Designers' Guide to en 1994-2_Eurocode 4_Design of Composite Steel and Concrete Structures

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DESIGNERS’ GUIDES TO THE EUROCODES

DESIGNERS’ GUIDE TO EN 1994-1-1 EUROCODE 4: DESIGN OF COMPOSITE STEEL AND CONCRETE STRUCTURES PART 1.1: GENERAL RULES AND RULES FOR BUILDINGS

Designers’ Guide to EN 1994-1-1 12 May 2004 11:52:25

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DESIGNERS’ GUIDE TO EN 1994-1-1 EUROCODE 4: DESIGN OF COMPOSITE STEEL AND CONCRETE STRUCTURES PART 1.1: GENERAL RULES AND RULES FOR BUILDINGS

R. P. JOHNSON and D. ANDERSON




Published by Thomas Telford Publishing, Thomas Telford Ltd, 1 Heron Quay, London E14 4JD URL: http://www.thomastelford.com

Distributors for Thomas Telford books are USA: ASCE Press, 1801 Alexander Bell Drive, Reston, VA 20191-4400 Japan: Maruzen Co. Ltd, Book Department, 3–10 Nihonbashi 2-chome, Chuo-ku, Tokyo 103 Australia: DA Books and Journals, 648 Whitehorse Road, Mitcham 3132, Victoria

First published 2004 Also available from Thomas Telford Books Designers’ Guide to EN 1990. Eurocode: Basis of Structural Design. H. Gulvanessian, J.-A.Calgaro and M. Holický. ISBN 0 7277 3011 8

A catalogue record for this book is available from the British Library

ISBN: 0 7277 3151 3

© The authors and Thomas Telford Limited 2004 All rights, including translation, reserved. Except as permitted by the Copyright, Designs and Patents Act 1988, no part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise, without the prior written permission of the Publishing Director, Thomas Telford Publishing, Thomas Telford Ltd, 1 Heron Quay, London E14 4JD This book is published on the understanding that the authors are solely responsible for the statements made and opinions expressed in it and that its publication does not necessarily imply that such statements and/or opinions are or reflect the views or opinions of the publishers. While every effort has been made to ensure that the statements made and the opinions expressed in this publication provide a safe and accurate guide, no liability or responsibility can be accepted in this respect by the authors or publishers Typeset by Helius, Brighton and Rochester Printed and bound in Great Britain by MPG Books, Bodmin




Preface EN 1994, also known as Eurocode 4, is one standard of the Eurocode suite and describes the principles and requirements for safety, serviceability and durability of composite steel and concrete structures. It is subdivided into three parts: • • •

Part 1.1: General Rules and Rules for Buildings Part 1.2: Structural Fire Design Part 2: Bridges.

It is intended to be used in conjunction with EN 1990, Basis of Structural Design, EN 1991, Actions on Structures, and the other design Eurocodes.

Aims and objectives of this guide The principal aim of this book is to provide the user with guidance on the interpretation and use of EN 1994-1-1 and to present worked examples. The guide explains the relationship with the other Eurocode parts to which it refers and with the relevant British codes. It also provides background information and references to enable users of Eurocode 4 to understand the origin and objectives of its provisions.

Layout of this guide EN 1994-1-1 has a foreword and nine sections, together with three annexes. This guide has an introduction which corresponds to the foreword of EN 1994-1-1, and Chapters 1 to 9 of the guide correspond to Sections 1 to 9 of the Eurocode. Chapters 10 and 11 correspond to Annexes A and B of the Eurocode, respectively. Appendices A to C of this guide include useful material from the draft Eurocode ENV 1994-1-1. The numbering and titles of the sections in this guide also correspond to those of the clauses of EN 1994-1-1. Some subsections are also numbered (e.g. 1.1.2). This implies correspondence with the subclause in EN 1994-1-1 of the same number. Their titles also correspond. There are extensive references to lower-level clause and paragraph numbers. The first significant reference is in bold italic type (e.g. clause 1.1.1(2)). These are in strict numerical sequence throughout the book, to help readers to find comments on particular provisions of the code. Some comments on clauses are necessarily out of sequence, but use of the index should enable these to be found. All cross-references in this guide to sections, clauses, subclauses, paragraphs, annexes, figures, tables and equations of EN 1994-1-1 are in italic type, which is also used where text from a clause in EN 1994-1-1 has been directly reproduced (conversely, cross-references to and quotations from other sources, including other Eurocodes, are in roman type).




DESIGNERS’ GUIDE TO EN 1994-1-1

Expressions repeated from EN 1994-1-1 retain their number; other expressions have numbers prefixed by D (for Designers’ Guide), e.g. equation (D6.1) in Chapter 6.

Acknowledgements The authors are deeply indebted to the other members of the four project teams for Eurocode 4 on which they have worked: Jean-Marie Aribert, Gerhard Hanswille, Bernt Johansson, Basil Kolias, Jean-Paul Lebet, Henri Mathieu, Michel Mele, Joel Raoul, Karl-Heinz Roik and Jan Stark; and also to the Liaison Engineers, National Technical Contacts, and others who prepared national comments. They thank the University of Warwick for the facilities provided for Eurocode work, and, especially, their wives Diana and Linda for their unfailing support. R. P. Johnson D. Anderson

vi




Contents Preface

Aims and objectives of this guide Layout of this guide Acknowledgements

Introduction

v v v vi 1

Chapter 1.

General 1.1. Scope 1.1.1. Scope of Eurocode 4 1.1.2. Scope of Part 1.1 of Eurocode 4 1.2. Normative references 1.2.1. General reference standards 1.2.2. Other reference standards 1.3. Assumptions 1.4. Distinction between principles and application rules 1.5. Definitions 1.5.1. General 1.5.2. Additional terms and definitions 1.6. Symbols

Chapter 2.

Basis of design 2.1. Requirements 2.2. Principles of limit states design 2.3. Basic variables 2.4. Verification by the partial factor method 2.4.1. Design values 2.4.2. Combination of actions 2.4.3. Verification of static equilibrium (EQU)

9 9 9 9 10 10 11 11

Chapter 3.

Materials 3.1. Concrete 3.2. Reinforcing steel 3.3. Structural steel 3.4. Connecting devices 3.4.1. General 3.4.2. Stud shear connectors 3.5. Profiled steel sheeting for composite slabs in buildings

13 13 15 16 16 16 17 17



3 3 3 3 5 5 5 5 5 6 6 6 6



Chapter 4.

Durability 4.1. General 4.2. Profiled steel sheeting for composite slabs in buildings

19 19 19

Chapter 5.

Structural analysis 5.1. Structural modelling for analysis 5.1.1. Structural modelling and basic assumptions 5.1.2. Joint modelling 5.2. Structural stability 5.2.1. Effects of deformed geometry of the structure 5.2.2. Methods of analysis for buildings 5.3. Imperfections 5.3.1. Basis 5.3.2. Imperfections in buildings 5.4. Calculation of action effects 5.4.1. Methods of global analysis Example 5.1: effective width of concrete flange 5.4.2. Linear elastic analysis 5.4.3. Non-linear global analysis 5.4.4. Linear elastic analysis with limited redistribution for buildings 5.4.5. Rigid plastic global analysis for buildings 5.5. Classification of cross-sections

21 21 21 21 22 22 23 24 24 24 27 27 29 29 33

Chapter 6.

Ultimate limit states 6.1. Beams 6.1.1. Beams for buildings 6.1.2. Effective width for verification of cross-sections 6.2. Resistances of cross-sections of beams 6.2.1. Bending resistance Example 6.1: resistance moment in hogging bending, with effective web 6.2.2. Resistance to vertical shear Example 6.2: resistance to bending and vertical shear 6.3. Resistance of cross-sections of beams for buildings with partial encasement 6.3.1. Scope 6.3.2. Resistance to bending 6.3.3–6.3.4. Resistance to vertical shear, and to bending and vertical shear 6.4. Lateral–torsional buckling of composite beams 6.4.1. General 6.4.2. Verification of lateral–torsional buckling of continuous composite beams with cross-sections in Class 1, 2 and 3 for buildings 6.4.3. Simplified verification for buildings without direct calculation Use of intermediate lateral bracing Flow charts for continuous beam Example 6.3: lateral–torsional buckling of two-span beam 6.5. Transverse forces on webs 6.6. Shear connection 6.6.1. General Example 6.4: arrangement of shear connectors

viii



34 36 37 41 41 41 43 43 44 50 54 55 57 57 57 57 58 58 58 61 63 64 66 66 67 67 69


CONTENTS

6.6.2. 6.6.3.

Longitudinal shear force in beams for buildings Headed stud connectors in solid slabs and concrete encasement 6.6.4. Design resistance of headed studs used with profiled steel sheeting in buildings Example 6.5: reduction factors for transverse sheeting 6.6.5. Detailing of the shear connection and influence of execution 6.6.6. Longitudinal shear in concrete slabs Example 6.6: transverse reinforcement for longitudinal shear Example 6.7: two-span beam with a composite slab – ultimate limit state Example 6.8: partial shear connection with non-ductile connectors Example 6.9: elastic resistance to bending, and influence of degree of shear connection and type of connector on bending resistance 6.7. Composite columns and composite compression members 6.7.1. General 6.7.2. General method of design 6.7.3. Simplified method of design 6.7.4. Shear connection and load introduction 6.7.5. Detailing provisions Example 6.10: composite column with bending about one or both axes Example 6.11: longitudinal shear outside areas of load introduction, for a composite column 6.8. Fatigue 6.8.1. General 6.8.2. Partial factors for fatigue assessment 6.8.3. Fatigue strength 6.8.4. Internal forces and fatigue loadings 6.8.5. Stresses 6.8.6. Stress ranges 6.8.7. Fatigue assessment based on nominal stress ranges Example 6.12: fatigue in reinforcement and shear connection

70

118 119 119 119 120 120 121 122 123 124

Chapter 7.

Serviceability limit states 7.1. General 7.2. Stresses 7.3. Deformations in buildings 7.3.1. Deflections 7.3.2. Vibration 7.4. Cracking of concrete 7.4.1. General 7.4.2. Minimum reinforcement 7.4.3. Control of cracking due to direct loading General comments on clause 7.4 Example 7.1: two-span beam (continued) – SLS

127 127 128 128 128 130 131 131 132 134 135 136

Chapter 8.

Composite joints in frames for buildings 8.1. Scope 8.2. Analysis, including modelling and classification 8.3. Design methods 8.4. Resistance of components Example 8.1: end-plate joints in a two-span beam in a braced frame

141 141 142 144 145 147

70 72 76 76 81 82 84 100 101 103 103 105 105 111 113 113

ix





Chapter 9.

Composite slabs with profiled steel sheeting for buildings 9.1. General 9.2. Detailing provisions 9.3. Actions and action effects 9.4. Analysis for internal forces and moments 9.5–9.6. Verification of profiled steel sheeting as shuttering 9.7. Verification of composite slabs for the ultimate limit states 9.7.1. Design criterion 9.7.2. Flexure 9.7.3. Longitudinal shear for slabs without end anchorage 9.7.4. Longitudinal shear for slabs with end anchorage 9.7.5. Vertical shear 9.7.6. Punching shear 9.8. Verification of composite slabs for serviceability limit states 9.8.1. Cracking of concrete 9.8.2. Deflection Example 9.1: two-span continuous composite slab

161 161 162 162 163 164 164 164 164 165 167 168 168 168 168 168 170

Chapter 10. Annex A (Informative). Stiffness of joint components in buildings A.1. Scope A.2. Stiffness coefficients A.3. Deformation of the shear connection Further comments on stiffness Example 10.1: elastic stiffness of an end-plate joint

179 179 179 181 181 181

Chapter 11. Annex B (Informative). Standard tests B.1. General B.2. Tests on shear connectors B.3. Testing of composite floor slabs Example 11.1: m–k tests on composite floor slabs Example 11.2: the partial-interaction method

187 187 188 191 194 198

Appendix A. Lateral–torsional buckling of composite beams for buildings Simplified expression for ‘cracked’ flexural stiffness of a composite slab Flexural stiffness of beam with encased web Maximum spacing of shear connectors for continuous U-frame action Top transverse reinforcement above an edge beam Derivation of the simplified expression for λLT Effect of web encasement on λLT Factor C4 for the distribution of bending moment Criteria for verification of lateral–torsional stability without direct calculation Web encasement

203

Appendix B. The effect of slab thickness on resistance of composite slabs to longitudinal shear Summary The model

x



203 204 204 206 206 208 209 209 210

211 211 211


CONTENTS

The m–k method The use of test results as predictors Shape of function y(x) Estimate of errors of prediction Conclusion for the m–k method The partial-connection method Conclusion for the partial-connection method

212 212 213 213 214 214 214

Appendix C. Simplified calculation method for the interaction curve for resistance of composite column cross-sections to compression and uniaxial bending Scope and method Resistance to compression Position of neutral axis Bending resistances Interaction with transverse shear Neutral axes and plastic section moduli of some cross-sections General Major-axis bending of encased I-sections Minor-axis bending of encased I-sections Concrete-filled circular and rectangular hollow sections Example C.1: N–M interaction polygon for a column cross-section

217 217 218 219 219 219 219 219 220 220 221 222

References

225

Index

231

xi




Introduction 1

The provisions of EN 1994-1-1 are preceded by a foreword, most of which is common to all Eurocodes. This Foreword contains clauses on: • • • • • •

the background to the Eurocode programme the status and field of application of the Eurocodes national standards implementing Eurocodes links between Eurocodes and harmonized technical specifications for products additional information specific to EN 1994-1-1 National Annex for EN 1994-1-1.

Guidance on the common text is provided in the introduction to the Designers’ Guide to 2 EN 1990, Eurocode: Basis of Structural Design, and only background information essential to users of EN 1994-1-1 is given here. 3 EN 1990 lists the following structural Eurocodes, each generally consisting of a number of parts which are in different stages of development at present: EN 1990 EN 1991 EN 1992 EN 1993 EN 1994 EN 1995 EN 1996 EN 1997 EN 1998 EN 1999

Eurocode: Basis of Structural Design Eurocode 1: Actions on Structures Eurocode 2: Design of Concrete Structures Eurocode 3: Design of Steel Structures Eurocode 4: Design of Composite Steel and Concrete Structures Eurocode 5: Design of Timber Structures Eurocode 6: Design of Masonry Structures Eurocode 7: Geotechnical Design Eurocode 8: Design of Structures for Earthquake Resistance Eurocode 9: Design of Aluminium Structures

The information specific to EN 1994-1-1 emphasizes that this standard is to be used with other Eurocodes. The standard includes many cross-references to particular clauses in 4 5 EN 1992 and EN 1993. Similarly, this guide is one of a series on Eurocodes, and is for use 6 7 with the guide for EN 1992-1-1 and the guide for EN 1993-1-1. It is the responsibility of each national standards body to implement each Eurocode part as a national standard. This will comprise, without any alterations, the full text of the Eurocode and its annexes as published by the European Committee for Standardization (CEN). This will usually be preceded by a National Title Page and a National Foreword, and may be followed by a National Annex. Each Eurocode recognizes the right of national regulatory authorities to determine values related to safety matters. Values, classes or methods to be chosen or determined at national level are referred to as Nationally Determined Parameters (NDPs), and are listed in the foreword to each Eurocode, in the clauses on National Annexes. NDPs are also indicated by





notes immediately after relevant clauses. Each National Annex will give or cross-refer to the NDPs to be used in the relevant country. Otherwise the National Annex may contain only the 8 following: • •

decisions on the application of informative annexes, and references to non-contradictory complementary information to assist the user in applying the Eurocode.

In EN 1994-1-1 the NDPs are principally the partial factors for material or product properties peculiar to this standard; for example, for the resistance of headed stud shear connectors, and of composite slabs to longitudinal shear. Other NDPs are values that may depend on climate, such as the free shrinkage of concrete.

2




CHAPTER 1

General This chapter is concerned with the general aspects of EN 1994-1-1, Eurocode 4: Design of Composite Steel and Concrete Structures, Part 1.1: General Rules and Rules for Buildings. The material described in this chapter is covered in Section 1, in the following clauses: • • • • • •

Scope Normative references Assumptions Distinction between principles and application rules Definitions Symbols

Clause 1.1 Clause 1.2 Clause 1.3 Clause 1.4 Clause 1.5 Clause 1.6

1.1. Scope 1.1.1. Scope of Eurocode 4

The scope of EN 1994 (all three parts) is outlined in clause 1.1.1. It is to be used with EN 1990, Eurocode: Basis of Structural Design, which is the head document of the Eurocode suite. Clause 1.1.1(2) emphasizes that the Eurocodes are concerned with structural behaviour and that other requirements, e.g. thermal and acoustic insulation, are not considered. The basis for verification of safety and serviceability is the partial factor method. EN 1990 recommends values for load factors and gives various possibilities for combinations of actions. The values and choice of combinations are to be set by the National Annex for the country in which the structure is to be constructed. Eurocode 4 is also to be used in conjunction with EN 1991, Eurocode 1: Actions on Structures9 and its National Annex, to determine characteristic or nominal loads. When a composite structure is to be built in a seismic region, account needs to be taken of EN 1998, Eurocode 8: Design of Structures for Earthquake Resistance.10 The Eurocodes are concerned with design and not execution, but minimum standards of workmanship are required to ensure that the design assumptions are valid. For this reason, clause 1.1.1(3) lists the European standards for the execution of steel structures and the execution of concrete structures. The former includes some requirements for composite construction, for example for the testing of welded stud shear connectors.

1.1.2. Scope of Part 1.1 of Eurocode 4

EN 1994-1-1 deals with aspects of design that are common to the principal types of composite structure, buildings and bridges. This results from the CEN requirement that a provision should not appear in more than one EN standard, as this can cause inconsistency when one standard is revised before another. For example, if the same rules for resistance to bending apply for a composite beam in a building as in a bridge (as most of them do), then



Clause 1.1.1 Clause 1.1.1(2)

Clause 1.1.1(3)



Clause 1.1.2(2)

those rules are ‘general’ and must appear in EN 1994-1-1 and not in EN 1994-2 (on bridges).11 This has been done even where most applications occur in bridges. For example, clause 6.8 (fatigue) is in Part 1.1, with a few additional provisions in Part 2. In EN 1994-1-1, all rules that are for buildings only are preceded by a heading that includes the word ‘buildings’, or, if an isolated paragraph, are placed at the end of the relevant clause, e.g. clauses 5.3.2 and 5.4.2.3(5). The coverage in this guide of the ‘general’ clauses of Part 1.1 is relevant to both buildings and bridges, except where noted otherwise. However, guidance provided by or related to the worked examples may be relevant only to applications in buildings. Clause 1.1.2(2) lists the titles of the sections of Part 1.1. Those for Sections 1–7 are the same as in the other material-dependent Eurocodes. The contents of Sections 1 and 2 similarly follow an agreed model. The provisions of Part 1.1 cover the design of the common composite members: • • • •

beams in which a steel section acts compositely with concrete composite slabs formed with profiled steel sheeting concrete-encased or filled composite columns joints between composite beams and steel or composite columns.

Sections 5 and 8 concern connected members. Section 5, ‘Structural analysis’, is needed particularly for a frame that is not of ‘simple’ construction. Unbraced frames and sway frames are within its scope. The provisions include the use of second-order global analysis and prestress by imposed deformations, and define imperfections. The scope of Part 1.1 extends to steel sections that are partially encased. The web of the steel section is encased by reinforced concrete, and shear connection is provided between the concrete and the steel. This is a well-established form of construction. The primary reason for its choice is improved resistance in fire. Fully encased composite beams are not included because: • •

no satisfactory model has been found for the ultimate strength in longitudinal shear of a beam without shear connectors it is not known to what extent some design rules (e.g. for moment–shear interaction and redistribution of moments) are applicable.

A fully encased beam with shear connectors can usually be designed as if partly encased or uncased, provided that care is taken to prevent premature spalling of encasement in compression. Part 2, Bridges, includes further provisions that may on occasion be useful for buildings, such as those on: • • • •

composite plates (where the steel member is a flat steel plate, not a profiled section) composite box girders tapered or non-uniform composite members structures that are prestressed by tendons.

The omission of application rules for a type of member or structure should not prevent its use, where appropriate. Some omissions are deliberate, to encourage the use of innovative design, based on specialized literature, the properties of materials, and the fundamentals of equilibrium and compatibility; and following the principles given in the relevant Eurocodes. This applies, for example, to: • • • • •

large holes in webs of beams types of shear connector other than welded studs base plates beneath composite columns shear heads in reinforced concrete framed structures, and many aspects of ‘mixed’ structures, as used in tall buildings.

4




CHAPTER 1. GENERAL

In addition to its nine normative sections, EN 1994-1-1 includes three informative annexes: • • •

Annex A, ‘Stiffness of joint components in buildings’ Annex B, ‘Standard tests’ Annex C, ‘Shrinkage of concrete for composite structures for buildings’.

The reasons for these annexes, additional to the normative provisions, are explained in the relevant chapters of this guide.

1.2. Normative references References are given only to other European standards, all of which are intended to be used as a package. Formally, the Standards of the International Organization for Standardization (ISO) apply only if given an EN ISO designation. National standards for design and for products do not apply if they conflict with a relevant EN standard. It is intended that, following a period of overlap, all competing national standards will be withdrawn by around 2010. As Eurocodes may not cross-refer to national standards, replacement of national standards for products by EN or ISO standards is in progress, with a time-scale similar to that for the Eurocodes. During the period of changeover to Eurocodes and EN standards it is likely that an EN referred to, or its National Annex, may not be complete. Designers who then seek guidance from national standards should take account of differences between the design philosophies and safety factors in the two sets of documents.

1.2.1. General reference standards

Some references here, and also in clause 1.2.2, appear to repeat references in clause 1.1.1. The difference is explained in clause 1.2. These ‘dated’ references define the issue of the standard that is referred to in detailed cross-references, given later in EN 1994-1-1.

1.2.2. Other reference standards

Eurocode 4 necessarily refers to EN 1992-1-1, Eurocode 2: Design of Concrete Structures, Part 1.1: General Rules and Rules for Buildings, and to several parts of EN 1993, Eurocode 3: Design of Steel Structures. In its application to buildings, EN 1994-1-1 is based on the concept of the initial erection of a steel frame, which may include prefabricated concrete-encased members. The placing of profiled steel sheeting or other shuttering follows. The addition of reinforcement and in situ concrete completes the composite structure. The presentation and content of EN 1994-1-1 therefore relate more closely to EN 1993-1-1 than to EN 1992-1-1.

1.3. Assumptions The general assumptions are those of EN 1990, EN 1992 and EN 1993. Commentary on them will be found in the relevant guides in this series.

1.4. Distinction between principles and application rules Clauses in the Eurocodes are set out as either principles or application rules. As defined by EN 1990: • • •

‘Principles comprise general statements for which there is no alternative and requirements and analytical models for which no alternative is permitted unless specifically stated’ ‘Principles are distinguished by the letter ‘P’ following the paragraph number’ ‘Application Rules are generally recognised rules which comply with the principles and satisfy their requirements’.

5





There are relatively few principles. It has been recognized that a requirement or analytical model for which ‘no alternative is permitted unless specifically stated’ can rarely include a numerical value, because most values are influenced by research and/or experience, and may change over the years. (Even the specified elastic modulus for structural steel is an approximate value.) Furthermore, a clause cannot be a principle if it requires the use of another clause that is an application rule; effectively that clause also would become a principle. It follows that, ideally, the principles in all the codes should form a consistent set, referring only to each other, and intelligible if all the application rules were deleted. This over-riding principle has strongly influenced the drafting of EN 1994.

1.5. Definitions 1.5.1. General

Clause 1.5.1

Clause 1.5.2

In accordance with the model for Section 1, reference is made to the definitions given in clauses 1.5 of EN 1990, EN 1992-1-1, and EN 1993-1-1. Many types of analysis are defined in clause 1.5.6 of EN 1990. It is important to note that an analysis based on the deformed geometry of a structure or element under load is termed ‘second order’ rather than ‘non-linear’. The latter term refers to the treatment of material properties in structural analysis. Thus, according to EN 1990 ‘non-linear analysis’ includes ‘rigid plastic’. This convention is not followed in EN 1994-1-1, where the heading ‘Non-linear global analysis’ (clause 5.4.3) does not include ‘rigid plastic global analysis’ (clause 5.4.5). References from clause 1.5.1 include clause 1.5.2 of EN 1992-1-1, which defines prestress as an action caused by the stressing of tendons. This applies to EN 1994-2 but not to EN 1994-1-1, as this type of prestress is outside its scope. Prestress by jacking at supports, which is outside the scope of EN 1992-1-1, is within the scope of EN 1994-1-1. The definitions in clauses 1.5.1 to 1.5.9 of EN 1993-1-1 apply where they occur in clauses in EN 1993 to which EN 1994 refers. None of them uses the word ‘steel’.

1.5.2. Additional terms and definitions

Most of the 13 definitions in clause 1.5.2 of EN 1994-1-1 include the word ‘composite’. The definition of ‘shear connection’ does not require the absence of separation or slip at the interface between steel and concrete. Separation is always assumed to be negligible, but explicit allowance may need to be made for effects of slip, e.g. in clauses 5.4.3, 7.2.1, 9.8.2(7) and A.3. The definition ‘composite frame’ is relevant to the use of Section 5. Where the behaviour is essentially that of a reinforced or prestressed concrete structure, with only a few composite members, global analysis should generally be in accordance with Eurocode 2. These lists of definitions are not exhaustive, because all the codes use terms with precise meanings that can be inferred from their contexts. Concerning use of words generally, there are significant differences from British codes. These arose from the use of English as the base language for the drafting process, and the need to improve precision of meaning and to facilitate translation into other European languages. In particular: • •

‘action’ means a load and/or an imposed deformation ‘action effect’ (clause 5.4) and ‘effect of action’ have the same meaning: any deformation or internal force or moment that results from an action.

1.6. Symbols The symbols in the Eurocodes are all based on ISO standard 3898: 1987.12 Each code has its own list, applicable within that code. Some symbols have more than one meaning, the particular meaning being stated in the clause.

6




CHAPTER 1. GENERAL

There are a few important changes from previous practice in the UK. For example, an x–x axis is along a member, a y–y axis is parallel to the flanges of a steel section (clause 1.7(2) of EN 1993-1-1), and a section modulus is W, with subscripts to denote elastic or plastic behaviour. Wherever possible, definitions in EN 1994-1-1 have been aligned with those in EN 1990, EN 1992 and EN 1993; but this should not be assumed without checking the list in clause 1.6. Some quite minor differences are significant. The symbol fy has different meanings in EN 1992-1-1 and EN 1993-1-1. It is retained in EN 1994-1-1 for the nominal yield strength of structural steel, though the generic subscript for that material is ‘a’, based on the French word for steel, ‘acier’. Subscript ‘a’ is not used in EN 1993-1-1, where the partial factor for steel is not γA, but γM; and this usage is followed in EN 1994-1-1. The characteristic yield strength of reinforcement is fsk, with partial factor γS.

7




CHAPTER 2

Basis of design The material described in this chapter is covered in Section 2 of EN 1994-1-1, in the following clauses: • • • •

Requirements Principles of limit states design Basic variables Verification by the partial factor method

Clause 2.1 Clause 2.2 Clause 2.3 Clause 2.4

The sequence follows that of EN 1990, Sections 2–4 and 6.

2.1. Requirements Design is to be in accordance with the general requirements of EN 1990. The purpose of Section 2 is to give supplementary provisions for composite structures. Clause 2.1(3) reminds the user again that design is based on actions and combinations of actions in accordance with EN 1991 and EN 1990, respectively. The use of partial safety factors for actions and resistances (the ‘partial factor method’) is expected but is not a requirement of Eurocodes. The method is presented in Section 6 of EN 1990 as one way of satisfying the basic requirements set out in Section 2 of that standard. This is why use of the partial factor method is given ‘deemed to satisfy’ status in clause 2.1(3). To establish that a design was in accordance with the Eurocodes, the user of any other method would normally have to demonstrate, to the satisfaction of the regulatory authority and/or the client, that the method satisfied the basic requirements of EN 1990.

Clause 2.1(3)

2.2. Principles of limit states design The clause provides a reminder that the influence of sequence of construction on action effects must be considered. It does not affect the bending resistance of beams that are in Class 1 or 2 (as defined in clause 5.5) or the resistance of a composite column, as these are determined by rigid plastic theory, but it does affect the resistances of beams in Class 3 or 4.

2.3. Basic variables The classification of effects of shrinkage and temperature in clause 2.3.3 into ‘primary’ and ‘secondary’ will be familiar to designers of continuous beams, especially for bridges. Secondary effects are to be treated as ‘indirect actions’, which are ‘sets of imposed deformations’ (clause 1.5.3.1 of EN 1990), not as action effects. This distinction appears to have no consequences in practice, for the use of EN 1994-1-1.



Clause 2.3.3



2.4. Verification by the partial factor method Clause 2.4.1.1 Clause 2.4.1.2

2.4.1. Design values

Clauses 2.4.1.1 and 2.4.1.2 illustrate the treatment of partial factors. Recommended values are given in Notes, in the hope of eventual convergence between the values for each partial factor that will be specified in the National Annexes. This process was adopted because the regulatory bodies in the member states of CEN, rather than CEN itself, are responsible for setting safety levels. The Notes are informative, not normative (i.e., not part of the preceding provision), so that there are no numerical values in the principles of clause 2.4.1.2, as explained earlier. The Notes also link the partial factors for concrete, reinforcing steel and structural steel to those recommended in EN 1992 and EN 1993. Design would be more difficult if the factors for these materials in composite structures differed from the values in reinforced concrete and steel structures. The remainder of EN 1994-1-1 normally refers to design strengths, rather than characteristic or nominal values with partial factors. The design strength for concrete is given by fcd = fck /γC

(2.1)

where fck is the characteristic cylinder strength. This definition is stated algebraically because it differs from that of EN 1992-1-1, in which an additional coefficient αcc is applied: fcd = αcc fck /γC

Clause 2.4.1.3 Clause 2.4.1.4

(D2.1)

The coefficient is explained by EN 1992-1-1 as taking account of long-term effects and of unfavourable effects resulting from the way the load is applied. The recommended value is 1.0, but a different value could be chosen in a National Annex. This possibility is not appropriate for EN 1994-1-1 because the coefficient has been taken as 1.0 in calibration of composite elements. Clause 2.4.1.3 refers to ‘product standards hEN’. The ‘h’ stands for ‘harmonized’. This term from the Construction Products Directive13 is explained in the Designers’ Guide to EN 1990.2 Clause 2.4.1.4, on design resistances, refers to expressions (6.6a) and (6.6c) given in clause 6.3.5 of EN 1990. Resistances in EN 1994-1-1 often need more than one partial factor, and so use expression (6.6a), which is Rd = R{(ηi Xk, i /γM, i); ad}

i≥1

(D2.2)

For example, clause 6.7.3.2(1) gives the plastic resistance to compression of a cross-section as the sum of terms for the structural steel, concrete and reinforcement: Npl, Rd = Aa fyd + 0.85Ac fcd + As fsd

(6.30)

In this case, there is no separate term ad based on geometrical data, because uncertainties in areas of cross-sections are allowed for in the γM factors. In terms of characteristic strengths, from clause 2.4.1.2, equation (6.30) becomes: Npl, Rd = Aa fy /γM + 0.85Ac fck/γC + As fsk/γS

(D2.3)

in which: • • •

the characteristic material strengths Xk, i are fy, fck and fsk the conversion factors, ηi in EN 1990, are 1.0 for steel and reinforcement and 0.85 for concrete the partial factors γM, i are γM, γC and γS.

Expression (6.6c) of EN 1990 is Rd = Rk/γM. It applies where characteristic properties and a single partial factor can be used; for example, in expressions for the shear resistance of a headed stud (clause 6.6.3.1).

10




CHAPTER 2. BASIS OF DESIGN

2.4.2. Combination of actions No comment is necessary.

2.4.3. Verification of static equilibrium (EQU)

The abbreviation EQU appears in EN 1990, where four types of ultimate limit state are defined in clause 6.4.1: • • • •

EQU, for loss of static equilibrium FAT, for fatigue failure GEO, for failure or excessive deformation of the ground STR, for internal failure or excessive deformation of the structure.

This guide covers ultimate limit states only of types STR and FAT. Use of type GEO arises in design of foundations to EN 1997.14

11




CHAPTER 3

Materials This chapter concerns the properties of materials needed for the design of composite structures. It corresponds to Section 3, which has the following clauses: • • • • •

Concrete Reinforcing steel Structural steel Connecting devices Profiled steel sheeting for composite slabs in buildings

Clause 3.1 Clause 3.2 Clause 3.3 Clause 3.4 Clause 3.5

Rather than repeating information given elsewhere, Section 3 consists mainly of crossreferences to other Eurocodes and EN standards. The following comments relate to provisions of particular significance for composite structures.

3.1. Concrete Clause 3.1(1) refers to EN 1992-1-1 for the properties of concrete. For lightweight-aggregate concrete, several properties are dependent on the oven-dry density, relative to 2200 kg/m3. Complex sets of time-dependent properties are given in its clause 3.1 for normal concrete and clause 11.3 for lightweight-aggregate concrete. For composite structures built unpropped, with several stages of construction, simplification is essential. Specific properties are now discussed. (For thermal expansion, see Section 3.3.)

Strength and stiffness Strength and deformation characteristics are summarized in EN 1992-1-1, Table 3.1 for normal concrete and Table 11.3.1 for lightweight-aggregate concrete. Strength classes for normal concrete are defined as Cx/y, where x and y are respectively the cylinder and cube compressive strengths in units of newtons per square millimetre. All compressive strengths in design rules in Eurocodes are cylinder strengths, so an unsafe error occurs if a specified cube strength is used in calculations. It should be replaced at the outset by the equivalent cylinder strength, using the relationships given by the strength classes. Classes for lightweight concrete are designated LCx/y. The relationships between cylinder and cube strengths differ from those of normal concrete. Except where prestressing by tendons is used (which is outside the scope of this guide), the tensile strength of concrete is rarely used in design calculations for composite members. The mean tensile strength fctm appears in the definitions of ‘cracked’ global analysis in clause 5.4.2.3(2), and in clause 7.4.2(1) on minimum reinforcement. Its value and the 5 and 95% fractile values are given in Tables 3.1 and 11.3.1 of EN 1992-1-1. The appropriate fractile value should be used in any limit state verification that relies on either an adverse or beneficial effect of the tensile strength of concrete.



Clause 3.1(1)



EN 1992-1-1: fcd = accfck/gC 0.8x

x

EN 1994-1-1: 0.85fcd, with fcd = fck/gC 0 Plastic neutral axis

0.0035

Compressive strain

0 Compressive stress

Fig. 3.1. Stress blocks for concrete at ultimate limit states

Values of the modulus of elasticity are given in Tables 3.1 and 11.3.1. Clause 3.1.3 of EN 1992-1-1 points out that these are indicative, for general applications. The short-term elastic modulus Ecm increases for ages greater than 28 days. The influence of this small change on the effective modulus is negligible compared with the uncertainties in the modelling of creep, so it should be ignored.

Stress/strain properties The design compressive strength of concrete, fcd, is defined in clause 3.1.6(1)P of EN 1992-1-1 as fcd = αcc fck/γC where αcc is the coefficient taking account of long term effects on the compressive strength and of unfavourable effects resulting from the way the load is applied. Note: The value of αcc for use in a Country should lie between 0.8 and 1.0 and may be found in its National Annex. The recommended value is 1.

The reference in clause 3.1(1) to EN 1992-1-1 for properties of concrete begins ‘unless otherwise given by Eurocode 4 ’ . Resistances of composite members given in EN 1994-1-1 are based on extensive calibration studies (e.g. see Johnson and Huang15,16). The numerical coefficients given in resistance formulae are consistent with the value αcc = 1.0 and the use of either elastic theory or the stress block defined in clause 6.2.1.2. Therefore, there is no reference in EN 1994-1-1 to a coefficient αcc or to a choice to be made in a National Annex. The symbol fcd always means fck/γC, and for beams and most columns is used with the coefficient 0.85, as in equation (6.30) in clause 6.7.3.2(1). An exception, in that clause, is when the value of 0.85 is replaced by 1.0 for concrete-filled column sections, based on calibration. The approximation made to the shape of the stress–strain curve is also relevant. Those given in clause 3.1 of EN 1992-1-1 are mainly curved or bilinear, but in clause 3.1.7(3) there is a simpler rectangular stress distribution, similar to the stress block given in the British Standard for the structural use of concrete, BS 8110.17 Its shape, for concrete strength classes up to C50/60, and the corresponding strain distribution are shown in Fig. 3.1. This stress block is inconvenient for use with composite cross-sections, because the region near the neutral axis assumed to be unstressed is often occupied by a steel flange, and algebraic expressions for resistance to bending become complex. In composite sections, the contribution from the steel section to the bending resistance reduces the significance of that from the concrete. It is thus possible18 for EN 1994 to allow the use of a rectangular stress block extending to the neutral axis, as shown in Fig. 3.1. For a member of unit width, the moment about the neutral axis of the EN 1992 stress block ranges from 0.38fckx2/γC to 0.48fckx2/γC, depending on the value chosen for αcc. The value for beams in EN 1994-1-1 is 0.425fckx2/γC. Calibration studies have shown that this overestimates

14




CHAPTER 3. MATERIALS

the bending resistance of cross-sections of columns, so a correction factor αM is given in clause 6.7.3.6(1). See also the comments on clauses 6.2.1.2(2) and 6.7.3.6. Clause 3.1(2) limits the scope of EN 1994-1-1 to the strength range C20/25 to C60/75 for normal concrete and from LC20/22 to LC60/66 for lightweight concrete. These ranges are narrower than those given in EN 1992-1-1 because there is limited knowledge and experience of the behaviour of composite members with weak or very strong concrete. This applies, for example, to the load/slip properties of shear connectors, the redistribution of moments in continuous beams and the resistance of columns. The use of rectangular stress blocks for resistance to bending (clause 6.2.1.2(d)) relies on the strain capacity of the materials. The relevant property of concrete, εcu3 in Table 3.1 of EN 1992-1-1, is –0.0035 for classes up to C50/60, but is only –0.0026 for class C90/105.

Shrinkage The shrinkage of concrete referred to in clause 3.1(3) is the drying shrinkage that occurs after setting. It does not include the plastic shrinkage that precedes setting, nor autogenous shrinkage. The latter develops during hardening of the concrete (clause 3.1.4(6) of EN 1992-1-1), and is that which occurs in enclosed or sealed concrete, as in a concrete-filled tube, where no loss of moisture occurs. Clause 3.1(4) permits its effect on stresses and deflections to be neglected, but does not refer to crack widths. It has little influence on cracking due to direct loading, and the rules for initial cracking (clause 7.4.2) take account of its effects. The shrinkage strains given in clause 3.1.4(6) of EN 1992-1-1 are significantly higher than those given in BS 8110. Taking grade C40/50 concrete as an example, with ‘dry’ environment (relative humidity 60%), the final drying shrinkage could be –400 × 10–6, plus autogenous shrinkage of –75 × 10–6. The value in ENV 1994-1-1 was –325 × 10–6, based on practice and experience. In the absence of adverse comment on the ENV, this value is repeated in Annex C (informative) of EN 1994-1-1, with a Note below clause 3.1 that permits other values to be given in National Annexes. In the absence of this Note, a design using a value from Annex C, confirmed in a National Annex, would not be in accordance with the Eurocodes. This is because normative clause 3.1.4(6) of EN 1992-1-1 takes precedence over an informative National Annex, and all variations in National Annexes have to be permitted in this way. In typical environments in the UK, the influence of shrinkage of normal-weight concrete on the design of composite structures for buildings is significant only in: • • •

Clause 3.1(2)

Clause 3.1(3)

Clause 3.1(4)

very tall structures very long structures without movement joints the prediction of deflections of beams with high span/depth ratios (clause 7.3.1 (8)). There is further comment on shrinkage in Chapter 5.

Creep The provisions of EN 1992-1-1 on creep of concrete can be simplified for composite structures for buildings, as discussed in comments on clause 5.4.2.2.

3.2. Reinforcing steel Clause 3.2(1) refers to EN 1992-1-1, which states in clause 3.2.2(3)P that its rules are valid for specified yield strengths fyk up to 600 N/mm2. The scope of clause 3.2 of EN 1992-1-1, and hence of EN 1994-1-1, is limited to reinforcement, including wire fabrics with a nominal bar diameter of 5 mm and above, that is, ‘ribbed’ (high bond) and weldable. There are three ductility classes, from A (the lowest) to C. The requirements include the characteristic strain at maximum force, rather than the

Clause 3.2(1)

15





Clause 3.2(2)

elongation at fracture used in past British standards. Clause 5.5.1(5) of EN 1994-1-1 excludes the use of Class A reinforcement in any composite cross-section in Class 1 or 2. The minimum ductility properties of wire fabric given in Table C.1 of EN 1992-1-1 may not be sufficient to satisfy clause 5.5.1(6) of EN 1994-1-1, as this requires demonstration of sufficient ductility to avoid fracture when built into a concrete slab.19 It has been found in tests on continuous composite beams with fabric in tension that the cross-wires initiate cracks in concrete, so that tensile strain becomes concentrated at the locations of the welds in the fabric. For simplicity, clause 3.2(2) permits the modulus of elasticity of reinforcement to be taken as 210 kN/mm2, the value given in EN 1993-1-1 for structural steel, rather than 200 kN/mm2, the value in EN 1992-1-1.

3.3. Structural steel Clause 3.3(1) Clause 3.3(2)

Clause 3.3(1) refers to EN 1993-1-1. This lists in its Table 3.1 steel grades with nominal yield strengths up to 460 N/mm2, and allows other steel products to be included in National Annexes. Clause 3.3(2) sets an upper limit of 460 N/mm2 for use with EN 1994-1-1. There has been extensive research20–23 on the use in composite members of structural steels with yield strengths exceeding 355 N/mm2. It has been found that some design rules need modification for use with steel grades higher than S355, to avoid premature crushing of concrete. This applies to: • • • •

redistribution of moments (clause 5.4.4(6)) rotation capacity (clause 5.4.5(4a)) plastic resistance moment (clause 6.2.1.2(2)) resistance of columns (clause 6.7.3.6(1)).

Thermal expansion For the coefficient of linear thermal expansion of steel, clause 3.2.6 of EN 1993-1-1 gives a value of 12 × 10–6 ‘per °C’ (also written in Eurocodes as /K or K–1). This is followed by a Note that for calculating the ‘structural effects of unequal temperatures’ in composite structures, the coefficient may be taken as 10 × 10–6 per °C, which is the value given for normal-weight concrete in clause 3.1.3(5) of EN 1992-1-1 ‘unless more accurate information is available’. Thermal expansion of reinforcement is not mentioned in EN 1992-1-1, presumably because it is assumed to be the same as that of normal-weight concrete. For reinforcement in composite structures the coefficient should be taken as 10 × 10–6 K–1. This was stated in ENV 1994-1-1, but is not in the EN. Coefficients of thermal expansion for lightweight-aggregate concretes can range from 4 × 10–6 to 14 × 10–6 K–1. Clause 11.3.2(2) of EN 1992-1-1 states that The differences between the coefficients of thermal expansion of steel and lightweight aggregate concrete need not be considered in design,

but ‘steel’ here means reinforcement, not structural steel. The effects of the difference from 10 × 10–6 K–1 should be considered in design of composite members for situations where the temperatures of the concrete and the structural steel could differ significantly.

3.4. Connecting devices 3.4.1. General

Reference is made to EN 1993, Eurocode 3: Design of Steel Structures, Part 1.8: Design of Joints,24 for information relating to fasteners, such as bolts, and welding consumables. Provisions for ‘other types of mechanical fastener’ are given in clause 3.3.2 of EN 1993-1-3.25 Commentary on joints is given in Chapters 8 and 10.

16





3.4.2. Stud shear connectors

Headed studs are the only type of shear connector for which detailed provisions are given in EN 1994-1-1, in clause 6.6.5.7. Any other method of connection must satisfy clause 6.6.1.1. The use of adhesives on a steel flange is unlikely to be suitable. Clause 3.4.2 refers to EN 13918, Welding – Studs and Ceramic Ferrules for Arc Stud Welding.26 This gives minimum dimensions for weld collars. Other methods of attaching studs, such as spinning, may not provide weld collars large enough for the resistances of studs given in clause 6.6.3.1(1) to be applicable. Shear connection between steel and concrete by bond or friction is permitted only in accordance with clause 6.7.4, for columns, and clauses 9.1.2.1 and 9.7, for composite slabs.

Clause 3.4.2

3.5. Profiled steel sheeting for composite slabs in buildings The title includes ‘in buildings’, as this clause and other provisions for composite slabs are not applicable to composite bridges. The materials for profiled steel sheeting must conform to the standards listed in clause 3.5. There are at present no EN standards for the wide range of profiled sheets available. Such standards should include tolerances on embossments and indentations, as these influence resistance to longitudinal shear. Tolerances on embossments, given for test specimens in clause B.3.3(2), provide guidance. The minimum bare metal thickness has been controversial, and in EN 1994-1-1 is subject to National Annexes, with a recommended minimum of 0.70 mm. The total thickness of zinc coating in accordance with clause 4.2(3) is about 0.05 mm.

Clause 3.5

17




CHAPTER 4

Durability This chapter concerns the durability of composite structures. It corresponds to Section 4, which has the following clauses: • •

General Profiled steel sheeting for composite slabs in buildings

Clause 4.1 Clause 4.2

4.1. General Almost all aspects of the durability of composite structures are covered by cross-references to EN 1990, EN 1992 and EN 1993. The material-independent provisions, in clause 2.4 of EN 1990, require the designer to take into account 10 factors. These include the foreseeable use of the structure, the expected environmental conditions, the design criteria, the performance of the materials, the particular protective measures, the quality of workmanship and the intended level of maintenance. Clauses 4.2 and 4.4.1 of EN 1992-1-1 define exposure classes and cover to reinforcement. A Note defines structural classes. These and the ‘acceptable deviations’ (tolerances) for cover may be modified in a National Annex. Clause 4.4.1.3 recommends an addition of 10 mm to the minimum cover to allow for the deviation. As an example, a concrete floor of a multi-storey car park will be subject to the action of chlorides in an environment consisting of cyclic wet and dry conditions. For these conditions (designated class XD3) the recommended structural class is 4, giving a minimum cover for a 50 year service life of 45 mm plus a tolerance of 10 mm. This total of 55 mm can be reduced, typically by 5 mm, where special quality assurance is in place. Section 4 of EN 1993-1-1 refers to execution of protective treatments for steelwork. If parts will be susceptible to corrosion, there is need for access for inspection and maintenance. This will not be possible for shear connectors, and clause 4.1(2) of EN 1994-1-1 refers to clause 6.6.5, which includes provisions for minimum cover.

4.2. Profiled steel sheeting for composite slabs in buildings For profiled steel sheeting, clause 4.2(1)P requires the corrosion protection to be adequate for its environment. Zinc coating to clause 4.2(3) is ‘sufficient for internal floors in a non-aggressive environment’. This implies that it may not provide sufficient durability for use in a multi-storey car park or near the sea.



Clause 4.2(1)P Clause 4.2(3)


CHAPTER 5

Structural analysis Structural analysis may be performed at three levels: global analysis, member analysis, and local analysis. This chapter concerns global analysis to determine deformations and internal forces and moments in beams and framed structures. It corresponds to Section 5, which has the following clauses: • • • • •

Structural modelling for analysis Structural stability Imperfections Calculation of action effects Classification of cross-sections


Wherever possible, analyses for serviceability and ultimate limit states use the same methods. It is generally more convenient, therefore, to specify them together in a single section, rather than to include them in Sections 6 and 7. For composite slabs, though, all provisions, including those for global analysis, are given in Section 9. The division of material between Section 5 and Section 6 (ultimate limit states) is not always obvious. Calculation of vertical shear is clearly ‘analysis’, but longitudinal shear is in Section 6. This is because its calculation for beams in buildings is dependent on the method used to determine the resistance to bending. However, for composite columns, methods of analysis and member imperfections are considered in clause 6.7.3.4. This separation of imperfections in frames from those in columns requires care, and receives detailed explanation after the comments on clause 5.4. The flow charts for global analysis (Fig. 5.1) include relevant provisions from Section 6.

5.1. Structural modelling for analysis 5.1.1. Structural modelling and basic assumptions

General provisions are given in EN 1990. The clause referred to says, in effect, that models shall be appropriate and based on established theory and practice and that the variables shall be relevant. Composite members and joints are commonly used in conjunction with others of structural steel. Clause 5.1.1(2) makes clear that this is the type of construction envisaged in Section 5, which is aligned with and cross-refers to Section 5 of EN 1993-1-1 wherever possible. Where there are significant differences between these two sections, they are referred to here.

5.1.2. Joint modelling

The three simplified joint models listed in clause 5.1.2(2) – simple, continuous and semi-continuous – are those given in EN 1993. The subject of joints in steelwork has its



Clause 5.1.1(2)

Clause 5.1.2(2)



Clause 5.1.2(3)

own Eurocode part, EN 1993-1-8.24 For composite joints, its provisions are modified and supplemented by Section 8 of EN 1994-1-1. The first two joint models are those commonly used for beam-to-column joints in steel frames. For each joint in the ‘simple’ model, the location of the nominal pin relative to the centre-line of the column, the ‘nominal eccentricity’, has to be chosen. This determines the effective span of each beam and the bending moments in each column. Practice varies across Europe, and neither EN 1993-1-1 nor EN 1994-1-1 gives values for nominal eccentricities. Guidance may be given in a National Annex, or in other literature. In reality, most joints in buildings are neither ‘simple’ (i.e. pinned) nor ‘continuous’. The third model, ‘semi-continuous’, is appropriate for a wide range of joints with moment– rotation behaviours intermediate between ‘simple’ and ‘continuous’. This model is rarely applicable to bridges, so the cross-reference to EN 1993-1-8 in clause 5.1.2(3) is ‘for buildings’. The provisions of EN 1993-1-8 are for joints ‘subjected to predominantly static loading’ (its clause 1.1(1)). They are applicable to wind loading on buildings, but not to fatigue loading, which is covered in EN 1993-1-9 and in clause 6.8. For composite beams, the need for continuity of slab reinforcement past the columns, to control cracking, causes joints to transmit moments. For the joint to ‘have no effect on the analysis’ (from the definition of a ‘continuous’ joint in clause 5.1.1(2) of EN 1993-1-8), so much reinforcement and stiffening of steelwork are needed that the design becomes uneconomic. Joints with some continuity are usually semi-continuous. Structural analysis then requires prior calculation of the properties of joints, except where they can be treated as ‘simple’ or ‘continuous’ on the basis of ‘significant experience of previous satisfactory performance in similar cases’ (clause 5.2.2.1(2) of EN 1993-1-8, referred to from clause 8.2.3(1)) or experimental evidence. Clause 5.1.2(2) refers to clause 5.1.1 of EN 1993-1-8, which gives the terminology for the semi-continuous joint model. For elastic analysis, the joint is ‘semi-rigid’. It has a rotational stiffness, and a design resistance which may be ‘partial strength’ or ‘full strength’, normally meaning less than or greater than the bending resistance of the connected beam. If the resistance of the joint is reached, then elastic–plastic or rigid plastic global analysis is required.

5.2. Structural stability The following comments refer mainly to beam-and-column frames, and assume that the global analyses will be based on elastic theory. The exceptions, in clauses 5.4.3, 5.4.4 and 5.4.5 are discussed later. All design methods must take account of errors in the initial positions of joints (global imperfections) and in the initial geometry of members (member imperfections); of the effects of cracking of concrete and of any semi-rigid joints; and of residual stresses in compression members. The stage at which each of these is considered or allowed for will depend on the software being used, which leads to some complexity in clauses 5.2 to 5.4.

5.2.1. Effects of deformed geometry of the structure

Clause 5.2.1(2)P Clause 5.2.1(3)

In its clause 1.5.6, EN 1990 defines types of analysis. ‘First-order’ analysis is performed on the initial geometry of the structure. ‘Second-order’ analysis takes account of the deformations of the structure, which are a function of its loading. Clearly, second-order analysis may always be applied. With appropriate software increasingly available, second-order analysis is the most straightforward approach. The criteria for neglect of second-order effects given in clauses 5.2.1(2)P and 5.2.1(3) need not be considered. The analysis allowing for second-order effects will usually be iterative but normally the iteration will take place within the software. Methods for second-order analysis are described in textbooks such as that by Trahair et al.27

22




CHAPTER 5. STRUCTURAL ANALYSIS

A disadvantage of second-order analysis is that, in general, the useful principle of superposition does not apply. Clause 5.2.1(3) provides a basis for the use of first-order analysis. The check is done for a particular load combination and arrangement. The provisions in this clause are similar to those for elastic analysis in the corresponding clause in EN 1993-1-1. In an elastic frame, second-order effects are dependent on the nearness of the design loads to the elastic critical load. This is the basis for expression (5.1), in which αcr is defined as ‘the factor … to cause elastic instability’. This may be taken as the load factor at which bifurcation of equilibrium occurs. For a conventional beam-and-column frame, it is assumed that the frame is perfect, and that only vertical loads are present, usually at their maximum design values. These are replaced by a set of loads which produces the same set of member axial forces without any bending. An eigenvalue analysis then gives the factor αcr, applied to the whole of the loading, at which the total frame stiffness vanishes, and elastic instability occurs. To sufficient accuracy, αcr may also be determined by a second-order load–deflection analysis. The non-linear load–deflection response approaches asymptotically to the elastic critical value. Normally, though, it is pointless to use this method, as it is simpler to use the same software to account for the second-order effects due to the design loads. A more useful method for αcr is given in clause 5.2.2(1). Unlike the corresponding clause in EN 1993-1-1, the check in clause 5.2.1(3) is not just for a sway mode. This is because clause 5.2.1 is relevant not only to complete frames but also to the design of individual composite columns (see clause 6.7.3.4). Such members may be held in position against sway but still be subject to significant second-order effects due to bowing. Clause 5.2.1(4)P is a reminder that the analysis needs to account for the reduction in stiffness arising from cracking and creep of concrete and from possible non-linear behaviour of the joints. Further remarks on how this should be done are made in the comments on clauses 5.4.2.2, 5.4.2.3 and 8.2.2, and the procedures are illustrated in Fig. 5.1(b)–(d). In general, such effects are dependent on the internal moments and forces, and iteration is therefore required. Manual intervention may be needed, to adjust stiffness values before repeating the analysis. It is expected, though, that advanced software will be written for EN 1994 to account automatically for these effects. The designer may of course make assumptions, although care is needed to ensure these are conservative. For example, assuming that joints have zero rotational stiffness (resulting in simply-supported composite beams) could lead to neglect of the reduction in beam stiffness due to cracking. The overall lateral stiffness would probably be a conservative value, but this is not certain. However, in a frame with stiff bracing it will be worth first calculating αcr, assuming joints are pinned and beams are steel section only; it may well be found that this value of αcr is sufficiently high for first-order global analysis to be used. Using elastic analysis, it is not considered necessary to account for slip (see clause 5.4.1.1(8)), provided that the shear connection is in accordance with clause 6.6.

5.2.2. Methods of analysis for buildings

Clause 5.2.2(1) refers to clause 5.2.1(4) of EN 1993-1-1 for a simpler check, applicable to many structures for buildings. This requires calculation of sway deflections due to horizontal loads only, and first-order analysis can be used to determine these deflections. It is assumed that any significant second-order effects will arise only from interaction of column forces with sway deflection. It follows that the check will only be valid if axial compression in beams is not significant. Fig. 5.1(e) illustrates the procedure. Even where second-order effects are significant, clause 5.2.2(2) allows these to be determined by amplifying the results from a first-order analysis. No further information is given, but clause 5.2.2(5) of EN 1993-1-1 describes a method for frames, provided that the conditions in its clause 5.2.2(6) are satisfied.

Clause 5.2.1(3)

Clause 5.2.1(4)P

Clause 5.2.2(1)

Clause 5.2.2(2)

23





Clause 5.2.2(3) Clause 5.2.2(4) Clause 5.2.2(5) Clause 5.2.2(6) Clause 5.2.2(7)

Clauses 5.2.2(3) to 5.2.2(7) concern the relationships between the analysis of the frame and the stability of individual members. A number of possibilities are presented. If relevant software is available, clause 5.2.2(3) provides a convenient route for composite columns, because column design to clause 6.7 generally requires a second-order analysis. Usually, though, the global analysis will not account for all local effects, and clause 5.2.2(4) describes in general terms how the designer should then proceed. Clause 5.2.2(5) refers to the methods of EN 1994-1-1 for lateral–torsional buckling, which allow for member imperfections. This applies also to local and shear buckling in beams, so imperfections in beams can usually be omitted from global analyses. In clause 5.2.2(6), ‘compression members’ are referred to as well as columns, to include composite members used in bracing systems and trusses. Further comments on clauses 5.2.2(3) to 5.2.2(7) are made in the sections of this guide dealing with clauses 5.5, 6.2.2.3, 6.4 and 6.7. Figure 5.1(a) illustrates how global and member analyses may be used, for a plane frame including composite columns.

5.3. Imperfections Clause 5.3.1(1)P Clause 5.3.1(2)

5.3.1. Basis

Clause 5.3.1(1)P lists possible sources of imperfection. Subsequent clauses (and also clause 5.2) describe how these should be allowed for. This may be by inclusion in the global analyses or in methods of checking resistance, as explained above. Clause 5.3.1(2) requires imperfections to be in the most unfavourable direction and form. The most unfavourable geometric imperfection normally has the same shape as the lowest buckling mode. This can sometimes be difficult to find; but it can be assumed that this condition is satisfied by the Eurocode methods for checking resistance that include effects of member imperfections (see comments on clause 5.2.2).

5.3.2. Imperfections in buildings

Clause 5.3.2.1(1)

Clause 5.3.2.1(2) Clause 5.3.2.1(3) Clause 5.3.2.1(4)

Generally, an explicit treatment of geometric imperfections is required for composite frames. In both EN 1993-1-1 and EN 1994-1-1 the values are equivalent rather than measured values (clause 5.3.2.1(1)), because they allow for effects such as residual stresses, in addition to imperfections of shape. The codes define both global sway imperfections for frames and local bow imperfections of individual members (meaning a span of a beam or the length of a column between storeys). The usual aim in global analysis is to determine the action effects at the ends of members. If necessary, a member analysis is performed subsequently, as illustrated in Fig. 5.1(a); for example to determine the local moments in a column due to transverse loading. Normally the action effects at members’ ends are affected by the global sway imperfections but not significantly by the local bow imperfections. In both EN 1993-1-1 and EN 1994-1-1 the effect of a bow imperfection on the end moments and forces may be neglected in global analysis if the design normal force NEd does not exceed 25% of the Euler buckling load for the pin-ended member (clause 5.3.2.1(2)). Clause 5.3.2.1(3) is a reminder that an explicit treatment of bow imperfections is always required for checking individual composite columns, because the resistance formulae are for cross-sections only and do not allow for action effects caused by these imperfections. The reference to EN 1993-1-1 in clause 5.3.2.1(4) leads to two alternative methods of allowing for imperfections in steel columns. One method includes all imperfections in the global analysis. Like the method just described for composite columns, no individual stability check is then necessary. The alternative approach is that familiar to most designers. Member imperfections are not accounted for in the global analysis. The stability of each member is then checked using end moments and forces from that analysis, with buckling formulae that take account of imperfections.

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Determine frame imperfections as equivalent horizontal forces, to clause 5.3.2.2, which refers to clause 5.3.2 of EC3. Neglect member imperfections (clause 5.3.2.1(2))

Go to Fig. 5.1(b), on creep

Go to Fig. 5.1(c), on cracking

Note: These flow charts are for a particular load combination and arrangement for ultimate limit states, for a beam-and-column type plane frame in its own plane, and for global analyses in which allowances may be needed for creep, cracking of concrete, and the behaviour of joints.

Go to Fig. 5.1(d), on joints

‘EC3’ means EN 1993-1-1

Determine appropriate stiffnesses, making allowance for cracking and creep of concrete and for behaviour of joints

Go to Fig. 5.1(e), on methods of global analysis

Is second-order analysis needed for global analysis?

No

Yes For each column, estimate NEd, find l-to clause 5.3.2.1(2). Determine member imperfection for each column (to clause 5.3.2.3) and where condition (2) of clause 5.3.2.1(2) is not satisfied, include these these imperfections in second-order analysis

Do first-order global analysis

Do second-order global analysis

Check beams for lateral–torsional buckling, using resistance formulae that include member imperfections (clauses 5.2.2(5) and 5.3.2.3(2))

Were member imperfections for columns included in the global analysis? No

Yes

Is the member in axial compression only? No

Note: for columns, more detail is given in Fig. 6.36

Do second-order analysis for each column, with end actioneffects from the global analysis, including member imperfections, from clause 5.2.2(6)

Verify column cross-sections to clause 6.7.3.6 or 6.7.3.7, from clause 5.2.2(6)

Yes

Use buckling curves that account for second-order effects and member imperfections to check the member (clause 6.7.3.5)

(a) Flow chart, global analysis of a plane frame with composite columns

Fig. 5.1. Global analysis of a plane frame

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Does clause 5.4.2.2(11), on use of a nominal modular ratio, apply? Yes

No For composite beams, determine modular ratios n0 for short-term loading and nL for permanent loads. For a combination of short-term and permanent loading, estimate proportions of loading and determine a modular ratio n from n0 and nL

For composite beams, assume an effective modulus (clause 5.4.2.2(11)) and determine a nominal modular ratio, n

For each composite column, estimate the proportion of permanent to total normal force, determine effective modulus Ec, eff (clause 6.7.3.3(4)), and hence the design effective stiffness, (EI)eff, II, from clause 6.7.3.4(2)

(b) Supplementary flow chart, creep

Determine cracked stiffness for each composite column, to clause 6.7.3.4

Do adjacent spans satisfy clause 5.4.2.3(3)? No

Yes No

Is the frame braced? Yes Are internal joints rigid?

No

Yes

Assume uncracked beams Make appropriate allowances for creep (clause 5.4.2.2) and flexibility of joints (clause 8.2.2) Analyse under characteristic combinations to determine internal forces and moments (clause 5.4.2.3 (2)) and determine cracked regions of beams

Assume cracked lengths for beams (clause 5.4.2.3(3)) Assign appropriate stiffnesses for beams

(c) Supplementary flow chart, cracking of concrete

Can the joint be classified on the basis of experimental evidence or significant experience of previous performance in similar cases? (See EN 1993-1-8 (clause 5.2.2.1(2)) No

Yes

Calculate initial rotational stiffness, Sj, ini (clause 8.3.3, Annex B and EN 1993-1-8 (clause 6.3))

Is the joint nominally-pinned or rigid? Yes

Determine classification by stiffness (clause 8.2.3 and EN 1993-1-8 (clause 5.2))

No Determine rotational stiffness ((clause 8.2.2 and EN 1993-1-8 (clause 5.1.2))

In the model for the frame, assign appropriate rotational stiffness to the joint

(d) Supplementary flow chart, stiffness of joints, for elastic global analysis only

Fig. 5.1. (Contd)

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Determine appropriate allowances for cracking and creep of concrete and for behaviour of joints, clause 5.2.1(4). Assign appropriate stiffnesses to the structure. (See Figs 5.1(b) to (d))

Is the structure a beam-and-column plane frame? Yes

No

Determine acr by use of appropriate software or from the literature

Is axial compression in the beams ‘not significant’? Yes

Determine acr for each storey by EN 1993-1-1 (clause 5.2.1(4)). Determine the minimum value Is acr ≥ 10? Yes First-order analysis is acceptable

No Second-order effects to be taken into account

(e) Supplementary flow chart, choice between first-order and second-order global analysis

Fig. 5.1. (Contd)

Global imperfections Clause 5.3.2.2(1) refers to clause 5.3.2 of EN 1993-1-1. This gives values for the global sway Clause 5.3.2.2(1) imperfections, describes how imperfections may be replaced by equivalent horizontal forces, and permits these to be disregarded if the real horizontal forces (e.g. due to wind) are significant relative to the design vertical load. Member imperfections Clause 5.3.2.3(1) refers to Table 6.5, which gives the amplitudes of the central bow of a Clause 5.3.2.3(1) member designed as straight. It makes little difference whether the curve is assumed to be a half sine wave or a circular arc. These single-curvature shapes are assumed irrespective of the shape of the bending-moment diagram for the member, but the designer has to decide on which side of the member the bow is present. Clauses 5.3.2.3(2) and 5.3.2.3(3) refer to member imperfections that need not be included Clause 5.3.2.3(2) in global analyses. If they are, only cross-section properties are required for checking Clause 5.3.2.3(3) resistances.

5.4. Calculation of action effects 5.4.1. Methods of global analysis

EN 1990 defines several types of analysis that may be appropriate for ultimate limit states. For global analysis of buildings, EN 1994-1-1 gives four methods: linear elastic analysis (with or without redistribution), non-linear analysis and rigid plastic analysis. Clause 5.4.1.1 gives Clause 5.4.1.1 guidance on matters common to more than one method. For reasons of economy, plastic (rectangular stress block) theory is preferred for checking the resistance of cross-sections. In such cases, clause 5.4.1.1(1) allows the action effects to Clause 5.4.1.1(1)

27





Clause 5.4.1.1(2)


Clause 5.4.1.1(7)

Clause 5.4.1.1(8)

be determined by elastic analysis; for composite structures this method has the widest application. Clause 5.4.1.1(2) makes clear that for serviceability limit states, elastic analysis should be used. Linear elastic analysis is based on linear stress/strain laws, but for composite structures, cracking of concrete needs to be considered (clause 5.4.2.3). Other possible non-linear effects include the flexibility of semi-continuous joints (Section 8). Methods for satisfying the principle of clause 5.4.1.1(4) are given for local buckling in clauses 5.4.1.1(5) and 5.4.1.1(6), and for shear lag in concrete in clause 5.4.1.2. Reference is made to the classification of cross-sections. This is the established method of taking account of local buckling of steel elements in compression. It determines the available methods of global analysis and the basis for resistance to bending. The classification system is defined in clause 5.5. There are several reasons28,29 why the apparent incompatibility between the methods used for analysis and for resistance is accepted, as stated in clause 5.4.1.1(1). There is no such incompatibility for Class 3 sections, as resistance is based on an elastic model. For Class 4 sections (those in which local buckling will occur before the attainment of yield), clause 5.4.1.1(6) refers to clause 2.2 of EN 1993-1-1, which gives a general reference to EN 1993-1-5 (‘plated structural elements’).30 This defines those situations in which the effects of shear lag and local buckling in steel plating can be ignored in global analyses. Clause 5.4.1.1(7) reflects a general concern about slip, shared by EN 1993-1-1. For composite joints, clause A.3 gives a method to account for deformation of the adjacent shear connectors. Composite beams have to be provided with shear connection in accordance with clause 6.6. Clause 5.4.1.1(8) therefore permits internal moments and forces to be determined assuming full interaction. For composite columns, clause 6.7.3.4(2) gives an effective flexural stiffness for use in global analysis.

Shear lag in concrete flanges, and effective width Accurate values for effective width of an uncracked elastic flange can be determined by numerical analysis. They are influenced by many parameters and vary significantly along each span. They are increased both by inelasticity and by cracking of concrete. For the bending resistance of a beam, underestimates are conservative, so values in codes have often been based on elastic values. Clause 5.4.1.2 The simplified values given in clause 5.4.1.2 of EN 1994-1-1 are very similar to those used in BS 5950: Part 3.1:199031 and BS ENV 1994-1-1:1994. The values are generally lower than those in EN 1992-1-1 for reinforced concrete T-beams. To adopt those would often increase the number of shear connectors. Without evidence that the greater effective widths are any more accurate, the established values for composite beams have mainly been retained. The effective width is based on the distance between points of contraflexure. In EN 1992-1-1, the sum of the distances for sagging and hogging regions equals the span of the beam. In reality, points of contraflexure are dependent on the load arrangement. EN 1994-1-1 therefore gives a larger effective width at an internal support, to reflect the critical load arrangement for this cross-section. In sagging regions, the assumed distances between points of contraflexure are the same in both codes. Although there are significant differences between effective widths for supports and Clause 5.4.1.2(4) mid-span regions, it is possible to ignore this in elastic global analysis (clause 5.4.1.2(4)). This is because shear lag has limited influence on the results. A small difference from earlier codes for buildings concerns the width of steel flange Clause 5.4.1.2(5) occupied by the shear connectors. Clause 5.4.1.2(5) allows this width to be included within the effective region. Alternatively, it may be ignored (clause 5.4.1.2(9)). Clause 5.4.1.2(8) Clause 5.4.1.2(8) is a reminder that Fig. 5.1 is based on continuous beams. Although clause 8.4.2.1(1) refers to it, clause 5.4.1.2 does not define the effective flange width adjacent to an external column. Figure 5.1 may be used as a guide, or the width may be taken as the width occupied by slab reinforcement that is anchored to the column (see Fig. 8.2).

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Example 5.1: effective width of concrete flange

The notation and method used are those of clause 5.4.1.2. A continuous beam of uniform section consists of two spans and a cantilever, as shown in Fig. 5.2. Values for beff are required for the mid-span regions AB and CD, for the support regions BC and DE, and for the support at A. The calculation is shown in Table 5.1. The result for support A is found from equations (5.4) and (5.5), as follows: beff = 0.2 + [0.55 + (0.025 × 6.8/0.4)] × 0.4 + [0.55 + (0.025 × 6.8/0.85)] × 0.85 = 1.23 m Global analysis may be based on stiffness calculated using the results for AB and CD, but the difference between them is so small that member ABCDE would be analysed as a beam of uniform section. 0.4 0.2

1.4

8

A

10

B

C

2

D

E

Fig. 5.2. Worked example: effective width Table 5.1. Effective width of the concrete flange of a composite T-beam Region

Le (from Fig. 5.2) (m) Le /8 (m) be1 (m) be2 (m) beff (m)

AB

BC

CD

DE

Support A

6.80 0.85 0.40 0.85 1.45

4.5 0.562 0.4 0.562 1.162

7.0 0.875 0.4 0.875 1.475

4.0 0.50 0.4 0.50 1.10

6.80 – 0.40 0.85 1.23

5.4.2. Linear elastic analysis

The restrictions on the use of rigid plastic global analysis (plastic hinge analysis), in clause 5.4.5, are such that linear–elastic global analysis will often be used for composite frames.

Creep and shrinkage There are some differences in clause 5.4.2.2 from previous practice in the UK. The elastic modulus for concrete under short-term loading, Ecm, is a function of the grade and density of the concrete. For normal-weight concrete, it ranges from 30 kN/mm2 for grade C20/25 to 39 kN/mm2 for grade C60/75. With Ea for structural steel given as 210 kN/mm2, the short-term modular ratio, given by n0 = Ea/Ecm, thus ranges from 7 to 5.3. Figure 5.1(b) illustrates the procedure to allow for creep in members of a composite frame. The proportion of loading that is permanent could be obtained by a preliminary global analysis, but in many cases this can be estimated by simpler calculations. For composite beams in structures for buildings where first-order global analysis is acceptable (the majority), clause 5.4.2.2(11) allows the modular ratio to be taken as 2n0 for both short-term and long-term loading – an important simplification, not given in BS 5950. The only exceptions are:

Clause 5.4.2.2

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sc/sc, 0

S 1.0 P

ID

0 Time

Fig. 5.3. Time-dependent compressive stress in concrete, for three types of loading

• • •

structures where second-order global analysis is required by clause 5.2 structures for buildings mainly intended for storage structures prestressed by ‘controlled imposed deformations’ – this would apply, for example, to the bending of steel beams by jacking before concrete is cast around one of the flanges.

Where the conditions of clause 5.4.2.2(11) do not apply, the modular ratio for use in analyses for the effects of long-term loading, nL, depends on the type of loading and the creep coefficient ϕt. This coefficient depends on both the age of the concrete on first loading, t0, and on the age at the time considered in the analysis, which is normally taken as ‘infinity’. The use of this method is excluded for members with both flanges composite; but as these occur mainly in bridges, no alternative is given in Part 1.1. Clause 5.4.2.2(3) Although clause 5.4.2.2(4) gives the age of loading by effects of shrinkage as 1 day, clause 5.4.2.2(3) allows one mean value of t0 to be assumed. If, for example, unpropped construction is used for floor slabs, this might be taken as the age at which they could be subjected to non-trivial imposed loads. These are likely to be construction loads. It makes quite a difference whether this age is assumed to be (for example) 2 weeks or 2 months. From clause 5.4.2.2(2), the values for normal-weight concrete are found from clause 3.1.4 of EN 1992-1-1. Suppose that normal cement is used for grade C25/30 concrete, that the building will be centrally heated, so ‘inside conditions’ apply, and that composite floor slabs with a mean concrete thickness of 100 mm are used. Only one side of the slabs is exposed to drying, so the notional size is 200 mm. The increase in t0 from 14 to 60 days reduces the creep coefficient from 3.0 to about 2.1. The effect of type of loading is introduced by the symbol ψL in the equation nL = n0(1 + ψLϕt)

(5.6)

The reason for taking account of it is illustrated in Fig. 5.3. This shows three schematic curves of the change of compressive stress in concrete with time. The top one, labelled S, is typical of stress caused by the increase of shrinkage with time. Concrete is more susceptible to creep when young, so there is less creep than for the more uniform stress caused by permanent loads (line P). The effects of imposed deformations can be significantly reduced by creep when the concrete is young, so the curve is of type ID. The creep multiplier ψL has the values 0.55, 1.1 and 1.5, respectively, for these three types of loading. The value for permanent loading on reinforced concrete is 1.0. It is increased to 1.1 for composite members because the steel component does not creep. Stress in concrete is reduced by creep less than it would be in a reinforced member, so there is more creep. These application rules are based mainly on extensive theoretical work on composite beams of many sizes and proportions,32 and find application more in design of composite bridges, than in buildings. Clause 5.4.2.2(6) The ‘time-dependent secondary effects due to creep’ referred to in clause 5.4.2.2(6) are most unlikely to be found in buildings. Their calculation is quite complex, and is explained, with an example, in Johnson and Hanswille.33

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For creep in columns, clause 5.4.2.2(9) refers to clause 6.7.3.4(2), which in turn refers to an effective modulus for concrete given in clause 6.7.3.3(4). If separate analyses are to be made for long-term and short-term effects, clause 6.7.3.3(4) can be used, assuming that the ratio of permanent to total load is 1.0 and 0, respectively.

Shrinkage of concrete For the determination of shrinkage strain, reference should be made to the commentary on clause 3.1. The effects in columns are unimportant, except in very tall structures. In beams with the slab above the steel member, shrinkage causes sagging curvature. This is its ‘primary effect’, which is reduced almost to zero where the concrete slab is cracked through its thickness. In continuous beams, the primary curvature is incompatible with the levels of the supports. It is counteracted by bending moments caused by changes in the support reactions, which increase at internal supports and reduce at end supports. The moments and the associated shear forces are the ‘secondary effects’ of shrinkage. Clause 5.4.2.2(7) allows both types of effect to be neglected at ultimate limit states in a Clause 5.4.2.2(7) beam with all cross-sections in Class 1 or 2, unless its resistance to bending is reduced by lateral–torsional buckling. This restriction can be significant. Clause 5.4.2.2(8) allows Clause 5.4.2.2(8) the option of neglecting primary curvature in cracked regions.34 This complicates the determination of the secondary effects, because the extent of the cracked regions has to be found, and the beam then has a non-uniform section. It may be simpler not to take the option, even though the secondary hogging bending moments at internal supports are then higher. These moments, being a permanent effect, enter into all load combinations, and may influence the design of what is often a critical region. The long-term effects of shrinkage are significantly reduced by creep. In the example above, on creep of concrete, ϕt = 3 for t0 = 14 days. For shrinkage, with t0 = 1 day, clause 3.1.4 of EN 1992-1-1 gives ϕt = 5, and equation (5.6) gives the modular ratio as: n = n0(1 + 0.55 × 5) = 3.7n0 Where it is necessary to consider shrinkage effects within the first year or so after casting, a value for the relevant free shrinkage strain can be obtained from clause 3.1.4(6) of EN 1992-1-1. The influence of shrinkage on serviceability verifications is dealt with in Chapter 7.

Effects of cracking of concrete Clause 5.4.2.3 is applicable to both serviceability and ultimate limit states. Figure 5.1(c) Clause 5.4.2.3 illustrates the procedure. In conventional composite beams with the slab above the steel section, cracking of concrete reduces the flexural stiffness in hogging moment regions, but not in sagging regions. The change in relative stiffness needs to be taken into account in elastic global analysis. This is unlike analysis of reinforced concrete structures, where cracking occurs in both hogging and sagging bending, and uncracked cross-sections can be assumed throughout. EN 1994-1-1 provides several different methods to allow for cracking in beams. This is because its scope is both ‘general’ and ‘buildings’. Clause 5.4.2.3(2) provides a general Clause 5.4.2.3(2) method. This is followed in clause 5.4.2.3(3) by a simplified approach of limited application. Clause 5.4.2.3(3) For buildings, a further method is given separately in clause 5.4.4. In the general method, the first step is to determine the expected extent of cracking in beams. The envelope of moments and shears is calculated for characteristic combinations of actions, assuming uncracked sections and including long-term effects. The section is assumed to crack if the extreme-fibre tensile stress in concrete exceeds twice the mean value of the axial tensile strength given by EN 1992-1-1. The cracked stiffness is then adopted for such sections, and the structure re-analysed. This requires the beams with cracked regions to be treated as beams of non-uniform section.

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Table 5.2. Combination factors for imposed load and temperature Action

y0

y1

y2

Imposed load, building in category B Temperature (non-fire) in buildings

0.7 0.6

0.5 0.5

0.3 0

The ‘uncracked’ and ‘cracked’ flexural stiffnesses EaI1 and EaI2 are defined in clause 1.5.2. Steel reinforcement is normally neglected in the calculation of I1. The reasons why stiffness is not reduced to the ‘cracked’ value until an extreme-fibre stress of twice the mean tensile strength of the concrete is reached, are as follows: • • • •

the concrete is likely to be stronger than specified reaching fctm at the surface may not cause the slab to crack right through, and even if it does, the effects of tension stiffening are significant at the stage of initial cracking until after yielding of the reinforcement, the stiffness of a cracked region is greater than EaI2, because of tension stiffening between the cracks the calculation uses an envelope of moments, for which regions of slab in tension are more extensive than they are for any particular loading.

Clause 5.4.2.3(3) Clauses 5.4.2.3(3) to 5.4.2.3(5) provide a non-iterative method, but one that is applicable Clause 5.4.2.3(4) only to some situations. These include conventional continuous composite beams, and Clause 5.4.2.3(5) beams in braced frames. The cracked regions could differ significantly from the assumed values in a frame that resists wind loading by bending. Where the conditions are not satisfied, the general method of clause 5.4.2.3 (2) should be used. Cracking affects the stiffness of a frame, and therefore needs to be considered in the criteria for use of first-order analysis (clauses 5.2.1(3) and 5.2.2(1)). For braced frames within the scope of clause 5.4.2.3(3), the cracked regions in beams are of fixed extent, and the effective stiffness of the columns is given by clause 6.7.3.4(2). The corresponding value of the elastic critical factor αcr can therefore be determined prior to analysis under the design loads. It is then worth checking if second-order effects can be neglected. For unbraced frames, the extent of the cracking can only be determined from analysis under the design loads. This analysis therefore needs to be carried out before the criteria can be checked. It is more straightforward to carry out a second-order analysis, without attempting to prove whether or not it is strictly necessary. Where second-order analysis is necessary, strictly the extent of cracking in beams should take account of the second-order effects. However, as this extent is based on the envelope of internal forces and moments for characteristic combinations, these effects may not be significant. The ‘encasement’ in clause 5.4.2.3(5) is a reference to the partially encased beams defined in clause 6.1.1(1)P. Fully encased beams are outside the scope of EN 1994-1-1. Temperature effects Clause 5.4.2.5(2) Clause 5.4.2.5(2) states that temperature effects, specified in EN 1991-1-5,35 may normally be neglected in analyses for certain situations. Its scope is narrow because it applies to all composite structures, not buildings only. It provides a further incentive to select steel sections for beams that are not weakened by lateral–torsional buckling. Study of the ψ factors of Annex A1 of EN 199036 for combinations of actions for buildings will show, for many projects, that temperature effects do not influence design. This is illustrated for the design action effects due to the combination of imposed load (Q) with temperature (T), for a building with floors in category B, office areas. Similar comments apply to other combinations of actions and types of building. The combination factors recommended in clause A1.2.2(1) of EN 1990 are given in Table 5.2. It is permitted to modify these values in a National Annex. For ultimate limit states, the combinations to be considered, in the usual notation and with the recommended γF factors, are

32





1.35Gk + 1.5(Qk + 0.6Tk)

and

1.35Gk + 1.5(Tk + 0.7Qk)

The second one, with T leading, governs only where Tk > 0.75Qk. Normally, action effects due to temperature are much smaller than those due to imposed load, and additional action effects resulting from the inclusion of T in the first combination are not significant. For serviceability limit states, much depends on the project. Note 2 to clause 3.4(1)P of EN 1990 states: ‘Usually the serviceability requirements are agreed for each individual project’. Similarly, clause A1.4.2(2) of Annex A1 of EN 1990 states, for buildings: ‘The serviceability criteria should be specified for each project and agreed with the client. Note: The serviceability criteria may be defined in the National Annex.’ There are three combinations of actions given in EN 1990 for serviceability limit states: characteristic, frequent and quasi-permanent. The first of these uses the same combination factors ψ0 as for ultimate limit states, and the comments made above therefore apply. The quasi-permanent combination is normally used for long-term effects, and temperature is therefore not included. For the frequent combination, the alternatives are: Gk + 0.5Qk

and

Gk + 0.5Tk + 0.3Qk

The second one governs only where Tk > 0.4Qk. This example suggests that unless there are members for which temperature is the most severe action, as can occur in some industrial structures, the effects T are unlikely to influence verifications for buildings.

Prestressing by controlled imposed deformations Clause 5.4.2.6(2) draws attention to the need to consider the effects of deviations of Clause 5.4.2.6(2) deformations and stiffnesses from their intended or expected values. If the deformations are controlled, clause 5.4.2.6(2) permits design values of internal forces and moments arising from this form of prestressing to be calculated from the characteristic or nominal value of the deformation, which will usually be the intended or measured value. The nature of the control required is not specified. It should take account of the sensitivity of the structure to any error in the deformation. Prestressing by jacking of supports is rarely used in buildings, as the subsequent loss of prestress can be high.

5.4.3. Non-linear global analysis

Clause 5.4.3 adds little to the corresponding clauses in EN 1992-1-1 and EN 1993-1-1, to which it refers. These clauses give provisions, mainly principles, that apply to any method of global analysis that does not conform to clause 5.4.2, 5.4.4 or 5.4.5. They are relevant, for example, to the use of finite-element methods. There is some inconsistency in the use of the term ‘non-linear’ in the Eurocodes. The notes to clauses 1.5.6.6 and 1.5.6.7 of EN 1990 make clear that all of the methods of global analysis defined in clauses 1.5.6.6 to 1.5.6.11 (which include ‘plastic’ methods) are ‘non-linear’ in Eurocode terminology. ‘Non-linear’ in these clauses refers to the deformation properties of the materials. Moderate geometrical non-linearity, such as can occur in composite structures, is allowed for by using analyses defined as ‘second-order’. The much larger deformations that can occur, for example, in some cable-stayed structures, need special treatment. In clause 5.4 of EN 1993-1-1, global analyses are either ‘elastic’ or ‘plastic’, and ‘plastic’ includes several types of non-linear analysis. The choice between these alternative methods should take account of the properties of composite joints given in Section 8 of EN 1994-1-1. Clause 5.7 in EN 1992-1-1 referred to from clause 5.4.3(1) is ‘Non-linear analysis’, which adds little new information. In EN 1994-1-1, non-linear analysis, clause 5.4.3, and rigid plastic analysis, clause 5.4.5, are treated as separate types of global analysis, so that clause 5.4.3 is not applicable where clause

Clause 5.4.3

Clause 5.4.3(1)

33





5.4.5 is being followed. The term ‘non-linear’ is used also for a type of resistance, in clause 6.2.1.4.

5.4.4. Linear elastic analysis with limited redistribution for buildings

Clause 5.4.4(1)

Clause 5.4.4(2)

The concept of redistribution of moments calculated by linear-elastic theory is well established in the design of concrete and composite framed structures. It makes limited allowance for inelastic behaviour, and enables the size of a design envelope of bending moments (from all relevant arrangements of variable loads) to be reduced. In composite beams it is generally easier to provide resistance to bending in mid-span regions than at internal supports. The flexural stiffness at mid-span is higher, sometimes much higher, than at internal supports, so that ‘uncracked’ global analyses overestimate hogging bending moments in continuous beams. A flow chart for this clause is given in Fig. 5.4. Clause 5.4.4(1) refers to redistribution in ‘continuous beams and frames’, but composite columns are not mentioned. At a beam–column intersection in a frame, there are usually bending moments in the column, arising from interaction with the beam. Redistribution for the beam may be done by assuming it to be continuous over simple supports. If the hogging moments are reduced, the bending moments in the column should be left unaltered. If hogging moments are increased, those in the column should be increased in proportion. The clause is applicable provided second-order effects are not significant. Inelastic behaviour results in loss of stiffness, but EN 1994-1-1 does not require this to be taken into account when determining whether the clause is applicable. Although there is considerable experience in using expression (5.1) as a criterion for rigid plastic global analysis of steel frames with full-strength joints, to allow for non-linear material properties, clause 5.2.1(3) of EN 1993-1-1 gives the more severe limit, αcr ≥ 15. This limit is a nationally determined parameter. EN 1994-1-1 retains the limit αcr ≥ 10, but account should be taken of cracking and creep of concrete and the behaviour of joints. One of the requirements of clause 5.4.4(2) is that redistribution should take account of ‘all types of buckling’. Where the shear resistance of a web is reduced to below the plastic value Vpl, Rd to allow for web buckling, and the cross-section is not in Class 4, it would be prudent either to design it for the vertical shear before redistribution, or to treat it as if in Class 4. Although the provisions of clause 5.4.4 appear similar to those of clause 5.2.3.1 of BS 5950-3-1, there are important differences. Some of these arise because the scope of the British standard is limited to conventional composite beams in normal building structures. The Eurocode provisions are not applicable where: (1) (2) (3) (4) (5)

second-order global analysis is required a serviceability or fatigue limit state is being verified the structure is an unbraced frame semi-rigid or partial-strength joints are used beams are partially encased, unless the rotation capacity is sufficient or encasement in compression is neglected (6) the depth of a beam varies within a span (7) a beam with steel of grade higher than S355 has cross-sections in Class 3 or 4 (8) the resistance of the beam is reduced to allow for lateral–torsional buckling. The reasons for these exclusions are now briefly explained: (1) redistribution arises from inelastic behaviour, which lessens the stiffness of the structure and threatens stability (2) fatigue verification is based on elastic analysis (3) the amounts of redistribution given in Table 5.1 have been established considering beams subjected only to gravity loading (4) the amounts of redistribution given in Table 5.1 allow for inelastic behaviour in composite beams, but not for the moment–rotation characteristics of semi-rigid or partial-strength joints

34





Is there a need for second-order global analysis? (Clause 5.4.4(1))

Do first-order elastic global analysis No

Yes No RoM from regions influenced by buckling

No RoM permitted

Yes

Are there any cross-sections or members where resistance is influenced by any form of buckling other than local buckling? (Clause 5.4.4(2)) No

Steel member. Follow clause 5.4(2) of EN 1993-1-1. (Clause 5.4.4(3)

Concrete member. Follow clause 5.5 of EN 1992-1-1. (Clause 5.4.4(3))

Composite member. Are there any partially encased beams without a concrete or composite slab? No

Yes

Does the structure analysed include any composite columns?

Follow the more restrictive of the rules for steel and for concrete members. (Clause 5.4.4(3))

No RoM that reduces moments in columns permitted, unless rotation capacity has been verified. (Inferred from clause 5.4.5(5))

Yes

No Either: check rotation capacity, or: ignore encasement in compression when finding MRd at sections where BM reduced. (Clause 5.4.4(4))

Note: The provisions from here onwards refer to ‘beams’ not to ‘frames’

Is the beam partially encased?

Yes

No Is the beam part of an unbraced frame? (Clause 5.4.4(4))

Yes

No Is the grade of the steel higher than S355? Yes

Yes

Are the conditions of clause 5.4.4(4) on joints and depth of member both satisfied?

No

Are all cross-sections in Class 1 or 2?

Are all cross-sections in Class 1 or 2?

No

Yes

No

Yes

No RoM permitted. (Clause 5.4.4(6)) (END)

No

No RoM permitted (END)

Treat Class 1 sections as Class 2 unless rotation capacity verified. (Clause 5.4.4(6))

No RoM of BMs applied to the steel member. (Clause 5.4.4(7))

Increase maximum hogging BMs by up to 10% for ‘cracked’ analyses and 20% for ‘uncracked’ analyses. (Clause 5.4.4(5))

Maximum hogging BMs may be reduced, by amounts within the limits given in Table 5.1 unless verified rotation capacity permits a higher value. (Clause 5.4.4(5)) (END)

Fig. 5.4. Flow chart for redistribution of moments. Note: this chart applies for verifications for limit states other than fatigue, based on linear-elastic global analysis (RoM, redistribution of bending moments; BM, bending moment)

(5) crushing of the concrete encasement in compression may limit the rotation capacity needed to achieve redistribution; limits to redistribution for partially encased beams can be determined by using the rules for steel members or concrete members, whichever is the more restrictive (clause 5.4.4(3)) (6) the amounts of redistribution given in Table 5.1 have been established only for beams of uniform section

Clause 5.4.4(3)

35





(7) the greater strains associated with higher grade steels may increase the rotation needed to achieve redistribution (8) lateral–torsional buckling may limit the rotation capacity available.

Clause 5.4.4(4) Clause 5.4.4(5) Clause 5.4.4(7)

The last condition can be restrictive, as it may apply where λLT > 0.2 (see clause 6.4.2(4)). However, for composite beams with rolled or equivalent welded steel sections, the value recommended in a note to clause 6.3.2.3(1) of EN 1993-1-1 is λLT > 0.4. The conditions above apply to the percentages given in Table 5.1, which applies only to beams. It should not be inferred that no redistribution is permitted in structures that fail to satisfy one or more of them. It would be necessary to show that any redistribution proposed satisfied clause 5.4.4(2). Under distributed loading, redistribution usually occurs from hogging to sagging regions of a beam (except of course at an end adjacent to a cantilever). The limits to this redistribution in Table 5.1 are based on extensive experience in the use of earlier codes and on research. They have been checked by parametric studies, based on Eurocode 4, of beams in Class 228 and Class 3.29 The limits given in Table 5.1 for ‘uncracked’ analyses are the same as in BS 5950,31 but are more restrictive by 5% for ‘cracked’ analyses of beams in Class 1 or 2. This reflects the finding37 that the difference caused by cracking, between moments calculated by elastic theory in such beams, is nearer 15% than 10%. No provision is made in EN 1994-1-1 for a ‘non-reinforced’ subdivision of Class 1, for which redistribution up to 50% is allowed in BS 5950-3-1. The use of such sections is not prevented by EN 1994-1-1. The requirements for minimum reinforcement given in clause 5.5.1(5) are applicable only if the calculated resistance moment takes account of composite action. The resistance of a non-reinforced cross-section is that of the steel member alone. The use of plastic resistance moments for action effects found by elastic global analysis (clause 5.4.1.1(1)) implies redistribution of moments, usually from internal supports to mid-span regions, additional to the degrees of redistribution permitted, but not required, by clause 5.4.4(4).38 Where there are heavy point loads, and in particular for adjacent spans of unequal length, there can be a need for redistribution from mid-span to supports. This is allowed, to a limited extent, by clause 5.4.4(5). The effects of sequence of construction should be considered where unpropped construction is used and the composite member is in Class 3 or 4. As Table 5.1 makes allowance for inelastic behaviour in a composite beam, clause 5.4.4(7) limits redistribution of moments to those arising after composite action is achieved. No such limitation applies to a cross-section in Class 1 or Class 2, as the moment resistance is determined by plastic analysis and is therefore independent of the loading sequence. The reference in clause 5.4.4(3)(b) to redistribution in steel members is to those that do not subsequently become composite.

5.4.5. Rigid plastic global analysis for buildings

Clause 5.4.5

Plastic hinge analysis, so well known in the UK, is referred to as ‘rigid plastic’ analysis because it is based on the assumption that the response of a member to bending moment is either rigid (no deformation) or plastic (rotation at constant bending moment). Other types of inelastic analysis defined in clause 1.5.6 of EN 1990 are covered in clause 5.4.3. Typical moment–curvature curves are shown in Fig. 5.5. No application rules are given for them because they require purpose-written computer programmes. These other methods are potentially more accurate than rigid plastic analysis, but only if the stress–strain curves are realistic and account is taken of longitudinal slip, as required by clause 5.4.3(2)P. Taking account of the cross-references in clause 5.4.5, the conditions under which use of rigid plastic global analysis is allowed extend over two pages, which need not be summarized here. Development of a collapse mechanism in a composite structure requires a greater

36





Rigid–plastic Bending moment Elastic–perfectly plastic Elasto-plastic

0

Curvature

Fig. 5.5. Moment–curvature curves for various types of global analysis

degree of redistribution of elastic moments than in most steel structures, because beams are usually much stronger at mid-span than at supports. The purpose of the conditions is to ensure that this redistribution, and the large in-plane deformations associated with it, can occur without loss of resistance caused by buckling, fracture of steel, or crushing of concrete. Rigid plastic analysis is applicable only if second-order effects are not significant. Comments on clause 5.4.1.1 referred to the use of expression (5.1) when material behaviour is non-linear. Care needs to be taken if plastic hinges are expected to form in partial-strength joints.39 These may be substantially weaker than the connected members and the plastic hinges may form at relatively low levels of load. It would be prudent to neglect the stiffness of such joints when determining αcr. Clause 5.4.5(1) requires cross-sections of ‘steel members’ to satisfy clause 5.6 of EN 1993-1-1. This applies during unpropped construction, but not to steel elements of composite members, except as provided in clause 5.4.5(6). The provision on rotation capacity in clause 5.6(2) of EN 1993-1-1 is replaced by the conditions of clause 5.4.5(4) of EN 1994-1-1. The rule on neutral axis depth in clause 5.4.5(4)(g) is discussed under clause 6.2.1.2.

Clause 5.4.5(1) Clause 5.4.5(4)

5.5. Classification of cross-sections Typical types of cross-section are shown in Fig. 6.1. The classification of cross-sections of composite beams is the established method of taking account in design of local buckling of steel elements in compression. It determines the available methods of global analysis and the basis for resistance to bending, in the same way as for steel members. Unlike the method in EN 1993-1-1, it does not apply to columns. A flow diagram for the provisions of clause 5.5 is given in Fig. 5.6. The clause numbers given are from EN 1994-1-1, unless noted otherwise. Clause 5.5.1(1)P refers to EN 1993-1-1 for definitions of the four classes and the slendernesses that define the class boundaries. Classes 1 to 4 correspond respectively to the terms ‘plastic’, ‘compact’, ‘ semi-compact’ and ‘slender’ that were formerly used in British codes. The limiting slendernesses are similar to those of BS 5950-3-1.31 The numbers appear different because the two definitions of flange breadth are different, and the coefficient that takes account of yield strength, ε, is defined as ÷(235/fy) in the Eurocodes, and as ÷(275/fy) in BS 5950. The scope of EN 1994-1-1 includes members where the cross-section of the steel component has no plane of symmetry parallel to the plane of its web (e.g. a channel section). Asymmetry of the concrete slab or its reinforcement is also acceptable. The classifications are done separately for steel flanges in compression and steel webs, but the methods interact, as described below. The class of the cross-section is the less favourable of the classes so found (clause 5.5.1(2)), with three exceptions. One is where a web is assumed

Clause 5.5 Clause 5.5.1(1)P

Clause 5.5.1(2)

37





Is steel compression flange restrained by shear connectors to clause 5.5.2(1)?

No

Is the web encased in concrete to clause 5.5.3(2)?

Note 1: ‘Flange’ means steel compression flange

No

Yes From clause 5.5.2(2), classify flange to Table 5.2 of EN 1993-1-1

Classify flange to Table 5.2 of EN 1994-1-1

Yes

Note: See Note 2, below

Compression flange is:

Class 1

Class 2

Class 3

Locate the plastic neutral axis, allowing for partial shear connection, if any

Locate the elastic neutral axis, assuming full shear connection, taking account of sequence of construction, creep, and shrinkage, to clause 5.5.1(4)

From clause 5.5.2(2), classify the web using the plastic stress distribution, to Table 5.2 of EN 1993-1-1

Web in Class 3

Class 4

From clause 5.5.2(2), classify the web using the elastic stress distribution, to Table 5.2 of EN 1993-1-1

Web not classified

Is web encased to clause 5.5.3(2)?

No

Web is Class 3

Yes

Is the compression flange in Class 1 or 2?

Yes

No

Replace web by effective web in Class 2, to clause 5.5.2(3)?

No

Yes Web is Class 1

Web is Class 2

Is the compression flange in Class 3?

Effective web is Class 2

Yes

Is the flange in Class 1? Yes Section is Class 1

No Section is Class 2

Effective section is Class 2

Web is Class 4

Section is Class 3

No

Section is Class 4

Note 2: Where elastic global analysis will be used, and the web will be assumed to resist shear force only, clause 5.5.4(6) of EN 1993-1-1 permits the section to be designed as Class 2, 3 or 4, depending only on the class of the flange

Fig. 5.6. Classification of a cross-section of a composite beam

to resist shear forces only (clause 5.5.2(12) of EN 1993-1-1). The others are the ‘hole-in-web’ option of clause 5.5.2(3) and the use of web encasement, both discussed later. Reference is sometimes made to a beam in a certain class. This means that none of its cross-sections is in a less favourable class than the one stated, and may imply a certain distribution of bending moment. Clause 5.5.1(2) warns that the class of a composite section depends on the sign of the bending moment (sagging or hogging), as it does for a steel section that is not symmetrical about its neutral axis for bending. Designers of structures for buildings normally select beams with steel sections such that the composite sections are in Class 1 or 2, for the following reasons: •

Rigid plastic global analysis is not excluded, provided that the sections at locations of plastic hinges are in Class 1.

38





• • •

Bending resistances of beams can be found using plastic theory. For composite sections, this gives resistances from 20 to 40% above the elastic resistance, whereas the increase for steel sections is about 15%. The limits to redistribution of moments are more favourable than for Classes 3 and 4. Where composite floor slabs are used, it may be difficult to provide full shear connection. Clause 6.6.1.1(14) permits partial connection, but only where all beam cross-sections are in Class 1 or 2.

Simply-supported composite beams in buildings are almost always in Class 1 or 2, because the depth of web in compression (if any) is small, and the connection to the concrete slab prevents local buckling of the adjacent steel flange. Clause 5.5.1(3) refers to this, and clause 5.5.2(1) refers to the more useful clause 6.6.5.5, which limits the spacing of the shear connectors required. Since the class of a web depends on the level of the neutral axis, and this is different for elastic and plastic bending, it is not obvious which stress distribution should be used for a section near the boundary between Classes 2 and 3. Clause 5.5.1(4) provides the answer, the plastic distribution. This is because the use of the elastic distribution could place a section in Class 2, for which the bending resistance would be based on the plastic distribution, which in turn could place the section in Class 3. Clause 5.5.1(5), on the minimum area of reinforcement for a concrete flange, appears here, rather than in Section 6, because it gives a further condition for a cross-section to be placed in Class 1 or 2. The reason is that these sections must maintain their bending resistance, without fracture of the reinforcement, while subjected to higher rotation than those in Class 3 or 4. This is ensured by disallowing the use of bars in ductility Class A (the lowest), and by requiring a minimum cross-sectional area, which depends on the tensile force in the slab just before it cracks.40 Clause 5.5.1(6), on welded mesh, has the same objective. Clause 5.5.1(7) draws attention to the use of unpropped construction, during which both the top flange and the web of a steel beam may be in a lower class until the member becomes composite.

The hole-in-web method This useful device first appeared in BS 5930-3-1.31 It is now in clause 6.2.2.4 of EN 1993-1-1, which is referred to from clause 5.5.2(3). In beams subjected to hogging bending, it often happens that the bottom flange is in Class 1 or 2, and the web is in Class 3. The initial effect of local buckling of the web would be a small reduction in the bending resistance of the section. The assumption that a defined depth of web, the ‘hole’, is not effective in bending enables the reduced section to be upgraded from Class 3 to Class 2, with the advantages for design that are listed above. The method is analogous to the use of effective areas for Class 4 sections, to allow for local buckling. There is a limitation to its scope that is not evident from the wording in EN 1993-1-1:

Clause 5.5.1(3)

Clause 5.5.1(4)

Clause 5.5.1(5)

Clause 5.5.1(6) Clause 5.5.1(7)

Clause 5.5.2(3)

The proportion of the web in compression should be replaced by a part of 20εtw adjacent to the compression flange, with another part of 20εtw adjacent to the plastic neutral axis of the effective cross-section.

It follows that for a design yield strength fyd the compressive force in the web is limited to 40εtw fyd. For a composite beam in hogging bending, the tensile force in the longitudinal reinforcement in the slab can exceed this value, especially where fyd is reduced to allow for vertical shear. The method is then not applicable, because the second ‘part of 20εtw’ is not adjacent to the plastic neutral axis, which lies within the top flange. The method, and this limitation, are illustrated in Examples 6.1 and 6.2.

Partially encased cross-sections Partially encased sections are defined in clause 6.1.1(1)P. Those illustrated there also have concrete flanges. The web encasement improves the resistance of both the web and the other

39





Clause 5.5.3(1)

flange to local buckling. A concrete flange is not essential, as shown in clause 5.5.3(1), which gives the increased slenderness ratios for compression flanges in Classes 2 and 3. The limit for Class 1 is unaltered. The rest of clause 5.5.3 specifies the encasement that enables a Class 3 web to be treated as Class 2, without loss of cross-section. Conditions under which the encasement contributes to the bending and shear resistance of the member are given in Section 6, where relevant comments will be found.

40




CHAPTER 6

Ultimate limit states This chapter corresponds to Section 6 of EN 1994-1-1, which has the following clauses: • • • • • • • •

Beams Resistances of cross-sections of beams Resistance of cross-sections of beams for buildings with partial encasement Lateral–torsional buckling of composite beams Transverse forces on webs Shear connection Composite columns and composite compression members Fatigue

Clause 6.1 Clause 6.2 Clause 6.3 Clause 6.4 Clause 6.5 Clause 6.6 Clause 6.7 Clause 6.8

Clauses 6.1 to 6.7 define resistances of cross-sections to static loading, for comparison with action effects determined by the methods of Section 5. The ultimate limit state considered is STR, defined in clause 6.4.1(1) of EN 1990 as: Internal failure or excessive deformation of the structure or structural members, … where the strength of constructional materials of the structure governs.

For lateral–torsional buckling of beams and for columns, the resistance is influenced by the properties of the whole member, and there is an implicit assumption that the member is of uniform cross-section, apart from variations arising from cracking of concrete and from detailing. The self-contained clause 6.8, ‘Fatigue’, covers steel, concrete and reinforcement by cross-reference to Eurocodes 2 and 3, and deals mainly with shear connection in beams. Most of the provisions of Section 6 are applicable to both buildings and bridges, but a number of clauses are headed ‘for buildings’, and are replaced by other clauses in EN 1994-2. Some of these differences arise from the different treatments of shear connection in the two codes, which are compared in comments on clause 6.1.1.

6.1. Beams 6.1.1. Beams for buildings

Figure 6.1 shows typical examples of beams for buildings within the scope of EN 1994-1-1. The details include web encasement, and profiled sheeting with spans at right angles to the span of the beam, and continuous or discontinuous over the beam. The top right-hand diagram represents a longitudinal haunch. Not shown (and not excluded) is the common situation in which profiled sheeting spans are parallel to the span of the beam. A re-entrant trough is shown in the bottom right-hand diagram. Sheeting with trapezoidal troughs is also within the scope of the code.



Clause 6.1.1



Fig. 6.1. Typical cross-sections of composite beams

The steel cross-section may be a rolled I- or H-section or may be a doubly-symmetrical or mono-symmetrical plate girder. Other possible types include any of those shown in sheet 1 of Table 5.2 of EN 1993-1-1; for example, rectangular hollow sections. Channel and angle sections should not be used unless the shear connection is designed to provide torsional restraint. Stub girders are not within the scope of EN 1994-1-1. There is an extensive literature on their design.41

Clause 6.1.1(4)P

Clause 6.1.1(5)

Shear connection In buildings, composite cross-sections are usually in Class 1 or 2, and the bending resistance is determined by plastic theory. At the plastic moment of resistance, the longitudinal force in a concrete flange is easily found, so design of shear connection for buildings is often based on the change in this force between two cross-sections where the force is known. This led to the concepts of critical cross-sections (clause 6.1.1(4)P to clause 6.1.1(6)) and critical lengths (clause 6.1.1(6)). These concepts are not used in bridge design. Cross-sections in Class 3 or 4 are common in bridges, and elastic methods are used. Longitudinal shear flows are therefore found from the well-known result from elastic theory, vL = VAy/I . Points of contraflexure are not critical cross-sections, partly because their location is different for each arrangement of variable load. A critical length in a continuous beam may therefore include both a sagging and a hogging region. Where connectors are uniformly spaced over this length, the number in the hogging region may not correspond to the force that has to be transferred from the longitudinal slab reinforcement. This does not matter, provided that the reinforcing bars are long enough to be anchored beyond the relevant connectors. The need for consistency between the spacing of connectors and curtailment of reinforcement is treated in clause 6.6.1.3(2)P. A sudden change in the cross-section of a member changes the longitudinal force in the concrete flange, even where the vertical shear is zero. In theory, shear connection to provide this change is needed. Clause 6.1.1(5) gives a criterion for deciding whether the change is sudden enough to be allowed for, and will normally show that changes in reinforcement can be ignored. Where the clause is applied, the new critical section has different forces in the flange on each side of it. It may not be clear which one to use. One method is to use the result that gives the greater change of force over the critical length being considered. An alternative is to locate critical cross-sections on both sides of the change point, not more than about two beam depths apart. The shear connection in the short

42




CHAPTER 6. ULTIMATE LIMIT STATES

critical length between these two sections, based on the longitudinal forces at those sections, needs to take account of the change of section. The application of clause 6.1.1(5) is clearer for a beam that is composite for only part of its length. The end of the composite region is then a critical section. A tapering member has a gradually changing cross-section. This can occur from variation in the thickness or effective width of the concrete flange, as well as from non-uniformity in the steel section. Where elastic theory is used, the equation vL, Ed = VEd A y /I should be replaced by ν L, Ed = VEd Ay/I + MEd

d(Ay/I ) dx

(D6.1)

where x is the coordinate along the member. For buildings, where resistances may be based on plastic theory, clause 6.1.1(6) enables the effect to be allowed for by using additional critical sections. It is applicable, for example, where the steel beam is haunched. The treatment of vertical shear then requires care, as part of it is resisted by the sloping steel flange. Provisions for composite floor slabs, using profiled steel sheeting, are given only for buildings. The space within the troughs available for the shear connection is often insufficient for the connectors needed to develop the ultimate compressive force in the concrete flange, and the resistance moment corresponding to that force is often more than is required, because of other constraints on design. This has led to the use of partial shear connection, which is defined in clause 6.1.1(7). It is applicable only where the critical cross-sections are in Class 1 or 2. Thus, in buildings, bending resistances are often limited to what is needed, i.e. to MEd, with shear connection based on bending resistances; see clause 6.6.2.2. Where bending resistances of cross-sections are based on an elastic model and limiting stresses, longitudinal shear flows can be found from vL, Ed = VEd A y /I. They are related to action effects, not to resistances. Shear connection designed in this way, which is usual in bridges, is ‘partial’ according to the definition in clause 6.1.1(7)P, because increasing it would increase the bending resistances in the vicinity – though not in a way that is easily calculated, because inelastic behaviour and partial interaction are involved. For these reasons, the concept ‘partial shear connection’ is confusing in bridge design and not relevant. Clauses in EN 1994-1-1 that refer to it are therefore labelled ‘for buildings’.

Clause 6.1.1(6)

Clause 6.1.1(7)

Effective cross-section of a beam with a composite slab Where the span of a composite slab is at right angles to that of the beam, as in the lower half of Fig. 6.1, the effective area of concrete does not include that within the ribs. Where the spans are parallel (θ = 0), the effective area includes the area within the depth of the ribs, but usually this is neglected. For ribs that run at an angle θ to the beam, the effective area of concrete within an effective width of flange may be taken as the full area above the ribs plus cos2 θ times the area of concrete within the ribs. Where θ > 60°, cos2 θ should be taken as zero. Service ducts in slabs can cause a significant loss of effective cross-section.

6.1.2. Effective width for verification of cross-sections

The variation of effective width along a span, as given by clause 5.4.1.2, is too complex for verification of cross-sections in beams for buildings. The simplification in clause 6.1.2(2) often enables checks on the bending resistance of continuous beams to be limited to the supports and the mid-span regions. This paragraph should not be confused with clause 5.4.1.2(4), which applies to global analysis.

Clause 6.1.2(2)

6.2. Resistances of cross-sections of beams This clause is for beams without partial or full encasement in concrete. Most of it is applicable to both buildings and bridges. Partial encasement is treated for buildings only, in clause 6.3. Full encasement is outside the scope of EN 1994.

43





No guidance is given in EN 1994-1-1, or in EN 1993, on the treatment of large holes in steel webs, but specialized literature is available.42,43 Bolt holes in steelwork should be treated in accordance with EN 1993-1-1, particularly clauses 6.2.2 to 6.2.6.

6.2.1. Bending resistance

In clause 6.2.1.1, three different approaches are given, based on rigid plastic theory, non-linear theory and elastic analysis. The ‘non-linear theory’ is that given in clause 6.2.1.4. This is not a reference to non-linear global analysis. Clause 6.2.1.1(3) The assumption that composite cross-sections remain plane is always permitted by clause 6.2.1.1(3), where elastic and non-linear theory are used, because the conditions set will be satisfied if the design is in accordance with EN 1994. The implication is that longitudinal slip is negligible. There is no requirement for slip to be determined. This would be difficult because the stiffness of shear connectors is not known accurately, especially where the slab is cracked. Wherever slip may not be negligible, the design methods of EN 1994-1-1 are intended to allow for its effects. For beams with curvature in plan sufficiently sharp for torsional moments not to be Clause 6.2.1.1(5) negligible, clause 6.2.1.1(5) gives no guidance on how to allow for the effects of curvature. In analysis from first principles, checks for beams in buildings can be made by assuming that the changing direction of the longitudinal force in a flange (and a web, if significant) creates a transverse load on that flange, which is then designed as a horizontal beam to resist that load. A steel bottom flange may require horizontal restraint at points within the span of the beam, and the shear connection should be designed for both longitudinal and transverse forces. Clause ‘Full interaction’ in clause 6.2.1.2(1)(a) means that no account need be taken of slip or 6.2.1.2(1)(a) separation at the steel–concrete interface.

Clause 6.2.1.2(1)(c)

Reinforcement in compression It is usual to neglect slab reinforcement in compression (clause 6.2.1.2(1)(c)). If it is included, and the concrete cover is little greater than the bar diameter, consideration should be given to possible buckling of the bars. Guidance is given in clause 9.6.3(1) of EN 1992-1-1 on reinforcement in concrete walls. The meaning is that the reinforcement in compression should not be the layer nearest to the free surface of the slab.

Small concrete flanges Where the concrete slab is in compression, the method of clause 6.2.1.2 is based on the assumption that the whole effective areas of steel and concrete can reach their design strengths before the concrete begins to crush. This may not be so if the concrete flange is small compared with the steel section. This lowers the plastic neutral axis, and so increases the maximum compressive strain at the top of the slab, for a given tensile strain in the steel bottom flange. A detailed study of the problem has been reported.44 Laboratory tests on beams show that strain hardening of steel usually occurs before crushing of concrete. The effect of this, and the low probability that the strength of both the steel and the concrete will be only at the design level, led to the conclusion that premature crushing can be neglected unless the grade Clause 6.2.1.2(2) of the structural steel is higher than S355. Clause 6.2.1.2(2) specifies a reduction in Mpl, Rd where the steel grade is S420 or S460 and the depth of the plastic neutral axis is high. This problem also affects the rotation capacity of plastic hinges. Extensive research45,46 led to the upper limit to neutral-axis depth given in clause 5.4.5(4)(g). For composite columns, the risk of premature crushing led to a reduction in the factor αM, given in clause 6.7.3.6(1), for S420 and S460 steels.

44





Fig. 6.2. Example of a composite beam with the slab in tension at mid-span

Ductility of reinforcement Reinforcement with insufficient ductility to satisfy clause 5.5.1(5), and welded mesh, should not be included within the effective section of beams in Class 1 or 2 (clause 6.2.1.2(3)). This is Clause 6.2.1.2(3) because laboratory tests on hogging moment regions have shown19 that some reinforcing bars, and most welded meshes, fracture before the moment–rotation curve for a typical double-cantilever specimen reaches a plateau. The problem with welded mesh is explained in comments on clause 3.2(1). Profiled steel sheeting The contribution of profiled steel sheeting in compression to the plastic moment of resistance of a beam is ignored (clause 6.2.1.2(4)) because at large strains its resistance can Clause 6.2.1.2(4) be much reduced by local buckling. Profiled sheeting with troughs that are not parallel to the span of a beam is ineffective in tension. This is because deformation could arise from change in shape of the profile rather than strain resulting from stress. Where the troughs are parallel, resistance to tension may still be difficult to achieve. For advantage to be taken Clause 6.2.1.2(5) of clause 6.2.1.2(5), the sheeting needs to be continuous, and interaction with other components of the cross-section has to be achieved. Beams with partial shear connection in buildings The background to the use of partial shear connection is explained in comments on clause 6.1.1. It is permitted only for the compressive force in the concrete slab (clause 6.2.1.3(1)). Clause 6.2.1.3(1) Where the slab is in tension the shear connection must be sufficient to ‘ensure yielding’ (clause 6.2.1.3(2)) of the reinforcement within the effective section. Full shear connection is Clause 6.2.1.3(2) required in hogging regions of composite beams for several reasons: • • • •

the bending moment may be larger than predicted because the concrete has not cracked or, if it has, because of tension stiffening the yield strength of the reinforcement exceeds fsd ( = fsk/γS) tests show that at high curvatures, strain hardening occurs in the reinforcement the design rules for lateral–torsional buckling do not allow for the effects of partial interaction.

It could be inferred from the definition of full shear connection in clause 6.1.1(7) that where the bending resistance is reduced below Mpl, Rd by the effects of lateral buckling, shear connection is required only for the reduced resistance. Clause 6.2.1.3(2) makes it clear that the inference is incorrect. Thus, clause 6.2.1.2 on the plastic resistance moment Mpl, Rd applies, amongst other cases, to all beams in Class 1 or 2 with tensile force in the slab. The words ‘hogging bending’ in clause 6.2.1.3 imply that the concrete slab is above the steel beam. This is an assumption implicit in much of the drafting of the provisions ‘for buildings’. In the ‘general’ clauses, phrases such as ‘regions where the slab is in tension’ are used instead, because in bridges this can occur in regions of sagging curvature (Fig. 6.2). The provisions referred to in clause 6.2.1.3(1) include clause 6.6.1.1(14), which begins ‘If all cross-sections are in Class 1 or Class 2 …’. This means all sections within the span considered. In practice, the use of an effective web in Class 2 (clause 5.5.2(3)) ensures that few Class 3 sections need be excluded.

45





Ductile connectors Clause 6.2.1.3(3) Clause 6.2.1.3(3) refers to ‘ductile connectors’. The basic condition for the use of partial shear connection is that the bending resistance must not fall below the design value until after the curvature has reached the minimum value relied upon in the method of global analysis used. Use of redistribution of moments, for example, relies upon curvatures beyond the elastic range. In other words: slip required (i.e. relied on in design) £ slip available

(D6.2) 47,48

that It has been shown by extensive numerical analyses, checked against test results, the slip required increases with the span of the beam and, of course, as the number of shear connectors is reduced. The latter parameter is represented by the ratio of the number of connectors provided within a critical length, n, to the number nf required for ‘full shear connection’ (defined in clause 6.1.1(7)P), i.e. the number that will transmit the force Nc, f to the slab (see Fig. 6.2). A reduced number, n, will transmit a reduced force, Nc. Thus, for connectors of a given shear strength, the ‘degree of shear connection’ is η = Nc /Nc, f = n/nf

(D6.3)

The application rules for general use are based on an available slip of 6 mm. Condition (D6.2) was then applied by defining combinations of η and span length such that the slip required did not exceed 6 mm. Headed studs regarded as ductile are defined in clause 6.6.1.2, where a flow chart (see Fig. 6.11) and further comments are given. For partial shear connection with non-ductile connectors, reference should be made to comments on clause 6.2.1.4.

Calculation models The calculation model given in clause 6.2.1.3(3) can be explained as follows. For a given cross-section, the force Nc, f can be found using clause 6.2.1.2. For η < 1, the concrete stress block has a reduced depth, and a neutral axis at its lower edge. For longitudinal equilibrium, part of the steel beam must also be in compression, so it too has a neutral axis. The model assumes no separation of the slab from the beam, so their curvatures must be the same. The strain distribution is thus as shown in Fig. 6.3, which is for the situation shown in Fig. 6.4 in EN 1994-1-1. At the interface between steel and concrete there is slip strain, i.e. rate of change of longitudinal slip. Neither the slip at this point nor the slip strain need be calculated in practice. Clause 6.2.1.3(4) Clauses 6.2.1.3(4) and 6.2.1.3(5) give two relationships between resistance moment MRd Clause 6.2.1.3(5) and degree of shear connection. Calculations using the method above give curve AHC in the upper part of Fig. 6.4(a), in which Mpl, a, Rd is the plastic resistance of the steel section. The line AC is a simpler and more conservative approximation to it. Their use is now illustrated. The lower half of Fig. 6.4(a) shows the limits to the use of partial shear connection given in clause 6.6.1.2.

0.85fcd

Nc = hNc, f

MRd fyd

Ma Na

fyd

0 Stresses

Strains

Fig. 6.3. Plastic stress and strain distributions under sagging bending for partial shear connection

46





C

Mpl, Rd MEd

H

G

C

Mpl, Rd

Mel, Rd

Q

K

B

Mpl, a, Rd A 0

J

0.4

U

Ma, Ed

N

V

Nc/Nc, f Nc, el /Nc, f

F

1.0

(b) Non-ductile connectors

Q

20

P

0

1.0

h (= Nc/Nc, f)

Le (m) 25

S Studs to clause 6.6.1.2(3), Ab = At

P

fy = 275

10 D

Other studs, Ab = At, from clause 6.6.1.2(1)

fy = 460

E

Other studs, Ab = 3At

R 0

T

MEd

M

0.4

0.6

0.8

1.0

n /n f (a) Ductile connectors

Fig. 6.4. Design methods for partial shear connection

Outline of a typical design procedure The example is a simply-supported beam of span L, with a Class 1 section at mid-span. With full shear connection the resistance is Mpl, Rd. The steel section has flanges of equal area, and the loading is uniformly distributed. Stud connectors are to be used. (1) Find the minimum shear connection, (n/nf)min, for which the connectors are ductile, from clause 6.6.1.2 (e.g. route DEF in Fig. 6.4(a)), and the corresponding resistance to bending (route FMB). (2) If that resistance exceeds the design moment MEd, this degree of shear connection is sufficient. Calculate the number of connectors for full shear connection, and then, from (n/nf)min, the number required. (3) If the resistance, point B, is much higher than MEd, it may be possible to reduce the number of connectors by using the method for non-ductile connectors, as explained in Example 6.4 below. (4) If the resistance, point B, is below MEd, as shown in Fig. 6.4(a), the interpolation method can be used (route GKN) to give the value of n/nf required. Alternatively, point H can be determined, as shown below, and hence point J. The higher of the values of η given by points J and F is the minimum degree of shear connection, and, hence, n can be found. (5) The spacing of the n connectors is now considered, along the length Lcr between the two relevant critical cross-sections (here, mid-span and a support). If the conditions of clause 6.6.1.3(3) are satisfied, as they are here if Mpl, Rd £ 2.5Ma, pl, Rd, the spacing may be

47





beff Nc = 0.85beffxcfcd xc

hc

hp hg

xa

tf

Nac = 2bxafyd

b G

Na = Asfyd

(b)

(a)

Fig. 6.5. Theory for force Nc. (a) Cross-section. (b) Longitudinal stresses

uniform. Otherwise, an intermediate critical section must be chosen, clause 6.6.1.3(4), or the spacing must be related to the elastic distribution of longitudinal shear, clause 6.6.1.3(5).

Determination of n for a given MEd by the equilibrium method The number of connectors needed to develop the moment MEd is n = Nc /PRd, where PRd is the design resistance of a connector and Nc is the force referred to in clause 6.2.1.3(3), for a moment MEd. Its calculation is tedious, but can be simplified a little, as shown here. Figure 6.5 shows a cross-section of a beam in sagging bending, where the compressive force in the concrete slab is Nc, less than Nc, f (equation (D6.3)), being limited by the strength of the shear connection. Following clause 6.2.1.3(3), the plastic stress blocks are as shown in Fig. 6.5. The neutral axis in the steel is at a depth xa below the interface. It is convenient to retain the known force Na acting at the centre of area G of the steel section, and to take the compressive strength over depth xa as 2fyd. This is because the stress in that area is to be changed from yield in tension to yield in compression, providing a compressive force Nac in the steel. We need to find out whether or not xa > tf. In most beams with full shear connection, Nc, f = Na, because the plastic neutral axis lies within the slab. It is required that Nc /Nc, f ≥ 0.4, from Fig. 6.4(a), so Nc ≥ 0.4Nc, f ≥ 0.4Na

(a)

From equilibrium, Nc + Nac = Na

(b)

so, from expression (a), Nac £ 0.6Na

(c)

When the neutral axis is at the underside of the steel top flange, of area Atop, Nac /Na = 2Atop /Aa

(d)

For most rolled I-sections, Atop ≥ 0.3Aa, so when xa = tf, from equation (d), Nac ≥ 0.6Na

(e)

From expressions (c) and (e), xa < tf when the preceding assumptions are valid. They usually are, so only the case xa < tf is considered. Force Nac is then as shown in Fig. 6.5, and acts at a depth hc + hp + 0.5xa below the top of the slab. For rolled sections, xa = hc + hp, so this depth can be taken as hc + hp. Taking moments about the top of the slab,

48





MEd = Na(hg + hp + hc) – ½xc Nc – Nac(hc + hp)

(D6.4)

Substitution for Nac from equation (b) and use of the expression for Nc in Fig. 6.5 gives MEd = Na hg + 0.85beff xc fcd (hc + hp – ½xc) This can be solved for xc, which gives Nc, and then n from n = Nc /PRd , since PRd is known for given connectors. In practice, it may be simpler to calculate Na, choose a convenient value for n, find Nc and then xc, calculate Nac from equation (b), and see whether or not equation (D6.4) gives a value that exceeds MEd. If it does not, n is increased and the process repeated. There is much interaction between clause 6.2.1.3 and clauses 6.6.1.2 and 6.6.1.3. The use of partial shear connection is illustrated in Example 6.7, which follows the comments on clause 6.6, in Examples 6.8 and 6.9, which are based on the same data, and in Fig. 6.11.

Non-ductile connectors These are connectors that do not satisfy the requirements for ductile connectors given in clauses 6.6.1.1 and 6.6.1.2. Plastic behaviour of the shear connection can no longer be assumed. Non-linear or elastic theory should now be used to determine resistance to bending. Provisions are given in clause 6.2.1.4 and clause 6.2.1.5. Clause 6.2.1.4 The effect of slip at the steel–concrete interface is to increase curvature, and usually to reduce longitudinal shear, for a given distribution of bending moment along a span. Where connectors are not ‘ductile’, slip must be kept small, so it is rational to neglect slip when calculating longitudinal shear. This is why clause 6.2.1.4(2) says that cross-sections should be Clause 6.2.1.4(2) assumed to remain plane. Non-linear resistance to bending There are two approaches, described in clause 6.2.1.4. With both, the calculations should be done at the critical sections for the design bending moments. The first approach, given in clause 6.2.1.4(1) to 6.2.1.4(5), enables the resistance of a section to be determined iteratively from the stress–strain relationships of the materials. A strain distribution is assumed for the cross-section, and the resulting stresses determined. Usually, the assumed strain distribution will have to be revised, to ensure that the stresses correspond to zero axial force on the section. Once this condition is satisfied, the bending moment is calculated from the stress distribution. This may show that the design bending moment does not exceed the resistance, in which case the calculation for bending resistance may be terminated. Otherwise, a general increase in strain should be made and the calculations repeated. For concrete, EN 1992-1-1 gives ultimate strains for concrete and reinforcement which eventually limit the moment resistance. Clearly, in practice this procedure requires the use of software. For sections in Class 1 or 2, a simplified approach is given in clause 6.2.1.4(6). This is based on three points on the curve relating longitudinal force in the slab, Nc, to design bending moment MEd that are easily determined. With reference to Fig. 6.4(b), which is based on Fig. 6.6, these points are: • • •

Clause 6.2.1.4(1) Clause 6.2.1.4(2) Clause 6.2.1.4(3) Clause 6.2.1.4(4) Clause 6.2.1.4(5)

Clause 6.2.1.4(6)

P, where the composite member resists no moment, so Nc = 0 Q, which is defined by the results of an elastic analysis of the section C, based on plastic analysis of the section.

Accurate calculation shows QC to be a convex-upwards curve, so the straight line QC is a conservative approximation. Clause 6.2.1.4(6) thus enables hand calculation to be used. For buildings, clause 6.2.1.4(7) refers to a simplified treatment of creep. Clause 6.2.1.4(7)

Shear connection to clause 6.2.1.4 Computations based on the stress–strain curves referred to in clause 6.2.1.4(3) to 6.2.1.4(5) lead to a complete moment-curvature curve for the cross-section, including a falling branch. The definition of partial shear connection in clause 6.1.1(7)P is unhelpful, because the

49





number of connectors for full shear connection can be found only from the stress distribution corresponding to the maximum moment. This is why Nc, f in Fig. 6.4(b) is based on plastic analysis to clause 6.2.1.2. This figure is approximately to scale for a typical composite section in sagging bending, for which Mpl, Rd /Mel, Rd = 1.33 Nc, el/Nc, f = 0.6 The position of line PQ depends on the method of construction. It is here assumed that the beam was unpropped, and that at the critical section for sagging bending, the moment Ma, Ed applied to the steel alone was 0.25Mpl, Rd. For a given design moment MEd (which must include Ma, Ed, as resistances of sections are found by plastic theory), the required ratio n/nf is given by route TUV in Fig. 6.4(b). No specific guidance is given for the spacing of shear connectors when moment resistance is determined by non-linear theory. The theory is based on plane cross-sections (clause 6.2.1.4(2)), so the spacing of connectors should ideally correspond to the variation of the force in the slab, Nc, along the member. Where non-ductile connectors are used, this should be done. For ductile connectors, clause 6.6.1.3(3), which permits uniform spacing, may be assumed to apply.

Elastic resistance to bending Clause 6.2.1.4(6) includes, almost incidentally, a definition of Mel, Rd that may seem strange. It is a peculiarity of composite structures that when unpropped construction is used, the elastic resistance to bending depends on the proportion of the total load that is applied before the member becomes composite. Let Ma, Ed and MEd be the design bending moments for the steel and composite sections, respectively, for a section in Class 3. Their total is typically less than the elastic resistance to bending, so to find Mel, Rd, one or both of them must be increased Clause 6.2.1.5(2) until one or more of the limiting stresses in clause 6.2.1.5(2) is reached. To enable a unique result to be obtained, clause 6.2.1.4(6) says that MEd is to be increased, and Ma, Ed left unchanged. This is because Ma, Ed is mainly from permanent actions, which are less uncertain than the variable actions whose effects comprise most of MEd. Unpropped construction normally proceeds by stages, which may have to be considered individually in bridge design. For simply-supported spans in buildings, it is usually sufficiently accurate to assume that the whole of the wet concrete is placed simultaneously on the bare steelwork. The weight of formwork is, in reality, applied to the steel structure and removed from the composite structure. This process leaves self-equilibrated residual stresses in composite cross-sections. For composite beams in buildings, these can usually be ignored in calculations for the final situation. Clause 6.2.1.5(5) One permanent action that influences MEd is shrinkage of concrete. Clause 6.2.1.5(5) enables the primary stresses to be neglected in cracked concrete, but the implication is that they should be included where the slab is in compression. This provision, which affects Mel, Rd, should not be confused with clause 5.4.2.2(8), which concerns global analysis to determine the secondary effects of shrinkage in statically-indeterminate structures. (Secondary effects are defined in clause 2.3.3.) These complications explain why, for ultimate limit states in buildings, design methods based on elastic behaviour are best avoided, as far as possible.

Example 6.1: resistance moment in hogging bending, with effective web

A typical cross-section near an internal support of a continuous composite beam is shown in Fig. 6.6(a). The plastic and elastic methods of calculation for the hogging moment of resistance are illustrated by preparing a graph that shows changes in this resistance as the effective area of longitudinal reinforcement in the slab, As, is increased from zero to

50





80

As

116 100

70

0

116

19

New pna

100

fyd

0 286

ht = 152 ac – 40te

D

20te

c = 360

26

pna ac

2(ac – 40te)

Hole

102

116 B

C

102

19

8.6

fyd fyd

142 (a)

228

222

228

328

hh = 4

199

20te

6.3

95

340

A

0

0

(b)

(c)

Fig. 6.6. Plastic resistance moment in hogging bending, for As = 267 mm2 (units: mm and kN). (a) Cross-section of Class 3 beam with hole in web. (b) Stress blocks for Mpl, Rd for Class 2 beam. (c) Stress blocks in web for Mpl, Rd for Class 3 beam

1800 mm2. The use of an effective web to clause 5.5.2(3), discussed in Chapter 5, is also illustrated. The depth of the ‘web’, as used here, is the depth between flange fillets (or welds), defined as c in Table 5.2 of EN 1993-1-1, not the clear depth between the flanges. Also from EN 1993-1-1, αc is the depth of the web in compression, for a section in Class 1 or 2, and ε is the correction factor for yield strength of the steel. The definition of the hole-in-web model, given in Fig. 6.3 of EN 1993-1-1, omits the depth of the hole and the location of the new plastic neutral axis (pna). They were given in ENV 1994-1-1,49 and are shown in Fig. 6.6(a). The depth of the hole, 2(αc – 40tε), includes a small approximation, which is now explained. In principle, the ‘hole’ should have zero depth when α, c, t and ε are such that the web is on the boundary between Class 2 and Class 3. When α > 0.5 (as is usual), and from Table 5.2 of EN 1993-1-1, this is when αc/tε = 456α/(13α – 1) As α increases from 0.5 to 1.0, the right-hand side of this equation reduces from 41.4 to 38. For α < 0.5, it is 41.5. For simplicity, it is taken as 40 for all values of α, so that the depth of web in compression can be defined as 40tε, in blocks of depth 20tε above and below the hole. The original depth in compression, αc, is reduced to 40tε. For equilibrium, the depth in tension must be reduced by αc – 40tε, so the plastic neutral axis moves up by this amount, as shown in Fig. 6.6(a), and the depth of the hole is thus 2(αc – 40tε). Other useful results for a symmetrical steel section are as follows. The depth of web in compression, 40tε, includes the depth needed to balance the tensile force in the reinforcement, which is hr = As fsd /tfyd

(D6.5)

The tension in the top flange, including web fillets, balances the compression in the bottom flange, so for longitudinal equilibrium the depth of web in tension is ht = 40tε – hr

(D6.6)

The total depth of the web is c, so the depth of the hole is hh = c – 40tε – ht = c – 80tε + hr

(D6.7)

51





M (kN m) 340

C

Mpl, Rd Class 2

B

320 H

Mpl, Rd Class 3, hole in web

300 A 280

Mel, Rd bottom flange yields

F

260 J 240

Mel, Rd reinforcement yields G

220

E

200 0

500

1000

1500

2000

As (mm2)

Fig. 6.7. Influence of longitudinal reinforcement on hogging moments of resistance

Data and results Data for the calculations, in addition to the dimensions shown in Fig. 6.6, are: • • •

structural steel: fy = 355 N/mm2, γM = 1.0, so fyd = 355 N/mm2 reinforcement: fsk = 500 N/mm2, γS = 1.15, so fsd = 435 N/mm2 steel section: 406 × 140 UB39 with Aa = 4940 mm2, Ia = 124.5 × 106 mm4.

The upper limit chosen for As corresponds to a reinforcement ratio of 1.5%, which is quite high for a beam in a building, in a flange with beff = 1.5 m. If there were no profiled sheeting (which plays no part in these calculations) the ratio for the 150 mm slab would be 0.8%, and there would be two layers of bars. For simplicity, one layer of bars is assumed here, with propped construction. The full results are shown in Fig. 6.7. Typical calculations only are given here.

Classification of the cross-section Clause 5.5.1(1)P refers to EN 1993-1-1, where Table 5.2 applies. For fy = 355 N/mm2, it gives ε = 0.81. For the bottom flange, c = (142 – 6.3)/2 – 10.2 = 57.6 mm c/tε = 57.6/(8.6 × 0.81) = 8.3 This is less than 9, so the flange is in Class 1, irrespective of the area of slab reinforcement. The web depth between fillets is c = 360 mm, so c/tε = 360/(6.3 × 0.81) = 70.5

52





This is less than 72, so when As = 0, the web is in Class 1. Addition of reinforcement increases the depth of web in compression, so its class depends on As. Here, α > 0.5, and from Table 5.2 in EN 1993-1-1 the limit for Class 2 is c/tε = 456/(13α – 1) For c/tε = 70.5 this gives α = 0.574. It will be shown that this corresponds to As = 267 mm2, by calculating Mpl, Rd by the methods for both a Class 2 and a Class 3 section.

Mpl, Rd for Class 2 section with As = 267 mm2 The stress blocks for the web are shown in Fig. 6.6(b). (1) Find Mpl, a, Rd for the steel section. For a rolled section, the plastic section modulus is usually found from tables. Here, Wpl = 0.721 × 106 mm3 so Mpl, a, Rd = 0.721 × 355 = 256 kN m (2) Find Fs, the force in the reinforcement at yield: Fs = 267 × 0.435 = 116 kN From equation (D6.5), the depth of web for this force is hr = 116/(6.3 × 0.355) = 52 mm To balance the force Fs, the stress in a depth of web hr/2 changes from +fyd to –fyd. This is shown as ABCD in Fig. 6.6. (3) The lever arm for the forces Fs is 286 mm, so taking moments, Mpl, Rd = 256 + 116 × 0.286 = 289 kN m

Mpl, Rd for Class 3 section with As = 267 mm2 The hole-in-web method is now used. Steps 1 and 2 are as above. (3) The contribution of the web is deducted from Mpl, a, Rd : Mpl, a, flanges = 256 – 0.362 × 6.3 × 355/4 = 183.5 kN m (This value is not calculated directly because of the complex shape of each ‘flange’ which includes the web fillets, or, for a plate girder, a small depth of web.) (4) From equation (D6.7), the depth of the hole is hh = 360 – 408 + 52 = 4 mm (5) From equation (D6.6), the depth of web in tension is ht = 204 – 52 = 152 mm and the force in it is Ft = 152 × 6.3 × 0.355 = 340 kN (6) The stress blocks in the web are shown in Fig. 6.6(c). Taking moments about the bottom of the slab, Mpl, Rd = 183.5 + 116 × 0.1 – 340 × 0.095 + 228(0.222 + 0.328) = 288 kN m This agrees with the result for the Class 2 member, point A in Fig. 6.7, because this section is at the class boundary. Hence, the depth of the hole is close to zero.

53





Mpl, Rd for As > 267 mm2 Similar calculations for higher values of As give curve AB in Fig. 6.7, with increasing depths of hole until, at As = 1048 mm2, the new plastic neutral axis reaches the top of the web, ht = 0, and the hole reaches its maximum depth. Further increase in As, ∆As say, causes changes of stress within the top flange only. Taking moments about the interface, it is evident that the plastic bending resistance is increased by ∆Mpl, Rd ª ∆As fyd hs where hs is the height of the reinforcement above the interface, 100 mm here. This is shown by line BC in Fig. 6.7. The bending resistance given by this method no longer approaches that given by elastic theory (as it should, as the slenderness of the web in compression approaches the Class 3/4 boundary). Use of the method with the new plastic neutral axis in the top flange is excluded by Eurocode 3, as explained in comments on clause 5.5.2(3). It is, in any case, not recommended because: • •

the authors are not aware of any experimental validation for this situation (which is uncommon in practice) Mpl, Rd is being calculated using a model where the compressive strain in the steel bottom flange is so high that the rotation capacity associated with a Class 2 section may not be available.

Equation (D6.6) shows that this restriction is equivalent to placing an upper limit of 40tε on hr. Any slab reinforcement that increases hr above this value moves the section back into Class 3. This can also be a consequence of vertical shear, as illustrated in Example 6.2. This point is relevant to the writing of software based on the code because software, once written, tends to be used blindly.

Elastic resistance to bending For simplicity, Mel, Rd has been calculated assuming propped construction. The stress in the reinforcement governs until As reaches 451 mm2 (point J), after which Mel, Rd is determined by yield of the bottom flange. This is shown by curves EF and GH in Fig. 6.7. Web in Class 4 The hole-in-web method is available only for webs in Class 3, so in principle a check should be made that the web is not in Class 4, using the elastic stress distribution. A Class 4 web can occur in a plate girder, but is most unlikely in a rolled I- or H-section. In this example, the Class 3/4 boundary is reached at As = 3720 mm2.

Clause 6.2.2

6.2.2. Resistance to vertical shear

Clause 6.2.2 is for beams without web encasement. The whole of the vertical shear is usually assumed to be resisted by the steel section, as in previous codes for composite beams. This enables the design rules of EN 1993-1-1, and EN 1993-1-530 where necessary, to be used. The assumption can be conservative where the slab is in compression. Even where it is in tension and cracked in flexure, consideration of equilibrium shows that the slab must make some contribution to shear resistance, except where the reinforcement has yielded. For solid slabs, the effect is significant where the depth of the steel beam is only twice that of the slab,50 but diminishes as this ratio increases. In composite plate girders with vertical stiffeners, the concrete slab can contribute to the anchorage of a tension field in the web,51 but the shear connectors must then be designed for Clause 6.2.2.3(2) vertical forces (clause 6.2.2.3(2)). The simpler alternative is to follow Eurocode 3, ignoring both the interaction with the slab and vertical tension across the interface.

54





VEd/VRd VRd is the lesser of Vpl, Rd and Vb, Rd

326

1.0

100 43 48

0.5 132

or Mpl, Rd (a)

213

102

Mel, Rd

118

Mpl, a, fl

180

0

Mb, Rd

213

102 108

Bending resistance

Mf, Rd

100

328

fyd (b)

Fig. 6.8. Resistance to bending and vertical shear (dimensions in mm)

Bending and vertical shear The methods of clause 6.2.2.4 are summarized in Fig. 6.8. Shear stress does not significantly Clause 6.2.2.4 reduce bending resistance unless the shear is quite high. For this reason, the interaction may Clause 6.2.2.4(1) be neglected until the shear force exceeds half of the shear resistance (clause 6.2.2.4(1)). Both EN 1993-1-1 and EN 1994-1-1 use a parabolic interaction curve. In clause 6.2.2.4(2) Clause 6.2.2.4(2) the reduction factor for the design yield strength of the web is (1 – ρ), where ρ = [(2VEd /VRd) – 1]2

(6.5)

and VRd is the resistance in shear. For a design shear force equal to VRd, the bending resistance is that provided by the flanges alone, denoted Mf, Rd. This is calculated in Example 6.2. The bending resistance at VEd = 0 may be the elastic or the plastic value, depending on the class of the cross-section. Where it is reduced to Mb, Rd by lateral–torsional buckling, interaction between bending and shear does not begin until a higher shear force than VRd /2 is present, as shown in Fig. 6.8(a). Where the shear resistance VRd is less than the plastic resistance to shear, Vpl, Rd, because of shear buckling, clause 6.2.2.4(2) replaces Vpl, Rd by the shear buckling resistance Vb, Rd. Where the design yield strength of the web is reduced to allow for vertical shear, the effect on a Class 3 section in hogging bending is to increase the depth of web in compression. If the change is small, the hole-in-web model can still be used, as shown in Example 6.2. For a higher shear force, the new plastic neutral axis may be within the top flange, and the hole-in-web method is inapplicable. The section is then treated as Class 3 or 4, and clause 6.2.2.4(3) applies. It refers to Clause 6.2.2.4(3) EN 1993-1-5. For beams, the rule given there is essentially MEd /MRd + (1 – Mf, Rd /Mpl, Rd)(2VEd /VRd – 1)2 £ 1

(D6.8)

These symbols relate to the steel section only. For a composite section, longitudinal stresses are found by elastic theory. These lead to values of MEd, NEd and VEd acting on the steel section, which is then checked to EN 1993-1-5. It is fairly easy to check if a given combination of these action effects can be resisted – but calculation of bending resistance for a given vertical shear is difficult.

Example 6.2: resistance to bending and vertical shear

Vertical shear is more likely to reduce resistance to bending in a continuous beam than in a simply-supported one, and it is instructive to consider its influence on a beam with bending resistance found by the hole-in-web method. Where the web is not susceptible to shear buckling, the application of clause 6.2.2.4 is straightforward. This example is therefore based on one of the few UB sections where web buckling can occur, if S355 steel

55





is used. It is the section used in Example 6.1, 406 × 140 UB39, shown in Fig. 6.6, with longitudinal reinforcement of area As = 750 mm2. The bending resistance will be calculated when the design vertical shear is VEd = 300 kN. All other data are as in Example 6.1. From EN 1993-1-1, resistance to shear buckling must be checked if hw/tw > 72ε/η, where η is a factor for which EN 1993-1-5 recommends the value 1.2. These clauses are usually applied to plate girders, for which hw is the clear depth between the flanges. Ignoring the corner fillets of this rolled section, hw = 381 mm, and for S355 steel, ε = 0.81, so hwη/twε = 381 × 1.2/(6.3 × 0.81) = 90.5 The resistance of this unstiffened web to shear buckling is found using clauses 5.2 and 5.3 of EN 1993-1-5, assuming a web of area hwtw, that there is no contribution from the flanges, and that there are transverse stiffeners at the supports. The result is Vb, Rd = 475 kN which is 85% of Vpl, Rd, as found by the method of EN 1993-1-1 for a rolled I-section. From clause 6.2.2.4(2), ρ = [(2VEd /VRd) – 1]2 = (600/475 – 1)2 = 0.068 The reduced yield strength of the web is (1 – 0.068) × 355 = 331 N/mm2. For hogging bending, the tensile force in the reinforcement is Fs = 750 × 0.5/1.15 = 326 kN

(D6.9)

From plastic theory, the depth of web in compression that is above the neutral axis of the I-section is 326/(2 × 0.331 × 6.3) = 78 mm This places the cross-section in Class 3, so the hole-in-web method is applied. From Fig. 6.6(a), the depth of the compressive stress-blocks in the web is 20tε. The value for ε should be based on the full yield strength of the web, not on the reduced yield strength, so each block is 102 mm deep, as in Fig. 6.6(c). The use of the reduced yield strength would increase ε and so reduce the depth of the ‘hole’ in the web, which would be unconservative. However, the force in each stress block should be found using the reduced yield strength, and so is now 228 × 331/355 = 213 kN The tensile force in the web is therefore 2 × 213 – 326 = 100 kN and the longitudinal forces are as shown in Fig. 6.8(b). From Example 6.1, Mpl, a, flanges = 183.5 kN m Taking moments about the bottom of the slab, Mpl, Rd = 183.5 + 326 × 0.1 + 213(0.118 + 0.328) – 100 × 0.043 Mpl, Rd = 307 kN m

(D6.10)

For this cross-section in bending only, the method of Example 6.1, with As = 750 mm2, gives Mpl, Rd = 314 kN m The alternative to this method would be to use elastic theory. The result would then depend on the method of construction.

56





6.3. Resistance of cross-sections of beams for buildings with partial encasement Encasement in concrete of the webs of steel beams is normally done before erection, casting one side at a time. This obviously increases the cost of fabrication, transport and erection, but when it satisfies clause 5.5.3(2) it has many advantages for design: • • • • • •

it provides complete fire resistance for the web and, with longitudinal reinforcement, compensation for the weakness of the bottom flange in fire, in accordance with EN 1994-1-252 it enables a Class 3 web to be upgraded to Class 2, and the slenderness limit for a Class 2 compression flange to be increased by 40% (clause 5.5.3) it widens significantly the range of steel sections that are not susceptible to lateral– torsional buckling (clause 6.4.3(1)(h)) it increases resistance to vertical shear, to clause 6.3.3(2) it improves resistance to combined bending and shear, to clause 6.3.4(2) it improves resistance to buckling in shear, clause 6.3.3(1).

6.3.1. Scope

To avoid shear buckling, clause 6.3.1(2) limits the slenderness of the encased web to d/tw £ 124ε. In practice, with encasement, steel sections in buildings are almost certain to be in Class 1 or 2. Clause 6.3 is applicable only to these.

6.3.2. Resistance to bending

The rules for resistance to bending, clause 6.3.2, correspond to those for uncased sections of the same class, except that lateral–torsional buckling is not mentioned in clause 6.3.2(2). Encasement greatly improves resistance to lateral–torsional buckling. Example 6.3, the comments on clause 6.4.2(7), and pp. 153–154 of Johnson and Molenstra,44 are relevant. However, it is possible for a beam within the scope of clause 6.3 to be susceptible; a web-encased IPE 450 section in S420 steel is an example. Continuous beams that do not satisfy clause 6.4.3 should therefore be checked. Partial shear connection is permitted for a concrete flange, but not for web encasement. The resistance to longitudinal shear provided by studs within the encasement is found in the usual way, but no guidance is given on the contribution from bars that pass through holes in the web or stirrups welded to the web, in accordance with clauses 5.5.3(2) and 6.3.3(2). They are provided to ensure the integrity of the encased section. The load–slip properties of different types of shear connection should be compatible (clause 6.6.1.1(6)P). It is known from research on ‘Perfobond’ shear connectors (longitudinal flange plates projecting into the slab, with holes through which bars pass) that these bars provide shear connection with good slip properties,53,54 so a contribution from them could be used here; but welds to stirrups may be too brittle.

6.3.3–6.3.4. Resistance to vertical shear, and to bending and vertical shear

The rules in clause 6.3.3 are based on the concept of superposition of resistances of composite and reinforced concrete members. This concept has been used in Japan for decades, in design of structures for earthquake resistance. The shear connection must be designed to ensure that the shear force is shared between the steel web and the concrete encasement. The references to EN 1992-1-1 are intended to ensure that the web encasement retains its shear resistance at a shear strain sufficient to cause yielding of the steel web. Clause 6.3.3(1) shows that no account need be taken of web buckling in shear. Moment– shear interaction is treated in clause 6.3.4 in a manner consistent with the rules for uncased sections.

Clause 6.3.1(2)


Clause 6.3.3

Clause 6.3.3(1) Clause 6.3.4

57





6.4. Lateral–torsional buckling of composite beams 6.4.1. General Clause 6.4.1(1)

Clause 6.4.1(2)

Clause 6.4.1(3)

It is assumed in this section that in completed structures for buildings, the steel top flanges of all composite beams will be stabilized laterally by connection to a concrete or composite slab (clause 6.4.1(1)). The rules on maximum spacing of connectors in clauses 6.6.5.5(1) and 6.6.5.5(2) relate to the classification of the top flange, and thus only to local buckling. For lateral–torsional buckling, the relevant rule, given in clause 6.6.5.5(3), is less restrictive. Any steel top flange in compression that is not so stabilized should be checked for lateral buckling (clause 6.4.1(2)) using clause 6.3.2 of EN 1993-1-1. This applies particularly during unpropped construction. In a long span, it may be necessary to check a steel beam that is composite along only part of its length. The general method of clause 6.4.2, based on the use of a computed value of the elastic critical moment Mcr, is applicable, but no detailed guidance on the calculation of Mcr is given in either EN 1993-1-1 or EN 1994-1-1. Buckling of web-encased beams without a concrete flange has been studied.55 However, for buildings, the construction phase is rarely critical in practice, because the loading is so much less than the design total load. Steel bottom flanges are in compression only in cantilevers and continuous beams. The length in compression may include most of the span, when that span is lightly loaded and both adjacent spans are fully loaded. Bottom flanges in compression should always be restrained laterally at supports (clause 6.4.3(1)(f) is relevant). It should not be assumed that a point of contraflexure is equivalent to a lateral restraint. In a composite beam, the concrete slab provides lateral restraint to the steel member, and also restrains its rotation about a longitudinal axis. Lateral buckling is always associated with distortion (change of shape) of the cross-section. Design methods for composite beams must take account of the bending of the web, Fig. 6.9(b). They differ in detail from the method of clause 6.3.2 of EN 1993-1-1, but the same imperfection factors and buckling curves are used, in the absence of any better-established alternatives. The reference in clause 6.4.1(3) to EN 1993-1-1 provides a general method for use where neither of the methods of EN 1994-1-1 are applicable (e.g. for a Class 4 beam).

6.4.2. Verification of lateral–torsional buckling of continuous composite beams with cross-sections in Class 1, 2 and 3 for buildings


This general method of design is written with distortional buckling of bottom flanges in mind. It would not apply, for example, to a mid-span cross-section of a beam with the slab at bottom-flange level (see Fig. 6.2). Although not stated, it is implied that the span concerned is of uniform composite section, excluding minor changes such as reinforcement details and effects of cracking of concrete. The use of this method for a two-span beam is illustrated in Example 6.7. The method is based closely on clause 6.3.2 of EN 1993-1-1. There is correspondence in the definitions of the reduction factor χLT (clause 6.4.2(1)) and the relative slenderness, λLT (clause 6.4.2(4)). The reduction factor is applied to the design resistance moment MRd, which is defined in clauses 6.4.2(1) to 6.4.2(3). Expressions for MRd include the design yield strength fyd. The reference in these clauses to the use of γM1 is provided because this is a check on instability. The recommended values for γM0 and γM1 are the same (1.0), but a National Annex could define different values. The determination of MRd for a Class 3 section differs from that of Mel, Rd in clause 6.2.1.4(6) only in that the limiting stress fcd for concrete in compression need not be considered. It is necessary to take account of the method of construction. The buckling resistance moment Mb, Rd given by equation (6.6) must exceed the highest applied moment MEd within the unbraced length of compression flange considered.

58





A

B

ds

hs

h tw

d

F

d

F

tf a (a) C

L

L1

D q0

L

L2

L1

0.8 ≥ L2 /L1 £ 1.25 (c)

(b)

Fig. 6.9. U-frame action and distortional lateral buckling

Lateral buckling for a Class 3 cross-section with unpropped construction The influence of method of construction on the verification of a Class 3 composite section for lateral buckling is as follows. From equation (6.4) MRd = Mel, Rd = Ma, Ed + kMc, Ed

(a)

where the subscript c indicates the action effect on the composite member. From equation (6.6) the verification is MEd = Ma, Ed + Mc, Ed £ χLT Mel, Rd

(b)

which is χLT ≥ (Ma, Ed + Mc, Ed)/Mel, Rd = MEd /Mel, Rd

(c)

The total hogging bending moment MEd may be almost independent of the method of construction. However, the stress limit that determines Mel, Rd may be different for propped and unpropped construction. If it is bottom-flange compression in both cases, then Mel, Rd is lower for unpropped construction, and the limit on χLT from equation (c) is more severe.

Elastic critical buckling moment Clause 6.4.2(4) requires the determination of the elastic critical buckling moment, taking account of the relevant restraints, so their stiffnesses have to be calculated. The lateral restraint from the slab can usually be assumed to be rigid. Where the structure is such that a pair of steel beams and a concrete flange attached to them can be modelled as an inverted-U frame (Fig. 6.11), continuous along the span, the rotational restraining stiffness at top-flange level, ks, can be found from clauses 6.4.2(5) to 6.4.2(7). Clause 6.4.2(5) gives conditions that define this frame. Analysis is based on its stiffness per unit length, ks, given by the ratio F/δ, where δ is the lateral displacement caused by a force F (Fig. 6.9(a)). The flexibility δ/F is the sum of flexibilities due to:

Clause 6.4.2(4)


59





• • •

bending of the slab, which may not be negligible: 1/k1 from equation (6.9) bending of the steel web, which predominates: 1/k2 from equation (6.10) flexibility of the shear connection.

It has been found56 that this last flexibility can be neglected in design to EN 1994-1-1. This leads to equation (6.8) for stiffness ks. This ‘continuous U-frame’ concept has long been used in the design of steel bridges.57 There is a similar ‘discrete U-frame’ concept, which would be relevant to composite beams if the steel sections had vertical web stiffeners. The shear connectors closest to those stiffeners would then have to transmit almost the whole of the bending moment Fh (Fig. 6.9(a)), where F is now a force on a discrete U-frame. The last of the three flexibilities listed above might then not be negligible, nor is it certain that the shear connection and the adjacent slab would be sufficiently strong.58 Where stiffeners are present, the resistance of the connection above each stiffener to repeated transverse bending should be established, as there is a risk of local shear failure within the slab. There is at present no simple method of verification. The words ‘may be unstiffened’ in clause 6.4.3(1)(f) are misleading, as the resistance model is based on both theory and research on unstiffened webs. It should, in the authors’ opinion, read ‘should be unstiffened’. The problem is avoided in bridge design by using transverse steel members, such as U- or H-frames (Fig. 6.10(a)). The conditions in clause 6.4.3(1) referred to from clause 6.4.2(5) are commonly satisfied in buildings by the beams that support composite slabs; but where these are secondary beams, the method is not applicable to the primary beams because condition (e) is not satisfied. Bottom flanges of primary beams can sometimes be stabilized by bracing from the secondary beams. The calculation of ks is straightforward, apart from finding (EI)2, the cracked flexural stiffness of a composite slab. An approximate method, used in Example 6.7, is derived in Appendix A.

Clause 6.4.2(7) Clause 6.4.2(9)

Concrete-encased web Clauses 6.4.2(7) and 6.4.2(9) allow for the additional stiffness provided by web encasement. This is significant: for rolled steel sections, k2 from equation (6.11) is from 10 to 40 times the value from equation (6.10), depending on the ratio of flange breadth to web thickness. Encasement will often remove any susceptibility to lateral–torsional buckling. The model used for equation (6.11) is explained in Appendix A. Theory for the continuous inverted-U frame model A formula for the elastic critical buckling moment for the U-frame model was given in Annex B of ENV 1994-1-1,49 but was removed from EN 1994-1-1, as it was considered to be ‘textbook material’. However, it is sufficiently unfamiliar to be worth giving here. Subject to conditions discussed below, the elastic critical buckling moment at an internal support of a continuous beam is Mcr = (kcC4/L)[(Ga Iat + ks L2/π2)Ea Iafz]1/2

(D6.11)

where: kc is a property of the composite section, given below, C4 is a property of the distribution of bending moment within length L, Ga is the shear modulus for steel (Ga = Ea/[2(1 + υ)] = 80.8 kN/mm2), Iat is the torsional moment of area of the steel section, ks is the rotational stiffness defined in clause 6.4.2(6), L is the length of the beam between points at which the bottom flange of the steel member is laterally restrained (typically, the span length), and Iafz is the minor-axis second moment of area of the steel bottom flange. Where the cross-section of the steel member is symmetrical about both axes, the factor kc is given by kc = (hsIy /Iay)/[(hs2/4 + ix2)/e + hs]

(D6.12)

with e = AIay /[Aa zc(A – Aa)]

60



(D6.13)



where: hs is the distance between the centres of the flanges of the steel section, Iy is the second moment of area for major-axis bending of the cracked composite section of area A, Iay is the corresponding second moment of area of the steel section, ix2 = (Iay + Iaz)/Aa, where Iaz and Aa are properties of the steel section, and zc is the distance between the centroid of the steel beam and the mid-depth of the slab. Four of the conditions for the use of these formulae are given in paragraphs (c) to (f) of clause 6.4.3(1). Three further conditions were given in ENV 1994-1-1. These related to the resistance of the slab part of the U-frame to hogging transverse bending in the plane of the U-frame, to its flexural stiffness, and to the spacing of shear connectors. It is now considered that other requirements are such that these will, in practice, be satisfied. Further explanation is given in Appendix A. The coefficient C4 was given in a set of tables, determined by numerical analyses, in which its range is 6.2–47.6. These values are given in Appendix A (see Figs A.3 and A.4). The coefficient accounts for the increased resistance to lateral buckling where the bending moment is not uniform along the member. When checking lateral stability, the distribution of bending moments corresponding to C4 must be used as the action effects, and not an equivalent uniform value. The calculation method is shown in Example 6.7.

Alternative theory for the elastic critical moment There is an analogy between the differential equations for distortional lateral buckling, taking account of restraint to warping, and those for a compressed member on an elastic foundation. This has led59 to an alternative expression for the elastic critical moment. Like equation (D6.11), its use requires computed values that depend on the bending-moment distribution and, for this method, also on the parameter ηB2 = ks L4/(EaIωD) where ks, L and Ea are as above, and IωD is the sectorial moment of inertia of the steel member related to the centre of the restrained steel flange. Four graphs of these values are given in Hanswille,59 and a more general set in Hanswille et al.60 Predictions of Mcr by this method and by equation (D6.11) were compared with results of finite-element analyses, for beams with IPE 500 and HEA 1000 rolled sections. This method was found to agree with the finite-element results for both internal and external spans. Equation (D6.11) was found to be satisfactory for internal spans, but to be less accurate generally for external spans, and unconservative by over 30% in some cases. This suggests that it needs further validation.

6.4.3. Simplified verification for buildings without direct calculation

As calculations for the U-frame model are quite extensive, a simplified method has been developed from it. Clause 6.4.3(1) defines continuous beams and cantilevers that may be designed without lateral bracing to the bottom flange, except at supports. Its Table 6.1 gives limits to the steel grade and overall depth of the steel member, provided that it is an IPE or HE rolled section. The contribution from partial encasement is allowed for in paragraph (h) of this clause. These results are derived from equation (D6.11) for the elastic critical buckling moment, making assumptions that further reduce the scope of the method. Accounts of its origin are available in both English61 and German.62 It is outlined in Johnson and Fan,56 and is similar to that used in the treatment of lateral–torsional buckling of haunched beams.63 The basis is that there shall be no reduction, due to lateral buckling, in the resistance of the beam to hogging bending. It is assumed that this is achieved when λLT £ 0.4. This value is given in a Note to clause 6.3.2.3 of EN 1993-1-1, which can be modified by a National Annex. Any National Annex that defines a lower limit should therefore also state if the method is permitted.

Clause 6.4.3(1)

61





The slenderness λLT is a function of the variation of bending moment along the span. This was studied using various loadings on continuous beams of the types shown in Fig. 6.9(c), and on beams with cantilevers. The limitations in paragraphs (a) and (b) of clause 6.4.3(1) on spans and loading result from this work.

Simplified expression for λ LT , and use of British UB rolled sections Table 6.1 applies only to IPE and HE rolled sections. Criteria for other rolled I- and H-sections are deduced in Appendix A. The basis for the method is as follows. For uncased beams that satisfy the conditions that apply to equation (D6.11) for Mcr, have a double symmetrical steel section, and are not concrete encased, the slenderness ratio for a Class 1 or Class 2 cross-section may conservatively be taken as Ê t h ˆÊh ˆ λLT = 5.0 Á 1 + w s ˜ Á s ˜ 4 bf tf ¯ Ë tw ¯ Ë

0.75

Ê tf ˆ ÁË b ˜¯ f

0.25

Ê fy ˆ ÁË E C ˜¯ a 4

0.5

(D6.14)

The derivation from equation (D6.11) is given in Appendix A. Most of the terms in equation (D6.14) define properties of the steel I-section; bf is the breadth of the bottom flange, and other symbols are as in Fig. 6.9(a). To check if a particular section qualifies for ‘simplified verification’, a section parameter F is calculated. From equation (D6.14), it is Ê t h ˆÊh ˆ F = Á1 + w s ˜ Á s ˜ 4 bf tf ¯ Ë tw ¯ Ë

0.75


0.25

(D6.15)

Limiting values of F, Flim say, are given in Appendix A (see Fig. A.5) for the nominal steel grades listed in Table 6.1. The horizontal ‘S’ line at or next above the plotted point F gives the highest grade of steel for which the method of clause 6.3.3 can be used for that section. Some examples are given in Table 6.1, with the values of Flim in the column headings. Many of the heavier wide-flange sections in S275 steel qualify for verification without direct calculation, but few UB sections in S355 steel do so. In ENV 1994-1-1, verification without direct calculation was permitted for hot-rolled sections of ‘similar shape’ to IPE and HE sections, that conformed to Table 6.1 and a geometrical condition similar to the limit on F. In EN 1994-1-1 this has been replaced by a reference to National Annexes.

Use of UB rolled sections with encased webs to clause 5.5.3(2) It is shown in Appendix A (equation (DA.4)) that the effect of web encasement is to increase Flim by at least 29%. All the sections shown in Table 6.1 now qualify for S275 steel, and all Table 6.1. Qualification of some UB rolled steel sections for verification of lateral–torsional stability, in a composite beam, without direct calculation

Section

Right-hand side of expression (D6.15)

S275 steel, uncased (13.9)

S355 steel, uncased (12.3)

S275 steel, cased (18.0)

S355 steel, cased (15.8)

457 × 152 UB52 457 × 152 UB67 457 × 191 UB67 457 × 191 UB98 533 × 210 UB82 533 × 210 UB122 610 × 229 UB125 610 × 229 UB140 610 × 305 UB149 610 × 305 UB238

16.4 14.9 13.6 11.8 14.4 12.5 14.1 13.5 12.2 9.83

No No Yes Yes No Yes No Yes Yes Yes

No No No Yes No No No No Yes Yes

Yes Yes Yes Yes Yes Yes Yes Yes Yes Yes

No Yes Yes Yes Yes Yes Yes Yes Yes Yes

62





(a)

(b)

Fig. 6.10. Laterally restrained bottom flanges

except one do so for S355 steel. Web encasement is thus an effective option for improving the lateral stability of a rolled steel section in a continuous composite beam. The method of clause 6.4.2 should be used where the steel section does not qualify.

Use of intermediate lateral bracing

Where the buckling resistance moment Mb, Rd as found by one of the preceding methods is significantly less than the design resistance moment MRd of the cross-sections concerned, it may be cost-effective to provide discrete lateral restraint to the steel bottom flange. Where the slab is composite, a steel cross-member may be needed (Fig. 6.10(a)), but for solid slabs, other solutions are possible (e.g. Fig. 6.10(b)). Clause 6.3.2.1(2) of EN 1993-1-1 refers to ‘beams with sufficient restraint to the compression flange’, but does not define ‘sufficient’. A Note to clause 6.3.2.4(3) of EN 1993-1-1 refers to Annex BB.3 in that EN standard for buckling of components of building structures ‘with restraints’. Clause BB.3.2(1) gives the minimum ‘stable length between lateral restraints’, but this is intended to apply to lateral–torsional buckling, and may not be appropriate for distortional buckling. Further provisions for steel structures are given in EN 1993-2.64 There is no guidance in EN 1994-1-1 on the minimum strength or stiffness that a lateral restraint must have. There is guidance in Lawson and Rackham,63 based on BS 5950: Part 1, clause 4.3.2.65 This states that a discrete restraint should be designed for 2% of the maximum compressive force in the flange. It is suggested63 that where discrete and continuous restraints act together, as in a composite beam, the design force can be reduced to 1% of the force in the flange. Provision of discrete bracing to make up a deficiency in continuous restraint is attractive in principle, but the relative stiffness of the two types of restraint must be such that they are effective in parallel. Another proposal is to relate the restraining force to the total compression in the flange and the web at the cross-section where the bracing is provided, to take advantage of the steep moment gradient in a region of hogging bending. In tests at the University of Warwick,66 bracings that could resist 1% of the total compression, defined in this way, were found to be effective. The calculation of this compressive force involves an elastic analysis of the section, not otherwise needed, and the rule is unsafe near points of contraflexure. There is at present no simple design method better than the 2% rule quoted above, which can be over-conservative. An elastic analysis can be avoided by taking the stress in the flange as the yield stress. The design methods based on equation (D6.11) for Mcr work well for complete spans, but become unsatisfactory (over-conservative) for short lengths of beam between lateral bracings. This is because the correct values of the factors C4 are functions of the length between lateral restraints. For simplicity, only the minimum values of C4 are given in the figures in Appendix B of this guide. They are applicable where the half-wavelength of a buckle is less than the length L in equation (D6.11). This is always so where L is a complete span, but where L is part of a span, the value given may be over-conservative. This is

63





Elevation of beam Classify sections at A and C, to clause 5.5 A Class 1 or 2

B

Use effective section in Class 2, to clause 5.5.2(3)?

Class 3

No

Yes Find Mel, Rd at A and C to clause 6.4.2(3)

Find MRd as Mpl, Rd , to clause 6.2.1.2

C Find Ma, Ed at A and C by elastic global analysis for relevant arrangements of actions on steel members; no redistribution

–

Find lLT at A and C, to clause 6.4. Is steel element a ‘rolled or equivalent welded’ section? No

Yes –

–

Is lLT > 0.4 for AB or BC? (Clause 6.3.2.3(1) of EN 1993-1-1)

Is lLT > 0.2 for AB or BC? Yes

No Ignore shrinkage (clause 5.4.2.2(7))

Yes

No

Is section Class 1 or 2?

Yes

Find cLT to clause 6.4, and hence find Mb, Rd

No

Use elastic global analysis, with relevant modular ratio, for actions on steel and composite members and relevant load arrangements to find MEd (= Ma, Ed + Mc, Ed) at A, B and C. – Is lLT > 0.2 (or 0.4) for either AB or BC? (See Notes (1) and (2), before Example 6.3) No

Yes

Redistribute hogging bending moments Mc, Ed to clause 5.4.4, to find new MEd at sections A, B and C. See Fig. 5.4

Is M Ed less than MRd or Mb, Rd, as appropriate at both of sections A and C? Yes

No

Classify cross-section B. Find MRd at B for full shear connection. Is MEd < MRd at B? Yes Re-design is required (END)

Go to Fig. 6.11(b) for design of shear connection

No Re-design is required (END)

Fig. 6.11. (a) Flow chart for design for ultimate limit state of an internal span of a continuous beam in a building, with uniform steel section and no cross-sections in Class 4

illustrated at the end of Example 6.7, where the effect of providing lateral bracing is examined.

Flow charts for continuous beam

The flow charts in Fig. 6.11 cover some aspects of design of an internal span of a continuous composite beam. Their scope is limited to cross-sections in Class 1, 2 or 3, an uncased web, a uniform steel section, and no flexural interaction with supporting members. It is assumed that for lateral–torsional buckling, the simplified method of clause 6.4.3 is not applicable. Figure 6.11(a) refers to the following notes:

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No

Are all critical sections in Class 1 or 2, to clause 6.6.1.1(14)?

Find longitudinal shear flows, ignoring tensile strength of concrete and tension stiffening, to clauses 6.6.2.1 and 6.6.2.2. See comments on clause 6.6.2

Yes Hogging region

Sagging region

Use full shear connection for Nc, f found from MRd, to clause 6.6.2.1(1) or 6.6.2.2(1)

Are connectors studs to clause 6.6.1.2(1) or (3)? Yes (advised)

No Has it been shown that proposed connectors satisfy clauses 6.6.1.1(4)P and 6.6.1.1(5)?

Find hmin to clause 6.6.1.2. Do you intend to use h ≥ hmin? Yes (advised)

No Find MRd to clauses 6.2.1.3(3) to 6.2.1.3(5)

(a) Choose h ≥ hmin so that studs are ‘ductile’

Yes (b) Choose h

No (c) Choose h and find MRd such that clause 6.6.1.1(3)P is satisfied

Find MRd to clause 6.2.1.4 or 6.2.1.5 Increase h. Return to (a), (b), or (c)

No

Is MRd ≥ MEd Yes

Find PRd for stud connectors to clause 6.6.3.1 or 6.6.4. For other connectors, find PRd that satisfies clause 6.6.1.1 and determine if ‘ductile’

Are all critical cross-sections in Class 1 or 2?

No

Yes

Sagging region. Find ns = hNc, f /PRd for region each side of B

Space connectors in accordance with the shear flow, to clause 6.6.1.3(5) (END)

Hogging region. Find nh = Nc, f /PRd for regions adjacent to A and C

Find n = ns + nh for critical lengths AB and BC. Are connectors ‘ductile’?

No

Yes Space n connectors uniformly over lengths AB (and CD) to clause 6.6.1.3(3)

Either

Or

No

Is Mpl, Rd > 2.5 Mpl, a, Rd at A, B, or C ? Yes Additional checks required to clause 6.6.1.3(4)

Space n connectors over AB (and CD) in accordance with longitudinal shear to clause 6.6.1.3(5)

Check spacing of connectors to clauses 6.6.5.5(3) and 6.6.5.7(4). Do they ensure the stability of any part of the member? Yes

Check spacing to clause 6.6.5.5(2) and revise if necessary

No

(END)

Fig. 6.11. (Contd) (b) Flow chart for design of an internal span of a continuous beam in a building – shear connection

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(1) The elastic global analysis is simpler if the ‘uncracked’ model is used. The limits to redistribution of hogging moments include allowance for the effects of cracking, but redistribution is not permitted where allowance for lateral–torsional buckling is required (clause 5.4.4(4)). ‘Cracked’ analysis may then be preferred, as it gives lower hogging moments at internal supports. (2) If fully propped construction is used, Ma, Ed may be zero at all cross-sections.

Example 6.3: lateral–torsional buckling of two-span beam

The design method of clause 6.4.2, with application of the theory in Appendix A, is illustrated in Example 6.7, which will be found after the comments on clause 6.6. The length of the method is such that the ‘simplified verification’ of clause 6.4.3 is used wherever possible. Limitations of the simplified method are now illustrated, with reference to the twospan beam treated in Example 6.7. This uses an IPE 450 steel section in grade S355 steel, and is continuous over two 12 m spans. Details are shown in Figs 6.23–6.25. Relevant results for hogging bending of the composite section at support B in Fig. 6.23(c) are as follows: Mpl, Rd = 781 kN m

λLT = 0.43

Mb, Rd = 687 kN m

The design ultimate loads per unit length of beam, from Table 6.2, are permanent: variable:

7.80 + 1.62 = 9.42 kN/m

26.25 kN/m

The steel section fails the condition in paragraph (g) of clause 6.4.3(1), which limits its depth to 400 mm. The beam satisfies all the other conditions except that in paragraph (b), for its ratio of permanent to total load is only 9.42/35.67 = 0.26, far below the specified minimum of 0.4. The simplest way to satisfy paragraph (g) would be to encase the web in concrete, which increases the depth limit to 600 mm, and the permanent load to 11.9 kN/m. The condition in paragraph (b) is quite severe. In this case, it is 11.9 ≥ 0.4(11.9 + qd) whence qd £ 17.8 kN/m This corresponds to a characteristic floor loading of 17.8/(1.5 × 2.5) = 4.75 kN/m2 which is a big reduction from the 7 kN/m2 specified, even though the required reduction in λLT (to £ 0.4) is less than 10%. This result illustrates a common feature of ‘simplified’ methods. They have to cover so wide a variety of situations that they are over-conservative for some of them.

6.5. Transverse forces on webs Clause 6.5

The local resistance of an unstiffened and unencased web to forces (typically, vertical forces) applied through a steel flange can be assumed to be the same in a composite member as in a steel member, so clause 6.5 consists mainly of references to EN 1993-1-5. The provisions of Section 8 of EN 1993-1-5 are not limited to rolled sections, so are applicable to webs where the neutral axis is not at mid-depth, as is usual in composite beams. In buildings, local yielding or buckling of a web may occur where a composite beam is continuous over a steel beam that supports it. Rolled I-sections in Class 1 or 2 may not be

66





susceptible, but a Class 3 web, treated as effective Class 2, should almost always be stiffened (clause 6.5.1(3)). A plate girder launched over roller supports cannot be stiffened at every section, so the erection condition may be critical; but this situation is rare in buildings. Flange-induced web buckling, clause 6.5.2, could occur where a large compression flange is restrained from buckling out of its plane by a weak or slender web. This is prevented by specifying a limit to the web slenderness, as a function of the ratio of flange area to web area. The slenderness limit is reduced if the flange is curved in elevation, to ensure that the web can resist the radial component of the force in the flange. This form of web buckling cannot occur with straight rolled steel I-sections, and needs to be checked only for plate girders of unusual proportions and for members sharply curved in elevation. The effect of sharp curvature is illustrated by reference to an IPE 400 section of grade S355 steel, which has been cold curved about its major axis, before erection. Assuming that its plastic resistance moment is to be used, the minimum permitted radius of curvature given by clause 8(2) of EN 1993-1-5 is 2.1 m. Flange-induced buckling is clearly a rare problem for hot-rolled sections.

Clause 6.5.1(3) Clause 6.5.2

6.6. Shear connection 6.6.1. General Basis of design Clause 6.6 is applicable to shear connection in composite beams. Clause 6.6.1.1(1) refers also to ‘other types of composite member’. Shear connection in composite columns is treated in clause 6.7.4, but reference is made to clause 6.6.3.1 for the design resistance of headed stud connectors. Similarly, headed studs used for end anchorage in composite slabs are treated in clause 9.7.4, but some provisions in clause 6.6 are also applicable. Although the uncertain effects of bond are excluded by clause 6.6.1.1(2)P, friction is not excluded. Its essential difference from bond is that there must be compressive force across the relevant surfaces. This usually arises from wedging action. Provisions for shear connection by friction are given in clauses 6.7.4.2(4) (columns) and 9.1.2.1 (composite slabs). ‘Inelastic redistribution of shear’ (clause 6.6.1.1(3)P) is relied on in the many provisions that permit uniform spacing of connectors. Clause 6.6.1.1(4)P uses the term ‘ductile’ for connectors that have deformation capacity sufficient to assume ideal plastic behaviour of the shear connection. Clause 6.6.1.1(5) quantifies this as a characteristic slip capacity of 6 mm.47 In practice, designers will not wish to calculate required and available slip capacities. Clause 6.6.1.2(1) enables such calculations to be avoided by limiting the extent of partial shear connection and by specifying the type and range of shear connectors. The need for compatibility of load/slip properties, clause 6.6.1.1(6)P, is one reason why neither bond nor adhesives can be used to supplement the shear resistance of studs. The combined use of studs and block-and-hoop connectors has been discouraged for the same reason, though there is little doubt that effectively rigid projections into the concrete slab, such as bolt heads and ends of flange plates, contribute to shear connection. ‘Separation’, in clause 6.6.1.1(7)P, means separation sufficient for the curvatures of the two elements to be different at a cross-section, or for there to be a risk of local corrosion. None of the design methods in EN 1994-1-1 takes account of differences of curvature, which can arise from a very small separation. Even where most of the load is applied by or above the slab, as is usual, tests on beams with unheaded studs show separation, especially after inelastic behaviour begins. This arises from local variations in the flexural stiffnesses of the concrete and steel elements, and from the tendency of the slab to ride up on the weld collars. The standard heads of stud connectors have been found to be large enough to control separation, and the rule in clause 6.6.1.1(8) is intended to ensure that other types of connector, with anchoring devices if necessary, can do so.

Clause 6.6.1.1(1)

Clause 6.6.1.1(2)P

Clause 6.6.1.1(3)P Clause 6.6.1.1(4)P Clause 6.6.1.1(5)

Clause 6.6.1.1(6)P

Clause 6.6.1.1(7)P

Clause 6.6.1.1(8)

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Clause 6.6.1.1(9)

Clause 6.6.1.1(10)P

Clause 6.6.1.1(12)

Clause 6.6.1.1(13)

Clause 6.6.1.1(14)P

Resistance to uplift is much influenced by the reinforcement near the bottom of the slab, so if the resistance of an anchor is to be checked by testing, reinforcement in accordance with clause 6.6.6 should be provided in the test specimens. Anchors are inevitably subjected also to shear. Clause 6.6.1.1(9) refers to ‘direct tension’. Load from a travelling crane hanging from the steel member is an example. In bridges, it can be caused by the differential deflexions of adjacent beams under certain patterns of imposed load. Where it is present, its design magnitude must be determined. Clause 6.6.1.1(10)P is a principle that has led to many application rules. The shear forces are inevitably ‘concentrated’. One research study67 found that 70% of the shear on a stud was resisted by its weld collar, and that the local (triaxial) stress in the concrete was several times its cube strength. Transverse reinforcement performs a dual role. It acts as horizontal shear reinforcement for the concrete flanges, and controls and limits splitting. Its detailing is critical where connectors are close to a free surface of the slab. Larger concentrated forces occur where precast slabs are used, and connectors are placed in groups in holes in the slabs. This influences the detailing of the reinforcement near these holes, and is referred to in Section 8 of EN 1994-2. Clause 6.6.1.1(12) is intended to permit the use of other types of connector. ENV 1994-1-1 included provisions for many types of connector other than studs: block connectors, anchors, hoops, angles, and friction-grip bolts. They have all been omitted because of their limited use and to shorten the code. Clause 6.6.1.1(13) says that connectors should resist at least the design shear force, meaning the action effect. Their design for Class 1 and 2 sections is based on the design bending resistances (see clause 6.6.2), and hence on a shear force that normally exceeds the action effect. The principle on partial shear connection, clause 6.6.1.1(14)P, leads to application rules in clause 6.2.1.3. The flow chart in Fig. 6.11(b) is for a beam without web encasement, and may assist in following the comments on clauses 6.2 and 6.6.

Limitation on the use of partial shear connection in beams for buildings As noted in comments on clause 6.2.1.3, the rules for partial shear connection are based on an available slip of 6 mm. Connectors defined as ‘ductile’ are those that had been shown to have (or were believed to have) a characteristic slip capacity (defined in clause B.2.5(4)) exceeding 6 mm. Prediction of slip capacity is difficult. Push tests on stud connectors have been reported in scores of publications, but few tests were continued for slips exceeding 3 mm. Slip capacity depends on the degree of containment of the connector by the concrete and its reinforcement, and hence on the location of free surfaces (e.g. in haunches or edge beams) as well as on the shape, size and spacing of the connectors. The information has been summarized.47,48 The conclusions led to the approval of certain stud connectors as ‘ductile’ Clause 6.6.1.2(1) (clause 6.6.1.2(1)), and also friction-grip bolts, which were within the scope of ENV 1994-1-1. These conclusions seem optimistic when compared with the results of some push tests using solid slabs, but connectors behave much better in beams reinforced as required by the Eurocode than in small push-test specimens, where splitting can cause premature failure. One might expect a lower available slip from studs in very strong concrete, but the 6 mm limit has been confirmed68 by four push tests with cylinder strengths fcm ª 86 N/mm2. For certain types of profiled sheeting, available slips were found to be greater than 6 mm.43,69 These results and other test data led to a relaxation of the limiting effective spans at which low degrees of shear connection can be used, as shown in the lower part of Fig. Clause 6.6.1.2(3) 6.4(a). This applies only where the conditions in paragraphs (a)–(e) of clause 6.6.1.2(3) are satisfied, because these are the situations for which test data are available. There is no validated theoretical model that includes all the many relevant variables, so this relaxation is

68





allowed only where the force Nc is determined by the more conservative of the two methods given in clause 6.2.1.3, the interpolation method. Research continues on the influence on slip capacity of profile shape and the detailing of studs in troughs. The limits to the use of partial shear connection in buildings are summarized in Fig. 6.4(a), where Le is the effective span. The span limits given by the lines PQ to RS are from the provisions of clause 6.6.1.2 for ‘ductile’ connectors. The design of a long-span beam with a low degree of shear connection is likely to be governed by the need to limit its deflection, unless it is propped during construction or is continuous. For composite beams in sagging bending, the steel top flange must be wide enough to resist lateral buckling during erection, and for attachment of the connectors, but can often be smaller than the bottom flange. A smaller flange lowers the plastic neutral axis of the composite section, and increases the slip required by the model for partial-interaction design. This is why clauses 6.6.1.2(1) and 6.6.1.2(2) give limits on the degree of shear connection that are less liberal than those for beams with equal steel flanges.

Spacing of shear connectors in beams for buildings Clause 6.6.1.3(1)P extends clause 6.6.1.1(2)P a little, by referring to ‘spacing’ of connectors Clause 6.6.1.3(1)P and an ‘appropriate distribution’ of longitudinal shear. The interpretation of ‘appropriate’ can depend on the method of analysis used and the ductility of the connectors. The principle may be assumed to be satisfied where connectors are spaced ‘elastically’ to clause 6.6.1.3(5), which has general applicability. The more convenient use of uniform spacing requires the connectors to satisfy clause 6.6.1.3(3), which implies (but does not require) the use of plastic resistance moments. The connectors must be ‘ductile’, as defined in clauses 6.6.1.1(4)P and 6.6.1.1(5). This is normally achieved by satisfying clause 6.6.1.2. Clauses 6.6.1.1(3)P to 6.6.1.1(5) provide an alternative, which enables research-based evidence to be used, but its use is not appropriate for routine design. In practice, it is possible to space connectors uniformly in most beams for buildings, so for continuous beams clause 6.6.1.3(2)P requires the tension reinforcement to be curtailed to Clause 6.6.1.3(2)P suit the spacing of the shear connectors. Clause 6.6.1.3(4) could apply to a simply-supported or a continuous beam with a large Clause 6.6.1.3(4) concrete slab and a relatively small steel top flange. Connectors spaced uniformly along a critical length might then have insufficient available slip. Use of an additional critical section would lead to a more suitable distribution.

Example 6.4: arrangement of shear connectors

As an example of the use of these rules, a simply-supported beam of span 10 m is considered. It has distributed loading, equal steel flanges, a uniform cross-section in Class 2, S355 steel, and stud connectors. At mid-span, MEd is much less than Mpl, Rd. The cross-section is such that the required resistance to bending can be provided using 40% of full shear connection (n/nf = 0.4). Clause 6.6.1.2(1) gives n/nf ≥ 0.55. However, if the slab is composite, and the other conditions of clause 6.6.1.2(3) are satisfied, n/nf = 0.4 may be used. Suppose now that the span of the beam is 12 m. The preceding limits to n/nf are increased to 0.61 and 0.48, respectively. One can either design using these limits, or go to clause 6.6.1.3(5), which refers to ‘longitudinal shear calculated by elastic theory’. This presumably means using vL = VEd A y /I, where VEd is the vertical shear on the composite section. This gives a triangular distribution of longitudinal shear, or separate distributions, to be superimposed, if creep is allowed for by using several modular ratios. The force in the slab at mid-span now depends on the proportion of MEd that is applied to the composite member, and on the proportions of the cross-section. Connectors corresponding to this force are then spaced accordingly, possibly with extra ones near mid-span to satisfy the rule of clause 6.6.5.5(3) on maximum spacing of studs.

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Strictly, the envelope of vertical shear should be used for VEd, which gives non-zero shear at mid-span. For distributed loading this increases the shear connection needed by only 2%, but the envelope should certainly be used for more complex variable loading. For continuous beams where MEd at mid-span is much less than Mpl, Rd, the method is more complex, as partial shear connection is permitted only where the slab is in compression. The envelope of design vertical shear from the global analysis should be used. For simplicity, longitudinal shear can be found using properties of the uncracked cross-section throughout, because this gives an overestimate in cracked regions. Examples 6.7 and 6.8 (see below) are relevant. It does not help, in the present example, to define additional critical sections within the 6 m shear span of the beam, because the limits of clause 6.6.1.2 are given in terms of effective span, not critical length.


6.6.2. Longitudinal shear force in beams for buildings

Clauses 6.6.2.1 and 6.6.2.2 say, in effect, that the design longitudinal shear force should be consistent with the bending resistances of the cross-sections at the ends of the critical length considered, not with the design vertical shear forces (the action effects). This is done for two reasons: • •

simplicity – for the design bending moments often lie between the elastic and plastic resistances, and calculation of longitudinal shear becomes complex robustness – for otherwise longitudinal shear failure, which may be more brittle than flexural failure, could occur first.

Beam with Class 3 sections at supports and a Class 1 or 2 section at mid-span Clause 6.6.2.1 applies because non-linear or elastic theory will have been ‘applied to cross-sections’. The longitudinal forces in the slab at the Class 3 sections are then calculated by elastic theory, based on the bending moments in the composite section. At mid-span, it is not clear whether clause 6.6.2.2 applies, because its heading does not say ‘resistance of all cross-sections’. The simpler and recommended method is to assume that it does apply, and to calculate the longitudinal force at mid-span based on MRd at that section, as that is consistent with the model used for bending. The total shear flow between a support and mid-span is the sum of the longitudinal forces at those points. The alternative would be to find the longitudinal force at mid-span by elastic theory for the moments applied to the composite section, even though the bending stresses could exceed the specified limits. Clause 6.6.2.2(3) This absence of ‘all’ from the heading is also relevant to the use of clause 6.6.2.2(3), on the use of partial shear connection. The design for a beam with Class 3 sections at internal supports limits the curvature of those regions, so the ultimate-load curvature at mid-span will be too low for the full-interaction bending resistance to be reached. The use of partial shear connection is then appropriate, with MRd less than Mpl, Rd.

6.6.3. Headed stud connectors in solid slabs and concrete encasement Resistance to longitudinal shear

In BS 5950-3-131 and in earlier UK codes, the characteristic shear resistances of studs are given in a table, applicable only when the stud material has particular properties. There was no theoretical model for the shear resistance. The Eurocodes must be applicable to a wider range of products, so design equations are Clause 6.6.3.1(1) essential. Those given in clause 6.6.3.1(1) are based on the model that a stud with a shank diameter d and an ultimate strength fu, set in concrete with a characteristic strength fck and a mean secant modulus Ecm, fails either in the steel alone or in the concrete alone. The concrete failure is found in tests to be influenced by both the stiffness and the strength of the concrete.

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PRd (kN)

100 Normal-density concrete

fu = 500 N/mm2

450 80 400

Density class 1.8 60

20

30

40

25

37

50

50 fck (N/mm2)

60

fcu (N/mm2)

Fig. 6.12. Design shear resistances of 19 mm studs with h/d ≥ 4 in solid slabs

This led to equations (6.18)–(6.21), in which the numerical constants and partial safety factor γV have been deduced from analyses of test data. In situations where the resistances from equations (6.18) and (6.19) are similar, tests show that interaction occurs between the two assumed modes of failure. An equation based on analyses of test data, but not on a defined model,70 PRd = k(πd2/4)fu(Ecm/Ea)0.4(fck/fu)0.35

(D6.16)

gives a curve with a shape that approximates better both to test data and to values tabulated in BS 5950. In the statistical analyses done for EN 1994-1-171,72 both of these methods were studied. Equation (D6.16) gave results with slightly less scatter, but the equations of clause 6.6.3.1(1) were preferred because of their clear basis and experience of their use in some countries. Here, and elsewhere in Section 6, coefficients from such analyses were modified slightly, to enable a single partial factor, denoted γV (‘V’ for shear), 1.25, to be recommended for all types of shear connection. This value has been used in draft Eurocodes for over 20 years. It was concluded from this study72 that the coefficient in equation (6.19) should be 0.26. This result was based on push tests, where the mean number of studs per specimen was only six, and where lateral restraint from the narrow test slabs was usually less stiff than in the concrete flange of a composite beam. Strength of studs in many beams is also increased by the presence of hogging transverse bending of the slab. For these reasons the coefficient was increased from 0.26 to 0.29, a value that is supported by a subsequent calibration study15 based on beams with partial shear connection. Design resistances of 19 mm stud connectors in solid slabs, given by clause 6.6.3.1, are shown in Fig. 6.12. It is assumed that the penalty for short studs, equation (6.20), does not apply. For any given values of fu and fck, the figure shows which failure mode governs. It can be used for this purpose for studs of other diameters, provided that h/d ≥ 4. The ‘overall nominal height’ of a stud, used in equations (6.20) and (6.21), is about 5 mm greater than the ‘length after welding’, a term which is also in use.

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Weld collars Clause 6.6.3.1(2) Clause 6.6.3.1(2) on weld collars refers to EN 13918,26 which gives ‘guide values’ for the height and diameter of collars, with the note that these may vary in through-deck stud welding. It is known that for studs with normal weld collars, a high proportion of the shear is transmitted through the collar.67 It should not be assumed that the shear resistances of clause 6.6.3.1 are applicable to studs without collars (e.g. where friction welding by high-speed spinning is used). A normal collar should be fused to the shank of the stud. Typical collars in the test specimens from which the design formulae were deduced had a diameter not less than 1.25d and a minimum height not less than 0.15d, where d is the diameter of the shank. The collars of studs welded through profiled sheeting can be of different shape from those for studs welded direct to steel flanges, and the shear strength may also depend on the effective diameter of the weld between the sheeting and the flange, about which little is known. The resistances of clause 6.6.3.1 are applicable where the welding is in accordance with EN ISO 14555,73 as is required. Studs with a diameter exceeding 20 mm are rarely used in buildings, as welding through sheeting becomes more difficult, and more powerful welding plant is required. Splitting of the slab Clause 6.6.3.1(3) Clause 6.6.3.1(3) refers to ‘splitting forces in the direction of the slab thickness’. These occur where the axis of a stud lies in a plane parallel to that of the concrete slab; for example, if studs are welded to the web of a steel T-section that projects into a concrete flange. These are referred to as ‘lying studs’ in published research74 on the local reinforcement needed to prevent or control splitting. There is an informative annex on this subject in EN 1994-2.11 A similar problem occurs in composite L-beams with studs close to a free edge of the slab. This is addressed in clause 6.6.5.3(2). Tension in studs Pressure under the head of a stud connector and friction on the shank normally causes the stud weld to be subjected to vertical tension before shear failure is reached. This is why clause 6.6.1.1(8) requires shear connectors to have a resistance to tension that is at least 10% of the Clause 6.6.3.2(2) shear resistance. Clause 6.6.3.2(2) therefore permits tensile forces that are less than this to be neglected. Resistance of studs to higher tensile forces has been found to depend on so many variables, especially the layout of local reinforcement, that no simple design rules could be given. It is usually possible to find other ways of resisting the vertical tension that occurs, e.g. where a travelling crane is supported from the steel element of a composite beam.

6.6.4. Design resistance of headed studs used with profiled steel sheeting in buildings

The load–slip behaviour of a stud connector in a trough (of sheeting) or rib (of concrete; both terms are used) (Fig. 6.13) is more complex than in a solid slab. It is influenced by • • • • •

the direction of the ribs relative to the span of the beam their mean breadth b0 and depth hp the diameter d and height hsc of the stud the number nr of the studs in one trough, and their spacing whether or not a stud is central within a trough, and, if not, by its eccentricity and the direction of the shear.

The interactions between these parameters have been explored by testing, with sheeting continuous across the beam. It is clear that the most significant are the ratios hsc/hp and b0/hp and, for ribs transverse to the supporting beams, nr and the eccentricity, if any. In EN 1994-1-1, as in earlier codes, reduction factors k (£ 1.0) are given, for application to the design resistances of studs in solid slabs. They are based entirely on testing and experience.

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40

hsc = 145 19 =

hp = 75 q

=

26

26 60

60

146, or b0 for continuous sheeting

Fig. 6.13. Details of a haunch, with parallel sheeting (dimensions in mm)

Sheeting with ribs parallel to the supporting beams (clause 6.6.4.1) There are two situations. The sheeting may be continuous across the beams – its side walls then provide lateral restraint to the concrete around the studs – or it may be discontinuous, as shown in Fig. 6.13, providing a haunch with a breadth that usually exceeds the breadth b0 of a trough. Edge fixings provided for erection may not provide much lateral restraint. The detailing rules for studs in unsheeted haunches, clause 6.6.5.4, then provide a guide to good practice, and may be more conservative than the rules of clause 6.6.4.1. Alternatively, the sheeting may be anchored to the beam. As practice varies, the means to achieve appropriate anchorage is a matter for the National Annex. A haunch that just complies with clause 6.6.5.4 is shown, to scale, in Fig. 6.13. The rules specify: • • • is

Clause 6.6.4.1

the angle θ (£ 45°) the concrete side cover to the connector (≥ 50 mm) the depth of the transverse reinforcement below the underside of the head (≥ 40 mm). The application of clause 6.6.4.1(2) to this haunch is now considered. The reduction factor kl = 0.6(b0 /hp)[(hsc /hp) – 1] £ 1

(6.22)

where hsc may not be taken as greater than hp + 75 mm. The equation is from Grant et al.,75 dating from 1977 because there is little recent research on ribs parallel to the beam.76 Here, it gives kl = 0.6(146/75)[(145/75) – 1] = 1.09 so there is no reduction. There would be, kl = 0.78, if the height of the stud were reduced to, say, 125 mm. In an unsheeted haunch it would then be necessary to provide transverse reinforcement at a lower level than in Fig. 6.13, which might be impracticable. Reasonable consistence is thus shown between clauses 6.6.4.1 and 6.6.5.4. There is no penalty for off-centre studs in either of these clauses, or for more than one at a cross-section. The rules of clause 6.6.4.1 for continuous or anchored sheeting, and of clause 6.6.5.4, for discontinuous un-anchored sheeting, should ensure good detailing.

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Where sheeting is continuous across a beam and through-deck stud welding is used, there may be shear transfer from the sheeting to the beam. No guidance is given on this complex situation, which may be ignored in design.

Clause 6.6.4.2

Sheeting with ribs transverse to the supporting beams (clause 6.6.4.2) In the preceding paragraphs, the likely mode of failure was loss of restraint to the base of the stud due to bursting (lateral expansion) of the haunch. A rib transverse to the beam, shown in Fig. 6.14 with typical dimensions, is more highly stressed, as it has to transfer much of the shear from the base of the stud to the continuous slab above. Three modes of failure are shown in Fig. 6.15(a)–(c): (a) failure surface above a stud that is too short (‘concrete pullout’) (b) haunch too slender, with plastic hinges in the stud (c) eccentricity on the ‘weak’ side of the centre, which reduces the effective breadth of the haunch,69 and causes ‘rib punching’ failure.77 In BS 5950-3-1,31 the reduction formula from Grant et al.75 is given. It is, essentially, kt = 0.85(b0 /hp)[(hsc/hp) – 1]/nr0.5 £ kt, max

(D6.17)

with kt, max falling from 1 to 0.6 as nr, the number of studs in a rib, increases from 1 to 3. Where a single stud is placed on the weak side of the centre of a rib, the breadth b0 is taken as 2e (Fig. 6.15(c)). In a study of this subject61 it was concluded that equation (D6.17) was inconsistent and could be unsafe. A new formula by Lawson78 gave better predictions, and was recommended. For the Eurocode, a review of Mottram and Johnson69 and Lawson,78 and their application to a wider range of profiles than is used in the UK, found that more test data were needed. It was initially decided72 to reduce the factor 0.85 in equation (D6.17) to 0.7, to eliminate situations where it had been found to be unsafe, with the conditions: nr not to be taken as greater than 2 in computations, b0 ≥ hp, and hp £ 85 mm, with kt, max as before. Where there is one stud per rib, Fig. 6.16 shows that: • •

where hsc ≥ 2hp and t > 1 mm, the reduction factor is usually 1.0 where hsc £ 1.5hp and/or t £ 1 mm, reductions in resistance can be large.

Most of the test data had been from sheeting exceeding 1.0 mm in thickness, with through-deck welding of studs up to 20 mm in diameter. Later work led to the reductions in kt, max for other situations, given in Table 6.2, for thinner sheeting and studs welded through

hc = 75 e = 30

hsc = 95 hp = 55

= =

50

b0 = 160

Fig. 6.14. Details of a haunch and positions of stud connectors, for trapezoidal sheeting transverse to the supporting beams (dimensions in mm)

74





e

(a)

(b)

(c)

(d)

(e)

Fig. 6.15. Failure modes and placing of studs, for troughs of profiled sheeting. The alternate favourable and unfavourable placing of pairs of studs is shown in (d), and the diagonal placing of pairs of studs in (e)

kt t > 1 mm 1.0

nr = 1 t £ 1 mm t > 1 mm

0.8

hsc/hp = 2

0.6

nr = 2 t £ 1 mm

nr = 1

hsc/hp = 1.5

1.0

nr = 2

2.0

3.0

b 0/ h p

Range common in practice

Fig. 6.16. Reduction factor kt for studs with d

20 mm and through-deck welding

holes. Research also found that the use of a reduction factor for the strength of studs in a solid slab is not appropriate where strong studs are placed in a relatively weak rib, so further limitations of scope were added: • •

stud diameter not to exceed the limits given in Table 6.2 ultimate strength of studs in sheeting not to be taken as greater than 450 N/mm2.

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Some trapezoidal sheetings have a small rib projecting above the main top surface; for example, as shown in Fig. 9.4. It is clear from clause 6.6.4.1(1) that in equation (6.22), hp is the depth including the rib. The words ‘overall depth’ are not repeated in clause 6.6.4.2, but should be assumed to apply, unless there is extensive test evidence that equation (6.23) gives safe predictions of kt when hp is taken as the depth excluding the rib. Crowding studs into a rib reduces ductility69 as well as strength. Ductility is needed most in long spans where, fortunately, longitudinal shears are usually low (in buildings) in relation to the width of steel top flange available for the placing of studs. This is relevant because the failures sketched in Fig. 6.15 are in reality three-dimensional, and resistances depend in a complex way on the arrangement of pairs of studs within a rib. Where sheeting has a small stiffening rib that prevents studs from being placed centrally within each trough, and only one stud per trough is required, application of the rule given in clause 6.6.5.8(3) is straightforward. Comments on that clause are given later. It is not clear from EN 1994-1-1 how two studs per trough should then be arranged. If there were no central rib, then two studs in line, spaced ≥ 5d apart (clause 6.6.5.7(4)) would be permitted. Small stiffening ribs, as shown in Fig. 6.14, should have little effect on shear resistance, so two-in-line is an acceptable layout. If the troughs are too narrow to permit this, then two studs side-by-side spaced ≥ 4d apart (clause 6.6.5.7(4)) fall within the scope of clause 6.6.5.8(3), and can be arranged as shown in plan in Fig. 6.15(d). There is limited evidence from tests76 that the diagonal layout of Fig. 6.15(e), to which the code does not refer, may be weaker than two studs in line.

Example 6.5: reduction factors for transverse sheeting

The sheeting is as shown in Fig. 6.14, with an assumed thickness t = 0.9 mm, including zinc coating. The overall depth is 55 mm. For one stud in the central location (shown by dashed lines in Fig. 6.14), b0 = 160 mm, hsc = 95 mm, and nr = 1. Equation (6.23) and Table 6.2 give kt = (0.7 × 160/55)(95/55 – 1) = 1.48 (but £ 0.85) For two studs per trough, placed centrally and side by side, kt = 1.48/1.41 = 1.05 (but £ 0.7) Further calculations of reduction factors are included in Example 6.7.

Clause 6.6.4.3

Biaxial loading of shear connectors The biaxial horizontal loading referred to in clause 6.6.4.3 occurs where stud connectors are used to provide end anchorage for composite slabs, as shown in Fig. 9.1(c). Clause 9.7.4(3) gives the design anchorage resistance as the lesser of that given by clause 6.6 and Ppb, Rd, the bearing resistance of the sheet, from equation (9.10). Even where 16 mm studs are used in lightweight concrete, the bearing resistance (typically about 14 kN) will be the lower of the two. These studs resist horizontal shear from both the slab and the beam. The interaction equation of clause 6.6.4.3(1) is based on vectorial addition of the two shear forces.

6.6.5. Detailing of the shear connection and influence of execution

It is rarely possible to prove the general validity of application rules for detailing, because they apply to so great a variety of situations. They are based partly on previous practice. An adverse experience causes the relevant rule to be made more restrictive. In research, existing rules are often violated when test specimens are designed, in the hope that extensive good experience may enable existing rules to be relaxed. Rules are often expressed in the form of limiting dimensions, even though most behaviour (excluding corrosion) is more influenced by ratios of dimensions than by a single value. Minimum dimensions that would be appropriate for an unusually large structural member

76





55 95

A

C 30

B

a

800

Fig. 6.17. Level of bottom transverse reinforcement (dimensions in mm)

could exceed those given in the code. Similarly, code maxima may be too large for use in a small member. Designers are unwise to follow detailing rules blindly, because no set of rules can be comprehensive.

Resistance to separation The object of clause 6.6.5.1(1) on resistance to separation is to ensure that failure surfaces in Clause 6.6.5.1(1) the concrete cannot pass above the connectors and below the reinforcement, intersecting neither. Tests have found that these surfaces may not be plane; the problem is threedimensional. A longitudinal section through a possible failure surface ABC is shown in Fig. 6.17. The studs are at the maximum spacing allowed by clause 6.6.5.5(3). Clause 6.6.5.1 defines only the highest level for the bottom reinforcement. Ideally, its longitudinal location relative to the studs should also be defined, because the objective is to prevent failure surfaces where the angle α (Fig. 6.17) is small. It is impracticable to specify a minimum angle α, or to link detailing rules for reinforcement with those for connectors. The angle α obviously depends on the spacing of the bottom bars, assumed to be 800 mm in Fig. 6.17. What is the maximum permitted value for this spacing? The answer is quite complex. Clause 6.6.6.3(1) refers to clause 9.2.2(5) in EN 1992-1-1, where a Note recommends a minimum reinforcement ratio. For fck = 30 N/mm2 and fsk = 500 N/mm2 this gives 0.088%, or 131 mm2/m for this 150 mm slab. However, if the slab is continuous across the beam, most of this could be in the top of the slab. The minimum bottom reinforcement depends on whether the slab is continuous or not. If it is simply-supported on the beam, the bottom bars are ‘principal reinforcement’ to clause 9.3.1.1(3) of EN 1992-1-1, where a Note gives their maximum spacing as 400 mm for this example, and 450 mm for ‘secondary’ reinforcement. The rules of clause 6.6.5.3(2) on splitting may also apply. If the slab is continuous over the beam, there may be no need for bottom flexural reinforcement to EN 1992-1-1. Let us assume that there is a single row of 19 mm studs with a design resistance, 91 kN per stud, which is possible in concrete of class C50/60 (Fig. 6.12). The bottom transverse reinforcement will then be determined by the rules of clause 6.6.6.1 for shear surfaces of type b–b or c–c. These show that 12 mm bars at 750 mm spacing are sufficient. There appears to be no rule that requires closer spacing, but it would be prudent to treat these bars as ‘secondary flexural’ bars to EN 1992-1-1, not least because it increases the angle α. Using its maximum spacing, 450 mm, 10 mm bars are sufficient. Cover to connectors Shear connectors must project significantly above profiled steel sheeting, because of the term (hsc /hp – 1) in equations (6.22) and (6.23) and the rule in clause 6.6.5.8(1) for projection of 2d above ‘the top of the steel deck’. The interpretation of this for profiles with a small additional top rib is not clear. If 2d is measured from this rib, and the rules of EN 1992-1-1 for cover are applied to the top of the connector, the resulting minimum thickness of a composite slab may govern its thickness. However, these slabs are normally used only in dry environments, where the concessions on minimum cover given in clause 6.6.5.2 are appropriate.

Clause 6.6.5.2

77





Loading of shear connection during execution As almost all relevant provisions on execution appear in standards for either steel or Clause 6.6.5.2(4) concrete structures, there is no section on execution in EN 1994-1-1. This is why clause 6.6.5.2(4) appears here. The method of construction of composite beams and slabs (i.e. whether propped or unpropped), and also the sequence of concreting, affects stresses calculated by elastic theory, and the magnitude of deflections. It is significant, therefore, in verifications of resistance of cross-sections in Class 3 or 4, and in serviceability checks. Where propped construction is used, it is usual to retain the props until the concrete has achieved a compressive strength of at least 75% of its design value. If this is not done, verifications at removal of props should be based on a reduced compressive strength. Clause 6.6.5.2(4) gives a lower limit for this reduction in concrete strength. It also relates to the staged casting of a concrete flange for an unpropped composite beam, setting, in effect, a minimum time interval between successive stages of casting. Local reinforcement in the slab Where shear connectors are close to a longitudinal edge of a concrete flange, use of U-bars is Clause 6.6.5.3(1) almost the only way of providing the full anchorage required by clause 6.6.5.3(1). The Clause 6.6.5.3(2) splitting referred to in clause 6.6.5.3(2) is a common mode of failure in push-test specimens with narrow slabs (e.g. 300 mm, which has long been the standard width in British codes). It was also found, in full-scale tests, to be the normal failure mode for composite L-beams constructed with precast slabs.79 Detailing rules are given in clause 6.6.5.3(2) for slabs where the edge distance e in Fig. 6.18 is less than 300 mm. The required area of bottom transverse reinforcement, Ab, per unit length of beam, should be found using clause 6.6.6. In the unhaunched slab shown in Fig. 6.18, failure surface b–b will be critical (unless the slab is very thick) because the shear on surface a–a is low in an L-beam with an asymmetrical concrete flange. To ensure that the reinforcement is fully anchored to the left of the line a–a, it is recommended that U-bars be used. These can be in a horizontal plane or, where top reinforcement is needed, in a vertical plane. No rules are given for the effectiveness as transverse reinforcement of profiled sheeting transverse to the supporting beams, with ribs that extend to a cantilever edge of the slab. The length needed to develop the full tensile resistance of the sheeting will be known from the design procedure for the composite slab. It is always greater than the minimum edge distance of 6d, and usually greater than 300 mm. Where it exceeds the length e available, bottom transverse reinforcement will be needed. The situation can be improved by placing all the connectors near the inner edge of the steel flange. Sheeting with ribs parallel to the free edge should not be assumed to resist longitudinal splitting of the slab. e (≥ 6 d ) a

b

d

a

b

Fig. 6.18. Longitudinal shear reinforcement in an L-beam

78



≥ 0.5 d



≥ 6d ≥ 6d (a)

(b)

Fig. 6.19. Examples of details susceptible to longitudinal splitting

A

Steel beam

B

Studs Slab

C Transverse reinforcement not shown

Fig. 6.20. Reinforcement at the end of a cantilever

Lying studs There is also a risk of splitting where the shank of a stud (a ‘lying stud’) is parallel and close to a free surface of the slab, as shown, for example, in Fig. 6.19. The U-bars should then be in a vertical plane. Research has found74 that the ‘6d rule’ for edge distance in clause 6.6.5.3(2) and the provision of hoops may not be sufficient to ensure that the design shear resistance of the stud is reached. The height of the stud and the longitudinal bars shown are also important. The slip capacity may be less than 6 mm, so the shear connection may not be ‘ductile’. The stud shown in Fig. 6.19(b) may be subjected to simultaneous axial tension and vertical and longitudinal shear, so details of this type should be avoided. Lying studs are outside the scope of EN 1994-1-1. There is an informative annex in EN 1994-2. Reinforcement at the end of a cantilever At the end of a composite cantilever, the force on the concrete from the connectors acts towards the nearest edge of the slab. The effects of shrinkage and temperature can add further stresses37 that tend to cause splitting in region B in Fig. 6.20, so reinforcement in this region needs careful detailing. Clause 6.6.5.3(3)P can be satisfied by providing ‘herring-bone’ Clause 6.6.5.3(3)P bottom reinforcement (ABC in Fig. 6.20) sufficient to anchor the force from the connectors into the slab, and to ensure that the longitudinal bars provided to resist that force are anchored beyond their intersection with ABC. The situation where a column is supported at point B is particularly critical. Incompatibility between the vertical stiffnesses of the steel beam and the slab can cause local shear failure of the slab, even where the steel beam alone is locally strong enough to carry the column.80 Haunches The detailing rules of clause 6.6.5.4 are based on limited test evidence, but are longestablished.81 In regions of high longitudinal shear, deep haunches should be used with caution because there may be little warning of failure. Haunches formed by profiled sheeting are considered under clause 6.6.4.1.

Clause 6.6.5.4

79





Maximum spacing of connectors Situations where the stability of a concrete slab is ensured by its connection to a steel beam are unlikely to occur in buildings. The converse situation, stabilization of the steel flange, is of interest only where the steel compression flange is not already in Class 1, so it rarely arises where rolled I-sections are used. Where the steel beam is a plate girder, there is unlikely to be any need for a top flange in Class 1. Its proportions can usually be chosen such that it is in Class 2, unless a wide thin flange is needed to avoid lateral buckling during construction. Clause 6.6.5.5(2) Clause 6.6.5.5(2) is not restrictive in practice. As an example, a plate girder is considered, in steel with fy = 355 N/mm2, where the top flange has tf = 20 mm, an overall breadth of 350 mm, and an outstand c of 165 mm. The ratio ε is 0.81 and the slenderness is c/tf ε = 165/(20 × 0.81) = 10.2 so, from Table 5.2 of EN 1993-1-1, the flange is in Class 3. From clause 6.6.5.5(2), it can be assumed to be in Class 1 if shear connectors are provided within 146 mm of each free edge, at a longitudinal spacing not exceeding 356 mm, for a solid slab. The ratio 22 in this clause is based on the assumption that the steel flange cannot buckle towards the slab. Where there are transverse ribs (e.g. due to the use of profiled sheeting), the assumption may not be correct, so the ratio is reduced to 15. The maximum spacing in this example is then 243 mm. The ratio 9 for edge distance, used in the formula 9tf(235/fy)0.5 is the same as in BS 5400: Part 5,82 and the ratio 22 for longitudinal spacing is the same as the ratio for staggered rows given in BS 5400. Clause 6.6.5.5(3) The maximum longitudinal spacing in buildings, given in clause 6.6.5.5(3), 6hc, £ 800 mm, is more liberal than the general rule of BS 5950-3-1,31 though that code permits a conditional increase to 8hc. The rule of EN 1994-1-1 is based on behaviour observed in tests, particularly those where composite slabs have studs in alternate ribs only. Some uplift then occurs at intermediate ribs. Spacing at 800 mm would in some situations allow shear connection only in every third rib, and there was a requirement in ENV 1994-1-1 for anchorage, but not necessarily shear connection, to be provided in every rib. It is expected that this requirement will be included in the forthcoming EN 1090 for execution of steel structures.

Detailing, for stud connectors The rules of clause 6.6.5.6 are intended to prevent local overstress of a steel flange near a Clause 6.6.5.6 shear connector and to avoid problems with stud welding. Application rules for minimum Clause 6.6.5.7 flange thickness are given in clause 6.6.5.7. Clauses 6.6.5.7(1) and 6.6.5.7(2) are concerned with resistance to uplift. Rules for resistance of studs, minimum cover and projection of studs above bottom reinforcement usually lead to the use of studs of height greater than 3d. Clause 6.6.5.7(3) The limit 1.5 for the ratio d/tf in clause 6.6.5.7(3) influences the design of shear connection for closed-top box girders in bridges. For buildings, the more liberal limit of clause 6.6.5.7(5) normally applies. It has appeared in several earlier codes. Clause 6.6.5.7(4) In clause 6.6.5.7(4), the minimum lateral spacing of studs in ‘solid slabs’ has been reduced to 2.5d, compared with the 4d of BS 5950-3-1. Although connection to precast slabs is outside the scope of EN 1994-1-1, this facilitates the use of large precast slabs supported on the edges of the steel flanges, with projecting U-bars that loop over pairs of studs. There is much experience, validated by tests,83 of this form of construction, especially in multistorey car parks. Closely spaced pairs of studs must be well confined laterally. The term ‘solid slabs’ should therefore be understood to exclude haunches. Further detailing rules for studs placed within troughs of profiled sheeting are given in Clause 6.6.5.8 clause 6.6.5.8. The first two concern resistance to uplift and compaction of concrete around studs.

80





Clause 6.6.5.8(3) relates to a common situation: where a small stiffening rib in sheeting Clause 6.6.5.8(3) prevents the placing of studs centrally within a trough. It would be prudent to locate studs on the ‘favourable’ side (see Fig. 6.15(a)), which, for symmetrical loading on a simply-supported span, is the side nearest the nearer support. This was not given as an application rule because it is difficult to ensure that the ‘favourable’ side would be correctly chosen on site. Instead, alternate-side placing is recommended, with alternate studs on the ‘unfavourable’ side (see Fig. 6.15(c)). However, research has found77 that the mean strength of pairs of studs placed on the two sides of a trough is about 5% less than if both were central, with a greater reduction for sheeting less than 1 mm thick. Sheeting profiles with troughs wide enough to have off-centre stiffening ribs are more suitable for composite slabs than those with central ribs. This clause does not refer to layouts that require two studs per trough. These are discussed in comments on clause 6.6.4.2

6.6.6. Longitudinal shear in concrete slabs

The subject of clause 6.6.6 is the avoidance of local failure of a concrete flange near the Clause 6.6.6 shear connection, by the provision of appropriate reinforcement. These bars enhance the resistance of a thin concrete slab to in-plane shear in the same way that stirrups strengthen a concrete web in vertical shear. Transverse reinforcement is also needed to control and limit the longitudinal splitting of the slab that can be caused by local forces from individual connectors. In this respect, the detailing problem is more acute than in the flanges of concrete T-beams, where the shear from the web is applied more uniformly. The principal change from earlier codes is that the equations for the required area of transverse reinforcement have been replaced by cross-reference to EN 1992-1-1. Its provisions are based on a truss analogy, as before, but a more general version of it, in which the angle between members of the truss can be chosen by the designer. It is an application of strut-and-tie modelling, which is widely used in EN 1992-1-1. The definitions of shear surfaces in clause 6.6.6.1(2)P and the basic design method are as Clause 6.6.6.1(2)P before. The method of presentation reflects the need to separate the ‘general’ provisions (clauses 6.6.6.1 to 6.6.6.3) from those restricted to ‘buildings’ (clause 6.6.6.4). Clause 6.6.6.1(4) requires the design longitudinal shear to be ‘consistent with’ that used for Clause 6.6.6.1(4) the design of the shear connectors. This means that the distribution along the beam of resistance to in-plane shear in the slab should be the same as that assumed for the design of the shear connection. For example, uniform resistance to longitudinal shear flow (vL) should be provided where the connectors are uniformly spaced, even if the vertical shear over the length considered is not constant. It does not mean, for example, that if, for reasons concerning detailing, vL, Rd = 1.3vL, Ed for the connectors, the transverse reinforcement must provide the same degree of over-strength. In applying clause 6.6.6.1(5), it is sufficiently accurate to assume that longitudinal bending Clause 6.6.6.1(5) stress in the concrete flange is constant across its effective width, and zero outside it. The clause is relevant, for example, to finding the shear on the plane a–a in the haunched beam shown in Fig. 6.15, which, for a symmetrical flange, is less than half of the shear resisted by the connectors.

Resistance of a concrete flange to longitudinal shear Clause 6.6.6.2(1) refers to clause 6.2.4 of EN 1992-1-1, which is written for a design Clause 6.6.6.2(1) longitudinal shear stress vEd acting on a cross-section of thickness hf. The clause requires the area of transverse reinforcement Asf at spacing sf to satisfy Asf fyd /sf > vEd hf /cot θf

(6.21 in EN 1992-1-1)

and the longitudinal shear stress to satisfy vEd < νfcd sin θf cos θf

(6.22 in EN 1992-1-1)

where ν = 0.6(1 – fck /250), with fck in units of newtons per square millimetre. (The Greek

81





letter ν used here in EN 1992-1-1 should not be confused with the Roman letter v, which is used for shear stress.) The angle θf between the diagonal strut and the axis of the beam is chosen (within limits) by the designer. The use of the method is illustrated in the following example.

Example 6.6: transverse reinforcement for longitudinal shear

Figure 6.21 shows a plan of an area ABCD of a concrete flange, assumed to be in longitudinal compression, with shear stress vEd and transverse reinforcing bars of area Asf at spacing sf. The shear force per transverse bar is Fv = vEd hf sf acting on side AB of the rectangle shown. It is transferred to side CD by a concrete strut AC at angle θf to AB, and with edges that pass through the mid-points of AB, etc., as shown, so that the width of the strut is sf sin θf. For equilibrium at A, the force in the strut is Fc = Fv sec θf

(D6.18)

For equilibrium at C, the force in the transverse bar BC is Ft = Fc sin θf = Fv tan θf

(D6.19)

For minimum area of transverse reinforcement, θf should be chosen to be as small as possible. For a flange in compression, the limits to θf given in clause 6.2.4(4) of EN 1992-1-1 are 45° ≥ θf ≥ 26.5°

(D6.20)

so the initial choice for θf is 26.5°. Then, from equation (D6.19), Ft = 0.5Fv

(D6.21)

From equation (6.22) above, vEd < 0.40νf cd If this inequality is satisfied, then the value chosen for θf is satisfactory. However, let us assume that the concrete strut is over-stressed, because vEd = 0.48νf cd To satisfy equation (6.22) Fv A

B qf

Ft Fc

D

C

Fv

sf sin qf sf

Fig. 6.21. Truss analogy for in-plane shear in a concrete flange

82





sin θf cos θf ≥ 0.48 whence θf ≥ 37° The designer increases θf to 40°, which satisfies expression (D6.20). From equation (D6.19), Ft = Fv tan 40° = 0.84Fv From equation (D6.21), the change in θf, made to limit the compressive stress in the concrete strut AC, increases the required area of transverse reinforcement by 68%. The lengths of the sides of the triangle ABC in Fig. 6.21 are proportional to the forces Fv, Ft and Fc. For given Fv and sf, increasing θf increases Fc, but for θf < 45° (the maximum value permitted), the increase in the width sf sin θf is greater, so the stress in the concrete is reduced.

Shear planes Clause 6.6.6.1(3) refers to ‘shear surfaces’. Those of types b–b, c–c, and d–d in Fig. 6.15 are different from the type a–a surface because they resist almost the whole longitudinal shear, not (typically) about half of it. The relevant reinforcement intersects them twice, as shown by the factor 2 in the table in Fig. 6.15. For a surface of type c–c in a beam with the steel section near one edge of the concrete flange, it is clearly wrong to assume that half of the shear crosses half of the surface c–c. However, in this situation the shear on the adjacent plane of type a–a will govern, so the method is not unsafe. Minimum transverse reinforcement Clause 6.6.6.3 on this subject is discussed under clause 6.6.5.1 on resistance to separation. Longitudinal shear in beams with composite slabs (clause 6.6.6.4) Design rules are given for sheeting with troughs that run either transverse to the span of the steel beam, or parallel to it. Where they intersect a beam at some other angle, one can either use the more adverse of the two sets of rules, or combine them in an appropriate way. The rule for thickness of concrete in clause 6.6.6.4(1) is independent of the direction of the sheeting. In clause 6.6.6.4(2), the ‘type b–b’ shear surface is as shown in Fig. 6.16, in which the labelling of the shear surfaces differs from that in Fig. 6.15. The contribution made by sheeting to resistance in longitudinal shear depends not only on its direction but also on whether the designer can determine the position of the ends of individual sheets, and whether these ends are attached to the steel beam by through-deck welding of stud connectors. If these decisions are made subsequently by the contractor, the designer may be unable to use transverse sheeting as reinforcement for shear in the plane of the slab. Its contribution to shear resistance is substantial where it is continuous across the beam (clause 6.6.6.4(4)) and is useful where through-deck welding is used (clause 6.6.6.4(5)).This is applied in Example 6.7. Sheets attached only by fixings used for erection and those with their span parallel to that of the beam are ineffective as transverse reinforcement. Through-deck welding of studs is also used to enhance the resistance of a composite slab to longitudinal shear. EN 1994-1-1 does not state whether both procedures can be used at once. The question is discussed with reference to Fig. 6.22, which shows an exploded view of the base of a stud welded to a steel flange through a layer of sheeting, which spans towards the right-hand side of the diagram. The concrete in the mid-span region of the composite slab


Clause 6.6.6.4(1) Clause 6.6.6.4(2)

Clause 6.6.6.4(4) Clause 6.6.6.4(5)

83





A

Stud

A A A+B

Sheeting

B B

B Flange

Fig. 6.22. A through-deck welded stud acting as an end anchorage for a composite slab, and also providing continuity for transverse reinforcement

applies the force A to the upper part of the stud, which enables it to anchor a tension A in the sheeting. The force B is due to the action of the steel top flange and the sheeting as equivalent transverse reinforcement. It is clear from Fig. 6.22 that forces A and B are not additive for the stud, because A exists above the sheeting and B below it, but they are additive for the sheeting. The model in clause 9.7.4(3) for the calculation of resistance Ppb, Rd is not shear failure of the stud but bearing failure of the sheeting, which depends on the distance from the stud to the edge of the sheet. This analysis shows that if the two procedures are used for the same stud, the available resistance Ppb, Rd should be split between them, in whatever ratio may be required.

Example 6.7: two-span beam with a composite slab – ultimate limit state

A floor structure for a department store consists of composite beams of uniform cross-section at 2.50 m spacing, fully continuous over two equal spans of 12.0 m. The floor consists of a composite slab of overall thickness 130 mm, spanning between the beams. The three supports for each beam may be treated as point supports, providing lateral and vertical restraint. A design is required for an internal beam, subjected to vertical loading only. The design service life is 50 years. This particular concept is unlikely to be used in practice. The design will be found to be governed by resistance at the internal support, and only a small part of the bending resistance at mid-span can be used. However, it illustrates the use of many of the provisions of EN 1994-1-1, and is chosen for that reason. The design of the composite slab, the checking of the beam for serviceability limit states, and the influence of semi-continuous beam-to-column connections are the subjects of subsequent worked examples in this and later chapters of this guide. It will be assumed initially that unpropped construction is used, and that the whole 24 m length of concrete flange is cast before significant composite action is developed in any of this length.

Loadings and materials The characteristic imposed load, including partitions, is qk = 7.0 kN/m2

γF = 1.50

where γF is a partial safety factor. The floor finish adds gk1 = 1.20 kN/m2

γF = 1.35

These are applied to the composite structure. For simplicity, the floor finish will be treated in global analyses as an imposed load, which is conservative.

84





E

beff = 1600 12∆ at 125

12∆ at 200 30

42

80 1.0

88

50

IPE 450

E

450

25

379

75

100

(b)

9.4 21

D

14.6 190 B

A

(a)

12 000

C

D 12 000

(c)

Fig. 6.23. Elevation and cross-sections of the two-span beam (dimensions in mm). (a) Section D–D. (b) Section E–E. (c) Elevation of beam Table 6.2. Loadings per unit length of beam Load (kN/m) Characteristic Composite slab Steel beam Total, on steel beam

Ultimate (minimum)

Ultimate (maximum)

5.02 0.76 5.78

6.78 1.02 7.80

6.78 1.02 7.80

Floor finishes Imposed load

1.20 17.50

1.62 0

1.62 26.25

Total, for composite beam

18.7

1.62

27.9

Total, for vertical shear

24.5

9.42

35.7

Preliminary studies have led to the choice of lightweight-aggregate concrete, of density class 1.8 and strength class LC25/28. For an oven-dry density not exceeding 1800 kg/m3, Table 11.1 in clause 11.3 of EN 1992-1-1 gives the design density of this reinforced concrete as 1950 kg/m3, which is now assumed to include the sheeting. The chosen steel section is IPE 450, with the dimensions shown in Fig. 6.23. Its weight is gk3 = 0.76 kN/m, with γF = 1.35. Table 6.2 gives the loadings for a beam spacing of 2.5 m. At this stage, the shape for the profiled sheeting is assumed to be as in Fig. 6.23. The mean thickness of the floor is 0.105 m, giving a characteristic load gk2 = 0.105 × 1.95 × 9.81 = 2.01 kN/m2 with a partial safety factor γF = 1.35.

Properties of materials Structural steel: grade S355, with γM = 1.0, so

85





fy = fyd = 355 N/mm2

(D6.22)

and ε = ÷(235/255) = 0.81 Concrete: fck = 25 N/mm2, γC = 1.5, 0.85fcd = 14.2 N/mm2

(D6.23)

From clause 11.3.1 of EN 1992-1-1, η1 = 0.4 + 0.6 × 1800/2200 = 0.891, so the mean tensile strength is flctm = 0.891 × 2.6 = 2.32 N/mm2

(D6.24) 2

2

From clause 11.3.2 of EN 1992-1-1, Elcm = 31 × (18/22) = 20.7 kN/mm , so the shortterm modular ratio is n0 = 210/20.7 = 10.1 From informative Annex C, and a ‘dry’ environment, the ‘nominal total final free shrinkage strain’ is εcs = –500 × 10–6 Reinforcement: fsk = 500 N/mm2, γS = 1.15, fsd = 435 N/mm2 Ductility: to be in Class B or C, from clause 5.5.1(5). Shear connectors: it is assumed that 19 mm studs will be used, welded through the steel sheeting, with ultimate tensile strength fu = 500 N/mm2 and γV = 1.25.

Durability The floor finish is assumed to be such that the top of the slab is exposed to ‘low air humidity’. From clause 4.2 of EN 1992-1-1, the exposure class is XC1. The minimum cover (clause 4.4.1) is then 15 mm for a service life of 50 years, plus a tolerance of between 5 and 10 mm that depends on the quality assurance system, and is here taken as 9 mm. At the internal support, the 12 mm longitudinal bars are placed above the transverse bars, with a top cover of 24 mm, as shown in Fig. 6.23. Properties of the IPE 450 cross-section From section tables: Area:

Aa = 9880 mm2

fillet radius: r = 21 mm

Second moments of area:

10–6Iay = 337.4 mm4

10–6Iaz = 16.8 mm4

Torsional moment of area:

10–6Iat = 0.659 mm4

Radii of gyration:

iy = 185 mm

Section moduli:

10–6Way = 1.50 mm3

Plastic section modulus:

10–6Wpl, a, y = 1.702 mm3

iz = 41.2 mm

ix = 190 mm

10–6Waz = 0.176 mm3 (D6.25)

The factor 10–6 is used to enable moments in kN m to be related to stresses in N/mm2 without further adjustment, and because it gives numbers of convenient size.

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Effective widths of concrete flange In Fig. 5.1 of clause 5.4.1.2, L1 = L2 = 12.0 m; Le for beff,1 = 10.2 m. Assume b0 = 0.1 m; then b1 = b2 = 2.5/2 – 0.05 = 1.20 m At mid-span, beff = 0.1 + 2 × 10.2/8 = 2.65 m (but £ 2.5 m) so bei = 2.4/2 = 1.2 m. At the internal support, Le for beff, 2 = 0.25 × 24 = 6 m; beff = 0.1 + 2 × 6/8 = 1.60 m At an end support, bei = 1.20 m; βi = 0.55 + 0.025 × 10.2/1.20 = 0.762 beff = 0.1 + 2 × 0.762 × 1.2 = 1.83 m

Classification of composite cross-section The class of the web is quite sensitive to the area of longitudinal reinforcement in the slab at the internal support. It is (inconveniently) necessary to assume a value for this before the checks that govern it can be made. Large-diameter bars may not give sufficient control of crack widths, so the reinforcement is assumed to be 12 mm bars at 125 mm, giving 13 bars within a 1.625 m width, so As = 1470 mm2 The effective area of concrete slab is 1.6 m by 80 mm, so As /Ac = ρs = 0.0113 The requirement of clause 5.5.1(5) for minimum ρs will be checked during the design for crack-width control; see Example 7.1. The force in these bars at yield is Fs, y = 1470 × 0.435 = 639 kN Assuming a Class 2 section, the stress distribution for Mpl, Rd is needed. Starting from the stress distribution for Mpl, a, Rd, the depth of steel web that changes from tension to compression is 639/(9.4 × 2 × 0.355) = 96 mm For classification, clause 5.5.1(1)P refers to clause 5.5.2 of EN 1993-1-1. Its Table 5.2 defines the depth of web, c, as that bounded by the root radii. Here, c = 379 mm, of which the depth in compression is 379/2 + 96 = 285 mm whence α = 285/379 = 0.75 This exceeds c/2, so from Table 5.2, for Class 2, c/t £ 456 ε/(13α – 1) = 42.2 The actual c/t = 381/9.4 = 40.5; so at the internal support, the web is Class 2. At mid-span, it is obviously in Class 1 or 2. For the compression flange, from Table 5.2, c = (190 – 9.4)/23 – 21 = 69.3 mm Hence

87





c/tε = 69.3/(14.6 × 0.81) = 5.86 The limit for a Class 1 flange is 9.0; so the bottom flange is Class 1. The member is Class 2 at the internal support and Class 1 or 2 at mid-span. It follows from clause 5.4.2.4(2) that for global analyses for ultimate limit states, (provided that lateral–torsional buckling does not govern) the whole of the loading may be assumed to act on the composite member; and from clause 6.2.1.1(1)P that rigid-plastic theory may be used for resistances to bending at all cross-sections of the beam. The use of partial shear connection (clause 6.2.1.3(1)) is limited to sagging regions, in accordance with clause 6.6.2.2, which is for ‘beams in which plastic theory is used for resistance of cross-sections’ (i.e. sections in Class 1 or 2). Design is thus much simpler when there are no beams with cross-sections in Class 3 or 4. This can usually be achieved in buildings, but rarely in multi-span bridges. It is notable that if the steel top flange were in Class 3, its connection to the slab would not enable it to be upgraded, because the condition of clause 6.6.5.5(2) is that the spacing of shear connectors does not exceed 15tfε, which is 15 × 14.6 × 0.81 = 177 mm. This would be impracticable with the sheeting profile shown in Fig. 6.23.

Plastic resistance to bending (clause 6.2.1.2) At the internal support, it has been found (above) that a 96 mm depth of the upper half of the web is in compression. The design plastic resistance to hogging bending is that of the steel section plus the effect of the reinforcing bars, shown in Fig. 6.24(a): Mpl, a, Rd = 1.702 × 0.355 = 604 kN m

(hogging and sagging)

Mpl, Rd = 604 + 639 × 0.277 = 781 kN m

(hogging)

The characteristic plastic resistance is also needed. With γS = 1 for the reinforcement, its force at yield increases to 639 × 1.15 = 735 kN; the depth of web to change stress is 110 mm; and, by the method used for Mpl, Rd, Mpl, Rk = 802 kN m

(hogging)

All of these resistances may need to be reduced to allow for lateral–torsional buckling or vertical shear. For sagging bending at mid-span, reinforcement in compression is ignored, and the available area of concrete is 2.5 m wide and 80 mm thick. If it is all stressed to 0.85fcd, the compressive force is Fc = 14.2 × 2.5 × 80 = 2840 kN

(D6.26)

If the whole steel section is at yield, the tensile force is 1600

2500 639 kN

100

2840 kN 90

93

277 225

225 96

5

667 kN

222

639 kN 3507 kN

604 kN m

(a)

(b)

Fig. 6.24. Plastic moments of resistance in (a) hogging and (b) sagging bending (dimensions in mm)

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Table 6.3. Elastic section properties of the composite cross-section Cross-section

Modular ratio

beff (m)

Neutral axis (mm)

10-6Iy (mm4)

10-6Wc, top (mm3)

(1) Support, cracked, reinforced (2) Support, uncracked (3) Support, uncracked (4) Mid-span, uncracked (5) Mid-span, uncracked (6) Mid-span, uncracked

– 10.1 20.2 10.1 20.2 28.7

1.6 1.6 1.6 2.5 2.5 2.5

42 177 123 210 158 130

467 894 718 996 828 741

– 50.7 62.5 69.5 84.7 94.5

Fa = 9880 × 0.355 = 3507 kN Assuming full shear connection, with the plastic neutral axis in the steel top flange, the thickness of flange to change from tension to compression is tf, c = (3507 – 2840)/(2 × 0.355 × 190) = 5.0 mm The longitudinal forces are then as shown in Fig. 6.24(b), and Mpl, Rd = 2840 × 0.315 + 667 × 0.222 = 1043 kN m

(sagging)

(D6.27)

This resistance will be reduced later by the use of partial shear connection.

Plastic resistance to vertical shear Clause 6.2.2.2 refers to clause 6.2.6 of EN 1993-1-1. This defines the shear area of a rolled I-section as Av = A – 2btf + (tw + 2r)tf = 9880 – 380 × 14.6 + 51.4 × 14.6 = 5082 mm2 and gives the design plastic shear resistance as Vpl, Rd = Av( fy/÷3)/γM0 = 5082 × 0.355/÷3 = 1042 kN For shear buckling, clause 6.2.2.3 refers to Section 5 of EN 1993-1-5. No buckling check is required, because hw/t for the steel web, based on the depth between the flanges, is 45, below the limit of 48.6.

Flexural properties of elastic cross-sections Several sets of elastic properties are needed, even where the steel beam is of uniform section, because of changes of modular ratio and effective width, and the use of ‘cracked’ and ‘uncracked’ sections. Here, slab reinforcement is ignored in the ‘uncracked’ analyses. The error is conservative (except for the shear connection) and usually very small. It is convenient to calculate all these properties at the outset (Table 6.3). From clause 5.4.2.2(11), creep may be allowed for by using a modular ratio n = 2n0 = 20.2 for both short-term and long-term loadings. Results for n = n0 are included in Table 6.3 for use in Chapter 7. Effects of shrinkage are unusually high in this example. For these, it helps to use the more accurate modular ratio 28.7, as explained later. The calculation for the uncracked properties at the internal support, with n = 10.1, is now given, as an example. In ‘steel’ units, the width of the slab is 1.6/10.1 = 0.158 m, so the composite section is as shown in Fig. 6.25. Its properties are: Area: A = 9880 + 158 × 80 = 9880 + 12640 = 22 520 mm2 Height of neutral axis above centre of steel section: zna = 12 640 × 315/22 520 = 177 mm

89





Second moment of area: 10–6Iy = 337.4 + 9880 × 0.1772 + 12 640 × 0.1382 = 894 mm4 Flexural stiffness: 10–6Ea Iy = 210 × 894 = 187 700 kN mm2 Section modulus, top of slab, in concrete units: 10–6Wc, top = 894 × 10.1/178 = 50.7 mm3 Similar calculations were done for the other elastic section properties required. The results are given in Table 6.3.

Global analysis The resistance to lateral–torsional buckling of the beam near the internal support, Mb, Rd, depends on the bending-moment distribution, so global analysis is done next. Wherever possible, the rules for global analysis in Section 5 apply to both ultimate and serviceability limit states. Where alternatives are permitted, both types of limit state should be considered before making a choice. For this beam, there is no need to take account of the flexibility of joints (clause 5.1.2). First-order elastic theory may be used (clauses 5.2.2(1) and 5.4.1.1(1)). Clause 5.2.2(4) on imperfections is satisfied, because lateral–torsional buckling is the only type of instability that need be considered, and its resistance formulae take account of imperfections. The simplest method of allowing for cracking, in clause 5.4.2.3(3), is applicable here, and will be used. Cracking almost always occurs in continuous beams. In this example, calculations for uncracked sections, using the longer method of clause 5.4.2.3(2), found that for the ultimate imposed loading on both spans, the tensile stress in the concrete at the internal support exceeded three times its tensile strength. This ignored shrinkage, which further increases the tension. For this beam, clause 5.4.2.3(3) requires the use of cracked section properties over a length of 1.8 m each side of the internal support. For global analysis, clause 5.4.1.2(4) permits the use of the mid-span effective width for the whole span; but here beff for the cracked region is taken as 1.6 m, because resistances are based on this width and reinforcement outside it may be quite light. The proposed use of n = 20.2 for the uncracked region merits discussion. Most of the permanent load is applied to the steel beam, which does not creep, but resistances are based on the whole of the load acting on the composite beam. Bending moment at the internal support governs. Creep increases this, and the long-term effects of shrinkage are so significant that the case t Æ • is more critical than t ª 0. 158

12 640 90

138

225 177 9880

Neutral axis

Fig. 6.25. Uncracked composite section at internal support, with n0 = 10.1 (dimensions in mm)

90





w

B

C

MEd, B I lI

0.15L

0.85L

Fig. 6.26. Elastic propped cantilever with change of section at 0.15L Table 6.4. Design ultimate bending moments at the fixed end of the propped cantilever Loading

w (kN/m)

n

10–6Iy (mm4)

λ

MEd, B (kN m)

Permanent Permanent Variable Variable

9.42 9.42 26.25 26.25

10.1 20.2 10.1 20.2

996 828 996 828

2.13 1.77 2.13 1.77

133 142 370 394

For the quasi-permanent combination, the coefficient ψ2 for variable load in a department store is given in clause A1.2.2(1) of EN 1990 as 0.6, and 0.6qk is 40% of 1.5qk, so some creep of the composite member is likely. In this example, n = 2n0 = 20.2 will be used for all action effects except shrinkage, for which a more accurate value is determined, as follows. Modular ratio for the effects of shrinkage From clause 5.4.2.2(4), the age of loading can be assumed to be 1 day. Creep coefficients for normal-weight concrete are given in Fig. 3.1 of EN 1992-1-1 in terms of h0, the notional size of the cross-section. For a slab with both surfaces exposed, this equals the slab thickness; but the slab here has one sealed surface, and h0 is twice its thickness. The mean thickness (see Fig. 6.23) is 105 mm, so h0 = 210 mm. For normally hardening cement and ‘inside conditions’, Fig. 3.1 of EN 1992-1-1 gives the creep coefficient ϕ(•, t0) as 5.0. For lightweight concrete, clause 11.3.3(1) of EN 1992-1-1 gives a correction factor, which in this case is (18/22)2, giving ϕ = 3.35. The creep multiplier ψL in clause 5.4.2.2(2) takes account of the shape of the stress–time curve for the effect considered, and is 0.55 for shrinkage. The modular ratio for shrinkage effects is n = n0(1 + ψLϕt) = 10.1(1 + 0.55 × 3.35) = 28.7 Bending moments Although, through cracking, the beam is of non-uniform section, calculation of bending moments algebraically is straightforward, because the two spans are equal. When both spans are fully loaded, only a propped cantilever need be considered (Fig. 6.26). For distributed loading w per unit length, and ratio of flexural stiffnesses λ, as shown, an equation for the elastic bending moment MEd at B is derived in the first edition of Johnson and Buckby37 (p. 375). It is MEd, B = (wL2/4)(0.110λ + 0.890)/(0.772λ + 1.228) The results are given in Table 6.4. Hence, the design bending moment for the internal support, excluding shrinkage effects, with n = 20.2, is MEd, B = 394 + 142 = 536 kN m The vertical shear at the internal support is

91





F

zsh R

Fzsh d

F

0.85L

0.30L

0.85L

(b)

(a)

P/2

Msh, B

P/2

d

P/2 d

0.15L

P

0.85L (d)

(c)

Fig. 6.27. Secondary effects of shrinkage

VEd, B = (9.42 + 26.25) × 6 + 536/12 = 259 kN The plastic shear resistance of the IPE 450 section, Vpl, Rd, was found earlier to be 1042 kN. From clause 6.2.2.4(1), bending resistance is not reduced by shear until VEd > Vpl, Rd /2, so there is no reduction here. Redistribution of moments is not used because clause 5.4.4(4) does not permit it if allowance for lateral–torsional buckling is required. Secondary effects of shrinkage Shrinkage of the slab causes sagging curvature and shortening of the composite member, the ‘primary effects’. In a continuous beam, the curvature causes bending moments and shear forces, the ‘secondary effects’. In regions assumed to be cracked, both the curvature and the stresses from the primary effects are neglected (clauses 5.4.2.2(8) and 6.2.1.5(5)). The important secondary effect in this beam, a hogging bending moment at the internal support, is now calculated. Shrinkage is a permanent action, and so is not reduced by a combination factor ψ0. The slab is imagined to be separated from the steel beam, Fig. 6.27(a). Its area Ac is that of the concrete above the sheeting. It shrinks. A force is applied to extend it to its original length. It is F = Ac(Ea/n)|εcs|

(D6.28)

This acts at the centre of the slab, at a distance zsh above the centroid of the composite section. The parts of the beam are reconnected. To restore equilibrium, an opposite force F and a sagging moment Fzsh are applied to the composite section. The radius of curvature of the uncracked part of the beam is R = EaλIy /Fzsh

(D6.29)

If the centre support is removed, the deflection δ at that point is δ = (0.85L)2/2R

(D6.30)

from the geometry of the circle (Fig. 6.27(b)).

92





It remains to calculate the force P, applied at that point, to reduce the deflection to zero, so that the centre support can be replaced (Fig. 6.27(d)). The secondary hogging bending moment at B is MEd, sh, B = PL/2

(D6.31)

and the vertical shear is P/2. Using slope and deflection coefficients for a cantilever (Fig. 6.27(c)), P can be found from the result δ = (P/2)L3(0.13λ + 0.20)/(EaIyλ)

(D6.32)

The calculation is as follows: Ac = 2.5 × 0.08 = 0.20 m2 Ea = 210 kN/mm2, n = 28.7 and εcs = –500 × 10 –6; so, from equation (D6.28), F = 732 kN. From Table 6.3, λIy = 741 × 106 mm4

zsh = 225 – 40 = 185 mm

so, from equation (D6.29), R = 1149 m, and, from equation (D6.30) with L = 12 m, δ = 45.3 mm. From Table 6.3, Iy = 467 × 106 mm4 so λ = 741/467 = 1.587 so, from equation (D6.32), P/2 = 10.0 kN and MEd, sh, B = 10 × 12 = 120 kN m Should shrinkage be neglected at ultimate load? This unusually high value, 120 kN m, results from the use of a material with high shrinkage and a continuous beam with two equal spans. It increases the ultimate design bending moment at B by 22%. Clause 5.4.2.2(7) permits this to be neglected if resistance is not influenced by lateral–torsional buckling. The reasoning is that as the bending resistances of the sections are determined by plastic theory, the ultimate condition approaches a collapse mechanism, in which elastic deformations (e.g. those from shrinkage) become negligible in comparison with total deformations. However, if the resistance at the internal support is governed by lateral–torsional buckling, and if the buckling moment (to be calculated) is far below the plastic moment, the inelastic behaviour may not be sufficient for the shrinkage effects to become negligible, before failure occurs at the internal support. Hence, the secondary shrinkage moment is not neglected at this stage, even though this beam happens to have a large reserve of strength at mid-span, and would not fail until the support section was far into the post-buckling phase.

Resistance to lateral–torsional buckling The top flange of the steel beam is restrained in both position and direction by the composite slab. Lateral buckling of the bottom flange near the internal support is accompanied by bending of the web, so the problem here is distortional lateral buckling. The provisions in EN 1994-1-1 headed ‘lateral–torsional buckling’ (clause 6.4), are in fact for distortional buckling. Clause 6.4.1(3) permits, as an alternative, use of the provisions in EN 1993-1-1 for steel beams. The method of clause 6.4.2 is used here. The

93





detailed comments on it, both in the main text and in Appendix A, should be referred to as required. The method requires the calculation of the elastic critical buckling moment at the internal support, Mcr, for which information is given in tables in ENV 1994-1-1 (reproduced as graphs in Appendix A (Figs A.3 and A.4)). The simple method of clause 6.4.3 is not available, as the loading does not conform to paragraph (b) of clause 6.4.3(1). Buckling near an internal support is often most critical in a span with minimum load that is adjacent to a fully loaded span. In this beam it was found that although the buckling moment Mb, Rd is increased when both spans are loaded (because the length of bottom flange in compression is reduced), the increase in the applied moment MEd is greater, so the both-spans-loaded case is more critical. This case is now considered, with n = 20.2 and all load assumed to act on the composite member. From Table 6.4, the bending moments in the beam are as shown in Fig. 6.28. Shrinkage is considered separately. The ‘simply-supported’ moment M0 is 35.7 × 122/8 = 643 kN m, so, from Fig. A.3 for C4 (see Appendix A), ψ = MB/M0 = 536/643 = 0.834

and

C4 = 28.3

The elastic critical buckling moment was given earlier as Mcr = (kcC4/L)[(GaIat + ksL2/π2)EaIafz]1/2

(D6.11)

where kc is a property of the composite section, given in Section 6.4.2, Ga is the shear modulus for steel, Ga = Ea/[2(1 + υa)] = 80.8 kN/mm2 Iat is the torsional moment of area of the steel section, Iat = 0.659 × 106 mm4 ks is defined in clause 6.4.2(6), and Iafz is the minor-axis second moment of area of the steel bottom flange, 10–6Iafz = 1.903 × 14.6/12 = 8.345 mm4 The stiffness ks is now found. It depends on the lesser of the ‘cracked’ flexural stiffnesses of the composite slab at a support and at mid-span, (EI)2. The value at the support governs. An approximation for this, derived in Appendix A, is (EI)2 = Ea[As Ae z2/(As + Ae) + Ae hp2/12] where Ae is the equivalent transformed area per unit width of concrete in compression, Ae = b0 hp /nbs 536

Full load on both spans Variable load on span AB only

339

0

A

C D

B

403 494

7.25 m

4.75 m

5.26 m

Fig. 6.28. Bending-moment distributions for ultimate limit state, excluding shrinkage

94





ds As z

hp/2 b0

bp

Fig. 6.29. Cross-section of the composite slab

where b0 is the mean width of the troughs, bs is the spacing of the troughs, hp is the depth of the sheeting, As is the area of top reinforcement per unit width of slab and z is the ‘lever arm’, as shown in Fig. A.1 of Appendix A. Calculation of (EI)2 for the composite slab, and ks It is assumed that the transverse reinforcement above the steel beam will be below the longitudinal bars and not less than 12 mm bars at 200 mm, giving As = 565 mm2/m and ds = 42 mm, whence z = 63 mm (Fig. 6.29). Assuming that buckling is caused by a short-term overload, n is taken as 10.1. From Fig. 6.23, b0/bs = 0.5; hp = 50 mm; so Ae = 2475 mm2/m. Hence, (EI)2 = 210[0.565 × 2.475 × 632/3040 + 2.475 × 502/12 000] = 491 kN m2/m From clause 6.4.2(6), for unit width of a slab continuous across the steel beams at spacing a, and assuming that the conditions for using α = 4 apply, k1 = 4(EI)2/a = 4 × 491/2.5 = 786 kN m/rad per metre width, and k2 = Eatw3/[4hs(1 – υa2)] where hs is the distance between the centres of the flanges of the IPE 450 section, 435 mm. Thus, k2 = 210 × 9.43/(4 × 435 × 0.91) = 110 kN/rad and ks = k1k2/(k1 + k2) = 786 × 110/896 = 96.4 kN/rad Calculation of kc For a doubly symmetrical steel section, from equations (D6.12) and (D6.13), kc = (hs Iy /Iay)/[(hs2/4 + ix2)/e + hs] with e = AIay /[Aazc(A – Aa)] In these expressions, the symbols are properties of the steel section, given earlier, except that A is the area of the cracked composite section, A = Aa + As = 11 350 mm2

95





and zc is the distance between the centroid of the steel beam and mid-depth of the slab. Here, ‘slab’ means the 130 mm-deep composite slab, not the 80 mm depth of concrete that contributes to the composite section. It is the stiffness of the composite slab in the transverse direction that prevents rotation of the steel top flange; so zc = 225 + 130/2 = 290 mm Hence, e = 11 350 × 337 × 106/(9880 × 290 × 1470) = 909 mm and kc = (435 × 467/337)/[(2182 + 1902)/909 + 435] = 1.15 Calculation of Mcr and Mb, Rd From equation (D6.11) for Mcr: Mcr = (1.15 × 28.3/12)[(80.8 × 0.659 + 96.4 ×122/π2) × 210 × 8.345]1/2 = 4340 kN m From clause 6.4.2(4), the relative slenderness depends also on the characteristic resistance moment, calculated earlier. The slenderness is λLT = ÷(MRk/Mcr) = (802/4340)1/2 = 0.430

For the reduction factor χLT for a rolled section, clause 6.4.2(1) refers to clause 6.3.2.3 of EN 1993-1-1, where buckling curve c is specified for this IPE section. This could be taken to mean the curve c plotted in Fig. 6.4 of EN 1993-1-1; but that curve is inconsistent with the equation for χLT in clause 6.3.2.3. This is understood to mean that the value of αLT given for curve c in Table 6.3 of EN 1993-1-1 should be used in calculating χLT. Its value depends on the parameters λLT, 0 and β, which can be given in a National Annex. The recommended values, 0.4 and 0.75 respectively, are used here. The calculation is ΦLT = 0.5[1 + αLT (λLT - λLT, 0 ) + βλLT 2 ] = 0.577 χLT =

1 [ΦLT + ÷ (ΦLT 2 - βλLT 2 )]

= 0.983

(The use of Fig. 6.4 gives the much lower result χLT = 0.88.) The buckling resistance is Mb, Rd = χLTMpl, Rd = 0.983 × 781 = 767 kN m This is well above the design ultimate moment with shrinkage included, MEd = 656 kN m. The result is quite sensitive to the values specified for λLT, 0 and β. Clause 6.3.2.2 of EN 1993-1-1 includes a rule that if MEd/Mcr £ 0.16, lateral–torsional buckling effects may be ignored. This applies here: MEd = 0.153Mcr. The rule appears to be linked to the use of λLT, 0 = 0.4. It is not clear whether it applies if a National Annex specifies a lower value. The provision of bracing to the bottom flange is considered at the end of this example.

Design for sagging bending The maximum sagging bending occurs in a span when the other span carries minimum load, and is reduced by creep, so n = 10.1 is assumed. Removal of variable load from one span halves the bending moment it causes at the internal support, so from Table 6.4 the bending moment at the internal support is MEd, B = 133 + 370/2 = 318 kN m For the span with loading 35.67 kN/m, the end reaction is

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VEd, A = 35.67 × 6 – 318/12 = 188 kN so the point of maximum moment is at a distance 188/35.67 = 5.26 m from the support, and the maximum sagging moment is MEd = 188 × 5.26/2 = 494 kN m This is so far below Mpl, Rd, 1043 kN m, that the least permitted degree of shear connection will be used. MEd is even below Mpl, a, Rd, which is 604 kN m. From clause 6.6.1.2(1) the span length in sagging bending may be taken as 0.85L, or 10.2 m here. From equation (6.12) with fy = 355 N/mm2, the minimum degree of shear connection is η = n/nf = 1 – (0.75 – 0.03 × 10.2) = 0.56 Clause 6.6.1.2(3) permits a lower value, subject to some conditions. It was found that one of these – that there should be only one stud connector per rib of sheeting – could not be satisfied. The minimum number of connectors in each half of the sagging region is therefore 0.56nf, where nf is the number for full shear connection. From Fig. 6.24(b), the compressive force in the slab is then not less than 2840 × 0.56 = 1590 kN. Recalculation of Mpl, Rd, by the method used before gives Mpl, Rd = 946 kN m which is almost twice the resistance required. It does not follow that the design of this beam as non-composite would be satisfactory. Example 7.1 shows that its deflection would probably be excessive.

Design of the shear connection The degree of shear connection used here enables the studs to be treated as ‘ductile’. An alternative design, using non-ductile connectors, is given in Example 6.8. From clause 6.6.5.8(1) the height of the 19 mm studs must be at least 50 + 2 × 19 = 88 mm. The standard height of 95 mm, after welding, satisfies this rule. The design shear resistance per stud is governed by equation (6.19) of clause 6.6.3.1(1): PRd = 0.29d2(fckEcm)1/2/γV = 0.29 × 192(25 × 20 700)1/2/(1000 × 1.25) = 60.25 kN This result is modified by a factor kt given in clause 6.6.4.2. It depends on the height of the stud, hsc, the dimensions of the trough in the sheeting (see Fig. 6.23), the thickness of the sheeting (assumed to be 1.0 mm) and the number of studs per trough, nr. For nr = 1:

kt = 0.7 × 100/50 × (95/50 – 1) = 1.26, but £ 0.85

For nr = 2:

kt = 1.26/÷2= 0.89, but £ 0.70

Provided that the studs are not also required to anchor the profiled sheeting, the resistances are therefore PRd, 1 = 0.85 × 60.25 = 51.2 kN

(D6.33)

PRd, 2 = 0.7 × 60.25 = 42.2 kN

(D6.34)

so that a trough with two studs provides the equivalent of 2 × 42.2/51.2 = 1.65 single studs. From Fig. 6.28, the studs for maximum sagging bending have to be provided within a length of 5.26 m. The troughs are spaced at 0.2 m, so 26 are available. The design compressive force in the slab is 1590 kN, so the design shear flow is 1590/5.2 = 306 kN/m The number of single studs required is

97





ns = 1590/51.2 = 31 Hence, two studs per trough have to be used over part of the span. For a shear span in a building, EN 1994-1-1 does not specify how non-uniform shear connection should be arranged. Slip is minimized when the density of the connection is related to the shear per unit length, so the regions with pairs of studs should be adjacent to the supports. If the minimum number of troughs with two studs is n2s, 1.65n2s + 26 – n2s = 31 whence n2s ≥ 7.8 For the cracked section in hogging bending, As fsd = 639 kN, which requires 12.5 single studs. When the hogging moment is a maximum, the distance from an internal support to the cross-section of maximum sagging moment is 7.25 m (Fig. 6.28), so 36 troughs are available for a total of 31 + 12.5 = 43.5 single studs. If the minimum number of troughs with two studs is n2h, 1.65n2h + 36 – n2h = 43.5 whence n2h ≥ 11.6 The design shear flow is (1590 + 639)/7.2 = 310 kN/m The arrangement of studs shown in Fig. 6.30 provides the equivalent of 31.2 and 43.8 studs, within the lengths 5.2 and 7.2 m, respectively. The maximum sagging and hogging bending moments are caused by different arrangements of imposed loading. This method of calculation takes account of this, with the result that two studs near mid-span are effective for both sagging and hogging resistance. It is quicker to assume that the maximum sagging and hogging bending moments are caused by a single loading. The resulting increase in the total number of studs is negligible. The disadvantage is that it is not clear how many of the troughs with two studs per trough should be near each end of the span.

nL (kN/m)

nL, Rd

422

nL, Ed for maximum sagging BM

nL, Ed for maximum hogging BM 310

306 256

100

A 1.6

4.8 5.2

9.6

8

16

2

22

12

13.2

16

2

22

19.8

Equivalent number of single studs

Fig. 6.30. Arrangement of stud connectors in one 12 m span

98


B 12.0


Distance from support A (m)

Number of troughs



a

45

50

a 35

40

20

Fig. 6.31. Detail of a stud welded through discontinuous profiled sheeting (dimensions in mm)

Design of the transverse reinforcement Clause 6.6.6.1(4) says that the design longitudinal shear for the concrete slab should be ‘consistent with the design and spacing of the shear connectors’. This is taken to mean that the resistance of the shear connection, rather than the design loading, determines the longitudinal shear. Its maximum occurs where there are two studs per trough, and is vL, Ed = 10 × 42.2 = 422 kN/m From clause 6.6.6.4(2), shear surfaces that pass closely around a stud need not be considered. The critical situation is thus where sheeting is not continuous across a beam. It is assumed to be anchored by a stud, as shown in Fig. 6.31. From symmetry, the critical shear plane, labelled a–a, is to be designed to resist 211 kN/m. The shear resisted by the sheeting is given in clause 6.6.6.4(5). For the design bearing resistance of the sheeting it refers to clause 9.7.4. For the detailing shown in Fig. 6.31, the end distance a is 40 mm. The diameter of the weld collar is taken as 1.1 × 19 = 20.9 mm whence kϕ in clause 9.7.4(3) is kϕ = 1 + 40/20.9 = 2.91 The thickness of the profiled sheeting is shown in Fig. 6.23 as 1.0 mm; but the composite slab has not yet been designed. Here, it is assumed to be at least 0.9 mm thick, with a yield strength of 350 N/mm2 and γM = 1.0. From equation (9.10), Ppb, Rd = kϕdd0 tfyp, d = 2.91 × 20.9 × 0.9 × 0.35 = 19.1 kN/stud From clause 6.6.6.4(5), with a stud spacing of 200 mm, the shear resistance provided by the sheeting is vL, pd, Rd = 19.1/0.2 = 95 kN/m This must not exceed the yield strength of the sheeting, Ap fyp, d, which for this sheeting is over 400 kN/m. The design shear for the concrete slab, of thickness 80 mm, is vL, Ed = 211 – 95 = 116 kN/m Clause 6.6.6.2(1) refers to clause 6.2.4 of EN 1992-1-1, for shear between web and flanges of T-sections. The method is a truss analogy, with some choice provided for the angle θf between the concrete diagonals of the truss and the longitudinal direction. The

99





reinforcement required increases with the angle θf. For simplicity, the minimum angle given in clause 6.2.4(4) of EN 1992-1-1 for a tension flange, 38.6°, will be used for the whole length of the span. The area of transverse reinforcement per unit length is given by Asf > vL, Ed /(fsd cot θf) = 116/(0. 435 × 1.25) = 213 mm2/m This is much less than the amount assumed in design for lateral buckling (565 mm2/m). It may also be affected by the need for crack control above the beam, which arises in the design of the composite slab.

Design of bracing to bottom flanges near the internal support It is now shown that the values of factor C4 given in Appendix A are not suitable for design based on Mcr where intermediate lateral bracing is provided. From Fig. 6.28, with full loading on both spans, the distance of a point of contraflexure from support B is 12 – 2 × 4.75 = 2.5 m. Let us suppose that lateral bracing is to be provided at this point, so that L in equation (D6.11) is reduced from 12 to 2.5 m. Assuming that the bending-moment distribution over this 2.5 m length is linear, Fig. A.4 in Appendix A gives C4 = 11.1. Substituting these values into equation (D.6.11), with other values unchanged, gives Mcr = 2285 kN m, which is much lower than the previous value, 4340 kN m. For this situation, it is necessary to find Mcr from elastic critical analysis by computer. Summary of Example 6.7 All important aspects of the design of this beam for persistent situations for ultimate limit states have now been considered. Serviceability checks are given in the worked example at the end of Chapter 7. The same beam with semi-continuous joints at support B is studied in Examples 8.1 and 10.1.

Example 6.8: partial shear connection with non-ductile connectors

The design bending-moment diagram for sagging bending in the preceding worked example is shown in Fig. 6.28. The shear connection for the length AD of span AB was designed for connectors that satisfied the definition of ‘ductile’ in clause 6.6.1.2. The result is shown in Fig. 6.30. This work is now repeated, using the same data, except that the proposed connectors are not ‘ductile’, to illustrate the use of clause 6.6.1.3(5). This requires the calculation of the shear flow vL, Ed by elastic theory. No ‘inelastic redistribution of shear’ is required, so clause 6.6.1.1(3)P on deformation capacity does not apply. Calculation of the resistance moment MRd according to clause 6.2.1.3(3) is not permitted, so stresses are calculated by elastic theory and checked against the limits in clause 6.2.1.5(2). From equations (D6.22) and (D6.23), these are fcd = fck/γC = 25/1.5 = 16.7 N/mm2

(D6.35)

and fyd = fyk = 355 N/mm2

(D6.36)

In this continuous beam, creep reduces stiffness at mid-span more than at the internal support, where the concrete is cracked, so the sagging bending moment and longitudinal shear (for constant loading) reduce over time. The short-term modular ratio, 10.1, is therefore used. Taking account of the use of unpropped construction, it is found for the loads in Table 6.2 that the maximum sagging bending moment acting on the composite section, Mc, Ed, occurs at 5.4 m from support A (Fig. 6.28), and is Mc, Ed = 404 kN m

with

Ma, Ed = 110 kN m

100





nL (kN/m) 422 For ‘ductile’ studs, from Example 6.7

nL, Rd 311 256 nL, Ed

For elastic design, from Example 6.8

128

0

No. of troughs

1.2

6

3.8

13

5.4

Distance from support A (m)

8

Fig. 6.32. Longitudinal shear and shear resistance for length AD of the beam in Fig. 6.28

Shrinkage of concrete in this beam reduces both the mid-span moments and the compressive stress in concrete. For simplicity and safety, its effects are ignored. Using the elastic section properties from row 4 of Table 6.3, the stresses are found to be well below the limits given above. The mean stress in the 80 mm thickness of the concrete slab, with beff = 2.5 m, is 4.20 N/mm2, giving the longitudinal force in the slab as Nc = 4.2 × 2.5 × 80 = 840 kN From equation (D6.26), the force for full shear connection is Nc, f = Fc = 2840 kN so the degree of shear connection needed is η = Nc/Nc, f = 840/2840 = 0.30 The elastic shear flow diagram is triangular, so the shear flow at support A is vL, Ed = 2 × 840/5.4 = 311 kN/m Stud connectors will be used, as before. They are not ‘ductile’ at this low degree of shear connection. Their resistances are given by equations (D6.33) and (D6.34). For the profiled sheeting used (Fig. 6.23(b)), there are five troughs per metre. One stud per trough (vL, Rd = 256 kN/m) is not sufficient near support A. Two studs per trough provide 422 kN/m. Near mid-span, it is possible to use one stud every other trough, (vL, Rd = 128 kN/m) as their spacing, 400 mm, is less than the limit set in clause 6.6.5.5(3). Details of a possible layout of studs are shown in Fig. 6.32. This also shows the resistance provided over this 5.4 m length from Fig. 6.30. That is higher because in the previous example, η = 0.56, not 0.30, and shear connection is provided for the whole load, not just that applied to the composite member. This is not a typical result, because the design sagging moment here is unusually low, in relation to the plastic resistance to sagging bending. It may become more typical in design to the Eurocodes, because of the influence of lateral–torsional buckling on accounting for shrinkage and the restrictions on redistribution of moments.

Example 6.9: elastic resistance to bending, and influence of degree of shear connection and type of connector on bending resistance

This example makes use of the properties of materials and cross-sections found in Example 6.7 on the two-span beam, and of the results from Example 6.8 on non-ductile

101





Bending moment (kN m) C

Mpl, Rd = 1043 D

MRd = 950

G

Mel, Rd = 720

B

Mpl, a, Rd = 604 A 500 E

Ma, Ed = 110 0

0.2

0.404

0.574

0.8

1.0 h = Nc/Nc, f

Fig. 6.33. Design methods for partial shear connection

connectors. The cross-section of maximum sagging bending moment, point D, was then found to be 5.4 m from support A in Fig. 6.28. For simplicity, the beam is now assumed to be simply-supported and of span 10.8 m, so that this cross-section is at mid-span, with unpropped construction as before, so that the bending moment in the steel beam at this section is Ma, Ed = 110 kN m. Shrinkage effects are beneficial, and are neglected. The purpose is to find the relationship between the degree of shear connection, η, and the bending resistance of the beam at mid-span, in accordance with clauses 6.2.1.3–6.2.1.5, 6.6.1.2 and 6.6.1.3. The result is shown in Fig. 6.33. At low degrees of shear connection, only elastic design is permitted, so the elastic bending resistance Mel, Rd is required, to clause 6.2.1.4(6). It depends on the modular ratio. It is found that the limiting stress (equations (D6.35) and (D6.36)) is reached first in the steel bottom flange, and is increased by creep, so n = 20.2 is assumed. For Ma, Ed = 110 kN m, the maximum stress in the steel beam is 73 N/mm2, leaving 355 – 73 = 282 N/mm2 for loading on the composite beam. Using the elastic section properties from row 5 of Table 6.3, the steel reaches yield when the bending moment on the composite section is 610 kN m, so Mel, Rd = 110 + 610 = 720 kN m The compressive force in the concrete slab is then Nc, el = 1148 kN, so that η = Nc, el/Nc, f = 1148/2840 = 0.404 Figure 6.33 is based on Figs 6.5 and 6.6(b) of EN 1994-1-1. The results above enable point B to be plotted. When MEd = 110 kN m, Nc = 0. From equation (6.2) in clause 6.2.1.4(6), line BE is drawn towards the point (0, 110) as shown. No lower limit to η is defined. It will in practice be determined by the detailing rules for the shear connection. For full shear connection,

102





Mpl, Rd = 1043 kN m from equation (D6.27). This fixes point C in Fig. 6.33, and line BC (from equation (6.3)) is drawn. Line EBC applies whether the connection is ‘ductile’ or not. For this beam, with fy = 355 N/mm2 and equal steel flanges, equation (6.12) in clause 6.6.1.2(1) gives η ≥ 1 – (0.75 – 0.03 × 10.8) = 0.574 Hence, Nc ≥ 0.574 × 2840 = 1630 kN Using the method of clause 6.2.1.3(3) for ductile connectors, illustrated in Fig. 6.3, the plastic bending resistance is MRd = 950 kN m This gives point D in Fig. 6.33. When η = 0, the plastic resistance is Mpl, a, Rd. From equation (D6.25) this is Mpl, a, Rd = 1.702 × 355 = 604 kN m which is point A in Fig. 6.33. Similar calculations for assumed degrees of shear connection give the curve ADC, which can, for simplicity, be replaced by the line AC (clause 6.2.1.3(5)). Both the curve and the line are valid only where η is high enough for the connection to be ‘ductile’. The design bending resistances for this example are therefore given by EBGDC in Fig. 6.33. Line BE gives the least shear connection that may be used when MEd < Mel, Rd, without restriction on the type of connector. For higher values of MEd, the required degree of shear connection for non-ductile connectors is given by the line BC. The bonus for using ductile connectors (as defined in clause 6.6.1.2) is the area GDC, where the position of the line GD is determined by the span of the beam, and moves to the right as the span increases. For the beam analysed here, with Ma, Ed = 110 kN m, a total bending moment MEd = 1000 kN m (for example) requires shear connection for about 2100 kN (η ª 0.74) if headed studs to clause 6.6.1.2 are used, but this increases to over 2600 kN for non-ductile connectors.

6.7. Composite columns and composite compression members 6.7.1. General Scope

A composite column is defined in clause 1.5.2.5 as ‘a composite member subjected mainly to compression or to compression and bending’. The title of clause 6.7 includes ‘compression members’, to make clear that its scope is not limited to vertical members but includes, for example, composite members in triangulated or Vierendeel girders. These girders may also have composite tension members, for which provisions are given in EN 1994-2. In this guide, references to ‘columns’ includes other composite compression members, unless noted otherwise, and for buildings, ‘column’ means a length of column between adjacent lateral restraints; typically, a storey height. Design rules for columns sometimes refer to ‘effective length’. That term is not generally used in clause 6.7. Instead, the ‘relative slenderness’ is defined, in clause 6.7.3.3(2), in terms of Ncr, ‘the elastic critical normal force for the relevant buckling mode’. This use of Ncr is explained in the comments on clause 6.7.3.3. Clause 6.7.1(1)P refers to Fig. 6.17, in which all the sections shown have double symmetry; but clause 6.7.1(6) makes clear that the scope of the general method of clause 6.7.2 includes members of non-symmetrical section.84

Clause 6.7.1(1)P

103





C

N

Npm, Rd/2

A D

gFNEk 0.8gFNEk

0

E B

Mpl, Rd

MRd

M

Fig. 6.34. Independent bending moment and normal force (not to scale)


Clause 6.7.1(4)

Clause 6.7.1(7)

The bending moment in a column depends on the assumed location of the line of action of the axial force, N. Where the cross-section has double symmetry, this is the intersection of the axes of symmetry. In other cases the choice, made in the modelling for global analysis, should be retained for the analysis of the cross-sections. A small degree of asymmetry (e.g. due to an embedded pipe) can be allowed for by ignoring in calculations concrete areas elsewhere, such that symmetry is restored. No shear connectors are shown in the cross-sections in Fig. 6.17, because within a column length the longitudinal shear is normally much lower than in a beam, and sufficient interaction may be provided by bond or friction. Shear connectors should be provided for load introduction, following clause 6.7.4. The minimum compression for a member to be regarded as a column, rather than a beam, is not stated. As shown in Example 6.11, the use of the cross-sections in Fig. 6.17 as beams without shear connectors is usually prevented by the low design shear strengths due to bond and friction (clause 6.7.4.3). The strengths of materials in clause 6.7.1(2)P are as for beams, except that class C60/75 and lightweight-aggregate concretes are excluded. For these, additional provisions (e.g. for creep, shrinkage, and strain capacity) would be required.85,86 Clause 6.7.1(3) and clause 5.1.1(2) both concern the scope of EN 1994-1-1. They appear to exclude composite columns in high-rise buildings with a reinforced concrete core. For these ‘mixed’ structures, additional consideration in the global analysis of the effects of shrinkage, creep, and column shortening may be needed. The steel contribution ratio (clause 6.7.1(4)), is essentially the proportion of the squash load of the section that is provided by the structural steel member. If it is outside the limits given, the member should be treated as reinforced concrete or as structural steel, as appropriate.

Independent action effects The interaction curve for the resistance of a column cross-section to combined axial force N and uniaxial bending M is shown in Fig. 6.19, and, as a polygon, in Fig. 6.38 of this guide. It has a region BD where an increase in NEd increases MRd. Clause 6.7.1(7) refers to a situation where at ultimate load the factored bending moment, γF MEk, could co-exist with an ‘independent’ axial force that was less than its design value, γF NEk. It says that verification should be based on the lower value, 0.8γF NEk. Discussion of this rule is illustrated in Fig. 6.34, which shows region BDC of the interaction curve in Fig. 6.19, which is symmetrical about the line AD. If

104





γF NEk < Npm, Rd/2

(D6.37)

then MRd should be found for an axial force of 0.8γF NEk, as shown by point E. The reduction in MRd is usually small. A simpler but more conservative rule is given in ENV 1994-1-1, with a clearer definition of ‘independent actions’: if MRd corresponding to γF NEk is found to exceed Mpl, Rd, MRd should be taken as Mpl, Rd. It is applicable unless the bending moment MEd is due solely to the eccentricity of the force NEd. Its effect is to replace the curve BDC in Fig. 6.34 with the line BC.

Local buckling The principle of clause 6.7.1(8)P is followed by its application rules. They ensure that the concrete (which will be reinforced in accordance with EN 1992-1-1) restrains the steel and prevents it from buckling even when yielded. For partly encased sections, the encasement prevents local buckling of the steel web, and prevents rotation of the steel flange at its junction with the web, so that a higher bf/t ratio may be used than for a bare steel section. Table 6.3 gives the limit as 44ε, compared with about 22ε (from EN 1993-1-1) for a Class 2 flange. (In EN 1994, as in EN 1993, ε = ÷(235/fy), in units of newtons per square millimetre.) For concrete-filled rectangular hollow steel sections (RHS), the limit of 52ε compares with about 41ε for a steel RHS. For a concrete-filled circular hollow section, the limiting d/t of 90ε2 compares with 70ε2 for Class 2 in EN 1993-1-1.

Clause 6.7.1(8)P

6.7.2. General method of design

Designers of composite columns will normally ensure that they fall within the scope of the simplified method of clause 6.7.3; but occasionally the need arises for a non-uniform or asymmetric member. The ‘general method’ of clause 6.7.2 is provided both for this reason, and to enable advanced software-based methods to be used. Clause 6.7.2 is more a set of principles than a design method. Development of software that satisfies these principles is a complex task. Clause 6.7.2(3)P refers to ‘internal forces’. These are the action effects within the column length, found from those acting on its ends, determined by global analysis to Section 5. At present, such an analysis is likely to exclude member imperfections and second-order effects within members, but comprehensive software may become available. Clause 6.7.2(3) also refers to ‘elastic-plastic analysis’. This is defined in clause 1.5.6.10 of EN 1990 as ‘structural analysis that uses stress/strain or moment/curvature relationships consisting of a linear elastic part followed by a plastic part with or without hardening.’ As the three materials in a composite section follow different non-linear relationships, direct analysis of cross-sections is not possible. One has first to assume the dimensions and materials of the member, and then determine the axial force N and bending moment M at a cross-section from assumed values of axial strain and curvature φ, using the relevant material properties. The M–N–φ relationship for each section can be found from many such calculations. This becomes even more complex where biaxial bending is present.87 Integration along the length of the column then leads to a non-linear member stiffness matrix that relates axial force and end moments to the axial change of length and end rotations.

Clause 6.7.2 Clause 6.7.2(3)P

6.7.3. Simplified method of design Scope of the simplified method

The method has been calibrated by comparison with test results.88 Its scope (clause 6.7.3.1) is Clause 6.7.3.1 limited mainly by the range of results available, which leads to the restriction λ £ 2 in clause Clause 6.7.3.1(1) 6.7.3.1(1). For most columns, the method requires second-order analysis in which explicit account is taken of imperfections. The use of strut curves is limited to axially loaded members.

105





0.85fcd

fsd fyd

–

Mpl, Rd +

+

fyd fsd Concrete

Steel

Reinforcement

Fig. 6.35. Stress distributions for resistance in bending (tension positive)

The restriction on unconnected steel sections in clause 6.7.3.1(1) is to prevent loss of stiffness due to slip, which would invalidate the formulae for EI of the column cross-section. Clause 6.7.3.1(2) The limits to concrete cover in clause 6.7.3.1(2) arise from concern over strain softening of concrete invalidating the interaction diagram (Fig. 6.19), and from the limited test data for columns with thicker covers. These provisions normally ensure that for each axis of bending, the flexural stiffness of the steel section makes a significant contribution to the total stiffness. Greater cover can be used by ignoring in calculation the concrete that exceeds the stated limits. Clause 6.7.3.1(3) The limit of 6% in clause 6.7.3.1(3) on the reinforcement used in calculation is more liberal than the 4% (except at laps) recommended in EN 1992-1-1. This limit and that on maximum slenderness are unlikely to be restrictive in practice. Clause 6.7.3.1(4) Clause 6.7.3.1(4) is intended to prevent the use of sections susceptible to lateral–torsional buckling. The reference to hc < bc arises because hc is defined as the overall depth in the direction normal to the major axis of the steel section (Fig. 6.17). The term ‘major axis’ can be misleading, because some column sections have Iz > Iy, even though Ia, y > Ia, z.

Resistance of cross-sections Calculations for composite sections, with three materials, are potentially more complex than for reinforced concrete, so simplifications to some provisions of EN 1992-1-1 are made in EN 1994-1-1. Reference to the partial safety factors for the materials is avoided by specifying resistances in terms of design values for strength, rather than characteristic values; for Clause 6.7.3.2(1) example in equation (6.30) for plastic resistance to compression in clause 6.7.3.2(1). This resistance, Npl, Rd, is the ultimate axial load that a short column can sustain, assuming that the structural steel and reinforcement are yielding and the concrete is crushing. For concrete-encased sections, the crushing stress is taken as 85% of the design cylinder strength, as explained in the comments on clause 3.1. For concrete-filled sections, the concrete component develops a higher strength because of the confinement from the steel section, and the 15% reduction is not made; see also the comments on clause 6.7.3.2(6). Resistance to combined compression and bending The bending resistance of a column cross-section, Mpl, Rd, is calculated as for a composite Clause 6.7.3.2(2) beam in Class 1 or 2 (clause 6.7.3.2(2)). Points on the interaction curve shown in Figs 6.18 and 6.19 represent limiting combinations of compressive axial load N and moment M which correspond to the plastic resistance of the cross-section. The resistance is found using rectangular stress blocks. For simplicity, that for the concrete extends to the neutral axis, as shown in Fig. 6.35 for resistance to bending (point B in Fig. 6.19 and Fig. 6.38). As explained in the comments on clause 3.1(1), this simplification is unconservative in comparison with stress/strain curves for concrete and the rules of EN 1992-1-1. To compensate for this, the plastic resistance moment for the column section is reduced by a factor αM in clause 6.7.3.6(1).

106





As axial compression increases, the neutral axis moves; for example, towards the lower edge of the section shown in Fig. 6.35, and then outside the section. The interaction curve is therefore determined by moving the neutral axis in increments across the section, and finding pairs of values of M and N from the corresponding stress blocks. This requires a computer program, unless the simplification given in clause 6.7.3.2(5) is used. Simplified expressions for the coordinates of points B, C and D on the interaction curve in Fig. 6.34 are given in Appendix C. Further comment is given in Examples 6.10 and C.1.

Influence of transverse shear Clauses 6.7.3.2(3) and 6.7.3.2(4), on the influence of transverse shear on the interaction Clause 6.7.3.2(3) curve, are generally the same as clause 6.2.2.4 on moment–shear interaction in beams. One Clause 6.7.3.2(4) assumes first that the shear VEd acts on the structural steel section alone. If it is less than 0.5Vpl, a, Rd, it has no effect. If it is greater, there is an option of sharing it between the steel and reinforced concrete sections, which may reduce that acting on the steel to below 0.5Vpl, a, Rd. If it does not, then a reduced design yield strength is used for the shear area, as for the web of a beam. In a column, however, the shear area depends on the plane of bending considered, and may consist of the flanges of the steel section. It is assumed that shear buckling does not occur. Simplified interaction curve Clause 6.7.3.2(5) explains the use of the polygonal diagram BDCA in Fig. 6.19 as an Clause 6.7.3.2(5) approximation to the interaction curve, suitable for hand calculation. The method applies to any cross-section with biaxial symmetry, not just to encased I-sections. First, the location of the neutral axis for pure bending is found, by equating the longitudinal forces from the stress blocks on either side of it. Let this be at distance hn from the centroid of the uncracked section, as shown in Fig. 6.19(B) and Fig. C.2 in Appendix C. It is shown in Appendix C that the neutral axis for point C on the interaction diagram is at distance hn on the other side of the centroid, and the neutral axis for point D passes through the centroid. The values of M and N at each point are easily found from the stress blocks shown in Fig. 6.19. For concrete-filled sections the factor 0.85 may be omitted. Concrete-filled tubes of circular or rectangular cross-section Clause 6.7.3.2(6) is based on the lateral expansion that occurs in concrete under axial Clause 6.7.3.2(6) compression. This causes circumferential tension in the steel tube and triaxial compression in the concrete. This increases the crushing strength of the concrete88 to an extent that outweighs the reduction in the effective yield strength of the steel in vertical compression. The coefficients ηa and ηc given in this clause allow for these effects. This containment effect is not present to the same extent in concrete-filled rectangular tubes because less circumferential tension can be developed. In all tubes the effects of containment reduce as bending moments are applied; this is because the mean compressive strain in the concrete and the associated lateral expansion are reduced. With increasing slenderness, bowing of the member under load increases the bending moment, and therefore the effectiveness of containment is further reduced. For these reasons, ηa and ηc are dependent on the eccentricity of loading and on the slenderness of the member. Properties of the column For columns in a frame, some properties of each column length are needed before or during global analysis of the frame: • the steel contribution ratio (clause 6.7.3.3(1)) • the relative slenderness λ (clause 6.7.3.3(2)) • the effective flexural stiffnesses (clauses 6.7.3.3(3) and 6.7.3.4(2)) • the creep coefficient and effective modulus for concrete (clause 6.7.3.3(4)).

Clause 6.7.3.3(1) Clause 6.7.3.3(2) Clause 6.7.3.3(3) Clause 6.7.3.3(4)

The steel contribution ratio is explained in the comments on clause 6.7.1(4).

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The relative slenderness λ is needed to check that the column is within the scope of the simplified method, clause 6.7.3.1(1). λ is calculated using characteristic values and the appropriate flexural stiffness is that given in clauses 6.7.3.3(3) and 6.7.3.3(4). The correction factor Ke is to allow for cracking. As λ depends on the elastic critical normal force for the relevant buckling mode, the behaviour of the surrounding members needs to be taken into account. This could require a calculation of the load factor αcr for elastic instability, following the procedure shown in Fig. 5.1(e). It will often be possible though to make simplifying assumptions to show that a proposed column is within the scope for the method. For example, in a frame with a high stiffness against sway it would be reasonable to calculate Ncr assuming the member to be pin-ended. In an unbraced continuous frame, the stiffness of each beam could be taken as that of the steel section alone, permitting Ncr to be determined from effective length charts that assume a beam to be of uniform stiffness. In any case, the upper limit on λ is somewhat arbitrary and does not justify great precision in Ncr. The creep coefficient ϕt influences the effective modulus Ec, eff (clause 6.7.3.3(4)), and hence the flexural stiffness of each column. It depends on the age at which concrete is stressed and the duration of the load. These will not be the same for all the columns in a frame. The effective modulus depends also on the proportion of the design axial load that is permanent. The design of a column is rarely sensitive to the influence of the creep coefficient on Ec, eff, so conservative assumptions can be made about uncertainties. Normally, a single value of effective modulus can be used for all the columns in the frame.

Clause 6.7.3.4

Verification of a column The flow chart of Fig. 6.36 shows a possible calculation route, intended to minimize iteration, for a column as part of a frame. It is used in Example 6.10. It is assumed that the column has cross-section details that can satisfy clauses 6.7.1(9), 6.7.3.1(2) to 6.7.3.1(4) and 6.7.5.2(1), and has already been shown to have λ £ 2, so that it is within the scope of the simplified method of clause 6.7.3. The relationship between the analysis of a frame and the stability of individual members is discussed in the comments on clauses 5.2.2(3) to 5.2.2(7). Conventionally, the stability of members is checked by analysis of individual members, using end moments and forces from global analysis of the frame. Figure 6.36 follows this procedure, giving in more detail the procedures outlined in the lower part of the flow chart for global analysis (see Fig. 5.1(a)). It is assumed that the slenderness of the column, determined according to clause 5.3.2.1(2), is such that member imperfections have been neglected in the global analysis. Comments are now given in the sequence of Fig. 6.36, rather than clause sequence. If bending is biaxial, the chart is followed for each axis in turn, as noted. It is assumed that loading is applied to the column only at its ends. The starting point is the output from the global analysis, listed at the top of Fig. 6.36. The design axial compression is the sum of the forces from the two frames of which the column is assumed to be a member. If, at an end of the column (e.g. the top) the joints to beams in these frames are at a different level, they could conservatively be assumed both to be at the upper level, if the difference is small. The flow chart does not cover situations where the difference is large (e.g. a storey height). The axial force NEd is normally almost constant along the column length. Where it varies, its maximum value can conservatively be assumed to be applied at the upper end. For most columns, the method requires a second-order analysis in which explicit account is taken of imperfections (clause 6.7.3.4). However, for a member subject only to end compression, clause 6.7.3.5(2) enables buckling curves to be used. For columns that qualify, this is a useful simplification because these curves allow for member imperfections. The reduction factor χ depends on the non-dimensional slenderness λ . The buckling curves are also useful as a preliminary check for columns with end moment; if the resistance to the normal force NEd is not sufficient, the proposed column is clearly inadequate.

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Known, for both the y and z axes: NEd, Npl, Rd, L, (EI)eff, II and the bending moments at both ends of the column. Find the axial compression: NEd = NEd, y + NEd, z

–

Yes

Find l to clause 6.7.3.3(2), then c to clause 6.7.3.5(2). Is NEd £ cNpl, Rd?

Yes

Is the column in axial compression only? (Clause 6.7.3.5(2))

Column verified (END)

No No

Column not strong enough (END)

Find Vpl, a, Rd to clause 6.2.2.2(2). Is VEd > 0.5Vpl, a, Rd? (Clause 6.7.3.2(3)) Yes

Note 1: Until Note 2, the chart is for both the y and z planes, separately, and the y or z subscript is not given

No

Find Mpl, a, Rd and Mpl, Rd, and hence Va, Ed and Vc, Ed from clause 6.7.3.2(4). Is Va, Ed > 0.5Vpl, a, Rd?

Find interaction curve or polygon for the cross-section (clauses 6.7.3.2(2) and 6.7.3.2(5))

No

Yes Find r from clause 6.2.2.4(2) and hence reduced fyd from clause 6.7.3.2(3)

Apply bow imperfection to clause 6.7.3.4(4); max. bow e0

Can first-order member analysis be used, because the member satisfies clause 5.2.1(3), based on stiffness to clause 6.7.3.4? No

Yes Either

Find MEd , the maximum firstorder bending moment within the column length. If MEd, 1 = MEd, 2 it is

MEd, max = MEd, 1 + NEde0

Note 2: For biaxial bending, repeat steps since Note 1 for the other axis

From clause 6.7.3.4(5): find

Ncr, eff = p (EI)eff, II/L 2

2

find b for end moments MEd, top and MEd, bot to Table 6.4, and hence k (= k1); find k2 for b = 1; find the design moment for the column,

Or

Find MEd, max by secondorder analysis of the pin-ended column length with force NEd and end moments MEd, 1 and MEd, 2

MEd, max = k1MEd + k2NEde0 ≥ MEd, top ≥ MEd, bot

From NEd and the interaction diagrams, find mdy and mdz from clause 6.7.3.7(1). Check that the cross-section can resist My, Ed, max and Mz, Ed, max from clause 6.7.3.7(2) (END)

Fig. 6.36. Flow chart for verification of a column length

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Comment has already been made on the calculation of λ when used to check the scope of the method. When λ is used as the basis for resistance, calculation of this parameter may still be simplified, provided the result is conservative. Columns without end moments are exceptional, and for most members the design process continues, as shown in Fig. 6.36. Columns with transverse shear exceeding half of the shear resistance of the steel element are rare; but shear is checked next, because if it is high the interaction curve for the cross-section may be affected. Comment has been given earlier on clauses 6.7.3.2(3) and 6.7.3.2(4), on shear, and on the interaction curve or polygon. Much of the remainder of the flow chart is concerned with finding the maximum bending moment to which the column will be subjected. In general, two calculations are necessary (clause 6.7.3.6(1)). The maximum bending moment may occur at one end, when the design moment equals to the larger of the two end moments; or the maximum bending moment may occur at an intermediate point along the member. This is because second-order effects and lateral load within the length of the member and initial bowing affect the bending moment. If member imperfections have been neglected in the global analysis, it is necessary to Clause 6.7.3.4(4) include them in the analysis of the column. Clause 6.7.3.4(4) gives the member imperfections in Table 6.5 as proportional to the length L of the column between lateral restraints. The imperfection is the lateral departure at mid-height of the column of its axis of symmetry from the line joining the centres of symmetry at the ends of the column. The values account principally for truly geometric imperfections and for the effects of residual stresses. They do not depend on the distribution of bending moment along the column. The curved shape is usually assumed to be sinusoidal, but use of a circular arc would be acceptable. The curve is assumed initially to lie in the plane of the frame being analysed. The next step is to ascertain whether second-order effects need to be considered within the member length. Clause 6.7.3.4(3) refers to clause 5.2.1(3). Second-order effects can therefore be neglected if the load factor αcr for elastic instability of the member exceeds 10. Possible sway effects will have been determined by global analysis, and will already be included within the values for the end moments and forces. To calculate αcr, the ends of the column are assumed to be pinned, and αcr is found using the Euler formula Ncr, eff = π2EI/L2, L being the physical length of the column. The flexural stiffness to use is (EI)eff, II (clause 6.7.3.4(2)), with the modulus of elasticity for concrete modified to take account of long-term effects (clause 6.7.3.3(4)). This flexural stiffness is lower than that defined in clause 6.7.3.3(3) because it is essentially a design value for ultimate limit states. The factor Ke, II allows for cracking. The factor Ko is from research-based calibration studies. The neglect of second-order effects does not mean that increase in bending moment caused by the member imperfection can also be ignored. The next box, on finding MEd, max, gives the example of a column in uniform single-curvature bending. If the end moments are dissimilar or of opposite sign, the maximum moment in the column, MEd, max, is likely to be the greater end moment. In practice, most columns are relatively slender, and second-order effects will usually need to be included. This can be done by second-order analysis of the member, treated as pin-ended but subject to the end moments and forces given by the global analysis. Any intermediate loads would also be applied. The analysis is to obtain the maximum moment in the column, which is taken as the design moment, MEd, max. Formulae may be obtained from Clause 6.7.3.4(5) the literature. Alternatively, use may be made of the factor k given by clause 6.7.3.4(5). It is assumed in Fig. 6.36 that the column is free from intermediate lateral loads. Two factors are used, written as k1 and k2, because two moment distributions must be considered. The first gives the equivalent moment k1MEd in the ‘perfect’ column, where MEd is the larger end moment given by the global analysis. The definitions of MEd in clause 6.7.3.4(5) and Table 6.4 may appear contradictory. In the text before equation (6.43), MEd is referred to as a first-order moment. This is because it does not include second-order effects arising within the column length. However, Table 6.4 makes clear that MEd is to be determined by either first-order or second-order global analysis; the choice would depend on clause 5.2.

110





The multiplier β (from Table 6.4) allows for the shape of the bending-moment diagram. The condition β ≥ 0.44 in the table is to ensure sufficient protection against snap-through buckling in double-curvature bending. The first-order moment from the imperfection, NEde0, has a distribution such that β = 1, from Table 6.4, so k2 normally differs from k1. The imperfection can be in any direction, so the equivalent moment k2NEd e0 always has the same sign as k1MEd, when the two are combined. Equation (6.43) states that k must be greater than or equal to unity, and this is correct for a single distribution of bending moment. However, for a combination of end moments and member imperfection, it could be conservative to limit all values of k in this way. At mid-length the component due to end moments depends on their ratio, r, and therefore could be small. The appropriate component is therefore k1MEd without the limit k ≥ 1.0, and the design moment within the column length is (k1MEd + k2NEd e0). In biaxial bending, the initial member imperfection may be neglected in the less critical plane (clause 6.7.3.7(1)). The limit k ≥ 1.0 is intended to ensure that the design moment is at least the larger end moment MEd. For uniaxial bending, the final step is to check that the cross-section can resist MEd, max with compression NEd. The interaction diagram gives a resistance µd Mpl, Rd with axial load NEd, as shown in Fig. 6.18. This is unconservative, being based on rectangular stress blocks, as explained in the comment on clause 3.1(1), so in clause 6.7.3.6(1) it is reduced, by the use of a Clause 6.7.3.6(1) factor αM that depends on the grade of structural steel. This factor allows for the increase in the compressive strain in the cross-section at yield of the steel (which is adverse for the concrete), when the yield strength of the steel is increased.

Biaxial bending Where values of MEd, max have been found for both axes, clause 6.7.3.7 applies, in which they are written as My, Ed and Mz, Ed. If one is much greater than the other, the relevant check for uniaxial bending, equation (6.46) will govern. Otherwise, the linear interaction given by equation (6.47) applies. If the member fails this biaxial condition by a small margin, it may be helpful to recalculate the less critical bending moment, omitting the member imperfection, as permitted by clause 6.7.3.7(1).

Clause 6.7.3.7

6.7.4. Shear connection and load introduction Load introduction

The provisions for the resistance of cross-sections of columns assume that no significant slip occurs at the interface between the concrete and structural steel components. Clauses Clause 6.7.4.1(1)P 6.7.4.1(1)P and 6.7.4.1(2)P give the principles for limiting slip to an ‘insignificant’ level in the Clause 6.7.4.1(2)P critical regions: those where axial load and/or bending moments are applied to the column. For any assumed ‘clearly defined load path’ it is possible to estimate stresses, including shear at the interface. In regions of load introduction, shear stress is likely to exceed the design shear strength from clause 6.7.4.3, and shear connection is then required (clause 6.7.4.2(1)). It is unlikely to be needed elsewhere, unless the shear strength τRd from Table 6.6 is very low, or the member is also acting as a beam, or has a high degree of double-curvature Clause 6.7.4.1(3) bending. Clause 6.7.4.1(3) refers to the special case of an axially loaded column. Few shear connectors reach their design shear strength until the slip is at least 1 mm; but this is not ‘significant’ slip for a resistance model based on plastic behaviour and rectangular stress blocks. However, a long load path implies greater slip, so the assumed path should not extend beyond the introduction length given in clause 6.7.4.2(2). Where axial load is applied through a joint attached only to the steel component, the force to be transferred to the concrete can be estimated from the relative axial loads in the two materials given by the resistance model. Accurate calculation is rarely practicable where the cross-section concerned does not govern the design of the column. In this partly plastic situation, the more adverse of the elastic and fully plastic models gives a safe result

111





Clause 6.7.4.2(1) (clause 6.7.4.2(1), last line). In practice, it may be simpler to provide shear connection based on a conservative (high) estimate of the force to be transferred. Where axial force is applied by a plate bearing on both materials or on concrete only, the proportion of the force resisted by the concrete gradually decreases, due to creep and shrinkage. It could be inferred from clause 6.7.4.2(1) that shear connection should be provided for a high proportion of the force applied. However, models based on elastic theory are over-conservative in this inherently stable situation, where large strains are acceptable. The application rules that follow are based mainly on tests. In a concrete-filled tube, shrinkage effects are low, for only the autogenous shrinkage strain occurs, with a long-term value below 10–4, from clause 3.1.4(6) of EN 1992-1-1. Radial shrinkage is outweighed by the lateral expansion of concrete in compression, for its inelastic Poisson’s ratio increases at high compressive stress. Friction then provides significant shear Clause 6.7.4.2(3) connection (clause 6.7.4.2(3)). Friction is also the basis for the enhanced resistance of stud Clause 6.7.4.2(4) connectors given in clause 6.7.4.2(4). Detailing at points of load introduction or change of cross-section is assisted by the high Clause 6.7.4.2(5) bearing stresses given in clauses 6.7.4.2(5) and 6.7.4.2(6). As an example, the following data Clause 6.7.4.2(6) are assumed for the detail shown in Fig. 6.22(B), with axial loading: steel tube with external diameter 300 mm and wall thickness 10 mm bearing plate 15 mm thick, with strength fy = fyd = 355 N/mm2 concrete with fck = 45 N/mm2 and γC = 1.5.

• • •

Then, Ac = π × 1402 = 61 600 mm2; A1 = 15 × 280 = 4200 mm2. From equation (6.48), σc, Rd /fcd = [1 + (4.9 × 10/300)(355/45)](14.7)0.5 = 8.8 and σc, Rd = 8.8 × 30 = 260 N/mm2 This bearing stress is so high that the fin plate would need to be at least 180 mm deep to have sufficient resistance to vertical shear. Clause 6.7.4.2(9) Figure 6.23 illustrates the requirement of clause 6.7.4.2(9) for transverse reinforcement, which must have a resistance equal to the force Nc1. If longitudinal reinforcement is ignored, this is given by Nc1 = Ac2/2nA where A is the transformed area of the cross-section 1–1 of the column in Fig. 6.23, given by A = As + (Ac1 + Ac2)/n and Ac1 and Ac2 are the unshaded and shaded areas of concrete, respectively, in section 1–1.

Transverse shear Clause 6.7.4.3 gives application rules (used in Example 6.11) relevant to the principle of clause 6.7.4.1(2), for columns with the longitudinal shear that arises from transverse shear. The design shear strengths τRd in Table 6.6 are far lower than the tensile strength of concrete. They rely on friction, not bond, and are related to the extent to which separation at the interface is prevented. For example, in partially encased I-sections, lateral expansion of the concrete creates pressure on the flanges, but not on the web, for which τRd = 0; and the highest shear strengths are for concrete within steel tubes. Where small steel I-sections are provided, mainly for erection, and the column is mainly Clause 6.7.4.3(4) concrete, clause 6.7.4.3(4) provides a useful increase to τRd, for covers up to 115 mm, more simply presented as

Clause 6.7.4.3

βc = 0.2 + cz/50 £ 2.5 Concern about the attachment of concrete to steel in partially encased I-sections appears Clause 6.7.4.3(5) again in clause 6.7.4.3(5), because under weak-axis bending, separation tends to develop between the encasement and the web.

112





6.7.5. Detailing provisions

If a steel I-section in an environment in class X0 to EN 1992-1-1 has links in contact with its flange (permitted by clause 6.7.5.2(3)), the cover to the steel section could be as low as 25 mm. For a wide steel flange, this thin layer of concrete would have little resistance to buckling outwards, so the minimum thickness is increased to 40 mm in clause 6.7.5.1(2). Clause 6.7.5.1(2) Minimum longitudinal reinforcement (clause 6.7.5.2(1)), is needed to control the width of Clause 6.7.5.2(1) cracks, which can be caused by shrinkage even in columns with concrete nominally in compression. Clause 6.7.5.2(4) refers to exposure class X0 of EN 1992-1-1. This is a ‘very dry’ Clause 6.7.5.2(4) environment, with ‘no risk of corrosion or attack’. Buildings with ‘very low air humidity’ are given as an example, so some buildings, or spaces in buildings, would not qualify. The minimum reinforcement provides robustness during construction.

Example 6.10: composite column with bending about one or both axes

A composite column of length 7.0 m has the uniform cross-section shown in Fig. 6.37. Its resistance to given action effects will be found. After checking that the column is within the scope of the simplified method, the calculations follow the sequence of the flow chart in Fig. 6.36. The properties of the steel member, 254 × 254 UC89, are taken from section tables, and given here in Eurocode notation. The properties of the materials, in the usual notation, are as follows: Structural steel: grade S355, fy = fyd = 355 N/mm2, Ea = 210 kN/mm2. Concrete: C25/30; fck = 25 N/mm2, fcd = 25/1.5 = 16.7 N/mm2, 0.85fcd = 14.2 N/mm2, Ecm = 31 kN/mm2, n0 = 210/31 = 6.77. Reinforcement: ribbed bars, fsk = 500 N/mm2, fsd = 500/1.15 = 435 N/mm2.

Geometrical properties of the cross-section In the notation of Fig. 6.17(a), bc = hc = 400 mm b = 256 mm h = 260 mm cz = 200 – 130 = 70 mm cy = 200 – 128 = 72 mm These satisfy the conditions of clauses 6.7.3.1(2), 6.7.3.1(4) and 6.7.5.1(2), so all the concrete casing is included in the calculations. Area of reinforcement = 4 × 36π = 446 mm2 Area of concrete = 4002 – 11 400 – 446 = 148 150 mm2

Steel section: 254 × 254 UC89

400 256

72

Aa = 11 400 mm2 10–6Ia, y = 143.1 mm4

70 17.3

10–6Ia, z = 48.5 mm4 10–6Wpa, y = 1.228 mm3

400

260 10–6Wpa, z = 0.575 mm3 10.5 12 ∆

Fig. 6.37. Cross-section and properties of a composite column (dimensions in mm)

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The reinforcement has area 0.301% of the concrete area, so clause 6.7.5.2(1) permits it to be included in calculations. For simplicity, its small contribution will be ignored, so the values are Aa = 11 400 mm2

Ac = 4002 – 11 400 = 148 600 mm2

As = 0

For the steel section: 10–6Ia, y = 143.1 mm4

10–6Ia, z = 48.5 mm4

Design action effects, ultimate limit state For the most critical load arrangement, global analysis gives these values: NEd = 1800 kN, of which NG, Ed = 1200 kN My, Ed, top = 380 kN m; Mz, Ed, top = 0 The bending moments at the lower end of the column and the lateral loading are zero. Later, the effect of adding Mz, Ed, top = 50 kN m is determined.

Properties of the column length From clause 6.7.3.2(1), Npl, Rd = Aa fyd + 0.85Ac fcd = 11.4 × 355 + 148.6 × 14.2 = 4047 + 2109 = 6156 kN From clause 6.7.3.3(1), the steel contribution ratio is δ = 4047/6156 = 0.657 which is within the limits of clause 6.7.1(4). For λ to clause 6.7.3.3(2), with γC = 1.5, Npl, Rk = 4047 + 1.5 × 2109 = 7210 kN Creep coefficient From clause 6.7.3.3(4), Ec = Ecm/[1 + (NG, Ed /NEd)ϕt]

(6.41)

The creep coefficient ϕt is ϕ(t, t0) to clause 5.4.2.2. The time t0 is taken as 30 days, and t is taken as ‘infinity’, as creep reduces the stiffness, and hence the stability, of a column. From clause 3.1.4(5) of EN 1992-1-1, the ‘perimeter exposed to drying’ is u = 2(bc + hc) = 1600 mm so h0 = 2Ac/u = 297 200/1600 = 186 mm Assuming ‘inside conditions’ and the use of normal cement, the graphs in Fig. 3.1 of EN 1992-1-1 give ϕ(•, 30) = 2.7 = ϕt The assumed ‘age at first loading’ has little influence on the result if it exceeds about 20 days. If, however, significant load were applied at age 10 days, ϕt would be increased to about 3.3. From equation (6.41), Ec, eff = 31/[1 + 2.7(1200/1800)] = 11.1 kN/mm2 Elastic critical load, with characteristic stiffness The minor axis is the more critical, so λ z is needed. From clause 6.7.3.3(3),

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(EI)eff = EaIa + KeEc, effIc

(6.40)

For the concrete, 10–6Ic, z = 0.42 × 4002/12 – 48.5 = 2085 mm4 From equation (6.40), 10–6(EI)eff, z = 210 × 48.5 + 0.6 × 11.1 × 2085 = 24 070 kN mm2 In this example, the end conditions for the column are assumed to be such that lateral restraint is provided, but no elastic rotational restraint, so the effective length is the actual length, 7.0 m, and Ncr, z = π2(EI)eff, z/L2 = 24 070π2/49 = 4848 kN From equation (6.39), λ z = (Npl, Rk/Ncr, z)0.5 = (7210/4848)0.5 = 1.22 Similar calculations for the y-axis give 10–6Ic, y = 1990 mm4 Ncr, y = 8715 kN

10–6(EI)eff, y = 43 270 kN mm2 λ y = 0.91

The non-dimensional slenderness does not exceed 2.0, so clause 6.7.3.1(1) is satisfied.

Resistance to axial load, z-axis From clause 6.7.3.5(2), buckling curve (c) is applicable. From Fig. 6.4 of EN 1993-1-1, for λ z = 1.22, χz = 0.43 From equation (6.44), NEd £ χz Npl, Rd = 0.43 × 6156 = 2647 kN This condition is satisfied.

Transverse shear For My, Ed, top = 380 kN m, the transverse shear is Vz, Ed = 380/7 = 54 kN This is obviously less than 0.5Vpl, a, Rd, so clause 6.7.3.2(3) does not apply.

Interaction curves The interaction polygons corresponding to Fig. 6.19 are determined in Appendix C (see Example C.1), and reproduced in Fig. 6.38. Clause 6.7.3.2(5) permits them to be used as approximations to the N–M interaction curves for the cross-section. First-order bending moments, y-axis The distribution of the external bending moment is shown in Fig. 6.39(a). From clause 6.7.3.4(4), the equivalent member imperfection is e0, z = L/200 = 35 mm The mid-length bending moment due to NEd is NEd e0, z = 1800 × 0.035 = 63 kN m Its distribution is shown in Fig. 6.39(b).

115





NRd (MN) 6

6.156 A

5

328

4

Major axis

414

504

Minor axis

3

534 C

2.482

2

1.80 1.241

D 1 559 B

0

200

400

600

333

MRd (kN m)

Fig. 6.38. Interaction polygons for bending about the major and minor axes

First order

My, Ed (kN m)

k1 = 0.87

380

M from initial bow (kN m) k2 = 1.32

0.66 × 380 0.87 × 380 83 63 0

3.5

7.0

0

3.5

x (m) (a)

7.0 x (m)

(b)

Fig. 6.39. Design second-order bending moments for a column of length 7.0 m

To check whether second-order moments can be neglected, a reduced value of Ncr is required, to clause 6.7.3.4(3). From equation (6.42), (EI)y, eff, II = 0.9(EaIa + 0.5EcIc) = 0.9 × 106(210 × 143.1 + 0.5 × 11.1 × 1990) = 3.70 × 1010 kN mm2 Hence, Ncr, y, eff = 37 000π2/72 = 7450 kN This is less than 10NEd, so second-order effects must be considered.

Second-order bending moments, y-axis From clause 6.7.3.4(5), Table 6.4, for the end moments, r = 0, β = 0.66, and from equation (6.43), k1 = β/(1 – NEd/Ncr, y, eff) = 0.66/(1 – 1800/7450) = 0.87

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This is not increased to 1.0, as it will be combined with the effect of imperfections, so the major-axis bending moments are as shown in Fig. 6.39(a). For the bending moment from the member imperfection, β = 1.0. From equation (6.43), k2 = 1/(1 – 1800/7450) = 1.32 thus increasing Nede0, z to 83 kN m (Fig. 6.39(b)). The total mid-length bending moment is 331 + 83 = 414 kN m This exceeds the greater end moment, 380 kN m, and so governs. The point (NEd, My, Ed, max) is (1800, 414) on Fig. 6.38. From the values shown in the figure, My, Rd = 504 + (2482 – 1800) × 55/(2482 – 1241) = 534 kN m This exceeds Mpl, y, Rd, so clause 6.7.3.6(2) is relevant. In this case, it makes no difference whether NEd and MEd are from independent actions or not, because the point My, Rd lies on line CD in Fig. 6.38, not on line BD, so the ‘additional verification’ to clause 6.7.1(7) would not alter the result. The ratio My, Ed, max /µd, yMpl, y, Rd = 414/534 = 0.78 This is below 0.9, so clause 6.7.3.6(1) is satisfied. The column is strong enough.

Biaxial bending The effect of adding a minor-axis bending moment Mz, Ed, top = 50 kN m is as follows. It is much smaller than My, Ed, so major-axis failure is assumed. From clause 6.7.3.7(1), there is assumed to be no bow imperfection in the x–y plane; but second-order effects have to be considered. From equation (6.42), (EI)z, eff, II = 0.9 × 106(210 × 48.5 + 0.5 × 11.1 × 2085) = 1.96 × 1010 kN mm2 Hence, Ncr, z, eff = 19 600π2/49 = 3948 kN m As before, β = 0.66, and from equation (6.43), k1 = 0.66/(1 – 1800/3948) = 1.21 so Mz, Ed, max = 1.21 × 50 = 60.5 kN m From Fig. 6.38, for NEd = 1800 kN, Mz, Rd = µd, zMpl, z, Rd = 330 kN m From clause 6.7.3.7(2), My, Ed, max /0.9My, Rd + Mz, Ed, max /0.9Mz, Rd = 414/(0.9 × 534) + 60.5/(0.9 × 330) = 0.861 + 0.204 = 1.07 This exceeds 1.0, so the check is thus not satisfied, and the column cannot resist the additional bending moment.

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Example 6.11: longitudinal shear outside areas of load introduction, for a composite column

All of the data for this example are given in Example 6.10 and Fig. 6.37, and are not repeated here. The design transverse shear was found to be Vz, Ed = 54 kN The maximum shear that can be resisted without provision of shear connection is now calculated. The critical cross-section is B–B in Fig. 6.40. Clause 6.7.4.3(2) permits the use of elastic analysis. Creep and cracking should be considered. Creep reduces the shear stress on plane B–B, so the modular ratio n0 = 6.77 is used. Uncracked section properties are used, and cracking is considered later. The cover cz = 70 mm, so from equation (6.49) in clause 6.7.4.3(4), βc = 1.60. From Table 6.6, τRd = 0.30 × 1.6 = 0.48 N/mm2 From Example 6.10, 10–6Ia, y = 143.1 mm4

10–6Ic, y = 1990 mm4

so for the uncracked section in ‘concrete’ units, 10–6Iy = 1990 + 143.1 × 6.77 = 2959 mm4 The ‘excluded area’ Aex = 400 × 70 = 28 000 mm2 and its centre of area is z = 200 – 35 = 165 mm from G in Fig. 6.40. Hence, Vz, Rd = τRd Iy bc/(Aex z) = 0.48 × 2959 × 400/(28 000 × 0.165) = 123 kN This exceeds Vz, Ed, so no shear connection is needed. The margin is so great that there is no need to consider the effect of cracking.

Column section used as a beam The shear strengths given in Table 6.6 are unlikely to be high enough to permit omission of shear connection from a column member with significant transverse loading. As an example, it is assumed that the column section of Fig. 6.40 is used as a simply-supported beam of span 8.0 m, with uniformly distributed loading. 400

cz = 70 B

B

400 G

Fig. 6.40 Longitudinal shear on plane B–B in a column cross-section (dimensions in mm)

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Using elastic analysis of the cracked cross-section with n0 = 6.77 as before, but no axial load, the results are 10–6Iy, cracked = 1441 mm4

Vz, Rd = 79 kN

This corresponds to a design ultimate load of 19.7 kN/m, which is only 4.4 times the unfactored weight of the member.

6.8. Fatigue 6.8.1. General

Although fatigue verification is mainly needed for bridges, these ‘general’ provisions find application in some buildings; for example, where travelling cranes or fork-lift trucks are used. They refer extensively to EN 1993-1-9,89 Fatigue Strength, which is also ‘general’. The Eurocode methods for fatigue are quite complex. There are supplementary provisions in EN 1994-2, Composite Bridges. The only complete set of provisions on fatigue in EN 1994-1-1 is for stud shear connectors. Fatigue in reinforcement, concrete, and structural steel is covered mainly by cross-reference to EN 1992 and EN 1993. Commentary will be found in the guides to those codes.6,7 Comments here are limited to design for a single cyclic loading: a defined number of cycles, NE, of a loading event for which can be calculated, at a given point, either: • •

a single range of stress, ∆σE(NE) or ∆τE(NE), or several stress ranges, that can be represented as N* cycles of a single ‘damage equivalent stress range’ (e.g. ∆σE, equ(N*)) by using the Palmgren–Miner rule for summation of fatigue damage.

The term ‘equivalent constant-amplitude stress range’, defined in clause 1.2.2.11 of EN 1993-1-9, has the same meaning as ‘damage equivalent stress range’, used here and in clause 6.8.5 of EN 1992-1-1. Damage equivalent factors (typically λ, as used in bridge design) are not considered here. Reference may be made to guides in this series to Part 2 of Eurocodes 2, 3 and 4 (e.g. Designers’ Guide to EN 199490). Fatigue damage is related mainly to the number and amplitude of the stress ranges. The peak of the stress range has a secondary influence that can be, and usually is, ignored in practice for peak stresses below about 60% of the characteristic strength. Ultimate loads are higher than peak fatigue loads, and the use of partial safety factors for ultimate-load design normally ensures that peak fatigue stresses are below this limit. This may not be so for buildings with partial shear connection, so clause 6.8.1(3) limits the force per stud to 0.75PRd, or 0.6PRk for γV = 1.25. Clause 6.8.1(4) gives guidance on the types of building where fatigue assessment may be required. By reference to EN 1993-1-1, these include buildings with members subject to wind-induced or crowd-induced oscillations. ‘Repeated stress cycles from vibrating machinery’ are also listed, but these should in practice be kept out of the structure by appropriate mountings.

Clause 6.8.1(3) Clause 6.8.1(4)

6.8.2. Partial factors for fatigue assessment

Resistance factors γMf may be given in National Annexes, so only the recommended values can be discussed here. For fatigue strength of concrete and reinforcement, clause 6.8.2(1) refers to EN 1992-1-1, which recommends the partial factors 1.5 and 1.15, respectively. For structural steel, EN 1993-1-9 recommends values ranging from 1.0 to 1.35, depending on the design concept and consequence of failure. These apply, as appropriate, for a fatigue failure of a steel flange caused by a stud weld.

Clause 6.8.2(1)

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Fatigue failure of a stud shear connector, not involving the flange, is covered by EN 1994-1-1. The value recommended in a note to clause 2.4.1.2(7)P, γMf, s = 1.0, corresponds to the value in EN 1993-1-9 for ‘damage tolerant design concept’ with ‘low consequence of failure’. From clause 3(2) of EN 1993-1-9, the use of the damage tolerant method should be satisfactory, provided that ‘a prescribed inspection and maintenance regime for detecting and correcting fatigue damage is implemented...’. A note to this clause states that the damage tolerant method may be applied where ‘in the event of fatigue damage occurring a load redistribution between components of structural elements can occur’. The second condition applies to stud connectors, but the first does not, for lack of access prevents detection of small cracks by any simple method of inspection. The recommendation of EN 1994-1-1 is based on other considerations, as follows. Fatigue failure results from a complex interaction between steel and concrete, commencing with powdering of the highly stressed concrete adjacent to the weld collar. This displaces upwards the line of action of the shear force, increasing the bending and shear in the shank just above the weld collar, and probably also altering the tension. Initial fatigue cracking further alters the relative stiffnesses and the local stresses. Research has found that the exponent that relates the cumulative damage to the stress range may be higher than the value, five, for other welds in shear. The value chosen for EN 1994-1-1, eight, is controversial, as discussed later. As may be expected from the involvement of a tiny volume of concrete, tests show a wide scatter in fatigue lives, which is allowed for in the design resistances. Studs are provided in large numbers, and are well able to redistribute shear between themselves. The strongest reason for not recommending a value more conservative than 1.0 comes from experience with bridges, where stud connectors have been used for almost 50 years. Whenever occasion has arisen in print or at a conference, the first author has stated that there is no known instance of fatigue failure of a stud in a bridge, other than a few clearly attributable to errors in design. This has not been challenged. Research has identified, but not yet quantified, many reasons for this remarkable experience.91,92 Most of them (e.g. slip, shear lag, permanent set, partial interaction, adventitious connection from bolt heads, and friction) lead to predicted stress ranges on studs lower than those assumed in design. With an eighth-power law, a 10% reduction in stress range more than doubles the fatigue life.

Clause 6.8.3(3)

6.8.3. Fatigue strength

The format of clause 6.8.3(3), as in EN 1993-1-9, uses a reference value of range of shear stress at 2 million cycles, ∆τC = 90 N/mm2. It defines the slope m of the line through this point on the log–log plot of range of stress ∆τR against number of constant-range stress cycles to failure, NR (Fig. 6.25). It is a complex matter to deduce a value for m from the mass of test data, which are often inconsistent. Many types of test specimen have been used, and the resulting scatter of results must be disentangled from that due to inherent variability. Values for m recommended in the literature range from 5 to 12, mostly based on linear-regression analyses. The method of regression used (x on y, or y on x) can alter the value found by up to three.91 The value eight, which was also used in BS 5400: Part 10, may be too high. In design for a loading spectrum, its practical effect is that the cumulative damage is governed by the highest-range components of the spectrum (e.g. by the small number of maximum-weight lifts made by a crane). A lower value, such as five, would give more weight to the much higher number of average-range components. While fatigue design methods for stud connectors continue to be conservative (for bridges and probably for buildings too) the precise value for m is of academic interest. Any future proposals for more accurate methods for prediction of stress ranges should be associated with re-examination of the value for m.

6.8.4. Internal forces and fatigue loadings

The object of a calculation is usually to find the range or ranges of stress in a given material at a chosen cross-section, caused by a defined event; for example, the passage of a vehicle along

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a beam. Loading other than the vehicle influences the extent of cracking in the concrete, and, hence, the stiffnesses of members. Cracking depends mainly on the heaviest previous loading, and so tends to increase with time. Clause 6.8.4(1) refers to a relevant clause in Eurocode 2. This defines the non-cyclic loading assumed to co-exist with the design value of the cyclic load, Qfat: it is the ‘frequent’ combination, represented by

ÂG j≥1

k, j

Clause 6.8.4(1)

+ P + ψ1, 1 Qk, 1 + Â ψ2, i Qk, i i>1

where the Qs are non-cyclic variable actions. Often, only one G and one Q are relevant, and there is no prestress. The design combination is then Gk + ψ1Qk + Qfat

(D6.38)

Clause 6.8.4(2) defines symbols that are used for bending moments in clause 6.8.5.4. The sign convention is evident from Fig. 6.26, which shows that MEd, max, f is the bending moment that causes the greatest tension in the slab, and is positive. Clause 6.8.4(2) also refers to internal forces, but does not give symbols. Analogous use of calculated tensile forces in a concrete slab (e.g. NEd, max, f, etc.) may sometimes be necessary.

Clause 6.8.4(2)

6.8.5. Stresses

Clause 6.8.5.1(1) refers to a list of action effects in clause 7.2.1(1)P to be taken into account Clause 6.8.5.1(1) ‘where relevant’. They are all relevant, in theory, to the extent of cracking. However, this can usually be represented by the same simplified model, chosen from clause 5.4.2.3, that is used for other global analyses. They also influence the maximum value of the fatigue stress range, which is limited for each material (e.g. the limit for shear connectors in clause 6.8.1(3)). It is unusual for any of these limits to be reached in design for a building; but if there are highly stressed cross-sections where most of the variable action is cyclic, the maximum value should be checked. The provisions for fatigue are based on the assumption that the stress range caused by a given fluctuation of loading, such as the passage of a vehicle of known weight, remains approximately constant after an initial shakedown period. ‘Shakedown’ here includes the changes due to cracking, shrinkage and creep of concrete that occur mainly within the first year or two. Most fatigue verifications are for load cycles with more than 104 repetitions. For this number in a 50 year life, the mean cycle time is less than 2 days. Thus, load fluctuations slow enough to cause creep (e.g. from the use of a tank for storing fuel oil) are unlikely to be numerous enough to cause fatigue damage. This may not be so for industrial processes with dynamic effects, such as forging, or for the charging floor for a blast furnace, but other uncertainties are likely to outweigh those from creep. The short-term modular ratio should therefore be used when finding stress ranges from the cyclic action Qfat. Where a peak stress is being checked, creep from permanent loading should be allowed for, if it increases the relevant stress. This would apply, for example, for most verifications for the structural steel in a composite beam. The effect of tension Clause 6.8.5.1(2)P stiffening (clauses 6.8.5.1(2)P and 6.8.5.1(3)), is illustrated in Example 6.12, below. The designer chooses a location where fatigue is most likely to govern. For a vehicle Clause 6.8.5.1(3) travelling along a continuous beam, the critical section for shear connection may be near mid-span; for reinforcement, it is near an internal support. The extent of a continuous structure that needs to be analysed depends on what is being checked. For a fatigue check on reinforcement it would be necessary to include at least two spans of the beam, and perhaps the two adjacent column lengths; but ranges of vertical shear, as the vehicle passes, are barely influenced by the rest of the structure, so for the shear connection it may be possible to consider the beam in isolation. For analysis, the linear-elastic method of Section 5 is used, from clause 6.8.4(1). Redistribution of moments is not permitted, clause 5.4.4(1). Calculation of range of stress, or of shear flow, from the action effects is based entirely on elastic theory, following clause 7.2.1.

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For stresses in structural steel, the effects of tension stiffening may be included or Clause 6.8.5.1(4) neglected (clause 6.8.5.1(4)).

Concrete Clause 6.8.5.2(1) For concrete, clause 6.8.5.2(1) refers to clause 6.8 of EN 1992-1-1, which provides (in clause 6.8.7) for concrete subjected to a damage equivalent stress range. For a building, fatigue of concrete is unlikely to influence design, so this clause is not discussed here. Structural steel Clause 6.8.5.3(1) Clause 6.8.5.3(1) repeats, in effect, the concession in clause 6.8.5.1(4). Where the words ‘or Clause 6.8.5.3(2) only MEd, min, f’ in clause 6.8.5.3(2) apply, MEd, max, f causes tension in the slab. The use of the uncracked section for MEd, max, f could then under-estimate the stress ranges in steel flanges. Reinforcement For reinforcement, clause 6.8.3(2) refers to EN 1992-1-1, where clause 6.8.4 gives the verification procedure. Its recommended value N* for straight bars is 106. This should not be confused with the corresponding value for structural steel in EN 1993-1-9, 2 × 106, denoted NC, which is used also for shear connectors (clause 6.8.6.2(1)). Using the γ values recommended in EN 1992-1-1, its equation (6.71) for verification of reinforcement becomes ∆σE, equ(N*) £ ∆σRsk(N*)/1.15 2

(D6.39) 6

with ∆σRsk = 162.5 N/mm for N* = 10 , from Table 6.4N. Where a range ∆σE(NE) has been determined, the resistance ∆σRsk(NE) can be found from the S–N curve for reinforcement, and the verification is ∆σE(NE) £ ∆σRsk(NE)/1.15

(D6.40)

Clause 6.8.5.4(1)

Clause 6.8.5.4(1) permits the use of the approximation to the effects of tension stiffening that is used for other limit states. It consists of adding to the maximum tensile stress in the ‘fully cracked’ section, σs, 0, an amount ∆σs that is independent of σs, 0 and of the limit state. Clause 6.8.5.4(2) Clauses 6.8.5.4(2) and 6.8.5.4(3) give simplified rules for calculating stresses, with Clause 6.8.5.4(3) reference to Fig. 6.26, which is discussed using Fig. 6.41. This has the same axes, and also shows a minimum bending moment that causes compression in the slab. A calculated value for the stress σs in reinforcement, that assumes concrete to be effective, would lie on line AOD. On initial cracking, the stress σs jumps from B to point E. Lines OBE are not shown in Fig. 6.26 because clause 7.2.1(5)P requires the tensile strength of concrete to be neglected in calculations for σs. This gives line OE. For moments exceeding Mcr, the stress σs follows route EFG on first loading. Calculation of σs using section property I2 gives line OC. At bending moment MEd, max, f the stress σs, 0 thus found is increased by ∆σs, from equation (7.5), as shown by line HJ. Tension stiffening tends to diminish with repeated loading,93 so clause 6.8.5.4 defines the unloading route from point J as JOA, on which the stress σs, min, f lies. Points K and L give two examples, for MEd, min, f causing tension and compression, respectively, in the slab. The fatigue stress ranges ∆σs, f for these two cases are shown.

Shear connection Clause 6.8.5.5(1)P The interpretation of clause 6.8.5.5(1)P is complex when tension stiffening is allowed for. Spacing of shear connectors near internal supports is unlikely to be governed by fatigue, so it is simplest to use uncracked section properties when calculating the range of shear flow from Clause 6.8.5.5(2) the range of vertical shear (clause 6.8.5.5(2)). These points are illustrated in Example 6.12. Clause 6.8.6.1

6.8.6. Stress ranges

Clause 6.8.6.1 is more relevant to the complex cyclic loadings that occur in bridges than to buildings, and relates to the provisions of EN 1992 and EN 1993. Where a spectrum

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ss (tension) G ss, max, f

J

Dss C H

ss, 0

Dss, f E

F K

ss, min, f

D 0 A

L

Mcr

B

MEd, min, f

MEd, max, f

M

MEd, min, f

Fig. 6.41. Stress ranges in reinforcement in cracked regions

of loading is specified, the maximum and minimum stresses, as discussed above for reinforcement, are modified by the damage equivalent factor λ, a property of the spectrum and the exponent m (clause 6.8.3(3)). Comment and guidance will be found in the relevant guides in this series. The need to combine global and local fatigue loading events (clause 6.8.6.1(3)), rarely Clause 6.8.6.1(3) occurs in buildings, and is outside of the scope of this guide. Where the design cyclic loading consists of a single load cycle repeated NE times, the damage equivalent factor λv used in clause 6.8.6.2 on shear connection can be found using the Clause 6.8.6.2 Palmgren–Miner rule, as follows. Let the load cycle cause a shear stress range ∆τ in a stud connector, for which m = 8. Then, (∆τE)8NE = (∆τE, 2)8NC where NC = 2 × 106 cycles. Hence, ∆τE, 2/∆τE = λv = (NE/NC)1/8

(D6.41)

6.8.7. Fatigue assessment based on nominal stress ranges

Comment on the methods referred to from clause 6.8.7.1 will be found in other guides in this Clause 6.8.7.1 series. For shear connectors, clause 6.8.7.2(1) introduces the partial factors. The recommended Clause 6.8.7.2(1) value of γMf, s is 1.0 (clause 6.8.2(1)). For γFf, EN 1990 refers to the other Eurocodes. The recommended value in EN 1992-1-1, clause 6.8.4(1), is 1.0. No value has been found in EN 1993-1-1 or EN 1993-1-9. Clause 9.3(1) of EN 1993-2 recommends 1.0 for bridges. Clause 6.8.7.2(2) covers interaction between the fatigue failures of a stud and of the steel Clause 6.8.7.2(2) flange to which it is welded, when the flange is in tension. The first of expressions (6.57) is the verification for the flange, from clause 8(1) of EN 1993-1-9, and the second is for the stud, copied from equation (6.55). The linear interaction condition is given in expression (6.56). It is necessary to calculate the longitudinal stress range in the steel flange that coexists with the stress range for the connectors. The load cycle that gives the maximum value of ∆σE, 2 in the flange will not, in general, be that which gives the maximum value of ∆τE, 2 in a

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shear connector, because the first is caused by flexure and the second by shear. Also, both ∆σE, 2 and ∆τE, 2 may be influenced by whether the concrete is cracked, or not. It thus appears that expression (6.56) may have to be checked four times. In practice, it is best to check first the conditions in expression (6.57). It should be obvious, for these, whether the ‘cracked’ or the ‘uncracked’ model is the more adverse. Usually, one or both of the left-hand sides is so far below 1.0 that no check to expression (6.56) is needed.

Example 6.12: fatigue in reinforcement and shear connection

It is assumed that the imposed floor load of 7.0 kN/m2 for the two-span beam in Examples 6.7 and 7.1 is partly replaced by a cyclic load. The resistance to fatigue of the reinforcement at the internal support, point B in Figs 6.23 and 6.28, and of the shear connection near the cyclic load are checked. All other data are as before.

Loading and global analysis The cyclic load is a four-wheeled vehicle with two characteristic axle loads of 35 kN each. It travels at right angles to beam ABC, on a fixed path that is 2.0 m wide and free from other variable loads. The axle spacing exceeds the beam spacing of 2.5 m, so each passage can be represented by two cycles of point load, 0–35–0 kN, applied at point D in Fig. 6.42(a). For a 25 year design life, 20 passages per hour for 5000 h/year gives NEd = 2.5 × 106 cycles of each point load. The partial factor γFf is taken as 1.0. In comparison with Example 6.7, the reduction in static characteristic imposed load is 7 × 2.5 × 2 = 35 kN, the same as the additional axle load, so previous global analyses for the characteristic combination can be used. These led to the bending moments MEk at support B given in the four rows of Table 7.2. Those in rows 2 and 4 are unchanged, and MEk, B = 263 kN m for loading qk. Analysis for the load Qfat alone, with 15% of each span cracked, gave the results in Fig. 6.42(b), with 31 kN m at support B. The frequent combination of non-cyclic imposed load is specified, for which ψ1 = 0.7. From Table 6.2, qk = 17.5 kN/m. Therefore, ψ1qk = 0.7 × 17.5 = 12.25 kN/m, acting on span AB and on 10 m only of span BC, giving MEk, B = 0.7 × (263 – 31) = 162 kN m Table 6.5 gives MEd, min, f = 18 + 162 + 120 = 300 kN m

31

12

4

11

35

6 2.6 + 23

2.6

12 84

Qfat = 35 kN

0.7qk A

B

D

(b) C

120

(a)

10.0

20.0

10.0

(c)

Fig. 6.42. Fatigue checks for a two-span beam. (a) Variable static and cyclic loads. (b) Design action effects, cyclic load. (c) Design action effects, shrinkage

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Table 6.5. Stresses in longitudinal reinforcement at support B Action

n

Load (kN/m)

Permanent, composite Variable, static (ψ1 = 0.7) Shrinkage Cyclic load

20.2 20.2 28.7 20.2

1.2 12.25 – –

Totals

MEk (kN m)

10–6Ws, cr (mm3)

18 162 120 31

1.65 1.65 1.65 1.65

331

σs, 0 (N/mm2) 11 98 73 19 201

From Fig. 6.42(b), MEd, max, f = 300 + 31 = 331 kN m

Verification for reinforcement at cross-section B From equation (D7.5), the allowance for tension stiffening is ∆σs = 52 N/mm2. From σs, 0 given in Table 6.5, σs, max, f = 201 + 52 = 253 N/mm2 From Fig. 6.41, σs, min, f = 253 × 300/331 = 229 N/mm2 so ∆σs, f = 253 – 229 = 24 N/mm2 From clause 6.8.4 of EN 1992-1-1, for reinforcement, m = 9 for NE > 106

N* = 106

∆σRsk(N*) = 162.5 N/mm2

By analogy with equation (D6.41), the damage equivalent stress range for NEd = 5 × 106 cycles of stress range 24 N/mm2 is given by 249 × 5 × 106 = (∆σE, equ)9 × 106 whence ∆σE, equ = 29 N/mm2 Using equation (D6.40), 29 £ 162.5/1.15 so the reinforcement is verified. If the axle loads had been unequal, say 35 and 30 kN, the stress range for the lighter axle would be a little higher than 24 × 30/35 = 20.6 N/mm2, because its line OJ in Fig. 6.41 would be steeper. Assuming this stress range to be 21 N/mm2, the cumulative damage check, for γMf = 1.15, would be 2.5 × 106 × (249 + 219) £ (162.5/1.15)9 × 106 which is 8.6 × 1018 £ 2.25 × 1025

Verification for shear connection near point D Vertical shear is higher on the left of point D in Fig. 6.42, than on the right. From Fig. 6.42(b), ∆VEd, f = 23 kN for each axle load.

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Table 6.6. Fatigue of shear connectors near cross-section D Action

10–6Iy (mm4)

10–3Ac/n (mm2)

z (mm)

VEd (kN)

VEdAcz/nIy (kN/m)

Permanent, composite Variable, static (ψ1 = 0.7) Shrinkage Cyclic load

828 828 741 828

9.90 9.90 6.97 9.90

157 157 185 157

2.7 23.8 10.0 23.0

5.0 44.7 17.4 43.2

59.5

110

Totals

The maximum vertical shear at this point, including 10.0 kN from the secondary effect of shrinkage, Fig. 6.42(c), is 59.5 kN, Table 6.6. The shear forces VEd are found from the loads and values MEk in Table 6.5. The resulting maximum longitudinal shear flow, for the uncracked unreinforced composite section, is 110 kN/m, of which the cyclic part is 43.2 kN/m. Clause 6.8.1(3) limits the shear per connector under the characteristic combination to 0.75PRd. For this combination, the shear flow from the non-cyclic variable action increases by 19 from 44.7 to 44.7/0.7 = 63.9 kN/m, so the new total is 110 + 19 = 129 kN/m. The shear connection (see Fig. 6.30), is 5 studs/m, with PRd = 51.2 kN/stud. Hence, PEk/PRd = 129/(5 × 51.2) = 0.50 which is below the limit 0.75 in clause 6.8.1(3). The range of shear stress is ∆τE = 43 200/(5π × 9.52) = 30.5 N/mm2 The concrete is in density class 1.8. From clause 6.8.3(4), ∆τc = 90 × (1.8/2.2)2 = 60 N/mm2 From equation (D6.41), ∆τE, 2 = 30.5[5 × 106/(2 × 106)]1/8 = 34.2 N/mm2 As γMf, s = 1.0, ∆τc, d = 60 N/mm2 so the shear connection is verified.

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CHAPTER 7

Serviceability limit states This chapter corresponds to Section 7 of EN 1994-1-1, which has the following clauses: • • • •

General Stresses Deformations in buildings Cracking of concrete


7.1. General Section 7 of EN 1994-1-1 is limited to provisions on serviceability that are specific to composite structures and are not in Sections 1, 2, 4, 5 (for global analysis) or 9 (for composite slabs), or in Eurocodes 1990, 1991, 1992 or 1993. Some of these other, more general provisions are briefly referred to here. Further comments on them are in other chapters of this book, or in other handbooks in this series. The initial design of a structure is usually based on the requirements for ultimate limit states, which are specific and leave little to the judgement of the designer. Serviceability is then checked. The consequences of unserviceability are less serious than those of reaching an ultimate limit state, and its occurrence is less easily defined. For example, a beam with an imposed-load deflection of span/300 may be acceptable in some situations, but in others the client may prefer to spend more on a stiffer beam. The drafting of the serviceability provisions of the EN Eurocodes is intended to give designers and clients greater freedom to take account of factors such as the intended use of a building and the nature of its finishes. The content of Section 7 was also influenced by the need to minimize calculations. Results already obtained for ultimate limit states are scaled or re-used wherever possible. Experienced designers know that many structural elements satisfy serviceability criteria by wide margins. For these, design checks must be simple, and it does not matter if they are conservative. For other elements, a longer but more accurate calculation may be justified. Some application rules therefore include alternative methods. Clauses 7.1(1)P and 7.1(2) refer to clause 3.4 of EN 1990. This gives criteria for placing a limit state within the ‘serviceability’ group, with reference to deformations (including vibration), durability, and the functioning of the structure.

Serviceability verification and criteria The requirement for a serviceability verification is given in clause 6.5.1(1)P of EN 1990 as Ed £ Cd






where Ed is the design value of the effects of the specified actions and the ‘relevant’ combination, and Cd is the limiting design value of the ‘relevant’ criterion. From clause 6.5.3 of EN 1990, the relevant combination is ‘normally’ the characteristic, frequent, or quasi-permanent combination, for serviceability limit states that are, respectively, irreversible, reversible or a consequence of long-term effects. The quasi-permanent combination is also relevant for the appearance of the structure. For buildings, these combinations are used with the partial safety factor 1.0, from clause A1.4.1 of EN 1990, ‘unless differently specified’ in another Eurocode. There are no departures from 1.0 in EN 1994-1-1. The same provision, with value 1.0, is given for partial safety factors for properties of materials, in clause 6.5.4(1) of EN 1990. Clause A1.4.2 of EN 1990 refers to serviceability criteria relevant for buildings. These may be defined in National Annexes, and should be specified for each project and agreed with the client. Clause A1.4.4 of EN 1990 says that possible sources of vibration and relevant aspects of vibration behaviour should be considered for each project and agreed with the client and/or the relevant authority. Further guidance may be found in the relevant Eurocode Part 2 (bridges) and in specialized literature. Comments on limits to crack width are given under clause 7.4. No serviceability limit state of ‘excessive slip of shear connection’ has been defined, but the effect of slip is recognized in clause 7.3.1(4) on deflection of beams. Generally, it is assumed that design of shear connection for ultimate limit states ensures satisfactory performance in service, but composite slabs can be an exception. Relevant rules are given in clause 9.8.2. No serviceability criteria are specified for composite columns, so, from here on, this chapter is referring to composite beams or, in some places, to composite frames.

7.2. Stresses Clause 7.2.1 Clause 7.2.2

Excessive stress is not itself a serviceability limit state, though stress calculations to clause 7.2.1 are required for some verifications for deformation and cracking. For most buildings, no checks on stresses are required, clause 7.2.2. No stress limits for buildings are given in the Eurocodes for concrete and steel structures, other than warnings in clause 7.2 of EN 1992-1-1, with recommended stress limits in notes. The ‘bridge’ parts of these Eurocodes include stress limits, which may be applicable for buildings that have prestressing or fatigue loading.

7.3. Deformations in buildings 7.3.1. Deflections Global analysis

Clause 7.3.1(1)

Deflections are influenced by the method of construction, and may govern design, especially where beams designed as simpl- supported are built unpropped. For propped construction, props to beams should not be removed until all of the concrete that would then be stressed has reached a strength equivalent to grade C20/25, from clause 6.6.5.2(4). Then, elastic global analysis to Section 5 is sufficient (clause 7.3.1(2)). Where unpropped construction is used and beams are not designed as simply supported, the analysis may be more complex than is revealed by the reference to EN 1993 in clause 7.3.1(1). In a continuous beam or a frame, the deflection of a beam depends on how much of the structure is already composite when the slab for each span is cast. A simple and usually conservative method is to assume that the whole of the steel frame is erected first. Then, all of the concrete for the composite members is cast at once, its whole weight being carried by the steelwork; but more realistic multi-stage analyses may be needed for a high-rise structure and for long-span beams.

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CHAPTER 7. SERVICEABILITY LIMIT STATES

Where falsework or re-usable formwork is supported from the steel beam, it will be removed after the member becomes composite. The small locked-in stresses that result can usually be ignored in buildings, but not always in bridges. Where first-order elastic global analysis was used for ultimate limit states (ULS), it may be possible to obtain some of the results needed for serviceability limit states (SLS) by simple scaling by the ratio of the relevant loads. This ratio will depend on the method of construction, and also on which of the three serviceability load combinations is being used for the limit state considered. As an example, suppose that for an unbraced frame at ULS, αcr = 8, so that second-order global analysis was used, from clause 5.2.1(3). If most of the load on columns is from suspended floors, and these loads for SLS are 60% of those for ULS, the elastic critical load will be little altered, so for SLS, αcr ª 8/0.6 = 13. This exceeds 10, so first-order analysis can be used. Redistribution of moments is permitted for most framed structures at SLS by clause 5.4.4(1), but the details in paragraphs (4) to (7) apply only to ULS. The relevant provisions in Section 7 are in clauses 7.3.1(6) and 7.3.1(7), discussed below.

Limits to deflection of beams The specification of a deflection limit for a long-span beam needs care, especially where construction is unpropped and/or the steel beam is pre-cambered. Reference should be made to the three components of deflection defined in clause A1.4.3 of EN 1990. Depending on circumstances, it may be necessary to set limits to any one of them, or to more than one, related to a defined load level. Prediction of long-term values should take account of creep of concrete, based on the quasi-permanent combination, and may need to allow for shrinkage. Where precast floor units are used, it must be decided whether they should be cambered to compensate for creep. Clause 7.3.1(3) relates to the use of false ceilings, which conceal the sagging of a beam due to dead loading. Longitudinal slip Clause 7.3.1(4) refers to the additional deflection caused by slip at the interface between steel and concrete. Its three conditions all apply. Condition (b) relates to the minimum value of the degree of shear connection, η, given as 0.4 in clause 6.6.1.2(1), and gives a higher limit, 0.5. For use where the design is such that 0.4 £ η < 0.5, ENV 1994-1-149 gave the following equation for the additional deflection due to partial interaction: δ = δc + α(δa – δc)(1 – η)

Clause 7.3.1(3)

Clause 7.3.1(4)

(D7.1)

where α = 0.5 for propped construction and 0.3 for unpropped construction, δa is the deflection of the steel beam acting alone, and δc is the deflection for the composite beam with complete interaction; both δa and δc are calculated for the design loading for the composite member. The method comes from a summary of pre-1975 research on this subject,94 which also gives results of relevant tests and parametric studies. Other methods are also available.95

Cracking in global analysis Apart from the different loading, global analysis for serviceability differs little from that for an ultimate limit state. Clause 5.4.1.1(2) requires ‘appropriate corrections’ for cracking of concrete, and clause 7.3.1(5) says that clause 5.4.2.3 applies. Clause 5.4.2.3(2) permits the use of the same distribution of beam stiffnesses at SLS as for ULS. Clauses 5.4.2.3(3) to 5.4.2.3(5) also apply, including the reference in clause 5.4.2.3(4) to a method given in Section 6 for the effect of cracking on the stiffness of composite columns. In the absence of cracking, continuous beams in buildings can often be assumed to be of uniform section within each span, which simplifies global analysis. Cracking reduces bending moments at internal supports to an extent that can be estimated by the method of clause

Clause 7.3.1(5)

Clause 7.3.1(6)

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7.3.1(6), based on Stark and van Hove.95 The maximum deflection of a given span normally occurs when no imposed load acts on adjacent spans. The conditions for the use of curve A in Fig. 7.1 are then not satisfied, and the method consists simply of reducing all ‘uncracked’ moments at internal supports by 40%. Using the new end moments, Mh1 and Mh2, say, the maximum deflection can be found either by elastic theory for the span, of uncracked flexural stiffness EaI1, or by an approximate method given in BS 5950-3-1.31 This consists of multiplying the deflection for the simplysupported span by the factor 1 – 0.6(Mh1 + Mh2)/M0

(D7.2)

where M0 is the maximum sagging moment in the beam when it is simply supported.

Clause 7.3.1(7)

Yielding of steel In continuous beams built unpropped, with steel beams in Class 1 or 2, it is possible that serviceability loading may cause yielding at internal supports. This is permitted for beams in buildings, but it causes additional deflection, which should be allowed for. Clause 7.3.1(7) provides a method. The bending moments at internal supports are found by elastic analysis, with allowance for effects of cracking. The two values given in the clause for factors f2 correspond to different checks. The first is for dead load only: wet concrete on a steel beam. According to the UK’s draft National Annex to EN 1990,96 the load combination to be used for the second check depends on the functioning of the structure. It may be the characteristic, frequent or quasi-permanent combination, with the load additional to that for the first check acting on the composite beam. For each analysis, appropriate assumptions are needed for the adjacent spans, on their loading and on the state of construction. Local buckling This does not influence stiffnesses for elastic analysis except for Class 4 sections. For these, clause 5.4.1.1(6) refers to clause 2.2 in EN 1993-1-5, which gives a design rule.

Clause 7.3.1(8)

Shrinkage In principle, shrinkage effects appear in all load combinations. For SLS, clause 5.4.2.2(7) refers to Section 7, where clause 7.3.1(8) enables effects of shrinkage on deflections of beams to be ignored for span/depth ratios up to 20. In more slender beams, shrinkage deflections are significantly reduced by provision of continuity at supports. Temperature Clause 5.4.2.5(2), on neglect of temperature effects, does not apply. For buildings, neither ψ0 nor ψ1 is given as zero in clause A1.2.2 of EN 1990 (nor in the UK’s draft National Annex to BS EN 199096), so if temperature effects are relevant at ULS, they should be included in all SLS combinations except quasi-permanent. Welded mesh Clause 5.5.1(6) gives conditions for the inclusion of welded mesh in the effective section, within the rules for classification of sections.

7.3.2. Vibration

Clause 7.3.2(1)

Limits to vibration in buildings are material-independent, and vibration is in clause A1.4.4 of EN 1990, not in EN 1994. Composite floor systems are lighter and have less inherent damping than their equivalents in reinforced concrete. During their design, dynamic behaviour should be checked against the criteria in EN 1990 referred to from clause 7.3.2(1). These are general, and advise that the lowest natural frequency of vibration of the structure or member should be kept above a value to be agreed with the client and/or the relevant authority. No values are given for either limiting frequencies or damping coefficients.

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More specific guidance can be found in EN 1991-1-1 and the extensive literature on this subject.97,98 These sources refer to several criteria that are likely to be specific to the individual project, and, with other aspects, should be agreed with the client. A note to clause 7.2.3 of EN 1993-1-1 says that limits to vibration of floors may be specified in a National Annex.

7.4. Cracking of concrete 7.4.1. General

In the early 1980s it was found44,99 that for composite beams in hogging bending, the long-established British methods for control of crack width were unreliable for initial cracks, which were wider than predicted. Before this, it had been found for reinforced concrete that the appropriate theoretical model for cracking caused by restraint of imposed deformation was different from that for cracking caused by applied loading. This has led to the presentation of design rules for control of cracking as two distinct procedures: • •

for minimum reinforcement, in clause 7.4.2, for all cross-sections that could be subjected to significant tension by imposed deformations (e.g. by effects of shrinkage, which cause higher stresses than in reinforced concrete, because of restraint from the steel beam) for reinforcement to control cracking due to direct loading (clause 7.4.3).

The rules given in EN 1994-1-1 are based on an extensive and quite complex theory, supported by testing on hogging regions of composite beams.99,100 Much of the original literature is in German, so a detailed account of the theory has recently been published in English,101 with comparisons with results of tests on composite beams, additional to those used originally. The paper includes derivations of the equations given in clause 7.4, comments on their scope and underlying assumptions, and procedures for estimating the mean width and spacing of cracks. These are tedious, and so are not in EN 1994-1-1. Its methods are simple: Tables 7.1 and 7.2 give maximum diameters and spacings of reinforcing bars for three design crack widths: 0.2, 0.3 and 0.4 mm. Tables 7.1 and 7.2 are for ‘high-bond’ bars only. This means ribbed bars with properties as in clause 3.2.2(2)P of EN 1992-1-1. The use of reinforcement other than ribbed is outside the scope of the Eurocodes. The references to EN 1992-1-1 in clause 7.4.1(1) give the surface crack-width limits required for design. Concrete in tension in a composite beam or slab for a building will usually be in exposure class XC3, for which the recommended limit is 0.3 mm; however, for spaces with low or very low air humidity, Tables 4.1 and 7.1N of EN 1992-1-1 recommend a limit of 0.4 mm. The limits are more severe for prestressed members, which are not discussed further. The severe environment for a floor of a multistorey car park is discussed in Chapter 4. All these limits may be modified in a National Annex. Clause 7.4.1(2) refers to ‘estimation’ of crack width, using EN 1992-1-1. This rather long procedure is rarely needed, and does not take full account of the following differences between the behaviours of composite beams and reinforced concrete T-beams. The steel member in a composite beam does not shrink or creep and has much greater flexural stiffness than the reinforcement in a concrete beam. Also, the steel member is attached to the concrete flange only by discrete connectors that are not effective until there is longitudinal slip, whereas in reinforced concrete there is monolithic connection. Clause 7.4.1(3) refers to the methods developed for composite members, which are easier to apply than the methods for reinforced concrete members.

Clause 7.4.1(1)

Clause 7.4.1(2)

Clause 7.4.1(3)

Uncontrolled cracking Clause 7.3.1(4) of EN 1992-1-1 (referred to from clause 7.4.1(1)) permits cracking of uncontrolled width in some circumstances; for example, beams designed as simply supported, with a concrete top flange that is continuous over ‘simple’ beam-to-column

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Clause 7.4.1(4)

connections. These are flexible in bending, and rotate about a point that cannot be predicted, as its position depends on tolerances and methods of erection of the steelwork. It may then be impossible to predict the widths of cracks. Where the environment is dry and the concrete surface is concealed by a flexible finish, such as carpeting, crack widths exceeding 0.4 mm may be acceptable. Even so, the minimum reinforcement required (for other reasons) by EN 1992-1-1 may be inadequate to prevent the fracture of small-diameter bars near internal supports, or the formation of very wide cracks. Minimum areas greater than those of EN 1992-1-1 are therefore specified in clause 7.4.1(4) and, for composite slabs, in clause 9.8.1(2). The maximum thicknesses of slab that can be reinforced by one layer of standard welded fabric, according to these rules, are given in Table 7.1. For composite slabs, the relevant thickness is that above the profiled steel sheeting. The maximum spacing of flexural reinforcement permitted by clause 9.3.1.1(3) of EN 1992-1-1 is 3h, but not exceeding 400 mm, where h is the total depth of the slab. This rule is for solid slabs. It is not intended for slabs formed with profiled steel sheeting, for which a more appropriate rule is that given in clause 9.2.1(5): spacing not exceeding 2h (and £ 350 mm) in both directions, where h is the overall thickness of the slab, including ribs of composite slabs.

7.4.2. Minimum reinforcement

Clause 7.4.2(1)

The only data needed when using Tables 7.1 and 7.2 are the tensile stresses in the reinforcement, σs. For minimum reinforcement, σs is the stress immediately after initial cracking. It is assumed that the curvature of the steel beam does not change, so all of the tensile force in the concrete just before cracking is transferred to the reinforcement, of area As. If the slab were in uniform tension, equation (7.1) in clause 7.4.2(1) would be Asσs = Act fct, eff The three correction factors in equation (7.1) are based on calibration work.102 These allow for the non-uniform stress distribution in the area Act of concrete assumed to crack. ‘Non-uniform self-equilibrating stresses’ arise from primary shrinkage and temperature effects, which cause curvature of the composite member. Slip of the shear connection also causes curvature and reduces the tensile force in the slab. The magnitude of these effects depends on the geometry of the uncracked composite section, as given by equation (7.2). With experience, calculation of kc can often be omitted, because it is less than 1.0 only where z0 < 1.2hc. Especially for beams supporting composite slabs, the depth of the ‘uncracked’ neutral axis below the bottom of the slab (excluding ribs) normally exceeds about 70% of the slab thickness, and then, kc = 1. For design, the design crack width and thickness of the slab, hc will be known. It will be evident whether there should be one layer of reinforcement or two. Two layers will often consist of bars of the same size and spacing, which satisfies clause 7.4.2(3). For a chosen bar diameter φ, Table 7.1 gives σs, and equation (7.1) gives the bar spacing. If this is too high or low, φ is changed.

Table 7.1. Use of steel fabric as minimum reinforcement, to clause 7.4.1(4)

Bar size and spacing

Cross-sectional area (mm2 per m width)

6 mm, 200 mm 7 mm, 200 mm 8 mm, 200 mm

142 193 252

132



Maximum thickness of slab (mm) Unpropped, 0.2%

Propped, 0.4%

71 96 126

– 48 63



hc = 100 mm 150

s (mm) 400

200 300

300 200

100

0

5

6

8

10

12

16

20 f (mm)

Fig. 7.1. Bar diameter and spacing for minimum reinforcement in two equal layers, for wk = 0.3 mm and fct, eff = 3.0 N/mm2

A typical relationship between slab thickness hc, bar spacing s and bar diameter φ is shown in Fig. 7.1. It is for two similar layers of bars, with kc = 1 and fct, eff = 3.0 N/mm2. Equation (7.1) then gives, for a slab of breadth b, (πφ2/4)(2b/s) = 0.72 × 3bhc/σs Hence, hc s = 0.727φ2σs

(D7.3)

For each bar diameter and a given crack width, Table 7.1 gives φ2σs, so the product hcs is known. This is plotted in Fig. 7.1, for wk = 0.3 mm, as curves of bar spacing for four slab thicknesses, which can of course also be read as slab thicknesses for four bar spacings. The shape of the curves results partly from the use of rounded values of σs in Table 7.1. The optional correction to minimum reinforcement given in clause 7.4.2(2) is negligible here, and has not been made. Figure 7.1 can be used for slabs with one layer of bars by halving the slab thickness. The weight of minimum reinforcement, per unit area of slab, is proportional to φ2/s, which is proportional to σs–1, from equation (D7.3). This increases with bar diameter, from Table 7.1, so the use of smaller bars reduces the weight of minimum reinforcement. This is because their greater surface area provides more bond strength. The method of clause 7.4.2(1) is not intended for the control of early thermal cracking, which can occur in concrete a few days old, if the temperature rise caused by the heat of hydration is excessive. The flanges of composite beams are usually too thin for this to occur. It would not be correct, therefore, to assume a very low value for fct, eff. The suggested value, 3 N/mm2, was probably rounded from the mean 28 day tensile strength of grade C30/37 concrete, given in EN 1992-1-1 as 2.9 N/mm2 – the value used as the basis for the optional correction given in clause 7.4.2(2). The difference between 2.9 and 3.0 is obviously negligible. If there is good reason to assume a value for fct, eff such that the correction is not negligible, it is best used by assuming a standard bar diameter φ, calculating φ*, and then finding σs by interpolation in Table 7.1.

Clause 7.4.2(2)

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N

N

Tensile strain es2 e s( x )

esm es1

e c( x )

ecm 0

x ecs

Le

Le

Fig. 7.2. Strain distributions near a crack in a reinforced concrete tension member

Clause 7.4.2(6)

Clause 7.4.3(2)

Clause 7.4.3(3)

Reinforcement for encasement of a steel web Clause 7.4.2(6) gives a minimum value of As/Act for encasement of the type shown in Fig. 6.1. The maximum bar size is not specified. Detailing of this reinforcement is usually determined by the requirements of clause 5.5.3(2) and of EN 1994-1-2, for fire resistance.

7.4.3. Control of cracking due to direct loading

Clause 7.4.3(2) specifies elastic global analysis to Section 5, allowing for the effects of cracking. The preceding comments on global analysis for deformations apply also to this analysis for bending moments in regions with concrete in tension. Paragraph (4) on loading should come next, but it is placed last in clause 7.4.3 because of the drafting rule that ‘general’ paragraphs precede those ‘for buildings’. It specifies the quasi-permanent combination. Except for storage areas, the values of factors ψ2 for floor loads in buildings are typically 0.3 or 0.6. The bending moments will then be much less than for the ultimate limit state, especially for cross-sections in Class 1 or 2 in beams built unpropped. There is no need to reduce the extent of the cracked regions below that assumed for global analysis, so the new bending moments for the composite members can be found by scaling values found for ultimate loadings. At each cross-section, the area of reinforcement will be already known: that required for ultimate loading or the specified minimum, if greater; so the stresses σs, 0 (clause 7.4.3(3)) can be found.

Tension stiffening A correction for tension stiffening is now required. At one time, these effects were not well understood. It was thought that, for a given tensile strain at the level of the reinforcement, the total extension must be the extension of the concrete plus the width of the cracks, so that allowing for the former reduced the latter. The true behaviour is more complex. The upper part of Fig. 7.2 shows a single crack in a concrete member with a central reinforcing bar. At the crack, the external tensile force N causes strain εs2 = N/AsEa in the bar, and the strain in the concrete is the free shrinkage strain εcs, which is negative, as shown. There is a transmission length Le each side of the crack, within which there is transfer of shear between the bar and the concrete. Outside this length, the strain in both the steel and the concrete is εs1, and the stress in the concrete is fractionally below its tensile strength.

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Within the length 2Le, the curves εs(x) and εc(x) give the strains in the two materials, with mean strains εsm in the bar and εcm in the concrete. It is now assumed that the graph represents the typical behaviour of a reinforcing bar in a cracked concrete flange of a composite beam, in a region of constant bending moment such that the crack spacing is 2Le. The curvature of the steel beam is determined by the mean stiffness of the slab, not the fully cracked stiffness, and is compatible with the mean longitudinal strain in the reinforcement, εsm. Midway between the cracks, the strain is the cracking strain of the concrete, corresponding to a stress less than 30 N/mm2 in the bar. Its peak strain, at the crack, is much greater than εsm, but less than the yield strain, if crack widths are not to exceed 0.4 mm. The crack width corresponds to this higher strain, not to the strain εsm that is compatible with the curvature, so a correction to the strain is needed. It is presented in clause 7.4.3(3) as a correction to the stress σs, 0 because that is easily calculated, and Tables 7.1 and 7.2 are based on stress. The strain correction cannot be shown in Fig. 7.1 because the stress σs, 0 is calculated using the ‘fully cracked’ stiffness, and so relates to a curvature greater than the true curvature. The derivation of the correction101 takes account of crack spacings less than 2Le, the bond properties of reinforcement, and other factors omitted from this simplified outline. The section properties needed for the calculation of the correction ∆σs will usually be known. For the composite section, A is needed to find I, which is used in calculating σs, 0, and Aa and Ia are standard properties of the steel section. The result is independent of the modular ratio. For simplicity, αst may conservatively be taken as 1.0, because AI > Aa Ia. When the stress σs at a crack has been found, the maximum bar diameter or the maximum spacing are found from Tables 7.1 and 7.2. Only one of these is needed, as the known area of reinforcement then gives the other. The correction of clause 7.4.2(2) does not apply.

Influence of profiled sheeting on the control of cracking The only references to profiled steel sheeting in clause 7.4 are in clause 7.4.1(4), ‘… no account should be taken of any profiled steel sheeting’, and in the definition of hc in clause 7.4.2(1), ‘… thickness … excluding any haunch or ribs’. The effects of the use of profiled sheeting for a slab that forms the top flange of a continuous composite beam are as follows: • • •

there is no need for control of crack widths at the lower surface of the slab where the sheeting spans in the transverse direction, there is at present no evidence that it contributes to the control of transverse cracks at the top surface of the slab where the sheeting spans parallel to the beam, it probably contributes to crack control, but no research on this subject is known to the authors.

For design, the definition of ‘effective tension area’ in clause 7.3.4(2) of EN 1992-1-1 should be noted. A layer of reinforcement at depth c + φ/2 below the top surface of the slab, where c is the cover, may be assumed to influence cracking over a depth 2.5(c + φ/2) of the slab. If the depth of the concrete above the top of the sheeting, hc, is greater than this, it would be reasonable to use the lower value, when calculating As from equation (7.1). This recognizes the ability of the sheeting to control cracking in the lower half of the slab, and has the effect of reducing the minimum amount of reinforcement required, for the thicker composite slabs

General comments on clause 7.4

In regions where tension in concrete may arise from shrinkage or temperature effects, but not from other actions, the minimum reinforcement required may exceed that provided in previous practice. Where unpropped construction is used for a continuous beam, the design loading for checking cracking is usually much less than that for the ultimate limit state, so that the quantity of reinforcement provided for resistance to load should be sufficient to control

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cracking. The main use of clause 7.4.3 is then to check that the spacing of the bars is not too great. Where propped construction is used, the disparity between the design loadings for the two limit states is smaller. If cracks are to be controlled to 0.3 mm, a check to clause 7.4.3 is more likely to influence the reinforcement required. For beams in frames, the preceding comments apply where semi-rigid or rigid connections are used. Where floors have brittle finishes or an adverse environment, simple beamto-column joints should not be used, because effective control of crack width may not be possible.

Example 7.1: two-span beam (continued) – SLS

Details of this beam are shown in Fig. 6.23. All of the design data and calculations for the ultimate limit state are given in Examples 6.7 to 6.12. For data and results required here, reference should be made to: • • • •

Table 6.2, for characteristic loads per unit length Table 6.3, for elastic properties of the cross-sections at the internal support (B in Fig. 6.23(c)) and at mid-span Table 6.4, for bending moments at support B for uniform loading on both spans Fig. 6.28, for bending-moment diagrams for design ultimate loadings, excluding the effects of shrinkage.

The secondary effects of shrinkage are significant in this beam, and cause a hogging bending moment at support B of 120 kN m (Example 6.7). Clause 7.3.1(8) does not permit shrinkage to be ignored for serviceability checks on this beam, because it does not refer to lightweight-aggregate concrete, which is used here.

Stresses From clause 7.2.2(1), there are no limitations on stress; but stresses in the steel beam need to be calculated, because if yielding occurs under service loads, account should be taken of the resulting increase in deflections, from clause 7.3.1(7). Yielding is irreversible, so, from a note to clause 6.5.3(2) of EN 1990, it should be checked for the characteristic load combination. However, the loading for checking deflections depends on the serviceability requirement.96 The maximum stress in the steel beam occurs in the bottom flange at support B. Results for the characteristic combination with variable load on both spans and 15% of each span cracked are given in Table 7.2. The permanent load, other than floor finishes, is assumed to act on the steel beam alone. Following clause 5.4.2.2(11), the modular ratio is taken as 20.2 for all of the loading except shrinkage.

Table 7.2. Hogging bending moments at support B and stresses in the steel bottom flange, for the characteristic load combination Loading (1) Permanent (on steel beam) (2) Permanent (on composite beam) (3) Variable (4) Shrinkage

Modular ratio

10–6Iy, B (mm4)

MEk, B (kN m)

10–6Wa, bot (mm3)

5.78

–

337

104

1.50

69

1.2

20.2

467

18

1.75

10

17.5 –

20.2 28.7

467 467

263 120

1.75 1.75

150 69

w (kN/m)

136



σa, bot (N/mm2)



Table 7.3. Deflections at 4.8 m from support A, for the frequent combination Load

Modular ratio

Deflection (mm)

Dead, on steel beam Dead, on composite beam Imposed, on composite beam Primary shrinkage Secondary shrinkage

– 20.2 20.2 27.9 27.9

9 1 14 33 –26

10.2

A

45.3

1.8

R

10 kN B

10 kN 26 mm

45.3 mm

33

E

A

E

B

B' 5.4

20 kN

(a)

(b)

Fig. 7.3. Sagging deflection at point E caused by shrinkage

The total bottom-flange compressive stress is σEk, bot, a = 298 N/mm2 (= 0.84fy) so no allowance is needed for yielding.

Deflections The maximum deflection of span AB of the beam will occur at about 4.8 m from A (40% of the span), when variable load acts on span AB only. The additional deflection caused by slip of the shear connection is ignored, as clause 7.3.1(4)(a) is satisfied. Calculated deflections at this point, with 15% of each span assumed to be cracked, are given in Table 7.3. The frequent combination is used, for which ψ1 = 0.7, so the variable loading is 0.7 × 17.5 = 12.3 kN/m The following method was used for the shrinkage deflections. From Example 6.7 and Fig. 6.27, the primary effect is uniform sagging curvature at radius R = 1149 m, with deflection δ = 45.3 mm at support B. The secondary reaction at B is 20 kN. From the geometry of the circle, the primary deflection at point E in Fig. 7.3(a) is δ1, E = 45.3 – 5.42 × 1000 × (2 × 1149) = 33 mm The upwards displacement at E caused by the 20 kN reaction at B that moves point B¢ back to B was found to be 26 mm by elastic analysis of the model shown in Fig. 7.3(b), with 15% of each span cracked. The total shrinkage deflection is only 7 mm, despite the high free shrinkage strain, but would not be negligible in a simply-supported span. The total deflection, 31 mm, is span/390. This ratio appears not to be excessive. However, the functioning of the floor may depend on its maximum deflection relative to the supporting columns. It is found in Example 9.1 that the deflection of the composite slab, if cast unpropped, is 15 mm for the frequent combination. This is relative to the supporting beams, so the maximum floor deflection is

137





31 + 15 = 46 mm or span/260. For the characteristic combination, this increases to 37 + 16 = 53 mm or span/230. The limiting deflections given in the UK’s draft National Annex to EN 199096 depend on the serviceability requirement. For floors with partitions, they range from span/300 to span/500. For this floor, some combination of using propped construction for the composite slabs and/or the beams, and cambering the beams, will be necessary. Reducing the modular ratio for imposed loading to 10.1 makes little difference: the value 14 mm in Table 7.3 becomes 12 mm. The extensive calculations for shrinkage lead to a net deflection of only 7 mm, because the secondary effect cancels out most of the primary effect. This benefit would not occur, of course, in a simply-supported span.

Control of crack width Clause 7.4 applies to reinforced concrete that forms part of a composite member. In the beam considered here, the relevant cracks are those near support B caused by hogging bending of the beam, and cracks along the beam caused by hogging bending of the composite slab that the beam supports. The latter are treated in Example 9.1 on a composite slab. Clause 7.4.1(1) refers to exposure classes. From clause 4.2(2) of EN 1992-1-1, Class XC3 is appropriate for concrete ‘inside buildings with moderate humidity’. For this class, a note to clause 7.3.1(5) of EN 1992-1-1 gives the design crack width as 0.3 mm. The method of clause 7.4.1(3) is followed, as clause 7.4.1(4) does not apply. Minimum reinforcement The relevant cross-section of the concrete flange is as shown in Fig. 6.23(a), except that effective widths up to 2.5 m should be considered. From clause 5.4.1.2, the effective width is assumed to increase from 1.6 m at support B to 2.5 m at sections more than 3 m from B. It may be difficult to show that sections 3 m from B are never ‘subjected to significant tension’ (clause 7.4.2(1)). Calculations are therefore done for both effective widths, assuming uncracked unreinforced concrete. From the definition of z0 in that clause, n0 = 10.1. It is found for both of these flange widths that z0 is such that kc > 1, so, from equation (7.2), it is taken as 1.0. A value is required for the strength of the concrete when cracks first occur. As unpropped construction is used, there is at first little load on the composite member, so from clause 7.4.2(1), conservatively, fct, eff = 3.0 N/mm2. Assuming that 10 mm bars are used for the minimum reinforcement, Table 7.1 gives σs = 320 N/mm2. Then, from equation (7.1), 100As/Act = 100 × 0.9 × 1 × 0.8 × 3.0/320 = 0.675% However, clause 5.5.1(5) also sets a limit, as a condition for the use of plastic resistance moments. For this concrete, flctm = 2.32 N/mm2, and fsk = 500 N/mm2. Hence, from equation (5.8) with kc = 1.0, 100ρs = 100 × (355/235)(2.32/500) = 0.70%

(D7.4)

This limit governs; so for a slab 80 mm thick above the sheeting the minimum reinforcement is As, min = 7 × 80 = 560 mm2/m One layer of 10 mm bars at 125 mm spacing provides 628 mm2/m.

138





Cracking due to direct loading Only the most critical cross-section, at support B, will be considered. Clause 7.4.3(4) permits the use of the quasi-permanent combination, for which the variable loading is ψ2qk, with ψ2 = 0.6, from clause A1.2.2(1) of EN 1990. From Table 7.2, the bending moment at B that stresses the reinforcement is ME, qp, B = 18 + 263 × 0.6 + 127 = 303 kN m The neutral axis for the cracked section is 313 mm below the top of the slab (Table 6.12), so the section modulus for reinforcement at depth 30 mm is 10–6Ws = 467/(313 – 30) = 1.65 mm3 Hence, from clause 7.4.3(3), σs, 0 = 303/1.65 = 184 N/mm2 The correction for tension stiffening, equation (7.5), is now calculated, assuming that the reinforcement used in Example 6.7, 12 mm bars at 125 mm spacing, will be satisfactory. This gives ρs = 0.0113. Using values obtained earlier, αst = AI/Aa Ia = 11 350 × 467/(9880 × 337) = 1.59 From equation (7.5), ∆σs = 0.4 × 2.32/(1.59 × 0.0113) = 52 N/mm2

(D7.5)

From equation (7.4), σs = 184 + 52 = 236 N/mm2 From Table 7.1, φs £ 16 mm. From Table 7.2, the bar spacing £ 200 mm. The use of 12 mm bars at 125 mm spacing at support B satisfies both conditions. Finding the cross-sections of the beam at which this reinforcement can be reduced to the minimum found above may require consideration of the bending-moment envelopes both for ultimate loads and for the quasi-permanent combination.

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CHAPTER 8

Composite joints in frames for buildings This chapter corresponds to Section 8 of EN 1994-1-1, which has the following clauses: • • • •

Scope Analysis, modelling and classification Design methods Resistance of components


8.1. Scope Section 8 is based on relatively recent research on beam-to-column and beam-to-beam joints of the types used in steel and composite frames for buildings, so its scope has been limited to ‘frames for buildings’. The definition of composite joints to which clause 8.1(1) refers includes joints with reinforced concrete members. These could occur, for example, in a tower block with a concrete core and composite floors. However, no application rules are given for such joints. As stated in clause 8.1(2), both Section 8 and Annex A are essentially extensions to the Eurocode for joints between steel members, EN 1993-1-8.24 It is assumed that a user will be familiar with this code, especially its Sections 5 and 6. The only steel members considered in detail are I- and H-sections, which may have concrete-encased webs. Plate girders are not excluded. The application rules of EN 1994-1-1 are limited to composite joints in which reinforcement is in tension and the lower part of the steel section is in compression (Fig. 8.1 and clause 8.4.1(1)). There are no application rules for joints where the axes of the members connected do not intersect, or do so at angles other than 90°; but the basic approach is more general than the procedures prescribed in detail, and is capable of application in a wider range of situations. Many types of joint are in use in steelwork, so that EN 1993-1-8 is around 130 pages long. The majority of the calculations needed for composite joints are specified there, and explained in the relevant guide in this series.103 The worked examples and much of the comment in the present guide are limited to a single type of joint – the double-sided configuration shown in Fig. 8.1 – but with an end plate, not a contact plate, and an uncased column, as shown on the left of Fig. 8.1 and in Fig. 8.8. Commentary on the design of this joint will be found, as appropriate, in this chapter and its examples, and in Chapter 10 on Annex A. Before the Eurocodes come into regular use, it is expected that tables of resistances and stiffnesses of a wide range of steel and composite joints will become available, based on



Clause 8.1(1)

Clause 8.1(2)



EN 1993 and EN 1994. The extensive calculations given here will rarely be needed. Many of them serve to show that a particular property of a joint does not govern its resistance. Experience will enable such checks to be omitted. Detailed guidance on the Eurocode methods for composite joints appeared in 1998, in the context of design to British codes.104 An explanation of the provisions and approximations in the Eurocodes, with worked examples, was then prepared, mainly by those who drafted the codes. Its first edition39 refers to the draft codes as they were in 1998, so some differences, mainly in symbols, will be found between it and the published EN Eurocodes. With over 200 pages, it provides much broader coverage than is possible here.

8.2. Analysis, including modelling and classification Clause 8.2.1(1)

Clause 8.2.2(1)

Clause 8.2.3(2) Clause 8.2.3(3)

Clause 8.2.1(1) refers to Section 5 of EN 1993-1-8, which covers the same subjects as clause 8.2. Table 5.1 in clause 5.1 of EN 1993-1-8 defines the links between the three types of global analysis, elastic, rigid plastic, and elastic–plastic, and the types of models used for joints. This enables the designer to determine whether the stiffness of the joint, its resistance, or both properties, are relevant to the analysis. Joints are classified in Section 5 by stiffness, as rigid, nominally pinned, or semi-rigid; and by strength as full-strength, nominally pinned, or partial-strength. This classification relates the property of the joint (stiffness or resistance) to that of the connected member, normally taken as the beam. This applies also to composite joints. The only modification, in clause 8.2.2(1), concerns the rotational stiffness of a joint, Sj. This is bending moment per unit rotation, shown in Fig. 8.2(b). The symbol φ is used for rotation, as well as for bar diameter. The initial elastic stiffness, Sj, ini, is reduced at high bending moments to allow for inelastic behaviour. For global analysis, it is divided by η, values of which, between 3.0 and 3.5, are tabulated in clause 5.1.2 of EN 1993-1-8 for various types of steel joint. These apply where the joint is composite. Clause 8.2.2(1) provides a further value, for contact-plate joints, as shown on the right of Fig. 8.1 and in Fig. 8.4. The classification of a composite joint may depend on the direction of the bending moment (e.g. sagging or hogging). This is unlikely in a steelwork joint, and so is referred to in clause 8.2.3(2). The reference in clause 8.2.3(3) to neglect of cracking and creep applies only to the classification of the joint according to stiffness. Its initial stiffness is to be compared with that of the connected beam, using Fig. 5.4 of EN 1993-1-8. The stiffer the beam the less likely it is that the joint can be classified as rigid. A more precise calculation of beam stiffness is permitted by the use of ‘may’ in clause 8.2.3(3). For example, a representative value of modular ratio, to clause 5.4.2.2(11), may be used. Account could also be taken of cracked and uncracked lengths within the beam

Nominally pinned joint Semi-rigid joint

A

B

C

Lb

Fig. 8.1. Model for a two-span beam in a frame

142




CHAPTER 8. COMPOSITE JOINTS IN FRAMES FOR BUILDINGS

Mj

Mj (1) Sj, ini ≥ 8EIb/Lb (1) Rigid

Mj, Rd Mj, Ed

(2) Semi-rigid (3) Sj, ini £ 0.5EIb/Lb

2Mj, Rd/3

(2) tan–1 Sj, ini

tan–1 Sj (Sj = Sj, ini/m) (3) Nominal pin 0

0 f (a)

fCd

f

(b)

Fig. 8.2. Moment–rotation relationships for joints

in accordance with clause 5.4.2.3, but the additional calculation would not normally be worthwhile.

Outline of modelling of joints for global analysis In global analysis, nominally pinned joints are represented by pins, and semi-rigid joints by rotational springs, as shown in Fig. 8.1 for a two-span beam of uniform depth, supported by three columns in a braced frame. Joints to external columns are usually designed as nominally pinned, to reduce bending moments in the columns. The use of partial-strength semi-rigid joints at point B, rather than nominally pinned joints, has advantages in design: • • •

possible reduction in the section sizes for beams reduction in the deflection of beams reduction in crack widths near support B.

In comparison with full-strength rigid joints, the advantages are: • • •

beams less susceptible to lateral–torsional buckling simpler construction and significant reduction in cost lower bending moments in columns.

The stiffness of a rotational spring, Sj, is the slope of the moment–rotation relationship for the joint (Fig. 8.2(a)). The stiffness class is determined by the ratio of the initial slope, Sj, ini, to the stiffness EaIb/Lb of the beam adjacent to the joint, as shown. The initial stiffness of a joint is assembled from the stiffnesses of its components, represented by elastic springs. Those for an end-plate joint with a single row of bolts in tension, between beams of equal depth are shown in Fig. 8.3, in which all elements except springs and pins are rigid. The notation for the spring stiffnesses ki is as in EN 1993-1-8 and in Examples 8.1 and 10.1, as follows: k1 k2 k3 k4 k5 k10

shear in column web compression of column web extension of column web bending of column flange, caused by tension from a single row of bolts bending of end plate, caused by tension from a single row of bolts extension of bolts, for a single row of bolts.

Stiffnesses in EN 1994-1-1, but not in EN 1993-1-8, are: extension of reinforcement (denoted k13 by ECCS TC1139) ks, r Ksc/Es slip of shear connection. Each spring has a finite strength, governed by yield or buckling of the steel. The design method ensures that non-ductile modes, such as fracture of bolts, do not govern.

143





For the tension region, the weakest of the springs numbered 3, 4, 5 and 10, and of the tension reinforcement, is found. This resistance is compared with the compressive resistance of spring 2. The product of the lower of these resistances and the effective lever arm gives the plastic bending resistance of the joint. The resistance can be increased by strengthening the weakest link; for example, by the addition of column-web stiffeners. Where the beams are of unequal depth, or MEd, 1 π MEd, 2 (Fig. 8.3), rotation at the joint is increased by shearing deformation of the column web. For beams of equal depth, this is the area ABCD in Fig. 8.3. Its deformation is resisted by the spring of stiffness k1. Depending on the out-of-balance moment |MEd, 1 – MEd, 2|, the column web panel may govern the resistance of the joint.

8.3. Design methods Clause 8.3.1(1)

Clause 8.3.1(1) refers to Section 6 of EN 1993-1-8, which is 40 pages long. It defines the ‘basic components’ of a steelwork joint, their strengths and their elastic stiffnesses. It is shown how these are assembled to obtain the resistances, rotational stiffness and rotation capacity of complete joints. A composite joint has these additional components: • • •


longitudinal slab reinforcement in tension concrete encasement, where present, of the column web steel contact plates, if used (not covered in EN 1993-1-8).

In addition, account is taken of the slip of shear connection, by modifying the stiffness of the reinforcement (Fig. 8.3). All the properties of components given in, or cross-referenced from, EN 1994-1-1 satisfy the condition of clause 8.3.1(2). The application of clause 8.3.1(3) to reinforcing bars is illustrated in Examples 8.1 and 10.1. None of the additional components listed above influences resistance to vertical shear, so this aspect of design is fully covered by EN 1993-1-8 (clause 8.3.2(1)).

ks, r Ksc/Es

MEd, 2

MEd, 1 A

B

k3 k4, 5, 10

k1 D

k2 C

Fig. 8.3. Model for an internal beam-to-column joint

144





Composite joints in framed structures for buildings are almost always in regions of hogging bending, for which full shear connection is normally required, to clauses 6.2.1.3(2) and 8.4.2.1(2). The reference to shear connection in clause 8.3.2(2) reminds the user that no provisions are given for composite joints in regions with partial shear connection. The use of Annex A (informative) for finding rotational stiffnesses satisfies clause 8.3.3(1), and is illustrated in Example 10.1. The coefficient ψ, referred to in clause 8.3.3(2), is used in clause 6.3.1(6) of EN 1993-1-8 to define the shape of the moment-rotation curve for a joint at bending moments Mj, Ed that exceed 2Mj, Rd /3, as follows. Let Sj, ini be the stiffness at low bending moments. For 2Mj, Rd /3 < Mj, Ed £ Mj, Rd, the stiffness is Sj = Sj, ini /µ


(D8.1)

(Fig. 8.2(b)), where µ = (1.5Mj, Ed /Mj, Rd)ψ

(D8.2)

This clause gives the value for ψ for a type of joint not included in Table 6.8 of EN 1993-1-8. The table is applicable to other types of composite joint. The rotation capacity of composite joints, φCd in Fig. 8.2(b), has been extensively researched.105 There are many relevant parameters. Analytical prediction is still difficult, and there are as yet no design rules sufficiently well established to be included in EN 1994-1-1. So-called ‘simple’ joints have been widely used in composite structures. Some of them will be found to qualify as ‘partial-strength’ when Eurocode methods are used. The experience referred to in clause 8.3.4(2) is then available. It is rarely necessary in design to calculate either the available rotation capacity or the rotation required of a composite joint. Further guidance is given by ECCS TC11.39

Clause 8.3.4(2)

8.4. Resistance of components This clause supplements clause 6.2 of EN 1993-1-8. The effective width of concrete flange in tension is the same at a joint as for the adjacent beam (clause 8.4.2.1(1)). Longitudinal bars above the beam should pass either side of the column. Clause 8.4.2.1(4) applies at an external column with a partial- or full-strength joint. The tensile force in the bars must be transferred to the column; for example, by being looped round it. This applies also at internal columns where there is a change in the tension in the bars (Fig. 8.2, clause 8.4.2.1(3) and Example 8.2). Clauses 8.4.2.2(1) and 8.4.3(1) permit the same 45° spread of force in a contact plate as used in EN 1993-1-8 for an end plate. The force is assumed in EN 1993-1-8 to spread at tan–1 2.5 (68°) through the flange and root radius of the column. Where the compressive force relied on in design exceeds the resistance of the steel bottom flange, the length of the contact plate should allow for this (Fig. 8.4). In EN 1994-1-1, the word ‘connection’ appears only in clause 8.4.3(1), clause A.2.3.2 and Table A.1. It means the set of components that connect a member to another member; for example, an end plate, its bolts and a column flange. Thus, a ‘connection’ is part of a ‘joint’. The model used in clause 8.4.4.1(2) is illustrated in Fig. 8.5. This figure shows an elevation of the concrete encasement of width h – 2tf (column depth less flange thicknesses) and depth z, the lever arm between the resultant horizontal forces from the beam. A shear force V is transferred through the encasement, which is of thickness bc – tw (column width less web thickness). The concrete strut ABDEFG has width 0.8(h – 2tf)cos θ, where tan θ = (h – 2tf)/z, so its area is Ac = 0.8(h – 2tf)cos θ (bc – tw)

Clause 8.4.2.1(1) Clause 8.4.2.1(4)

Clause 8.4.2.2(1) Clause 8.4.3(1)

Clause 8.4.4.1(2)

(8.2)

145





Beam

Column

45˚

68˚

Contact plate

Fig. 8.4. Detail of a contact plate between a beam bottom flange and a column

h – 2 tf 0.4(h – 2tf)

V

A q

B q D

z G

0.8(h – 2tf)cos q

C

F

V

E

NEd

Fig. 8.5. Strut model for the shear resistance of the concrete encasement to a column web r tp Column Beam

kwc, c 2.0

teff, c

tf, b 1.3

45˚

68˚

End plate

tf, c

(a)

0

0.212

scom, c, Ed /fcd

(b)

Fig. 8.6. Model for resistance to compression of the concrete encasement to a column web

146





Its compressive strength is 0.85νfcd, giving the force C in Fig. 8.5. For horizontal equilibrium at B and F, C sin θ = V. These are equations (8.1) to (8.3) in clause 8.4.4.1. The shear strength of concrete is increased by compression. This is allowed for by the factor ν in clause 8.4.4.1(3), which ranges from 0.55 for zero axial compression to 1.1 for Clause 8.4.4.1(3) NEd ≥ 0.55Npl, Rd. The contribution of concrete encasement to the resistance of a column web to horizontal Clause 8.4.4.2 compression is given in clause 8.4.4.2. For an end-plate joint to a column flange, the depth of encasement assumed to resist compression, teff, c, is shown in Fig. 8.6(a), with the 2.5:1 dispersion, referred to above, extending through the root radius r. The horizontal compressive strength of the concrete is 0.85 kwc, c fcd, where kwc, c depends on the vertical compressive stress in the column, σcom, c, Ed, as shown in Fig. 8.6(b).

Example 8.1: end-plate joints in a two-span beam in a braced frame

In development work that followed the publication of ENV 1994-1-1, a set of application rules for composite joints was prepared, more detailed than those now given in Section 8 and Annex A of EN 1994-1-1. These are published by ECCS39 as a model annex J. They provide useful guidance in this example, and are referred to, for example, as ‘clause J.1.1 of ECCS TC11’.

Data The subject of Examples 6.7 and 7.1 is a two-span beam ABC continuous over its central support (see Figs 6.23–6.28). It is now assumed that this beam is one of several similar beams in a multistorey braced frame (Fig. 8.7). Its joints with the external columns are nominal pins. The spans of the composite-slab floors are 2.5 m, as before. For simplicity, in the work on beams AB and BC, column EBF will be treated as fixed at nodes E and F. These beams are attached to the column at B by the end-plate connections shown in Fig. 8.8, which also gives dimensions of the column section. Its other properties are as follows: HEB 240 cross-section, Aa = 10 600 mm2, fy = 355 N/mm2, 10–6Iy = 112.6 mm4, 10–6Iz = 39.23 mm4. The end plates are of mild steel, fy = 275 N/mm2, and relatively thin, 12 mm, to provide the plastic behaviour required. They are attached to the beam by 10 mm fillet welds to the flanges, and 8 mm welds to the web. They are each attached to the column by four Grade 8.8 M20 bolts with properties: fub = 800 N/mm2, fyb = 640 N/mm2, net area at root of thread As, b = 245 mm2 per bolt. The only other change from the data used in Examples 6.7 and 7.1 on geometry, materials and loadings concerns the reinforcement in the slab. Longitudinal reinforcement at support B These partial-strength joints need rotation capacity. Its value cannot be found at this stage, but it is known to increase with both the diameter of the reinforcing bars in the slab

E 3 A

B

D

C

3

F 12

12

Fig. 8.7. Model for a two-span beam ABC, with an internal column EBF

147





240

200

25

60

A 160

m2 = 35.4 + 2 450

383 9.4

12

14.6

10

60

=

= 25

A

90

(b) Section A–A

(a) Elevation

30 16 dia.

100 21 240 10

130

230

270 800

17 100 mm

(c) Column section

(d) Slab reinforcement

Fig. 8.8. Details of the beam-to-column end-plate connections

and the area of reinforcement provided. However, the amount of top reinforcement should be limited, so that the whole of the compressive force across the joint can be resisted by the beam bottom flange and the unstiffened column web. Detailed guidance is given in Couchman and Way.104 For a steel beam of depth 450 mm in S355 steel, the recommended minimum areas are 3000 mm2 for bars with 5% elongation and 860 mm2 for bars with 10% elongation. The recommended maximum amount depends on the size of the column and the details of the bolts in tension, and is about 1200 mm2 for this example. The recommended bar diameters are 16 and 20 mm. For these reasons, the previous reinforcement (13 No. 12 mm bars, As = 1470 mm2) is replaced by six No. 16 mm hot-formed bars (minimum elongation 10%): As = 1206 mm2, fsk = 500 N/mm2.

Classification of the joints It is assumed initially that flexural failure of a joint will occur in a ductile manner, by yielding of the reinforcement in tension and the end-plate or column flange in bending;

148





and that at bottom-flange level the compressive resistance of the column web will be sufficient. As the spans are equal, it is unlikely that shear of the column web will be critical. The joint is expected to be ‘partial-strength’. This can be checked by comparing the tension resistance of the top two bolts, from Table 3.4 of EN 1993-1-8, with the force to yield the beam top flange: FT, Rd, bolts = 2(k2 fub As/γM2) = 2 × 0.9 × 0.8 × 245/1.25 = 282 kN

(D8.3)

FRd, flange = bf tf fyd = 190 × 14.6 × 0.355 = 985 kN Thus, the resistance moment Mj, Rd for the joint will be much less than Mpl, Rd for the beam, and lateral buckling will be less critical than before. There is no need to find the stiffness of the joint at this stage, because it is clearly either ‘rigid’ or ‘semi-rigid’. Either type may be treated as ‘semi-rigid’.

Approximate global analysis Tables in Appendix B of Couchman and Way104 enable a rough check to be made on this initial design, without much calculation. They give resistances Mj, Rd in terms of the cross-section of the steel beam, its yield strength, the thickness and grade of the end plate, the number and size of bolts in tension, and the area of reinforcement. Even though the beam used here is an IPE section, it can be deduced that Mj, Rd is about 400 kN m. For both spans fully loaded, it was found in Example 6.7 that MEd at B was 536 kN m from loading (see Fig. 6.28) plus 120 kN m from shrinkage. The flexural stiffness of the joint is not yet known, but it will be between zero and ‘fully rigid’. If fully rigid, the joint will obviously be ‘plastic’ under ultimate loading, and there will then be no secondary shrinkage moment. At mid-span, for the total load of 35.7 kN/m (see Table 6.2), the sagging bending moment is then 357 ×122/8 – 400/2 = 443 kN m If the joint acts as a pin, the mid-span moment is 443 + 200 = 643 kN m It is recommended in Couchman and Way104 that mid-span resistances should be taken as 0.85Mpl, Rd, to limit the rotation required at the joints. From Example 6.7, Mpl, Rd with full shear connection is 1043 kN m, so the bending resistance of the beam is obviously sufficient.

Vertical shear For MEd = 400 kN m at B, the vertical shear at B is Fv, Ed, B = 35.7 × 6 + 400/12 = 247 kN The shear resistance of the four M20 bolts is now found, using Table 3.4 of EN 1993-1-8. Two of the bolts may be at yield in tension. The shear applied to these bolts must satisfy Fv, Ed /Fv, Rd + Ft, Ed /(1.4Ft, Rd) £ 1.0

(D8.4) 2

The net shear area of each bolt is As, b = 245 mm , so from Table 3.4 of EN 1993-1-8, Fv, Rd = 0.6fub As, b/γM2 = 0.6 × 800 × 0.245/1.25 = 94.1 kN From equation (D8.3) with Ft, Ed = Ft, Rd, Fv, Ed £ (1 – 1/1.4)Fv, Rd = 27 kN For four bolts, Fv, Rd = 2 × (94.1 + 27) = 242 kN

(D8.5)

149





This shows that it may be necessary to add a second pair of bolts in the compression region of the joint.

Bending resistance of the joint, excluding reinforcement Unpropped construction was used in Example 6.7. From Table 6.2, the design ultimate load for the steel beam is 7.8 kN/m. For the construction phase, this is increased to 9.15 kN/m, to allow for the higher density of fresh concrete and the construction imposed loading. For rigid joints at the internal support between two 12 m spans, MEd, B = wL2/8 = 9.15 × 122/8 = 165 kN m The plastic resistance of the joint during construction is required. An upper limit is easily obtained. The lever arm from the top bolts to the centre of the bottom flange is zbolts = 450 – 60 – 7.3 = 383 mm

(D8.6)

The resistance cannot exceed FT, Rd, bolts zbolts = 282 × 0.383 = 108 kN m

(D8.7)

so the previous hogging bending moment of 165 kN m cannot be reached. It is assumed that the stiffness of the joint is sufficient for its plastic resistance, found later to be 83 kN m, to be reached under the factored construction loading. Resistance of T-stubs and bolts in tension The calculation of the bending resistance consists of finding the ‘weakest links’ in both tension and compression. In tension, the column flange and the end plate are each modelled as T-stubs, and prying action may occur. Some of the dimensions required are shown in Fig. 8.9. From Fig. 6.8 of EN 1993-1-8, the dimensions m overlap with 20% of the corner fillet or weld. Thus, in Fig. 8.9(a), for the end plate: m = 45 – 4.7 – 0.8 × 8 = 33.9 mm

(D.8.8)

It is evident from the geometry shown in Fig. 8.9 that the end plate is weaker than the column flange, so its resistance is now found.

m = 23.2

0.8rc = 16.8

e = 75 tf = 17

tp = 12

rc = 21

55

m = 33.9

m2 = 37.4

8

(a) Plan details of T-stubs

55

(b) Yield line pattern in end plate

Fig. 8.9. Dimensions of T-stubs, and the yield line pattern

150



33.9



Q/2 Ft, Rd

q

n

m

FT, 2, Rd

Fig. 8.10. Plan of a T-stub, showing failure mode 2

Clause 6.2.4.1 of EN 1993-1-8 gives three possible failure modes: (1) yielding of the plate (2) a combination of (1) and (3) (3) failure of the bolts in tension. Yield line theory is used for bending of the plate. The critical mechanism in this case will be either that shown in Fig. 8.9(b) or a circular fan, for which the perimeter is leff, cp = 2πm from Table 6.6 in clause 6.2.6.5 of EN 1993-1-8. For the non-circular pattern, dimension m2 in Fig. 8.8(a) is also relevant, and leff, nc = αm

£ 2πm

where α is given by Fig. 6.11 in EN 1993-1-8 or in Fig. 4.9 of Couchman and Way.104 In this case, α = 6.8, so the circular pattern governs for mode (1), and leff, 1 = 2πm = 6.28 × 33.9 = 213 mm The plastic resistance per unit length of plate is mpl, Rd = 0.25tf2 fy /γM0 = 0.25 × 122 × 0.275/1.0 = 9.90 kN m/m

(D8.9)

From equation (D8.3), the tensile resistance of a pair of bolts is 282 kN. Mode 1. For mode 1, yielding is confined to the plate. From Table 6.2 of EN 1993-1-8, the equation for this mode is FT, 1, Rd = 4Mpl, 1, Rd /m with Mpl, 1, Rd = 0.25leff, 1tf2 fy /γM0 = leff, 1mpl, Rd = 0.213 × 9.9 = 2.11 kN m From equation (D8.8) for m, FT, 1, Rd = 4 × 2.11/0.0339 = 249 kN Mode 2. This mode is more complex. The equation for the tension resistance FT, 2, Rd is now explained. The effective length of the perimeter of the mechanism is leff, 2, and the work done for a rotation θ at its perimeter is 2mpl, Rdleff, 2θ, from yield line theory. With each bolt failing in tension, the work equation is

151





FT, 2, Rd(m + n)θ = n

SF

t, Rd

θ + 2mpl, Rd leff, 2θ

(D8.10)

where n is shown in Fig. 8.10. In Table 6.2 of EN 1993-1-8 there is the further condition that n £ 1.25m = 1.25 × 33.9 = 42.4 mm The effective length leff, 2 = αm = 6.8 × 33.9 = 231 mm from Table 6.6 of EN 1993-1-8. For two bolts, Substituting in equation (D8.10):

SF

t, Rd

= 282 kN, from equation (D8.3).

S

FT, 2, Rd = (2mpl, Rdleff, 2 + n Ft, Rd)/(m + n) = (2 × 9.9 × 231 + 42.4 × 282)/76.3 = 217 kN

(D8.11)

Mode 3. Failure of the bolts – mode 3 – has FT, 3, Rd = 282 kN from equation (D8.3), so mode 2 governs. From Fig. 8.10, the prying force is Q = 282 – 217 = 65 kN Beam web in tension The equivalent T-stub in Fig. 8.10 applies a tensile force of 217 kN to the web of the beam. Its resistance is given in clause 6.2.6.8 of EN 1993-1-8 as Ft, wb, Rd = beff, t, wb twb fy, wb /γM0 (equation (6.22) of EN 1993-1-8), and beff, t, wb is taken as the effective length of the T-stub, leff, 2 = 231 mm. Hence, Ft, wb, Rd = 231 × 9.4 × 0.355/1.0 = 771 kN

(D8.12)

so this does not govern. Column web in tension The effective width of the column web in tension, to clause 6.2.6.3 of EN 1993-1-8, is the length of the T-stub representing the column flange. The resistance is Ft, wc, Rd = ωbeff, t, wc twc fy, wc /γM0 where ω is a reduction factor to allow for shear in the column web. In this case, the shear is zero, and ω = 1. The column web is thicker than the beam web, so from result (D8.12), its resistance does not govern. Column web in transverse compression The resistance is given in clause 6.2.6.2 of EN 1993-1-8. It depends on the plate slenderness λp and the width of the column web in compression, which is beff, c, wc = tf, b + 2÷2ap + 5(tfc + s) + sp (equation (6.11) of EN 1993-1-8), where ap is the throat thickness of the bottom-flange welds, so ÷2ap = 10 mm here; sp allows for 45° dispersion through the end plate, and is 24 mm here; and s is the root radius of the column section (s = rc = 21 mm). Hence, beff, c, wc = 14.6 + 20 + 5 × (17 + 21) + 24 = 248 mm For web buckling, the effective compressed length is dwc = hc – 2(tfc + rc) = 240 – 2 × (17 + 21) = 164 mm

152





The plate slenderness is λ p = 0.932(beff, c, wcdwc fy, wc /Eatwc2)0.5 = 0.932 × [248 × 164 × 0.355/(210 × 100)]0.5 = 0.773 The reduction factor for plate buckling is ρ = (λ p – 0.2)/λ p 2 = 0.573/0.7732 = 0.96 The factor ω for web shear is 1.0, as before. It is assumed that the maximum longitudinal compressive stress in the column is less than 0.7fy, wc, so from clause 6.2.6.2(2) of EN 1993-1-8, the reduction factor for this, kwc, is 1.0. From equation (6.9) in EN 1993-1-8, Fc, wc, Rd = ωkwcρbeff, c, wctwc fy, wc /γM1 = 0.96 × 248 ×10 × 0.355/1.0 = 845 kN

(D8.13)

Clearly, the tensile force of 217 kN governs the resistance of the steel connection. Bending resistance of the steel joint, for both beams fully loaded From equation (D8.5), the lever arm is 383 mm, so the resistance, excluding the reinforcement, is Mj, Rd, steel = 217 × 0.383 = 83 kN m

(D8.14)

governed by bending of the end plate. The critical mode 2 includes failure of the top row of bolts in tension. However, the joint is closely based on a type given in Couchman and Way,104 which is confirmed by ECCS TC1139 as having ‘ductile’ behaviour. From Example 6.7, the plastic bending resistance of the steel beam, an IPE 450 section, is Mpl, a, Rd = 1.702 × 355 = 604 kN m This exceeds four times Mj, Rd, so clause 5.2.3.2(3) of EN 1993-1-8 permits this joint to be classified as ‘nominally pinned’ for the construction stage.

Resistance of the composite joint For the composite joint, the reinforcement is at yield in tension. Its resistance is Ft, s, Rd = 1206 × 0.500/1.15 = 524 kN This increases the total compressive force to Fc = 217 + 524 = 741 kN

(D8.15)

This is less than the compressive resistance of 845 kN, found above. The bars act at a lever arm of 543 mm (Fig. 8.8(a)), so the bending resistance of the composite joint is Mj, Rd, comp = 83 + 524 × 0.543 = 83 + 284 = 367 kN m

(D8.16)

Check on vertical shear For the maximum design beam load of 35.7 kN/m and a hogging resistance moment at B of 367 kN m, the vertical shear in each beam at B is 244 kN, which just exceeds the shear resistance found earlier, 242 kN. It will probably be found from elastic–plastic global analysis that the vertical shear at B is reduced by the flexibility of the joints. If necessary, two extra M20 bolts can be added in the lower half of each end plate. This has no effect on the preceding results for resistance to bending.

Maximum load on span BC, with minimum load on span AB This loading causes maximum shear in the column web. There is an abrupt change in the tension in the slab reinforcement at B. The load acting on the steel members is equal for

153





the two spans, and is assumed to cause a hogging bending moment at node B equal to the resistance of the joints, 83 kN m from equation (D8.11). The ultimate loads on the composite member are 1.62 kN/m on AB and 27.9 kN/m on BC (see Table 6.2). The flexibility of the joints and cracking of concrete both reduce hogging bending moments, so both are neglected in these checks on shear in the column web and anchorage of the reinforcement. The moment on the composite joint at B in span BC is taken as the additional resistance provided by the slab reinforcement, which is 284 kN m (equation (D8.16)). For the other three members meeting at node B, elastic analysis gives the bending moments shown in Fig. 8.11(a). The total bending moments at B, including construction, are shown in Fig. 8.11(b). The shear forces in columns DB and BE are (75 + 37.5)/3 = 37.5 kN If the end plate in span AB is plastic under ultimate construction loading, the whole of the difference between the beam moments shown is caused by change of tension in the reinforcement. Thus, the relevant lever arm, z, is 543 mm. From clause 5.3(3) of EN 1993-1-8, the shear force on the web panel is Vwp, Ed = (Mb1, Ed – Mb2, Ed)/z – (Vc1, Ed – Vc2, Ed)/2 = (367 – 217)/0.543 – [37.5 – (–37.5)]/2 = 276.2 – 37.5 = 239 kN The sign convention used here is given in Fig. 5.6 of EN 1993-1-8. It is evident from Fig. 8.11(b) and the equation above that the web shear from the beams, 276 kN, is reduced by the shear forces in the column. The change of force in the reinforcement is 276 kN. Shear resistance of the column web From clause 6.2.6.1(1) of EN 1993-1-8, the shear resistance of an unstiffened column web panel is Vwp, Rd = 0.9fy, wc Avc/(÷3γM0) where Avc is the shear area of the column web. This is given in EN 1993-1-1, and is 3324 mm2 here. Hence, Vwp, Rd = 0.9 × 0.355 × 3324/(÷3× 1.0) = 613 kN

(> Vwp, Ed)

This load arrangement therefore does not govern the design of the joint.

75 37.5 D 134 + 83 284

3

134

1.62 kN/m A

75

27.9 kN/m

z 284 + 83

B

C

37.5 E 12

37.5 75

12

(a) Bending-moment diagram (kN m)

(b) Action affects on joint (kN and kN m)

Fig. 8.11. Analyses for unequal design loadings (ultimate limit state) on spans AB and BC

154





A

138 kN

184 kN 122 kN 163 420

D B

3 No. 12 bars at 200 mm

300

C

120

50

120

Fig. 8.12. Strut-and-tie model for anchorage of unbalanced tension in slab reinforcement

Table 8.1. Initial stiffnesses of joints, Sj, ini (kN m/mrad)

Steel joint Composite joint, elastic Composite joint, no top bolts

No shear

Joint BA, with shear

Joint BC, with shear

61.1 146 110

– 118 118

– 48.8 38.6

Anchorage of the force from the reinforcement The force of 276 kN (above) has to be anchored in the column. The strut-and-tie model shown in Fig. 8.2 requires transverse reinforcement to resist the force Ftq shown in the figure, and the force depends on the directions chosen for the struts. The mean distance of the three 16 mm bars shown in Fig. 8.8(d) from the centre-line of the column is 420 mm. The two concrete struts AB and AD shown in Fig. 8.12 can, for calculation, be replaced by line AC. Resolution of forces at point A gives the strut force as 184 kN. The depth of concrete available is 80 mm. With fck = 25 N/mm2, the total width of the struts is bc = (184 × 1.5)/(0.08 × 0.85 × 25) = 163 mm This width is shown to scale in Fig. 8.12, and is obviously available. The existing transverse reinforcement (Example 6.7) is 12 mm bars at 200 mm spacing. Insertion of three more bars (As = 339 mm2) at 200 mm spacing provides an extra resistance TRd = 339 × 0.5/1.15 = 147 kN which exceeds the tie force of 122 kN shown in Fig. 8.12. Three extra bars are provided on each side of the column. The available area of column flange to resist bearing stress is 80 × (120 + 115) = 18 800 mm2 so the mean stress is

155





(276/2)/18.8 = 7.34 N/mm2 well below the design compressive strength of the concrete.

Stiffness of the joints and rotation capacity, ultimate limit state The initial stiffnesses of these joints are calculated in Example 10.1, and are given in Table 8.1. In the preceding global analyses, the joints were assumed to be rigid until their resistance Mj, Rd was reached, and to act as hinges for further loading. This neglect of the elastic rotation of the joints at moments below Mj, Rd leads to overestimation of the hogging bending moments at the joints. The mid-span moments therefore exceed those calculated. This method is safe for the verification of the joints, and is appropriate where there is ample bending resistance at mid-span, as in this example. There is little inelastic curvature in the regions of sagging moment, so the rotation at the joints is much less than that required for the development of mid-span plastic hinges. The typical joint details given in Couchman and Way,104 which were used here, were shown to have adequate rotation capacity by a calculation method19 supported by tests (clause 8.3.4(3)) or are those ‘which experience has proved have adequate properties’ (clause 8.3.4(2)), so no further verification of rotation capacity is required. Serviceability checks The preceding analyses are inadequate for serviceability checks on deflections or crack width. Account must be taken of the flexibility of each joint, as given by equations (D8.1) and (D8.2). From clause 6.3.1(6) of EN 1993-1-8 or clause J.4.1(5) of ECCS TC11,39 Ψ = 2.7 for bolted end-plate joints. Thus, where Mj, Ed = Mj, Rd, from equation (D8.2), µ = 1.52.7 = 2.99

(D8.17)

This dependence of the joint stiffness Sj on Mj, Ed leads to iterative analysis, so simplifications are given in clause 5.1.2 of EN 1993-1-8 and in section 9.5 of ECCS TC1139 for use in elastic frame analysis, as follows: •

•

for a composite beam-to-column joint with a flush end-plate connection, and ‘usual cases’, a nominal stiffness Sj = Sj, ini/2 may be used, for moments up to Mj, Rd (i.e. µ = 2) where joints are required to behave within their elastic range, the stiffness Sj, ini should be used, and Mj, Ed should not exceed 2Mj, Rd/3.

Here, the joints are included in conventional elastic analyses as follows. From Fig. 8.8, a pair of joints (see Fig. 8.1) has an overall length of 240 + 2 × 12 = 264 mm, so each joint is represented by a beam-type member of length Lj = 132 mm and second moment of area Ij. For a bending moment Mj, the rotation is φ = Mj/Sj For the ‘beam’, it is φ = MjLj/EaIj Eliminating φ/Mj, Ij = (Lj/Ea)Sj Hence, 10–6Ij = (132/210)Sj = 0.63Sj with Ij in units of mm4 and Sj in units of kN m/mrad.

156



(D8.18)



5.78 kN/m

Mj,Ek 19.2

337.4 10–6I

B 132

C 11 868

Fig. 8.13. Analysis for steel beams, serviceability limit state

10–6I 113 13.5 kN/m

1.2 kN/m

A 10–6I

828

467

74 12.2

467

132

1668

12 000

828

C

10 200

Fig. 8.14. Analysis for composite beams, serviceability limit state

Maximum deflection, with imposed load on span BC only Separate calculations are required for the steel joints and the composite joints. The load at the end of construction, 5.78 kN/m, is applied to both spans. From Table 8.1 and equation (D8.18), with µ = 2, 10–6Ij = 0.63 × 61.1/2 = 19.2 mm4 For the beam, 10–6Iay = 337.4 mm4 so the calculation model for the steel joints is as shown in Fig. 8.13, with I in units of mm4. The results are: • •

bending moment in the connection in span BC: 67.7 kN m, which is 82% of Mj, Rd, steel maximum deflection of span BC: 13.8 mm.

The calculation model for the composite phase takes account of cracking, and uses the modular ratio n = 20.2. The floor finishes, 1.2 kN/m, act on both spans, and the frequent value of the imposed load, 0.7 × 17.5 = 12.3 kN/m, acts on span BC only. For the joint in span BC, µ = 2. There is a little unused tensile resistance from the steel connection. If this is neglected, then for the composite joint in span BC, Sj, ini = 39.6 kN m/mrad (Table 8.1) and 10–6Ij, BC = 0.63 × 39.6/2 = 12.5 mm4 (The effect of including the full stiffness of the steel joint (Sj, ini = 48.8 kN m/mrad, not 38.6 kN m/mrad) is small; it reduced the deflection by less than 1 mm.) The moment in the connection in span BA is low, so µ = 1. From Table 8.1, 10–6Ij, BA = 0.63 × 118 = 74 mm4 For the column, 10–6Iy = 113 mm4 The calculation model is shown in Fig. 8.14, with I in units of mm4. The results are:

157





R 3.8 mrad C 15 mm

B

1800

d 10 200

Fig. 8.15. Primary shrinkage deformation of span BC

• •

bending moment at B in span BC: 70.2 kN m maximum deflection of span BC: 17.9 mm.

The total deflection of span BC is therefore 13.8 + 17.9 = 31.7 mm or span/380. These two components exceed the values found for the fully continuous beam (see Table 7.3) by 4.8 and 2.9 mm, respectively. The imposed-load deflection is about 17 mm, or span/700, for the frequent loading. The modular ratio used allows for some creep. The additional deflection of the composite slabs, relative to that of the beams, is discussed in Example 7.1. The effect of shrinkage of concrete. Accurate calculation is difficult where semi-rigid joints are used, but estimates can be made, as follows. The primary shrinkage deformations for the 24 m length of beam are shown in Fig. 6.27(b). Calculations in Example 6.7 found that R = 1149 m and δ = 45.3 mm. The most conservative assumption is that the flexibility of the joints reduces the secondary shrinkage moment (which reduces deflections) to zero. Fig. 8.15 shows the same primary curvature as in Fig. 6.27(b), but with compatibility restored by rotation of the joint at B, rather than by secondary bending of the beam. From the geometry of Fig. 8.15, this rotation is found to be 3.8 mrad for each joint. The mid-span shrinkage deflection is then about 15 mm. The direction shown for δ is consistent with that in Fig. 6.1(b), and is not important, as these angles of slope are very small. The preceding calculation for the composite joint found Mj, Ek = 70 kN m, and used Sj = 38.6/2 = 19.3 kN m/mrad, giving a rotation of 70/19.3 = 3.6 mrad. Thus, shrinkage imposes a significant increase of rotation on each joint, and the resulting increase of hogging moment at B, while far less than the 120 kN m found for the fully continuous beam, will decrease the 15 mm deflection found above. The result for the fully continuous beam was 7 mm (see Table 7.3). It is concluded that the shrinkage deflection lies in the range 10–12 mm, additional to the 31.7 mm found above. Cracking of concrete Elastic analysis of the composite frame for service (frequent) loading acting on both spans, otherwise similar to those outlined above, found the hogging bending moment at B to be 133 kN m. The lever arm for the reinforcement, of area 1206 mm2, is 543 mm (Fig. 8.8(a)), so the tensile stress is σs = 133/(0.543 × 1.206) = 203 N/mm2

158





The mean spacing of the 16 mm bars is about 250 mm (see Fig. 8.8(d)), which happens to be the limiting value for a crack width of 0.3 mm given in Table 7.2 of EN 1994-1-1. The alternative condition in Table 7.1 is satisfied by a wide margin. However, the strain field in the slab is disturbed locally by the column and by the concentrated rotations associated with the joints. The values in Tables 7.1 and 7.2 take no account of this situation. It can be concluded that the top reinforcement is unlikely to yield in service, and that very wide cracks will not occur.

159




CHAPTER 9

Composite slabs with profiled steel sheeting for buildings This chapter corresponds to Section 9 of EN 1994-1-1, which has the following clauses: • • • • • • • •

General Detailing provisions Actions and action effects Analysis for internal forces and moments Verification of profiled steel sheeting as shuttering for ultimate limit states Verification of profiled steel sheeting as shuttering for serviceability limit states Verification of composite slabs for ultimate limit states Verification of composite slabs for serviceability limit states

Clause 9.1 Clause 9.2 Clause 9.3 Clause 9.4 Clause 9.5 Clause 9.6 Clause 9.7 Clause 9.8

9.1. General Scope The form of construction and the scope of Section 9 are defined in clause 9.1.1. The shape of the steel profile, with ribs running in one direction, and its action as tensile reinforcement for the finished floor, result in a system that effectively spans in one direction only. The slab can also act as the concrete flange of a composite beam spanning in any direction relative to that of the ribs. Provision is made for this in the clauses on design of beams in Sections 5, 6 and 7. The ratio of the gap between webs to the web spacing, br/bs in clause 9.1.1(2)P, is an important property of a composite slab. This notation is as in Fig. 9.2 and Fig. A.1 in Appendix A. If the troughs are too narrow, the shear strength of stud connectors placed within them is reduced (clause 6.6.4), and there may be insufficient resistance to vertical shear. If the web spacing is too wide, the ability of the slab to spread loads across several webs may be inadequate, especially if the thickness of the slab above the sheeting is minimized, to save weight. Such a wide range of profiles is in use that it was necessary to permit the limit to br/bs to be determined nationally. It should probably be a function of the thickness of the slab above the sheeting. As a guide, it should normally be less than about 0.6. No account is taken of any contribution from the top flange of the sheeting to resistance to transverse bending. The design methods for composite slabs given in Section 9 are based on test procedures described in clause B.3. Although the initial loading is cyclic, the test to failure is under static loading. Thus, if dynamic effects are expected, the detailed design for the particular project



Clause 9.1.1

Clause 9.1.1(2)P



Clause 9.1.1(3)P Clause 9.1.1(4)P Clause 9.1.1(5)

Clause 9.1.2.1

Clause 9.1.2.2

must ensure that the integrity of the composite action is maintained (clauses 9.1.1(3)P and 9.1.1(4)P). Guidance on the degree of lateral restraint provided to steel beams (clause 9.1.1(5)) is available in EN 1993-1-1 and elsewhere.106 Inverted U-frame action relies also on flexural restraint. This subject is covered in comments on clause 6.4.2. Because of the wide range of profiles used, resistance to longitudinal shear has always been based on tests. Slabs made with some profiles have a brittle mode of failure, which is penalized in clause B.3.5(1).

Types of shear connection As for other types of composite member, bond is not accepted in clause 9.1.2.1 as a reliable method of shear connection. Sheeting without local deformations of profile is permitted where the profile is such that some lateral pressure will arise from shrinkage of the concrete (Fig. 9.1(b)). Here, the distinction between ‘frictional interlock’ and ‘bond’ is, in effect, that the former is what remains after the 5000 cycles of loading specified in clause B.3.4. The quality of mechanical interlock is sensitive to the height or depth of the small local deformations of the sheeting, so tight tolerances (clause B.3.3(2)) should be maintained on these during manufacture, with occasional checking on site. These two standard forms of interlock are sometimes insufficient to provide full shear connection, as defined in clause 9.1.2.2. They can be augmented by anchorages at the ends of each sheet, as shown in Fig. 9.1, or design can be based on partial shear connection.

9.2. Detailing provisions Clause 9.2.1.(1)P The limits to thickness given in clauses 9.2.1(1)P and 9.2.1(2)P are based on satisfactory Clause 9.2.1(2)P experience of floors with these dimensions. No limits are given for the depth of the profiled sheeting. Its minimum depth will be governed by deflection. For a slab acting compositely with a beam, the minimum depths are increased (clause 9.2.1(2)P) to suit the detailing rules for stud connectors, such as the length of stud that extends above the sheeting and the concrete cover. A slab used as a diaphragm is treated similarly. Where a slab spans onto a hogging moment region of a composite beam, the minimum reinforcement transverse to its span is governed by the rules for the flange of the beam (e.g. Clause 9.2.1(4) Table 7.1), not by the lower amount given in clause 9.2.1(4). Clause 9.2.3 The minimum bearing lengths (clause 9.2.3) are based on accepted good practice. The lengths for bearing onto steel or concrete are identical to those given in BS 5950: Part 4.107

9.3. Actions and action effects Clause 9.3.1(2)P

Clause 9.3.2(1)

Profiled sheeting Where props are used for profiled sheeting (clause 9.3.1(2)P), care should be taken to set these at the correct level, taking account of any expected deflection of the surface that supports them. If verification relies on the redistribution of moments in the sheeting due to local buckling or yielding, this must be allowed for in the subsequent check on deflection of the completed floor; but this is, of course, less likely to be critical where propping is used. For the loading on the profiled sheeting, clause 9.3.2(1) refers to clause 4.11 of EN 1991-1-6.108 For working personnel and small site equipment, a note to clause 4.11.1(3) proposes a characteristic distributed load of 1 kN/m2. Further guidance may be given in the National Annex. For the weight density of normal-weight concrete, Annex A of EN 1991-1-19 recommends 24 kN/m3, increased by 1 kN/m3 for ‘normal’ reinforcement and by another 1 kN/m3 for unhardened concrete. In addition to self-weight, clause 4.11 of EN 1991-1-6 specifies an imposed load of 10% of the weight of the concrete, but not less than 0.75 kN/m2 (which

162




CHAPTER 9. COMPOSITE SLABS WITH PROFILED STEEL SHEETING FOR BUILDINGS

0.125wL2

0.070wL2

L

L

Fig. 9.1. Bending moments for a two-span beam or slab for uniform loading; elastic theory without redistribution

usually governs), applied to a working area of 3 × 3 m, and 0.75 kN/m2 outside this area. This corresponds to a layer of normal-weight concrete about 35 mm thick, to allow for the mounding that occurs during delivery of fresh concrete. Guidance on the avoidance of overload during construction is available elsewhere.109 Partial factors for ultimate limit states are recommended in Table A1.2(B) of EN 1990, as 1.35 for permanent actions and 1.5 for variable actions. It would be reasonable to use 1.35 for the whole of the weight density of 26 kN/m3, explained above, even though the extra 1 kN/m3 for unhardened concrete is not strictly ‘permanent’. Sometimes, to increase the speed of construction, the profiled sheeting is not propped. It then carries all these loads. This condition, or the check on the deflection of the finished floor, normally governs its design. For the serviceability limit state, the deflection of the sheeting when the concrete hardens is important, for use when checking the total deflection of the floor in service. The construction load and the extra loading from mounding are not present at this time, so the deflection is from permanent load only, and the ψ factors for serviceability, given in Table A.1 of EN 1991-1-6, are not required. Clause 9.3.2(1) refers to ‘ponding’, and clause 9.3.2(2) gives a condition for its effects to be ignored. Where profiled sheeting is continuous over several supports, this check should be made using the most critical arrangement of imposed load.

Composite slab The resistances of composite slabs are determined by plastic theory or by empirical factors based on tests in which all of the loading is resisted by the composite section (clause B.3.3(6)). This permits design checks for the ultimate limit state to be made under the whole of the loading (clause 9.3.3(2)).

Clause 9.3.2(2)

Clause 9.3.3(2)

9.4. Analysis for internal forces and moments Profiled steel sheeting Clause 9.4.1(1) refers to EN 1993-1-3,25 which gives no guidance on global analysis of continuous members of light-gauge steel. Clause 9.4.1(2) rules out plastic redistribution where propping is used, but not where the sheeting extends over more than one span, as is usual. Subsequent flexure over a permanent support will be in the same direction (hogging) as during construction, whereas at the location of a prop it will be in the opposite direction. Elastic global analysis can be used, because a safe lower bound to the ultimate resistance is obtained. Elastic moments calculated for uniform stiffness are normally greatest at internal supports, as shown in Fig. 9.1 for a two-span slab under distributed loading. The reduction in stiffness due to parts of the cross-section yielding in compression will be greatest in these regions, which will cause redistribution of moment from the supports to mid-span. In a technical note from 1984,110 and in a note to clause 5.2 of BS 5950-4,107 the redistribution is given as between 5 and 15%. This suggests that redistribution exceeding about 10% should not be used in absence of supporting evidence from tests.

Clause 9.4.1(1) Clause 9.4.1(2)

163





Clause 9.4.2(3) Clause 9.4.2(4)

Clause 9.4.3

Clause 9.4.3(5)

Composite slab As the steel sheets are normally continuous over more than one span, and the concrete is cast over this length without joints, the composite slab is in reality continuous. If elastic global analysis is used based on the uncracked stiffness, the resulting moments at internal supports are high, as in the example in Fig. 9.1. To resist these moments may require heavy reinforcement. This can be avoided by designing the slab as a series of simply-supported spans (clause 9.4.2(5)), provided that crack-width control is not a problem. Other approaches that reduce the quantity of hogging reinforcement needed are the use of redistribution of moments (clause 9.4.2(3)), and of plastic analysis (clause 9.4.2(4)). Numerical and experimental research on continuous slabs has been reported.111 With typical relative values of moment resistance at internal supports and at mid-span, the maximum design loads calculated by elastic analysis with limited redistribution were found to be less than those obtained by treating each span as simply supported. This arises because the large resistance to sagging moment is not fully utilized. If the slab is to be treated as continuous, plastic analysis is more advantageous. The studies showed that no check on rotation capacity need be made provided the conditions given in clause 9.4.2(4) are satisfied. Effective width for concentrated point and line loads The ability of composite slabs to carry masonry walls or other heavy local loads is limited. The rules of clause 9.4.3 for the effective widths bm, bem and bev are important in practice. They are based on a mixture of simplified analysis, test data and experience,107 and are further discussed, with a worked example, in Johnson.81 The effective width depends on the ratio between the longitudinal and transverse flexural stiffnesses of the slab. The nature of these slabs results in effective widths narrower than those given in BS 811017 for solid reinforced concrete slabs. The nominal transverse reinforcement given in clause 9.4.3(5) is not generous for a point load of 7.5 kN, and should not be assumed to apply for the ‘largely repetitive’ loads to which clause 9.1.1(3)P refers.

9.5–9.6. Verification of profiled steel sheeting as shuttering Clause 9.5(1)

Clause 9.6(2)

The design checks before composite action is established are done to EN 1993-1-3. Clause 9.5(1) refers to the loss of effective cross-section that may be caused by deep deformations of the sheeting. This loss and the effects of local buckling are both difficult to determine theoretically. Design recommendations provided by manufacturers are based in part on the results of loading tests on the sheeting concerned. The maximum deflection of L/180 given in the note to clause 9.6(2) is accepted good practice. Deflection from ponding of wet concrete is covered in clause 9.3.2(2).

9.7. Verification of composite slabs for the ultimate limit states 9.7.1. Design criterion No comment is needed.

Clause 9.7.2(3)

9.7.2. Flexure

The rules in clause 9.7.2 are based on research reported in Stark and Brekelmans.111 Clause 9.7.2(3) says that deformed areas of sheeting should be ignored in calculations of section properties, unless tests show otherwise. No guidance on relevant testing is given. Test results are also influenced by local buckling within the flat parts of the steel profile, and by the enhanced yield strength at cold-formed corners.

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For a composite slab in sagging bending, tests can be done in which the shear span is long enough, or the end anchorage is sufficient, for flexural failure to occur. If the strengths of the materials are known, the effective area of the sheeting, when in tension, can be calculated from the moment resisted, and may be close to the gross area of the sheeting. For a composite slab in hogging bending, the contribution from the sheeting is usually ignored, because it may not be continuous. Where it is continuous, the area of tensile reinforcement is usually small compared with the effective area of the sheeting, so that a conservative estimate of the latter (e.g. excluding embossed areas) may reduce only slightly the calculated resistance to bending. Alternatively, a value found from a bending test on the sheeting alone could be used. The effective widths in clause 9.7.2(4) for local buckling take account of the restraint provided to one side of the sheeting by the concrete. Bending resistances of composite slabs are based on rectangular stress blocks (clauses 9.7.2(5) to 9.7.2(7)). In design of reinforced concrete beams, the compressive strain in concrete is limited, to prevent premature crushing of the concrete before the reinforcement yields. There is no similar restriction for composite slabs. The design yield strength of the profiled sheeting, typically between 280 and 420 N/mm2 (lower than that of reinforcement), and its own bending resistance make composite slabs less sensitive to premature crushing of concrete. However, it could be a problem where stronger sheeting is used. For stress in concrete, the 0.85 factor is included, as discussed in comments on clause 3.1(1). When the neutral axis is within the sheeting, theory based on stress blocks as in Fig. 9.6 becomes very complex for some profiles so simplified equations are given in clause 9.7.2(6). Their derivation is on record.111 As shown in Fig. 9.6, the bending resistance is MRd = Mpr + Nc, f z

Clause 9.7.2(4) Clause 9.7.2(5)

Clause 9.7.2(6)

(D9.1)

where z and Mpr are given by equations (9.5) and (9.6). The concrete in compression within the trough is neglected. When Nc, f = Ape fyp, d and the sheeting is entirely in tension, the neutral axis is at its top edge. Equation (9.5) then correctly gives the lever arm as z = h – hc/2 – e.

9.7.3. Longitudinal shear for slabs without end anchorage

Design of composite slabs for longitudinal shear is based on the results of tests. The specification for these involves a compromise between exploring interactions between the many relevant parameters and limiting the cost of testing to a level such that the use of new profiles is not prevented. The tests on composite slabs are defined in clause B.3, on which comments are given in Chapter 11 of this guide. The tests are suitable for finding the design resistance to longitudinal shear by either of the methods referred to in clause 9.7.3(2).

Clause 9.7.3(2)

The m–k method, and existing tests An empirical ‘m–k method’ has long been established. It is difficult to predict the effect of changes from test conditions using this method, because of the lack of analytical models, especially for slabs with ‘non-ductile’ behaviour. A model for ductile behaviour is given in Appendix B. The m–k test is included in EN 1994-1-1 in a modified form, to provide continuity with earlier practice; but values of m and k, determined in accordance with codes such as BS 5950: Part 4,107 cannot be used in design to EN 1994-1-1, as explained in comments on clause B.3.5. Subject to sufficient test data being available, it may be possible to convert the former values to ‘Eurocode’ values. A study for the European Convention for Constructional Steelwork112 found many differences between methods of testing used. Conversion can be difficult, both for the m–k

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L/4

L/2

L/4

A B

L0

C

D

Ls

Fig. 9.2. Shear spans for a composite slab with two-point loading

Clause 9.7.3(4)

Clause 9.7.3(5)

Clause 9.7.3(6)

Clause 9.7.3(7)

method and for the partial-connection method, and the range of applicability of the values found may be uncertain. Further details are given in Examples 11.1 and 11.2. For the m–k method, resistance to longitudinal shear is presented in clause 9.7.3(4) in terms of vertical shear because of the way in which m and k are defined. Shear–bond failure is characterized by the formation of a major crack in the slab at between one-quarter and one-third of the span from a support, as shown in Fig. 9.2. It is over the length AC from this point to the end of the sheeting that significant slip occurs between the concrete and the sheeting, so this is the length over which the resistance to longitudinal shear is mobilized. The location of point C is unknown, and the finite widths of both the applied load and the support are further complications. The definition of ‘shear span’ Ls for use with the m–k method is treated in clause 9.7.3(5), which gives values for common load arrangements. For the two-point loading in Fig. 9.2, it is the length BD. For the partial-interaction method, calculation of the mean shear strength τu is based on the length Ls + L0. In clauses 9.7.3(3) and 9.7.3(5), L is the span of a simply-supported test specimen, which differs from its use in clause 9.7.3(6). In clause 9.7.3(6), the ‘isostatic span’ is the approximate length between points of contraflexure in a continuous span of length L between supports.

The partial connection method To avoid the risk of sudden failure, profiled sheeting should have ductile behaviour in longitudinal shear. For plain sheeting, the ultimate shear resistance is not significantly greater than that for initial slip. Such behaviour is not ‘ductile’ to clause 9.7.3(3), and is referred to as ‘brittle’ in clause B.3.5(1). The partial shear connection method is not applicable to slabs with this behaviour (clause 9.7.3(2)). The m–k method may still be used, but with an additional partial safety factor of 1.25, expressed by the reduction factor 0.8 in clause B.3.5(1). For profiles with deformations, the expected ultimate behaviour involves a combination of friction and mechanical interlock after initial slip, giving a relationship between load and deflection that should satisfy the definition of ‘ductile’ in clause 9.7.3(3). Design data from test results should be found by the partial connection method of clauses 9.7.3(7) to 9.7.3(10). The analytical model, now given, is similar to that used for composite beams, clause 6.2.1.3, and has been verified for slabs by full-scale tests.113,114 It is used in Example 11.2. For an assumed flexural failure at a cross-section at a distance Lx from the nearest support, the compressive force Nc in a slab of breadth b is assumed to be given (equation (9.8)) by Nc = τu, Rd bLx

(D9.2)

where the design shear strength τu, Rd is found by testing, to clause B.3. Its derivation takes account of the difference between the perimeter of a cross-section of sheeting and its overall breadth, b. By its definition in equation (9.8), force Nc cannot exceed the force for full interaction, Nc, f. Hence, there are two neutral axes, one of which is within the steel profile. In clause B.3.6(2) the degree of shear connection is defined as

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η = Nc/Nc, f

(D9.3)

The longitudinal forces that determine the partial-interaction bending resistance MRd are all known, for given η, but calculation of the resistance moment by the method used for partial interaction in composite beams is difficult. The line of action of the longitudinal force in the sheeting (shown in Fig. 11.5) depends on its complex geometry. A simplified method is given in clause 9.7.3(8). It consists of using equation (9.5) of clause 9.7.2(6), with Nc, f replaced by ηNc, f and 0.5hc replaced by 0.5xpl, to determine the lever arm z: z = h – 0.5xpl – ep + (ep – e)ηNc, f/Ape fyp, d

Clause 9.7.3(8)

(9.9)

The reason for these changes is that where xpl is much less than hc, the method gives too low a value for MRd, because equation (9.5) assumes that the line of action of the force Nc in the slab is at depth hc/2. The correct value is ηxpl/2, where xpl is the full-interaction value as shown in Fig. 9.5. It is not clear in EN 1994-1-1 whether the symbol xpl in equation (9.9) means the full-interaction value, or the reduced value, which in this guide is written ηxpl. The reduced value corresponds to the model used, gives the higher value for z, and is recommended. The value of the last two terms in equation (9.9) increases from –ep to –e as η is increased from 0 to 1. For profiled sheeting, the plastic neutral axis is usually above the centroidal axis (i.e. ep > e), and (ep – e) = z. It can then be assumed, for simplicity, that the force Nc in the sheeting acts at height ep above its bottom fibre, giving a lever arm that is correct for η = 0, and slightly too low for η < 1. With these changes, equation (9.9) becomes z = h – 0.5ηxpl – ep

(D9.4)

The bending resistance MRd of the slab, not stated in clause 9.7.3, can be deduced from Fig. 9.6 (and Fig. 11.5 in this guide). It is MRd = Mpr + Nc z

(D9.5)

Calculations for a range of values of Lx thus give the curve relating resistance MRd to the distance to the nearest support. The design is satisfactory for longitudinal shear if the corresponding curve for MEd lies entirely within the one for MRd, as shown in Example 9.1 and Fig. 9.12. In tests, the resistance to longitudinal shear is increased by the friction associated with the reaction at the adjacent end support. If this is allowed for when calculating τu, Rd, a lower value is obtained. Clause 9.7.3(9) provides compensation by using the same effect in the structure being designed, µREd, to contribute to the shear resistance required. The value recommended for the coefficient of friction µ is based on tests.

Additional reinforcement Reinforcing bars may be provided in the troughs of the profiled sheeting, and this reinforcement may be taken into account when calculating the resistance of the slab by the partial connection method, clause 9.7.3(10). The analytical model assumes that the total resistance is that from composite action of the concrete with both the sheeting and the bars, as for reinforced concrete, determined by plastic analysis of the cross-section. The value of τu, Rd is obtained, as before, by testing of specimens without additional reinforcement, clause B.3.2(7). If the m–k method were to be used, reinforcement in troughs would be an additional variable, which would require a separate test series (clause B.3.1(3)), with evaluation based on measured strength of the reinforcement.

9.7.4. Longitudinal shear for slabs with end anchorage

Clause 9.7.4 refers to the two types of end anchorage defined in clause 9.1.2.1. The anchorage provided by studs is ductile. It is preferable to that provided by deformed ribs of re-entrant profiles, where poor compaction of concrete may occur.

Clause 9.7.3(9)

Clause 9.7.3(10)

Clause 9.7.4

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Clause 9.7.4(2) Clause 9.7.4(3)

Clause 9.7.5

Clause 9.7.6(1)

In the partial-connection method, the anchorage force is used as a contribution to the total force Nc (clause 9.7.4(2)). For through-deck-welded studs it can be calculated, to clause 9.7.4(3). The model is based on the weld collar pulling through the end of the sheeting. Further comments are given in clause 6.6.6.4. It is not clear how a contribution from deformed ribs should be determined, as end anchorage is not used in the test specimens (clause B.3.2(7)). For the m–k method, end anchorage of either type may be included in the test specimens, but each type is an additional variable (clause B.3.1(3)), and would require a separate test series.

9.7.5. Vertical shear

Clause 9.7.5 refers to EN 1992-1-1, where resistance to vertical shear depends on the effective depth d of the section. In a composite slab, where the sheeting is the reinforcement, d is the distance dp to the centroid of the profile, shown in Fig. 9.6.

9.7.6. Punching shear

The critical perimeter for punching shear (clause 9.7.6(1)) has rounded corners, as does that used in EN 1992-1-1. It is therefore shorter than the rectangular perimeter used in BS 5950: Part 4. It is based on dispersion at 45° to the centroidal axis of the sheeting in the direction parallel to the ribs, but only to the top of the sheeting in the less stiff transverse direction. Clause 6.4.4 of EN 1992-1-1 gives the shear resistance as a stress, so one needs to know the depth of slab on which this stress is assumed to act. For a concrete slab this would be the appropriate effective depth in each direction. For a composite slab, no guidance is given. BS 5950-4107 takes the effective depth in both directions as that of the concrete above the top of the profiled steel sheeting, to be used in conjunction with BS 8110.17

9.8. Verification of composite slabs for serviceability limit states Clause 9.8.1(1)

9.8.1. Cracking of concrete

Clause 9.8.1(1) refers to EN 1992-1-1, where, from clause 7.3.1(4), Cracks may be permitted to form without any attempt to control their width, provided that they do not impair the functioning of the structure.

Clause 9.8.1(2)

Clause 9.8.2(1) Clause 9.8.2(2)

For this situation, the provisions for minimum reinforcement above internal supports of composite slabs (clause 9.8.1(2)) are the same as those for beams in clause 7.4.1(4), where further comment is given. Crack widths should always be controlled above supports of slabs subjected to travelling loads. The methods of EN 1992-1-1 are applicable, neglecting the presence of the sheeting.

9.8.2. Deflection

Clause 9.8.2(1) refers to EN 1990, which lists basic criteria for the verification of deformations. For the construction phase, clause 9.8.2(2) refers to EN 1993-1-3, where clause 7.3(2) says that elastic theory should be used, with the characteristic load combination (clause 7.1(3)). This corresponds to an ‘irreversible’ limit state, which is appropriate for the deflection of sheeting due to the weight of the finished slab. Where it can be assumed that when the slab hardens, the imposed load on it is negligible, the remaining deflection of the sheeting due to construction loads should also be negligible; but this assumption, if made, should be based on relevant experience. The prediction of construction loads may also be difficult. If in doubt, the deflection of the sheeting at this time should be determined for the characteristic combination, even though the less adverse frequent combination is permitted for reversible deflections.

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Clause 7.3.(4) of EN 1993-1-3 refers to a clause in EN 1993-1-1 which says that limits to deflection of sheeting ‘should be agreed’, and may be given in a National Annex. Further comment is given under clause 9.3.2, and in Examples 7.1 and 9.1. For the composite member, clause 9.8.2(3) refers to Section 5 for global analysis, where the comments made on continuous beams apply. Restraint of bending of slabs from the torsional stiffness of supporting members is usually ignored, but situations arise where it could cause cracking. The rules in clause 9.8.2(4) permit calculation of deflections to be omitted if two conditions are satisfied. The first refers to limits to the ratio of span to effective depth given in EN 1992-1-1. These are 20 for a simply-supported slab, 26 for an external span of a continuous slab and 30 for an internal span. The effective depth should be that given in the comment on clause 9.7.5. The second condition, clause 9.8.2(6), applies to external spans only, and relates to the initial slip load found in tests – information which may not be available to the designer. Two-point loading is used for testing, so the ‘design service load’ must be converted to a point load. This can be done by assuming that each point load equals the end reaction for a simply-supported span under service loading. Where the initial slip load in the tests is below the limit given in clause 9.8.2(6), clause 9.8.2(7) provides a choice. Either the slip should be allowed for, which means estimating the deflection from the test results, or ‘end anchors should be provided’. There is no guidance on how much end anchorage is required. Its effectiveness would have to be found by testing and analysis by the m–k method, because clause B.3.2(7) does not permit end anchors to be used in tests for the partial-connection method.

Tied-arch model The situation that is covered by clause 9.8.2(8) seems likely to arise only where: • • •

Clause 9.8.2(3)

Clause 9.8.2(4)

Clause 9.8.2(6)

Clause 9.8.2(7)

Clause 9.8.2(8)

a high proportion of the shear connection is provided by welded studs its amount is established by calculation to clause 9.7.4 no test data are available.

If the end anchorage is provided by deformed ribs, then experimental verification is essential, and the slip behaviour will be known. In the tied-arch model proposed, the whole of the shear connection is provided by end anchorage. Accurate calculation of deflection is difficult, but the arch will be so shallow that the following simplified method is quite accurate. The tie shown in Fig. 9.3(a) consists of the effective area of the steel sheeting. Line AB denotes its centroid. The thickness hc of the arch rib must be such that at the relevant bending moment its compressive stress at mid-span is not excessive. There is interaction between its assumed thickness hc and the lever arm ha, so a little trial-and-error is needed. Longitudinal forces at mid-span are now known, and the strains εc and εt in the concrete and steel members can be found, taking appropriate account of creep. In Fig. 9.3(b), curve CB represents the arch rib. The ratio L/ha is unlikely to be less than 20, so both the curve length and the chord length CB can be taken as L/2, and cosec α ª cot α = L/2ha As shown in the figure, the changes in length of the members are et = εtL/2 and ec = εcL/2 so the deflection δ is given by

169





hc

et cosec a

C C d

ha

B

a A

B

et cot a

A

ec

et

L/2 (a)

(b)

Fig. 9.3. Tied-arch model for deflection of a composite slab with an end anchorage

δ/L = (εt + εc)(L/4ha)

(D9.6)

with both strains taken as positive numbers.

Example 9.1: two-span continuous composite slab Data

Details of the geometry assumed for this composite slab are shown in Fig. 9.4(a). The properties of the concrete and reinforcement are as used in the worked example on a two-span beam in Chapters 6 and 7, where further information on them is given. The spacing of the supporting composite beams is different, 3.0 m rather than 2.5 m. Their top-flange width is 190 mm. The cross-section assumed for the sheeting satisfies the condition of clause 9.1.1 for ‘narrowly spaced webs’. It provides shear connection by embossments, in accordance with clause 9.1.2.1(a). Its dimensions hc and hp in Fig. 9.2 are defined in clause 1.6. From Fig. 9.4(a), they are hc = 75 mm (concrete above the ‘main flat surface’ of the sheeting) and hp = 70 mm (‘overall depth’ of the sheeting). Thus, for profiles of this type, the thickness of the slab, 130 mm here, is less than hc + hp. For flexure of the composite slab, hc is the appropriate thickness, but for flexure of the composite beam supporting the sheeting, or for in-plane shear in the slab, the relevant thickness is, for this sheeting, hc – 15 mm. Some of the design data used here have been taken from the relevant manufacturer’s brochure. For simplicity, it is assumed that the reinforcement above the supporting beams, which affects the properties of the composite slab in hogging bending, is not less than that determined in Example 6.7 on the two-span beam, as follows: • •

in regions where the beam resists hogging bending, as determined by resistance to distortional lateral buckling: As = 565 mm2/m and 12 mm bars at 200 mm spacing in other regions, as determined by resistance to longitudinal shear: As ≥ 213 mm2/m.

It is assumed that A252 mesh is provided (As = 252 mm2/m), resting on the sheeting. This has 8 mm bars at 200 mm spacing, both ways. These details satisfy all the requirements of clause 9.2.1. At the outset, two assumptions have to be made that affect the verification of a composite slab: • •

whether construction will be unpropped or propped whether the spans are modelled as simply supported or continuous.

Here, it will be assumed that unpropped construction will be used wherever possible. The sheeting is provided in 6 m lengths, so no 3 m span can have sheeting continuous at both ends. For a building 9 m wide, there will also be spans where 3 m lengths of sheeting are used, so several end-of-span conditions will be considered.

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200

A252 mesh

48

60

0.85fcd 29

15

412 kN 0.9 (nom.)

55 26

30.3

26 164

136

(a)

(b)

49

48

252

27

x 100 51

51.4

27

x

28 1178

27

e = 30

162 (d)

(c)

Fig. 9.4. Composite slab. (a) Dimensions. (b) Mpl, Rd, sagging. (c) Second moment of area, sagging. (d) Second moment of area, hogging

Some of the design checks on a composite slab are usually satisfied by wide margins. These checks will be simplified here by making conservative assumptions.

Properties of materials and profiled sheeting Lightweight-aggregate concrete: grade LC25/28; ρ £ 1800 kg/m3; 0.85fcd = 14.2 N/mm2; Elctm = 20.7 kN/mm2; n0 = 10.1; flctm = 2.32 N/mm2. Creep is allowed for by using n = 20.2 for all loading (clause 5.4.2.2(11)). Reinforcement: fsk = 500 N/mm2, fsd = 435 N/mm2. Profiled sheeting: nominal thickness including zinc coating, 0.9 mm; bare-metal thickness, 0.86 mm; area, Ap = 1178 mm2/m; weight, 0.10 kN/m2; second moment of area, 10–6Iy, p = 0.548 mm4/m; plastic neutral axis, ep = 33 mm above bottom of section (see Fig. 9.6); centre of area, e = 30.3 mm above bottom of section; Ea = 210 kN/mm2; yield strength, fyk, p = 350 N/mm2, γM, p = 1.0; plastic moment of resistance in hogging and sagging bending, Mpa = 6.18 kN m/m – this value is assumed to take account of the effect of embossments (clause 9.5(1)). Composite slab: 130 mm thick; volume of concrete, 0.105 m3/m2. Loading for profiled sheeting From clause 11.3 of EN 1992-1-1, the design density of the reinforced concrete is 1950 kg/m3, which is assumed to include the sheeting. The dead weight of the floor is gk1 = 1.95 × 9.81 × 0.105 = 2.01 kN/m2 From Note 2 to clause 4.11.1(7) of EN 1991-1-6:108 for fresh concrete the weight density should be increased by 1 kN/m3, so for initial loading on the sheeting, gk1 = 2.01 × 20.5/19.5 = 2.11 kN/m2

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Table 9.1. Loadings per unit area of composite slab (kN/m2) Type of load

Characteristic, maximum

Characteristic, minimum

During concreting Sheeting and concrete Imposed load

2.11 1.00

0.10 0

2.85 1.50

0.13 0

3.11

0.10

4.35

0.13

2.01 + 0.48 7.00

2.49 0

3.36 10.5

3.36 0

9.49

2.49

13.9

3.36

0.7 × 7 + 0.48 = 5.38

0.48

NA

NA

Total For composite slab Slab and floor finish Imposed load Total For deflection of composite slab (frequent)

Ultimate, maximum

Ultimate, minimum

For construction loading, clause 9.3.2(1) refers to clause 4.11.2 of EN 1991-1-6. Clause 4.11.2(1) of EN 1991-1-6 gives the actions from personnel and equipment, qk, as ‘10% of the self weight of the concrete, but not less than 0.75 and not more than 1.5 kN/m2’. EN 12812115 may also be relevant. The mounding of concrete during placing is referred to in clause 9.3.2(1), but not in EN 1991-1-6, although its clause 4.11.1(2) refers to ‘storage of materials’. Here, to demonstrate the method of calculation, qk will be taken as 1.0 kN/m2. The National Annex may specify a different value.

Loading for composite slab Floor finish (permanent): gk2 = 0.48 kN/m2. Variable load (including partitions, services, etc.): qk = 7.0 kN/m2. The loads per unit area are summarized in Table 9.1, using γF, g = 1.35 and γF, q = 1.5.

Verification of sheeting Ultimate limit state From clause 9.2.3(2), the minimum width of bearing of the sheeting on a steel top flange is 50 mm. Assuming an effective support at the centre of this width, and a 190 mm steel flange, the effective length of a simply-supported span is 3.0 – 0.19 + 0.05 = 2.86 m Hence, MEd = 4.35 × 2.862/8 = 4.45 kN m/m This is only 72% of Mpl, Rd, so there is no need to consider bending moments in the continuous slab, or to check rotation capacity to clause 9.4.2.1(1)(b). Excluding effects of continuity, for vertical shear: VEd = 4.35 × 1.43 = 6.22 kN/m There are 6.7 webs of depth 61 mm per metre width of sheeting, at an angle cos–1 55/61 to the vertical. In the absence of buckling, their shear resistance is given by EN 1993-1-3 as (61/55)VRd = 6.7 × 61 × 0.9 × 0.350/÷3 whence VRd = 74.3 kN/m

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The slenderness of each web is 61/0.9 = 68, which is close to the limit at which buckling must be considered; but VEd is so far below VRd, as is usual for the construction stage, that no calculation is needed. Deflection A note to clause 9.6(2) recommends that the deflection should not exceed span/180. The worst case is where a span is simply-supported. Then, δ = 5wL4/(384EaIy) = 5 × 2.01 × 2.864 × 1000/(384 × 210 × 0.548) = 15.2 mm For fresh concrete, this is increased to 15.2 × 20.5/19.5 = 16.0 mm From clause 9.3.2(1), allowance should be made for ponding if the deflection exceeds 1/10th of the slab thickness (13 mm), as it does here. The specified thickness of the additional concrete is 0.7δ. Its weight, for fresh concrete, is 0.7 × 0.016 × 20.5 = 0.23 kN/m2 This increases the deflection to 16 × 2.34/2.01 = 17.7 mm which is span/161, and so exceeds span/180. It follows that where the sheeting is not continuous at either end of a 3 m span, propping should be used during construction. The effects of continuity at one end of a span are now considered. The most adverse condition occurs when the concrete in one span hardens (with no construction load present) before the other span is cast. The loadings are then 2.01 kN/m on one span and 0.10 kN/m on the other. The spans should be taken as slightly longer than 2.84 m, to allow for the hogging curvature over the width of the central support. The appropriate length is (2.86 + 3.0)/2 = 2.93 m It can be shown by elastic analysis of a continuous beam of uniform section, with uniformly distributed loading, that the deflection δ at the centre of a span is δ = δ0[1 – 0.6 (M1 + M2)/M0)] where the hogging end moments are respectively M1 and M2, and δ0 and M0 are the deflection and mid-span moment of the span when the end moments are zero. From elastic analysis for a span of 2.93 m, M0 = 2.16 kN m, M1 = 1.13 kN m, M2 = 0 and δ0 = 16.7 mm. Hence, δ = 16.7[1 – 0.6 × (1.13/2.16)] = 11.5 mm When the other span is cast, this deflection is reduced; but if the concrete in the first span has already hardened, the reduction is small. The deflection is less than span/180.

Properties of the composite slab Plastic resistance moment For sagging bending, the plastic neutral axis is likely to be above the sheeting, so clause 9.7.2(5) applies. The tensile force in the sheeting, when at yield, is Fy, p = Ap fyp, d = 1178 × 0.35 = 412 kN/m

(D9.7)

The depth of slab in compression is 412/14.2 = 29 mm

(D9.8)

The lever arm is

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130 – 30.3 – 29/2 = 85 mm (Fig. 9.4(b)), so Mpl, Rd = 412 × 0.085 = 35 kN m/m

(D9.9)

Second moments of area For sagging bending, the transformed width of concrete is 1000/20.2 = 49 mm/m. For an elastic neutral axis at depth x (Fig. 9.4(c)), first moments of area give 1178(100 – x) = 49x2/2 whence x = 49 mm. Hence 10–6Iy = 0.548 + 1178 × 0.0512 + 49 × 0.492/3= 5.53 mm4/m For hogging bending with sheeting present, each trough in the sheeting is replaced by a rectangle of width 162 mm. There is one trough per 300 mm, so the transformed width of concrete is 49 × 162/300 = 27 mm/m For a neutral axis at height x above the bottom of the slab (Fig. 9.4(d)), 252(82 – x) = 1178(x – 30) + 27 x (x/2) whence x = 30.6 mm. Thus, the neutral axis almost coincides with the centre of area of the sheeting, and 10–6Iy = 0.548 + (252 × 51.42 + 27 × 30.63/3) × 10–6 = 1.47 mm4/m

Verification of the composite slab Deflection This is considered first, as it sometimes governs the design. Clause 9.8.2(4) gives conditions under which a check on deflection may be omitted. It refers to clause 7.4 of EN 1992-1-1. Table 7.4 in that clause gives the limiting ratio of span to effective depth for an end span as 26. The depth to the centroid of the sheeting is 100 mm, so the ratio is 2.93/0.1 = 29.3, and the condition is not satisfied. Deflection is a reversible limit state, for which a note to clause 6.5.3(2) of EN 1990 recommends use of the frequent combination. The ψ1 factor for this combination depends on the floor loading category. From Table A1.1 in EN 1990, it ranges from 0.5 to 0.9, and is here taken as 0.7, the value for ‘shopping’ or ‘congregation’ areas. From Table 9.1, the maximum and minimum loadings are 5.38 and 0.48 kN/m, respectively. For a simply-supported loaded span of 2.93 m, with Iy = 5.53 × 106 mm4/m, the deflection is 4.4 mm. The region above the central support is likely to be cracked. A more accurate calculation for the two-span slab, assuming 15% of each span to be cracked, and with 0.48 kN/m on the other span, gives the deflection of the fully loaded span as 3.5 mm. Hence, the total deflection of the slab, if cast unpropped, is δtotal = 11.5 + 3.5 = 15 mm which is span/195. This value is not the total deflection, as it is relative to the levels of the supporting beams. In considering whether it is acceptable, account should also be taken of their deflection (Example 7.1). If it is found to be excessive, propped construction should be used for the composite slab. Ultimate limit states: flexure From Table 9.1, the maximum loading is 13.9 kN/m. For a simply-supported span, MEd = 13.9 × 2.932/8 = 14.9 kN m

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This is so far below the plastic moment of resistance, 35 kN m/m, that there is no need, for this check, to consider continuity or the resistance to hogging bending. Where it is necessary to consider continuity, clause 9.4.2 on global analysis and clause 9.7.2(4) on resistance to hogging bending are applicable. Longitudinal shear by the m–k method For longitudinal shear, it is assumed that there is no end anchorage, so that clause 9.7.3 is applicable. The shear properties of this sheeting are determined and discussed in Examples 11.1 and 11.2 and Appendix B. It is assumed that tests have shown that the sheeting provides ‘ductile’ shear connection to clause 9.7.3(3), and that the values for use in the m–k method are m = 184 N/mm2 and k = 0.0530 N/mm2. From clause 9.7.3(4), the design shear resistance is Vl, Rd = bdp /γVS[(mAp/bLs) + k] where dp is the depth to the centroidal axis of the sheeting, 100 mm; Ap is the cross-sectional area of breadth b of the sheeting, 1178 mm2/m; Ls is span/4, or 0.73 m, from clause 9.7.3(5); and γVS is the partial safety factor, with a recommended value of 1.25. These values give: Vl, Rd = 28.0 kN/m, which must not be exceeded by the vertical shear in the slab. For a simply-supported span, VEd ª 1.5 × 13.9 = 20.9 kN/m For the two-span layout, it will be a little higher at the internal support, but clearly will not exceed 28 kN/m. Vertical shear Clause 9.7.5 refers to clause 6.2.2 of EN 1992-1-1. This gives a formula for the resistance in terms of the ‘area of tensile reinforcement’, which is required to extend a certain distance beyond the section considered. The sheeting is unlikely to satisfy this condition at an end support, but its anchorage has already been confirmed by the check on longitudinal shear. Treating the sheeting as the ‘reinforcement’, the clause gives the resistance to vertical shear as 49 kN/m, which far exceeds VEd, found above. Serviceability limit state – cracking Clause 9.8.1(1) is for ‘continuous’ slabs. This slab can be assumed to satisfy the condition of clause 9.8.1(2): to have been designed as simply supported in accordance with clause 9.4.2(5). To control cracking above intermediate supports, that clause requires the provision of reinforcement to clause 9.8.1. The amount required, for unpropped construction, is 0.2% of the area of concrete ‘on top of the steel sheet’. For this purpose, the mean concrete thickness is relevant. This is close to 75 mm, so the area required is 150 mm2/m, and the A252 mesh used here is sufficient. However, if any spans are constructed propped, to reduce deflections, then the required area is doubled, and A252 mesh is not sufficient.

Longitudinal shear resistance by the partial-connection method In Example 11.2 it is deduced from tests on slabs with the sheeting used here that its design shear strength was τu, Rd = 0.144 N/mm2

(D9.10)

Other conditions for the use of this result in design to EN 1994-1-1, based on clause B.3.1(4), are now compared with the data for this example, which are given in parentheses, with 3 indicating compliance:

175





M (kN m/m) B 20 E D C

MRd

MEd

10

Mpa = 6.18

1.27MEd A

0

0.5

1.0

1.5

x (m)

Fig. 9.5. Design partial-interaction diagram

(a) (b) (c) (d)

thickness of sheeting ≥ 0.9 mm, including coating (t = 0.86 mm, plus coating; 3) concrete strength, fck ≥ 0.8fcm = 0.8 × 29.8 = 23.8 N/mm2 (C25/30 concrete; 3) steel yield strength, fyp ≥ 0.8 fyp, m = 0.8 × 376 = 301 N/mm2 (fyp = 350 N/mm2; 3) concrete density, measured 1.5 h after mixing: 1944 kg/m3 (design density £ 1800 kg/m3; ?) (e) slab thickness, h = 170 mm, as in tests 1 to 4 (h = 130 mm; 7).

Clause B.3.1(3) defines concrete density and slab thickness as ‘variables to be investigated’. Section 9 gives no guidance on what allowance should be made, if any, for differences between test and design values of these variables. The density, measured at an early age, takes no account of subsequent loss of moisture. The concrete strength, item (b), is acceptable, and the difference of density, item (d), is small, so its effects can be ignored. The difference in slab thickness, item (e), is significant. Its effect is discussed more generally in Appendix B. The test results on thinner slabs led to a higher value of τu, Rd, for reasons explained in Example 11.2. It is assumed that the value for τu, Rd given in equation (D9.7) can be used here. Design partial-interaction diagram To satisfy clause 9.7.3(7) it is necessary to show that throughout the span (coordinate x) the curve of design bending moment, MEd(x), nowhere lies above the curve of design resistance, MRd(x), which is a function of the degree of shear connection, η(x). These two curves are now constructed. From Table 9.1, the design ultimate load for the slab is 13.9 kN/m2, assumed to act on the composite member. The verification for flexure assumed simply-supported spans of 2.93 m, which gives MEd, max as 14.9 kN m/m, at mid-span. The parabolic bending-moment diagram is plotted, for a half span, in Fig. 9.5. From equation (9.8) in clause 9.7.3(8), the compressive force in the slab at distance x m from an end support (i.e. with Lx = x) is Nc = τu, Rdbx = 0.144 × 1000x = 144x kN/m From equation (D9.4), Nc, f = Ap fyp, d = 412 kN/m The length of shear span needed for full interaction is therefore

176





Ls, f = 412/144 = 2.86 m This exceeds span/2 (1.47 m), so full interaction is not achieved in a span of this length. From equation (D9.5), the depth of the full-interaction stress block in the slab is xpl = 29 mm. With partial interaction, this value gives a slightly conservative result for z, and is used here for simplicity. With h = 130 mm, ep = 33 mm and e = 30.3 mm, as before, equation (9.9) for the lever arm gives z = 130 – 29/2 – 33 + (33 – 30.3) × 144x/412 = 82 + 1.05x mm The reduced bending resistance of the composite slab is given by equation (9.6) with Nc, f replaced by Nc and Mpa = 6.18 kN m/m, as before: Mpr = 1.25 × 6.18 × (1 – 144x/412) = 7.72 – 2.7x £ 6.18 so x ≥ 1.54/2.7 = 0.570 m From Fig. 9.6, the plastic resistance is MRd = Ncz + Mpr = 0.144x × (82 + 1.05x) + 7.72 – 2.7x = 7.72 + 9.11x + 0.151x2 for 0.57 £ x £ 1.47 m For x < 0.57 m, MRd = 0.144x × (82 + 1.05x) + 6.18 = 6.18 + 11.8x + 0.151x2 The curve MRd (x) is plotted as AB in Fig. 9.5. It lies above the curve 0C for MEd at all cross-sections, showing that there is sufficient resistance to longitudinal shear. This result can be compared with that from the m–k method, as follows. Curve 0C is scaled up until it touches curve AB. The scale factor is found to be 1.27 (curve 0DE), with contact at x = 1.0 m. Shear failure thus occurs along a length of 1.0 m adjacent to an end support. The vertical reaction at that support is then VEd = 1.27 × 13.9 × 2.93/2 = 25.9 kN/m This is 8% lower than the 28 kN/m found by the m–k method. It is concluded in Example 11.2 that its result for τu, Rd is probably too low because the test span was too long. The two methods therefore give consistent results, for this example. No general comparison of them is possible, because the partial shear-connection method involves the bendingmoment distribution, while the m–k method does not. The calculations summarized in this example illustrate provisions of EN 1994-1-1 that are unlikely to be needed for routine design. They are relevant to the preparation of design charts or tables for composite slabs using sheeting of a particular thickness and profile. These are normally prepared by specialists working on behalf of the manufacturer.

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CHAPTER 10

Annex A (Informative). Stiffness of joint components in buildings This chapter corresponds to Annex A in EN 1994-1-1, which has the following clauses: • • •

Scope Stiffness coefficients Deformation of the shear connection

Clause A.1 Clause A.2 Clause A.3

Annex A is needed for the application of clause 8. It is ‘informative’ because the ‘component’ approach to the design of steel and composite joints continues to be developed. Its content is based on the best available research, much of which is recent. It is informed by and generally consistent with two reports prepared on behalf of the steel industry.39,104

A.1. Scope This annex supplements the provisions on stiffness of steel joints in clause 6.3 of EN 1993-1-8.24 Its scope is limited (clause A.1(1)). It covers conventional joints in regions where the longitudinal slab reinforcement is in tension, and the use of steel contact plates in compression. As in EN 1993-1-8, stiffness coefficients ki are determined (clause A.1(2)), with dimensions of length, such that when multiplied by Young’s elastic modulus for steel, the result is a conventional stiffness (force per unit extension or compression). For a known lever arm z between the tensile and compressive forces across the joint, the rotational stiffness is easily found from these coefficients. As in EN 1993-1-8, stiffnesses of components are combined in the usual way, for example: k = k1 + k2

(for two components in parallel)

1/k = 1/k1 + 1/k2

(for two components in series)

Clause A.1(1) Clause A.1(2)

(D10.1)

Stiffness coefficients k1 to k16 are defined in Table 6.11 of EN 1993-1-8. Those relevant to composite joints are listed in Section 8.2 of this guide. Of these, only k1 and k2 are modified here, to allow for steel contact plates and for the encasement of a column web in concrete (clause A.1(3)).

Clause A.1(3)

A.2. Stiffness coefficients The background to this clause is available.116 The coefficients have been calibrated against test results. For longitudinal reinforcement in tension, clause A.2.1.1(1) gives formulae for Clause A.2.1.1(1) the coefficient ks, r in terms of the bending moments MEd, 1 and MEd, 2, shown in Fig. A.1 and





l

b

For precise b

2

For approx. b

l1

b1 and b2, approx.

5

b1 and b2, precise

3.6 1 0.5 0 –1

0

1

(a)

1

M2 / M1

l2

M2 / M1 (b)

Fig. 10.1. Flexibility of reinforcement. (a) Transformation parameter, β. (b) Flexibility of reinforcement, represented by λ1 and λ2, for M1 ≥ M2

Fig. 8.3. Subscripts 1 and 2 are used here also for the properties of the connections on which these moments act. The concrete slab is assumed to be fully cracked. The transformation parameters β1 and β2 allow for the effects of unequal bending moments applied to the pair of connections on either side of a column. They are given in clause 5.3(9) of EN 1993-1-8, with simplified values in clause 5.3(8). The latter are discontinuous functions of MEd, 2/MEd, 1, as shown in Fig. 10.1(a). The sign convention in EN 1993-1-8 is that both moments are positive when hogging, with MEd, 1 ≥ MEd, 2. The range covered by EN 1993-1-8 is –1 £ MEd, 2/MEd, 1 £ 1 However, the stiffness coefficients ks, r in Table A.1 are based on reinforcement in tension, and composite joints with MEd, 2 < 0 are outside the scope of EN 1994-1-1. A typical stiffness coefficient is ks, r = As, r/λh

Clause A.2.1.2 Clause A.2.2.1 Clause A.2.2.2

(D10.2)

where h is the depth of the steel section of the column, and λ1 and λ2 (for the connections on sides 1 and 2 of the column) are functions of β1 and β2, respectively. Based on both the precise and the simplified values for the βs given in EN 1993-1-8, they are plotted against the moment ratio M2/M1 (using this notation for MEd, 2 and MEd, 1) in Fig. 10.1(b). For the simplified βs, they are discontinuous at M2/M1 = 0 and 1. It is the extension of a length λh of reinforcement that is assumed to contribute to the flexibility of the joint. The figure shows that when M2 = M1, this length for connection 1 is h/2, increasing to 3.6h when M2 = 0. This is the value in Table A.1 for a single-sided joint. The flexibility 1/ks, r for connection 2 becomes negative for M2 < 0.5M1. The ‘infinite’ stiffness of a steel contact plate (clause A.2.1.2) simply means that one term in an equation of type (D10.1) is zero. In clause A.2.2.1, the stiffness of a column web panel in shear is reduced below the value in EN 1993-1-8 because the force applied by a contact plate may be more concentrated than would occur with other types of end-plate connection. Similarly, the stiffness for a web in transverse compression has been reduced in clause A.2.2.2, where the value 0.2 replaces 0.7 in EN 1993-1-8.

Concrete encasement Encasement in concrete increases the stiffness of the column web in shear, which is given for an uncased web in EN 1993-1-8 as k1 = 0.38Avc/βz

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CHAPTER 10. ANNEX A (INFORMATIVE). STIFFNESS OF JOINT COMPONENTS IN BUILDINGS

Clause A.2.3.1(1)

The addition to k1 given in clause A.2.3.1(1) has a similar form: k1, c = 0.06(Ecm/Ea)bc hc/βz where Ecm/Ea is the modular ratio and bc hc is the area of concrete. For the column web in compression (clause A.2.3.2), the relationship with the stiffness of the steel web is similar to that for the web in shear. The coefficient in the additional stiffness k2, c is 0.5 for an end plate, but reduces to 0.13 for a contact plate, because of the more concentrated force.

Clause A.2.3.2

A.3. Deformation of the shear connection The background to the rather complex provisions of clause A.3 is given in Appendix 3 of ECCS TC1139 and in COST-C1.117 They are based on linear partial-interaction theory for the shear connection. Equations (A.6) to (A.8) are derived as equations (7.26) and (7.28) in Aribert.118 Equation (A.5) in clause A.3(2) can be rearranged as follows:

Clause A.3

1/kslip ks, r = (Ksc + Es ks, r)/Ksc ks, r = Es/Ksc + 1/ks, r This format shows that the flexibility Es/Ksc has been added to that of the reinforcement, 1/ks, r, to give the combined stiffness, kslip ks, r. It also shows that, unlike the ki, Ksc is a conventional stiffness, force per unit extension. The definition of the stiffness of a shear connector in clause A.3(3) assumes that the mean load per connector will be a little below the design strength, PRk/γVS (typically 0.8PRk). Where the slab is composite, the tests should ideally be ‘specific push tests’ to clause B.2.2(3). In practice, the approximate value given in clause A.3(4) for 19 mm studs, 100 kN/mm, may be preferred. Its use is limited to slabs in which the reduction factor kt (clause 6.6.4.2) is unity. It may not apply, therefore, to pairs of studs in each trough, as used in Example 6.7. In fact, this stiffness, the connector modulus, can vary widely. Johnson and Buckby37 refer to a range from 60 kN/mm for 16 mm studs to 700 kN/mm for 25 mm square bar connectors; and an example in Johnson81 uses 150 kN/mm for 19 mm studs. The value 100 kN/mm is of the correct magnitude, but designs that are sensitive to its accuracy should be avoided.

Clause A.3(3)

Further comments on stiffness

These are found in Chapter 8 and in the following example.

Example 10.1: elastic stiffness of an end-plate joint

The rotational stiffness Sj, ini, modified to give Sj, is needed for elastic or elastic–plastic global analysis, both for finding the required rotation of the joint, and for checking deflection of the beams. The stiffnesses ki shown in Fig. 8.3 are now calculated. Formulae for the steel components are given in Table 6.11 of EN 1993-1-8. Dimensions are shown in Figs 8.8 and 8.9. Column web in shear From Table 6.11 of EN 1993-1-8, k1 = 0.38Avc/βz It was shown in Example 8.1 that the relevant lever arm, z, is 543 mm. This stiffness is relevant only for unequal beam loading, for which the transformation parameter β is 1.0 (Fig. 10.1). The shear area of the column web is 3324 mm2, so k1 = 0.38 × 3324/543 = 2.33 mm

(D10.3)

181





Column web in compression, unstiffened From Table 6.11 of EN 1993-1-8, k2 = 0.7beff, c, wc twc/dc From Example 8.1, the width of the web, beff, c, wc, is 248 mm; twc = 10 mm, and dc = 240 – 2 × (21 + 17) = 164 mm so k2 = 0.7 × 248 × 10/164 = 10.6 mm Column web in tension, unstiffened From Table 6.11 of EN 1993-1-8, k3 = 0.7beff, t, wc twc /dc where beff, t, wc is the smallest of the effective lengths leff found for the column T-stub. These are given in Table 6.4 of EN 1993-1-8. The smallest is 2πm, with m = 23.2 mm (Fig. 8.9), so leff = 146 mm. Hence, k3 = 0.7 × 146 × 10/164 = 6.22 mm Column flange in bending From Table 6.11 of EN 1993-1-8, k4 = 0.9leff (tfc /m)3 where leff = 2πm as above, so k4 = 0.9 × 146 × (17/23.2)3 = 51.7 mm End plate in bending From Table 6.11 of EN 1993-1-8, k5 = 0.9leff (tp /m)3 The smallest leff for the end plate was 213 mm (Example 8.1), and from Fig. 8.9, m = 33.9 mm. Hence, k5 = 0.9 × 213 × (12/33.9)3 = 8.50 mm Bolt in tension From Table 6.11 of EN 1993-1-8, k10 = 1.6As /Lb where Lb is the grip length (29 mm) plus an allowance for the bolt head and nut; total 44 mm. The tensile stress area is 245 mm2, so k10 = 1.6 × 245/44 = 8.91 mm per row of two bolts.

Slab reinforcement in tension Let λh be the length of reinforcing bar assumed to contribute to the flexibility of the joint. There are two cases: (a) beams equally loaded, for which λ1 = λ2 = 0.5, from Fig. 10.1 (b) beams unequally loaded, for which λ1 = 3.6 and λ2 = 0. From equation (D10.2),

182





ks, r = As, r /λh where As, r = 1206 mm2 and h = 240 mm. It follows that ks, r(a) = 1206/(0.5 × 240) = 10.05 mm for both joints, ks, r(b) = 1206/(3.6 × 240) = 1.40 mm for the joint in span BC and ks, r(b) Æ • for the joint in span AB. This result is revised later. Deformation of the shear connection In the notation of clause A.3: • • • •

hogging length of beam: l = 0.15 × 12 = 1.80 m distance of bars above centre of compression: hs = 543 mm from Fig. 8.8(a) distance of bars above centroid of steel beam: ds = 325 mm from Fig. 8.8 second moment of area of steel beam: 10–6Ia = 337 mm4.

Allowance will be made for the possible reduced stiffness of pairs of studs in a trough of sheeting by taking N, the number in length l, as the equivalent number of single studs. In Example 6.7, 19.8 equivalent studs were spread along a 2.4 m length of beam each side of support B (see Fig. 6.30). In Example 8.1, the area of tension reinforcement was reduced from 1470 to 1206 mm2, but the shear connection is now assumed to be as before. For l = 1.8 m, N = 19.8 × 1.8/2.4 = 14.85 studs From equation (A.8), ξ = Ea Ia /ds2Es As = 210 × 337/(0.3252× 200 × 1206) = 2.778 From equation (A.7), ν = [(1 + ξ)Nksclds2/EaIa]1/2 = [3.778 × 14.85 × 0.100 × 1.8 × 3252/(210 × 337)]1/2 = 3.88 From equation (A.6), Ksc = Nksc /[ν – (ν – 1)(hs/ds)/(1 + ξ)] = 14.85 × 100/[3.88 – 2.88 × (543/325)/3.778] = 570 kN/mm From equation (A.5), the reduction factor to be applied to ks, r is kslip = 1/(1 + Es ks, r/Ksc) = 1/(1 + 210ks, r /570) The symbol ks, red is used for the reduced value. For ks, r = 10.05 mm, kslip = 0.213, whence ks, red = ks, r kslip = 2.14 mm for beams equally loaded. For ks, r = 1.40 mm, kslip = 0.660, whence ks, red = 0.924 mm

(D10.4)

for joint 1, unequal loading. For ks, r Æ •, kslip = 0. Hence, ks, red is indeterminate. This is an anomaly in the code. Research has found that the steel tension zone of the joint that resists the lower bending moment should be treated as rigid, so in this case, for joint BA, ks, red Æ ∞

(D10.5)

183





Stiffness of joints, for both beams fully loaded The rules for assembly of stiffnesses in Table 6.10 of EN 1993-1-8 are extended on p. B3.7 of ECCS TC1139 to allow for the slab reinforcement. Stiffness in tension Stiffnesses 3, 4, 5 and 10 are in series (see Fig. 8.3). For these, 1/kt = 1/6.22 + 1/51.7 + 1/8.5 + 1/8.91 = 0.410 mm–1 kt = 2.44 mm The lever arm for kt is z2 = 0.383 m. The lever arm for the reinforcement is z1= 0.543 m, and ks, red = 2.14 mm. The equivalent lever arm is zeq = (ks, red z12 + kt z22)/(ks, red z1 + kt z2) = 0.989/2.096 = 0.472 m

(D10.6)

The equivalent stiffness in tension is keq = (ks, red z1 + kt z2)/zeq = 2.096/0.472 = 4.44 mm

(D10.7)

Stiffness in compression Only the stiffness for the column web is required: k2 = 10.6 mm. Stiffness of joints From clause 6.3.1 of EN 1993-1-8, the initial stiffness of each composite joint is Sj, ini = Ea zeq2/(1/keq+ 1/k2) = 210 × 0.4722/(1/4.44 + 1/10.6) = 146 kN m/mrad

(D10.8)

If, during construction, the end plates yield in tension, the stiffness represented by kt is ineffective for actions on the composite member. Then, for the stiffness of the composite joint, kt = 0, zeq = z1 and keq = ks, red = 2.14 mm, and equation (D10.8) becomes Sj, ini = Ea z12/(1/ks, red + 1/k2) = 210 × 0.5432/(1/2.14 + 1/10.6) = 110 kN m/mrad

(D10.9)

For each joint during construction, ks, red = 0, z = z2 and keq = kt, so Sj, ini, steel = 210 × 0.3832/(1/2.44 + 1/10.6) = 61.1 kN m/mrad

(D10.10)

Stiffness of joints, for imposed load on span BC only The flexibility of the column web in shear, 1/k1 (equation (D10.3)), must now be included, and the values of ks, red are different for the two joints. There are two cases: either the end plate is elastic, or it has yielded in the tension zone, so that kt = 0. For the connection in span BC, with z1 and z2 as above, from equations (D10.6) and (D10.7), zeq = (0.924 × 0.5432 + 2.44 × 0.3832)/(0.924 × 0.543 + 2.44 × 0.383) = 0.630/1.436 = 0.439 m keq = 1.436/0.439 = 3.27 mm Including 1/k1 in equation (D10.8), Sj, ini, BC = 210 × 0.4392/(1/2.33 + 1/10.6 +1/3.27) = 48.8 kN m/mrad

(D10.11)

If kt = 0, then from equations (D10.6) and (D10.7), zeq = z1, and keq = ks, red = 0.924 mm. From equation (D10.8), Sj, ini, BC = 210 × 0.5432/(1/2.33 + 1/10.6 +1/0.924) = 38.6 kN m/mrad

184



(D10.12)



For the joint in span AB, with ks, red Æ •, then zeq = z1 and keq Æ •, and equation (D10.8) gives Sj, ini, BA = 210 × 0.5432/(1/2.33 + 1/10.6) = 118 kN m/mrad

(D10.13)

This use of equation (D10.8) where there is shear in the web panel includes an approximation that leads to a small overestimate of the deflection of span BC. The shear deformation of the column web panel, of stiffness k1 (2.33 mm here), causes clockwise rotation of end B of span AB, and hence increases the hogging bending moment at this point. The effective stiffness Sj, ini, BA therefore exceeds 118 kN m/rad. The resulting increase in the hogging moment at B in span BC is small, and so is the associated decrease in the mid-span deflection. Results (D10.8) to (D10.13) are repeated in Table 8.1, for use in Example 8.1.

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CHAPTER 11

Annex B (Informative). Standard tests B.1. General Annex B is ‘informative’, not ‘normative’, because test procedures for products are strictly outside the scope of a design code. From the note to clause B.1(1), they should be given in a European standard or in guidelines for European technical approvals, which are not yet available. One of the objectives of standard tests is to provide guidance to designers in situations where calculation models are not sufficient. This commonly occurs for two components of composite structures: shear connectors and profiled steel sheeting. Existing design rules for both shear connection and composite slabs are based mainly on test data obtained over many decades using various procedures and types of test specimen, for which there has been no international standard. There are many national standards, and there is some international consensus on details of the m–k test for resistance of composite slabs to longitudinal shear. However, evidence has accumulated over the past 20 years that both this test and the UK’s version of the push test for shear connectors have significant weaknesses. These restrict the development of new products, typically by giving results that are over-conservative (push test) or misleading (m–k test). A full set of m–k tests for a new profile is also expensive and time-consuming. When the specification for an existing test is changed, past practice should, in principle, be re-evaluated. This is one reason why the UK’s push test has survived so long in its present form. The new push test (clause B.2) has been in drafts of Eurocode 4 for 15 years, and was based on research work in the preceding decade. Almost all non-commercial push tests since that time have used slabs wider than the 300 mm of the specimen that is defined, for example, in BS 5400: Part 5.82 The new test generally gives higher results, and so does not raise questions about past practice. It costs more, but gives results that are more consistent and relevant to the behaviour of connectors in composite beams and columns. For profiled sheeting, most research workers have concluded that the empirical m–k test procedure should be phased out. This method, as given in BS 5950: Part 4,107 does not distinguish sufficiently between profiles that fail in a ductile manner and those that fail suddenly, and does not exploit the use of end anchorage or the ability of many modern profiles to provide partial shear connection.113,119 However, its use has been the principal basis world-wide for the design of composite slabs for longitudinal shear. Re-testing of the scores (if not hundreds) of types of composite slabs used in existing structures is impracticable.





Clause B.3 therefore sets out in detail a method of testing that can be used with what is essentially the existing m–k procedure, though with a tighter specification relating to the mode of failure, and other changes based on recent experience of its use. The test method is intended also for use with the more rational partial-interaction design of composite slabs, which is to be preferred for reasons given in comments on clauses 9.7 and B.3.

Properties of materials Ideally, the strengths of the materials in a test specimen should equal the characteristic values specified for the application concerned. This is rarely possible in composite test specimens that include three different materials. It is therefore necessary to adjust resistances found by testing, or to limit the range of acceptable measured strengths of a material so that any adjustment would be negligible. Relevant provisions are given within clauses B.2 and B.3. The influence of cracking of concrete may be assumed to be allowed for in the test procedures. Tests that fully reproduce the effects of shrinkage and creep of concrete are rarely practicable; but these effects can normally be predicted once the behaviour in a short-term test has been established, and have little influence on ultimate strength except in slender composite columns.

B.2. Tests on shear connectors

Clause B.2.2(1) Clause B.2.2(2)

General The property of a shear connector that is needed for design is a curve that relates longitudinal slip, δ, to shear force per connector, P, of the type shown in Fig. B.2. No reliable method has been found for deducing such curves from the results of tests on composite beams, mainly because bending resistance is insensitive to the degree of shear connection, as shown by curve CH in Fig. 6.4(a). Almost all the load-slip curves on which current practice is based were obtained from push tests, which were first standardized in the UK in 1965, in CP117: Part 1. A metricated version of this test is given in BS 5400: Part 5,82 and referred to in clause 5.4.3 of BS 5950: Part 3.131 without comment on the need to modify it when profiled sheeting is present. This test has two variants, because the slab and reinforcement ‘should be either as given in [the code] ... or as in the beams for which the test is designed’. This distinction is maintained in EN 1994-1-1 (clause B.2.2(1)). A ‘standard’ specimen is specified in clause B.2.2(2) and Fig. B.1, and a specimen for ‘specific push tests’ is defined in more general terms in clause B.2.2(3). The principal differences between the ‘standard’ tests of EN 1994-1-1 and BS 5400 (the ‘BS test’) are summarized below, with reference to Fig. 11.1, and reasons for the changes are given. The standard test is intended for use where the shear connectors are used in T-beams with a concrete slab of uniform thickness, or with haunches complying with 6.6.5.4. … In other cases, specific push tests should be used.

It can be inferred from clause B.2.2(1) that separate ‘specific’ push tests should be done to determine the resistance of connectors in columns and in L-beams, which commonly occur at external walls of buildings and adjacent to large internal holes in floors. This is rarely, if ever, done, although a connector very close to a free edge of a slab is likely to be weaker and have less slip capacity than one in a T-beam.79 This problem can be avoided by appropriate detailing, and is the reason for the requirements of clauses 6.6.5.3 (on longitudinal splitting) and 6.6.5.4 (on the dimensions of haunches).

Push tests to clause B.2, compared with the BS test Welded headed studs are the only type of shear connector for which large numbers of tests have been done in many countries, so all reported studies of push testing (e.g. see Johnson and Oehlers,67 Stark and van Hove72 and Oehlers120) are based on these tests. It has been

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CHAPTER 11. ANNEX B (INFORMATIVE). STANDARD TESTS

P

180

180

180

Cover 15 35 150

150 150

250

150 250

100 35

Recess optional

30 200 150

260

150

200

200

Bedded in mortar or gypsum

100

600 Reinforcement: ribbed bars, ∆ 10 mm, resulting in a high bond with 450 £ fsk £ 550 N/mm2 Steel section: HE 260 B or 254 × 254 × 89 kg U.C.

Fig. 11.1. Test specimen for the standard push test (dimensions in mm)

found that the results of the tests are widely scattered.76 To obtain realistic characteristic values, it is necessary to separate inherent variability from that due to differences in the test specimens, the methods of casting and testing, and the ultimate tensile strength of the connectors. The BS specimen was probably designed to give results at the lower edge of the band of uncertainty that existed 40 years ago, because it has very small slabs, prone to split longitudinally because the mild steel reinforcement is light and poorly anchored. It has connectors at only one level, which in effect prevents redistribution of load from one slab to the other92 and so gives the resistance of the weaker of the two pairs of connectors. The changes from this test are as follows: (1) The slabs have the same thickness, but are larger (650 × 600 mm, cf. 460 × 300 mm). This enables reinforcement to be better anchored, and so avoids low results due to splitting. The bond properties of the reinforcement are more important than the yield strength, which has little influence on the result. Limits are given in Fig. B.1. (2) The transverse reinforcement is 10 high-yield ribbed bars per slab, instead of four mild steel bars of the same diameter, 10 mm, so the transverse stiffness provided by the bars is at least 2.5 times the previous value. In T-beams, the transverse restraint from the in-plane stiffness of the slab is greater than in a push specimen. The reinforcement is intended to simulate this restraint, not to reproduce the reinforcement provided in a beam. (3) Shear connectors are placed at two levels in each slab. This enables redistribution of load to occur, so that the test gives the mean resistance of eight stud connectors, and better simulates the redistribution that occurs within the shear span of a beam. (4) The flange of the steel section is wider (> 250 mm, cf. 146 mm), which enables wider block or angle connectors to be tested; and the lateral spacing of pairs of studs is standardized. The HE 260B section (Fig. B.1) is 260 × 260 mm, 93 kg/m. (5) Each concrete slab must be cast in the horizontal position, as it would be in practice (clause B.2.3(1)). In the past, many specimens were cast with the slabs vertical, with the risk that the concrete just below the connectors would be poorly compacted.

Clause B.2.3(1)

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(6) Unlike the BS test, details of concrete curing are specified, in clauses B.2.3(3) and B.2.3(4). (7) The strength of the concrete measured at the time the push test is done must satisfy 0.6 £ fcm /fck £ 0.8

(D11.1)

where fck is the specified strength in practice (clause B.2.3(5)). The corresponding rule for the BS test is 0.86 £ fcm /fcu £ 1.2 where fcu is ‘the cube strength of the concrete in the beams’. For both codes, the two strength tests must be done using the same type of specimen, cylinder or cube. Condition (D11.1) is now explained. It is essentially fcm = 0.7fck The resistance of a stud is usually found from equation (6.19): PRd = 0.29αd 2(fck Ecm)0.5/γV In the push test, fck is in effect replaced by 0.7fck. Then, PRd = 0.29αd2(0.7fck Ecm)0.5/γV = 0.29αd 2( fck Ecm)0.5/1.5

Clause B.2.4(1)


Clause B.2.5(1)

(D11.2)

when γV = 1.25. This shows that the purpose of condition (D11.1) is to compensate for the use of a γV factor of 1.25, lower than the value 1.5 normally used for concrete, and the likelihood that the quality of the concrete in the laboratory may be higher than on site. (8) The loading is cycled 25 times between 5 and 40% of the expected failure load (clause B.2.4(1)). The BS test does not require this. Stresses in concrete adjacent to shear connectors are so high that, even at 40% of the failure load, significant local cracking and inelastic behaviour could occur. This repeated loading ensures that if the connector tested is susceptible to progressive slip, this will become evident. (9) Longitudinal slip and transverse separation are measured (clauses B.2.4(3) and B.2.4(4)), to enable the characteristic slip and uplift to be determined, as explained below. The BS test does not require this.

Evaluation of results of push tests Normally, three tests are conducted on nominally identical specimens to determine the characteristic resistance PRk for concrete and connector material of specified strengths fck and fu, respectively. Let Pm be the mean and Pmin the lowest of the three measured resistances per connector, and fut be the measured ultimate strength of the connector material. If all three results are within 10% of Pm, then, from clause B.2.5(1), PRk = 0.9Pmin

(D11.3)

Clause B.2.5(2) refers to Annex D of EN 1990 (Informative) for the procedure to be followed if the scatter of results exceeds the 10% limit. A method to clause D.8 of EN 1990 for the deduction of a characteristic value from a small number of test results, which took no account of prior knowledge, would severely penalize a three-test series. It is necessary to rely also on the extensive past experience of push testing. Clause D.8.4 is relevant. For three tests it sets the condition that all results must be within 10% of the mean, Pm. This appears in clause B.2.5(1). Clause D.8.4 then gives the characteristic resistance as a function of Pm and of Vr, ‘the maximum coefficient of variation observed in previous tests’, in which the ‘10% from the mean’ condition was satisfied. Most of the previous results were from research programmes, with many different types of test specimen. The results for studs in profiled sheeting, for example, have been found to be samples from seven different statistical populations.76 It has not been possible to establish the value of Vr. The method of clause B.2.5(1), of reducing the lowest of the three results by 10%, is mainly based on previous practice. It can be deduced from clause D.8.4 that for a set

190





of three results with the lowest 10% below the mean, the method of clause B.2 implies that Vr = 11%. Clause B.2.5(1) gives a penalty that applies when fut > fu. This is appropriate where the resistance of a connector is governed by its own material, usually steel, but in practice the resistance of a connector can depend mainly on the strength of the concrete, especially where lightweight aggregate is used. The correction then seems over-conservative, because fu is limited to 500 N/mm2 by clause 6.6.3.1(1), and the strength of the material can exceed 600 N/mm2 for studs. In the BS test, a ‘nominal’ strength Pu is calculated from Pu = (fck /fc)Pmin and, then, PRd = Pu /1.4 It so happens that 1.25/0.9 = 1.4, so from equation (D11.3) the two methods give a similar relationship between Pmin and PRd, except that the Eurocode result is corrected for the strength of the steel, and the BS result is corrected for the strength of the concrete. This is probably because the results of the BS test are rarely governed by the strength of the steel, as the slabs are so likely to split. Clause B.2.5(3) finds application for connectors such as blocks with hoops, where the block resists most of the shear, and the hoop resists most of the uplift. The classification of a connector as ductile (clause 6.6.1.1(5)) depends on its characteristic slip capacity, which is defined in clause B.2.5(4). From the definition of PRk (clause B.2.5(1)), all three test specimens will have reached a higher load, so the slips δu in Fig. B.2 are all taken from the falling branches of the load–slip curves. It follows that a push test should not be terminated as soon as the maximum load is reached.


B.3. Testing of composite floor slabs General The most usual mode of failure of a composite slab is by longitudinal shear, loss of interlock occurring at the steel–concrete interface. Resistance to longitudinal shear is difficult to predict theoretically. The pattern and height of indentations or embossments and the shape of the sheeting profile all have significant effect. There is no established method to calculate this resistance, so the methods of EN 1994-1-1 rely on testing. Tests are needed for each new shape of profiled sheet. They are normally done by or for the manufacturer, who will naturally be concerned to minimize their cost. Their purpose (clause B.3.1(1)) is to provide values for either the factors m and k for the ‘m–k method’, or the longitudinal shear strength required for the partial shear connection method. These procedures for verifying resistance to longitudinal shear are given in clauses 9.7.3 and 9.7.4. Comments on them are relevant here. The tests also determine whether the shear connection is brittle or ductile (clause B.3.1(2)). There is a 20% penalty for brittle behaviour (clause B.3.5(1)). In view of the purpose of the tests, failure must be in longitudinal shear (clauses B.3.2(6) and B.3.2(7)). Number of tests The list of relevant variables in clause B.3.1(3) and the concessions in clause B.3.1(4) define the number of tests required. As an example, it is assumed that a manufacturer seeks to determine shear resistance for a new profile, as the basis for design data for a range of sheet thicknesses, slab thicknesses and spans, and concrete strengths, with both lightweight and normal-weight concrete. How many tests are required? In view of the penalty for brittle behaviour (clause B.3.5(1)), it is assumed that the new profile is found to satisfy clause 9.7.3(3) on ductility. The partial-connection method is more

Clause B.3.1(1)

Clause B.3.1(2)


191





versatile than the m–k method, and is recommended. Its calculations are straightforward on a spreadsheet. From clause B.3.2(7), its tests are done in groups of four. The variables are now considered in turn, to find the minimum number of values needed for each one, and, hence, the number of tests needed for a full set. (a) Thickness of sheeting: test the thinnest sheeting. As interlock is dependent on the local bending of individual plate elements in the sheeting profile, the results may not be applied to thinner or significantly weaker sheets, which would be more flexible . . . (1) (b) Type of sheeting, meaning the profile, including any overlap details, and the specification of embossments and their tolerances. Ensure that the embossments on sheets tested satisfy clause B.3.3(2), and standardize the other details . . . . . . . . . . . . . . . . (1) (c) Steel grade: test the highest and lowest grades to be used. The materials standards listed in clause 3.5 include several nominal yield strengths . . . . . . . . . . . . . . . . . . (2) (d) Coating: this should be standardized, if possible . . . . . . . . . . . . . . . . . . . . (1) (e) Density of concrete: test the lowest and highest densities. . . . . . . . . . . . . . . . (2) (f) Grade of concrete: test with a mean strength not exceeding 1.25 times the lowest value of fck to be specified (see clause B.3.1(4)). The results will be slightly conservative for stronger concretes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) (g) Slab thickness: test the thinnest and thickest slabs . . . . . . . . . . . . . . . . . . . (2) The use of a single thickness is not permitted because the effectiveness of shear connection may depend on the stiffness of the concrete component. Conclusions from a theoretical model for the effect of slab thickness, given in Appendix B of this guide, are summarized below. (h) Shear span: account is taken of this in the provisions for use of the test results. This gives a total of 4 × 14 × 23 = 32 tests If it is suspected that the results for parameters (a) and (g) will be over-conservative for thick sheets and strong concrete, respectively, even more tests would be needed. If an alternative coating is to be offered, it should be possible to compare its performance with that of the standard coating in a few tests, rather than another full set. For the m–k method, tests are in groups of six, so the number rises from 32 to 48. The main conclusions from Appendix B of this guide are as follows; • •

for the partial-interaction method, interpolation between results from tests on slabs of the same shear span and two thicknesses is valid for slabs of intermediate thickness for the m–k method, results from tests on two shear spans are applicable for shear spans between those tested.

The status of Annex B, and use of fewer tests Annex B is informative. From notes to clauses 9.7.3(4) and 9.7.3(8), its test methods ‘may be assumed to meet the basic requirements’ of the relevant design method for longitudinal shear. These requirements are not defined; they have to be inferred, mainly from Annex B. This implies that where the testing does not conform to the extensive scheme outlined above, some independent body, such as the relevant regulatory authority or its nominee, must be persuaded that the evidence presented does satisfy the ‘basic requirements’ of one or both of the two design methods. Where this is done, the design can presumably claim to be in accordance with Eurocode 4, in this respect; but that would not apply internationally. Annex B may eventually be superseded by a European standard on the determination of the shear resistance of composite slabs. Until then, the situation is unsatisfactory. Development of better theoretical models would help. At present, research workers cannot validate these because manufacturers rarely release their detailed test results. Where a new profile is a development from an existing range, it should be possible to use the results of earlier tests to predict the influence of some of the parameters, and so reduce the number of new tests required.

192





Testing arrangement Loading is applied symmetrically to a simply-supported slab of span L, at points distant L/4 from each support (clause B.3.2(2)). Crack inducers are placed beneath the loads (clause B.3.3(3)), to reduce the effect of local variations in the tensile strength of the concrete. The failure loading is much heavier than the slabs, so the shear spans Ls (Fig. B.3) are subjected to almost constant vertical shear. This differs from the test details in BS 5950: Part 4, in which vertical shear is not constant over the length between a crack inducer and the nearer support. As the shear span is a fixed proportion of the span, specimens for regions such as A and B for the m–k method (clause B.3.2(6) and Fig. B.4) are obtained by altering the span L. As this method is empirical, it is good practice to ensure that the tests also encompass the range of spans required for use in practice.107 For sheeting with ductile behaviour and where design will use the partial-interaction method, the number of tests in a series, for specimens of given thickness ht, can be reduced from six to four (clause B.3.2(7)). Clauses B.3.3(1) and B.3.3(2) are intended to minimize the differences between the profiled sheeting used in the tests and that used in practice. The depth of embossments has been found to have a significant effect on the resistance. Propped construction increases the longitudinal shear. It is required for the test specimens to enable the results to be used with or without propping (clause B.3.3(6)). From clause B.3.3(8), the specimens for the determination of concrete strength, defined in clause B.3.3(9), should be cured under the same conditions as the test slabs. This cannot of course be the curing under water normally used for standard cubes and cylinders. When deviation of strength from the mean is significant, the concrete strength is taken as the maximum value (clause B.3.3(9)).This causes the applicability of the test results to be more restricted (clause B.3.1(4)). The test procedure for the strength of the profiled sheeting (clause B.3.3(10)) is given elsewhere.121 The initial loading test is cyclic (clauses B.3.4(3) and B.3.4(4)), to destroy any chemical bond between the sheeting and the concrete, so that the subsequent test to failure gives a true indication of the long-term resistance to variations of longitudinal shear. The number of cycles, 5000, is fewer than that required by BS 5940: Part 4, but has been judged to be adequate for these purposes. Design values for m and k Clause B.3.5(1) gives a design rule for the possibility that the two end reactions may differ slightly, and applies an additional factor of safety of 1.25 to compensate for brittle behaviour, in the form of a reduction factor of 0.8. The method of clause B.3.5(2) is applicable to any set of six or more test results, irrespective of their scatter. An ‘appropriate statistical model’ will penalize both the scatter and the number of results, if small, and may be that given in EN 1990, to which clause B.3.5(3) refers. Where a series consists of six tests and the results are consistent, clause B.3.5(3) provides a simple method for finding the design line shown in Fig. B.4, and hence values for m and k. These are in units of N/mm2. These methods differ from that of BS 5950: Part 4, both in the determination of the line that gives the values of m and k (see Fig. 11.3) and in their definition. In BS 5950, k is proportional to the square root of the concrete strength, which causes complications with units, and the two sets of three results could be from slabs with different concrete strengths. It has been found that the deliberate use of very different strengths for the specimens in regions A and B in Fig. 11.3 can lead to unsafe applications of the method, when m and k are defined as in BS 5950: Part 4. All the results in a diagram such as Fig. B.4 are required by clause B.3.3(8) to be from specimens with nominally identical concrete, so there is no need to include concrete strength in the functions plotted in this figure.

Clause B.3.2(2)

Clause B.3.2(6)

Clause B.3.2(7) Clause B.3.3(1) Clause B.3.3(2) Clause B.3.3(6) Clause B.3.3(8) Clause B.3.3(9)

Clause B.3.3(10) Clause B.3.4(3) Clause B.3.4(4)

Clause B.3.5(1) Clause B.3.5(2) Clause B.3.5(3)

193





The formula for vertical shear in clause 9.7.3(4) and the definitions of m and k in clause B.3.5 are dimensionally correct, and can be used with any consistent set of units. However, analyses of a given set of test results by the Eurocode method give values for m and k different to those found by the BS method, and k even has different dimensions. The conversion of BS values to Eurocode values is illustrated in Example 11.1. The applicability of sets of test results not in accordance with Annex B are discussed in Appendix B of this guide.

Clause B.3.6(1)

Design values for τu, Rd Clause B.3.6(1) refers to the partial-interaction curve shown in Fig. B.5. This is for sagging bending, and is determined for a group of tests on specimens with nominally identical cross-sections as follows: (a) The measured values of the required dimensions and strengths of materials are determined, and used to calculate the full-interaction plastic moment of resistance of a test specimen, Mp, Rm and the corresponding compressive force in the concrete slab, Nc, f. (b) A value is chosen for η (= Nc/Nc, f), which determines a value Nc, the partial-interaction compressive force in the slab at the section where flexural failure is assumed to occur. The corresponding value of the bending resistance M is then calculated from M = Mpr + Nc z with Mpr from equation (9.6), with Nc replacing Nc, f, and z from equation (9.9). This gives a single point on the curve in Fig. B.5, which assumes that an undefined slip can occur at the interface between the sheeting and the concrete. This necessitates ductile behaviour. (c) By repeating step (b) with different values of η, sufficient points are found to define the partial-interaction curve.

Clause B.3.6(2)

Clause B.3.6(3)

Clause B.3.6(4)

From clause B.3.6(2), a bending moment M is found from each test. This is Mtest in Fig. B.5, and leads to a value ηtest, and hence τu from equation (B.2). For the test arrangement shown in Fig. B.3, there will be an overhang L0 beyond the shear span Ls, along which slip will occur. This is allowed for in equation (B.2), which assumes that the shear strength is uniform along the total length (Ls + L0). In reality, the strength includes a contribution from friction at the interface between the sheeting and the concrete, arising from the transmission of the vertical load across the interface to the support. In clause B.3.6(2), this effect is included in the value found for τu. A more accurate equation for τu is given in clause B.3.6(3), for use with the alternative method of clause 9.7.3(9), where relevant comment is given. From clause B.3.2(7), a group of four tests on specimens of given span and slab thickness gives three values of τu, and evidence on ductility (clause B.3.2(7)). All the values of τu are used in a single calculation of the lower 5% fractile value, to clause B.3.6(4). This is divided by γVS to obtain the design value used in clause 9.7.3(8). The shape of the partial-interaction curve depends on the slab thickness, so a separate one is needed for each thickness. The need for tests at different thicknesses is discussed at the end of Appendix B of this guide.

Example 11.1: m–k tests on composite floor slabs

In this example, values of m and k are determined from a set of tests not in accordance with clause B.3, ‘Testing of composite floor slabs’. These tests, done in accordance with the relevant Netherlands standard, RSBV 1990, were similar to ‘specific tests’ as specified in clause 10.3.2 of ENV 1994-1-1, which has been omitted from EN 1994-1-1. This set of tests on eight simply-supported composite slabs has been fully reported.122 The cross-section of the profiled sheeting is shown in Fig. 9.4. The values of m and k

194





L =

=

W/4

=

=

=

W/4

=

=

W/4

=

W/4

Crack inducers

W/2 Vertical shear due to applied loading 0

Ls

Fig. 11.2. Loading used in the tests on composite slabs

determined here are used in Example 9.1, ‘Two-span continuous composite slab’, which includes the results of tests on the non-composite sheeting.

Test specimens and procedure All of the composite slabs had the same breadth, 915 mm, which satisfies clause B.3.3(5), and were cast using the same mix of lightweight-aggregate concrete, with propped construction (clause B.3.3(6)). (Where a clause number is given without comment, as here, it means that the clause was complied with.) The tests to failure were done between the ages of 27 and 43 days, when the strengths of test specimens stored with slabs (clause B.3.3(8)) were as given in Table 11.1. Steel mesh with 6 mm bars at 200 mm spacing was provided in each slab (clause B.3.3(7)). For specimens 1–4, the overall thickness and span were ht = 170 mm and L = 4500 mm, where the notation is as in Fig. B.3. For specimens 5–8, ht = 120 mm and L = 2000 mm. The distance between the centre-line of each support and the adjacent end of each slab was 100 mm (clause B.3.2(4)). The surfaces of the sheeting were not degreased (clause B.3.3(1)). The tensile strength and the yield strength of the sheeting were found from coupons cut from its top and bottom flanges (clause B.3.3(10)). The mean values were fu = 417 N/mm2 and fy, 0.2 = 376 N/mm2. The stress 376 N/mm2, measured at the 0.2% proof strain, is significantly higher than the yield strength found in another series of tests,123 which was 320 N/mm2. The nominal yield strength for this sheeting, now 350 N/mm2, was then 280 N/mm2, which is only 74% of 376 N/mm2, so clause B.3.1(4) is not complied with. This is accepted, because small changes of yield strength have little influence on resistance to longitudinal shear. There is a similar ‘80%’ rule in clause B.3.1(4) for the strength of the concrete. Applying it to the mean cube strength for this series, 34.4 N/mm2, gives a cube strength of 27.5 N/mm2, which is below the value used in Example 9.1, so the rule is satisfied. The slabs were tested under four-point loading, as shown in Fig. 11.2, and crack inducers were provided at distances L/8 each side of mid-span. This is not in accordance with Fig. B.3; clause B.3.2(3) specifies two-point loading. An appropriate value of Ls has to be found, for use in the determination of m and k. A shear–force diagram for two-point loading is found (the dashed line in Fig. 11.2) that has the same area as the

195





actual shear–force diagram, and the same maximum vertical shear. Here, Ls = L/4, as it is in Annex B.

Cyclic loading Clause B.3.4 specifies 5000 cycles of loading between 0.2Wt and 0.6Wt, where Wt is a static failure load. In these tests, Wt for specimen 1 was 75.5 kN, and the range of loading in tests 2–4, for 10 000 cycles, was from 0.13Wt to 0.40Wt. For specimens 5–8, with a mean failure load of 94.4 kN, the fatigue loading in tests 6–8 ranged from 0.19Wt to 0.57Wt, close to the range specified. These divergences are not significant. Results of tests to failure To satisfy clause 9.7.3(3) on ductility, it is necessary to record the total load on the specimen, including its weight, at a recorded end slip of 0.1 mm, at a deflection (δ) of span/50, and at maximum load. These are given in Table 11.1. The failure load, as defined in clause 9.7.3(3), is for all these tests the value when L/50. These loads all exceed the load at a slip of 0.1 mm by more than 10%, the least excess being 13%. All failures are therefore ‘ductile’. From clause B.3.5(1), the representative vertical shear force Vt is taken as half the failure load. Determination of m and k In the original report122 the axes used for plotting the results were similar to those in BS 5950: Part 4 (Fig. 11.3(a)). They are X = Ap /[bLs(0.8 fcu)0.5]

Y = Vt /[bdp(0.8fcu)0.5]

(D11.4)

where fcu is the measured cube strength. The other symbols are in Eurocode notation. Values of X and Y were calculated from the results and are given in Table 11.1. From these, the following values were determined: m = 178 N/mm2

k = 0.0125 N0.5 mm

(D11.5)

For m and k as defined in EN 1994-1-1, the relevant axes (Fig. 11.3(b)) are x = Ap /bLs

y = Vt /bdp

so that x = X(0.8fcu)0.5

y = Y(0.8fcu)0.5

(D11.6)

Approximate values for m and k to EN 1994-1-1 can be found by assuming that m to BS 5950: Part 4 is unchanged, and k is (0.8fcu, m)0.5 times the BS value, given above, where fcu, m is the mean cube strength for the series. These values are Table 11.1. Results of tests on composite floor slabs Test No.

fcu (N/mm2)

Load at 0.1 mm slip (kN)

Load at δ = L/50 (kN)

103Y 103x Maximum 103X 0.5 0.5 load (kN) ((1/N )mm) (N mm)

103y (N/mm2)

1 2 3 4 5 6 7 8

31.4 30.4 32.1 34.2 35.3 37.4 37.2 36.9

63.3 65.3 64.3 66.8 52.2 54.2 52.2 56.2

75.5 75.5 73.9 75.4 94.0 90.9 91.7 94.7

75.5 75.5 73.9 75.4 94.2 94.0 93.9 95.4

294 294 287 293 575 574 573 582

0.222 0.226 0.220 0.213 0.472 0.459 0.460 0.462

196



58.6 59.6 56.7 56.1 108.2 104.9 105.0 107.2

1.115 1.115 1.115 1.115 2.51 2.51 2.51 2.51



Vt/bdpfcm0.5

Vt/bdp B

Shear-bond failure regression line

A Design line (minimum values reduced by 10%)

Design line (reduction of 15%)

Ap/bLsfcm0.5

Ap/bLs

(a)

(b)

Fig. 11.3. Evaluation of the results of tests on composite slabs. (a) BS 5950: Part 4. (b) EN 1994-1-1

V/bdp (N/mm2) 0.6

B D

0.4

m = 184 N/mm2 A 1.0 C

0.2

k = 0.053 N/mm2 0

0.001

0.002

0.003

Ap/bLs

Fig. 11.4. Determination of m and k

m = 178 N/mm2

k = 0.066 N/mm2

(D11.7)

These results are approximate because fcu is different for each test, and the procedure of Annex B for finding characteristic values differs from the BS procedure. The correct method is to calculate x and y for each test, plot a new diagram, and determine m and k from it in accordance with clause B.3.5, as follows. The values of x and y for these tests, from equations (D11.6), are given in Table 11.1. The differences within each group of four are so small that, at the scale of Fig. 11.4, each plots as a single point: A and B. Clause B.3.5(3), on variation within each group, is satisfied. Using the simplified method of that clause, the characteristic line is taken to be the line through points C and D, which have y coordinates 10% below the values for specimens 3 and 7, respectively. This line gives the results m = 184 N/mm2

k = 0.0530 N/mm2

(D11.8)

which are used in Example 9.1. For these tests, the approximate method gives a small error in m, –3%, and a larger error in k, +25%. From clause 9.7.3(4), the design shear resistance is

197





Vl, Rd = (bdp /γVS)[(mAp /bLs) + k]

(D11.9)

The second term in square brackets is much smaller than the first, and in this example the errors in m and k almost cancel out. In Example 9.1, Vl, Rd was found to be 28.0 kN/m. Using the approximate values of equations (D11.7) for m and k changes it only to 28.3 kN/m. Similar comparisons are needed with other sets of test results, before any general conclusion can be drawn about the accuracy of the approximate method of calculation.

Note on the partial-interaction method A composite slab using sheeting of the type tested was designed in Example 9.1. The m–k method was used for the verification for longitudinal shear, with values of m and k calculated from test results, as shown in Example 11.1. To illustrate the partial-interaction method of clauses 9.7.3(7) to 9.7.3(9), an attempt will be made in the following Example 11.2 to use it in an alternative verification of the same composite slab, using the same set of test results.122 It is assumed that the reader is familiar with the two examples referred to. Example 11.2 illustrates potential problems in using this method with existing test data, shows that a procedure given in ENV 1994-1-1,49 and omitted from EN 1994-1-1, can give unconservative results, and proposes a method to replace it.

Example 11.2: the partial-interaction method

It is shown in Example 11.1 that in the eight tests reported,122 the behaviour of the slabs was ‘ductile’ to clause 9.7.3(3). For the partial-interaction method, clause B.3.2(7) then requires a minimum of four tests on specimens of the same overall depth ht: three with a long shear span, to determine τu, and one with a short shear span, but not less than 3ht. The tests available satisfy the 3ht condition, but not that for uniform depth. The four long-span slabs were all thicker than the four short-span slabs. The purpose of the single short-span test is to verify ductility, which is satisfactory here. The maximum recorded end slips in the long-span tests (Nos. 1 to 4), only 0.3 mm, reveal a problem. These tests were discontinued when the deflections reached span/50, so it is unlikely that the maximum longitudinal shear was reached. The specimens had the high span/depth ratio of 26.5, and probably failed in flexure, not longitudinal shear. This is confirmed by the results that follow: the shear strength from tests 1–4 is about 30% lower than that from the short-span tests 5–8, where the end slips at maximum load were from 1 to 2 mm and the span/depth ratio was 16.7. Clause B.3.2(7) requires the shear strength to be determined from the results for the long-span slabs. Its condition for a shear span ‘as long as possible while still providing failure in longitudinal shear’ was probably not satisfied here, so that the final design value τu, Rd is lower than it would have been if a shorter span had been used for tests 1–4. In the absence of guidance from previous tests, this condition is difficult to satisfy when planning tests. Other aspects of these tests are compared with the provisions of Annex B in Example 11.1.

The partial-interaction diagram Measured cube strengths and maximum loads for the eight tests are given in Table 11.1. The mean measured yield strength and cross-sectional area of the sheeting were 376 N/mm2 and 1145 mm2/m, respectively.122 For full shear connection, the plastic neutral axes are above the sheeting, so clause 9.7.2(5) applies. The longitudinal force for full interaction is Nc, f = Ap fyp = 1145 × 0.376 = 431 kN/m

198



(D11.10)



8.5h 8.5

Nc = 431

Ncf = 431

65

z

81.5

Nc Np

55 431

e = 30 (a)

zp

Np (c)

(b)

Fig. 11.5. Stress blocks for bending resistance of composite slab with partial interaction (dimensions in mm)

and the stress blocks are as shown in Fig. 11.5(b). The mean cube strength for specimens 5–8, 36.7 N/mm2, corresponds to a cylinder strength fcm of 29.8 N/mm2, so that the depth of the concrete stress block is xpl = 431/(0.85 × 29.8) = 17.0 mm

(D11.11)

With full interaction, the force Nc, f in the sheeting acts at its centre of area, 30 mm above its bottom surface, so the lever arm is z = 120 – 17/2 – 30 = 81.5 mm and Mp, Rm = 431 × 0.0815 = 35.1 kN m/m The method of calculation for the partial-interaction diagram is now considered, following clause B.3.6(1). The stress blocks in Fig. B.5 correspond to those used in clauses 9.7.2(5) and 9.7.2(6) modified by clause 9.7.3(8). It follows from equation (D11.11) that for any degree of shear connection η (= Nc/Nc, f), the stress block depth is 17η mm, with a line of action 8.5η mm below the top of the slab (Fig. 11.5(c)). For any assumed value for η, the lever arm z is given by equation (9.9). For typical trapezoidal sheeting, where the profile is such that ep > e (these symbols are shown in Fig. 9.6), the simplification given in equation (D9.4) should for this purpose be replaced by z = ht – 0.5η xpl – e

(D11.12)

where ht is the thickness of the slab tested. The use of e in place of ep is because an approximation to the mean resistance MRm should over-estimate it. This moves curve FG in Fig. 11.6 upwards. For a given test resistance M, following the route ABC then gives a lower value for ηtest, and, hence, a lower predicted τu, from equation (B.2). Curve FG is found by calculations for a set of values for η that covers the range of M/Mp, Rm found in the tests. The mean bending resistance is given by MRm = Mpr + ηNc, f z

(D11.13)

(based on equation (D9.5)). The reduced plastic resistance of the sheeting, Mpr, which equals Np zp in Fig. 11.5(c), is found from equation (9.6).

Calculations for the partial-interaction diagram and τu For η ≥ 0.2, equation (9.6) and equation (D11.13) give MRm = ηNc, f z + 1.25Mp, a(1 – η)

(D11.14)

Assuming η = 0.7, for example, equation (D11.12) gives for specimens 5–8 z = 120 – 0.7 × 8.5 – 30 = 84.0 mm

199





From the tests,122 Mp, a = 5.65 kN m/m, with ep = 33 mm, so e – ep = 3 mm, much less than z. From equation (D11.14), MRm = 0.7 × 431 × 0.084 + 1.25 × 5.65 × 0.3 = 27.48 kN m/m and MRm /Mp, Rm = 27.48/35.1 = 0.783 For specimens 1–4, ht = 170 mm and Mp, Rm = 56.7 kN m/m. For η = 0.7, z = 170 – 0.7 × 8.5 – 30 = 134 mm MRm = 0.7 × 431 × 0.134 + 1.25 × 5.65 × 0.3 = 42.5 kN m/m MRm /Mp, Rm = 42.5/56.7 = 0.750 Similar calculations for other degrees of shear connection give curves DE for the shortspan slabs 5–8 and FG for slabs 1–4. In clause B.3.6(2), the bending moment M is defined as being ‘at the cross-section under the point load’, on the assumption that two-point loading is used in the tests. Here, four-point loading was used (see Fig. 11.2). At failure, there was significant slip throughout the length of 3L/8 between each inner point load and the nearer support, so, for these tests, M was determined at an inner point load, and Ls was taken as 3L/8. The calculation of Mtest for specimen 5 is now explained. From Table 11.1, the maximum load was 94.2 kN. This included 2.2 kN that was, in effect, applied to the composite slab by the removal of the prop that was present at mid-span during concreting.122 The loads on the composite member were thus as shown inset on Fig. 11.6, and the bending moment at point J was Mtest = 47.1 × 0.75 – 23.0 × 0.5 = 23.83 kN m This is for a slab of width 0.915 m, so that Mtest/Mp, Rm = 23.83/(0.915 × 35.1) = 0.742 From Fig.11.6, ηtest = 0.646. M/Mp, Rm G 0.9 0.86

E

A

46.0 B 2.2

Tests 5 to 8 0.8 Tests 1 to 4

47.1

250

F

0.742

500 Tests 5 to 8

D

0.7

htest = 0.83 htest = 0.646 C

0.6 0.6

0.7

0.8

0.9 h = Nc/Ncf

Fig. 11.6. Partial-interaction diagram from tests (units: mm and kN)

200


J

H


250



Table 11.2. Degree of shear connection, from tests on composite slabs Test No.

Maximum load (kN)

Mtest (kN m)

Mtest /Mp, Rm

η

1 2 3 4 5

75.5 75.5 73.9 75.4 94.2

44.8 44.8 43.9 44.7 23.83

0.863 0.863 0.846 0.862 0.742

0.835 0.835 0.814 0.833 0.646

Corresponding results for tests 1–4 are given in Table 11.2. Clause B.3.6(4) defines τu, Rk as the 5% lower fractile, based on the results for tests 1–4. There may be other evidence on the variance of such results. Here, it is assumed that this enables a value 10% below the mean to be used, as for the m–k method. The values of η for tests 1–4 are so close that τu can be found from their mean value, 0.83. From clause B.3.6(2), τu = ηNc, f /[b(Ls + L0)]

(D11.15)

For tests 1–4, Ls = 3 × 4.5/8 = 1.69 m. For all tests, Ncf = 431 kN/m, L0 = 0.1 m, and b is taken as 1.0 m. Hence, τu = 0.83 × 431/1790 = 0.200 N/mm2 From clause B.3.6(6), with γVS taken as 1.25, τu, Rd = 0.9 × 0.200/1.25 = 0.144 N/mm2 The interaction curve DE for specimens 5–8 is slightly higher than FG in Fig. 11.6. Its value at η = 0, Mp, a /Mp, Rm, is higher because, for these thinner slabs (ht = 120 mm), Mp, Rm is lower, at 35 kN m/m. Using the preceding method for these results gives τu, Rd = 0.24 N/mm2. This much higher result confirms the suspicion, noted above, that longitudinal shear failure was not reached in specimens 1–4.

Comments Where the test data are in accordance with the specification in Annex B, determination of τu, Rd is straightforward, as values of η can be found by replacing the graphical method (used here for illustration) by direct calculation. However, where tests are being planned, or other data are being used, as here, the work requires understanding of the basis of the provisions of Annex B. It is of particular importance to ensure that longitudinal shear failures occur in the tests.

201




APPENDIX A

Lateral–torsional buckling of composite beams for buildings This appendix supplements the comments on clause 6.4.

Simplified expression for ‘cracked’ flexural stiffness of a composite slab The ‘cracked’ stiffness per unit width of a composite slab is defined in clause 6.4.2(6) as the lower of the values at mid-span and at a support. The latter usually governs, because the profiled sheeting may be discontinuous at a support. It is now determined for the cross-section shown in Fig. A.1 with the sheeting neglected. It is assumed that only the concrete within the troughs is in compression. Its transformed area in ‘steel’ units is Ae = b0 hp /nbs

(a)

where n is the modular ratio. The position of the elastic neutral axis is defined by the dimensions a and c, so that Aec = Asa

and

a+c=z

(b)

where As is the area of top reinforcement per unit width of slab, and z = h – ds – hp /2

(c)

br a

ds As h

z c

hp/2

b0 bs

Fig. A.1. Model for stiffness of a composite slab in hogging bending





Assuming that each trough is rectangular, the second moment of area per unit width is I = As a2 + Ae(c2 + hp2/12)

(d)

Using equations (b) to (d), the flexural stiffness is (EI)2 = Ea [As Ae z2/(As + Ae) + Ae hp2/12]

(DA.1)

This result is used in Example 6.7.

Flexural stiffness of beam with encased web For a partially encased beam the model used for the derivation of equation (6.11) for the flexural stiffness k2 is as shown in Fig. A.2(a). A lateral force F applied to the steel bottom flange causes displacement δ. The rotation of line AB is φ = δ/hs. It is caused by a bending moment Fhs acting about A. The stiffness is k2 = M/φ = Fhs2/δ The force F is assumed to be resisted by vertical tension in the steel web and compression in a concrete strut BC, of width bc/4. Elastic analysis gives equation (6.11).

Maximum spacing of shear connectors for continuous U-frame action A rule given in ENV 1994-1-1 is derived. It is assumed that stud connectors are provided at spacing s in a single row along the centre of the steel top flange (Fig. A.2(b)). The tendency of the bottom flange to buckle laterally causes a transverse moment Mt per unit length, which is resisted by a tensile force T in each stud. From Fig. A.2, Mt s = 0.4bT

(a)

The initial inclination from the vertical of the web, θ0 in Fig. 6.9(b), due to the tendency of the bottom flange to buckle sideways, would be resisted by a moment ksθ0, from the definition of ks in clause 6.4.2(6). For a design longitudinal moment MEd at the adjacent internal support, θ0 is assumed to be increased to

bc/4

0.4b

T A C

T

bc/4 hs

B

F

Mt

d (a)

(b)

Fig. A.2. Resistance to transverse bending in an inverted U-frame. (a) Flexural stiffness of encased web. (b) Spacing of shear connectors

204




APPENDIX A. LATERAL–TORSIONAL BUCKLING OF COMPOSITE BEAMS FOR BUILDINGS

Table A.1. Maximum spacings for 19 mm studs, and minimum top reinforcement Serial size

Mass (kg/m)

Web thickness (mm)

smax (mm)

100As, min/ds

762 × 267 UB 610 × 305 UB 610 × 229 UB IPE 600 HEA 700

197 238 101 122 204

15.6 18.6 10.6 12.0 14.5

362 204 767 509 452

0.06 0.12 0.02 0.03 0.05

θ0[(MEd /Mcr)/(1 – MEd /Mcr)] where Mcr is the elastic critical buckling moment. This deformation causes a transverse bending moment per unit length Mt = ksθ0[(MEd /Mcr)/(1 – MEd /Mcr)]

(b)

where ks is the stiffness defined in clause 6.4.2(6). The design procedure of clause 6.4.2(1) is such that MEd £ χLT MRd. Here, MRd is taken as approximately equal to the characteristic resistance MRk. From clause 6.4.2(4), λLT 2 = MRk/Mcr, so that equation (b) becomes Mt = ksθ0[(χLT λLT 2 )/(1 – χLT λLT 2 )]

(c)

It is assumed that the resistance of the studs to longitudinal shear, PRd, must not be reduced, and that this is achieved if (d)

T £ 0.1PRd

The initial slope θ0 is taken as L/400 h, where h is the depth of the steel section. A typical L/h ratio is 20, giving θ0 = 0.05. From these results, s 0.4T 0.04 PRd (1 - χLT λLT 2 ) = £ b Mt 0.05 ks χLT λLT 2

(DA.2)

This upper limit to the spacing of studs reduces as the slenderness λLT increases. It can be evaluated where the conditions of clause 6.4.3 for simplified verification are satisfied, because the value λLT = 0.4 can be assumed. From Table 6.5 of EN 1993-1-1, buckling curve c should be used for rolled I-sections with a depth/breadth ratio exceeding 2.0. For λLT = 0.4 it gives χLT = 0.90. For a typical 19 mm stud, the resistance PRd is about 75 kN. The combined stiffness of the slab and the web, ks, depends mainly on the stiffness of the web, and is here taken as 0.9k2 where k2, the stiffness of the web, is given by equation (6.10) as k2 = Eatw3/[4(1 – νa2)hs]

(DA.3)

For a typical I-section, hs ª 0.97 h. With Ea = 210 kN/mm2 and ν = 0.3, substitution into equation (DA.2) gives s £ 6.66(b/tw)(h/tw)(1/tw)

(e)

For rolled sections, the closest stud spacing is thus required for relatively thick webs. Examples are given in Table A.1. For studs in two rows, these spacings can be doubled, because the assumed lever arm for the moment Mt would increase from 0.4b (Fig. A.2) to about 0.8b. For web-encased beams, ENV 1994-1-1 required the maximum spacings to be halved. This check is not required by EN 1994-1-1. The results show that it would not govern in normal practice, but could do so where there was a need for wide spacing of studs (e.g. because precast concrete floor slabs were being used) on a beam with a relatively thick web, or where web encasement was used.

205





Top transverse reinforcement above an edge beam

Where the concrete flange of a beam is continuous on one side only, as in Fig. 6.9(a), top transverse reinforcement (AB) is required to prevent lateral buckling by anticlockwise rotation of the steel section, in the plane of the diagram. The preceding results for spacing of stud connectors can be used to estimate the amount required. The reinforcement will be light, so the lever arm for transverse bending can be taken as 0.9ds (notation as in Fig. 6.9(a)), even where concrete in the lower half of the slab is present only in the troughs of sheeting. From expression (d) above, the force T per unit length is 0.1PRd/s; so from equation (a) the transverse bending moment is Mt = 0.4bT/s = 0.04bPRd /s = As fsd(0.9ds) where As is the area of top transverse reinforcement per unit length along the beam, at its design yield stress fsd. Using expression (e) for s, 0.9As fsd ds ≥ 0.0060PRd tw3/h Assuming PRd = 75 kN and fsd = 500/1.15 = 435 N/mm2: 100As /ds ≥ 115tw3/ds2h The area As is thus greatest for a thin slab, so an effective depth ds = 100 mm is assumed, giving 100As /ds ≥ 0.0115tw3/h

(f)

These values are given in the last row of Table A.1. They show that although top reinforcement is required for U-frame action, the amount is small. Clause 6.6.5.3, on local reinforcement in the slab, does not refer to this subject, and could be satisfied by bottom reinforcement only. The requirements for minimum reinforcement of clause 9.2.1(4) and of EN 1992-1-1 could also be satisfied by bottom reinforcement, whereas some should be placed near the upper surface.

Derivation of the simplified expression for λ LT

The notation is as in the comments on clause 6.4 and in this appendix, and is not redefined here. Repeating equation (D6.11): Mcr = (kcC4/L)[(Ga Iat + ks L2/π2)EaIafz]1/2

(D6.11)

From clause 6.4.2(4), λLT = (MRk /Mcr)0.5

(a)

It is on the safe side to neglect the term GaIat in equation (D6.11), which in practice is usually less than 0.1ksL2/π2. Hence, Mcr = (kcC4 /π)(ks Ea Iafz)1/2

(b)

It is assumed that the stiffness of the concrete slab k1 is at least 2.3 times the stiffness of the steel web, k2. The combined stiffness ks, given by equation (6.8) in clause 6.4.2, then always exceeds 0.7k2, so ks in equation (b) can be replaced by 0.7k2. This replacement would not be valid for an encased web, so these are excluded. For a steel flange of breadth bf and thickness tf, Iafz = bf3tf /12

(c)

The stiffness k2 is given by equation (6.10) in clause 6.4.2. Using it and the equations above, 2

λLT

4

Ê M ˆ 48π2 hs (1 - υa 2 ) = Á Rk ˜ Ë kc ¯ 0.7Ea 2 tw 3 bf 3 tf C4 2

206



(d)



For sections in Class 1 or Class 2, MRk = Mpl, Rk. It can be shown that Mpl, Rk is given approximately by Mpl, Rk = kc Mpl, a, Rk(1 + tw hs /4bf tf)

(e)

For double-symmetrical steel I-sections, the plastic resistance to bending is given approximately by Mpl, a, Rk = fy hs bf tf (1 + tw hs /4bf tf)

(f)

From equations (d) to (f), with υa = 0.3, Ê t h ˆÊh ˆ λLT = 5.0 Á 1 + w s ˜ Á s ˜ 4 bf tf ¯ Ë tw ¯ Ë

0.75


0.25

Ê fy ˆ ÁË E C ˜¯ a 4

0.5

(D6.14)

as given in Annex B of ENV 1994-1-1.

C4

yM 0

y M0

0.5yM0

30

M0 is the bending moment at midspan when both ends are simply supported

yM0 0.75yM0 yM 0

yM 0

yM 0

20 y M0

yM 0

yÆ• 10

yÆ•

4 0.4

1.0

2.0

3.0

y

Fig. A.3. Values of the factor C4 for uniformly distributed and centre point loading

207





Effect of web encasement on λ LT

The reduction in relative slenderness achieved by encasing a steel web to clause 5.5.3(2) can be estimated as follows. The subscript e is used for properties of the section after encasement. From equations (6.10) and (6.11), k2, e k2

=

(1 - υa 2 )bf 2 4(1 + 4 ntw /bf )tw 2

The modular ratio n rarely exceeds 12, and bf /tw is at least 15 for rolled or welded I-sections. With these values, and νa = 0.3, k2, e /k2 = 12.2 Assuming, as above, that k1 > 2.3k2 and using equation (6.8), the change in ks is ks, e ks

=

k2, e k1 + k2, e

k1 + k2 12.2 ¥ (2.3 + 1) > = 2.78 k2 2.3 + 12.2

It was found above that ks can be replaced by 0.7k2, so ks, e is now replaced by 2.78 × 0.7k2 = 1.95k2. Thus, the divisor 0.7 in equation (d) above is replaced by 1.95. Hence,

C4

Uniform loading No transverse loading 30

Lc/L = 0.25 yM0

M0

L

Lc

20

M

yM

Lc/L = 0.50 10

Lc/L = 0.75

Lc/L = 1.00 M 4

0

0.5

yM 1.0

y

Fig. A.4. Values of the factor C4 for cantilevers, and for spans without transverse loading

208





F 16 457 × 152 UB

533 × 210 UB

Flim for uncased section, S 235

610 × 229 UB 14

S 275 457 × 191 UB

S 355

IPE sections 12

610 × 305 UB S 420, S 460 10

Does not qualify for grades S 420 and S 460

HEA sections 8

300

400

500

600

700

800

900 Depth, h (mm)

Fig. A.5. Property F (equation (D6.15)) for some IPE, HEA and UB steel sections

λLT, e /λLT = (0.7/1.95)0.25 = 0.77

(DA.4)

Factor C4 for the distribution of bending moment

The tables that were given in ENV 199449 relate to distributions of bending moment between points at which the steel bottom flange is laterally restrained, not necessarily to complete spans. The more commonly used values for distributed loading on internal spans are plotted in Fig. A.3. For values of ψ exceeding 3.0, values corresponding to ψ Æ • can conservatively be used. These are also shown. The dashed lines in Fig. A.3 are for point loads at mid-span. Two other sets of values are plotted in Fig. A.4. The solid lines apply for lateral buckling of a cantilever of length Lc, where both it and the adjacent span of length L have the same intensity of distributed loading. The dashed lines are for an unloaded span with one or both ends continuous.

Criteria for verification of lateral–torsional stability without direct calculation Unlike UB steel sections, the basic sets of IPE and HEA sections have only one size for each overall depth, h. Plots of their section properties F, from equation (D6.15), against h lie on straight lines, as shown in Fig. A.5. This enables limits on F to be presented in Table 6.1 of clause 6.4.3 as limits to overall depth. From equation (D6.14), Flim for given λLT is proportional to fy–0.5. From this, and the qualifying sections, it can be deduced that the values of Flim used in EN 1994-1-1 for the various grades of steel are as given in Table A.2.

209





Table A.2. Limiting section parameter Flim, for uncased and web-encased sections Nominal steel grade

S235

S275

S355

S420 and S460

Flim, uncased Flim, encased web

15.1 19.5

13.9 18.0

12.3 15.8

10.8 13.9

The IPE and HEA sections shown qualify for all steel grades that have Flim above their plotted value of F. The only exception is HEA 550, which plots just below the S420 and S460 line, but does not qualify according to Table 6.1. For UB sections, crosses in Fig. A.5 represent the 10 sections listed in Table 6.1. The entries ‘Yes’ in that table correspond to the condition F £ Flim. It is not possible to give a qualifying condition in terms of depth only; equation (D6.15) should be used.

Web encasement

From equation (DA.4), the effect of web encasement is to increase Flim by a factor of at least 1/0.77 = 1.29. These values are given in Tables 6.1 and A.2. The additional depths permitted by clause 6.4.3(1)(h) are a little more conservative than this result.

210




APPENDIX B

The effect of slab thickness on resistance of composite slabs to longitudinal shear Summary A mechanical model based on ductile shear connection has been applied to the m–k and partial connection methods of Section 9 for design of composite slabs for longitudinal shear. For the m–k method it was shown124 that: • • • • •

where the assumptions of the model apply, the two sets of tests from which m and k are derived can be done on sets of slabs of different thickness but similar concrete strength two widely different shear spans should be used predictions by the m–k method of EN 1994-1-1 for degrees of shear connection between those corresponding to the shear spans tested, are conservative predictions for degrees of shear connection outside this range are unconservative the percentage errors can be estimated. For the partial-connection method it was found124 that:

• •

where tests are done on slabs of one thickness only, the model gives no help in predicting the effect of slab thickness on ultimate shear strength τu slabs of at least two different thicknesses should be tested, preferably with the same shear span.

The model The notation and assumptions are generally those of clauses 9.7.3 and B.3, to which reference should be made. Figure B.1 shows the left-hand shear span of a composite slab of breadth b and effective depth dp, at failure in a test in accordance with clause B.3. The self-weight of the slab is neglected in comparison with Vt, the value of each of the two point loads at failure. The shear connection is assumed to be ductile, as defined in clause 9.7.3(3), with ultimate shear strength τu, as would be found by the procedure of clause B.3.6 (except that all values here are mean values, with no partial safety factors). The sheeting and the slab are shown separated in Fig. B.1, and the longitudinal shear force between them is





ηNc, f = τubLs

(DB.1)

where η is the degree of shear connection, £ 1. The value of τu is assumed to be independent of the shear span, so that Ls = ηLsf

(DB.2)

where Lsf is the length of shear span at which the longitudinal force Nc, f equals the tensile strength of the sheeting, Npl. In the absence of a partial safety factor for strength of concrete, the rectangular stress block is quite shallow, and is assumed to lie within the concrete slab. Its depth is ηxpl, where xpl is the depth for full shear connection. Let the plastic bending resistance of the sheeting be Mp, a, reduced to Mpr in the presence of an axial force N, as shown by the stress blocks in Fig. B.1. The resistance Mpr is assumed to be given by Mpr = (1 – η2)Mp, a

(DB.3)

The bilinear relationship given in clause 9.7.2(6) is an approximation to this equation, which is also approximate, but accurate enough for the present work. For equilibrium of the length Ls of composite slab, Vt Ls = ηNcf (dp – ηxpl /2) + Mpr Hence, Vt = [ηNcf (dp – ηxpl /2) + (1 – η2)Mp, a]/(ηLsf)

(DB.4)

For a typical profiled sheeting, Mp, a ª 0.3hpNpl. This is assumed here. The conclusions do not depend on the accuracy of the factor 0.3. Hence, Vt = (Ncf /ηLsf)[ηdp – η2 xpl /2 + 0.3 hp(1 – η2)]

(DB.5)

For a particular sheeting and strength of concrete, it may be assumed that Ncf, Lsf, xpl and hp are constant. The independent variables are the slab thickness, represented by dp, and the shear span in a test, represented by the degree of shear connection, η. The dependent variable is the vertical shear resistance, Vt.

The m–k method The use of test results as predictors

The properties m and k are determined from the graph shown in Fig. B.2, by drawing a line through two test results. The line is y = mx + k For a single result, (x1, y1), say, y1 = mx1 + k

hxpl

dp

hNcf

Vt

dp – hxpl/2 hNcf

Mpr

hp e Vt

hNcf

L = hLsf

Fig. B.1. Shear span of composite slab, and stress blocks at failure in longitudinal shear

212




APPENDIX B. SLAB THICKNESS AND RESISTANCE TO LONGITUDINAL SHEAR

y = Vt/bdp

Vtrue Vpred

0

x = Ap/bhLsf

Fig. B.2. Determination of m and k from two sets of test results

From the definitions of x and y, shown in Fig. B.2, and for this test result, Vt1 = bdp1(mAp /bLs1 + k) This is equation (9.7) in clause 9.7.3(4).Thus, the m–k method predicts exactly the first test result. This is so also for the second test result, even if the slab thickness and shear span are different. Two test results ‘predict’ themselves. The basic assumption of the m–k method is that other results can be predicted by assuming a straight line through the two known results. It has been shown that it is in fact curved, so that other predictions are subject to error.

Shape of function y(x)

The slope of the curve y(x) was found for a set of tests done with different shear spans and constant slab thickness. It was shown, by differentiation of equation (DB.5), that the curve through the two points found in tests is convex upwards, as shown in Fig. B.2. For some degree of shear connection between those used in two tests on slabs of the same thickness, from which the m–k line was predicted, the method gives the result Vpred, shown in Fig. B.2. This is less than the resistance given by equation (DB.5), Vtrue, so the method is safe, according to the model. For a degree of shear connection outside this range, the m–k method is unsafe.

Estimate of errors of prediction

As an example, suppose that for a set of tests with dp /hp = 2.0 the sheeting and concrete are such that when the shear span is Lsf, the depth xpl of the concrete stress block is given by xpl /hp = 0.4. Equation (DB.5) then becomes Vt = (Nc, f hp /Lsf)(0.3/η + 2 – 0.5η)

(DB.6)

Suppose that four otherwise identical tests are done with shear spans such that η = 0.4, 0.5, 0.7 and 1.0. The (assumed) true results for Vt Lsf /Nc, f hp are calculated and plotted against 1/η. They lie on a convex-upwards curve, as expected. By drawing lines through any two of the points, values for the other two tests predicted by the m–k method are obtained. Comparison with the plotted points gives the error from the m–k method, as a percentage. Typical results are given in Table B.1, for which the m–k line is drawn through the results for η1 and η2, and used to predict the shear resistance for a slab with η = η3. The values in columns 4 and 5 of the table are proportional to Vt, so the percentage values are correct.

213





Table B.1. Errors in prediction of Vt by the m–k method η1

η2

η3

Prediction from m–k line

Plotted value

Error of prediction (%)

0.4 0.4 0.7

1.0 0.5 1.0

0.7 1.0 0.4

2.00 1.98 2.78

2.08 1.80 2.55

–4 +10 +9

Conclusion for the m–k method

Two sets of tests should be done, with shear spans as widely different as possible, subject to obtaining the correct failure mode. The slab thickness should be roughly central within the range of application, and can differ in the two sets; but the concrete strengths should be the same. The m–k results are applicable within the range of shear spans tested, and probably for a short distance outside it.

The partial-connection method From Fig. B.1, the ultimate bending moment at the end of the shear span is M = VtηLsf

(DB.7)

The overhang L0 (clause B.3.6(2) and Fig. 9.2) is much less than Lsf. For simplicity it is assumed that Lsf + L0 ª Lsf, so that equation (B.2) in clause B.3.6 becomes τu = ηNc, f /bLsf

(DB.8)

From equations (DB.5) and (DB.7), M/Mp, Rm = (Nc, f /Mp, Rm)[ηdp – η2xpl /2 + 0.3hp(1 – η2)]

(DB.9)

where Mp, Rm is the plastic resistance moment with full shear connection. For any assumed value for η, an ultimate bending moment M can be calculated from equation (DB.9). Thus, an M–η curve (Fig. B.5 of EN 1994-1-1) can be found. If safety factors are omitted, the same curve is used to find ηtest from a measured value Mtest, and hence τu from equation (DB.8). Equation (DB.9) is independent of shear span because ductile behaviour is assumed. It gives no information on the rate of change of η with slab thickness or shear span, so a single group of four tests gives no basis for predicting τu for slabs of different thickness from those tested. Assuming that the fourth test, with a short shear span (clause B.3.2(7)), shows ductile behaviour, the effect of thickness can be deduced from results of a further group of three tests. The specimens and shear span should be identical with those in tests 1–3 in the first group, except for slab thickness. The thicknesses for the two groups should be near the ends of the range to be used in practice. Let the ratios dp/hp for these two series be denoted v1 and v2, with v2 > v1, leading to the corresponding degrees of shear connection η1 and η2. It is likely that η2 > η1, because the longitudinal strain across the depth of the embossments will be more uniform in the thicker slabs; but the difference may be small. Assuming that over this range the relationship between test bending resistance M and ratio v is linear, it can be shown124 that the η–v curve is convex upwards. Hence, interpolation for η is conservative for slab thicknesses between those tested, and may be unconservative outside this range.

Conclusion for the partial-connection method

For this method, further tests at constant thickness but different shear span would only provide a check on the presence of ductile behaviour. Information on the effect of slab

214




APPENDIX B. SLAB THICKNESS AND RESISTANCE TO LONGITUDINAL SHEAR

thickness is best obtained from tests at constant shear span and two thicknesses. It can be shown that values of degree of shear connection for intermediate thicknesses can be obtained by linear interpolation between the values for the thicknesses tested.

215




APPENDIX C

Simplified calculation method for the interaction curve for resistance of composite column cross-sections to compression and uniaxial bending Scope and method Equations are given for the coordinates of points B, C and D in Fig. 6.19, also shown in Fig. C.1. They are applicable to cross-sections of columns where the structural steel, concrete and reinforcement are all doubly symmetric about a single pair of axes. The steel section should be an I- or H-section or a rectangular or circular hollow section. Examples are shown in Fig. 6.17. N Npl, Rd A

Npm, Rd

C

Npm, Rd/2

0

D

B Mpl, Rd Mmax, Rd

Fig. C.1. Polygonal interaction curve



M



bc b

cy

cy cz

(1)

tf

B

B hn (2)

D

G

hn

C

h

y D

ez

hc

tw

C

cz

(1)

ey (a)

z

(b)

Fig. C.2. Composite cross-sections symmetrical about two axes

Plastic analysis is used, with rectangular stress blocks for structural steel, reinforcement, and concrete in accordance with clauses 6.7.3.2(2) to 6.7.3.2(6). For filled tubes of circular section, the coefficient ηc in clause 6.7.3.2(6) has conservatively been taken as zero. In this annex, the compressive stress in concrete in a rectangular stress block is denoted fcc, where generally, fcc = 0.85fcd. However, for concrete-filled steel sections, the coefficient 0.85 may be replaced by 1.0, following clause 6.7.3.2(1).

Resistance to compression The plastic resistance Npl, Rd is given by clause 6.7.3.2. The resistance Npm, Rd is calculated as follows. Figure C.2(a) represents a generalized cross-section of structural steel and reinforcement (shaded area), and of concrete, symmetrical about two axes through its centre of area G. For bending only (point B) the neutral axis is line BB which defines region (1) of the cross-section, within which concrete is in compression. The line CC at the same distance hn on the other side of G is the neutral axis for point C in Fig. C.1. This is because the areas of structural steel, concrete and reinforcement in region (2) are all symmetrical about G, so that the changes of stress when the axis moves from BB to CC add up to the resistance Npm, Rd, and the bending resistance is unchanged. Using subscripts 1 to 3 to indicate regions (1) to (3), Npm, Rd = Rc2 + 2|Ra2|

(C.1)

where Rc2 is the resistance of the concrete in region (2), and Ra2 is the resistance of the steel in region (2). In the notation of clause 6.7.3.2(1), Rc2 = Ac2 fcc Ra2 = Aa2 fyd + As2 fsd where compressive forces and strengths fcc, fsd and fyd are taken as positive. From symmetry, Ra1 = |Ra3|

(C.2)

Rc1 = Rc3

218




APPENDIX C. RESISTANCE OF COMPOSITE COLUMNS TO COMPRESSION AND UNIAXIAL BENDING

When the neutral axis is at BB, N = 0, so that Ra1 + Rc1 = |Ra2| + |Ra3|

(C.3)

From eqs (C.2) and (C.3), |Ra2| = Rc1 = Rc3. Substituting in equation(C.l), Npm, Rd = Rc2 + Rc1 + Rc3 = Rc

(C.4)

where Rc is the compressive resistance of the whole area of concrete, which is easily calculated.

Position of neutral axis Equations for hn depend on the axis of bending, the type of cross section and the cross section properties. The equations are derived from equations (C.1) and (C.4), and are given below for some cross sections.

Bending resistances The axial resistance at point D in Fig. C.1 is half that at point C, so the neutral axis for point D is line DD in Fig. C.2(a). The bending resistance at point D is Mmax, Rd = Wpa fyd + Wps fsd + Wpc fcc /2

(C.5)

where Wpa, Wps and Wpc are the plastic section moduli for the structural steel, the reinforcement and the concrete part of the section (for the calculation of Wpc the concrete is assumed to be uncracked), and fyd, fsd and fcc are the design strengths for the structural steel, the reinforcement and the concrete. The bending resistance at point B is Mpl, Rd = Mmax, Rd – Mn, Rd

(C.6)

with Mn, Rd = Wpa, n fyd + Wps, n fsd + Wpc, n fcc /2

(C.7)

where Wpa, n, Wps, n and Wpc, n are the plastic section moduli for the structural steel, the reinforcement and the concrete parts of the section within region (2) of Fig. C.2(a). Equations for the plastic section moduli of some cross-sections are given below.

Interaction with transverse shear If the shear force to be resisted by the structural steel is considered according to clause 6.7.3.2(4) the appropriate areas of steel should be assumed to resist shear alone. The method given here can be applied using the remaining areas.

Neutral axes and plastic section moduli of some cross-sections General

The compressive resistance of the whole area of concrete is Npm, Rd = Ac fcc

(C.8)

The value of the plastic section modulus of the total reinforcement is given by

219





n

Wps = Â|As, i ei|

(C.9)

i =1

where ei are the distances of the reinforcement bars of area As, i to the relevant middle line (y-axis or z-axis). The equations for the position of the neutral axis hn are given for selected positions in the cross-sections. The resulting value hn should lie within the limits of the assumed region.

Major-axis bending of encased I-sections

The plastic section modulus of the structural steel may be taken from tables, or be calculated from Wpa =

( h - 2 tf )2 tw + btf ( h - tf ) 4

(C.10)

bc hc 2 - Wpa - Wps 4

(C.11)

and Wpc =

For the different positions of the neutral axes, hn and Wpa, n are given by: (a) Neutral axis in the web, hn £ h/2 – tf: hn =

N pm, Rd - Asn (2fsd - fcc ) 2 bc fcc + 2 tw (2fyd - fcc )

Wpa, n = tw hn2

(C.12) (C.13)

where Asn is the sum of the area of reinforcing bars within the region of depth 2hn. (b) Neutral axis in the flange, h/2 – tf < hn < h/2: hn =

N pm, Rd - Asn (2fsd - fcc ) + ( b - tw )( h - 2 tf )(2fyd - fcc ) 2 bc fcc + 2 b(2fyd - fcc )

( b - tw )( h - 2 tf )2 4 (c) Neutral axis outside the steel section, h/2 £ hn £ hc/2: Wpa, n = bhn 2 -

hn =

N pm, Rd - Asn (2fsd - fcc ) - Aa (2fyd - fcc ) 2 bc fcc

Wpa, n = Wpa

(C.14) (C.15)

(C.16) (C.17)

The plastic modulus of the concrete in the region of depth from 2hn then results from Wpc, n = bc hn2 – Wpa, n – Wps, n

(C.18)

with n

Wps, n = Â|Asn, i ez, i|

(C.19)

i=1

where Asn, i are the areas of reinforcing bars within the region of depth 2hn, and ez, i are the distances from the middle line.

Minor-axis bending of encased I-sections

The notation is given in Fig. C.2(b). The plastic section modulus of the structural steel may be taken from tables or be calculated from

220





Wpa =

( h - 2 tf )tw 2 2 tf b2 + 4 4

(C.20)

and hc bc 2 - Wpa - Wps 4 For the different positions of the neutral axes, hn and Wpa, n are given by: Wpc =

(C.21)

(a) Neutral axis in the web, hn £ tw/2: hn =

N pm, Rd - Asn (2fsd - fcc )

(C.22)

2 hc fcc + 2 h(2fyd - fcc )

Wpa, n = hhn2

(C.23)

(b) Neutral axis in the flanges, tw/2 < hn < b/2: hn =

N pm, Rd - Asn (2fsd - fcc ) + tw (2 tf - h)(2fyd - fcc ) 2 hc fcc + 4 tf (2fyd - fcc )

( h - 2 tf )tw 2 4 (c) Neutral axis outside the steel section, b/2 £ hn £ bc/2 Wpa, n = 2 tf hn 2 +

hn =

N pm, Rd - Asn (2fsd - fcc ) - Aa (2fyd - fcc ) 2 hc fcc

Wpa, n = Wpa

(C.24) (C.25)

(C.26) (C.27)

The plastic modulus of the concrete in the region of depth 2hn then results from Wpc, n = hc hn2 – Wpa, n – Wps, n

(C.28)

with Wps, n according to equation (C.19), changing the subscript z to y.

Concrete-filled circular and rectangular hollow sections The following equations are derived for rectangular hollow sections with bending about the y-axis of the section (see Fig. C.3). For bending about the z-axis the dimensions h and b are to be exchanged as well as the subscripts z and y. Equations (C.29) to (C.33) may be used for circular hollow sections with good approximation by substituting h=b=d

and

r = d/2 – t

( b - 2 t )( h - 2 t )2 2 3 2 - r - r (4 - π)(0.5 h - t - r ) - Wps 4 3 with Wps according to equation (C.9). Wpa may be taken from tables, or be calculated from Wpc =

bh2 2 - ( r + t )3 - ( r + t )2 (4 - π)(0.5 h - t - r ) - Wpc - Wps 4 3 N pm, Rd - Asn (2fsd - fcc )

Wpa = hn =

2 bfcc + 4 t(2fyd - fcc )

Wpc, n = (b – 2t)hn2 – Wps, n

(C.29)

(C.30) (C.31) (C.32)

2

Wpa, n = bhn – Wpc, n – Wps, n

(C.33)

with Wps, n according to equation (C.19).

221





b ey

ey

t

t

r

h y

y

ez

ez

z

z (b) (a)

Fig. C.3. Concrete-filled (a) rectangular and (b) circular hollow sections, with notation

Example C.1: N–M interaction polygon for a column cross-section

The method of Appendix C is used to obtain the interaction polygon given in Fig. 6.38 for the concrete-encased H section shown in Fig. 6.37. The small area of longitudinal reinforcement is neglected. The data and symbols are as in Example 6.10 and Figs 6.37, C.1 and C.2. Design strengths of the materials: fyd = 355 N/mm2; fcd = 16.7 N/mm2. Other data: Aa = 11 400 mm2; Ac = 148 600 mm2; tf = 17.3 mm; tw = 10.5 mm; bc = hc = 400 mm; b = 256 mm; h = 260 mm; 10 –6Wpa, y = 1.228 mm 3; 10–6Wpa, z = 0.575 mm 3; Npl, Rd = 6156 kN.

Major-axis bending From equation (C.8), Npm, Rd = 148.6 × 16.7 = 2482 kN From equation (C.12), hn = 2482/[0.8 × 16.7 + 0.021 × (710 – 16.7)] = 89 mm so the neutral axis is in the web (Fig. C.4(a)), as assumed. From equation (C.11), the plastic section modulus for the whole area of concrete is 10–6Wpc = 43/4 – 1.228 = 14.77 mm3 From equation (C.13), 10–6Wpa, n = 10.5 × 0.0892 = 0.083 mm3 From equation (C.18), 10–6Wpc, n = 400 × 0.0892 – 0.083 = 3.085 mm3 From equation (C.5), Mmax, Rd = 1.228 × 355 + 14.77 × 16.7/2 = 559 kN m From equations (C.6) and (C.7), Mpl, Rd = 559 – (0.083 × 355 + 3.085 × 16.7/2) = 504 kN m The results shown above in bold type are plotted on Fig. 6.38.

222





D B 13.7 C

B

B 89

D

D 89

C

C

B 13.7 (b)

(a)

D

C

Fig. C.4. Neutral axes at points B, C and D on the interaction polygons

Minor-axis bending From equation (C.4), Npm, Rd is the same for both axes of bending. Thus, Npm, Rd = 2482 kN Assuming that the neutral axis B–B intersects the flanges, from equation (C.24), hn = [2482 – 0.0105 × (260 – 34.6) × (710 – 16.7)]/[0.8 × 16.7 + 0.0692 × (710 – 16.7)] = 13.7 mm so axis B–B does intersect the flanges (Fig. C.4(b)). From equation (C.21), 10–6Wpc = 43/4 – 0.575 = 15.42 mm3 From equation (C.25), 10–6Wpa, n = 34.6 × 0.01372 + 0.01052 × (260 – 34.6)/4 = 0.0127 mm3 From equation (C.28), 10–6Wpc, n = 400 × 0.01372 – 0.0127 = 0.0624 mm3 From equation (C.5), Mmax, Rd = 0.575 × 355 + 15.42 × 16.7/2 = 333 kN m From equation (C.7), Mn, Rd = 0.0127 × 355 + 0.0624 × 16.7/2 = 5.03 kN m From equation (C.6), Mpl, Rd = 333 – 5 = 328 kN m These results are plotted on Fig. 6.38, and used in Example 6.10.

223




References 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16.

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230




Index

Note: references to ‘beams’ and to ‘columns’ are to composite members

action effect see actions, effects of actions 6 combinations of 32–3, 128, 130 characteristic 33, 124–5, 136 frequent 33, 121, 124, 174 quasi-permanent 33, 91, 129, 134, 139 concentrated 36, 209 effects of 6 independent 104, 117 primary 9, 92 secondary 9, 92 second-order 22–3, 34 horizontal 27 indirect 9 see also fatigue adhesives 17 analysis, elastic, of cross-sections see beams; columns; etc. analysis, global 6, 21–40, 121, 156 cracked 13, 36 elastic 29–36, 90–1, 121, 128, 134, 156–7 elastic–plastic 105 first-order 22–3, 90 for composite slabs 21, 164, 169 for profiled sheeting 163 non-linear 33–4 of frames 23–4, 149 rigid-plastic 33–4, 36–8 second-order 22–4, 30, 32–4, 105, 108, 110 uncracked 36 see also cracking of concrete analysis, types of 6, 27–8 anchorage 76, 162, 167–9, 187 see also reinforcement Annex, National see National Annex application rules 5 axes 7, 106


beams 41–67 bending resistance of 39, 44–54 hogging 45, 50–4, 88 sagging 44–50, 88–9, 96–7 Class 1 or 2 9, 36, 38–9, 43, 45, 49, 68, 88, 130, 134, 207 Class 3 9, 36, 39, 50, 56, 59, 67, 70 Class 4 9, 28, 34, 36, 54, 58, 130 concrete-encased 4, 32, 34, 43, 57, 180–1 cross-sections of 41–2, 44 asymmetric 37–8 classification of 28, 37–40, 52–3, 87–8 critical 42–3, 69 effective 43, 90 elastic analysis of 44, 50, 54, 89–90 non-uniform 43 sudden change in 42 see also slabs, composite; slabs, concrete curved in plan 44 design procedure for 47–8 effective width of flanges of 28–9, 87, 138, 145 flexural stiffness of 32 haunched 73–4, 79, 81 in frames 32 L-section 72 of non-uniform section 4, 31, 34 shear connection for see shear connection shear resistance of 89 stresses in 136–7 see also analysis; buckling; cantilevers; cracking of concrete; deflections; fire, resistance to; imperfections; interaction; shear …; vibration; webs bending moments elastic critical 58–65, 94–6 in columns 34, 105, 110–11, 115 redistribution of 15, 34–7, 39, 66, 129, 163–4




bolts, fracture of 143 bolts, holes for 44 bolts, stiffness of 182 bond see shear connection box girders 4, 80 bracing to bottom flanges 60, 63, 100 breadth of flange, effective see beams; slabs, composite bridges, beams in 68, 120 British Standards 5 BS 5400 80, 120, 187–8 BS 5950 34, 36–7, 39, 71, 74, 80, 130, 162–3, 165, 168, 187–8, 193, 196 BS 8110 14–15, 164 buckling 22–4 in columns 23, 105–6, 108, 113 lateral–torsional 24, 34–6, 41, 45, 55, 57–66, 90, 93–6, 203–10 local 24, 28, 37, 39–40, 67, 80, 130, 164 see also beams, cross-sections of of reinforcement 44 of webs in shear 24, 34, 55–7, 89 see also loads, elastic critical buildings, high-rise 104, 128 calibration 14–15, 71, 132 cantilevers 58, 79, 93, 209 CEN (Comité Européen Normalisation), 1, 3, 10 Class of section see Beams, Class … Codes of Practice see British Standards columns 103–119, 128 bi-axial bending in 105, 108, 111, 117 concrete-encased 4, 105–6, 112, 147, 220–3 concrete-filled 14, 105–7, 112, 218, 221 creep in 31, 114, 118 cross-sections of interaction diagram for 104, 106–7, 115, 217–23 non-symmetrical 103–5 section moduli for 219–21 design method for 105–11 eccentricity of loading for 107 effective length of 103 effective stiffness of 28, 107, 110 high-strength steel in 44 load introduction in 104, 111–12 moment–shear interaction in 107, 110, 115 second-order effects in 110, 116–17 shear in 112, 219 slenderness of 107–8, 115 squash load of 106 steel contribution ratio for 104, 107, 114 transverse loading on 24, 118 see also buckling; bending moments; imperfections; loads, elastic critical; shear connection; stresses, residual compression members see columns

concrete lightweight-aggregate 13, 15–16, 85, 91, 104, 136, 191, 195 precast 80, 129, 205 properties of 13–16, 119, 147 strength classes for 13 strength of 10 stress block for 14–15 see also cracking of concrete; creep of concrete; elasticity, modulus of; shrinkage of concrete connections see joints connector modulus see shear connection, stiffness of construction 3, 78, 168 loads 30, 162–3, 168, 172 methods of 9 propped 78, 136, 138, 193 unpropped 13, 36–7, 39, 50, 59, 84, 128, 135 Construction Products Directive 10 contraflexure, points of 42 cover 19, 73, 77, 86, 106, 113 crack inducers 193, 195 cracking of concrete 121, 131–6, 158–9, 168 and global analysis 31–2, 90, 129–30 control of load-induced 134–5, 139 restraint-induced 132–3, 138 early thermal 133 uncontrolled 131–2 creep of concrete 15, 29–31, 49, 89, 91, 121 in columns 108 see also modular ratio; elasticity, modulus of critical length 42 cross-sections see beams; columns curvature of beams 49 decking, metal see sheeting, profiled steel definitions 6 deflections 128–30, 137–8 due to shrinkage 130, 137 due to slip 137 limits to 129, 137–8 of beams 128, 157–8 of composite slabs 168–70, 174 of profiled sheeting 163–4, 173 deformation, imposed 9, 131 design, basis of 9–11 design, methods of see beams; columns; slabs Designers’ Guides 1, 10 ductility see reinforcement, fracture of durability 19, 86 effective length see columns effective width see beams; slabs, composite effect of action see actions, effects of eigenvalue see load, elastic critical elasticity, modulus of for concrete 14, 29, 108 for steels 16, 29

232




INDEX

EN 1090 80 EN 1990 3, 91, 127–9, 131, 174, 190, 193 EN 1991 3, 32, 131, 172 EN 1992-1-1 5 and detailing 77, 132 and materials 13–16, 49, 91 and resistances 28, 81–2, 122, 128, 168, 175 and serviceability 131, 174 EN 1993-1-1 21, 23, 39, 58, 63, 205 EN 1993-1-3 16, 163–4, 168–9 EN 1993-1-5 28, 55 EN 1993-1-8 16, 22, 141–5, 150–4, 179–80 EN 1993-1-9 119–20, 122 EN 1993-2 123 EN 1994-1-1, scope of 3–5, 15 EN 1994-1-2 57, 134 EN 1994-2 4, 72, 103, 119 EN 1998 3 EN ISO 14555 72 ENV 1994-1-1 15, 60–2, 68, 80, 105, 194, 198, 204, 209 equilibrium, static 11 Eurocodes 1 European Standard 5 see also EN … examples bending and vertical shear 55–6 composite beam, continuous 84–100, 136–9 composite column 113–9, 222–3 composite joint 147–59, 181–5 composite slab 170–7, 194–201 effective width 29 elastic resistance to bending 101–3 fatigue 124–6 lateral–torsional buckling 66 reduction factor for strength of stud 76 resistance to hogging bending 50–4 shear connection 69–70, 100–1 transverse reinforcement 82 execution see construction exposure classes 19, 86, 113, 131 factors, combination 32 factors, conversion 10 factors, damage equivalent 123 factors, partial 2, 10, 119–20, 163, 190 factors, reduction 58, 72–6, 96, 106, 108, 181 fatigue 11, 34, 119–26 finite-element methods 33 fire, resistance to 57 flanges, concrete see beams; slabs flow charts 25–7, 35, 38, 64–5, 109 forces, internal 105 formwork, re-usable 129 see also sheeting, profiled steel foundations 11 frame, inverted-U, 59–61, 162, 204–6 frames, composite 6 braced 32, 147

unbraced 4, 32, 34, 108 see also analysis, global; buckling; imperfections geometrical data 10 see also imperfections haunches see beams, haunched hole-in-web method 38–9, 51–4 imperfections 21–2, 24–7, 90, 110, 115, 117 interaction, full 44 interaction, partial 43, 45–9, 129, 176–7 see also shear connection ISO standards 5–6 joints 16, 21–2, 28 beam-to-column 108, 141–59 bending resistance of 150–4 classification of 142, 148–9 contact-plate 142, 144–5, 180 end-plate 141, 143, 147–59, 181 full-strength 22, 143, 145 modelling of 143–7 nominally pinned 22, 136, 143, 145, 147, 153 partial-strength 2, 34, 37, 143, 145, 149 rigid 136, 142 rotational stiffness of 142, 144–5, 156, 179–85 rotation capacity of 144–5, 147, 156 semi-continuous 22 semi-rigid 22, 34, 136, 143, 149 simple see nominally pinned L-beams 78, 188 length, effective 103 limit states serviceability 28, 33, 127–39 ultimate 41–126 loads dynamic 161–2 elastic critical 23, 32, 103, 108, 114–15 imposed, for fatigue 120 see also actions m–k method 165–9, 175, 177, 191–4, 211–14 see also slabs, composite, tests on materials, properties of 13–17 see also concrete; steel mesh, welded see reinforcement, welded mesh modular ratio 29–31, 89–91, 121 modulus of elasticity see elasticity, modulus of moments see bending moments; torsion nationally determined parameter 1, 34 National Annexes 1–3, 5, 19, 22, 73, 128 and actions 32, 130, 162, 172 and beams 61, 96 and materials 14–17

233





and partial factors 10, 58, 119 and serviceability 131, 138, 169 notation see symbols notes, in Eurocodes 2, 10 partial factors 2, 9, 128 γM, for materials and resistances 7, 10, 123, 175 γF, for actions 123 plastic theory see analysis, global, rigid-plastic plate girders 80, 141 see also beams plates, composite 4 prestressing 4, 6, 13, 30, 33 principles 5–6 propping see construction, methods of prying 150, 152 push tests see shear connectors, tests on redistribution see bending moments; shear, longitudinal reference standards 5 reinforcement 131, 182–3 anchorage of 42, 78–9, 155 at joints 147–8 fracture of, in joints 39, 45, 132 in beams 52, 139 minimum area of 13, 36, 39, 131–5, 138 spacing of 132–3 transverse 68, 77, 81–3, 99–100, 162, 206 in columns 106, 112–13 in compression 44 in haunches see beams, haunched strain hardening in 45 welded mesh (fabric), 15–16, 39, 45, 130 see also cover; fatigue; slabs, composite reinforcing steel 15–16, 119 resistances 10, 27, 34 see also beams, bending resistance of; etc. rotation capacity 34–5, 37, 44, 164 see also joints safety factors see partial factors section modulus 7 sections see beams; columns; etc. separation 6, 44, 67–8, 77, 80, 112, 190 serviceability see limit states shakedown 121 shear see columns; shear, longitudinal; shear, vertical; etc. shear–bond test see m–k test shear connection 42–3, 67–81, 122, 183 and execution 78 by bond or friction 17, 67, 104, 112, 162, 166–7, 193–4 degree of 46, 69, 97, 101–3, 129, 166–7, 176 design of 44 detailing of 76–81

for composite slabs 165–8 full 45, 48, 145 in columns 104, 111–2, 118, 188 partial 43, 45–50, 57, 68–70, 119, 145 equilibrium method for 48–9 interpolation method for 47 see also anchorage; fatigue; reinforcement, in beams, transverse; shear connectors; slab, composite; slip, longitudinal shear connectors 17, 57 bi-axial loading of 76 ductile 46, 50, 67–9, 79, 97, 103, 191 flexibility of see stiffness of non-ductile 47, 49–50, 100–1, 103 spacing of 47–8, 50, 58, 69, 77, 80, 98, 101, 204–5 stiffness of 44, 60, 181 tests on 181, 187–91 types of 68 see also studs, welded shear flow 42–3, 81, 122 shear heads 4 shear lag see slabs, composite, effective width of shear, longitudinal 21, 70–1, 81–4 in columns 112 see also shear connection; slabs, composite shear, punching see slabs, composite shear, vertical 21, 54, 91–2, 144, 149, 153–4 and bending moment 55–7 see also buckling; slabs, composite sheeting, profiled steel 17, 72–6, 85, 161–2, 171–3, 212 and cracking 135 and shear connection 68, 80–1, 192 as transverse reinforcement 83–4, 99 bearing length for 162, 172 depth of 162 design of 162–4, 172–3 effective area of 45 embossments on (dimples in), 17, 162, 164, 191–3 fixing of 83 in compression 45 loading on 162–3 properties of 171, 193, 195 see also buckling; deflection; durability; slabs, composite shrinkage of concrete 15 and cracking 15, 131–2, 134 effects of 9, 30–1, 50, 89–93, 101, 112, 136–8, 158 see also deflections slabs, composite 161–77 as diaphragms 162 bending resistance of 165, 167, 173–4, 177 brittle failure of 162, 165–6, 187, 191, 193 concentrated loads on 164 cracking in 168, 175 effective thickness of 43, 170, 175

234




INDEX

effective width of 43, 164–5 flexural stiffness of 203–4 longitudinal shear in 165–8, 191, 211–15 partial-connection design of 166–8, 175–7, 188, 191–4, 198 partial shear connection in 39, 191–201, 211–12, 214–15 punching shear in 168 reinforcement in 132, 164, 167–8 serviceability of 168–70 shear span for 166 tests on 165–7, 169, 187–8, 191–8, 201 thickness of 132, 162, 176, 211–15 vertical shear in 166, 168, 172–3, 175 see also anchorage; deflections; global analysis; m–k method; sheeting, profiled steel slabs, concrete reinforcement in 68, 79 splitting in 72, 78–9, 81, 189 see also concrete, precast; slabs, composite slabs, form-reinforced see slabs, composite slenderness ratios, limiting see beams, cross-sections of slenderness, relative 58, 96, 103, 107–8, 206–9 slip, longitudinal 6, 23, 36, 44, 46, 49, 106, 111, 128, 144, 166, 188 and deflections 129, 132 available 46 capacity 67–9, 190–1 slip strain 46 software for EN 1994 23, 49, 105 squash load see columns, squash load of standards see British Standards; EN … standards, harmonized 10 steel see reinforcing steel; structural steels; yielding of steel steel contribution ratio see columns steelwork, protection of see durability stiffness coefficient 179–185 stiffness, flexural see beams; columns; etc. strength 10 see also resistance stresses excessive 128

fatigue 121–4 residual 22, 24 see also beams, stresses in stress range, damage equivalent 119, 122 stress resultant see actions, effects of structural steels 16, 34, 44, 111 stub girders 42 studs, lying 79 studs, welded 46 length after welding 71 resistance of in composite slabs 72–6, 81, 97–8, 161 in solid slabs 70–2 tension in 72 weld collar of 17, 68, 72, 99 see also fatigue; shear connection, detailing of; shear connectors subscripts 7 supports, friction at see shear connection sway frames see frames, unbraced symbols 6–7, 121, 143 temperature, effects of 9, 16, 32–3, 130, 132 tension stiffening 45, 122, 125, 134–5, 139 testing see shear connectors; slabs, composite through-deck welding see welding, through-deck torsion 44, 169 trusses 24 tubes, steel see columns, concrete-filled U-frame see frame, inverted-U uplift see separation vibration 128, 130–1 webs 66–7, 180–4 encased 4, 38–40, 60–1, 66, 134, 141, 204–10 holes in 4, 44 see also hole-in-web method; shear, vertical web stiffeners 60 welding, through-deck 72, 74, 83, 168 width, effective see beams; slabs, composite worked examples see examples yielding of steel and deflections 130, 136–7 yield line theory 151–2

235




DESIGNERS’ GUIDE TO EN 1994-2 EUROCODE 4: DESIGN OF STEEL AND COMPOSITE STRUCTURES PART 2: GENERAL RULES AND RULES FOR BRIDGES

Eurocode Designers’ Guide Series Designers’ Guide to EN 1990. Eurocode: Basis of Structural Design. H. Gulvanessian, J.-A. Calgaro and M. Holicky´. 0 7277 3011 8. Published 2002. Designers’ Guide to EN 1994-1-1. Eurocode 4: Design of Composite Steel and Concrete Structures. Part 1.1: General Rules and Rules for Buildings. R. P. Johnson and D. Anderson. 0 7277 3151 3. Published 2004. Designers’ Guide to EN 1997-1. Eurocode 7: Geotechnical Design – General Rules. R. Frank, C. Bauduin, R. Driscoll, M. Kavvadas, N. Krebs Ovesen, T. Orr and B. Schuppener. 0 7277 3154 8. Published 2004. Designers’ Guide to EN 1993-1-1. Eurocode 3: Design of Steel Structures. General Rules and Rules for Buildings. L. Gardner and D. Nethercot. 0 7277 3163 7. Published 2004. Designers’ Guide to EN 1992-1-1 and EN 1992-1-2. Eurocode 2: Design of Concrete Structures. General Rules and Rules for Buildings and Structural Fire Design. A.W. Beeby and R. S. Narayanan. 0 7277 3105 X. Published 2005. Designers’ Guide to EN 1998-1 and EN 1998-5. Eurocode 8: Design of Structures for Earthquake Resistance. General Rules, Seismic Actions, Design Rules for Buildings, Foundations and Retaining Structures. M. Fardis, E. Carvalho, A. Elnashai, E. Faccioli, P. Pinto and A. Plumier. 0 7277 3348 6. Published 2005. Designers’ Guide to EN 1995-1-1. Eurocode 5: Design of Timber Structures. Common Rules and for Rules and Buildings. C. Mettem. 0 7277 3162 9. Forthcoming: 2007 (provisional). Designers’ Guide to EN 1991-4. Eurocode 1: Actions on Structures. Wind Actions. N. Cook. 0 7277 3152 1. Forthcoming: 2007 (provisional). Designers’ Guide to EN 1996. Eurocode 6: Part 1.1: Design of Masonry Structures. J. Morton. 0 7277 3155 6. Forthcoming: 2007 (provisional). Designers’ Guide to EN 1991-1-2, 1992-1-2, 1993-1-2 and EN 1994-1-2. Eurocode 1: Actions on Structures. Eurocode 3: Design of Steel Structures. Eurocode 4: Design of Composite Steel and Concrete Structures. Fire Engineering (Actions on Steel and Composite Structures). Y. Wang, C. Bailey, T. Lennon and D. Moore. 0 7277 3157 2. Forthcoming: 2007 (provisional). Designers’ Guide to EN 1992-2. Eurocode 2: Design of Concrete Structures. Bridges. D. Smith and C. Hendy. 0 7277 3159 9. Forthcoming: 2007 (provisional). Designers’ Guide to EN 1993-2. Eurocode 3: Design of Steel Structures. Bridges. C. Murphy and C. Hendy. 0 7277 3160 2. Forthcoming: 2007 (provisional). Designers’ Guide to EN 1991-2, 1991-1-1, 1991-1-3 and 1991-1-5 to 1-7. Eurocode 1: Actions on Structures. Traffic Loads and Other Actions on Bridges. J.-A. Calgaro, M. Tschumi, H. Gulvanessian and N. Shetty. 0 7277 3156 4. Forthcoming: 2007 (provisional). Designers’ Guide to EN 1991-1-1, EN 1991-1-3 and 1991-1-5 to 1-7. Eurocode 1: Actions on Structures. General Rules and Actions on Buildings (not Wind). H. Gulvanessian, J.-A. Calgaro, P. Formichi and G. Harding. 0 7277 3158 0. Forthcoming: 2007 (provisional).

www.eurocodes.co.uk


DESIGNERS’ GUIDE TO EN 1994-2 EUROCODE 4: DESIGN OF STEEL AND COMPOSITE STRUCTURES PART 2: GENERAL RULES AND RULES FOR BRIDGES

C. R. HENDY and R. P. JOHNSON

Published by Thomas Telford Publishing, Thomas Telford Ltd, 1 Heron Quay, London E14 4JD URL: www.thomastelford.com

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First published 2006

Eurocodes Expert Structural Eurocodes offer the opportunity of harmonized design standards for the European construction market and the rest of the world. To achieve this, the construction industry needs to become acquainted with the Eurocodes so that the maximum advantage can be taken of these opportunities Eurocodes Expert is a new ICE and Thomas Telford initiative set up to assist in creating a greater awareness of the impact and implementation of the Eurocodes within the UK construction industry Eurocodes Expert provides a range of products and services to aid and support the transition to Eurocodes. For comprehensive and useful information on the adoption of the Eurocodes and their implementation process please visit our website or email [email protected]

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All rights, including translation, reserved. Except as permitted by the Copyright, Designs and Patents Act 1988, no part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise, without the prior written permission of the Publishing Director, Thomas Telford Publishing, Thomas Telford Ltd, 1 Heron Quay, London E14 4JD. This book is published on the understanding that the authors are solely responsible for the statements made and opinions expressed in it and that its publication does not necessarily imply that such statements and/or opinions are or reflect the views or opinions of the publishers. While every effort has been made to ensure that the statements made and the opinions expressed in this publication provide a safe and accurate guide, no liability or responsibility can be accepted in this respect by the authors or publishers.

Typeset by Academic þ Technical, Bristol Printed and bound in Great Britain by MPG Books, Bodmin

Preface EN 1994, also known as Eurocode 4 or EC4, is one standard of the Eurocode suite and describes the principles and requirements for safety, serviceability and durability of composite steel and concrete structures. It is subdivided into three parts: . . .

Part 1.1: General Rules and Rules for Buildings Part 1.2: Structural Fire Design Part 2: General Rules and Rules for Bridges.

It is used in conjunction with EN 1990, Basis of Structural Design; EN 1991, Actions on Structures; and the other design Eurocodes.

Aims and objectives of this guide The principal aim of this book is to provide the user with guidance on the interpretation and use of EN 1994-2 and to present worked examples. It covers topics that will be encountered in typical steel and concrete composite bridge designs, and explains the relationship between EN 1994-1-1, EN 1994-2 and the other Eurocodes. It refers extensively to EN 1992 (Design of Concrete Structures) and EN 1993 (Design of Steel Structures), and includes the application of their provisions in composite structures. Further guidance on these and other Eurocodes will be found in other Guides in this series.17 This book also provides background information and references to enable users of Eurocode 4 to understand the origin and objectives of its provisions. The need to use many Eurocode parts can initially make it a daunting task to locate information in the sequence required for a real design. To assist with this process, flow charts are provided for selected topics. They are not intended to give detailed procedural information for a specific design.

Layout of this guide EN 1994-2 has a foreword, nine sections, and an annex. This guide has an introduction which corresponds to the foreword of EN 1994-2, Chapters 1 to 9 which correspond to Sections 1 to 9 of the Eurocode, and Chapter 10 which refers to Annexes A and B of EN 1994-1-1 and covers Annex C of EN 1994-2. Commentary on Annexes A and B is given in the Guide by Johnson and Anderson.5 The numbering and titles of the sections and second-level clauses in this guide also correspond to those of the clauses of EN 1994-2. Some third-level clauses are also numbered (for example, 1.1.2). This implies correspondence with the sub-clause in EN 1994-2 of the same number. Their titles also correspond. There are extensive references to lower-level clause and paragraph numbers. The first significant reference is in bold italic type (e.g. clause 1.1.1(2)).

DESIGNERS’ GUIDE TO EN 1994-2

These are in strict numerical sequence throughout the book, to help readers find comments on particular provisions of the code. Some comments on clauses are necessarily out of sequence, but use of the index should enable these to be found. All cross-references in this guide to sections, clauses, sub-clauses, paragraphs, annexes, figures, tables and expressions of EN 1994-2 are in italic type, and do not include ‘EN 1994-2’. Italic is also used where text from a clause in EN 1994-2 has been directly reproduced. Cross-references to, and quotations and expressions from, other Eurocodes are in roman type. Clause references include the EN number; for example, ‘clause 3.1.4 of EN 1992-1-1’ (a reference in clause 5.4.2.2(2)). All other quotations are in roman type. Expressions repeated from EN 1994-2 retain their number. The authors’ expressions have numbers prefixed by D (for Designers’ Guide); for example, equation (D6.1) in Chapter 6. Abbreviated terms are sometimes used for parts of Eurocodes (e.g. EC4-1-1 for EN 19941-18 ) and for limit states (e.g. ULS for ultimate limit state).

Acknowledgements The first author would like to thank his wife, Wendy, and two boys, Peter Edwin Hendy and Matthew Philip Hendy, for their patience and tolerance of his pleas to finish ‘just one more paragraph’. He thanks his employer, Atkins, for providing both facilities and time for the production of this guide, and the members of BSI B525/10 Working Group 2 who provided comment on many of the Eurocode clauses. The second author is deeply indebted to the other members of the project and editorial teams for Eurocode 4 on which he has worked: David Anderson, Gerhard Hanswille, Bernt Johansson, Basil Kolias, Jean-Paul Lebet, Henri Mathieu, Michel Mele, Joel Raoul, Karl-Heinz Roik and Jan Stark; and also to the Liaison Engineers, National Technical Contacts, and others who prepared national comments. He thanks the University of Warwick for facilities provided for Eurocode work, and, especially, his wife Diana for her unfailing support. Chris Hendy Roger Johnson

vi

Contents Preface Aims and objectives of this guide Layout of this guide Acknowledgements

v v v vi

Additional information specific to EN 1994-2

1 2

Chapter 1.

General 1.1. Scope 1.1.1. Scope of Eurocode 4 1.1.2. Scope of Part 1.1 of Eurocode 4 1.1.3. Scope of Part 2 of Eurocode 4 1.2. Normative references 1.3. Assumptions 1.4. Distinction between principles and application rules 1.5. Definitions 1.5.1. General 1.5.2. Additional terms and definitions 1.6. Symbols

3 3 3 3 4 5 7 7 8 8 8 8

Chapter 2.

Basis of design 2.1. Requirements 2.2. Principles of limit states design 2.3. Basic variables 2.4. Verification by the partial factor method 2.4.1. Design values 2.4.2. Combination of actions 2.4.3. Verification of static equilibrium (EQU)

11 11 12 12 12 12 15 15

Chapter 3.

Materials 3.1. Concrete 3.2. Reinforcing steel for bridges 3.3. Structural steel for bridges 3.4. Connecting devices 3.4.1. General 3.4.2. Headed stud shear connectors

17 17 19 21 22 22 22

Introduction


viii

3.5. Prestressing steel and devices 3.6. Tension components in steel

23 23

Chapter 4.

Durability 4.1. General 4.2. Corrosion protection at the steel–concrete interface in bridges

25 25 27

Chapter 5.

Structural analysis 5.1. Structural modelling for analysis 5.1.1. Structural modelling and basic assumptions 5.1.2. Joint modelling 5.1.3. Ground–structure interaction 5.2. Structural stability 5.2.1. Effects of deformed geometry of the structure 5.2.2. Methods of analysis for bridges 5.3. Imperfections 5.3.1. Basis 5.3.2. Imperfections for bridges 5.4. Calculation of action effects 5.4.1. Methods of global analysis Example 5.1: effective widths of concrete flange for shear lag 5.4.2. Linear elastic analysis Example 5.2: modular ratios for long-term loading and for shrinkage Example 5.3: primary effects of shrinkage 5.4.3. Non-linear global analysis for bridges 5.4.4. Combination of global and local action effects 5.5. Classification of cross-sections Example 5.4: classification of composite beam section in hogging bending Flow charts for global analysis

29 29 29 30 30 30 31 33 34 34 35 36 36 41 42 53 54 56 56 57 60 62

Chapter 6.

Ultimate limit states 6.1. Beams 6.1.1. Beams in bridges – general 6.1.2. Effective width for verification of cross-sections 6.2. Resistances of cross-sections of beams 6.2.1. Bending resistance Example 6.1: plastic resistance moment in sagging bending Example 6.2: resistance to hogging bending at an internal support Example 6.3: elastic bending resistance of a Class 4 cross-section 6.2.2. Resistance to vertical shear Example 6.4: resistance of a Class 4 section to hogging bending and vertical shear Example 6.5: addition of axial compression to a Class 4 cross-section 6.3. Filler beam decks 6.3.1. Scope 6.3.2. General 6.3.3. Bending moments 6.3.4. Vertical shear 6.3.5. Resistance and stability of steel beams during execution 6.4. Lateral–torsional buckling of composite beams 6.4.1. General 6.4.2. Beams in bridges with uniform cross-sections in Class 1, 2 and 3

67 67 67 68 68 69 72 73 77 79 85 86 89 89 90 90 91 91 91 91 92

CONTENTS

Chapter 7.

Chapter 8.

6.4.3. General methods for buckling of members and frames Example 6.6: bending and shear in a continuous composite beam Example 6.7: stiffness and required resistance of cross-bracing 6.5. Transverse forces on webs 6.6. Shear connection 6.6.1. General Example 6.8: shear resistance of a block connector with a hoop 6.6.2. Longitudinal shear force in beams for bridges 6.6.3. Headed stud connectors in solid slabs and concrete encasement 6.6.4. Headed studs that cause splitting in the direction of the slab thickness 6.6.5. Detailing of the shear connection and influence of execution 6.6.6. Longitudinal shear in concrete slabs Example 6.9: transverse reinforcement for longitudinal shear Example 6.10: longitudinal shear checks Example 6.11: influence of in-plane shear in a compressed flange on bending resistances of a beam 6.7. Composite columns and composite compression members 6.7.1. General 6.7.2. General method of design 6.7.3. Simplified method of design 6.7.4. Shear connection and load introduction 6.7.5. Detailing provisions Example 6.12: concrete-filled tube of circular cross-section 6.8. Fatigue 6.8.1. General 6.8.2. Partial factors for fatigue assessment of bridges 6.8.3. Fatigue strength 6.8.4. Internal forces and fatigue loadings 6.8.5. Stresses 6.8.6. Stress ranges 6.8.7. Fatigue assessment based on nominal stress ranges Example 6.13: fatigue verification of studs and reinforcement 6.9. Tension members in composite bridges

93 104 111 113 114 114 116 118

Serviceability limit states 7.1. General 7.2. Stresses 7.3. Deformations in bridges 7.3.1. Deflections 7.3.2. Vibrations 7.4. Cracking of concrete 7.4.1. General 7.4.2. Minimum reinforcement 7.4.3. Control of cracking due to direct loading 7.5. Filler beam decks Example 7.1: checks on serviceability stresses, and control of cracking

163 163 164 166 166 166 167 167 168 169 173

Precast concrete slabs in composite bridges 8.1. General 8.2. Actions

179 179 180

121 123 124 127 130 131 134 136 136 137 138 144 145 145 150 150 151 152 152 153 155 156 157 161

173

ix


Chapter 9.

Chapter 10.

x

8.3. Design, analysis and detailing of the bridge slab 8.4. Interface between steel beam and concrete slab

180 181

Composite plates in bridges 9.1. General 9.2. Design for local effects 9.3. Design for global effects 9.4. Design of shear connectors Example 9.1: design of shear connection for global effects at the serviceability limit state

183 183 183 184 185

Annex C (informative). Headed studs that cause splitting forces in the direction of the slab thickness C.1. Design resistance and detailing C.2. Fatigue strength Applicability of Annex C Example 10.1: design of lying studs

187

189 190 191 191 192

References

195

Index

201

Introduction The provisions of EN 1994-29 are preceded by a foreword, most of which is common to all Eurocodes. This Foreword contains clauses on: . . . . . .

the background to the Eurocode programme the status and field of application of the Eurocodes national standards implementing Eurocodes links between Eurocodes and harmonized technical specifications for products additional information specific to EN 1994-2 National Annex for EN 1994-2.

Guidance on the common text is provided in the introduction to the Designers’ Guide to EN 1990. Eurocode: Basis of Structural Design,1 and only background information relevant to users of EN 1994-2 is given here. It is the responsibility of each national standards body to implement each Eurocode part as a national standard. This will comprise, without any alterations, the full text of the Eurocode and its annexes as published by the European Committee for Standardisation, CEN (from its title in French). This will usually be preceded by a National Title Page and a National Foreword, and may be followed by a National Annex. Each Eurocode recognizes the right of national regulatory authorities to determine values related to safety matters. Values, classes or methods to be chosen or determined at national level are referred to as Nationally Determined Parameters (NDPs). Clauses in which these occur are listed in the Foreword. NDPs are also indicated by notes immediately after relevant clauses. These Notes give recommended values. Many of the values in EN 1994-2 have been in the draft code for over a decade. It is expected that most of the 28 Member States of CEN (listed in the Foreword) will specify the recommended values, as their use was assumed in the many calibration studies done during drafting. They are used in this guide, as the National Annex for the UK was not available at the time of writing. Each National Annex will give or cross-refer to the NDPs to be used in the relevant country. Otherwise the National Annex may contain only the following:10 . .

decisions on the use of informative annexes, and references to non-contradictory complementary information to assist the user to apply the Eurocode.

Each national standards body that is a member of CEN is required, as a condition of membership, to withdraw all ‘conflicting national standards’ by a given date, that is at present March 2010. The Eurocodes will supersede the British bridge code, BS 5400,11 which should therefore be withdrawn. This will lead to extensive revision of many sets of supplementary design rules, such as those published by the Highways Agency in the UK. Some countries have already adopted Eurocode methods for bridge design; for example, Germany in 2003.12

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Additional information specific to EN 1994-2 The information specific to EN 1994-2 emphasises that this standard is to be used with other Eurocodes. The standard includes many cross-references to particular clauses in EN 1990,13 EN 1991,14 EN 199215 and EN 1993.16 Similarly, this guide is one of a series on Eurocodes, and is for use with other guides, particularly those for EN 1991,2 EN 1992-1-1,6 EN 1993-11,7 EN 1992-23 and EN 1993-2.4 The Foreword refers to a difference between EN 1994-2 and the ‘bridge’ parts of the other Eurocodes. In Eurocode 4, the ‘general’ provisions of Part 1-1 are repeated word for word in Part 2, with identical numbering of clauses, paragraphs, equations, etc. Such repetition breaks a rule of CEN, and was permitted, for this code only, to shorten chains of crossreferences, mainly to Eurocodes 2 and 3. This determined the numbering and location of the provisions for bridges, and led to a few gaps in the sequences of numbers. The same policy has been followed in the guides on Eurocode 4. Where material in the Designers’ Guide to EN 1994-1-15 is as relevant to bridges as to buildings, it is repeated here, so this guide is self-contained, in respect of composite bridges, as is EN 1994-2. A very few ‘General’ clauses in EN 1994-1-1 are not applicable to bridges. They have been replaced in EN 1994-2 by clearly labelled ‘bridge’ clauses; for example, clause 3.2, ‘Reinforcing steel for bridges’. The Foreword lists the 15 clauses of EN 1994-2 in which national choice is permitted. Five of these relate to values for partial factors, three to shear connection, and seven to provision of ‘further guidance’. Elsewhere, there are cross-references to clauses with NDPs in other codes; for example, partial factors for steel and concrete, and values that may depend on climate, such as the free shrinkage of concrete. Otherwise, the Normative rules in the code must be followed, if the design is to be ‘in accordance with the Eurocodes’. In EN 1994-2, Sections 1 to 9 are Normative. Only its Annex C is ‘Informative’, because it is based on quite recent research. A National Annex may make it normative in the country concerned, and is itself normative in that country, but not elsewhere. The ‘non-contradictory complementary information’ referred to above could include, for example, reference to a document based on provisions of BS 5400 on matters not treated in the Eurocodes. Each country can do this, so some aspects of the design of a bridge will continue to depend on where it is to be built.

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CHAPTER 1

General This chapter is concerned with the general aspects of EN 1994-2, Eurocode 4: Design of Composite Steel and Concrete Structures, Part 2: General Rules and Rules for Bridges. The material described in this chapter is covered in Section 1, in the following clauses: . . . . . .

Scope Normative references Assumptions Distinction between principles and application rules Definitions Symbols


1.1. Scope 1.1.1. Scope of Eurocode 4 The scope of EN 1994 (all three Parts) is outlined in clause 1.1.1. It is to be used with EN 1990, Eurocode: Basis of Structural Design, which is the head document of the Eurocode suite, and has an Annex A2, ‘Application for bridges’. Clause 1.1.1(2) emphasizes that the Eurocodes are concerned with structural behaviour and that other requirements, e.g. thermal and acoustic insulation, are not considered. The basis for verification of safety and serviceability is the partial factor method. EN 1990 recommends values for load factors and gives various possibilities for combinations of actions. The values and choice of combinations are set by the National Annex for the country in which the structure is to be constructed. Eurocode 4 is also to be used in conjunction with EN 1991, Eurocode 1: Actions on Structures14 and its National Annex, to determine characteristic or nominal loads. When a composite structure is to be built in a seismic region, account needs to be taken of EN 1998, Eurocode 8: Design of Structures for Earthquake Resistance.17 Clause 1.1.1(3), as a statement of intention, gives undated references. It supplements the normative rules on dated reference standards, given in clause 1.2, where the distinction between dated and undated standards is explained. The Eurocodes are concerned with design and not execution, but minimum standards of workmanship are required to ensure that the design assumptions are valid. For this reason, clause 1.1.1(3) lists the European standards for the execution of steel structures and the execution of concrete structures. The standard for steel structures includes some requirements for composite construction – for example, for the testing of welded stud shear connectors.


Clause 1.1.1(3)

1.1.2. Scope of Part 1.1 of Eurocode 4 The general rules referred to in clause 1.1.2(1) appear also in EN 1994-2, so there is (in general) no need for it to cross-refer to Part 1-1, though it does refer (in clause 6.6.3.1(4))

Clause 1.1.2(1)


Clause 1.1.2(2)

to Annex B of Part 1-1. The list of the titles of sections in clause 1.1.2(2) is identical to that in clause 1.1.3, except for those of Sections 8 and 9. In Sections 1–7 of EN 1994-2, all ‘for buildings’ clauses of EN 1994-1-1 are omitted, and ‘for bridges’ clauses are added.

1.1.3. Scope of Part 2 of Eurocode 4 Clause 1.1.3(1)

Clause 1.1.3(2)

Clause 1.1.3(1) refers to the partial coverage of design of cable-stayed bridges. This is the only reference to them in EN 1994-2. It was considered here, and in EC2 and EC3, that for this rapidly evolving type of bridge, it was premature to codify much more than the design of their components (e.g. cables, in EN 1993-1-11), although EN 1993-1-11 does contain some requirements for global analysis. Composite construction is attractive for cable-stayed bridges, because the concrete deck is well able to resist longitudinal compression. There is an elegant example in central Johannesburg.18 Clause 1.1.3(2) lists the titles of the sections of Part 2. Those for Sections 1–7 are the same as in all the other material-dependent Eurocodes. The contents of Sections 1 and 2 similarly follow an agreed model. The provisions of Part 2 cover the design of the following: . . . . . .

beams in which a steel section acts compositely with concrete concrete-encased or concrete-filled composite columns composite plates (where the steel member is a flat steel plate, not a profiled section) composite box girders tapered or non-uniform composite members structures that are prestressed by imposed deformations or by tendons.

Joints in composite beams and between beams and steel or composite columns appear in clause 5.1.2, Joint modelling, which refers to EN 1993-1-8.19 There is little detailed coverage, because the main clauses on joints in Part 1-1 are ‘for buildings’. Section 5, Structural analysis concerns connected members and frames, both unbraced and braced. The provisions define their imperfections and include the use of second-order global analysis and prestress by imposed deformations. The scope of Part 2 includes double composite action, and also steel sections that are partially encased. The web of the steel section is encased by reinforced concrete, and shear connection is provided between the concrete and the steel. This is a well-established form of construction in buildings. The primary reason for its choice is improved resistance in fire. Fully-encased composite beams are not included because: .

.

no satisfactory model has been found for the ultimate strength in longitudinal shear of a beam without shear connectors it is not known to what extent some design rules (e.g. for moment–shear interaction and redistribution of moments) are applicable.

A fully-encased beam with shear connectors can usually be designed as if partly encased or uncased, provided that care is taken to prevent premature spalling of encasement in compression. Prestressing of composite members by tendons is rarely used, and is not treated in detail. Transverse prestress of a deck slab is covered in EN 1992-2.3 The omission of application rules for a type of member or structure should not prevent its use, where appropriate. Some omissions are deliberate, to encourage the use of innovative design, based on specialised literature, the properties of materials, and the fundamentals of equilibrium and compatibility. However, the principles given in the relevant Eurocodes must still be followed. This applies, for example, to: . .

members of non-uniform section, or curved in plan types of shear connector other than welded headed studs.

EN 1994-2 has a single Informative annex, considered in Chapter 10 of this book. The three annexes in EN 1994-1-1 were not copied into EN 1994-2 because they are ‘Informative’ and, except for tests on shear connectors, are for buildings. They are:

4

CHAPTER 1. GENERAL

. . .

Annex A, Stiffness of joint components in buildings Annex B, Standard tests (for shear connectors and for composite slabs) Annex C, Shrinkage of concrete for composite structures for buildings.

In ENV 1994-1-1,20 design rules for many types of shear connector were given. All except those for welded headed studs were omitted, clause 1.1.3(3), mainly in response to requests for a shorter code. The Note to this clause enables national annexes to refer to rules for any type of shear connector. In the UK, this is being done for block connectors with hoops and for channels, and in France for angle connectors, based on the rules in ENV 1994-1-1. Research on older types of connector and the development of new connectors continues.21ÿ25

Clause 1.1.3(3)

1.2. Normative references References are given only to other European standards, all of which are intended to be used as a package. Formally, the Standards of the International Organization for Standardization (ISO) apply only if given an EN ISO designation. National standards for design and for products do not apply if they conflict with a relevant EN standard. As Eurocodes may not cross-refer to national standards, replacement of national standards for products by EN or ISO standards is in progress, with a timescale similar to that for the Eurocodes. During the period of changeover to Eurocodes and EN standards it is possible that an EN referred to, or its national annex, may not be complete. Designers who then seek guidance from national standards should take account of differences between the design philosophies and safety factors in the two sets of documents. The lists in clause 1.2 are limited to standards referred to in the text of EN 1994-1-1 or 1994-2. The distinction between dated and undated references should be noted. Any relevant provision of the general reference standards, clause 1.2.1, should be assumed to apply. EN 1994-2 is based on the concept of the initial erection of structural steel members, which may include prefabricated concrete-encased members. The placing of formwork (which may or may not become part of the finished structure) follows. The addition of reinforcement and in situ concrete completes the composite structure. The presentation and content of EN 19942 therefore relate more closely to EN 1993 than to EN 1992. This may explain why this list includes execution of steel structures, but not EN 13670, on execution of concrete structures, which is listed in clause 1.1.1.

Clause 1.2 Clause 1.2.1

Table 1.1. References to EN 1992, Eurocode 2: Design of Concrete Structures Title of Part

Subjects referred to from EN 1994-2

EN 1992-1-1, General Rules and Rules for Buildings

Properties of concrete, reinforcement, and tendons General design of reinforced and prestressed concrete Partial factors M , including values for fatigue Resistance of reinforced concrete cross-sections to bending and shear Bond, anchorage, cover, and detailing of reinforcement Minimum areas of reinforcement; crack widths in concrete Limiting stresses in concrete, reinforcement and tendons Combination of actions for global analysis for fatigue Fatigue strengths of concrete, reinforcement and tendons Reinforced concrete and composite tension members Transverse reinforcement in composite columns Vertical shear and second-order effects in composite plates Effective areas for load introduction into concrete

EN 1992-2, Rules for Bridges

Many subjects with references also to EN 1992-1-1 (above) Environmental classes; exposure classes Limitation of crack widths Vertical shear in a concrete flange Exemptions from fatigue assessment for reinforcement and concrete Verification for fatigue; damage equivalent factors

5


Clause 1.2.2

Clause 1.2.3

The ‘other reference standards’ in clause 1.2.2 receive both general references, as in clause 2.3.2(1) (to EN 1992-1-115 ), and specific references to clauses, as in clause 3.1(1), which refers to EN 1992-1-1, 3.1. For composite bridges, further standards, of either type, are listed in clause 1.2.3. For actions, the main reference is in clause 2.3.1(1), to ‘the relevant parts of EN 1991’, which include those for unit weights of materials, wind loads, snow loads, thermal actions, and actions during execution. The only references in clause 1.2 are to EN 1991-2, ‘Traffic loads on bridges’,26 and to Annex A2 of EN 1990, which gives combination rules and recommended values for partial factors and combination factors for actions for bridges. EN 1990 is also referred to for modelling of structures for analysis, and general provisions on serviceability limit states and their verification.

Cross-references from EN 1994-2 to EN 1992 and EN 1993 The parts of EN 1992 and EN 1993 most likely to be referred to in the design of a steel and concrete composite bridge are listed in Tables 1.1 and 1.2, with the relevant aspects of design. Table 1.2. References to EN 1993, Eurocode 3: Design of Steel Structures

6

Title of Part

Subjects referred to from EN 1994-2

EN 1993-1-1, General Rules and Rules for Buildings

Stress–strain properties of steel; M for steel General design of unstiffened steelwork Classification of cross-sections Resistance of composite sections to vertical shear Buckling of members and frames; column buckling curves

EN 1993-1-5, Plated Structural Elements

Design of cross-sections in slenderness Class 3 or 4 Effects of shear lag in steel plate elements Design of beams before a concrete flange hardens Design where transverse, longitudinal, or bearing stiffeners are present Transverse distribution of stresses in a wide flange Shear buckling; flange-induced web buckling In-plane transverse forces on webs

EN 1993-1-8, Design of Joints

Modelling of flexible joints in analysis Design of joints and splices in steel and composite members Design using structural hollow sections Fasteners and welding consumables

EN 1993-1-9, Fatigue Strength of Steel Structures

Fatigue loading Classification of details into fatigue categories Limiting stress ranges for damage-equivalent stress verification Fatigue verification in welds and connectors

EN 1993-1-10, Material Toughness and Through-thickness Properties

For selection of steel grade (Charpy test, and Z quality)

EN 1993-1-11, Design of Structures with Tension Components

Design of bridges with external prestressing or cable support, such as cable-stayed bridges

EN 1993-2, Rules for Bridges

Global analysis; imperfections Buckling of members and frames Design of beams before a concrete flange hardens Limiting slenderness of web plates Distortion in box girders M for fatigue strength; F for fatigue loading Damage equivalent factors Limiting stresses in steel; fatigue in structural steel Limits to deformations Vibration

CHAPTER 1. GENERAL

Many references to EN 1992-227 and EN 1993-228 lead to references from them to EN 1992-1-1 and EN 1993-1-1, respectively. Unfortunately, the method of drafting of these two bridge parts was not harmonised. For many subjects, some of the clauses needed are ‘general’ and so are located in Part 1-1, and others are ‘for bridges’ and will be found in Part 2. There are examples in clauses 3.2(1), 7.2.2(2) and 7.4.1(1). Other Eurocode parts that may be applicable are: EN EN EN EN EN

1993-1-7 1993-1-12 1997 1998 1999

Strength and Stability of Planar Plated Structures Transversely Loaded Supplementary Rules for High Strength Steel Geotechnical Design, Parts 1 and 2 Design of Structures for Earthquake Resistance Design of Aluminium Structures.

1.3. Assumptions It is assumed in EN 1994-2 that the general assumptions of ENs 1990, 1992, and 1993 will be followed. Commentary on them will be found in the relevant Guides of this series. Various clauses in EN 1994-2 assume that EN 1090 will be followed in the fabrication and erection of the steelwork. This is important for the design of slender elements, where the methods of analysis and buckling resistance formulae rely on imperfections from fabrication and erection being limited to the levels in EN 1090. EN 1994-2 should therefore not be used for design of bridges that will be fabricated or erected to specifications other than EN 1090, without careful comparison of the respective requirements for tolerances and workmanship. Similarly, the requirements of EN 13670 for execution of concrete structures should be complied with in the construction of reinforced or prestressed concrete elements.

1.4. Distinction between principles and application rules Clauses in the Eurocodes are set out as either Principles or Application Rules. As defined by EN 1990: .

. .

‘Principles comprise general statements for which there is no alternative and requirements and analytical models for which no alternative is permitted unless specifically stated.’ ‘Principles are distinguished by the letter ‘‘P’’ following the paragraph number.’ ‘Application Rules are generally recognised rules which comply with the principles and satisfy their requirements.’

There may be other ways to comply with the Principles, that are at least equivalent to the Application Rules in respect of safety, serviceability, and durability. However, if these are substituted, the design cannot be deemed to be fully in accordance with the Eurocodes. Eurocodes 2, 3 and 4 are consistent in using the verbal form ‘shall’ only for a Principle. Application rules generally use ‘should’ or ‘may’, but this is not fully consistent. There are relatively few Principles in Parts 1.1 and 2 of ENs 1992 and 1994. Almost all of those in EN 1993-1-1 and EN 1993-2 were replaced by Application Rules at a late stage of drafting. It has been recognized that a requirement or analytical model for which ‘no alternative is permitted unless specifically stated’ can rarely include a numerical value, because most values are influenced by research and/or experience, and may change over the years. (Even the specified elastic modulus for structural steel is an approximate value.) Furthermore, a clause cannot be a Principle if it requires the use of another clause that is an Application Rule; effectively that clause also would become a Principle. It follows that, ideally, the Principles in all the codes should form a consistent set, referring only to each other, and intelligible if all the Application Rules were deleted. This overriding principle strongly influenced the drafting of EN 1994.

7


1.5. Definitions 1.5.1. General

Clause 1.5.1(1)

In accordance with the model specified for Section 1, reference is made to the definitions given in clauses 1.5 of EN 1990, EN 1992-1-1, and EN 1993-1-1. Many types of analysis are defined in clause 1.5.6 of EN 1990. It should be noted that an analysis based on the deformed geometry of a structure or element under load is termed ‘second-order’, rather than ‘non-linear’. The latter term refers to the treatment of material properties in structural analysis. Thus, according to EN 1990, ‘non-linear analysis’ includes ‘rigid-plastic’. This convention is not followed in EN 1994-2, where the heading ‘Non-linear global analysis for bridges’ (clause 5.4.3) does not include ‘rigid-plastic global analysis’. There is no provision for use of the latter in bridges, so relevant rules are found in the ‘buildings’ clause 5.4.5 of EN 1994-1-1. References from clause 1.5.1(1) include clause 1.5.2 of EN 1992-1-1, which defines prestress as an action caused by the stressing of tendons. This is not sufficient for EN 1994-2, because prestress by jacking at supports, which is outside the scope of EN 1992-1-1, is within the scope of EN 1994-2. The definitions in clauses 1.5.1 to 1.5.9 of EN 1993-1-1 apply where they occur in clauses in EN 1993 to which EN 1994 refers. None of them uses the word ‘steel’.

1.5.2. Additional terms and definitions Clause 1.5.2

Most of the 15 definitions in clause 1.5.2 include the word ‘composite’. The definition of ‘shear connection’ does not require the absence of separation or slip at the interface between steel and concrete. Separation is always assumed to be negligible, but explicit allowance may need to be made for effects of slip, for example in clauses 5.4.3, 6.6.2.3 and 7.2.1. The definition of ‘composite frame’ is relevant to the use of Section 5. Where the behaviour is essentially that of a reinforced or prestressed concrete structure, with only a few composite members, global analysis should be generally in accordance with EN 1992. These lists of definitions are not exhaustive, because all the codes use terms with precise meanings that can be inferred from their contexts. Concerning use of words generally, there are significant differences from British codes. These arose from the use of English as the base language for the drafting process, and the resulting need to improve precision of meaning, to facilitate translation into other European languages. In particular: . .

‘action’ means a load and/or an imposed deformation ‘action effect’ (clause 5.4) and ‘effect of action’ have the same meaning: any deformation or internal force or moment that results from an action.

1.6. Symbols The symbols in the Eurocodes are all based on ISO standard 3898.29 Each code has its own list, applicable within that code. Some symbols have more than one meaning, the particular meaning being stated in the clause. A few rarely-used symbols are defined only in clauses where they appear (e.g. Ac;eff in 7.5.3(1)). There are a few important changes from previous practice in the UK. For example, an x–x axis is along a member, a y–y axis is parallel to the flanges of a steel section (clause 1.7(2) of EN 1993-1-1), and a section modulus is W, with subscripts to denote elastic or plastic behaviour. This convention for member axes is more compatible with most commercially available analysis packages than that used in previous British bridge codes. The y–y axis generally represents the major principal axis, as shown in Fig. 1.1(a) and (b). Where this is not a principal axis, the major and minor principal axes are denoted u–u and v–v, as shown in Fig. 1.1(c). It is possible for the major axis of a composite cross-section to be the minor axis of its structural steel component.

8

CHAPTER 1. GENERAL

z

z

z v y

y u

y

y

y

y

u z (a)

z (b)

z

v

(c)

Fig. 1.1. Sign convention for axes of members

Wherever possible, definitions in EN 1994-2 have been aligned with those in ENs 1990, 1992 and 1993; but this should not be assumed without checking the list in clause 1.6. Some quite minor differences are significant. The symbol fy has different meanings in ENs 1992 and 1993. It is retained in EN 1994-2 for the nominal yield strength of structural steel, though the generic subscript for that material is ‘a’, based on the French word for steel, ‘acier’. Subscript ‘a’ is not used in EN 1993, where the partial factor for steel is not A , but M . The symbol M is also used in EN 1994-2. The characteristic yield strength of reinforcement is fsk , with partial factor S . The use of upper-case subscripts for factors for materials implies that the values given allow for two types of uncertainty: in the properties of the material and in the resistance model used.

Clause 1.6

9

CHAPTER 2

Basis of design The material described in this chapter is covered in Section 2 of EN 1994-2, in the following clauses: . . . .

Requirements Principles of limit states design Basic variables Verification by the partial factor method


The sequence follows that of EN 1990, Sections 2 to 4 and 6.

2.1. Requirements Design is to be in accordance with the general requirements of EN 1990. The purpose of Section 2 is to give supplementary provisions for composite structures. Clause 2.1(3) reminds the user again that design is based on actions in accordance with EN 1991, combinations of actions and load factors at the various limit states in accordance with EN 1990 (Annex A2), and the resistances, durability and serviceability provisions of EN 1994 (through extensive references to EC2 and EC3). The use of partial safety factors for actions and resistances (the ‘partial factor method’) is expected but is not a requirement of Eurocodes. The method is presented in Section 6 of EN 1990 as one way of satisfying the basic requirements set out in Section 2 of that standard. This is why use of the partial factor method is given ‘deemed to satisfy’ status in clause 2.1(3). To establish that a design was in accordance with the Eurocodes, the user of any other method would normally have to demonstrate, to the satisfaction of the regulatory authority and/or the client, that the method satisfied the basic requirements of EN 1990. The design working life for bridges and components of bridges is also given in EN 1990. This predominantly affects calculations on fatigue. Temporary structures (that will not be dismantled and reused) have an indicative design life of 10 years, while bearings have a life of 10–25 years and a permanent bridge has an indicative design life of 100 years. The design lives of temporary bridges and permanent bridges can be varied in project specifications and the National Annex respectively. For political reasons, the design life for permanent bridges in the UK may be maintained at 120 years. To achieve the design working life, bridges and bridge components should be designed against corrosion, fatigue and wear and should be regularly inspected and maintained. Where components cannot be designed for the full working life of the bridge, they need to be replaceable. Further detail is given in Chapter 4 of this guide.

Clause 2.1(3)


2.2. Principles of limit states design The clause provides a reminder that it is important to check strength and stability throughout all stages of construction in addition to the final condition. The strength of bare steel beams during pouring of the deck slab must be checked, as the restraint to the top flange provided by the completed deck slab is absent in this condition. A beam that is in Class 1 or 2 when completed may be in Class 3 or 4 during construction, if a greater depth of web is in compression. Its stresses must then be built up allowing for the construction history. For cross-sections that are in Class 1 or 2 when completed, final verifications of resistances can be based on accumulation of bending moments and shear forces, rather than stresses, as plastic bending resistances can be used. The serviceability checks would still necessitate consideration of the staged construction. All resistance formulae for composite members assume that the specified requirements for materials, such as ductility, fracture toughness and through-thickness properties, are met.

2.3. Basic variables Clause 2.3.1

Clause 2.3.2(1)

Clause 2.3.3

Clause 2.3.1 on actions refers only to EN 1991. Its Part 2, ‘Traffic loads on bridges’, defines load patterns and leaves clients, or designers, much choice over intensity of loading. Loads during construction are specified in EN 1991-1-6, ‘Actions during execution’.30 Actions include imposed deformations, such as settlement or jacking of supports, and effects of temperature and shrinkage. Further information is given in comments on clause 2.3.3. Clause 2.3.2(1) refers to EN 1992-1-1 for shrinkage and creep of concrete, where detailed and quite complex rules are given for prediction of free shrinkage strain and creep coefficients. These are discussed in comments on clauses 3.1 and 5.4.2.2. Effects of creep of concrete are not normally treated as imposed deformations. An exception arises in clause 5.4.2.2(6). The classification of effects of shrinkage and temperature in clause 2.3.3 into ‘primary’ and ‘secondary’ will be familiar to designers of continuous beams. Secondary effects are to be treated as ‘indirect actions’, which are ‘sets of imposed deformations’ (clause 1.5.3.1 of EN 1990), not as action effects. This distinction is relevant in clause 5.4.2.2(7), where indirect actions may be neglected in analyses for some verifications of composite members with all cross-sections in Class 1 or 2. This is because resistances are based on plastic analysis and there is therefore adequate rotation capacity to permit the effects of imposed deformations to be released.

2.4. Verification by the partial factor method 2.4.1. Design values Clause 2.4.1

Clause 2.4.1.1(1)

Clause 2.4.1.2

12

Clause 2.4.1 illustrates the treatment of partial factors. Recommended values are given in Notes, in the hope of eventual convergence between the values for each partial factor that will be specified in the national annexes. This process was adopted because the regulatory bodies in the member states of CEN, rather than CEN itself, are responsible for setting safety levels. The Notes are informative, not normative (i.e. not part of the preceding provision), so that there are no numerical values in the principles, as explained earlier. The Note below clause 2.4.1.1(1) recommends P ¼ 1:0 (where subscript ‘P’ represents prestress) for controlled imposed deformations. Examples of these include jacking up at supports or jacking down by the removal of packing plates. The latter might be done to increase the reaction at an adjacent end support where there is a risk of uplift occurring. The Notes to clause 2.4.1.2 link the partial factors for concrete, reinforcing steel and structural steel to those recommended in EN 1992-1-1 and EN 1993. Design would be more difficult if the factors for these materials in composite structures differed from the values in reinforced concrete and steel structures. The reference to EN 1993, as distinct from EN 1993-1-1, is required because some M factors differ for bridges and buildings.


εcu3

0.85fck/γC x

Plastic neutral axis

αcc fck/γC 0.8x

fyd

Strain

Stress to EN 1994

fyd

Stress to EN 1992, for fck ≤ 50 N/mm2

Fig. 2.1. Rectangular stress blocks for concrete in compression at ultimate limit states

The remainder of EN 1994-2 normally refers to design strengths, rather than to characteristic or nominal values with partial factors. Characteristic values are 5% lower fractiles for an infinite test series, predicted from experience and a smaller number of tests. Nominal values (e.g. the yield strength of structural steel) are used where distributions of test results cannot be predicted statistically. They are chosen to correspond to characteristic values. The design strength for concrete is given by: fcd ¼ fck =C

ð2:1Þ

where fck is the characteristic cylinder strength. This definition is stated algebraically because it differs from that of EN 1992-2, in which an additional coefficient cc is applied: fcd ¼ cc fck =C

ðD2:1Þ

The coefficient is explained in EN 1992-2 as taking account of long-term effects and of unfavourable effects resulting from the way the load is applied. The value for cc is to be given in national annexes to EN 1992-2, and ‘should lie between 0.80 and 1.00’. The value 1.00 has been used in EN 1994-2, without permitting national choice, for several reasons: .

.

.

.

The plastic stress block for use in resistance of composite sections, defined in clause 6.2.1.2, consists of a stress 0.85fcd extending to the neutral axis, as shown in Fig. 2.1. The depth of the stress block in EN 1992-2 is only 80% of this distance. The factor 0.85 is not fully equivalent to cc ; it allows also for the difference between the stress blocks. Predictions using the stress block of EN 1994 have been verified against test results for composite members conducted independently from verifications for concrete bridges. The EN 1994 block is easier to apply. The Eurocode 2 rule was not used in Eurocode 4 because resistance formulae become complex where the neutral axis is close to or within the steel flange adjacent to the concrete slab. Resistance formulae for composite elements given in EN 1994 are based on calibrations using its stress block, with cc ¼ 1:0.

The definition of fcd in equation (2.1) is applicable to verifications of all composite crosssections, but not where the section is reinforced concrete only; for example, in-plane shear in a concrete flange of a composite beam. For reinforced concrete, EN 1992-2 applies, with cc in equation (D2.1) as given in the National Annex. It is expected that the rules in the UK’s Annex will include: cc ¼ 0:85 for flexure and axial compression This is consistent with EN 1994-2, as the coefficient 0.85 appears in the resistance formulae in clauses 6.2.1.2 and 6.7.3.2. In these cases, the values 0.85fcd in EN 1994-2 and fcd in EN 1992-2 are equal, so the values of symbols fcd are not equal. There is a risk of error when switching between calculations for composite sections and for reinforced concrete elements such as a deck slab both for this reason and because of the different depth of stress block.

13


Table 2.1. Partial factors from EN 1992-2 for materials, for ultimate limit states

Clause 2.4.1.3

Clause 2.4.1.4

Design situations

C , for concrete

S , reinforcing steel

S , prestressing steel

Persistent and transient Accidental

1.5 1.2

1.15 1.0

1.15 1.0

Care is needed also with symbols for steels. The design strengths in EN 1994 are fyd for structural steel and fsd for reinforcement, but reinforcement in EN 1992 has fyd , not fsd . The recommended partial factors given in EN 1992-2 (referring to EN 1992-1-1) for materials for ultimate limit states other than fatigue are repeated in Table 2.1. For serviceability limit states, the recommended value is generally 1.0, from clause 2.4.2.4(2). The M values for structural steel are denoted M0 to M7 in clause 6.1 of EN 1993-2. Those for ultimate limit states other than fatigue are given in Table 2.2. Further values are given in clauses on fatigue. No distinction is made between persistent, transient, and accidental design situations, though it could be, in a national annex. For simplicity, M for resistances of shear connectors (denoted V ), given in a Note to clause 6.6.3.1(1), was standardised at 1.25, because this is the recommended value for most joints in steelwork. Where calibration led to a different value, a coefficient in the resistance formula was modified to enable 1.25 to be used. Clause 2.4.1.3 refers to ‘product standards hEN’ and to ‘nominal values’. The ‘h’ stands for ‘harmonised’. This term from the Construction Products Directive31 is explained in the Designers’ Guide to EN 1990.1 Generally, global analysis and resistances of cross-sections may be based on the ‘nominal’ values of dimensions, which are given on the project drawings or quoted in product standards. Geometrical tolerances as well as structural imperfections (such as welding residual stresses) are accounted for in the methods specified for global analyses and for buckling checks of individual structural elements. These subjects are discussed further in sections 5.2 and 5.3, respectively, of this guide. Clause 2.4.1.4, on design resistances to particular action effects, refers to expressions (6.6a) and (6.6c) given in clause 6.3.5 of EN 1990. Resistances in EN 1994-2 often need more than one partial factor, and so use expression (6.6a) which is: Rd ¼ Rfði Xk;i =M;i Þ; ad g i 1

ðD2:2Þ

Table 2.2. Partial factors from EN 1993-2 for materials, for ultimate limit states Resistance type

Factor

Recommended value

Resistance of members and cross-sections . Resistance of cross-sections to excessive yielding including local buckling . Resistance of members to instability assessed by member checks . Resistance to fracture of cross-sections in tension

M0 M1 M2

1.00 1.10 1.25

M2 M2

1.25 1.25

M3 M3;ser M4 M5 M6;ser M7

1.25 1.10 1.10 1.10 1.00 1.10

Resistance of joints . Resistance of bolts, rivets, pins and welds . Resistance of plates in bearing . Slip resistance: – at an ultimate limit state – at a serviceability limit state . Bearing resistance of an injection bolt . Resistance of joints in hollow section lattice girders . Resistance of pins at serviceability limit state . Pre-load of high-strength bolts

14


For example, clause 6.7.3.2(1) gives the plastic resistance to compression of a crosssection as the sum of terms for the structural steel, concrete and reinforcement: Npl;Rd ¼ Aa fyd þ 0:85Ac fcd þ As fsd

ð6:30Þ

In this case, there is no separate term ad for the influence of geometrical data on resistance, because uncertainties in areas of cross-sections are allowed for in the M factors. In terms of characteristic strengths, from clause 2.4.1.2, equation (6.30) becomes: Npl;Rd ¼ Aa fy =M þ 0:85Ac fck =C þ As fsk =S

ðD2:3Þ

where: – the characteristic material strengths Xk;i are fy , fck and fsk ; – the conversion factors, i in EN 1990, are 1.0 for steel and reinforcement and 0.85 for concrete. These factors enable allowance to be made for the difference between the material property obtained from tests and its in situ contribution to the particular resistance considered. In general, it is also permissible to allow for this effect in the values of M;i ; – the partial factors M;i are written M , C and S in EN 1994-2. Expression (6.6c) of EN 1990 is: Rd ¼ Rk =M It applies where characteristic properties and a single partial factor can be used; for example, in expressions for the shear resistance of a headed stud (clause 6.6.3.1). It is widely used in EN 1993, where only one material, steel, contributes to a resistance.

2.4.2. Combination of actions Clause 2.4.2 refers to the combinations of actions given in EN 1990. As in current practice, variable actions are included in a combination only in regions where they contribute to the total action effect considered. For permanent actions and ultimate limit states, the situation is more complex. Normally the same factor F (favourable or unfavourable as appropriate) is applied throughout the structure, irrespective of whether both favourable and unfavourable loading regions exist. Additionally, the characteristic action is a mean (50% fractile) value. Exceptions are covered by clause 6.4.3.1(4)P of EN 1990:

Clause 2.4.2

‘Where the results of a verification are very sensitive to variations of the magnitude of a permanent action from place to place in the structure, the unfavourable and the favourable parts of this action shall be considered as individual actions.’

A design permanent action is then Ed;min Gk;min in a ‘favourable’ region, and Ed;max Gk;max in an ‘unfavourable’ region. Recommendations on the choice of these values and the application of this principle are given in EN 1990, with guidance in the Designers’ Guide to EN 1990.1

2.4.3. Verification of static equilibrium (EQU) The preceding quotation from EN 1990 evidently applies to checks on static equilibrium, clause 2.4.3(1). It draws attention to the role of anchors and bearings in ensuring static equilibrium. The abbreviation EQU in this clause comes from EN 1990, where four types of ultimate limit state are defined in clause 6.4.1: . .

Clause 2.4.3(1)

EQU for loss of static equilibrium FAT for fatigue failure

15


. .

GEO for failure or excessive deformation of the ground STR for internal failure or excessive deformation of the structure.

As explained above, the main feature of EQU is that, unlike STR, the partial factor F for permanent actions is not uniform over the whole structure. It is higher for destabilizing actions than for those relied on for stability. This guide mainly covers ultimate limit states of types STR and FAT. Use of type GEO arises in design of foundations to EN 1997.32

16

CHAPTER 3

Materials This chapter concerns the properties of materials needed for the design of composite structures. It corresponds to Section 3, which has the following clauses: . . . . . .

Concrete Reinforcing steel for bridges Structural steel for bridges Connecting devices Prestressing steel and devices Tension components in steel


Rather than repeating information given elsewhere, Section 3 consists mainly of cross-references to other Eurocodes and EN standards. The following comments relate to provisions of particular significance for composite structures.

3.1. Concrete Clause 3.1(1) refers to EN 1992-1-1 for the properties of concrete. For lightweight-aggregate concrete, several properties are dependent on the oven-dry density, relative to 2200 kg/m3 . Comprehensive sets of time-dependent properties are given in its clause 3.1 for normal concrete and clause 11.3 for lightweight-aggregate concrete. For composite structures built unpropped, with several stages of construction, simplification may be needed. A simplification for considerations of creep is provided in clause 5.4.2.2(2). Specific properties are now discussed. (For thermal expansion, see Section 3.3 below.)

Compressive strength Strength and deformation characteristics are summarized in EN 1992-1-1, Table 3.1 for normal concrete and Table 11.3.1 for lightweight-aggregate concrete. Strength classes for normal concrete are defined as Cx/y, where x and y are respectively the cylinder and cube compressive strengths in N/mm2 units, determined at age 28 days. All compressive strengths in design rules in Eurocodes are cylinder strengths, so an unsafe error occurs if a specified cube strength is used in calculations. It should be replaced at the outset by the equivalent cylinder strength, using the relationships given by the strength classes. Most cube strengths in Table 3.1 are rounded to 5 N/mm2 . The ratios fck =fck;cube range from 0.78 to 0.83, for grades up to C70/85. Classes for lightweight concrete are designated LCx/y. The relationships between cylinder and cube strengths differ from those of normal concrete; for example, C40/50 and LC40/44. The ratios fck =fck;cube for the LC grades range from 0.89 to 0.92. Thus, cylinder strengths are about 80% of cube strengths for normal-weight concrete and 90% for lightweight concrete.

Clause 3.1(1)


Comment on the design compressive strength, fcd ¼ fck =C , is given at clause 2.4.1.2.

Tensile strength EN 1992 defines concrete tensile strength as the highest stress reached under concentric tensile loading. Values for the mean axial tensile strength of normal-weight concrete at 28 days, fctm , are given in Table 3.1 of EN 1992-1-1. They are based on the following formulae, in N/mm2 units: fctm ¼ 0:30ð fck Þ2=3 ;

fck C50=60

fctm ¼ 2:12 ln½1 þ ð fcm =10Þ;

fck > C50=60

ðD3:1Þ ðD3:2Þ

This table also gives the 5% and 95% fractile values for tensile strength. The appropriate fractile value should be used in any limit state verification that relies on either an adverse or beneficial effect of the tensile strength of concrete. Tensile strengths for lightweight concrete are given in Table 11.3.1 of EN 1992-1-1. Mean tensile stress, fctm , is used in several places in EN 1994-2 where the effects of tension stiffening are considered to be important. These include: . .

.

.

.

clause 5.4.2.3(2): rules on allowing for cracking in global analysis clause 5.4.2.8(6): calculation of internal forces in concrete tension members in bowstring arches clause 5.5.1(5): minimum area of reinforcement required in concrete tension flanges of composite beams clause 7.4.2(1): rules on minimum reinforcement to ensure that cracking does not cause yielding of reinforcement in the cracked region clause 7.4.3(3): rules on crack width calculation to allow for the increase in stress in reinforcement caused by tension stiffening.

Elastic deformation All properties of concrete are influenced by its composition. The values for the mean shortterm modulus of elasticity in Tables 3.1 and 11.3.1 of EN 1992-1-1 are given with a warning that they are ‘indicative’ and ‘should be specifically assessed if the structure is likely to be sensitive to deviations from these general values’. The values are for concrete with quartzite aggregates. Corrections for other types of aggregate are given in EN 1992-1-1, clause 3.1.3(2). All these are secant values; typically, 0.4fcm /(strain at 0.4fcm ), and so are slightly lower than the initial tangent modulus, because stress–strain curves for concrete are non-linear from the origin. Table 3.1 in EN 1992-1-1 gives the analytical relation: Ecm ¼ 22½ð fck þ 8Þ=100:3

Clause 3.1(2)

18

with Ecm in GPa or kN/mm2 units, and fck in N/mm2 . For fck ¼ 30, this gives Ecm ¼ 32:8 kN/mm2 , whereas the entry in the table is rounded to 33 kN/mm2 . A formula for the increase of Ecm with time, in clause 3.1.3(3) of EN 1992-1-1, gives the two-year value as 6% above Ecm at 28 days. The influence in a composite structure of so small a change is likely to be negligible compared with the uncertainties in the modelling of creep. Clause 3.1(2) limits the scope of EN 1994-2 to the strength range C20/25 to C60/75 for normal concrete and from LC20/22 to LC60/66 for lightweight concrete. The upper limits to these ranges are lower than that given in EN 1992-2 (C70/85) because there is limited knowledge and experience of the behaviour of composite members with very strong concrete. This applies, for example, to the load/slip properties of shear connectors, the redistribution of moments in continuous beams and the resistance of columns. The use of rectangular stress blocks for resistance to bending (clause 6.2.1.2(d)) relies on the strain capacity of the materials. The relevant property of concrete in compression, "cu3 in Table 3.1 of EN 1992-1-1, is 0.0035 for classes up to C50/60, but then falls, and is only 0.0026 for class C90/105.


Shrinkage The shrinkage of concrete referred to in clause 3.1(3) is (presumably) both the drying shrinkage that occurs after setting and the autogenous shrinkage, but not the plastic shrinkage that precedes setting. Drying shrinkage is associated with movement of water through and out of the concrete and therefore depends on relative humidity and effective section thickness as well as on the concrete mix. It takes several years to be substantially complete. The mean drying shrinkage strain (for unreinforced concrete) is given in clause 3.1.4(6) of EN 1992-1-1 as a function of grade of concrete, ambient relative humidity, effective thickness of the concrete cross-section, and elapsed time since the end of curing. It is stated that actual values have a coefficient of variation of about 30%. This implies a 16% probability that the shrinkage will exceed the prediction by at least 30%. A slightly better predictor is given in Annex B of EN 1992-1-1, as the type of cement is included as an additional parameter. Autogenous shrinkage develops during the hydration and hardening of concrete. It is that which occurs in enclosed or sealed concrete, as in a concrete-filled steel tube, where no loss of moisture occurs. This shrinkage strain depends only on the strength of the concrete, and is substantially complete in a few months. It is given in clause 3.1.4(6) of EN 1992-1-1 as a function of concrete grade and the age of the concrete in days. The time coefficient given is ½1 expð0:2t0:5 Þ, so this shrinkage should be 90% complete at age 19 weeks. The 90% shrinkage strain for a grade C40/50 concrete is given as 67 106 . It has little influence on cracking due to direct loading, and the rules for minimum reinforcement (clause 7.4.2) take account of its effects. The rules in EN 1992-1-1 become less accurate at high concrete strengths, especially if the mix includes silica fume. Data for shrinkage for concrete grades C55/67 and above are given in informative Annex B of EN 1992-2. Section 11 of EN 1992-2 gives supplementary requirements for lightweight concretes. The shrinkage of reinforced concrete is lower than the ‘free’ shrinkage, to an extent that depends on the reinforcement ratio. The difference is easily calculated by elastic theory, if the concrete is in compression. In steel–concrete composite bridges, restraint of reinforced concrete shrinkage by the structural steel leads to locked-in stresses in the composite section. In indeterminate bridges, secondary moments and forces from restraint to the free deflections also occur. Shrinkage, being a permanent action, occurs in every combination of actions. It increases hogging moments at internal supports, often a critical region, and so can influence design. The specified shrinkage strains will typically be found to be greater than that used in previous UK practice, but the recommended partial load factor, in clause 2.4.2.1 of EN 1992-1-1, is SH ¼ 1:0, lower than the value of 1.2 used in BS 5400. There is further comment on shrinkage in Chapter 5.

Clause 3.1(3)

Creep In EN 1994-2, the effects of creep are generally accounted for using an effective modulus of elasticity for the concrete, rather than by explicit calculation of creep deformation. However, it is still necessary to determine the creep coefficient ðt; t0 Þ (denoted t in EN 1994) from clause 3.1.4 of EN 1992-1-1. Guidance on deriving modular ratios is given in section 5.4.2 of this guide.

3.2. Reinforcing steel for bridges For properties of reinforcement, clause 3.2(1) refers to clause 3.2 of EN 1992-1-1, which in turn refers to its normative Annex C for bond characteristics. EN 1992 allows the use of bars, de-coiled rods and welded fabric as suitable reinforcement. Its rules are applicable to ribbed and weldable reinforcement only, and therefore cannot be used for plain round bars. The rules are valid for characteristic yield strengths between 400 N/mm2 and 600 N/mm2 . Wire fabrics with nominal bar size 5 mm and above are included. Exceptions to the rules for

Clause 3.2(1)

19


Table 3.1. Ductility classes for reinforcement Class

Characteristic strain at maximum force, "uk (%)

Minimum value of k ¼ ð ft =fy Þk

A B C

2.5 5 7.5

1.05 1.08 1.15, <1.35

fatigue of reinforcement may be given in the National Annex, and could refer to the use of wire fabric. In this section 3.2, symbols fyk and fyd are used for the yield strengths of reinforcement, as in EN 1992, although fsk and fsd are used in EN 1994, to distinguish reinforcement from structural steel. The grade of reinforcement denotes the specified characteristic yield strength, fyk . This is obtained by dividing the characteristic yield load by the nominal cross-sectional area of the bar. Alternatively, for products without a pronounced yield stress, the 0.2% proof stress, f0:2k may be used in place of the yield stress.

Clause 3.2(2)

Clause 3.2(3)

Elastic deformation For simplicity, clause 3.2(2) permits the modulus of elasticity of reinforcement to be taken as 210 kN/mm2 , the value given in EN 1993-1-1 for structural steel, rather than 200 kN/mm2 , the value in EN 1992-1-1. This simplification means that it is not necessary to ‘transform’ reinforcement into structural steel or vice versa when calculating cracked section properties of composite beams. Ductility Clause 3.2(3) refers to clause 3.2.4 of EN 1992-2; but provisions on ductility in Annex C of EN 1992-1-1 also apply. Reinforcement shall have adequate ductility, defined by the ratio of tensile strength to the yield stress, ð ft =fy Þk , and the strain at maximum force, "uk . The requirements for the three classes for ductility are given in Table 3.1, from EN 1992-1-1. Clause 3.2.4(101)P of EN 1992-2 recommends that Class A reinforcement is not used for bridges, although this is subject to variation in the National Annex. The reason is that high strain can occur in reinforcement in a reinforced concrete section in flexure before the concrete crushes. Clause 5.5.1(5) prohibits the use of Class A reinforcement in composite beams which are designed as either Class 1 or 2 for a similar reason: namely, that very high strains in reinforcement are possible due to plastification of the whole composite section. Class 3 and 4 sections are limited to first yield in the structural steel and so the reinforcement strain is limited to a relatively low value. The recommendations of EN 1992 and EN 1994 lead to some ambiguity with respect to ductility requirements for bars in reinforced concrete deck slabs forming part of a composite bridge with Class 3 or 4 beams. Where main longitudinal bars in the deck slab of a composite section are significantly stressed by local loading, it would be advisable to follow the recommendations of EN 1992 and not to use Class A reinforcement. Stress–strain curves The characteristic stress–strain diagram and the two alternative design diagrams defined in clause 3.2.7 of EN 1992-1-1 are shown in Fig. 3.1. The design diagrams (labelled B in Fig. 3.1) have: (a) an inclined top branch with a strain limit of "ud and a maximum stress of kfyk =S at "uk (for symbols k and "uk , see Table 3.1), and (b) a horizontal top branch without strain limit. A value for "ud may be found in the National Annex to EN 1992-1-1, and is recommended as 0.9"uk .

20


σ

A

kfyk

kfyk

kfyk /γS

fyk fyd = fyk /γS B k = ( ft / fy)k A – idealised B – design

0

fyd/Es

εud

εuk

ε

Fig. 3.1. Characteristic and design stress–strain diagrams for reinforcement (tension and compression)

From clause 6.2.1.4, reinforcement diagram (a) is only relevant when the non-linear method for bending resistance is used. Elastic and plastic bending resistances assume that the reinforcement stress is limited to the design yield strength. The minimum ductility properties of wire fabric given in Table C.1 of EN 1992-1-1 may not be sufficient to satisfy clause 5.5.1(6), as this requires demonstration of sufficient ductility to avoid fracture when built into a concrete slab. It has been found in tests on continuous composite beams with fabric in tension that the cross-wires initiate cracks in concrete, so that tensile strain becomes concentrated at the locations of the welds in the fabric.33

3.3. Structural steel for bridges Clause 3.3(1) refers to EN 1993-2, which in turn refers to EN 1993-1-1. This lists in its Table 3.1 steel grades with nominal yield strengths up to 460 N/mm2 , and allows other steel products to be included in national annexes. The nominal values of material properties have to be adopted as characteristic values in all design calculations. Two options for selecting material strength are provided. Either the yield strength and ultimate strength should be obtained from the relevant product standard or the simplified values provided in Table 3.1 of EN 1993-1-1 should be used. The National Annex for EC3-1-1 may make this choice. In either case, the strength varies with thickness, and the appropriate thickness must be used when determining the strength. The elastic constants for steel, given in clause 3.2.6 of EN 1993-1-1, are familiar values. In the notation of EN 1994, they are: Ea ¼ 210 kN/mm2 , Ga ¼ 81 kN/mm2 , and a ¼ 0:3. Moduli of elasticity for tension rods and cables of different types are not covered by this clause and are given in EN 1993-1-11. Clause 3.3(2) sets the same upper limit to nominal yield strength as in EN 1993-1-1, namely 460 N/mm2 , for use in composite bridges. EN 1993-1-12 covers steels up to grade S700. A comprehensive report on high-performance steels appeared in 2005,34 and there has been extensive research on the use in composite members of structural steels with yield strengths exceeding 355 N/mm2 .3537 It was found that some design rules need modification for use with steel grades higher than S355, to avoid premature crushing of concrete. This applies to: . .

Clause 3.3(1)

Clause 3.3(2)

plastic resistance moment (clause 6.2.1.2(2)), and resistance of columns (clause 6.7.3.6(1)).

Ductility Many design clauses in EN 1994 rely on the ductile behaviour of structural steel after yield. Ductility is covered by the references in clause 3.3(1) to EN 1993. The ductility characteristics required by clause 3.2.2 of EN 1993-1-1 are for a minimum ratio fu =fy of the specified values; a minimum elongation; and a minimum strain at the

21


specified ultimate tensile strength, fu . Recommended values are given, all of which can be modified in the National Annex. The steel grades in Table 3.1 of EN 1993-1-1 all provide the recommended level of ductility. It follows that the drafting of this part of a national annex to EN 1993-1-1 should consider both steel and composite structures.

Thermal expansion For the coefficient of linear thermal expansion of steel, clause 3.2.6 of EN 1993-1-1 gives a value of 12 106 ‘per 8C’ (also written in Eurocodes as /K or K1 ). This is followed by a Note that for calculating the ‘structural effects of unequal temperatures’ in composite structures, the coefficient may be taken as 10 106 per 8C, which is the value given for normal-weight concrete in clause 3.1.3(5) of EN 1992-1-1. This avoids the need to calculate the internal restraint stresses from uniform temperature change, which would result from different coefficients of thermal expansion for steel and concrete. Movement due to change of uniform temperature (or force due to restraint of movement) should however be calculated using ¼ 12 106 per 8C for all the structural materials (clause 5.4.2.5(3)). Thermal expansion of reinforcement is not mentioned in EN 1992-1-1, presumably because it is assumed to be the same as that of normal-weight concrete. For reinforcement in composite members the coefficient should be taken as 10 106 per 8C. This is not in EN 1994. Coefficients of thermal expansion for lightweight-aggregate concretes can range from 4 106 to 14 106 per 8C. Clause 11.3.2(2) of EN 1992-1-1 states that: ‘The differences between the coefficients of thermal expansion of steel and lightweight aggregate concrete need not be considered in design’, but ‘steel’ here means reinforcement, not structural steel. The effects of the difference from 10 106 per 8C should be considered in design of composite members for situations where the temperatures of the concrete and the structural steel could differ significantly.

3.4. Connecting devices 3.4.1. General Reference is made to EN 1993, Eurocode 3: Design of Steel Structures, Part 1-8: Design of Joints19 for information relating to fasteners, such as bolts, and welding consumables. Provisions for ‘other types of mechanical fastener’ are given in clause 3.3 of EN 1993-1-3.38

Composite joints Composite joints are defined in clause 1.5.2.8. In bridges, they are essentially steelwork joints across which a reinforced or prestressed concrete slab is continuous, and cannot be ignored. Composite joints are covered in Section 8 and Annex A of EN 1994-1-1, with extensive reference to EN 1993-1-8. These clauses are written ‘for buildings’, and so are not copied into EN 1994-2, though many of them are relevant. Commentary on them will be found in Chapters 8 and 10 of the Designers’ Guide to EN 1994-1-1.5 The joints classified as ‘rigid’ or ‘full-strength’ occur also in bridge construction. Where bending resistances of beams in Class 1 or 2 are determined by plastic theory, joints in regions of high bending moment must either have sufficient rotation capacity, or be stronger than the weaker of the members joined. The rotation capacity needed in bridges, where elastic global analysis is always used, is lower than in buildings. Tests, mainly on beam-to-column joints, have found that reinforcing bars of diameter up to 12 mm may fracture. Clause 5.5.1 gives rules for minimum reinforcement that apply also to joints, but does not exclude small-diameter bars.

3.4.2. Headed stud shear connectors Headed studs are the only type of shear connector for which detailed provisions are given in EN 1994-2, throughout clause 6.6. Their use is referred to elsewhere; for example, in clause 6.7.4.2(4). Their performance has been validated for diameters up to 25 mm.39 Research on

22


larger studs is in progress. Studs attached to steel top flanges present a hazard during construction, and other types of connector are sometimes used.23 These must satisfy clause 6.6.1.1, which gives the basis of design for shear connection. Research on perforated plate connectors (known initially as ‘Perfobond’) of S355 and S460 steel in grade C50/60 concrete has found slip capacities from 8–15 mm, which is better than the 6 mm found for 22-mm studs.25 The use of adhesives on a steel flange is unlikely to be suitable. See also the comment on clause 1.1.3(3). Clause 3.4.2 refers to EN 13918 Welding – Studs and Ceramic Ferrules for Arc Stud Welding.40 This gives minimum dimensions for weld collars. Other methods of attaching studs, such as spinning, may not provide weld collars large enough for the resistances of studs given in clause 6.6.3.1(1) to be applicable. Shear connection between steel and concrete by bond or friction is permitted only in accordance with clause 6.7.4, for columns.

Clause 3.4.2

3.5. Prestressing steel and devices Properties of materials for prestressing tendons and requirements for anchorage and coupling of tendons are covered in clauses 3.3 and 3.4, respectively, of EN 1992-1-1. Prestressing by tendons is rarely used for steel and concrete composite members and is not discussed further.

3.6. Tension components in steel The scope of EN 1993-1-11 is limited to bridges with adjustable and replaceable steel tension components. It identifies three generic groups: tension rod systems, ropes, and bundles of parallel wires or strands; and provides information on stiffness and other material properties. The analysis of cable-supported bridges, including treatment of load combinations and nonlinear effects, is also covered. These are not discussed further here but some discussion can be found in the Designers’ Guide to EN 1993-2.4

23

CHAPTER 4

Durability This chapter corresponds to Section 4, which has the following clauses: . .

General Corrosion protection at the steel–concrete interface in bridges

Clause 4.1 Clause 4.2

4.1. General Almost all aspects of the durability of composite structures are covered by cross-references in clause 4.1(1) to ENs 1990, 1992 and 1993. Bridges must be sufficiently durable to remain serviceable throughout their design life. Clause 2.4 of EN 1990 lists ten factors to be taken into account, and gives the following general requirement: ‘The structure shall be designed such that deterioration over its design working life does not impair the performance of the structure below that intended, having due regard to its environment and the anticipated level of maintenance.’

The specific provisions given in EN 1992 and EN 1993 focus on corrosion protection to reinforcement, tendons and structural steel.

Reinforced concrete The main durability provision in EN 1992 is the specification of concrete cover as a defence against corrosion of reinforcement and tendons. The following outline of the procedure is for reinforcement only. In addition to the durability aspect, adequate concrete cover is essential for the transmission of bond forces and for providing sufficient fire resistance (which is of less significance for bridge design). The minimum cover cmin to satisfy the durability requirements is defined in clause 4.4.1.2 of EN 1992-1-1 by the following expression: cmin ¼ maxfcmin;b ; cmin;dur þ cdur; cdur;st cdur;add ; 10 mmg

ðD4:1Þ

is the minimum cover due to bond requirements and is defined in Table 4.2 of EN 1992-1-1. For aggregate sizes up to 32 mm it is equal to the bar diameter (or equivalent bar diameter for bundled bars), cmin;dur is the minimum cover required for the environmental conditions, cdur; is an additional safety element which EC2 recommends to be 0 mm, cdur;st is a reduction of minimum cover for the use of stainless steel, which, if adopted, should be applied to all design calculations, including bond. The recommended value in EC2 without further specification is 0 mm, cdur;add is a reduction of minimum cover for the use of additional protection. This could cover coatings to the concrete surface or reinforcement (such as epoxy coating). EC2 recommends taking a value of 0 mm.

where: cmin;b

Clause 4.1(1)


Table 4.1. Minimum cover cmin;dur for reinforcement. (Source: based on Table 4.4N of EN 1992-1-115 ) Environmental Requirements for cmin (mm) Exposure Class (from Table 4.1 of EN 1992-1-1) Structural Class

X0

XC1

XC2/XC3

XC4

XD1/XS1

XD2/XS2

XD3/XS3

1 2 3 4 5 6

10 10 10 10 15 20

10 10 10 15 20 25

10 15 20 25 30 35

15 20 25 30 35 40

20 25 30 35 40 45

25 30 35 40 45 50

30 35 40 45 50 55

The minimum cover for durability requirements, cmin;dur , depends on the relevant ‘exposure class’ taken from Table 4.1 of EN 1992-1-1. There are 18 exposure classes, ranging from X0, ‘no risk of corrosion’, to XA3, ‘highly aggressive chemical environment’. It should be noted that a particular element may have more than one exposure class, e.g. XD3 and XF4. The XF and XA designations affect the minimum required concrete grade (via EN 1992-1-1 Annex E) and the chemical composition of the concrete. The XC and XD designations affect minimum cover and crack width requirements, and XD, XF and XS affect a stress limit for concrete under the characteristic combination, from clause 7.2(102) of EN 1992-2. The exposure classes most likely to be appropriate for composite bridge decks are: .

. . .

XC3 for a deck slab protected by waterproofing (recommended in clause 4.2(105) of EN 1992-2) XC3 for a deck slab soffit protected from the rain by adjacent girders XC4 for other parts of the deck slab exposed to cyclic wetting and drying XD3 for parapet edge beams in the splash zone of water contaminated with de-icing salts; and also XF2 or XF4 if exposed to both freeze–thaw and de-icing agents (recommended in clause 4.2(106) of EN 1992-2).

Informative Annex E of EN 1992-1-1 gives ‘indicative strength classes’ (e.g. C30/37) for each exposure class, for corrosion of reinforcement and for damage to concrete. The cover cmin;dur is given in Table 4.4N of EN 1992-1-1 in terms of the exposure class and the structural class, and the structural class is found from Table 4.3N. These are reproduced here as Tables 4.1 and 4.2, respectively. Table 4.2 gives modifications to the initial structural class, which is recommended (in a Note to clause 4.4.1.2(5) of EN 1992-1-1) to be class 4, assuming a service life of 50 years and concrete of the indicative strength. Taking exposure class XC4 as an example, the indicative strength class is C30/37. Starting with Structural Class 4, and using Tables 4.1 and 4.2: . . .

for 100-year life, increase by 2 to Class 6 for use of C40/50 concrete, reduce by 1 to Class 5 where the position of the reinforcement is not affected by the construction process, reduce by 1 to Class 4.

‘Special quality control’ (Table 4.2) is not defined, but clues are given in the Notes to Table 4.3N of EN 1992-1-1. Assuming that it will not be provided, the Class is 4, and Table 4.1 gives cmin;dur ¼ 30 mm. Using the recommendations that follow equation (D4.1), cmin ¼ 30 mm The cover to be specified on the drawings, cnom , shall include a further allowance for deviation (cdev ) according to clause 4.4.1.3(1)P of EN 1992-1-1, such that: cnom ¼ cmin þ cdev

26

CHAPTER 4. DURABILITY

Table 4.2. Recommended structural classification. (Source: based on Table 4.3N of EN 1992-1-115 ) Structural Class Exposure Class (from Table 4.1 of EN 1992-1-1) Criterion

X0

XC1

XC2/XC3 XC4

XD1

XD2/XS1 XD3/XS2/XS3

Service life of 100 years Increase Increase Increase Increase Increase Increase Increase class by 2 class by 2 class by 2 class by 2 class by 2 class by 2 class by 2 Strength Class (see notes 1 and 2)

C30/37 C30/37 C35/45 C40/50 C40/50 C40/50 C45/55 Reduce Reduce Reduce Reduce Reduce Reduce Reduce class by 1 class by 1 class by 1 class by 1 class by 1 class by 1 class by 1

Reduce Reduce Reduce Reduce Reduce Reduce Reduce Member with slab class by 1 class by 1 class by 1 class by 1 class by 1 class by 1 class by 1 geometry (position of reinforcement not affected by construction process) Special Quality Control Reduce Reduce Reduce Reduce Reduce Reduce Reduce of the concrete ensured class by 1 class by 1 class by 1 class by 1 class by 1 class by 1 class by 1 Note 1: The strength class and water/cement ratio are considered to be related values. The relationship is subject to a national code. A special composition (type of cement, w/c value, fine fillers) with the intent to produce low permeability may be considered. Note 2: The limit may be reduced by one strength class if air entrainment of more than 4% is applied.

The value of cdev for buildings and bridges is defined in the National Annex and is recommended in clause 4.4.1.3(2) of EN 1992-1-1 to be taken as 10 mm. This value may be reduced in situations where accurate measurements of cover achieved can be taken and non-conforming elements rejected. This could apply to precast units. Almost all the provisions on cover, but not the process to be followed, can be modified in the National Annex to EN 1992-1-1.

Structural steel The rules in Section 4 of EN 1993-1-1 cover the need for access for in-service inspection, maintenance, and possible reconstruction of parts susceptible to corrosion, wear or fatigue. Further provisions relevant to fatigue are given in Section 4 of EN 1993-2, and a list is given of parts that may need to be replaceable. Corrosion allowances for inaccessible surfaces may be given in the National Annex. Further discussion on durability of structural steel is presented in the Designers’ Guide to EN 1993-2.4 Access to shear connectors is not possible, so they must be protected from corrosion. Clause 4.1(2) refers to clause 6.6.5, which includes relevant detailing rules, for cover and for haunches.

Clause 4.1(2)

4.2. Corrosion protection at the steel–concrete interface in bridges The side cover to stud connectors must be at least 50 mm (clause 6.6.5.4(2)). Clause 4.2(1) requires provision of a minimum of 50 mm of corrosion protection to each edge of a steel flange at an interface with concrete. This does not imply that the connectors must be protected. For precast deck slabs, the reference to Section 8 is to clause 8.4.2, which requires greater corrosion protection to a steel flange that supports a precast slab without bedding. Normal UK practice when using ‘Omnia’ planks has been to extend the corrosion protection a minimum of 25 mm beyond the plank edge and its seating material, with due allowance

Clause 4.2(1)

27


for placing tolerance. The connectors are not mentioned. They are usually surrounded by in situ concrete, whether bedding is used (as is usual) or not. Corrosion protection to the connectors is not normally required. It is possible that a thick coating could reduce their stiffness in shear.

28

CHAPTER 5

Structural analysis This chapter corresponds to Section 5 of EN 1994-2, which has the following clauses: . . . . .

Structural modelling for analysis Structural stability Imperfections Calculation of action effects Classification of cross-sections


Structural analysis is performed at three levels: global analysis, member analysis and local analysis. Section 5 of EN 1994-2 covers the structural idealization of bridges and the methods of global analysis required in different situations to determine deformations and internal forces and moments. It also covers classification of cross-sections of members, for use in determining resistances by methods given in Sections 6 of EN 1993-2 and EN 1994-2. Much reference has to be made to other parts of EC3, especially EN 1993-1-541 for the effects of shear lag and plate buckling. Wherever possible, analyses for serviceability and ultimate limit states use the same methods. It is therefore more convenient to specify them in a single section, rather than to include them in Sections 6 and 7. The division of material between Section 5 and Section 6 (Ultimate limit states) is not always obvious. Calculation of vertical shear is clearly ‘analysis’, but longitudinal shear is in Section 6. For composite columns, ‘Methods of analysis and member imperfections’ is in clause 6.7.3.4. This separation of imperfections in frames from those in columns requires care, and receives detailed explanation in the Designers’ Guide to EN 1994-1-1.5 Two flow charts for global analysis, Figs 5.15 and 5.16, are given, with comments, at the end of this chapter. They include relevant provisions from Section 6.

5.1. Structural modelling for analysis 5.1.1. Structural modelling and basic assumptions The clause of EN 1990 referred to in clause 5.1.1(1)P says, in effect, that models shall be appropriate and based on established theory and practice and that the variables shall be relevant. The basic requirement is that analysis should realistically model the expected behaviour of the bridge and its constituent elements. For composite bridges, important factors in analysis are the effects on stiffness of shear lag and concrete cracking. For composite members, different rules for shear lag apply for concrete flanges and for the steel parts. The former is dealt with in clause 5.4.1.2 and the latter in Section 3 of EN 1993-1-5. They are discussed in this Guide under clause 5.4.1.2. The effects of cracking of concrete can be taken into account either by using cracked section properties in accordance with clause 5.4.2.3 or, for filler-beam decks only, by

Clause 5.1.1(1)P


Clause 5.1.1(2)

redistributing the moments determined from an uncracked analysis away from the cracked sections in accordance with clause 5.4.2.9. For Class 4 sections, plate buckling effects, which have to be considered in accordance with clause 2.2 of EN 1993-1-5, can also lead to a reduction in stiffness of cross-sections. This is discussed in this Guide under clause 5.4.1.1. Global analysis can be significantly affected by flexibility at connections and by interaction of the bridge structure with the soil, particularly in fully integral bridges. Guidance on modelling joints and ground–structure interaction are given in clauses 5.1.2 and 5.1.3, respectively. Composite members and joints are commonly used in conjunction with others of structural steel. Clause 5.1.1(2) makes clear that this is the type of construction envisaged in Section 5. Significant differences between Sections 5 of EC3 and EC4 are referred to in this chapter.

5.1.2. Joint modelling Clause 5.1.2(1)

Clause 5.1.2(2)

Clause 5.1.2(3)

In analysis of bridges, it is generally possible to treat joints as either rigid or pinned, as appropriate. Clause 5.1.2(1) refers to ‘semi-continuous’ joints as an exception. They are neither ‘rigid’ nor ‘pinned’, and have sufficient flexibility to influence the bending moment transmitted. This could occur, for example, from the flexure of thin end-plates in a bolted end-plate connection. The three simplified joint models listed in clause 5.1.2(2) – simple, continuous and semicontinuous – are those given in EN 1993. Joints in steelwork have their own Eurocode part, EN 1993-1-8.19 Its design methods are for joints ‘subjected to predominantly static loading’ (its clause 1.1(1)). Resistance to fatigue is covered in EN 1993-1-942 and in clause 6.8. Clause 5.1.2(3) prohibits the use of semi-continuous composite joints (defined in clause 1.5.2.8) in bridges. An example of such a prohibited joint might be a composite main beam joined together through end-plate connections. Semi-continuous non-composite joints should also be avoided where possible, so that fatigue can be assessed using the detail categories in EN 1993-1-9. Semi-continuous joints may, in some situations, be unavoidable, such as end-plate connections between composite cross-beams and main beam webs in some U-frame bridges, but these would not be composite joints due to the lack of continuity of the slab reinforcement. The flexibility of such a joint would have to be considered in deriving the restraint provided to the compression flange by the U-frame. Design rules are given in EN 1993-1-8 and in EN 1994-1-1. Another apparent exception to the above rule concerns the slip of bolts. This is discussed under clause 5.4.1.1(7).

5.1.3. Ground–structure interaction Clause 5.1.3(1)P

Clause 5.1.3(3)

Clause 5.1.3(1)P refers to ‘deformation of supports’, so the stiffness of the bearings, piers, abutments and ground have to be taken into account in analysis. This also includes consideration of stiffness in determining effective lengths for buckling or resistance to buckling by analysis. For further guidance on this, see Section 5.2 below. The effects of differential settlement must also be included in analysis, although from clause 5.1.3(3) they may be neglected in ultimate limit state checks. Similar considerations apply to other indirect actions, such as differential temperature and differential creep. They are discussed in this Guide under clause 5.4.2.2(6).

5.2. Structural stability The following comments refer to both entire bridges and isolated members. They assume that the global analyses will be based on elastic theory. The exception in clause 5.4.3 is discussed later. All design methods must take account of: .

30

errors in the initial positions of joints (global geometric imperfections) and in the initial geometry of members (member geometric imperfections)


. .

the effects of cracking of concrete and of any semi-rigid or nominally pinned joints residual stresses in compression members (structural imperfections).

The stage at which each of these is considered or allowed for can be selected by the designer, which leads to some complexity in clauses 5.2 to 5.4.

5.2.1. Effects of deformed geometry of the structure In its clause 1.5.6, EN 1990 defines types of analysis. ‘First-order’ analysis is performed on the initial geometry of the structure. ‘Second-order’ analysis takes account of the deformations of the structure, which are a function of its loading. Clearly, second-order analysis may always be applied. With appropriate software increasingly available, second-order analysis is now relatively straightforward to perform. The criteria for neglect of second-order effects given in clauses 5.2.1(2)P and 5.2.1(3) need not then be considered. The analysis allowing for second-order effects will usually be iterative but normally the iteration will take place within the software. Methods for second-order analysis are described in text books such as that by Trahair et al.43 A disadvantage of second-order analysis is that the principle of superposition does not apply and entire load combinations must be applied to the bridge model. In this case, the critical load combinations can still first be estimated using first-order analysis, influence lines (or surfaces) and superposition of load cases. Second-order effects apply to both in-plane and out-of-plane modes of buckling, including lateral–torsional buckling. The latter behaviour is more complex and requires a finiteelement analysis using shell elements to model properly second-order effects and instability. A method of checking beams for out-of-plane instability while modelling only in-plane second-order effects is given in clause 6.3.4 of EN 1993-1-1 and discussed in section 6.4.3 of this guide. Out-of-plane second-order effects can only be neglected in bridge beams where there is sufficient lateral bracing present. In-plane second-order effects in the beams will usually be negligible and lateral–torsional buckling may be checked using one of the simplified methods permitted in clause 6.4. Integral bridges, with high axial load in the beams caused by earth pressure, may be an exception. Clause 5.2.1(3) provides a basis for the use of first-order analysis. The check is done for a particular load combination and arrangement. The provisions in this clause are similar to those for elastic analysis in the corresponding clause in EN 1993-2. Clause 5.2.1(3) is not just for a sway mode. This is because clause 5.2.1 is relevant not only to complete frames but also to the design of individual columns (see clause 6.7.3.4 for composite columns, and comments on it). Such members may be held in position against sway but still be subject to significant second-order effects due to bowing. Second-order effects in local and global modes are illustrated in Fig. 5.1. In an elastic frame, second-order effects are dependent on the proximity of the design loads to the elastic critical buckling load. This is the basis for expression (5.1), in which cr is defined as ‘the factor . . . to cause elastic instability’. This may be taken as the load factor at which bifurcation of equilibrium occurs. For a column or frame, it is assumed that there are no member imperfections, and that only vertical loads are present, usually at their maximum design values. These are replaced by a set of loads that produce the same set of member axial forces without any bending. An eigenvalue analysis then gives the factor

A

B

C

E


D

F (a)

(b)

Fig. 5.1. Examples of local and global instability: (a) local second-order effects; (b) global second-order effects

31


cr , applied to the whole of the loading, at which the system stiffness vanishes and elastic instability occurs. To sufficient accuracy, cr may also be determined by a second-order load–deflection analysis. The non-linear load–deflection response approaches asymptotically to the elastic critical value. This may be useful as some software will perform this analysis but not an elastic critical buckling analysis. The use of expression (5.1) is one way of determining if first-order analysis will suffice. Clause 5.2.1(3) also states that second-order effects may be ignored where the increases in internal actions due to the deformations from first-order analysis are less than 10%. Hence, for members braced against lateral buckling: MI =MI 10

ðD5:1Þ

where MI is the moment from first-order analysis, including the effects of initial imperfections, and MI is the increase in bending moments calculated from the deflections obtained from first-order analysis (the P– moments). By convention, the symbols or are used for deformations. They should not be confused with , as used here in MI . Application of this criterion, in principle, avoids the need for elastic critical buckling analysis but its use has some problems as discussed below. For the case of a pin-ended strut with sinusoidal bow of magnitude a0 , expression (D5.1) is the same as expression (5.1). This can be shown as follows. The extra deflection from a first-order analysis can easily be shown to be given by: a ¼ a0 FEd =Fcr

ðD5:2Þ

where FEd is the applied axial load and Fcr is the elastic critical buckling load. It follows that the extra moment from the first-order deflection is: MI ¼ FEd ða0 FEd =Fcr Þ

ðD5:3Þ

Putting equation (D5.3) into equation (D5.1) gives expression (5.1): MI =MI ¼

FEd a0 F ¼ cr ¼ cr 10 FEd ða0 FEd =Fcr Þ FEd

This direct equivalence is only valid for a pin-ended strut with a sinusoidal bow and hence sinusoidal curvature but it generally remains sufficiently accurate. (Note: It is found for a strut with equal end moments that: 8 Fcr MI =MI ¼ 2 FEd

Clause 5.2.1(4)P

32

For anything other than a pin-ended strut or statically determinate structure, it will not be easy to determine MI from the deflections found by first-order analysis. This is because in indeterminate structures, the extra moment cannot be calculated at all sections directly from the local ‘P–’ because of the need to maintain compatibility. In the example shown in Fig. 5.2, it would be conservative to assume that at mid-height, MI ¼ N. (This is similar to secondary effects of prestressing in prestressed structures.) A more accurate value could be found from a further first-order analysis that models the firstorder deflected shape found by the previous analysis. To avoid the problem that low ratios MI =MI can be obtained near points of contraflexure, the condition MI =MI 10 should be applied only at the peak moment positions between each adjacent point of contraflexure. The maximum P– bending moment in the member can again be used as a conservative estimate of MI . Clause 5.2.1(4)P is a reminder that the analysis shall account for the reductions in stiffness arising from cracking and creep of concrete and from possible non-linear behaviour of the joints. In general, such effects are dependent on the internal moments and forces, so calculation is iterative. Simplified methods are therefore given in clauses 5.4.2.2 and 5.4.2.3, where further comment is given. Manual intervention may be needed, to adjust stiffness values before repeating an analysis. It is expected however that advanced software will be written for EN 1994 to account


N

N

∆

N

N (a)

(b)

(c)

Fig. 5.2. Extra bending moments from deflection: (a) first-order moment due to imperfections; (b) firstorder deflection; (c) additional moment from deflection

automatically for these effects. The designer may of course make assumptions, although care is needed to ensure these are conservative. For example, assuming that joints have zero rotational stiffness (resulting in simply-supported composite beams) could lead to neglect of the reduction in beam stiffness due to cracking. The overall lateral stiffness would probably be a conservative value, but this is not certain. Clause 6.7.3.4(2) gives an effective flexural stiffness for doubly symmetric columns which may be used to determine cr (clause 6.7.3.4(3)) and which makes allowance for the stiffness of the concrete, including the effects of cracking, and the reinforcement. The use of this stiffness in checking composite columns is discussed in section 6.7.3 of this guide. For asymmetric composite compression members in general, such as a composite bridge deck beam in an integral bridge, the effective stiffness usually depends on the direction of bowing of the member. This is influenced by the initial camber and by the deflection under the loading considered. The deflection under design ultimate load and after creep usually exceeds the initial camber. The direction of bow is then downwards. A conservative possibility for determining cr is to ignore completely the contribution of the concrete to the flexural stiffness, including reinforcement only; this is done in Example 6.6 to determine the elastic critical buckling load under axial force. An even more conservative possibility is to base the flexural stiffness on the steel section alone. If second-order analysis is necessary, this simplification will not usually be satisfactory as the use of cracked properties throughout the structure, irrespective of the sign of the axial force in the concrete, would not satisfy the requirements of clause 5.4.2.3 regarding cracking. Generally the results of a firstorder analysis can be used to determine which areas of the structure are cracked and the section properties for second-order analysis can then be modified as necessary. The stiffness of cracked areas can be based on the above simplification. The procedure can be iterative if the extent of cracked zones is significantly altered by the second-order analysis. An effective modulus of elasticity for compressed concrete is also required to calculate the flexural stiffness of uncracked areas. Clause 6.7.3.3(4) provides a formula.

5.2.2. Methods of analysis for bridges Where it is necessary to take second-order effects and imperfections into account, EN 1993-2 clause 5.2.2 provides the following three alternative methods by reference to EN 1993-1-1 clause 5.2.2(3). .

Use of second-order analysis including both ‘global’ system imperfections and ‘local’ member imperfections as discussed in section 5.3 below. If this method is followed, no individual checks of member stability are required and members are checked for crosssection resistance only. An alternative method for bare steel members is discussed under clause 5.3.1(2). For each composite member, it is necessary to use an appropriate flexural stiffness covering the effects of cracking and creep as discussed in section 5.2.1 above.

33


.

If lateral–torsional buckling is to be covered totally by second-order analysis, appropriate finite-element analysis capable of modelling the behaviour will be required. Use of second-order analysis including ‘global’ system imperfections only. For individual bare steel members, stability checks are then required according to EN 1993-2 clause 6.3. Since the member end forces and moments include second-order effects from global behaviour, the effective length of individual members is then based on the member length, rather than a greater effective length that includes the effects of global sway deformations. Note that when clause 6.3.3 of EN 1993-1-1 is used for member checks of bare steel members, the member moments will be further amplified by the ‘kij ’ parameters. Since the second-order analysis will already have amplified these moments, providing sufficient nodes have been included along the member in the analysis model, this is conservative and it would be permissible to set the ‘kij ’ parameters equal to unity where they exceed unity. However, the imperfections within the members have not been considered or amplified by the second-order analysis. These are included via the first term in the equations in this clause: NEd NRk =M1

.

For composite members in compression and bending, buckling resistance curves cannot be used, and the moments from member imperfections in the member length should be added. Second-order effects within the member are accounted for by magnifying the resulting moments from the local imperfections within the length of the member according to clauses 6.7.3.4(4) and 6.7.3.4(5) using an effective length based on the member length, and then checking the resistance of cross-sections. Only the local member imperfections need to be amplified if sufficient nodes have been included along the member in the analysis model, as all other moments will then have been amplified by the second-order global analysis. Further comment and a flow chart are given under clause 6.7.3.4. Use of first-order analysis without modelled imperfections. For bare steel members, the verification can be made using clause 6.3 of EN 1993-2 with appropriate effective lengths. All second-order effects are then included in the relevant resistance formulae. This latter method will be most familiar to bridge engineers in the UK, as tables of effective lengths for members with varying end conditions of rotational and positional fixity have commonly been used. The use of effective lengths for this method is discussed in the Designers’ Guide to EN 1993-2.4 For composite compression members, this approach is generally not appropriate. The method of clause 6.7.3 is based on calculation of second-order effects within members, followed by checks on resistance of cross-sections. No buckling resistance curves are provided. Composite beams in bending alone can however be checked for lateral– torsional buckling satisfactorily following this method.

Second-order analysis itself can be done either by direct computer analysis that accounts for the deformed geometry or by amplification of the moments from a first-order analysis (including the effects of imperfections) using clause 5.2.2(5) of EN 1993-2. Where either approach is used, it should only be performed by experienced engineers because the guidance on the use of imperfections in terms of shapes, combinations and directions of application is not comprehensive in EC3 and EC4 and judgement must be exercised.

5.3. Imperfections 5.3.1. Basis Clause 5.3.1(1)P

34

Clause 5.3.1(1)P lists possible sources of imperfection. Subsequent clauses (and also clause 5.2) describe how these should be allowed for. This may be by inclusion in the global analyses or in methods of checking resistance, as explained above. Imperfections comprise geometric imperfections and residual stresses. The term ‘geometric imperfection’ is used to describe departures from the intended centreline setting-out


dimensions found on drawings, which occur during fabrication and erection. This is inevitable as construction work can only be executed to certain tolerances. Geometric imperfections include lack of verticality, lack of straightness, lack of fit and minor joint eccentricities. The behaviour of members under load is also affected by residual stresses within the members. Residual stresses can lead to yielding of steel occurring locally at lower applied external load than predicted from stress analysis ignoring such effects. The effects of residual stresses can be modelled by additional geometric imperfections. The equivalent geometric imperfections given in EC3 and EC4 cover both geometric imperfections and residual stresses. Clause 5.3.1(2) requires imperfections to be in the most unfavourable direction and form. The most unfavourable geometric imperfection normally has the same shape as the lowest buckling mode. This can sometimes be difficult to find, but it can be assumed that this condition is satisfied by the Eurocode methods for checking resistance that include effects of member imperfections (see comments on clause 5.2.2). Clause 5.3.2(11) of EN 1993-1-1 covers the use of a unique global and local system imperfection based on the lowest buckling mode. This can generally only be used for bare steel members as the imperfection parameter is required and this is not provided for composite members. The method is discussed in the Designers’ Guide to EN 1993-2.4

Clause 5.3.1(2)

5.3.2. Imperfections for bridges Generally, an explicit treatment of geometric imperfections is required for composite frames. In both EN 1993-1-1 and EN 1994-1-1 the values are equivalent rather than measured values (clause 5.3.2(1)) because they allow for effects such as residual stresses, in addition to imperfections of shape. Clause 5.3.2(2) covers bracing design. In composite bridges, the deck slab acts as plan bracing. Compression flanges that require bracing occur in hogging regions of beam-andslab bridges and in sagging regions of half-through bridges, bowstring arches and similar structures. The bracing of compression flanges in sagging regions differs little from that in all-steel bridges, and is discussed in the Designers’ Guide to EN 1993-2.4 Steel bottom flanges in hogging regions of composite bridges are usually restrained laterally by continuous or discrete transverse frames. For deep main beams, plan bracing at bottom-flange level may also be used. Where the main beams are rolled I-sections, their webs may be stiff enough to serve as the vertical members of continuous inverted-U frames, which are completed by the shear connection and the deck slab. These systems are discussed under clause 6.4.2. Discrete U-frame bracing can be provided at the location of vertical web stiffeners. These frames need transverse steel members. If these are provided just below the concrete deck, they should be designed as composite. Otherwise, design for shrinkage and temperature effects in the transverse direction becomes difficult. This problem is often avoided by placing the steel cross-member at lower level, so creating an H-frame. Both types of frame provide elastic lateral restraint at bottom-flange level, with a spring stiffness that is easily calculated. The design transverse forces for these frames, or for plan bracing, arise from lateral imperfections in the compressed flanges. For these imperfections, clause 5.3.2(2) refers to EN 1993-2, which in turn refers to clauses 5.3.2 to 5.3.4 of EN 1993-1-1. The design transverse forces, FEd , and a design method are given in clause 6.3.4.2 of EN 1993-2, though it refers specifically only to U-frame restraints. Comments on these clauses are in the relevant Guides in this series.4;7 The relevant imperfections for analysis of the bracing system are not necessarily the same as those for the bridge beams themselves. In hogging regions of continuous beam-and-slab bridges, distortional lateral buckling is usually the critical mode. It should not be assumed that a point of contraflexure is a lateral restraint, for the buckling half-wavelength can exceed the length of flange in compression.44

Clause 5.3.2(1) Clause 5.3.2(2)

35


Where the restraint forces are to be transmitted to end supports by a system of plan bracing, this system should be designed to resist the more onerous of the transverse forces FEd from each restraint within a length equal to the half wavelength of buckling, and the forces generated by an overall flange bow in each flange according to clause 5.3.3 of EN 1993-1-1. For the latter case, the overall bow is given as e0 ¼ m L=500, where m is the reduction factor for the number of interconnected beams (m ¼ 0:866 for two beams), and L is the span. The plan bracing may be designed for an equivalent uniformly-distributed force per beam of 8NEd ðe0 þ q Þ=L, where q is the deflection of the bracing, and NEd is the maximum compressive force in the flange. For very stiff bracing, the total design lateral force for the bracing is: X X 8 NEd =L ðm L=500Þ ¼ NEd m =62:5 Clause 5.3.2(2) should also be used for system imperfections for composite columns, although its scope is given as ‘stabilizing transverse frames’. Its reference to clause 5.3 of EN 1993-2 leads to relevant clauses in EN 1993-1-1, as follows. Initial out-of-plumb of a column is given in clause 5.3.2(3) of EN 1993-1-1 which, although worded for ‘frames’, is applicable to a single column or row of columns. Where a steel column is very slender and has a moment-resisting joint at one or both ends, clause 5.3.2(6) of EN 1993-1-1 requires its local bow imperfection to be included in the secondorder global analysis used to determine the action effects at its ends. ‘Very slender’ is defined as: qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi > 0:5 Afy =NEd

Clause 5.3.2(3)

Clause 5.3.2(4)

It is advised that this rule should be used also for composite columns, in the form cr < 4, with cr as defined in clause 5.2.1(3). This is obtained by replacing Afy by Npl . Clause 5.3.2(3) covers imperfections in composite columns and compression members (e.g. in trusses), which must be considered explicitly. It refers to material in clause 6.7.3, which appears to be limited, by clause 6.7.3(1), to uniform members of doubly symmetrical cross-section. Clause 6.7.2(9), which is of general applicability, also refers to Table 6.5 of clause 6.7.3 for member imperfections; but the table only covers typical cross-sections of columns. Imperfections in compressed beams, which occur in integral bridges, appear to be outside the scope of EN 1994. The imperfections for buckling curve d in Table 5.1 of EN 1993-1-1 could conservatively be used for second-order effects in the plane of bending. For composite bridges with the deck slab on top of the main beams, lateral buckling effects can subsequently be included by a check of the compression flange using the member resistance formulae in clause 6.3 of EN 1993-1-1. Guidance on verifying beams in integral bridges in bending and axial load is discussed in section 6.4 of this guide. Clause 5.3.2(4) covers global and local imperfections in steel compression members, by reference to EN 1993-2. Imperfections for arches are covered in Annex D of EN 1993-2.

5.4. Calculation of action effects 5.4.1. Methods of global analysis

Clause 5.4.1.1(1)

36

EN 1990 defines several types of analysis that may be appropriate for ultimate limit states. For global analysis of bridges, EN 1994-2 gives three methods: linear elastic analysis, with or without corrections for cracking of concrete, and non-linear analysis. The latter is discussed in section 5.4.3 below, and is rarely used in practice. Clause 5.4.1.1(1) permits the use of elastic global analysis even where plastic (rectangularstress-block) theory is used for checking resistances of cross-sections. For resistance to flexure, these sections are in Class 1 or 2, and commonly occur in mid-span regions. There are several reasons4547 why the apparent incompatibility between the methods used for analysis and for resistance is accepted. It is essentially consistent with UK practice, but


Moments with bending moment at C reduced to Mel,C P B

C

D Moments from elastic analysis Mel,C Mpl,C

Fig. 5.3. Effect of mixing section classes, and an approximate method for checking bending moments at internal supports

care should be taken with mixing section classes within a bridge when elastic analysis is used. An example is a continuous bridge, with a mid-span section designed in bending as Class 2 and the section at an internal support as Class 3. The Class 3 section may become overstressed by the elastic moments shed from mid-span while the plastic section resistance develops there and stiffness is lost. There is no such incompatibility for Class 3 or 4 sections, as resistance is based on elastic models. Mixed-class design has rarely been found to be a problem, as the load cases producing maximum moment at mid-span and at a support rarely coexist, except where adjacent spans are very short compared to the span considered. A relevant design rule is given in clause 6.2.1.3(2). If redistribution is required to be checked, the conservative method illustrated in Fig. 5.3 may be used. In this example there is a Class 2 section at mid-span of the central span, and the support sections are Class 3. A simplified load case that produces maximum sagging moment is shown. Elastic analysis for the load P gives a bending moment at cross-section C that exceeds the elastic resistance moment, Mel;C . The excess moment is redistributed from section C, giving the distribution shown by the dashed line. In reality, the moment at C continues to increase, at a reduced rate, after the elastic value Mel;C is reached, so the true distribution lies between those shown in Fig. 5.3. The upper distribution therefore provides a safe estimate of the moments at supports B and D, and can be used to check that the elastic resistance moment is not exceeded at these points. Elastic global analysis is required for serviceability limit states (clause 5.4.1.1(2)) to enable Clause 5.4.1.1(2) yielding of steel to be avoided. Linear elastic analysis is based on linear stress–strain laws, so for composite structures, ‘appropriate corrections for . . . cracking of concrete’ are required. These are given in clause 5.4.2.3, and apply also for ultimate limit states. Clause 5.4.1.1(3) requires elastic analysis for fatigue, to enable realistic ranges of fatigue Clause 5.4.1.1(3) stress to be predicted. The effects of shear lag, local buckling of steel elements and slip of bolts must also be considered where they significantly influence the global analysis. Shear lag and local buckling effects can reduce member stiffness, while slip in bolt holes causes a localized loss of stiffness. Shear lag is discussed under clause 5.4.1.2, and plate buckling and bolt slip are discussed below. Methods for satisfying the principle of clause 5.4.1.1(4)P are given for local buckling in Clause 5.4.1.1(4)P clauses 5.4.1.1(5) and (6). These refer to the classification of cross-sections, the established Clause 5.4.1.1(5) method of allowing for local buckling of steel flanges and webs in compression. It determines Clause 5.4.1.1(6) the available methods of global analysis and the basis for resistance to bending. The classification system is defined in clause 5.5.

Plate buckling In Class 4 sections (those in which local buckling will occur before the attainment of yield), plate buckling can lead to a reduction of stiffness. The in-plane stiffness of perfectly flat plates suddenly reduces when the elastic critical buckling load is reached. In ‘real’ plates that have imperfections, there is an immediate reduction in stiffness from that expected from the gross

37


Fig. 5.4. Stress distribution across width of slender plate

plate area because of the growth of the bow imperfections under load. This stiffness continues to reduce with increasing load. This arises because non-uniform stress develops across the width of the plate as shown in Fig. 5.4. The non-uniform stress arises because the development of the buckle along the centre of the plate leads to a greater developed length of the plate along its centreline than along its edges. Thus the shortening due to membrane stress, and hence the membrane stress itself, is less along the centreline of the plate. This loss of stiffness must be considered in the global analysis where significant. It can be represented by an effective area or width of plate, determined from clause 2.2 of EN 1993-1-5. This area or width is greater than that used for resistance, which is given in clause 4.3 of EN 1993-1-5. The loss of stiffness may be ignored when the ratio of effective area to gross cross-sectional area exceeds a certain value. This ratio may be given in the National Annex. The recommended value, given in a Note to clause 2.2(5) of EN 1993-1-5, is 0.5. This should ensure that plate buckling effects rarely need to be considered in the global analysis. It is only likely to be of relevance for the determination of pre-camber of box girders under self-weight and wet concrete loads. After the deck slab has been cast, buckling of the steel flange plate will be prevented by its connection to the concrete flange via the shear connection.

Clause 5.4.1.1(7)

Clause 5.4.1.1(8)

38

Effects of slip at bolt holes and shear connectors Clause 5.4.1.1(7) requires consideration of ‘slip in bolt holes and similar deformations of connecting devices’. This applies to both first- and second-order analyses. There is a similar rule in clause 5.2.1(6) of EN 1993-1-1. No specific guidance is given in EN 1993-2 or EN 1994-2. Generally, bolt slip will have little effect in global analysis. It has often been practice in the UK to design bolts in main beam splices to slip at ultimate limit states (Category B to clause 3.4.1 of EN 1993-1-8). Although slip could alter the moment distribution in the beam, this is justifiable. Splices are usually near to the point of contraflexure, so that slip will not significantly alter the distribution of bending moment. Also, the loading that gives maximum moment at the splice will not be fully coexistent with that for either the maximum hogging moment or maximum sagging moment in adjacent regions. It is advised that bolt slip should be taken into account for bracing members in the analysis of braced systems. This is because a sudden loss of stiffness arising from bolt slip gives an increase in deflection of the main member and an increased force on the bracing member, which could lead to overall failure. Ideally, therefore, bolts in bracing members should be designed as non-slip at ultimate limit state (Category C to EN 1993-1-8). The ‘similar deformations’ quoted above could refer to slip at an interface between steel and concrete, caused by the flexibility of shear connectors. The provisions on shear connection in EN 1994 are intended to ensure that slip is too small to affect the results of elastic global analysis or the resistance of cross-sections. Clause 5.4.1.1(8) therefore permits


internal moments and forces to be determined assuming full interaction where shear connection is provided in accordance with EN 1994. Slip of shear connectors can also affect the flexural stiffness of a composite joint. A relevant design method is given in clause A.3 of EN 1994-1-1. It is mainly applicable to semi-continuous joints, and so is not included in EN 1994-2. An exception to the rules on allowing for cracking of concrete is given in clause 5.4.1.1(9), for the analysis of transient situations during erection stages. This permits uncracked global analysis to be used, for simplicity.

Effective width of flanges for shear lag Shear lag is defined in clause 5.4.1.2(1) with reference to the ‘flexibility’ of flanges due to in-plane shear. Shear lag in wide flanges causes the longitudinal bending stress adjacent to the web to exceed that expected from analysis with gross cross-sections, while the stress in the flange remote from the web is much lower than expected. This shear lag also leads to an apparent loss of stiffness of a section in bending which can be important in determining realistic distributions of moments in analysis. The determination of the actual distribution of stress is a complex problem. The Eurocodes account for both the loss of stiffness and localized increase in flange stresses by the use of an effective width of flange which is less than the actual available flange width. The effective flange width concept is artificial but, when used with engineering bending theory, leads to uniform stresses across the whole reduced flange width that are equivalent to the peak values adjacent to the webs in the true situation. The rules that follow provide effective widths for resistance of cross-sections, and simpler values for use in global analyses. The rules use the word ‘may’ because clause 5.4.1.2(1) permits ‘rigorous analysis’ as an alternative. This is not defined, but should take account of the many relevant influences, such as the cracking of concrete. Steel flanges For ‘steel plate elements’ clause 5.4.1.2(2) refers to EN 1993-1-1. This permits shear lag to be neglected in rolled sections and welded sections ‘with similar dimensions’, and refers to EN 1993-1-5 for more slender flanges. In these, the stress distribution depends on the stiffening to the flanges and any plasticity occurring for ultimate limit state behaviour. The elastic stress distribution can be modelled using finite-element analysis with appropriate shell elements. The rules in EN 1993-1-5 are not discussed further in this guide but are covered in the Designers’ Guide to EN 1993-2.4 Different values of effective width apply for cross-section design for serviceability and ultimate limit states, and the value appropriate to the location of the section along the beam should be used. Simplified effective widths, taken as constant throughout a span, are allowed in the global analysis.

Clause 5.4.1.1(9)

Clause 5.4.1.2(1)

Clause 5.4.1.2(2)

Concrete flanges Effective width of concrete flanges is covered in clauses 5.4.1.2(3) to (7). The behaviour is Clauses 5.4.1.2(3) complex, being influenced by the loading configuration, and by the extent of cracking and of to (7) yielding of the longitudinal reinforcement, both of which help to redistribute the stress across the cross-section. The ability of the transverse reinforcement to distribute the forces is also relevant. The ultimate behaviour in shear of wide flanges is modelled by a truss analogy similar to that for the web of a deep concrete beam. The values for effective width given in this clause are simpler than those in BS 5400:Part 5, and similar to those in BS 5950:Part 3.1:1990.48 The effective width at mid-span and internal supports is given by equation (5.3): X beff ¼ b0 þ bei ð5:3Þ where b0 is the distance between the outer shear connectors and bei is either be1 or be2 , as shown in Fig. 5.5, or the available width b1 or b2 , if lower.

39


Le = 0.25(L1 + L2) for beff,2

Le = 2L3 for beff,2

Le = 0.85L1

Le = 0.70L2

for beff,1

for beff,1

L2

L1

L1/4

L1/2

beff,1

beff,0

L1/4

beff,2

L2/4

L2/2

beff,1

L3

L2/4

beff,2

b0 b1

b2

be1

be2

Centreline

Fig. 5.5. Symbols and equivalent spans, for effective width of concrete flange (Source: based on Fig. 5.1 of EN 1994-2)

Clause 5.4.1.2(8)

40

Each width bei is limited to Le =8, where Le is the assumed distance between points of zero bending moment. It depends on the region of the beam considered and on whether the bending moment is hogging or sagging. This is shown in Fig. 5.5, which is based on Fig. 5.1. The values are generally lower than those in EN 1992-1-1 for reinforced concrete T-beams. To adopt those would often increase the number of shear connectors. Without evidence that the greater effective widths are any more accurate, the established values for composite beams have mainly been retained. In EN 1992-1-1, the sum of the lengths Le for sagging and hogging regions equals the span of the beam. In reality, points of contraflexure are dependent on the load arrangement. EN 1994, like EN 1993, therefore gives a larger effective width at an internal support. In sagging regions, the assumed distances between points of contraflexure are the same in all three codes. Although there are significant differences between effective widths for supports and midspan regions, it is possible to ignore this in elastic global analysis (clause 5.4.1.2(4)). This is because shear lag has limited influence on the results. There can however be some small advantage to be gained by modelling in analysis the distribution of effective width along the members given in Fig. 5.5 or Fig. 5.1, as this will tend to shed some moment from the hogging regions into the span. It would also be appropriate to model the distribution of effective widths more accurately in cable-stayed structures, but Fig. 5.1 does not cover these. Example 5.1 below illustrates the calculation of effective width. Some limitations on span length ratios when using Fig. 5.1 should be made so that the bending-moment distribution within a span conforms with the assumptions in the figure. It is suggested that the limitations given in EN 1992 and EN 1993 are adopted. These limit the use to cases where adjacent spans do not differ by more than 50% of the shorter span and a cantilever is not longer than half the adjacent span. For other span ratios or moment distributions, the distance between points of zero bending moment, Le , should be calculated from the moment distribution found from an initial analysis. Where it is necessary to determine a more realistic distribution of longitudinal stress across the width of the flange, clause 5.4.1.2(8) refers to clause 3.2.2 of EN 1993-1-5. This might be necessary, for example, in checking a deck slab at a transverse diaphragm between main


beams at a support, where the deck slab is in tension under global bending and also subjected to a local hogging moment from wheel loads. The use of EN 1993-1-5 can be beneficial here, as often the greatest local effects in a slab occur in the middle of the slab between webs where the global longitudinal stresses are lowest.

Composite plate flanges Clause 5.4.1.2(8) recommends the use of its stress distribution for both concrete and steel flanges. Where the flange is a composite plate, shear connection is usually concentrated near the webs, so this stress distribution is applicable. Effective widths of composite plates in bridges are based on clause 5.4.1.2, but with a different definition of b0 , given in clause 9.1(3). Composite trusses Clause 5.4.1.2(9) applies where a longitudinal composite beam is also a component of a larger structural system, such as a composite truss. For loading applied to it, the beam is continuous over spans equal to the spacing of the nodes of the truss. For the axial force in the beam, the relevant span is that of the truss.

Clause 5.4.1.2(9)

Example 5.1: effective widths of concrete flange for shear lag A composite bridge has the span layout and cross-section shown in Fig. 5.6. The effective width of top flange for external and internal beams is determined for the mid-span regions BC and DE, and the region CD above an internal support. A

B

D

C

L1 = 19.000 m

E

L2 = 31.000 m

L3 = 19.000 m

250 250

b2

b1

be1

C/L

be2

C/L

3100 2000

3100

3100

3100

2000

Fig. 5.6. Elevation and typical cross-section of bridge for Example 5.1

The effective spans Le , from Fig. 5.5, and the lengths Le =8 are shown in Table 5.1. For an external beam, the available widths on each side of the shear connection are: b1 ¼ 1:875 m; b2 ¼ 1:425 m: Table 5.1. Effective width of concrete flange of composite T-beam External beam

Internal beam

Region

BC

CD

DE

BC

CD

DE

Le (m) Le =8 (m) beff (m)

16.15 2.019 3.550

12.50 1.563 3.238

21.70 2.712 3.550

16.15 2.019 3.100

12.50 1.563 3.100

21.70 2.712 3.100

41


The effective widths are the lower of these values and Le =8, plus the width of the shear connection, as follows: for BC and DE;

beff ¼ 1:875 þ 1:425 þ 0:25 ¼ 3:550 m

for CD;

beff ¼ 1:563 þ 1:425 þ 0:25 ¼ 3:238 m

For an internal beam, the available widths on each side of the shear connection are: b1 ¼ b2 ¼ 1:425 m These available widths are both less than Le =8, and so govern. The effective widths are: for BC, CD, and DE, beff ¼ 2 1:425 þ 0:25 ¼ 3:100 m

5.4.2. Linear elastic analysis Clause 5.4.2.1(1)

Clause 5.4.2.2

Clause 5.4.2.2(2)

Cracking, creep, shrinkage, sequence of construction, and prestressing, listed in clause 5.4.2.1(1), can all affect the distribution of action affects in continuous beams and frames. This is always important for serviceability limit states, but can in some situations be ignored at ultimate limit states, as discussed under clause 5.4.2.2(6). Cracking of concrete is covered in clause 5.4.2.3.

Creep and shrinkage of concrete The rules provided in clause 5.4.2.2 allow creep to be taken into account using a modular ratio nL , that depends on the type of loading, and on the concrete composition and age at loading. This modular ratio is used both for global analysis and for elastic section analysis. It is defined in clause 5.4.2.2(2) by: nL ¼ n0 ð1 þ

L ðt; t0 ÞÞ

ð5:6Þ

where n0 is the modular ratio for short-term loading, Ea =Ecm . The concrete modulus Ecm is obtained from EN 1992 as discussed in section 3.1 of this guide. The creep coefficient ðt; t0 Þ is also obtained from EN 1992. The creep multiplier L takes account of the type of loading. Its values are given in clause 5.4.2.2(2) as follows: . .

.

for permanent load, L ¼ 1:1 for the primary and secondary effects of shrinkage (and also the secondary effects of creep, clause 5.4.2.2(6)), L ¼ 0:55 for imposed deformations, L ¼ 1:5.

The reason for the factor L is illustrated in Fig. 5.7. This shows three schematic curves of the change of compressive stress in concrete with time. The top one, labelled S, is typical of stress caused by the increase of shrinkage with time. Concrete is more susceptible to creep when young, so there is less creep ( L ¼ 0:55) than for the more uniform stress caused by permanent loads (line P). The effects of imposed deformations can be significantly reduced by creep when the concrete is young, so the curve is of type ID, with L ¼ 1:5. The value for permanent loading on reinforced concrete is 1.0. It is increased to 1.1 for composite

σc/σc,0

S P

1.0

ID

0

Time

Fig. 5.7. Time-dependent compressive stress in concrete, for three types of loading

42


Table 5.2. Values of 0 ¼ ð1; t0 Þ and modular ratio nL

t0 ¼ 7 days t0 ¼ 28 days

h0 ¼ 250 mm

h0 ¼ 500 mm

2.48, 23.7 1.90, 19.6

2.30, 22.4 1.78, 18.8

members because the steel component does not creep. Stress in concrete is reduced by creep less than it would be in a reinforced member, so there is more creep. These values are based mainly on extensive theoretical work on composite beams of many sizes and proportions.49 The factor L performs a similar function to the ageing coefficient found in Annex KK of EN 1992-2 and in the calculation for loss of prestress in clause 5.10.6 of EN 1992-1-1. The creep factor ðt; t0 Þ depends on the age of the concrete, t, at which the modular ratio is being calculated (usually taken as infinity) and the age of the concrete at first loading, t0 . For age t0 , clauses 5.4.2.2(3) and (4) make recommendations for permanent load and shrinkage, respectively. Since most bridges will follow a concrete pour sequence rather than have all the concrete placed in one go, this age at first loading could vary throughout the bridge. Clause 5.4.2.2(3) permits an assumed ‘mean’ value of t0 to be used throughout. This simplification is almost a necessity as it is rare for the designer to have sufficient knowledge of the construction phasing at the design stage to be more accurate than this, but some estimate of the expected timings is still required. ‘First loading’ could occur at an age as low as a week, for example, from erection of precast parapets, but the mean age for a multi-span bridge is unlikely to be less than a month. The creep coefficient depends also on the effective thickness of the concrete element considered, h0 . There is no moisture loss through sealed surfaces, so these are assumed to be at midthickness of the member. After striking of formwork, a deck slab of thickness, say, 250 mm, has two free surfaces, and an effective thickness of 250 mm. The application of waterproofing to the top surface increases this thickness to 500 mm, which reduces subsequent creep. The designer will not know the age(s) of the deck when waterproofed, and so must make assumptions on the safe side. Fortunately, the modular ratio is not sensitive to either the age of loading or the effective thickness. As resistances are checked for the structure at an early age, it is on the safe side for the long-term checks to overestimate creep. As an example, let us suppose that the short-term modular ratio is n0 ¼ 6:36 (as found in a subsequent example), and that a concrete deck slab has a mean thickness of 250 mm, with waterproofing on one surface. The long-term modular ratio is calculated for t0 ¼ 7 days and 28 days, and for h0 ¼ 250 mm and 500 mm. For ‘outdoor’ conditions with relative humidity 70%, the values of ð1; t0 Þ given by Annex B of EN 1992-1-1 with L ¼ 1:1 are as shown in Table 5.2. The resulting range of values of the modular ratio nL is from 18.8 to 23.7. A difference of this size has little effect on the results of a global analysis of continuous beams with all spans composite, and far less than the effect of the difference between n ¼ 6:4 for imposed load and around 20 for permanent load. For stresses at cross-sections of slab-on-top decks, the modular ratio has no influence in regions where the slab is in tension. In mid-span regions, compression in concrete is rarely critical, and maximum values occur at a low age, where creep is irrelevant. In steel, bottom-flange tension is the important outcome, and is increased by creep. From Table 5.2, h0 has little effect, and the choice of the low value of 7 days for age at first loading is on the safe side. Modular ratios are calculated in Example 5.2 below.

Shrinkage modified by creep For shrinkage, the advice in clause 5.4.2.2(4) to assume t0 ¼ 1 day rarely leads to a modular ratio higher than that for permanent actions, because of the factor L ¼ 0:55. Both the

Clause 5.4.2.2(3)

Clause 5.4.2.2(4)

43


Table 5.3. Effects of shrinkage

Clause 5.4.2.2(5)

Clause 5.4.2.2(6)

h0 (mm)

RH (%)

106 "sh

nL

250 500 250

70 70 75

340 304 305

18.8 18.0 19.2

long-term shrinkage strain and the creep coefficient are influenced by the assumed effective thickness h0 . For the preceding example, the 1-day rule gives the values in rows 1 and 2 of Table 5.3. It shows that doubling h0 has negligible effect on nL , but reduces shrinkage strain by 10%. Increasing the assumed mean relative humidity (RH) by only 5% has the same effect on shrinkage strain as doubling h0 , and negligible effect on nL . The error in an assumed RH may well exceed 5%. For this example, the ‘safe’ choices for shrinkage effects are h0 ¼ 250 mm, and an estimate for RH on the low side. As the concession in clause 5.4.2.2(3) (the use of a single time t0 for all creep coefficients) refers to ‘loads’, not to ‘actions’, it is not clear if shrinkage may be included. It is conservative to do so, because when t0 is assumed to exceed 1 day, the relief of shrinkage effects by creep is reduced. Hence, a single value of nL ð1; t0 Þ may usually be used in analyses for permanent actions, except perhaps in special situations, to which clause 5.4.2.2(5) refers.

Secondary effects of creep Where creep deflections cause a change in the support reactions, this leads to the development of secondary moments. This might occur, for example, where there are mixtures of reinforced concrete and steel–composite spans in a continuous structure. The redistribution arises because the ‘free’ creep deflections are not proportional everywhere to the initial elastic deflections and therefore the ‘free’ creep deflection would lead to some non-zero deflection at the supports. Other construction sequences could produce a similar effect but this does not affect normal steel–composite bridges to any significant extent. Clause 5.4.2.2(6) is however a prompt that the effects should be considered in the more unusual situations. Calculation of creep redistribution is more complex than for purely concrete structures, and is explained, with an example, in Ref. 50. The redistribution effects develop slowly with time, so L ¼ 0:55. Cross-sections in Class 1 or 2 Clause 5.4.2.2(6) is one of several places in EN 1994-2 where, in certain global analyses, various ‘indirect actions’, that impose displacements and/or rotations, are permitted to be ignored where all cross-sections are either Class 1 or 2. Large plastic strains are possible for beams where cross-sections are Class 1. Class 2 sections exhibit sufficient plastic strain to attain the plastic section capacity but have limited rotation capacity beyond this point. This is however normally considered adequate to relieve the effects of imposed deformations derived from elastic analysis, and EN 1994 therefore permits such relief to be taken. The corresponding clause 5.4.2(2) in EN 1993-2 only permits the effects of imposed deformations to be ignored where all sections are Class 1, so there is an inconsistency at present. In EN 1994, the effects which can be neglected in analyses for ultimate limit states other than fatigue, provided that all sections are Class 1 or 2, are as follows: . . . . .

44

differential settlement: clause 5.1.3(3) secondary creep redistribution of moments: clause 5.4.2.2(6) primary and secondary shrinkage and creep: clause 5.4.2.2(7) effects of staged construction: clause 5.4.2.4(2) differential temperature: clause 5.4.2.5(2).


The further condition that there should not be any reduction of resistance due to lateral– torsional buckling is imposed in all of these clauses, and is discussed under clause 6.4.2(1).

Primary and secondary effects of creep and shrinkage Clause 5.4.2.2(7) requires ‘appropriate’ account to be taken of both the primary and Clause 5.4.2.2(7) secondary effects of creep and shrinkage of the concrete. The recommended partial factor for shrinkage effects at ultimate limit states is SH ¼ 1, from clause 2.4.2.1 of EN 1992-1-1. In a fully-restrained member with the slab above the steel beam, shrinkage effects can be split into a hogging bending moment, an axial tensile force, and a set of self-equilibrated longitudinal stresses, as shown in Example 5.3 below. Where bearings permit axial shortening, there is no tensile force. In a statically determinate system, the hogging bending moment is released, causing sagging curvature. These, with the locked-in stresses, are the primary effects. They are reduced almost to zero where the concrete slab is cracked through its thickness. In a statically indeterminate system, such as a continuous beam, the primary shrinkage curvature is incompatible with the levels of the supports. It is counteracted by bending moments caused by changes in the support reactions, which increase at internal supports and reduce at end supports. The moments and the associated shear forces are the secondary effects of shrinkage. Clause 5.4.2.2(7) permits both types of effect to be neglected in some checks for ultimate limit states. This is discussed under clause 5.4.2.2(6). Clause 5.4.2.2(8) allows the option of neglecting primary shrinkage curvature in cracked Clause 5.4.2.2(8) regions.51 It would be reasonable to base this cracked zone on the same 15% of the span allowed by clause 5.4.2.3(3), where this is applicable. The use of this option reduces the secondary hogging bending at supports. These moments, being a permanent effect, enter into all load combinations, and may influence design of what is often a critical region. The long-term effects of shrinkage are significantly reduced by creep, as illustrated in clause 5.4.2.2(4). Where it is necessary to consider shrinkage effects within the first year or so after casting, a value for the relevant free shrinkage strain can be obtained from clause 3.1.4(6) of EN 1992-1-1. Primary effects of shrinkage are calculated in Example 5.3 below. The influence of shrinkage on serviceability verifications is dealt with in Chapter 7. For creep in columns, clause 5.4.2.2(9) refers to clause 6.7.3.4(2), which in turn refers to Clause 5.4.2.2(9) an effective modulus for concrete given in clause 6.7.3.3(4). If separate analyses are to be made for long-term and short-term effects, clause 6.7.3.3(4) can be used assuming ratios of permanent to total load of 1.0 and zero, respectively. Shrinkage effects in columns are unimportant, except in very tall structures. Clause 5.4.2.2(10) excludes the use of the preceding simplified methods for members Clause 5.4.2.2(10) with both flanges composite and uncracked. The ‘uncracked’ condition is omitted from clause 5.4.2.2(2), which is probably an oversight. This exclusion is not very restrictive, as new designs of this type are unusual in the UK. It may occur in strengthening schemes where the resistance of the compression flange is increased by making it composite over a short length. Torsional stiffness of box girders For box girders with a composite top flange or with a concrete flange closing the top of an open U-section, the torsional stiffness is usually calculated by reducing the thickness of the gross concrete flange on the basis of the appropriate long- or short-term modular ratio, and maintaining the centroid of the transformed flange in the same position as that of the gross concrete flange. From clause 5.4.2.2(11), the short-term modular ratio should be based on Clause 5.4.2.2(11) the ratio of shear moduli, n0;G ¼ Ga =Gcm , where for steel, Ga ¼ 81:0 kN/mm2 from clause 3.2.6 of EN 1993-1-1 and, for concrete: Gcm ¼ Ecm =½2ð1 þ c Þ

45


Transformed thickness: uncracked: hc /nL,G cracked: 0.5hc /nL,G

hc

Torsional perimeter

Fig. 5.8. Torsional stiffness of composite box girder

Clause 3.1.3(4) of EN 1992-1-1 gives Poisson’s ratio ( c ) as 0.2 or zero, depending on whether the concrete is uncracked or cracked. For this application it is accurate enough to assume c ¼ 0:2 everywhere. The method of clause 5.4.2.2(2) should be used for the modular ratio: nL;G ¼ n0;G ð1 þ

L ðt; t0 ÞÞ

The calculation of the torsional second moment of area (in ‘steel’ units) then follows the usual procedure such that: 4A2 IT ¼ þ 0 ds tðsÞ where A0 is the area enclosed by the torsional perimeter running through the centreline of the box walls. This is shown in Fig. 5.8. For closed steel boxes, the location of the centroid of the composite flange can, for simplicity, be located on the basis of first moment of area. Þ The integral ðds=tðsÞÞ is the summation of the lengths of each part of the perimeter divided by their respective thicknesses. It is usual to treat the parts of the web projection into the flange as having the thickness of the steel web. The torsional stiffness is also influenced by flexural cracking, which can cause a significant reduction in the in-plane shear stiffness of the concrete flange. To allow for this in regions where the slab is assumed to be cracked, clause 5.4.2.3(6) recommends a 50% reduction in the effective thickness of the flange.

Clause 5.4.2.3

Clause 5.4.2.3(2)

46

Effects of cracking of concrete Clause 5.4.2.3 is applicable to beams, at both serviceability and ultimate limit states. The flow chart of Fig. 5.15 below illustrates the procedure. In conventional composite beams with the slab above the steel section, cracking of concrete reduces the flexural stiffness in hogging moment regions, but not in sagging regions. The change in relative stiffness needs to be taken into account in elastic global analysis. This is unlike analysis of reinforced concrete beams, where cracking occurs in both hogging and sagging bending, and uncracked cross-sections can be assumed throughout. A draft of EN 1994-2 permitted allowance for cracking by redistribution of hogging moments from ‘uncracked’ analysis by up to 10%. Following detailed examination of its effects,52 this provision was deleted. Clause 5.4.2.3(2) provides a general method. This is followed in clause 5.4.2.3(3) by a simplified approach of limited application. Both methods refer to the ‘uncracked’ and ‘cracked’ flexural stiffnesses Ea I1 and Ea I2 , which are defined in clause 1.5.2. The flexural rigidity Ea I1 can usually be based on the gross concrete area excluding reinforcement with acceptable accuracy. In the general method, the first step is to determine the expected extent of cracking in beams. The envelope of moments and shears is calculated for characteristic combinations of actions, assuming uncracked sections and including long-term effects. The section is assumed to crack if the extreme-fibre tensile stress in concrete exceeds twice the mean value of the axial tensile strength given by EN 1992-1-1.


The reasons for ‘twice’ in this assumption are as follows: .

.

.

.

.

The concrete is likely to be stronger than specified, although this is partly catered for by the use of a mean rather than characteristic tensile strength. Test results for tensile strength show a wide scatter when plotted against compressive strength. Reaching fctm at the surface may not cause the slab to crack right through, and even if it does, the effects of tension stiffening are significant at the stage of initial cracking. Until after yielding of the reinforcement, the stiffness of a cracked region is greater than Ea I2 , because of tension stiffening between the cracks. The calculation uses an envelope of moments, for which regions of slab in tension are more extensive than they are for any particular loading.

The global model is then modified to reduce the beam stiffness to the cracked flexural rigidity, Ea I2 , over this region, and the structure is reanalysed. Clause 5.4.2.3(3) provides a non-iterative method, but one that is applicable only to some situations. These include conventional continuous composite beams, and beams in braced frames. The cracked regions could differ significantly from the assumed values in a bridge with highly unequal span lengths. Where the conditions are not satisfied, the general method of clause 5.4.2.3(2) should be used. The influence of cracking on the analysis of braced and unbraced frames is discussed in the Designers’ Guide to EN 1994-1-1.5 For composite columns, clause 5.4.2.3(4) makes reference to clause 6.7.3.4 for the calculation of cracked stiffness. The scope of the latter clause is limited (to double symmetry, etc.) by clause 6.7.3.1(1), where further comment is given. The reduced value of EI referred to here is intended for verifications for ultimate limit states, and may be inappropriate for analyses for serviceability. For column cross-sections without double symmetry, cracking and tension in columns are referred to in clause 6.7.2(1)P and clause 6.7.2(5)P, respectively; but there is no guidance on the extent of cracking to be assumed in global analysis. The assumption in clause 5.4.2.3(5), that effects of cracking in transverse composite members may be neglected, does not extend to decks with only two main beams. Their behaviour is influenced by the length of cantilever cross-beams, if any, and the torsional stiffness of the main beams. The method of clause 5.4.2.3(2) is applicable. Clause 5.4.2.3(6) supplements clause 5.4.2.2(11), where comment on clause 5.4.2.3(6) is given. For the effects of cracking on the design longitudinal shear for the shear connection at ultimate limit states, clause 5.4.2.3(7) refers to clause 6.6.2.1(2). This requires uncracked section properties to be used for uncracked members and for members assumed to be cracked in flexure where the effects of tension stiffening have been ignored in global analysis. Where tension stiffening and possible over-strength of the concrete (using upper characteristic values of the tensile strength) have been explicitly considered in the global analysis, then the same assumptions may be made in the determination of longitudinal shear flow. The reason for this is that tension stiffening can lead to a greater force being attracted to the shear connection than would be found from a fully cracked section analysis. The simplest and most conservative way to consider this effect is to determine the longitudinal shear with an uncracked concrete flange. The same approach is required for fatigue where tension stiffening could again elevate fatigue loads on the studs according to clauses 6.8.5.4(1) and 6.8.5.5(2). For longitudinal shear at serviceability limit states, clause 5.4.2.3(8) gives, in effect, the same rules as for ultimate limit states, explained above.

Stages and sequence of construction The need to consider staged construction is discussed in section 2.2 of this guide. The reason for allowing staged construction to be ignored at the ultimate limit state, if the conditions of clause 5.4.2.4(2) are met, is discussed under clause 5.4.2.2(6). However, it would not be

Clause 5.4.2.3(3)

Clause 5.4.2.3(4)

Clause 5.4.2.3(5)

Clause 5.4.2.3(6)

Clause 5.4.2.3(7)

Clause 5.4.2.3(8)

Clause 5.4.2.4(2)

47


common to do this, as a separate analysis considering the staged construction would then be required for the serviceability limit state.

Clause 5.4.2.5(1)

Clause 5.4.2.5(2)

Temperature effects Clause 5.4.2.5(1) refers to EN 1991-1-553 for temperature actions. These are uniform temperature change and temperature gradient through a beam, often referred to as differential temperature. Differential temperature produces primary and secondary effects in a similar way to shrinkage. The reason for allowing temperature to be ignored at the ultimate limit state, if the conditions of clause 5.4.2.5(2) are met, is discussed under clause 5.4.2.2(6). Recommended combination factors for temperature effects are given in Tables A2.1 to A2.4 of Annex A2 of EN 1990. If they are confirmed in the National Annex (as the further comments assume), temperature will be included in all combinations of actions for persistent and transient design situations. In this respect, design to Eurocodes will differ from previous practice in the UK. However, the tables for road bridges and footbridges have a Note which recommends that 0 for thermal actions ‘may in most cases be reduced to zero for ultimate limit states EQU, STR and GEO’. Only FAT (fatigue) is omitted. It is unlikely that temperature will have much influence on fatigue life. The table for railway bridges refers to EN 1991-1-5. The purpose may be to draw attention to its rules for simultaneity of uniform and temperature difference components, and the need to consider differences of temperature between the deck and the rails. With 0 ¼ 0, temperature effects appear in the ultimate and characteristic combinations only where temperature is the leading variable action. It will usually be evident which members and cross-sections need to be checked for these combinations. The factors 1 and 2 are required for the frequent and quasi-permanent combinations used for certain serviceability verifications. The recommended values are 1 ¼ 0:6, 2 ¼ 0:5. Temperature will rarely be the leading variable action, as the following example shows. For the effects of differential temperature, EN 1991-1-5 gives two approaches, from which the National Annex can select. The ‘Normal Procedure’ in Approach 2 is equivalent to the procedure in BS 5400.11 If this is used, both heating and cooling differential temperature cases tend to produce secondary sagging moments at internal supports where crack widths are checked in continuous beams. These effects of temperature will not normally add to other effects. Combinations of actions that involve temperature A cross-section is considered where the characteristic temperature action effect is Tk , and the action effect from traffic load model 1 (LM1) for road bridges is Qk . The recommended combination factors are given in Table 5.4. The frequent combinations of variable action effects are: . .

with load model 1 leading: with temperature leading:

þ T þ 1 k 1 Qk

¼ ð0:75 or 0:40ÞQk þ 0:5Tk Q ¼ 0:6Tk 2 k 2 Tk

The second of these governs only where 0:1Tk >(0.75 or 0.4)Qk . Thus, temperature should be taken as the leading variable action only where its action effect is at least 7.5 times (for TS) or 4 times (for UD) that from traffic load model 1.

Table 5.4. Recommended combination factors for traffic load and temperature according to EN 1990 Annex A2 Action effect from: Tandem system (TS), from LM1 Uniform loading (UD), from LM1 Temperature (non-fire)

48

0

0.75 0.40 0.6 or zero

1

0.75 0.40 0.6

2

0 0 0.5


The uncertainty about which variable action leads does not arise in quasi-permanent combinations, because the combination factor is always 2 . Temperature and construction loads are the only variable actions for which 2 > 0 is recommended. Both the values and the combination to be used can be changed in the National Annex.

Prestressing by controlled imposed deformations As a principle, clause 5.4.2.6(1) requires that possible deviations from the intended amount of imposed deflection be considered, such as might occur due to the tolerance achievable with the specified jacking equipment. The effect of variations in material properties on the action effects developed must also be considered. However, clause 5.4.2.6(2) permits these effects to be determined using characteristic or nominal values, ‘if the imposed deformations are controlled’. The nature of the control required is not specified. It should take account of the sensitivity of the structure to any error in the deformation. At the ultimate limit state, clause 2.4.1.1 recommends a load factor of 1.0 for imposed deformations, regardless of whether effects are favourable or unfavourable. It is recommended here that where a structure is particularly sensitive to departures from the intended amount of imposed deformation, tolerances should be determined for the proposed method of applying these deformations and upper and lower bound values considered in the analysis.

Clause 5.4.2.6(1)

Clause 5.4.2.6(2)

Prestressing by tendons Prestressing composite bridges of steel and concrete is uncommon in the UK and is therefore not covered in detail here. Clause 5.4.2.6(1) refers to EN 1992 for the treatment of prestress forces in analysis. This is generally sufficient, although EN 1994 itself emphasises, in clause 5.4.2.6(2), the distinction between bonded and unbonded tendons. Essentially, this is that while the force in bonded tendons increases everywhere in proportion to the local increase of strain in the adjacent concrete, the force in unbonded tendons changes in accordance with the overall deformation of the structure; that is, the change of strain in the adjacent concrete averaged over the length of the tendon. Tension members in composite bridges The purpose of the definitions (a) and (b) in clause 5.4.2.8(1) is to distinguish between the two types of structure shown in Fig. 5.9, and to define the terms in italic print. In Fig. 5.9(a), the concrete tension member AB is shear-connected to the steel structure, represented by member CD, only at its ends (‘concrete’ here means reinforced concrete). No design rules are given for a concrete member where cracking is prevented by prestressing. Figure 5.9(b) shows a composite tension member, which has normal shear connection. Its concrete flange is the concrete tension member. In both cases, there is a tensile force N from the rest of the structure, shared between the concrete and steel components. A member spanning between nodes in a sagging region of a composite truss with a deck at bottom-chord level could be of either type. The difference between members of types (a) and (b) is similar to that between unbonded and bonded tendons in prestressed concrete. Clause 5.4.2.8(2) lists the properties of concrete that should be considered in global analyses. These influence the stiffness of the concrete component, and hence the magnitude Ac

Clause 5.4.2.8(1)

Clause 5.4.2.8(2)

As Ms

A

B

M d

N

zs

Ns Ma

N N

za

Na

D

C Aa (a)

(b)

(c)

Fig. 5.9. Two types of tension member, and forces in the steel and concrete parts: (a) concrete tension member; (b) composite tension member; (c) action effects equivalent to N and M

49


Clause 5.4.2.8(3)


of the force N, and the proportions of it resisted by the two components. Force N is assumed to be a significant action effect. There will normally be others, arising from transverse loading on the member. The distribution of tension between the steel and concrete parts is greatly influenced by tension stiffening in the concrete (which is in turn affected by over-strength of the concrete). It is therefore important that an accurate representation of stiffness is made. This clause allows a rigorous non-linear method to be used. It could be based on Annex L of ENV 1994-2, ‘Effects of tension stiffening in composite bridges’.54 This annex was omitted from EN 1994-2, as being ‘text-book material’. Further information on the theory of tension stiffening and its basis in tests is given in Ref. 55, in its references, and below. The effects of over-strength of concrete in tension can in principle be allowed for by using the upper 5% fractile of tensile strength, fctk;0:95 . This is given in Table 3.1 of EN 1992-1-1 as 30% above the mean value, fctm . However, tension is caused by shrinkage, transverse loading, etc., as well as by force N, so simplified rules are given in clauses 5.4.2.8(5) to (7). Clause 5.4.2.8(3) requires effects of shrinkage to be included in ‘calculations of the internal forces and moments’ in a cracked concrete tension member. This means the axial force and bending moment, which are shown as Ns and Ms in Fig. 5.9(c). The simplification given here overestimates the mean shrinkage strain, and ‘should be used’ for the secondary effects. This clause is an exception to clause 5.4.2.2(8), which permits shrinkage in cracked regions to be ignored. Clause 5.4.2.8(4) refers to simplified methods. The simplest of these, clause 5.4.2.8(5), which requires both ‘uncracked’ and ‘cracked’ global analyses, can be quite conservative. Clause 5.4.2.8(6) gives a more accurate method for members of type (a) in Fig. 5.9. The longitudinal stiffness of the concrete tension member for use in global analysis is given by equation (5.6-1): ðEAs Þeff ¼ Es As =½1 0:35=ð1 þ n0 s Þ where:

ð5:6-1Þ

As is the reinforcement in the tension member, Ac is the effective cross-sectional area of the concrete, s ¼ As =Ac , and n0 is the short-term modular ratio.

This equation is derived from the model of Annex L of ENV 1994-2 for tension stiffening, shown in Fig. 5.10. The figure relates mean tensile strain, ", to tensile force N, in a concrete tension member with properties As , Ac , s and n0 , defined above. Lines 0A and 0B represent uncracked and fully cracked behaviour, respectively. Cracking first occurs at force Nc;cr , when the strain is "sr1 . The strain at the crack at once increases to "sr2 , but the mean strain hardly changes. As further cracks occur, the mean strain follows the line CD. If the local variations in the tensile strength of concrete are neglected, this becomes line CE. The effective stiffness within this ‘stage of single cracking’ is the slope of a line from 0 to some point within CE. After cracking has stabilized, the stiffness is given by a line such as 0F. The strain difference "sr remains constant until

β∆εsr N

F

A D C

Nc,cr

B E

(EAs)eff 0

εsr1

εcr

εsr2

ε

∆εsr

Fig. 5.10. Normal force and mean strain for a reinforced concrete tension member

50


the reinforcement yields. It represents tension stiffening, the term used for the stiffness of the concrete between the cracks. It has been found that in bridges, the post-cracking stiffness is given with sufficient accuracy by the slope of line 0E, with 0:35. This slope, equation (5.6-1), can be derived using Fig. 5.10, as follows. At a force of Nc;cr the following strains are obtained: fully cracked strain:

"sr2 ¼ Nc;cr =Es As

uncracked strain:

"sr1 ¼ Nc;cr =ðEs As þ Ec Ac Þ

Introducing s ¼ As =Ac and the short-term modular ratio n0 gives: "sr1 ¼ Nc;cr n0 s =½Es As ð1 þ n0 s Þ From Fig. 5.10, the strain at point E is: Nc;cr n0 s "cr ¼ "sr2 ð"sr2 "sr1 Þ ¼ 1 1 Es As ð1 þ n0 s Þ This can also be expressed in terms of effective stiffness as "cr ¼ Nc;cr =ðEs As Þeff Eliminating "cr from the last two equations and dividing by Nc;cr gives: 1=ðEs As Þeff ¼ 1=ðEs As Þ ½1 n0 s =ð1 þ n0 s Þ=ðEs As Þ ¼ ½1 0:35=ð1 þ n0 s Þ=ðEs As Þ which is equation (5.6-1). In Ref. 56, a study was made of the forces predicted in the tension members of a truss using a very similar factor to that in equation (5.6-1). Comparison was made against predictions from a non-linear analysis using the tension field model proposed in Annex L of ENV 19942.54 The two methods generally gave good agreement, with most results being closer to those from a fully cracked analysis than from an uncracked analysis. The forces given by global analysis using stiffness ðEAs Þeff are used for the design of the steel structure, but not the concrete tension member. The tension in the latter is usually highest just before cracking. For an axially loaded member it is, in theory: NEd ¼ Ac fct;eff ð1 þ n0 s Þ

ðD5:4Þ

with fct;eff being the tensile strength of the concrete when it cracks and other notation as above. Usually, there is also tensile stress in the member from local loading or shrinkage. This is allowed for by the assumption that fct;eff ¼ 0:7fctm , given in clause 5.4.2.8(6). Equation (D5.4) is given in this clause with partial factors 1.15 and 1.45 for serviceability and ultimate limit states, respectively. These allow for approximations in the method. Thus, for ultimate limit states: NEd;ult ¼ 1:45Ac ð0:7fctm Þð1 þ n0 s Þ ¼ 1:02fctm ½Ac þ ðEs =Ec ÞAs

ðD5:5Þ

which is the design tensile force at cracking at stress fctm . Clause 5.4.2.8(7) covers composite members of type (b) in Fig. 5.9. The cross-section properties are found using equation (5.6-1) for the stiffness of the cracked concrete flange, and are used in global analyses. As an example, it is assumed that an analysis for an ultimate limit state gives a tensile force N and a sagging moment M, as shown in Fig. 5.9(b). These are equivalent to action effects Na and Ma in the steel component plus Ns and Ms in the concrete component, as shown. This clause requires the normal force Ns to be calculated. Equations for this de-composition of N and M can be derived from elastic section analysis, neglecting slip, as follows. The crosses in Fig. 5.9(b) indicate the centres of area of the cross-sections of the concrete flange (or the reinforcement, for a cracked flange) and the structural steel section. Let zs ,

Clause 5.4.2.8(7)

51


za and d be as shown in Fig. 5.9(c) and I, Ia and Is be the second moments of area of the composite section, the steel component, and the concrete flange, respectively. For M ¼ 0; Na;N þ Ns;N ¼ N; and Ns;N zs ¼ Na;N za ; whence Ns;N ¼ Nðza =dÞ

ðaÞ

For N ¼ 0;

ðbÞ

Ns;M ¼ Na;M

Equating curvatures; For equilibrium;

M=I ¼ Ms =Is ¼ Ma =Ia

M ¼ Ms þ Ma þ Na;M za Ns;M zs

ðcÞ ðdÞ

From equations (b) to (d), with za þ zs ¼ d, M ¼ MðIs =I þ Ia =IÞ Ns;M d; whence Ns;M d ¼ MðIs þ Ia IÞ=I

ðeÞ

For N and M together, from equations (a), (c) and (e): Ns ¼ Nðza =dÞ MðI Ia Is Þ=Id

ðD5:6Þ

Ms ¼ MIs =I

ðD5:7Þ

In practice, Is Ia , so Is , and hence Ms , can often be taken as zero. The area of reinforcement in the concrete tension member (As , not As;eff ) must be sufficient to resist the greater of force Ns (plus Ms , if not negligible) and the force NEd given by equation (D5.5).

Clause 5.4.2.9(1)

Clause 5.4.2.9(2) Clause 5.4.2.9(3)

Clause 5.4.2.9(4)

52

Filler beam decks for bridges There are a great number of geometric, material and workmanship-related restrictions which have to be met in order to use the application rules for the design of filler beams. These restrictions are discussed in section 6.3, which deals with the resistances of filler beams, and are necessary because these clauses are based mainly in existing practice in the UK. There is very little relevant research. The same restrictions apply in the use of clause 5.4.2.9(1), which allows the effects of slip at the concrete–steel interface and shear lag to be neglected in global analysis only if these conditions are met. One significant difference from previous practice in the UK is that fully-encased filler beams are not covered by EN 1994. This is because there are no widely accepted design rules for longitudinal shear in fully-encased beams without shear connectors. Clause 5.4.2.9(2) covers the transverse distribution of imposed loading. Its option of assuming rigid behaviour in the transverse direction may be applicable to a small singletrack railway bridge, but generally, one of the methods of clause 5.4.2.9(3) will be used. These assume that there are no transverse steel members within the span. It is therefore essential that continuous transverse reinforcement in both top and bottom faces of the concrete is provided in accordance with the requirements of clause 6.3. Clause 5.4.2.9(3) permits global analysis by non-linear methods to clause 5.4.3, but normally orthotropic plate or grillage analysis will be used. For the longitudinal flexural stiffness, ‘smearing of the steel beams’ involves calculating the stiffness of the whole width of the deck, and hence finding a mean stiffness per unit width. It is inferred from clause 5.4.2.9(4) that cracking may be neglected, though clause 5.4.2.9(7) provides an alternative for some analyses for serviceability. The flexural stiffness per unit width in the transverse direction is calculated for the uncracked concrete slab, neglecting reinforcement. The result is a plate with different properties in orthogonal directions, i.e. orthogonally anisotropic or orthotropic for short. For grillage analysis, uncracked section properties should generally be used (as required by clause 5.4.2.9(4)) but it is permissible to account for the loss of stiffness in the transverse direction caused by cracking, by reducing the torsional and flexural stiffnesses of the transverse concrete members by 50%. This can be advantageous, as it reduces the transverse moments, and hence the stresses in the reinforcement. The longitudinal moments obtained from elastic analysis of an orthotropic slab or grillage may not be redistributed to allow for cracking. This is because cracking can occur in both hogging and sagging regions, and there is insufficient test evidence on which to base


design rules. However, clause 5.4.2.9(5) permits, for some analyses, up to 15% redistribution of hogging moments for beams in Class 1 at internal supports. This is less liberal than it may appear, because clause 5.5.3, which covers classification, does not relax the normal rules for Class 1 webs or flanges to allow for restraint from encasement. The concrete does, however, reduce the depth of web in compression. There are no provisions for creep of concrete at ultimate limit states, so clause 5.4.2.2 applies in the longitudinal direction. In the transverse direction, clause 3.1 of EN 1992-1-1 presumably applies. The modular ratios in the two directions may be found to be different, because of the L factors in EN 1994. These comments on creep also apply for deformations, from clause 5.4.2.9(6). Shrinkage can be neglected, because it causes little curvature where there is little difference between the levels of the centroids of the steel and concrete cross-sections. Clause 5.4.2.9(7) gives a simplified rule for the effects of cracking of concrete on deflections and camber. Clause 5.4.2.9(8) permits temperature effects to be ignored, except in certain railway bridges.

Clause 5.4.2.9(5)

Clause 5.4.2.9(6)

Clause 5.4.2.9(7) Clause 5.4.2.9(8)

Example 5.2: modular ratios for long-term loading and for shrinkage For the bridge of Example 5.1 (Fig. 5.6), modular ratios are calculated for: . . .

imposed load (short-term loading) superimposed dead load (long-term loading) effects of shrinkage (long-term).

Modular ratios for long-term loading depend on an assumed effective thickness for the concrete deck slab and a mean ‘age at first loading’. The choice of these values is discussed in comments on clause 5.4.2.2. Here, it is assumed that superimposed dead load is applied when the concrete has an average age of 7 days, and that the effective thickness of the deck slab is the value before application of waterproofing, which is its actual thickness, 250 mm. The deck concrete is grade C30/37 and the relative humidity is 70%.

Live load From Table 3.1 of EN 1992-1-1,

Ecm ¼ 33 kN/mm2

From clause 3.2.6 of EN 1993-1-1, for structural steel,

Ea ¼ 210 N/mm2

The short-term modular ratio,

n0 ¼ Ea =Ecm ¼ 210=33 ¼ 6:36

Superimposed dead load (long term), from Annex B of EN 1992-1-1 From equation (B.1) of EN 1992-1-1, the creep coefficient ðt; t0 Þ ¼ 0 c ðt; t0 Þ, where c ðt; t0 Þ is a factor that describes the amount of creep that occurs at time t. When t ! 1, c ¼ 1. The total creep is therefore given by: 0 ¼ RH ð fcm Þðt0 Þ

(B.2) in EN 1992-1-1

where RH is a factor to allow for the effect of relative humidity on the notional creep coefficient. Two expressions are given, depending on the size of fcm . From Table 3.1 of EN 1992-1-1, fcm ¼ fck þ 8 ¼ 30 þ 8 ¼ 38 N/mm2 1 RH=100 pffiffiffiffiffi 1 2 For fcm > 35 MPa; RH ¼ 1 þ (B.3b) in EN 1992-1-1 0:1 3 h0 RH is the relative humidity of the ambient environment in percentage terms; here, 70%. The factors 1 and 2 allow for the influence of the concrete strength: 0:7 35 0:7 35 1 ¼ ¼ ¼ 0:944 fcm 38

53


35 2 ¼ fcm

0:2

35 ¼ 38

0:2 ¼ 0:984

(B.8c) in EN 1992-1-1

From equation (B.3b) of EN 1992-1-1: 1 70=100 pffiffiffiffiffiffiffiffi 0:944 0:984 ¼ 1:43 RH ¼ 1 þ 0:1 3 250 From equation (B.4) of EN 1992-1-1, the factor ð fcm Þ, that allows for the effect of concrete strength on the notional creep coefficient, is: 16:8 16:8 ð fcm Þ ¼ pffiffiffiffiffiffiffi ¼ pffiffiffiffiffi ¼ 2:73 38 fcm The effect of the age of the concrete at first loading on the notional creep coefficient is given by the factor ðt0 Þ according to equation (B.5) of EN 1992-1-1. For loading at 7 days this gives: ðt0 Þ ¼

1 1 ¼ ¼ 0:63 0:20 ð0:1 þ t0 Þ ð0:1 þ 70:20 Þ

This expression is only valid as written for normal or rapid-hardening cements. The final creep coefficient from equation (B.2) of EN 1992-1-1 is then: ð1; t0 Þ ¼ 0 ¼ RH ð fcm Þðt0 Þ ¼ 1:43 2:73 0:63 ¼ 2:48 From equation (5.6), the modular ratio is given by: nL ¼ n0 ð1 þ where

L

L ðt; t0 ÞÞ

¼ 6:36ð1 þ 1:1 2:48Þ ¼ 23:7

¼ 1:1 for permanent load.

Effects of shrinkage, from Annex B of EN 1992-1-1 The calculation of the creep factor is as above, but the age at loading is assumed to be one day, from clause 5.4.2.2. For the factor ðt0 Þ this gives: ðt0 Þ ¼

1 1 ¼ ¼ 0:91 Þ ð0:1 þ t0:20 ð0:1 þ 10:20 Þ 0

The final creep coefficient from equation (B.2) of EN 1992-1-1 is then: ð1; t0 Þ ¼ 0 ¼ RH ð fcm Þðt0 Þ ¼ 1:43 2:73 0:91 ¼ 3:55 From equation (5.6), the modular ratio is given by: nL ¼ n0 ð1 þ where

L

L ðt; t0 ÞÞ

¼ 6:36ð1 þ 0:55 3:55Þ ¼ 18:8

¼ 0:55 for the primary and secondary effects of shrinkage.

Example 5.3: primary effects of shrinkage For the bridge in Example 5.1 (Fig. 5.6) the plate thicknesses of an internal beam at midspan of the main span are as shown in Fig. 5.11. The primary effects of shrinkage at this cross-section are calculated. The deck concrete is grade C30/37 and the relative humidity is 70%. It is assumed that, for the majority of the shrinkage, the length of continuous concrete deck is such that shear lag effects are negligible. The effective area of the concrete flange is taken as the actual area, for both the shrinkage and its primary effects. The secondary effects arise from changes in the reactions at the supports. For these, the effective widths should be those used for the other permanent actions. The free shrinkage strain is found first, from clause 3.1.4 of EN 1992-1-1. By interpolation, Table 3.2 of EN 1992-1-1 gives the drying shrinkage as "cd;0 ¼ 352 106 .

54


The factor kh allows for the influence of the shape and size of the concrete cross-section. From Example 5.2, h0 ¼ 250 mm. From Table 3.3 of EN 1992-1-1, kh ¼ 0:80. The long-term drying shrinkage strain is 0:80 352 106 ¼ 282 106 . The long-term autogenous shrinkage strain is: "ca ð1Þ ¼ 2:5ð fck 10Þ 106 ¼ 2:5 ð30 10Þ 106 ¼ 50 106 3100 0.35

25 haunch 400 × 20

49

T

250 1.08

C

NA 1225

1175 × 12.5

998

400 × 30

12

T

Fig. 5.11. Cross-section of beam for Example 5.3, and primary shrinkage stresses with nL ¼ 18:8 (T: tension; C: compression)

From clause 2.4.2.1 of EN 1992-1-1, the recommended partial factor for shrinkage is SH ¼ 1:0, so the design shrinkage strain, for both serviceability and ultimate limit states, is: esh ¼ 1:0 ð282 þ 50Þ 106 ¼ 332 106 From Example 5.2, the modular ratio for shrinkage is nL ¼ 18:8. The tensile force Fc to restore the slab to its length before shrinkage applies to the concrete a tensile stress: 332 106 210 103 =18:8 ¼ 3:71 N=mm2 The area of the concrete cross-section is: Ac ¼ 3:1 0:25 þ 0:4 0:025 ¼ 0:785 m2 therefore Fc ¼ "sh Ac Ea =nL ¼ 332 0:785 210=18:8 ¼ 2913 kN For nL ¼ 18:8, the location of the neutral axis of the uncracked unreinforced section is as shown in Fig. 5.11. This is 375 mm below the centroid of the concrete area. Relevant properties of this cross-section are: I ¼ 21 890 106 mm4 and A ¼ 76 443 mm2 The total external force is zero, so force Fc is balanced by applying a compressive force of 2913 kN and a sagging moment of 2913 0:375 ¼ 1092 kNm to the composite section. The long-term primary shrinkage stresses in the cross-section are as follows, with compression positive: at the top of the slab, 2913 103 1092 502 1 ¼ 0:35 N=mm2 þ ¼ 3:71 þ 21 890 18:8 76 443 at the interface, in concrete, 2913 103 1092 227 1 ¼ 1:08 N=mm2 þ ¼ 3:71 þ 21 890 18:8 76 443

55


at the interface, in steel, 2913 103 1092 227 ¼ þ49:4 N=mm2 þ 21 890 76 443 at the bottom of the steel beam, ¼þ

¼þ

2913 103 1092 998 ¼ 11:7 N=mm2 þ 21 890 76 443

5.4.3. Non-linear global analysis for bridges Clause 5.4.3

The provisions of EN 1992 and EN 1993 on non-linear analysis are clearly relevant, but are not referred to from clause 5.4.3. It gives Principles, but no Application Rules. It is stated in EN 1992-2 that clause 5.7(4)P of EN 1992-1-1 applies. It requires stiffnesses to be represented ‘in a realistic way’ taking account of the ‘uncertainties of failure’, and concludes ‘Only those design formats which are valid within the relevant fields of application shall be used’. Clause 5.7 of EN 1992-2, ‘Non-linear analysis’, consists mainly of Notes that give recommendations to national annexes. The majority of the provisions can be varied in the National Annex as agreement could not be obtained at the time of drafting over the use of the safety format proposed. The properties of materials specified in the Notes have been derived so that a single safety factor can be applied to all materials in the verification. Further comment is given under clause 6.7.2.8. Clause 5.4.1 of EN 1993-2 requires the use of an elastic analysis for ‘all persistent and transient design situations’. It has a Note that refers to the use of ‘plastic global analysis’ for accidental design situations, and to the relevant provisions of EN 1993-1-1. These include clause 5.4, which defines three types of non-linear analysis, all of which refer to ‘plastic’ behaviour. One of them, ‘rigid-plastic analysis’, should not be considered for composite bridge structures. This is evident from the omission from EN 1994-2 of a clause corresponding to clause 5.4.5 of EN 1994-1-1, ‘Rigid plastic global analysis for buildings’. This method of drafting arises from Notes to clauses 1.5.6.6 and 1.5.6.7 of EN 1990, which make clear that all of the methods of global analysis defined in clauses 1.5.6.6 to 1.5.6.11 (which include ‘plastic’ methods) are ‘non-linear’ in Eurocode terminology. ‘Non-linear’ in these clauses of EN 1990 refers to the deformation properties of the materials, and not to geometrical non-linearity (second-order effects), although these have to be considered when significant, as discussed in section 5.2 above. Non-linear analysis must satisfy both equilibrium and compatibility of deformations when using non-linear material properties. These broad requirements are given to enable methods more advanced than linear-elastic analysis to be developed and used within the scope of the Eurocodes. Unlike clause 5.4.1 of EN 1993-2, EN 1994-2 makes no reference to the use of plastic analysis for accidental situations, such as vehicular impact on a bridge pier or impact on a parapet. The National Annex to EN 1993-2 will give guidance. It is recommended that this be followed also for composite design. Further guidance on non-linear analysis is given in EN 1993-1-5, Annex C, on finiteelement modelling of steel plates.

5.4.4. Combination of global and local action effects

Clause 5.4.4(1)

56

A typical local action is a wheel load on a highway bridge. It is expected that the National Annex for the UK will require the effects of such loads to be combined with the global effects of coexisting actions for serviceability verifications, but not for checks for ultimate limit states. This is consistent with current practice to BS 5400.11 The Note to clause 5.4.4(1) refers to Normative Annex E of EN 1993-2. This annex was written for all-steel decks, where local stresses in welds can be significant and where local and global stresses always combine unfavourably. It recommends a combination factor for local and global effects that depends on the span and ranges from 0.7 to 1.0. The application


of this rule to reinforced concrete decks that satisfy the serviceability requirements for the combined effects is believed to be over-conservative, because of the beneficial local effects of membrane and arching action. By contrast, if the EN 1993 rules are adopted, global compression in the slab is usually favourable so consideration of 70% of the maximum compressive global stress when checking local effects may actually be unconservative.

5.5. Classification of cross-sections The classification of cross-sections of composite beams is the established method of taking account in design of local buckling of plane steel elements in compression. It determines the available methods of global analysis and the basis for resistance to bending, in the same way as for steel members. Unlike the method in EN 1993-1-1, it does not apply to columns. The Class of a steel element (a flange or a web) depends on its edge support conditions, b/t ratio, distribution of longitudinal stress across its width, yield strength, and in composite sections, the restraint provided against buckling by any attached concrete or concrete encasement. A flow diagram for the provisions of clause 5.5 is given in Fig. 5.12. The clause numbers given are from EN 1994-2, unless noted otherwise. Is steel compression flange restrained by shear connectors to clause 5.5.2(1)?

Note 1: ‘Flange’ means steel compression flange

No

Yes From clause 5.5.2(2), classify flange to Table 5.2 of EC3-1-1

Classify flange to Table 5.2 of EC4-2

Yes

Compression flange is:

Is the section a filler beam to clause 6.3?

No

Class 1

Class 2

Class 3

Locate the plastic neutral axis, using gross web and effective flanges, to clause 5.5.1(4)

Web is Class 3

Class 4

Locate the elastic neutral axis, taking account of sequence of construction, creep, and shrinkage, to clause 5.5.1(4)

From clause 5.5.2(2), classify the web using the plastic stress distribution, to Table 5.2 of EC3-1-1

From clause 5.5.2(2), classify the web using the elastic stress distribution, to Table 5.2 of EC3-1-1

Web is Class 3

Web not classified

Is web encased to clause 5.5.3(2)?

No

Yes

Yes

No No

Yes Web is Class 2

Web is Class 4

Is the compression flange in Class 1 or 2?

Replace web by effective web in Class 2, to clause 5.5.2(3)?

Web is Class 1

Clause 5.5

Is the compression flange in Class 3?

Effective web is Class 2

Yes

Is the flange in Class 1?

No

No

Yes Section is Class 1

Section is Class 2

Effective section is equivalent Class 2

Section is Class 3

Section is Class 4 (see Note 2)

Note 2. Where the stress levels in a Class 4 section satisfy the limit given in EC3-1-1/5.5.2(9), it can be treated as a Class 3 section, from clause 5.5.1(1)P.

Fig. 5.12. Classification of a cross-section of a composite beam

57


M

M

Mpl

Mpl

Mel

Mel

Class 1

Typical curve from a test

θ

θ

Class 2 M

M Mpl Mel

Mpl Mel

Class 3

θ

Class 4

θ

Fig. 5.13. Idealized moment–rotation relationships for sections in Classes 1 to 4

Clause 5.5.1(1)P

Clause 5.5.1(2)

Clause 5.5.1(1)P refers to EN 1993-1-1 for definitions of the four Classes and the slendernesses that define the Class boundaries. Classes 1 to 4 correspond respectively to the terms ‘plastic’, ‘compact’, ‘semi-compact’ and ‘slender’ that were formerly used in BS 5950.48 The classifications are done separately for steel flanges in compression and steel webs. The Class of the cross-section is the less favourable of the Classes so found, clause 5.5.1(2), with one exception: the ‘hole-in-web’ option of clause 5.5.2(3). Idealized moment–rotation curves for members in the four Classes are shown in Fig. 5.13. In reality, curves for sections in Class 1 or 2 depart from linearity as soon as (or even before) the yield moment is reached, and strain-hardening leads to a peak bending moment higher than Mpl , as shown. The following notes supplement the definitions given in clause 5.5.2(1) of EN 1993-1-1: .

.

.

.

58

Class 1 cross-sections can form a plastic hinge and tolerate a large plastic rotation without loss of resistance. It is a requirement of EN 1993-1-1 for the use of rigidplastic global analysis that the cross-sections at all plastic hinges are in Class 1. For composite bridges, EN 1994-2 does not permit rigid-plastic analysis. A Note to clause 5.4.1(1) of EN 1993-2 enables its use to be permitted, in a National Annex, for certain accidental design situations for steel bridges. Class 2 cross-sections can develop their plastic moment resistance, Mpl;Rd , but have limited rotation capacity after reaching it because of local buckling. Regions of sagging bending in composite beams are usually in Class 1 or 2. The resistance Mpl;Rd exceeds the resistance at first yield, Mel;Rd , by between 20% and 40%, compared with about 15% for steel beams. Some restrictions are necessary on the use of Mpl;Rd in combination with elastic global analysis, to limit the post-yield shedding of bending moment to adjacent cross-sections in Class 3 or 4. These are given in clauses 6.2.1.2(2) and 6.2.1.3(2). Class 3 cross-sections become susceptible to local buckling before development of the plastic moment of resistance. In clause 6.2.1.5(2) their bending resistance is defined as the ‘elastic resistance’, governed by stress limits for all three materials. A limit may be reached when the compressive stress in all restrained steel elements is below yield. Some rotation capacity then remains, but it is impracticable to take advantage of it in design. Class 4 cross-sections are those in which local buckling will occur before the attainment of yield stress in one or more parts of the cross-section. This is assumed in EN 1993 and EN 1994 to be an ultimate limit state. The effective cross-section should be derived in accordance with EN 1993-1-5. Guidance is given in comments on clause 6.2.1.5(7), which defines the procedure, and in the Designers’ Guide to EN 1993-2.4


The Class of a cross-section is determined from the width-to-thickness ratios given in Table 5.2 of EN 1993-1-1 for webs and flanges in compression. The numbers appear different 11 from those in BS 5400:Part that takes account of yield p 3:2000 because the coefficient p strength, ", is defined as ð235=fy Þ in the Eurocodes, and as ð355=fy Þ in BS 5400. After allowing for this, the limits for webs at the Class 2/3 boundary agree closely with those in BS 5400, but there are differences for flanges. For outstand flanges, EN 1993 is more liberal at the Class 2/3 boundary, and slightly more severe at the Class 3/4 boundary. For internal flanges of boxes, EN 1993 is considerably more liberal for all Classes. Reference is sometimes made to a beam in a certain Class. This may imply a certain distribution of bending moment. Clause 5.5.1(2) warns that the Class of a composite section depends on the sign of the bending moment (sagging or hogging), as it does for a steel section that is not symmetrical about its neutral axis for bending. Clause 5.5.1(3) permits account to be taken of restraint from concrete in determining the classification of elements, providing that the benefit has been established. Further comment is given at clause 5.5.2(1) on spacing of shear connectors. Since the Class of a web depends on the level of the neutral axis, which is different for elastic and plastic bending, it may not be obvious which stress distribution should be used for a section near the boundary between Classes 2 and 3. Clause 5.5.1(4) provides the answer: the plastic distribution. This is because the use of the elastic distribution could place a section in Class 2, for which the bending resistance would be based on the plastic distribution, which in turn could place the section in Class 3. Elastic stress distributions should be built up by taking the construction sequence into account, together with the effects of creep (generally through the use of different modular ratios for the different load types) and shrinkage. Where a steel element is longitudinally stiffened, it should be placed in Class 4 unless it can be classified in a higher Class by ignoring the longitudinal stiffeners. Where both axial load and moment are present, these should be combined when deriving the plastic stress block. Alternatively, the web Class can conservatively be determined on the basis of compressive axial load alone. Clause 5.5.1(5), on the minimum area of reinforcement for a concrete flange, appears here, rather than in Section 6, because it gives a further condition for a cross-section to be placed in Class 1 or 2. The reason is that these sections must maintain their bending resistance, without fracture of the reinforcement, while subjected to higher rotation than those in Class 3 or 4. This is ensured by disallowing the use of bars in ductility Class A (the lowest), and by requiring a minimum cross-sectional area, which depends on the tensile force in the slab just before it cracks.55 Clause 5.5.1(6), on welded mesh, has the same objective. Clause 3.2.4 of EN 1992-2 does not recommend the use of Class A reinforcement for bridges in any case, but this recommendation can be modified in a National Annex. During the construction of a composite bridge, it is quite likely that a beam will change its section Class, because the addition of the deck slab both prevents local buckling of the top flange and significantly shifts the neutral axis of the section. Typically, a mid-span section could be in Class 1 or 2 after casting the slab but in Class 3 or 4 prior to this. Clause 5.5.1(7) requires strength checks at intermediate stages of construction to be based on the relevant classification at the stage being checked. The words ‘without concrete encasement’ in the title of clause 5.5.2 are there because this clause is copied from EN 1994-1-1, where it is followed by a clause on beams with web encasement. These are outside the scope of EN 1994-2. Clause 5.5.2(1) is an application of clause 5.5.1(3). The spacing rules to which it refers may be restrictive where full-thickness precast deck slabs are used. Clause 5.5.2(2) adds little to clause 5.5.1.

The hole-in-web method This useful device first appeared in BS 5950-3-1.48 It is now in clause 6.2.2.4 of EN 1993-1-1, which is referred to from clause 5.5.2(3).

Clause 5.5.1(3)

Clause 5.5.1(4)

Clause 5.5.1(5)

Clause 5.5.1(6)

Clause 5.5.1(7)

Clause 5.5.2

Clause 5.5.2(1) Clause 5.5.2(2)

Clause 5.5.2(3)

59


In beams subjected to hogging bending, it often happens that the bottom flange is in Class 1 or 2, and the web is in Class 3. The initial effect of local buckling of the web would be a small reduction in the bending resistance of the section. The assumption that a defined depth of web, the ‘hole’, is not effective in bending enables the reduced section to be upgraded from Class 3 to Class 2, and removes the sudden change in the bending resistance that would otherwise occur. The method is analogous to the use of effective areas for Class 4 sections, to allow for local buckling. There is a limitation to its scope that is not evident in the following wording, from EN 1993-1-1: The proportion of the web in compression should be replaced by a part of 20"tw adjacent to the compression flange, with another part of 20"tw adjacent to the plastic neutral axis of the effective cross-section.

Clause 5.5.3(2)

It follows that for a design yield strength fyd , the compressive force in the web is limited to 40"tw fyd . For a composite beam in hogging bending, the tensile force in the longitudinal reinforcement in the slab can exceed this value, especially where fyd is reduced to allow for vertical shear. The method is then not applicable, because the second ‘element of 20"tw ’ is not adjacent to the plastic neutral axis, which lies within the top flange. The method, and this limitation, are illustrated in Examples in the Designers’ Guide to EN 1994-1-1.5 It should be noted that if a Class 3 cross-section is treated as an equivalent Class 2 crosssection for section design, it should still be treated as Class 3 when considering the actions to consider in its design. Indirect actions, such as differential settlement, which may be neglected for true Class 2 sections, should not be ignored for effective Class 2 sections. The primary self-equilibrating stresses could reasonably be neglected, but not the secondary effects. Clause 5.5.3(2) and Table 5.2 give allowable width-to-thickness ratios for the outstands of exposed flanges of filler beams. Those for Class 2 and 3 are greater than those from Table 5.2 of EN 1993-1-1. This is because even though a flange outstand can buckle away from the concrete, rotation of the flange at the junction with the web is prevented (or at least the rotational stiffness is greatly increased) by the presence of the concrete.

Example 5.4: classification of composite beam section in hogging bending The classification of the cross-section shown in Fig. 5.14 is determined for hogging bending moments. The effective flange width is 3.1 m. The top layer of reinforcement comprises pairs of 20 mm bars at 150 mm centres. The bottom layer comprises single 20 mm bars at 150 mm centres. All reinforcement has fsk ¼ 500 N/mm2 and S ¼ 1:15. These bars are shown in assumed locations, which would in practice depend on the specified covers and the diameter of the transverse bars. 3100

60 70 8.47 MN

250 25

400 × 25

Plastic NA 1225

3.45 MN

209

1160 × 25

Elastic NA

400 × 40

775

1.80 MN 10.01 MN

8.21 MN 5.52 MN

Fig. 5.14. Cross-section of beam for Example 5.4

The yield strength of structural steel is thickness-dependent and is 345 N/mm2 from EN 10025 for steel between 16 mm and 40 mm thick. (If Table 3.1 of EN 1993-1-1 is

60


used, the yield strength can be taken as 355 N/mm2 for steel up to 40 mm thick. The choice of method can be specified in the National Annex. It is likely that this will require the value from theprelevant ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiproduct standard to be used.) Hence, " ¼ 235=345 ¼ 0:825.

Steel bottom flange Ignoring the web-to-flange welds, the flange outstand c ¼ ð400 25Þ=2 ¼ 187:5 mm. From Table 5.2 of EN 1993-1-1, the condition for Class 1 is: c=t < 9" ¼ 9 0:825 ¼ 7:43 For the flange, c=t ¼ 187:5=40 ¼ 4:7, so the flange is Class 1.

Steel web To check if the web is in Class 1 or 2, the plastic neutral axis must be determined, using design material strengths. The total area of reinforcement is: As ¼ 3 100 3100=150 ¼ 19 480 mm2 so its tensile force at yield is: 19:48 0:5=1:15 ¼ 8:47 MN Similarly, the forces in the structural steel elements at yield are found to be as shown in Fig. 5.14. The total longitudinal force is 27.45 MN, so at Mpl;Rd the compressive force is 27:45=2 ¼ 13:73 MN. The plastic neutral axis is evidently within the web. The depth of web in compression is: 1160 ð13:73 5:52Þ=10:01 ¼ 951 mm From Table 5.2 of EN 1993-1-1, for a ‘part subject to bending and compression’, and ignoring the depth of the web-to-flange welds: ¼ 951=1160 ¼ 0:82; >0:5 The web is in Class 2 if: c=t 456"=ð13 1Þ ¼ 456 0:825=9:66 ¼ 39:0 Its ratio c=t ¼ 1160=25 ¼ 46:4, so it is not in Class 2. By inspection, it is in Class 3, but the check at the Class 3/4 boundary is given, as an example. It should be based on the built-up elastic stresses, which may not be available. It is conservative to assume that all stresses are applied to the composite section as this gives the greatest depth of web in compression. The location of the elastic neutral axis in hogging bending must be found, for the cracked reinforced composite section. There is no need to use a modular ratio for reinforcement as its modulus may be taken equal to that for structural steel, according to clause 3.2(2). The usual ‘first moment of area’ calculation finds the neutral axis to be as shown in Fig. 5.14, so the depth of web in compression is 775 40 ¼ 735 mm. Again neglecting the welds, from Table 5.2 of EN 1993-1-1 the stress ratio is ¼ ð1160 735Þ=735 ¼ 0:58 and the condition for a Class 3 web is: c=t 42"=ð0:67 þ 0:33 Þ ¼ 42 0:825=ð0:67 0:19Þ ¼ 72:4 The actual c=t ¼ 1160=25 ¼ 46:4, so the composite section is in Class 3 for hogging bending.

61


Flow charts for global analysis The flow charts given in Figs 5.15 and 5.16 are for a bridge with the general layout shown in Fig. 5.1. Figure 5.15, for the superstructure, provides design forces and displacements for the beams and at the four sets of bearings shown. Figure 5.16, for the columns, takes account of system instability shown in Fig. 5.1(b). Instability of members in compression is covered in comments on clause 6.7.3.4. For simplicity, the scope of these charts is limited by assumptions, as follows: . . . . . .

. .

.

.

. .

Fatigue, vibration, and settlement are excluded. Axial force in the superstructure (e.g. from friction at bearings) is negligible. The main imposed loading is traffic Load Model 1, from EN 1991-2. Only persistent design situations are included. The limit states considered are ULS (STR) and SLS (deformation and crack width). The superstructure consists of several parallel continuous non-hybrid plate girders without longitudinal stiffeners, composite with a reinforced normal-density-concrete deck slab. There are no structural steel transverse members at deck-slab level. The only steel cross-sections that may be in Class 4 are the webs near internal supports. The depth of web in compression is influenced by the ratio of non-composite to composite bending moment and the area of reinforcement in the slab. The Class is therefore difficult to predict until some analyses have been done. The deck is constructed unpropped, and all structural deck concrete is assumed to be in place before any of the members become composite. The formwork is structurally participating precast concrete planks. They are assumed (for simplicity here) to have the same creep and shrinkage properties as the in situ concrete of the deck. All joints except bearings are assumed to be continuous (clause 5.1.2). Bearings are ‘simple’ joints, with or without longitudinal sliding, as shown in Fig. 5.1. Transverse sliding cannot occur.

In the charts, creep and shrinkage effects are considered only as ‘long-term’ values (t ! 1). The values of all Nationally Determined Parameters, such as and factors, are assumed to be those recommended in the Notes in the Eurocodes. The following data are assumed to be available, based on preliminary analyses and the strengths of the materials to be used, fy , fsk and fck (converted from an assumed fcu ): . .

. .

dimensions of the flanges and webs of the plate girders dimensions of the cross-sections of the concrete deck and the two supporting systems, BE and CF in Fig. 5.1(a) details and weight of the superimposed dead load (finishes, parapets, etc.) estimated areas of longitudinal slab reinforcement above internal supports.

Assumptions relevant to out-of-plane system instability are as follows. The deck transmits most of the lateral wind loading to supports A and D, with negligible restraint from the two sets of internal supports. The lateral deflections of nodes B and C influence the design of the columns, but stiffnesses are such that wind-induced system instability is not possible. The following abbreviations are used: . . .

.

‘EC2’ means EN 1992-1-1 and/or EN 1992-2; similarly for ‘EC3’. A clause in EN 1994-2 is referred to as, for example, ‘5.4.2.2’. Symbols gk1 and gk2 are used for characteristic dead loads on the steelwork, and on composite members, respectively. Superimposed dead load is gk3 . Shrinkage is gsh . Characteristic imposed loads are denoted qk (traffic), wk (wind) and tk (temperature).

There is not space to list on Fig. 5.15 the combinations of actions required. The notation in the lists that follow is that each symbol, such as gk2 , represents the sets of action effects (MEd , VEd , deformations, etc.) resulting from the application of the arrangement of the action gk2 that is most adverse for the action effect considered.

62


Determine elastic properties of materials: Ea,Es (taken as Ea), Ecm, etc., from 3.1 to 3.3. Assume a mean value t0 for permanent actions, to 5.4.2.2(3). Determine creep coefficients ϕ(∞, t0); hence find modular ratios n0, nL,s, and nL,p for short-term, shrinkage, and permanent actions, respectively (5.4.2.2(2)). Determine free shrinkage strain, εcs(∞, 1 day), to 3.1(3) and 5.4.2.2(4)

Select exposure (environmental) class(es) for concrete surfaces, to 7.1(3). Hence find minimum covers to concrete from EC2/4.4, and the locations of main tensile reinforcement in composite beams. These are required for cracked section properties

Determine shear lag effective widths for global analyses: – for concrete flanges, at mid-span and internal supports, to 5.4.1.2 – for steel flanges, find beff for SLS and ULS, to EC3-1-5/3.1 to 3.3

Estimate distributions of longitudinal stress in steel webs at internal supports, and classify them, to 5.5. If any are in Class 4, thicken them or find effective properties to EC3-1-5/2.2

Flexural stiffnesses of cross-sections. Determine EaI1 for all uncracked composite and concrete sections, for modular ratios n0, nL,s and nL,p, using effective widths at mid-span, or at supports for cantilevers (1.5.2.11). Reinforcement may be included or omitted. Determine EaI2 for cracked reinforced longitudinal composite sections in hogging bending, with effective widths as above (1.5.2.12 ). Represent bearings by appropriate degrees of freedom

Imperfections. Determine system imperfections to 5.3.2(1). Out-of-plumb of supports BE and CF are relevant for these members, but are assumed not to affect this flow chart. For imperfections of column members, see flow chart of Fig. 6.44

Stability. If no buckling mode can be envisaged, as here for the superstructure, then αcr ≥ 10 can be assumed, and global analyses can be first-order, to 5.2.1(3)

Global analyses. Are No all span ratios ≥0.6? Yes

Use stiffnesses to 5.4.2.3(3) (‘15%’ rule) for global analyses

5.4.2.3(2) applies. Use uncracked stiffnesses and modular ratios n0, nL,s and nL,p. Analyse for load cases: g k2, g k3, gsh. Find moment and shear envelopes for q k and tk. Find the highest extreme-fibre tensile stresses in concrete, fct,max, for the combinations listed in expression D(5.8), with effective widths to 5.4.1.2(7) Find regions of longitudinal members where fct,max > 2fctm, and reduce stiffnesses of these regions to EaI2

Do global analyses for the load cases gk1 (on the steel structure), g k2, g k3, gsh, q k, wk and tk using both short-term and long-term modular ratios for the variable actions and for g k2, g k3 and long-term for gsh. Refer to expressions (D5.9) etc., for the combinations required. (END)

Fig. 5.15. Flow chart for global analysis for superstructure of three-span bridge

For the variable actions qk and wk , different arrangements govern at different crosssections, so envelopes are required. This may apply also for tk , as several sets of temperature actions are specified. For finding the ‘cracked’ regions of longitudinal members, it is assumed that the shortterm values are critical, because creep may reduce tensile stress in concrete more than

63


shrinkage increases it. From clause 5.4.2.3(2), the following characteristic combinations are required for finding ‘cracked’ regions: . . .

with traffic leading: with wind leading: with temperature leading:

gk2 þ gk3 þ qk þ 0;w wk gk2 þ gk3 þ 0;q qk þ wk gk2 þ gk3 þ 0;q qk þ 0;w wk þ tk

(D5.8)

In practice, of course, it will usually be evident which combination governs. Then, only regions in tension corresponding to that combination need be determined. For finding the most adverse action effects for the limit state ULS (STR), all combinations include the design permanent action effects: G ðgk1 þ gk2 þ gk3 Þ þ gsh using the more adverse of the long-term or short-term values. To these are added, in turn, the following combinations of variable action effects: . . .


1:35qk þ 1:5 0;w wk 1:35 0;q qk þ 1:5wk 1:35 0;q qk þ 1:5ð 0;w wk þ tk Þ

(D5.9)

For serviceability limit states, deformation is checked for frequent combinations. The combination for crack width is for national choice, and ‘frequent’ is assumed here. These combinations all include the permanent action effects as follows, again using the more adverse of short-term and long-term values: gk1 þ gk2 þ gk3 þ gsh To these are added, in turn, the following combinations of variable action effects: . . .


þ w 1;w k þ 1;t tk

1;q qk

2;t tk 2;t tk

because because

2;w 2;q

¼0 ¼0 (D5.10)

As before, it will usually be evident, for each action effect and location, which combination governs.

Flow chart for supporting systems at internal supports At each of points B and C in Fig. 5.1, it is assumed that the plate girders are supported on at least two bearings, mounted on a cross-head that is supported by a composite frame or by two or more composite columns, fixed at points E and F. Each bearing acts as a spherical pin. Design action effects and displacements (six per bearing) are known, for each limit state, from analyses of the superstructure. Preliminary cross-sections for all the members have been chosen. Composite columns are assumed to be within the scope of clause 6.7.3 (doubly symmetrical, uniform, etc.). The flow chart of Fig. 5.16 is for a single composite column, and is applicable to composite columns generally. For ultimate limit states, only long-term behaviour is considered, as this usually governs. Notes on Fig. 5.16 (1) For the elastic critical buckling force Ncr , the effective length for an unbraced column, as in Fig. 5.1(b), is at least 2L, where L is the actual length. If the foundation cannot be assumed to be ‘rigid’, its rotational stiffness should be included in an elastic critical analysis, as the effective length then exceeds 2L. In many cases, will be much less than 2, and cr will far exceed 10. These checks can then be done approximately, by simple hand calculation. Other methods of checking if second-order global analysis is required are discussed under clause 5.2.1. Here, it is assumed that for the transverse direction, cr > 10. No assumption is made for the plane shown in Fig. 5.1. The flow chart of Fig. 5.16, which is for a single column, includes second-order system effects in this plane.

64


Determine elastic properties of materials, mean value t0, creep coefficients ϕ(∞, t0), modular ratios n0, nL,s and nL,p, and shrinkage strain, εcs(∞, 1 day), as in Fig. 5.15. Select exposure class(es) for concrete, and find minimum covers, all as in Fig. 5.15. All effective widths are assumed to be actual widths. Check that concrete covers satisfy 6.7.5.1, and reinforcement satisfies 6.7.5.2. Modify if necessary

For a concrete-encased section, find concrete cross-section for use in calculations from 6.7.3.1(2) on excessive cover. Check that 6.7.3.1(3) on longitudinal reinforcement and 6.7.3.1(4) on shape of section are satisfied

Find Npl,Rd from eq. (6.30), steel contribution ratio δ from eq. (6.38), and check that 0.2 ≤ δ ≤ 0.9, to 6.7.1(4). Find Npl,Rk to 6.7.3.3(2)

Note: Until the final box, below, the chart is for both y and z planes, separately, and the y or z subscript is not given

Estimate ratio of permanent to total design normal force (axial compression) at ULS Hence find Ec,eff from eq. (6.41), to allow for creep. Find the characteristic and design flexural stiffnesses, (EI )eff and (EI )eff,II, from eqs (6.40) and (6.42)

Estimate maximum design axial force, NEd. Find αcr = Ncr/NEd. Is αcr ≥ 10? See Note 1

Determine Ncr and then λ, to 6.7.3.3(2). Check that λ ≤ 2, to 6.7.3.1(1). See Note 1 in main text No

Yes

Is αcr < 4? Yes Use second-order global analysis, including both system and member imperfections 5.3.2(2) → EC3-2 → EC3-1-1/5.3.2(6)

No Use method (i) or (ii) (see comment on clause 5.2.2) for global analysis

First-order global analysis permitted by5.2.1(3), with system imperfections

Find system imperfection φ to EC3-1-1/5.3.2(3) and hence notional transverse forces NEdφ at ends of column, and add to coexisting transverse forces (e.g. from wind loading) See Note 2 in the main text

Perform global analysis, to find action effects NEd, MEd and VEd at each end of column

Repeat for the other plane of bending. Include member imperfection only in the plane where its effect is more adverse; 6.7.3.7(1) (END)

Note: For verification of the column length, see flow chart of Fig. 6.44

Fig. 5.16. Flow chart for global analysis of a composite column

(2) Out-of-plumb of columns, a system imperfection, should be allowed for as follows. Figure 5.17(a) shows a nominally vertical column of length L, with design action effects M, N and V from a preliminary global analysis. The top-end moment M could represent an off-centre bearing. The design out-of-plumb angle for the column, , is found from clause 5.3.2(3) of EN 1993-1-1. Notional horizontal forces N are applied, as shown in Fig. 5.17(b). Second-order global analysis for the whole structure, including base flexibility, if any, then gives the deformations of the ends of the column and the action effects NEd , etc.,

65


MEd,1 NEd M

M

VEd,1

N

N

V

V + Nφ

wEd

φ L φ0 VEd,2

NEd MEd,2

(a)

(b)

(c)

Fig. 5.17. System imperfection and global analysis for a column

needed for verification of the column length. In Fig. 5.17(c), 0 is the rotation of the column base, and wEd represents the transverse loading, which may be negligible. For the determination of the bending moments within a column length and its verification, reference should be made to the flow chart of Fig. 6.36 of Ref. 5.

66

CHAPTER 6

Ultimate limit states This chapter corresponds to Section 6 of EN 1994-2, which has the following clauses: . . . . . . . . .

Beams Resistances of cross-sections of beams Filler beam decks Lateral–torsional buckling of composite beams Transverse forces on webs Shear connection Composite columns and composite compression members Fatigue Tension members in composite bridges

Clause 6.1 Clause 6.2 Clause 6.3 Clause 6.4 Clause 6.5 Clause 6.6 Clause 6.7 Clause 6.8 Clause 6.9

Clauses 6.1 to 6.7 define resistances of cross-sections to static loading, for comparison with action effects determined by the methods of Section 5. The ultimate limit state considered is STR, defined in clause 6.4.1(1) of EN 1990 as: Internal failure or excessive deformation of the structure or structural members . . . where the strength of constructional materials of the structure governs.

The self-contained clause 6.8, Fatigue, covers steel, concrete, and reinforcement by crossreference to Eurocodes 2 and 3. Requirements are given for shear connection. Clause 6.9 does not appear in EN 1994-1-1 and has been added in EN 1994-2 to cover concrete and composite tension members such as may be found in tied arch bridges and truss bridges.

6.1. Beams 6.1.1. Beams in bridges – general Clause 6.1.1(1) serves as a summary of the checks that should be performed on the beams themselves (excluding related elements such as bracing and diaphragms). The checks listed are as follows: .

.

Resistance of cross-sections to bending and shear – clauses 6.2 and 6.3. In the Eurocodes, local buckling in Class 4 members, due to direct stress, is covered under the heading of ‘cross-section’ resistance, even though this buckling resistance is derived considering a finite length of the beam. In Eurocode 3, shear buckling is similarly covered under the heading of ‘cross-section’ resistance, but this is separately itemized below. A check of the interaction between shear and bending is required in clause 6.2.2.4. Resistance to lateral–torsional buckling – clause 6.4. For lateral–torsional buckling, the resistance is influenced by the properties of the whole member. The rules of Eurocode 4 assume that the member is of uniform cross-section, apart from variations arising from

Clause 6.1.1(1)


.

.

.

cracking of concrete and from detailing. The resistance of non-uniform members is covered in clause 6.3.4 of EN 1993-2. Resistance to shear buckling and in-plane forces applied to webs – clauses 6.2.2 and 6.5 respectively. As discussed above, shear buckling resistance is treated as a property of a cross-section. Resistance to longitudinal shear – clause 6.6. According to clause 1.1.3(3), provisions for shear connection are given only for welded headed studs. This is misleading, for much of clause 6.6 is more widely applicable, as discussed under clause 6.6.1. Resistance to fatigue – clause 6.8.

The above checks are not exhaustive. Further checks that may be required include the following: .

.

. .

.

.

.

Interaction with axial force. Axial force is not included in the checks above as clause 1.5.2.4 defines a composite beam as ‘a composite member subjected mainly to bending’. Axial force does however occur in the beams of composite integral bridges.57 This is discussed in section 6.4 of this guide. Addition of stresses in webs and flanges generated from plan curvature, although this is identified in clause 6.2.1.1(5). No method of combining (or calculating) these effects is provided in Eurocodes 3 or 4. The Designers’ Guide to EN 1993-24 provides some guidance, as do the comments on clause 6.2.1.1(5). Flange-induced buckling of the web – clause 6.5.2 refers. Torsion in box girders, which adds to the shear in the webs and necessitates a further check on the flange – Section 7 of EN 1993-1-5 refers. The need to consider combinations of torsion and bending is mentioned in clause 6.2.1.3(1). Distortion of box girders, which causes both in-plane and out-of-plane bending in the box walls – clause 6.2.7 of EN 1993-2 refers and the Designers’ Guide to EN 1993-2 provides some guidance. Torsion of bare steel beams during construction, which often arises with the use of cantilever forms to construct the deck edge cantilevers. This usually involves a consideration of both St Venant torsion and warping torsion. Design of transverse stiffeners – Section 9 of EN 1993-1-5 refers.

Steel cross-sections may be rolled I- or H-section or doubly-symmetrical or monosymmetrical plate girders. Other possible types include any of those shown in sheet 1 of Table 5.2 of EN 1993-1-1; this includes box girders. Channel and angle sections should not be used unless the shear connection is designed to provide torsional restraint or there is adequate torsional bracing between beams.

6.1.2. Effective width for verification of cross-sections Effective widths for shear lag are discussed in section 5.4.1.2 of this guide. Unlike in global analysis, the effective width appropriate to the cross-section under consideration must be used in calculation of resistance to bending. Distributions of effective width along a span are given in Figure 5.1.

6.2. Resistances of cross-sections of beams This clause is for beams without partial or full encasement in concrete. Filler beams with partial encasement are treated in clause 6.3. Full encasement is outside the scope of EN 1994. No guidance is given in EN 1994, or in EN 1993, on the treatment of large holes in steel webs without recourse to finite-element modelling (following the requirements of EN 1993-15), but specialized literature is available.58;59 Bolt holes in steelwork should be treated in accordance with EN 1993-1-1, particularly clauses 6.2.2 to 6.2.6.

68


6.2.1. Bending resistance 6.2.1.1. General In clause 6.2.1.1, three different approaches are given, based on rigid-plastic theory, nonlinear theory, and elastic theory. The ‘non-linear theory’ is that given in clause 6.2.1.4. This is not a reference to non-linear global analysis. Clause 6.2.1.1(1) only permits rigid-plastic theory to be used where cross-sections are in Class 1 or 2 and where prestressing by tendons is not used. This is because no explicit check of yielding of bonded tendons is given and therefore non-linear resistance calculation is more appropriate. Comment on this use of plastic resistance with elastic analysis is given under clause 5.4.1.1(1). Clause 6.2.1.1(2) permits non-linear theory and elastic theory to be used for all crosssections. If unbonded tendons are used, the tendon forces used in section analysis should however be derived in accordance with clause 5.4.2.7(2). The assumption that composite cross-sections remain plane is always permitted by clause 6.2.1.1(3) where elastic and non-linear theory are used, because the conditions set will be satisfied if the design is in accordance with EN 1994. The implication is that longitudinal slip is negligible. There is no requirement for slip to be determined. This would be difficult because the stiffness of shear connectors is not known accurately, especially where the slab is cracked. Wherever slip may not be negligible, the design methods of EN 1994-2 are intended to allow for its effects. For beams with curvature in plan, clause 6.2.1.1(5) gives no guidance on how to allow for the torsional moments induced or how to assess their significance. Normal practice is to treat the changing direction of the longitudinal force in a flange (and a web, if significant) as a transverse load applied to that flange, which is then designed as a horizontal beam spanning between transverse restraints. It is common to use elastic section resistance in such circumstances to avoid the complexity of producing a plastic stress block for the combined local and global loading. The shear connection and bracing system should be designed for the additional transverse forces. Similar calculation should be carried out where curvature is achieved using a series of straight sections, except that the transverse forces will be concentrated at the splices between adjacent lengths. Particular care is needed with detailing the splices. Transverse stiffeners and bracing will usually be needed close to each splice to limit the bending in the flange. In box girders, the torsion from curvature will also tend to produce distortion of the box. This must be considered in the design of both the cross-section of the box and its internal restraints. Bending in transverse planes can also be induced in flanges by curvature of the flange in a vertical plane, and should be considered. Clause 6.5 covers the transverse forces on webs that this causes, but not the transverse bending in the flange. The latter is covered in the Designers’ Guide to EN 1993-2.4

6.2.1.2. Plastic resistance moment Mpl;Rd of a composite cross-section ‘Full interaction’ in clause 6.2.1.2(1)(a) means that no account need be taken of slip or separation at the steel–concrete interface. ‘Full interaction’ should not be confused with ‘full shear connection’. That concept is used only in the rules for buildings, and is explained in clause 6.1.1(7)P of EN 1994-1-1 as follows:

Clause 6.2.1.1(1)

Clause 6.2.1.1(2)

Clause 6.2.1.1(3)

Clause 6.2.1.1(5)

Clause 6.2.1.2(1)(a)

A span of a beam . . . has full shear connection when increase in the number of shear connectors would not increase the design bending resistance of the member.

This link of shear connection to bending resistance differs from the method of EN 1994-2, where shear connection is related to action effects, both static and fatigue. Shear connection to Part 2 is not necessarily ‘full’ according to the above definition (which should strictly read ‘. . . number of shear connectors within a critical length . . .’). It would be confusing to refer to it as ‘partial’, so this term is never used in Part 2.

69


Clause 6.2.1.2(1)(c)

Reinforcement in compression It is usual to neglect slab reinforcement in compression (clause 6.2.1.2(1)(c)). Its effect on the bending resistance of the composite section is negligible unless the slab is unusually small. If it is included, and the concrete cover is little greater than the bar diameter, consideration should be given to possible buckling of the bars. Guidance on detailing is given in clauses 9.5.3(6) and 9.6.3(1) of EN 1992-1-1 for reinforcement in concrete columns and walls respectively. The former requires that no bar within a compression zone should be further than 150 mm from a ‘restrained’ bar, but ‘restrained’ is not defined. This could be interpreted as requiring all compression bars in an outer layer to be within 150 mm of a bar held in place by transverse reinforcement. This would usually require link reinforcement in the flange. This interpretation was used in BS 5400 Part 411 for compression bars assumed to contribute to the resistance of the section. If the compression flange is classed as a wall, clause 9.6.3 of EN 1992-1-1 requires only that the longitudinal bars are placed inside horizontal (i.e. transverse) reinforcement unless the reinforcement in compression exceeds 2% of the gross concrete area. In the latter case, transverse reinforcement must be provided in accordance with the column rules. Stress/strain properties for concrete The design compressive strength of concrete, fcd , is defined in clause 3.1.6(1)P of EN 1992-1-1 as: fcd ¼ cc fck =C where: ‘cc is the coefficient taking account of long term effects on the compressive strength and of unfavourable effects resulting from the way the load is applied. ‘Note: The value of cc for use in a country should lie between 0.8 and 1.0 and may be found in its National Annex. The recommended value is 1.’ The reference in clause 3.1(1) to EN 1992-1-1 for properties of concrete begins ‘unless otherwise given by Eurocode 4’. Resistances of composite members given in EN 1994-2 are based on extensive calibration studies (e.g. Refs 60, 61). The numerical coefficients given in resistance formulae are consistent with the value cc ¼ 1:0 and the use of either elastic theory or the stress block defined in clause 6.2.1.2. Therefore, there is no reference in EN 1994-2 to a coefficient cc or to a choice to be made in a National Annex. The symbol fcd always means fck =C , and for beams and most columns is used with the coefficient 0.85, as in equation (6.30) in clause 6.7.3.2(1). An exception, in that clause, is that the 0.85 is replaced by 1.0 for concrete-filled column sections, based on calibration. The approximation made to the shape of the stress–strain curve is also relevant. Those given in clause 3.1 of EN 1992-1-1 are mainly curved or bilinear, but in clause 3.1.7(3) there is a simpler rectangular stress distribution, similar to the stress block given in the British Standard for the structural use of concrete, BS 8110.62 Its shape, for concrete strength classes up to C50/60, and the corresponding strain distribution are shown in Fig. 6.1 below. This stress block is inconvenient for use with composite cross-sections, because the region near the neutral axis assumed to be unstressed is often occupied by a steel flange, and algebraic expressions for resistance to bending become complex. EN 1992-1-1: fcd = αcc fck /γC x

0.8x EN 1994-1-1: 0.85fcd, with fcd = fck /γC 0


0.0035

Compressive strain

0 Compressive stress

Fig. 6.1. Stress blocks for concrete at ultimate limit states

70


In composite sections, the contribution from the steel section to the bending resistance reduces the significance of that from the concrete. It is thus possible63 for EN 1994 to allow the use of a rectangular stress block extending to the neutral axis, as shown in Fig. 6.1. For a member of unit width, the moment about the neutral axis of the EN 1992 stress block ranges from 0.38fck x2 =C to 0.48fck x2 =C , depending on the value chosen for cc . The value for beams in EN 1994-2 is 0.425fck x2 =C . Calibration studies have shown that this overestimates the bending resistance of cross-sections of columns, so a correction factor M is given in clause 6.7.3.6(1). See also the comments on clause 6.7.3.6.

Small concrete flanges Where the concrete slab is in compression, the method of clause 6.2.1.2 is based on the assumption that the whole effective areas of steel and concrete can reach their design strengths before the concrete begins to crush. This may not be so if the concrete flange is small compared with the steel section. This lowers the plastic neutral axis, and so increases the maximum compressive strain at the top of the slab, for a given tensile strain in the steel bottom flange. A detailed study of the problem has been reported.64 Laboratory tests on beams show that strain hardening of steel usually occurs before crushing of concrete. The effect of this, and the low probability that the strength of both the steel and the concrete will be only at the design level, led to the conclusion that premature crushing can be neglected unless the grade of the structural steel is higher than S355. Clause 6.2.1.2(2) specifies a reduction in Mpl;Rd where the steel grade is S420 or S460 and the depth of the plastic neutral axis is high. For composite columns, the risk of premature crushing led to a reduction in the factor M , given in clause 6.7.3.6(1), for S420 and S460 steels. Ductility of reinforcement Reinforcement with insufficient ductility to satisfy clause 5.5.1(5), and welded mesh, should not be included within the effective section of beams in Class 1 or 2 (clause 6.2.1.2(3)). This is because laboratory tests on hogging moment regions have shown33 that some reinforcing bars, and most welded meshes, fracture before the moment–rotation curve for a typical double-cantilever specimen reaches a plateau. The problem with welded mesh is explained in comments on clause 3.2(3). 6.2.1.3. Additional rules for beams in bridges Clause 6.2.1.3(1) is a reminder that composite beams need to be checked for possible combinations of internal actions that are not specifically covered in EN 1994. The combinations given are biaxial bending, bending and torsion and local and global effects, with a reference to clause 6.2.1(5) of EN 1993-1-1. Significant bending about a vertical axis is rare in composite bridges so biaxial bending is rarely a concern. Despite the reference to EN 1993-1-1, there is little of direct relevance for biaxial bending in composite beams therein. For Class 1 and 2 cross-sections, the interaction of EN 1993-1-1 expression (6.2) could be used for resistance of cross-sections. Rather than computing the resultant plastic stress block for axial load and biaxial bending, a linear interaction is provided: NEd My;Ed Mz;Ed þ þ 1:0 NRd My;Rd Mz;Rd

Clause 6.2.1.2(2)

Clause 6.2.1.2(3)

Clause 6.2.1.3(1)

where NRd , My;Rd and Mz;Rd are the design resistances for each effect acting individually, with reductions for shear where the shear force is sufficiently large. In theory, it is still necessary to derive the resultant plastic stress block to check whether the cross-section is either Class 1 or 2. This complexity can be avoided by performing the classification under axial compression only. The same expression can be applied to Class 3 and 4 cross-sections or the stresses can be summed using elastic section analysis. Care is needed where the sign of the stress in the slab is different for each constituent action.

71


In hogging zones of integral bridges, where there is usually a moderate coexistent axial load induced by temperature or soil pressure, it is common to do calculations on the basis of the fully cracked section. The non-linear method of clause 6.2.1.4 could also be used but this is likely to require the use of computer software. Buckling needs to be checked separately – see section 6.4. Significant torsion is unlikely to be encountered in most composite I-girder bridges due to the low St Venant torsional stiffness of the steel beams. There are some exceptions including: . . .

Clause 6.2.1.3(2)

torsion in curved beams as discussed in comments on clause 6.2.1.1(5) torsion in skew decks at end trimmers torsion of bare steel beams where formwork for deck cantilevers is clamped to the outer girders.

EN 1993-1-1 clause 6.2.7(7) permits St Venant torsion to be ignored at ultimate limit states provided that all the torsion is carried by resistance to warping. This is usually the most efficient model and avoids a further interaction with shear stress from vertical shear in the web. If the torque is resisted by opposing bending in the flanges, they can be designed for this bending combined with their axial force. If the length between restraints should be long, then the warping bending stresses would become large and the section would try to resist the torsion predominantly through St Venant shear flow. In that case it might be better to derive the separate contributions from St Venant and warping torsion. Further guidance on shear, torsion and bending is provided in the Designers’ Guide to EN 1993-2.4 Pure torsion in box beams is treated simply by a modification to the shear stress in the webs and flanges, and the design is checked using clause 7.1 of EN 1993-1-5. Pure torsion is however rare and most boxes will also suffer some distortion. This leads to both inplane warping and out-of-plane bending of the box walls as discussed in Ref. 4. The reference in clause 6.2.1.3(1) to combined local and global effects relates to the steel beam only, because this combination in a concrete deck is a matter for Eurocode 2, unless the deck is a composite plate, when clause 9.3 applies. Such combinations include bending, shear and transverse load (from wheel loads) according to clause 6.2.8(6) of EN 1993-1-1 and other combinations of local and global load. The Von Mises equivalent stress criterion of EN 19931-1 expression (6.1) should be used in the absence of test-based interaction equations for resistances. Clause 6.2.1.3(2) relates to the use of plastic resistances in bending, which implies shedding of bending moments, typically from mid-span regions to adjacent supports. Non-linear global analysis allows for this, but linear-elastic analysis does not. The reasons for permitting linear analysis, and for the limitations given in the present clause, are explained in comments on clause 5.4.1.1(1). A method for making use of the limited ductility of support regions has been proposed.65

Example 6.1: plastic resistance moment in sagging bending For the bridge in Example 5.1 (Fig. 5.6), the resistance moment for an internal beam at mid-span with the cross-section in Fig. 5.11 is determined. The deck concrete is Grade C30/37 and the structural steel is S355 J2 G3. The cross-section is shown in Fig. 6.2 with relevant dimensions, stress blocks and longitudinal forces. The notation is as in Fig. 6.2 and the concrete stress block as in Fig. 6.1. Using the partial factors recommended in EN 1992-2 and EN 1993-2, the design strengths are: fyd ¼ 345=1:0 ¼ 345 N=mm2 ; fcd ¼ 30=1:5 ¼ 20 N=mm2 The steel yield stress has been taken here as 345 N/mm2 throughout. This is given in EN 10025 for thicknesses between 16 mm and 40 mm. For the web, 355 N/mm2 could have been used.

72


3100 0.85fcd

227

11.97 MN

250

172 NA

25 haunch 400 × 20

2.76 MN 769

1225 1175 × 12.5

5.07 MN 4.14 MN

400 × 30

1372 fyd

Fig. 6.2. Plastic resistance of cross-section to sagging bending, for Example 6.1

Ignoring the haunch and any slab reinforcement, the available compressive force in the concrete is: Nc;f ¼ 3:1 0:25 0:85 20 ¼ 13:18 MN The available forces in the steel beam are: in the top flange,

Na;top ¼ 0:4 0:02 345 ¼ 2:76 MN

in the web,

Na;web ¼ 1:175 0:0125 345 ¼ 5:07 MN

in the bottom flange,

Na;bot ¼ 0:4 0:03 345 ¼ 4:14 MN

The total is 11.97 MN. As this is less than 13.18 MN, the neutral axis lies in the concrete slab, at a depth: zna ¼ 250ð11:97=13:18Þ ¼ 227 mm The distances below force Nc of the lines of action of the three forces Na are shown in Fig. 6.2. Hence: Mpl;Rd ¼ 2:76 0:172 þ 5:07 0:769 þ 4:14 1:372 ¼ 10:05 MNm The cross-section at the adjacent support is in Class 3, so potentially clause 6.2.1.3(2) applies. Since the ratio of adjacent spans is 0.61, greater than the limit of 0.6, there is no need to restrict the bending resistance at mid-span to 0.90Mpl;Rd .

Example 6.2: resistance to hogging bending at an internal support For the bridge shown in Fig. 5.6, and materials as in Example 6.1, the resistance to hogging bending of the cross-section shown in Fig. 5.14 is studied. It was found in Example 5.4 that the flanges are in Class 1 and the web is in Class 3. It appears from clause 5.5.2(3) that an effective section in Class 2 could be used, to clause 6.2.2.4 of EN 1993-1-1. However, the wording of that clause implies, and its Fig. 6.3 shows, that the plastic neutral axis of the effective section should lie within the web. It is concluded in the Designers’ Guide to EN 1994-1-15 that the hole-in-web approximation should only be used when this is so. As the area of longitudinal reinforcement in the present cross-section is quite high, this condition is checked first. From clause 6.2.2.4 of EN 1993-1-1, the effective area of web in compression is 40t2w ". Using rectangular stress blocks, the condition is: As fsd þ Aa;top fyd < 40t2w " fyd þ Aa;bot fyd Hence: As < ð40t2w " þ Aa;bot Aa;top Þ fyd =fsd

ðD6:1Þ

73


From Example 5.4: As ¼ 19 480 mm2 ;

and

"¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 235=345 ¼ 0:825

From Fig. 5.14: Aa;bot Aa;top ¼ 400 15 ¼ 6000 mm2 ; tw ¼ 25 mm For the reinforcement: fsd ¼ 500=1:15 ¼ 435 N=mm2 From expression (D6.1): As < ð40 252 0:825 þ 6000Þ 345=435 ¼ 21 120 mm2 Thus, expression (D6.1) is satisfied, so Mpl;Rd will be found by the hole-in-web method. The longitudinal forces in the reinforcement and the two steel flanges are found as in Example 6.1, and are shown in Fig. 6.3. 5.645 250

zweb = 83

400 × 25 Plastic NA

120

3.45

227.5

0.715

412

281.5 529

3.56

25

1160

2.82

Hole in web

253

400 × 40

412

Forces in MN units 3.56 1194 1420 5.52

Fig. 6.3. Plastic resistance of cross-section to hogging bending, for Example 6.2

The depth of each of the two compressive stress blocks in the web is: 20tw " ¼ 20 25 0:825 ¼ 412 mm The force in each of them is 20t2w " fyd ¼ 0:412 0:025 345 ¼ 3:56 MN The location of the plastic neutral axis is found from longitudinal equilibrium. The tensile force in the web is: Tw ¼ 3:56 2 þ 5:52 5:645 2:82 3:45 ¼ 0:715 MN ¼ 715 kN This force requires a depth of web: zw ¼ 715=ð25 0:345Þ ¼ 83 mm This leads to the depth of the ‘hole’ in the web, 253 mm, and the distances of the various forces below the level of the top reinforcement, shown in Fig. 6.3. Taking moments about this level gives the bending resistance, which is: M pl;Rd ¼ 12:64 MNm

Clause 6.2.1.4(1) to (5)

74

6.2.1.4. Non-linear resistance to bending There are two approaches, described in clause 6.2.1.4. With both, the calculations should be done at the critical sections for the design bending moments. The first approach, given in clause 6.2.1.4(1) to (5), enables the resistance of a section to be determined iteratively from the stress–strain relationships of the materials.


C

Mpl,Rd Mel,Rd

B

Ma,Ed A Nc /Nc,f 0

Nc,el /Nc,f

1.0

Fig. 6.4. Non-linear resistance to bending for Class 1 and 2 cross-sections

A curvature (strain gradient) and neutral-axis position are assumed, the stresses determined from the strains, and the neutral axis moved until the stresses correspond to the external longitudinal force, if any. The assumed strain distributions should allow for the shrinkage strain of the concrete and any strain and/or difference of curvature between steel and concrete caused by temperature. The bending resistance is calculated from this stress distribution. If it exceeds the external moment MEd , the calculation is terminated. If not, the assumed curvature is increased and the process repeated until a value MRd is found that exceeds MEd . If one of the ultimate strains given in EN 1992-1-1 for concrete and reinforcement is reached first, the cross-section has insufficient resistance. For Class 3 cross-sections or Class 4 effective cross-sections, the compressive strain in the structural steel must not exceed that at first yield. Clearly, in practice this procedure requires the use of software. For sections in Class 1 or 2, a simplified approach is given in clause 6.2.1.4(6). This is based on three points on the curve relating longitudinal force in the slab, Nc , to design bending moment MEd that are easily determined. With reference to Fig. 6.4, which is based on Fig. 6.6, these points are: . . .

Clause 6.2.1.4(6)

A, where the composite member resists no moment, so Nc ¼ 0 B, which is defined by the results of an elastic analysis of the section, and C, based on plastic analysis of the section.

Accurate calculation shows BC to be a convex-upwards curve, so the straight line BC is a conservative approximation. Clause 6.2.1.4(6) thus enables hand calculation to be used. The elastic analysis gives the resistance Mel;Rd , which is calculated according to equation (6.4). The moment acting on the composite section will generally comprise both shortterm and permanent actions and in calculating the stresses from these, appropriate modular ratios should be used in accordance with clause 5.4.2.2(2). Clause 6.2.1.4(7) makes reference to EN 1993-1-1 for the stress–strain relationship to be used for prestressing steel. The prestrain (which is the initial tendon strain after all losses, calculated in accordance with EN 1992-1-1 clause 5.10.8) must be taken into account in the section design. For bonded tendons, this can be done by displacing the origin of the stress–strain curve along the strain axis by an amount equal to the design prestrain and assuming that the strain change in the tendon is the same as that in the surrounding concrete. For unbonded tendons, the prestress should be treated as a constant force equal to the applied force after all losses. The general method of section analysis for composite columns in clause 6.7.2 would then be more appropriate.

Clause 6.2.1.4(7)

6.2.1.5. Elastic resistance to bending Clause 6.2.1.4(6) includes, almost incidentally, a definition of Mel;Rd that may seem strange. It is a peculiarity of composite structures that when unpropped construction is used, the elastic resistance to bending depends on the proportion of the total load that is applied before the member becomes composite. Let Ma;Ed and Mc;Ed be the design bending

75


Clause 6.2.1.5(2)

Clause 6.2.1.5(5)

Clause 6.2.1.5(6)

Clause 6.2.1.5(7)

moments for the steel and composite sections, respectively, for a section in Class 3. Their total is typically less than the elastic resistance to bending, so to find Mel;Rd , one or both of them must be increased until one or more of the limiting stresses in clause 6.2.1.5(2) is reached. To enable a unique result to be obtained, clause 6.2.1.4(6) says that Mc;Ed is to be increased, and Ma;Ed left unchanged. This is because Ma;Ed is mainly from permanent actions, which are less uncertain than the variable actions whose effects comprise most of Mc;Ed . Unpropped construction normally proceeds by stages, which may have to be considered individually in bridge design. While the sequence of erection of the beams is often known in the design stage, the concrete pour sequence is rarely known. Typically, either a range of possible pour sequences is considered or it is assumed that the whole of the wet concrete is placed simultaneously on the bare steelwork, and the resulting design is rechecked when the pour sequence is known. The weight of formwork is, in reality, applied to the steel structure and removed from the composite structure. This process leaves self-equilibrated residual stresses in composite cross-sections. Whether or not this is considered in the final situation is a matter for judgement, depending on the significance of the weight of the formwork. One permanent action that influences Mel;Rd is shrinkage of concrete. Clause 6.2.1.5(5) enables the primary stresses to be neglected in cracked concrete, but the implication is that they should be included where the slab is in compression. This provision should not be confused with clause 5.4.2.2(8), although it is consistent with it. The self-equilibrating stresses from the primary effects of shrinkage do not cause any moment but they can give rise to stress. In checking the beam section, if these stresses are adverse, they should be added to those from Ma;Ed and Mc;Ed when verifying stresses against the limits in clause 6.2.1.5(2). If it is necessary to determine the actual elastic resistance moment, Mel;Rd , the shrinkage stresses should be added to the stresses from Ma;Ed and kMc;Ed when determining k and hence Mel;Rd . If this addition increases Mel;Rd , it could be omitted, but this is not a requirement, because shrinkage is classified as a permanent action. Clause 6.2.1.5(6) is a reminder that lateral–torsional buckling should also be checked, which applies equally to the other methods of cross-section design. The calculation of Mel;Rd is relevant for Class 3 cross-sections if the method of clause 6.4.2 is used, but the above problem with shrinkage does not occur as the slab will be in tension in the critical region. Additional guidance is required for Class 4 cross-sections since the effectiveness of the Class 4 elements (usually only the web for composite I-beams) depends on the stress distributions within them. The loss of effectiveness for local buckling is dealt with by the use of effective widths according to EN 1993-1-5. For staged construction, there is the additional problem that the stress distribution changes during construction and therefore the size and location of the effective part of the element also change at each stage. To avoid the complexity of summing stresses from different effective cross-sections, clause 6.2.1.5(7) provides a simplified pragmatic rule. This requires that the stress distribution at any stage is built up using gross-section properties. The reference to ‘gross’ sections is not intended to mean that shear lag can be neglected; it refers only to the neglect of plate buckling. The stress distribution so derived is used to determine an effective web which is then used to determine section properties and stresses at all stages up to the one considered. The Note to clause 4.4(3) of EN 1993-1-5 provides almost identical guidance, but clarifies that an effective flange should be used together with the gross web to determine the initial stress distribution. ‘Effective’ in this sense includes the effects of both shear lag and plate buckling. Plate buckling for flanges is likely to be relevant only for box girders. Example 6.3 illustrates the method. Clause 6.2.1.5(7) refers to some clauses in EN 1993-1-5 that permit mid-plane stresses in steel plates to be used in verifications. For compression parts in Class 3, EN 1993-2 follows clause 6.2.1(9) of EN 1993-1-1. This says: Compressive stresses should be limited to the yield strength at the extreme fibres.

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It is followed by the Note: The extreme fibres may be assumed at the midplane of the flanges for ULS checks. For fatigue see EN 1993-1-9.

This concession can be assumed to apply also to composite beams. An assumption in the effective section method is that there is sufficient post-buckling strength to achieve the necessary redistribution of stress to allow all components to be stressed to their individual resistances. This approach is therefore not permitted (and is not appropriate) in a number of situations where there may not be sufficient post-buckling strength or where the geometry of the member is outside prescribed limits. These exceptions are given in EN 1993-1-5, clause 2.3(1). Where prestressing is used, clause 6.2.1.5(8) limits the stress in tendons to the elastic range and makes reference to clause 5.10.8 of EN 1992-1-1 for guidance on initial prestrain. The latter covers both bonded and unbonded prestress. Clause 6.2.1.5(9) provides an alternative method of treating Class 4 cross-sections using Section 10 of EN 1993-1-5. This method can be used where the conditions of EN 1993-1-5 clause 2.3(1) are not met. Section 10 requires that all stresses are calculated on gross sections and buckling checks are then carried out on the component plates of the cross-section. There is usually economic disadvantage in using this method because the beneficial load shedding of stress around the cross-section implicit in the effective section method does not occur. Additionally, the benefit of using test-based interactions between shear and bending is lost. If the whole member is prone to overall buckling instability, such as flexural or lateral– torsional buckling, these effects must either be calculated by second-order analysis and the additional stresses included when checking panels or by using a limiting stress limit in member buckling checks. For flexural buckling, limit can be calculated based on the lowest compressive value of axial stress x;Ed acting on its own, required to cause buckling failure in the weakest sub-panel or an entire panel, according to the verification formula in Section 10 of EN 1993-1-5. This value of limit is then used to replace fy in the member buckling check. It is conservative, particularly when the critical panel used to determine limit is not at the extreme compression fibre of the section where the greatest stress increase during buckling occurs. For lateral–torsional buckling, limit can be determined as the bending stress at the extreme compression fibre needed to cause buckling in the weakest panel. This would however again be very conservative where limit was determined from buckling of a web panel which was not at the extreme fibre, as the direct stress in a web panel would not increase much during lateral–torsional buckling. A detailed discussion of the use of Section 10 of EN 1993-1-5 is given in the Designers’ Guide to EN 1993-2.4

Clause 6.2.1.5(8)

Clause 6.2.1.5(9)

Example 6.3: elastic bending resistance of a Class 4 cross-section For the bridge in Example 5.1 (Fig. 5.6), the mid-span section of the internal beam in Fig. 5.11 continues to the splice adjacent to each pier. The top and bottom layers of reinforcement comprise 16 mm bars at 150 mm centres. There are 20 mm transverse bars, with top and bottom covers of 40 mm and 45 mm respectively, so the locations of the 16 mm bars are as shown in Fig. 6.5. All reinforcement has fsk ¼ 500 N/mm2 and S ¼ 1:15. The steel yield stress is taken as 345 N/mm2 throughout. The cross-section is checked for the ultimate limit state hogging moments adjacent to the splice, which are as follows: steel beam only:

Ma;Ed ¼ 150 kNm

cracked composite beam:

Mc;Ed ¼ 2600 kNm (including secondary effects of shrinkage)

By inspection, the cross-section is not in Class 1 or 2 so its classification is checked at the Class 3/4 boundary using elastic stresses.

77


4155 mm2

4155 mm2

68

250

109

25

73

400 × 20

Elastic neutral axis, effective composite section

1175 × 12.5 1225 389 47 400 × 30

247

Fig. 6.5. Effective section and reinforcement for Example 6.3

Steel bottom flange Ignoring the web-to-flange welds, the flange outstand c ¼ ð400 12:5Þ=2 ¼ 193:8 mm. From Table 5.2 of EN 1993-1-1, the condition for Class 1 is: c=t < 9" ¼ 9 0:825 ¼ 7:43 For the flange, c=t ¼ 193:8=30 ¼ 6:46, so the flange is Class 1.

Clause 3.2(2)

Steel web The area of each layer of reinforcement is As ¼ 64 3100=150 ¼ 4155 mm2 . It would be conservative to assume that all stresses are applied to the composite section as this gives the greatest depth of web in compression. The stresses below are however based on the built-up elastic stresses. The elastic modulus for the reinforcement is taken as equal to that for structural steel, from clause 3.2(2). The elastic section moduli for the gross cross-section are given in rows 1 and 3 of Table 6.1. The extreme-fibre stresses for the steel section are: a;top ¼ 150=12:87 þ 2600=25:85 ¼ 112:2 N/mm2 tension a;bot ¼ 150=15:96 þ 2600=18:91 ¼ 146:9 N/mm2 compression Table 6.1. Section moduli for hogging bending of the cross-section of Fig. 6.5, in 106 mm3 units, and height of neutral axis above bottom of section Section modulus

Gross steel section Effective steel section Gross composite section Effective composite section

Top layer of bars

Top of steel section

Bottom of steel section

Height of neutral axis (mm)

— — 18.47 18.47

12.87 12.90 25.85 25.94

15.96 15.77 18.91 18.63

547 551 707 713

Primary shrinkage stresses are neglected because the deck slab is assumed to be cracked. Using the stresses at the extreme fibres of the web, from Table 5.2 of EN 1993-1-1, the stress ratio is: ¼ 108=140:6 ¼ 0:768 and the condition for a Class 3 web is: c=t 42"=ð0:67 þ 0:33 Þ ¼ 42 0:825=ð0:67 0:253Þ ¼ 83:1 Neglecting the widths of the fillet welds, the actual c=t ¼ 1175=12:5 ¼ 94, so the composite section is in Class 4 for hogging bending. An effective section must therefore be

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derived for the web in accordance with clause 6.2.1.5(7) using the built-up stresses calculated on the gross cross-section above. From Table 4.1 of EN 1993-1-5: k ¼ 7:81 6:29 þ 9:78

2

¼ 18:4 for

¼ 0:768:

From clause 4.4(2) of EN 1993-1-5: p ¼

b=t 1175=12:5 pffiffiffiffiffiffiffiffiffi ¼ 0:935 > 0:673 pffiffiffiffiffi ¼ 28:4" k 28:4 0:825 18:4

so the reduction factor is: ¼

p 0:055ð3 þ Þ 2 p

¼

0:935 0:055ð3 0:768Þ ¼ 0:929 0:9352

beff ¼ b=ð1 Þ ¼ 617 mm; be1 ¼ 0:4 617 ¼ 247 mm; be2 ¼ 0:6 617 ¼ 370 mm Including the ‘hole’, the depth of web in compression is beff =, which is 664 mm, so the width of the hole is 664 617 ¼ 47 mm. The stress ratio for the web now differs from that for the gross section, but the effect of this on the properties of the net section can be neglected. It is clear from clause 4.4(3) of EN 1993-1-5 that (and hence and beff ) need not be recalculated. The level of the elastic neutral axis for this net section is found to be as shown in Table 6.1; consequently, the depth be2 is in fact 389 mm, not 370 mm. The new section moduli are given in rows 2 and 4 of Table 6.1. The effective section is as shown in Fig. 6.5. The final stresses are as follows: a;top ¼ 150=12:90 þ 2600=25:94 ¼ 111:8 N/mm2 tension < 345 N/mm2 a;bot ¼ 150=15:77 þ 2600=18:63 ¼ 149:1 N/mm2 compression < 345 N/mm2 s;top ¼ 2600=18:47 ¼ 140:8 N/mm2 < 500=1:15 ¼ 435 N/mm2 The stress change caused by the small reduction in web area is negligible in this case.

Elastic resistance to bending From clause 6.2.1.4(6), Mel;Rd is found by scaling up Mc;Ed by a factor k until a stress limit is reached. By inspection of the final stresses, the bottom flange will probably govern. In fact, it does, and: 150=15:77 þ ð2600=18:63Þk ¼ 345 whence k ¼ 2:40, and the elastic resistance moment is: Mel;Rd ¼ 150 þ 2:40 2600 ¼ 6390 kNm (provided that Ma;Ed ¼ 150 kNm)

6.2.2. Resistance to vertical shear Clause 6.2.2 is for beams without web encasement. The whole of the vertical shear is usually assumed to be resisted by the steel section, as in previous codes for composite beams. This enables the design rules of EN 1993-1-1 and EN 1993-1-5 to be used. The assumption can be conservative where the slab is in compression. Even where it is in tension and cracked in flexure, consideration of equilibrium shows that the slab must make some contribution to shear resistance, except where the reinforcement has yielded. For solid slabs, the effect is significant where the depth of the steel beam is only twice that of the slab,66 but diminishes as this ratio increases. In composite plate girders with vertical stiffeners, the concrete slab can contribute to the anchorage of a tension field in the web,67 but the shear connectors must then be designed for vertical forces (clause 6.2.2.3(2)). The tension field model used in EN 1993-1-5 is discussed in the Designers’ Guide to EN 1993-2.4 Since the additional tension field supported by the flanges must be anchored at both upper and lower surfaces of the web, the weaker flange

Clause 6.2.2

Clause 6.2.2.3(2)

79


VEd

Shear resistance Vb,Rd to EN 1993-1-5/5

Interaction to clause 6.2.2.4(2)

VEd Vpl,Rd

Vbw,Rd

0.5Vpl,Rd

0.5Vbw,Rd

MEd Mf,Rd (a)

MEd

Mpl,Rd

Mf,Rd

Mpl,Rd

(b)

Fig. 6.6. Shear–moment interaction for Class 1 and 2 cross-sections (a) with shear buckling and (b) without shear buckling

will govern the contribution of the flanges, Vbf;Rd , to shear resistance. Comment given later on clause 6.2.2.5(1) is relevant here.

Clause 6.2.2.4(1) Clause 6.2.2.4(2)

Bending and vertical shear – beams in Class 1 or 2 Shear stress does not significantly reduce bending resistance unless the shear is quite high. For this reason, the interaction may be neglected until the shear force exceeds half the shear resistance (clause 6.2.2.4(1)). Both EN 1993-1-1 and EN 1994-2 use a parabolic interaction curve. Clause 6.2.2.4(2) covers the case of Class 1 or 2 cross-sections where the reduction factor for the design yield strength of the web is ð1 Þ, where: ¼ ½ð2VEd =VRd Þ 12

ð6:5Þ

and VRd is the resistance in shear (which is either the plastic shear resistance or the shear buckling resistance if lower). The interactions for Class 1 and 2 cross-sections with and without shear buckling are shown in Fig. 6.6. For a web where the shear buckling resistance is less than the plastic shear resistance and MEd < Mf;Rd , the flanges may make a contribution Vbf;Rd to the shear resistance according to EN 1993-1-5 clause 5.4(1). For moments exceeding Mf;Rd (the plastic bending resistance ignoring the web), this contribution is zero as at least one flange is fully utilized for bending. VRd is then equal to Vbw;Rd . For moments less than Mf;Rd , VRd is equal to Vbw;Rd þ Vbf;Rd . This definition of VRd leads to some inconsistency in clause 6.2.2.4(2) as the resistance in bending produced therein can never be less than Mf;Rd . Where there is shear buckling therefore, it is best to consider that the interaction with bending and shear according to clause 6.2.2.4(2) is valid for moments in excess of Mf;Rd only. For lower moments, the interaction with shear is covered entirely by the shear check to EN 1993-1-5 clause 5.4(1). Where a Class 3 cross-section is treated as an equivalent Class 2 section and the design yield strength of the web is reduced to allow for vertical shear, the effect on a section in hogging bending is to increase the depth of web in compression. If the change is small, the hole-in-web model can still be used. For a higher shear force, the new plastic neutral axis may be within the top flange, and the hole-in-web method is inapplicable. The section should then be treated as a Class 3 section.

Clause 6.2.2.4(3)

80

Bending and vertical shear – beams in Class 3 or 4 If the cross-section is either Class 3 or Class 4, then clause 6.2.2.4(3) applies and the interaction should be checked using EN 1993-1-5 clause 7.1. This clause is similar to that for Class 1 and 2 sections but an interaction equation is provided. This allows the designer to neglect the interaction between shear and bending moment when the design shear force is less than 50% of the shear buckling resistance based on the web contribution alone. Where the design


VEd

Shear resistance Vb,Rd to EC3-1-5/5

Vbw,Rd (1834)

Interaction to EC3-1-5/7.1

B (2900), etc. Values from Example 6.4

A (1100) Vbw,Rd /2

(2900)

MEd Mf,Rd Mel,Rd Mpl,Rd (5568) (6390) (8089)

Fig. 6.7. Shear–moment interaction for Class 3 and 4 cross-sections to clause 7.1 of EN 1993-1-5

shear force exceeds this value and MEd Mf;Rd , the condition to be satisfied is: Mf;Rd 1 þ 1 ð23 1Þ2 1:0 ð7:1Þ in EN 1993-1-5 Mpl;Rd where 3 is the ratio VEd =Vbw;Rd and 1 is a usage factor for bending, MEd =Mpl;Rd , based on the plastic moment resistance of the section. Mf;Rd is the design plastic bending resistance based on a section comprising the flanges only. The definition of Mf;Rd is discussed under clause 6.2.2.5(2) below. For Class 4 sections, the calculation of Mf;Rd and Mpl;Rd must consider effective widths for flanges, allowing for plate buckling. Mpl;Rd is however calculated using the gross web, regardless of any reduction that might be required for local buckling under direct stress. If axial force is present, EN 1993-1-5 clause 7.1(4) requires appropriate reduction to be made to Mf;Rd and Mpl;Rd . Discussion of axial force is given before Example 6.4. The interaction for Class 3 and 4 beams is illustrated in Fig. 6.7. The full contribution to the shear resistance from the web, Vbw;Rd , is obtained at a moment of Mf;Rd . For smaller moments, the coexisting shear can increase further due to the flange shear contribution, Vbf;Rd , from clause 5.4 of EN 1993-1-5, provided that the web contribution is less than the plastic resistance. The applied bending moment must additionally not exceed the elastic bending resistance; that is, the accumulated stress must not exceed one of the limits in clause 6.2.1.5(2). This truncates the interaction diagram in Fig. 6.7 at a moment of Mel;Rd . The moment must also not exceed that for lateral–torsional buckling. The value of MEd for use in the interaction with Class 3 and 4 cross-sections is not clearly defined. Clause 6.2.2.4(3) states only that EN 1993-1-5 clause 7.1 is applicable ‘using the calculated stresses of the composite section’. These stresses are dependent on the sequence of construction and can include self-equilibrating stresses such as those from shrinkage which contribute no net moment. There was no problem with interpretation in earlier drafts as 1 , the accumulated stress divided by the appropriate stress limit, was used in the interaction rather than 1 . For compatibility with the use of Mpl;Rd in the interaction expression (based on the crosssection at the time considered) it is recommended here that MEd is taken as the greatest value of (i ÞW, where i is the total accumulated stress at an extreme fibre and W is the elastic modulus of the effective section at the same fibre at the time considered. This bending moment, when applied to the cross-section at the time considered, produces stresses at the extreme fibres which are at least as great as those accumulated. The reason for the use of plastic bending properties in the interaction for Class 3 and Class 4 beams needs some explanation. Test results on symmetric bare steel beams with Class 3 and Class 4 webs68 and also computer simulations on composite bridge beams with unequal flanges69 showed very weak interaction with shear. The former physical tests showed

81


virtually no interaction at all and the latter typically showed some minor interaction only after 80% of the shear resistance had been reached. The use of a plastic resistance moment in the interaction helps to force this observed behaviour as seen in Fig. 6.7. No distinction is made for beams with longitudinally stiffened webs, which can have less post-buckling strength when overall web panel buckling is critical. There are limited test results for such beams and the approach leads to an interaction with shear only at very high percentages of the web shear resistance. A safe option is to replace 1 by 1 in the interaction expression. For composite beams with longitudinally stiffened webs, 1 can be interpreted as the usage factor based on accumulated stress and the stress limits in clause 6.2.1.5(2). Various theories for post-critical behaviour in shear of webs in Class 3 or 4 under combined bending and vertical shear have been compared with 22 test results from composite beams.69 It was found that the method of EN 1993-1-5 gives good predictions for web panels of width/depth ratio exceeding 1.5, and is conservative for shorter panels. Checks of bare steel flanges of box girders are covered in the Designers’ Guide to EN 1993-2.4 For open steel boxes, clause 7.1(5) of EN 1993-1-5 clearly does not apply to the reinforced concrete top flange. For composite flanges, this clause should be applied to the steel part of the composite flange, but the effective area of the steel part may be taken as the gross area (reduced for shear lag if applicable) for all loads applied after the concrete flange has been cast, provided that the shear connectors are spaced in accordance with Table 9.1. Shear buckling need not be considered in the calculation of 3 . Since most continuous box-girder bridges will be in Class 3 or 4 at supports, the restriction to elastic bending resistance forced by clause 7.1(5) of EN 1993-1-5 should not be unduly conservative. The use of elastic analysis also facilitates addition of any distortional warping and transverse distortional bending stresses developed.

Clause 6.2.2.4(4)

Clause 6.2.2.5(1)

Clause 6.2.2.5(2)

82

Bending and vertical shear – all Classes Clause 6.2.2.4(4) confirms that when the depth of web in compression is increased to allow for shear, the resulting change in the plastic neutral axis should be ignored when classifying the web. The reduction of steel strength to represent the effect on bending resistance of shear is only a model to match test results. To add the sophistication of reclassifying the crosssection would be an unjustified complexity. The scatter of data for section classification further makes reclassification unjustified. The issue of reclassification does not arise when using EN 1993-1-5 clause 7.1 as the interaction with shear is given by an interaction expression. The movement of the neutral axis is never determined. Clause 6.2.2.5(1) refers to the contribution of flanges to the resistance of the web to buckling in shear. It permits the contribution of the flange in EN 1993-1-5 clause 5.4(1) to be based on the bare steel flange even if it has the larger plastic moment resistance. It implies that where this is done, the weaker flange is being assisted by the concrete slab in anchoring the tension field. From clause 6.2.2.3(2), the shear connection should then be designed for the relevant vertical force. This additional check can be avoided by neglecting the concrete contribution in calculating Vbf;Rd . The plastic bending resistance of the flanges, Mf;Rd , is defined in clause 6.2.2.5(2) for composite sections as the design plastic resistance of the effective section excluding the steel web. This implies a plastic neutral axis within the stronger flange (usually the composite one). Clause 7.1(3) of EN 1993-1-5 allows Mf;Rd to be taken as the product of the strength of the weaker flange and ‘the distance between the centroids of the flanges’. This gives a slightly lower result for a composite beam than application of the rule in EN 1994-2. The definition in EN 1994-2 is in fact also used in EN 1993-1-5, clauses 5.4(2) and 7.1(1). It is stated in clause 7.1(1) of EN 1993-1-5 that the interaction expression for bending and shear is valid only where 1 Mf;Rd =Mpl;Rd . From the definition of 1 , this condition is MEd Mf;Rd . Where it is not satisfied (as in Example 6.5), the bending moment MEd can be resisted entirely by the flanges of the section. The web is not involved, so there is no interaction between bending and shear unless the shear resistance is to be enhanced by the flange contribution in EN 1993-1-5, clause 5.4. In such cases, the check on interaction


between bending and shear is effectively carried out using that clause as illustrated in Figs 6.6 and 6.7. No such condition is stated for 1 , so it should not be applied when 1 is replaced by 1 when required by clause 7.1(5) of EN 1993-1-5.

Effect of compressive axial force Clause 6.2.2.5(1) makes clear that ‘axial force’ means a force NEd acting on the composite cross-section, or an axial force Na;Ed applied to the steel element before the member becomes composite. It is not the axial force in the steel element that contributes to the bending resistance of a composite beam. For Class 1 or 2 cross-sections, the resistance to bending, shear and axial force should be determined by first reducing the design yield strength of the web in accordance with clause 6.2.2.4(2) and then checking the resulting cross-section under bending and axial force. For Class 3 or 4 cross-sections, clause 7.1(4) of EN 1993-1-5 is also relevant. This effectively requires the plastic bending resistance Mpl;Rd in the interaction expression of EN 1993-1-5 clause 7.1(1) to be reduced to Mpl;N;Rd (using the notation of clause 6.7.3.6) where axial force is present. The resistance Mf;Rd should be reduced by the factor in clause 5.4(2) of EN 1993-1-5, which is as follows: NEd 1 ð5:9Þ in EN 1993-1-5 ðAf1 þ Af2 Þ fyf =M0 This is written for bare steel beams and Af1 and Af2 are the areas of the steel flanges. These are assumed here to be resisting the whole of the force NEd , presumably because in this tensionfield model, the web is fully used already. For composite beams in hogging zones, equation (5.9) above could be replaced by: NEd 1 ðD6:2Þ ðAf1 þ Af2 Þ fyf =M0 þ As fsd where As is the area of the longitudinal reinforcement in the top slab. For sagging bending, the shear force is unlikely to be high enough to reduce the resistance to axial force and bending. On the assumption that the axial force is applied only to the composite section, the value of MEd to use in the interaction expression can be derived from the accumulated stresses as suggested above for checking combined bending and shear, but the uniform stress component from the axial force should not be considered in calculating i . If the axial force, determined as acting at the elastic centroid of the composite section, acts at another level in the model used for the resistance, the moment arising from this change in its line of action should be included in MEd . This is illustrated in Example 6.5. Clause 7.1(4) and (5) of EN 1993-1-5 requires that where axial force is present such that the whole web is in compression, Mf;Rd should be taken as zero in the interaction expression, and 1 should be replaced by 1 (which is defined in EN 1993-1-5 clause 4.6). This leads to the interaction diagram shown in Fig. 6.8. The limit 1 Mf;Rd =Mpl;Rd for validity of expression η3

1.0

0.5

η1

1.0 Fig. 6.8. Shear–moment interaction for Class 3 and 4 cross-sections with webs fully in compression

83


(7.1) given in EN 1993-1-5 clause 7.1(1) is not applicable in this case. The expression is applicable where 1 0. The application of this requirement is unclear for beams built in stages. These could have axial load applied separately to the bare steel section and to the composite section. A safe interpretation, given the relatively small amount of testing on asymmetric sections, would be to take Mf;Rd as zero wherever the whole web is in compression under the built-up stresses. For composite bridges, 1 can be interpreted as the usage factor based on accumulated stress and the stress limits in clause 6.2.1.5(2). However, this is likely to be conservative at high shear, given the weak interaction between bending and shear found in the tests on composite beams discussed above.

Clause 6.2.2.5(3)

Vertical shear in a concrete flange Clause 6.2.2.5(3) gives the resistance to vertical shear in a concrete flange of a composite beam (represented here by a design shear strength, vRd;c ) by reference to clause 6.2.2 of EN 1992-2. That clause is intended mainly to enable higher shear strengths to be used in the presence of in-plane prestress. A Note in EN 1992-2 recommends values for its three nationally determined parameters (NDPs). Where the flange is in tension, as in a continuous composite beam, the reduced strengths obtained can be over-conservative. In EN 1994-2, the Note recommends different NDPs, based on recent research.70 With these values, and for effective slab depths d of at least 200 mm and C ¼ 1:5, the rules are: rffiffiffiffiffiffiffiffi! 200 ðaÞ vRd;c ¼ 0:10 1 þ ð100 fck Þ1=3 þ 0:12cp d and vRd;c

rffiffiffiffiffiffiffiffi!3=2 200 1=2 0:035 1 þ fck þ 0:12cp d

ðbÞ

where: ¼ As =bd 0:02 cp ¼ NEd =Ac < 0:2fcd (compression positive)

ðcÞ

and NEd is the in-plane axial force (negative if tensile) in the slab of breadth b and with tensile reinforcement As , and fck is in N/mm2 units. It can be inferred from Fig. 6.3 in EN 1992-2 that As is the reinforcement in tension under the loading ‘which produces the shear force considered’ (a wording that is used in clause 5.3.3.2 of BS 5400-4). Thus, for shear from a wheel load, only one layer of reinforcement (top or bottom, as appropriate) is relevant, even though both layers may be resisting global tension. It thus appears from equation (a) that the shear strength depends on the tensile force in the slab. This awkward interaction is usually avoided, because EN 1994-2 gives a further research-based recommendation, that where cp is tensile, it should not be taken as greater than 1.85 N/mm2 . The effect of this is now illustrated, with d ¼ 200 mm. Let the reinforcement ratios be 1 ¼ 0.010 for the ‘tensile reinforcement’, 2 ¼ 0.005 for the other layer, with fck ¼ 40 N/mm2 , and cp ¼ 1.85 N/mm2 . From equation (a) above: vRd;c ¼ 0:1 2ð1:0 40Þ1=3 0:12 1:85 ¼ 0:68 0:22 ¼ 0:46 N=mm2 and equation (b) does not govern. From equation (c) with values of cp , NEd and fs all negative: cp ¼ NEd =Ac ¼ ð fs As Þ=Ac ¼ fs ð0:01 þ 0:005Þbd=bh ¼ 0:015fs d=h where the summation is for both layers of reinforcement, because cp is the mean tensile stress in the slab if uncracked and unreinforced. For h ¼ 250 mm, the stress cp then reaches 1.85 N/mm2 when the mean tensile stress in the reinforcement is 154 N/mm2 , which is a low value in practice. At higher values, vRd;c is independent of the tensile force in the slab, though the resulting shear strength is usually lower than that from BS 5400-4. In the transverse direction, NEd is zero unless there is composite action in both directions, so for checking punching shear, two different shear strengths may be relevant.

84


Example 6.4: resistance of a Class 4 section to hogging bending and vertical shear The cross-section in Example 6.3 (Fig. 6.5) is checked for resistance to a vertical shear force of 1100 kN, combined with bending moments Ma;Ed ¼ 150 kNm and Mc;Ed ¼ 2600 kNm From clause 6.2.6(6) of EN 1993-1-1, resistance to shear buckling should be checked if: hw =tw > 72"= where is a factor for which a Note to clause 5.1.2 of EN 1993-1-5 recommends the value 1.2. For S355 steel and a 12.5 mm thick web plate, " ¼ 0:81, so: hw =tw " ¼ 1175 1:2=ð12:5 0:81Þ ¼ 139 > 72 The resistance of this unstiffened web to shear buckling is found using clauses 5.2 and 5.3 of EN 1993-1-5. The transverse stiffeners provided at the cross-bracings are conservatively ignored, as is the contribution from the flanges. For stiffeners at supports only, the slenderness is obtained from EN 1993-1-5 equation (5.5): w ¼

hw 1175 ¼ 1:343 ¼ 86:4t" 86:4 12:5 0:81

Away from an end support, the column ‘rigid end post’ in Table 5.1 of EN 1993-1-5 applies, so: w ¼

1:37 1:37 ¼ 0:67 ¼ 0:7 þ w 0:7 þ 1:343

From EN 1993-1-5 equation (5.2): w fyw hw t 0:67 355 1175 12:5 pffiffiffi Vbw;Rd ¼ pffiffiffi ¼ ¼ 1834 kN 3M1 3 1:1 The shear ratio 3 ¼ 1100=1834 ¼ 0:60. This exceeds 0.5 so interaction with the hogging bending moment must be considered using EN 1993-1-5, clause 7.1. The resistances Mf;Rd and Mpl;Rd are first determined for the composite section. Clause 6.2.2.5(2) requires Mf;Rd to be calculated for the composite section neglecting the web. Mpl;Rd is calculated using the gross web, regardless of the reduction for local buckling under direct stress. From section analysis, using fyd ¼ 345 N/mm2 throughout: Mf;Rd ¼ 5568 kNm Mpl;Rd ¼ 8089 kNm The applied bending moment MEd is taken as the greatest value of (i )W, as explained earlier. Using stresses from Example 6.3 and section moduli from Table 6.1, the values of MEd for the extreme fibres of the steel beam are as follows: top flange: bottom flange:

MEd ¼ 111:8 25:94 ¼ 2900 kNm, which governs MEd ¼ 149:1 18:63 ¼ 2778 kNm

The bending ratio 1 ¼ 2900=8089 ¼ 0:359. This is less than the ratio Mf;Rd =Mpl;Rd , which is 0.69, so there is no interaction between bending and shear. To illustrate the use of interaction expression (7.1) of EN 1993-1-5, let us assume that MEd is increased, such that 1 ¼ 0:75, with 3 ¼ 0:60 as before. Then: Mf;Rd 5568 2 1 þ 1 ð23 1Þ ¼ 0:75 þ 1 ðD6:3Þ ð2 0:60 1Þ2 ¼ 0:762 8089 Mpl;Rd The original action effects are shown as point A in Fig. 6.7. This lies below point B, showing that in this case, the vertical shear does not reduce the resistance to bending.

85


The check of 1 also required by the reference in clause 7.1(1) of EN 1993-1-5 to its clause 4.6 is covered in Example 6.3. The ratio 1 is the greatest usage based on accumulated stress, which is 149:5=345 ¼ 0:43 at the bottom flange. If the moment–shear interaction above is checked conservatively using 1 , equation (D6.3) becomes: MEd 5568 þ 1 ð2 0:60 1Þ2 1; or MEd 0:988Mel;Rd 8089 Mel;Rd thus giving a slight reduction to the bending resistance. This reduction will always occur when 3 > 0:5 is used in this interaction expression as can be seen from Fig. 6.8.

Example 6.5: addition of axial compression to a Class 4 cross-section The effect of adding an axial compression NEd ¼ 4.8 MN to the composite cross-section studied in Examples 6.3 and 6.4 is now calculated, using the method explained earlier. It is assumed that in the global analysis, the beam was located at the level of the long-term elastic neutral axis of the uncracked unreinforced section at mid-span, with nL ¼ 23.7. This is 951 mm (denoted zelu ) above the bottom of the 1500 mm deep section. Where the neutral axis of the cross-section being verified is at some lower level, z, for example, with NEd acting at that level, the coexisting hogging bending moment should be reduced by NEd (zelu z), to allow for the change in the level of NEd . The composite section, shown in Fig. 6.5, is also subjected, as before, to hogging bending moments Ma;Ed ¼ 150 kNm and Mc;Ed ¼ 2600 kNm, and to vertical shear VEd ¼ 1100 kN. The concrete slab is assumed to be fully cracked. It was found in Example 6.4 that for the vertical shear, 3 ¼ 0:60, so the interaction factor (2 3 1Þ2 is only 0.04. The first check is therefore on the elastic resistance of the net section to NEd plus MEd . If NEd is assumed to act at the centroid of the net elastic Class 4 section, allowing for the hole, its line of action, and hence MEd , change at each iteration. It will be found that the whole of the effective web is in compression, so that the interaction with shear should be based on 1 (accumulated stresses), not on 1 (action effects and resistances). This enables stresses from NEd and MEd to be added, and a non-iterative method to be used, based on separate effective cross-sections for axial force and for bending. This method can always be used as a conservative approach.

Stresses from axial compression For the gross cracked cross-section, A ¼ 43 000 mm2 , so: a ¼ 4800=43:0 ¼ 112 N/mm2 compression ¼ 1 and k ¼ 4.0

From Table 4.1 in EN 1993-1-5, From Example 6.3, for the web,

b=t ¼ c=t ¼ 94

Assuming fy ¼ 345 N/mm2 , as in Example 6.1, then " ¼ 0:825. From EN 1993-1-5 clause 4.4(2) and Table 4.1: p ¼ ¼

b=t 94 pffiffiffiffiffiffiffi ¼ 2:01 pffiffiffiffiffi ¼ 28:4 0:825 4:0 28:4" k

p 0:055ð3 þ Þ 2 p

¼

2:01 0:055ð3 þ 1Þ ¼ 0:444 2:012

beff ¼ b ¼ 0:444 1175 ¼ 522 mm;

so

be1 ¼ be2 ¼ 261 mm

The depth of the hole in the web is 1175 522 ¼ 653 mm.

86


The net cross-section for axial compression is shown in Fig. 6.9(a). Its net area is: Aeff;N ¼ 43 000 653 12:5 ¼ 34 840 mm2 The compressive stress from NEd is a;N ¼ 4800=34:84 ¼ 138 N/mm2 . The elastic neutral axis of the net section is 729 mm above the bottom, so the change in neutral axis is z ¼ 951 729 ¼ 222 mm2 , and NEd z ¼ 1066 kNm. 4155 mm2

4155 mm2 109 98

400 × 20 12.5

400 × 20

261

Elastic NA

12.5

493

Elastic NA 396

653 712

729 400 × 30

400 × 30

261

(a)

39 247

(b)

Fig. 6.9. Net cross-sections for (a) axial compression and (b) hogging bending

Stresses from bending moment In Example 6.4, Mc;Ed ¼ 2600 kNm, hogging. The line of action of NEd has been moved downwards, so NEd z is sagging, and now Mc;Ed ¼ 2600 1066 ¼ 1534 kNm, with Ma;Ed ¼ 150 kNm, as before. Using section moduli for the gross cross-section from Example 6.3, the stresses at the edge of the web are: a;top ¼ 68.4 N/mm2 tension, a;bot ¼ 86.5 N/mm2 compression Hence, ¼ 68:4=86:5 ¼ 0:790 Proceeding as in Example 6.3, the results are as follows: k ¼ 18:9, p ¼ 0:923, ¼ 0:941, beff ¼ 618 mm, be1 ¼ 247 mm, be2 ¼ 371 mm The hole in the web is 39 mm deep. The effective cross-section is shown in Fig. 6.9(b). The extreme-fibre stresses from Ma;Ed , Mc;Ed and NEd respectively are: a;top ¼ ð11:6 þ 59:1Þ þ 138 ¼ 67 N=mm2 a;bot ¼ þð9:5 þ 82:4Þ þ 138 ¼ 230 N=mm2 It follows that there is some compression in the deck slab, which has been neglected, for simplicity. The reinforcement is assumed here to carry all the compressive force in the slab.

Interaction with vertical shear The whole of the web is in compression, so from clause 7.1(5) of EN 1993-1-5, Mf;Rd ¼ 0 and 1 , not 1 , is used. For the steel bottom flange, which governs, 1 ¼ 230=345 ¼ 0:67 From equation (7.1) of EN 1993-1-5, with 3 = 0.60 and Mf;Rd ¼ 0, 3 1Þ2 ¼ 0:67 þ 0:04 ¼ 0:71 1 þ ð2 This is less than 1.0, so the cross-section is verified.

87


Method when the web is partly in tension This example is now repeated with the axial compression reduced to NEd ¼ 2.5 MN and all other data as before, to illustrate the method where 1 is used and Mf;Rd is not zero. From clause 7.1(4) of EN 1993-1-5, Mpl;Rd is reduced to allow for NEd , as follows. For the gross cross-section Mpl;Rd ¼ 8089 kNm, and the plastic neutral axis is 876 mm above the bottom. Force NEd is assumed to act at its level. The depth of web needed to resist it is hw;N ¼ 2500=ð12:5 0:345Þ ¼ 580 mm. This depth is centred on the neutral axis as shown in Fig. 6.10 and remains wholly within the web. Its contribution to Mpl;Rd was: ð fyd th2w;N Þ=4 ¼ 345 12:5 0:582 =4 ¼ 363 kNm As its depth is centred on the plastic neutral axis for bending alone, M pl;N;Rd ¼ 8089 363 ¼ 7726 kNm 1.805 1.805

250 45

1175

Tension

2.76

400 × 20 290


0.17

39

NEd 290

Forces in MN units

2.50

12.5 556 400 × 30

2.40 Compression 4.14

Fig. 6.10. Plastic resistance of Class 4 section to hogging bending and axial compression

From Example 6.4, Mf;Rd ¼ 5568 kNm. From equation (D6.2) and using forces from Fig. 6.10, the reduction factor that allows for NEd is: NEd 2:50 ¼ 0:762 1 ¼1 2:76 þ 4:14 þ 2 1:805 ðAf1 þ Af2 Þ fyf =M0 þ As fsd Hence, Mf;N;Rd ¼ 0:762 5568 ¼ 4243 kNm The bending moment MEd is found from accumulated stresses, using the Class 4 crosssection for bending. Its elastic properties should strictly be determined with a value for MEd such that the line of action of NEd is at the neutral axis of the effective crosssection. This requires iteration. Finally, the line of action of NEd should be moved to the neutral axis of the plastic section used for the calculation of Mpl;Rd . This requires a further correction to MEd . The simpler and sufficiently accurate method used here is to find the Class 4 section properties ignoring any moment from the axial force, to use these to scale up Ma;Ed , and then to add the correction from NEd . The change in the line of action of NEd is: 951 876 ¼ 75 mm, so Mc;Ed ¼ 2600 2500 0:075 ¼ 2413 kNm Calculation similar to that in the section ‘Stresses from bending moment’, above, finds the depth of the hole in the web to be 322 mm. The top flange is found to govern, and MEd ¼ Ma;Ed ðWc;top =Wa;top Þ þ Mc;Ed ¼ 150 26:6=13:0 þ 2413 ¼ 2720 kNm This bending moment is less than Mf;N;Rd , so it can be resisted entirely by the flanges, and there is no need to consider interaction with shear in accordance with clause 7.1(1) of EN 1993-1-5.

88


Stability of the span in the vertical plane Buckling of the whole span in the vertical plane is possible. Its elastic critical axial force is now estimated, treating it as pin-ended, and assuming sagging bending. The modular ratio nL ¼ 18.8 is used, intermediate between the values for short- and long-term loading. The presence of a stiffer cross-section near the supports is ignored. The result is Ncr ¼ 47 MN. The ratio cr ¼ Ncr =NEd ¼ 8.1 (<10), so from clause 5.2.1(3), second-order effects should be included in MEd . There appears to be a sufficient margin for these, but this has not been checked.

6.3. Filler beam decks 6.3.1. Scope The encasement of steel bridge beams in concrete provides several advantages for design: .

. . .

It enables a Class 3 web to be upgraded to Class 2, and the slenderness limit for a Class 2 compression flange to be increased by 40% (clause 5.5.3). It prevents lateral–torsional buckling. It prevents shear buckling (clause 6.3.4(1)). It greatly increases the resistance of the bridge deck to vehicular impact or terrorist attack.

These design advantages may not however lead to the most economic solution. The use of longitudinal filler beams in new construction is not common at present. There are a great number of geometric, material and workmanship-related restrictions given in clauses 6.3.1(1) to (4) which have to be met in order to use the application rules for the design of filler beams. These are necessary because the rules derive mainly from existing practice in the UK and from clause 8 of BS 5400:Part 5.11 No explicit check of the shear connection between steel beams and concrete (provided by friction and bond only) is required. Clause 6.3.1(1) excludes fully-encased filler beams from the scope of clause 6.3. This is because there are no widely-accepted design rules for longitudinal shear in fully-encased beams without shear connectors. Clause 6.3.1(2) requires the beams to be of uniform cross-section and to have a web depth and flange width within the ranges found for rolled H- or I-sections. This is due to the lack of existing examples of filler beams with cross-sections other than these. There is no requirement for the beams to be H- or I-sections, but hollow sections would be outside the scope of clause 6.3. Clause 6.3.1(3) permits spans to be either simply supported or continuous with square or skew supports. This clarification is based on existing practice, and takes account of the many other restrictions. Clause 6.3.1(4) contains the majority of the restrictions which relate mainly to ensuring the adequacy of the bond between steel beam and concrete, as follows. .

.

.

.

.

Clause 6.3.1(1)

Clause 6.3.1(2)

Clause 6.3.1(3)

Clause 6.3.1(4)

Steel beams should not be curved in plan. This is because the torsion produced would lead to additional bond stresses between the structural steel and concrete, for which no application rules are available. The deck skew should not exceed 308. This limits the magnitude of torsional moments, which can become large with high skew. The nominal depth, h, of the beam should lie between 210 mm and 1100 mm. This is because anything less than 210 mm should be treated as reinforced concrete, and there could in future be rolled sections deeper than 1100 mm. A maximum spacing of the steel beams is set: the lesser of h=3 þ 600 mm and 750 mm. This reflects existing practice and limits the longitudinal shear flow (and bond stresses) between the concrete and the steel beam. The minimum concrete cover to the top of the steel beams is restricted to 70 mm. A larger value may however be necessary to provide adequate cover to the reinforcement. The

89


.

.

.

.

.

maximum cover is limited to the lesser of 150 mm and h/3, based on existing practice and to limit the longitudinal shear stress developed. A further restriction is given such that the plastic neutral axis for sagging bending remains below the level of the bottom of the top flange, since cracking of the concrete in the vicinity of the top flange could reduce the bond stress developed. This rule could only govern where the steel beams were unusually small. The side cover to the top flange should be at least 80 mm. The clear distance between top flanges should not be less than 150 mm so that the concrete can be adequately compacted. This is essential to ensure that the required bond to the steel is obtained. Bottom transverse reinforcement should be provided (through holes in the beam webs) such that transverse moments developed can be carried. A minimum bar size and maximum spacing are specified. Minimum reinforcement, here and elsewhere, should also satisfy the requirements of EN 1992. Normal-density concrete should be used. This is because there is little experience of filler-beam construction with concrete other than normal-density, where the bond characteristics could be affected. The flange should be de-scaled. This again is to ensure good bond between the concrete and the steel beam. For road and railway bridges the holes in steel webs should be drilled. This is discussed under clause 6.3.2(2).

6.3.2. General Clause 6.3.2(1)

Clause 6.3.2(2)

Clause 6.3.2(3) Clause 6.3.2(4)

Clause 6.3.2(1) refers to other clauses for the cross-section checks, which should be conducted at ultimate and serviceability limit states. These references do not require a check of torsion as discussed below. Clause 6.3.2(2) requires beams with bolted connections or welding to be checked against fatigue. The implication is that filler beams without these need not be checked for fatigue, even though they will contain stress-raising holes through which the transverse reinforcement passes. For road and railway bridges, where fatigue loading is significant, clause 6.3.1(4) requires that all holes in webs are drilled (rather than punched), which improves the fatigue category of the detail. Clause 6.3.2(3) is a reminder to refer to the relaxations for cross-section Class in clause 5.5.3. Mechanical shear connection need not be provided for filler beams (clause 6.3.2(4)). This reliance on bond improves the relative economy of filler-beam construction but leads to many of the restrictions noted above under clause 6.3.1.

6.3.3. Bending moments Clause 6.3.3(1)

Clause 6.3.3(2)

90

The resistance of cross-sections to bending, clause 6.3.3(1), is determined in the same way as for uncased sections of the same Class, with Class determined in accordance with clause 5.5.3. The relaxations in clause 5.5.3 should generally ensure that beams can be designed plastically and thus imposed deformations generally need not be considered at ultimate limit states (the comments made under clause 5.4.2.2(6) refer). Lateral–torsional buckling is not mentioned in clause 6.3.3(1) because a filler-beam deck is inherently stable against lateral–torsional buckling in its completed state due to its large transverse stiffness. The steel beams are likely to be susceptible during construction and the title of clause 6.3.5 provides a warning. For the influence of vertical shear on resistance to bending, reference is made to the rules for uncased beams. The shear resistance of filler-beam decks is high, so interaction is unlikely, but it should be checked for continuous spans. In the transverse direction, a filler-beam deck behaves as a reinforced concrete slab. Clause 6.3.3(2) therefore makes reference to EN 1992-2 for the bending resistance in the transverse


direction. A Note to clause 9.1(103) of EN 1992-2 makes minimum reinforcement a nationally determined parameter. No requirement is given for a check on torsion, which will be produced to some degree in both longitudinal and transverse directions of the global analysis models allowed by clause 5.4.2.9(3). Neglect of torsion is justified by the limits imposed on geometry in clause 6.3.1(4), particularly the limit on skew angle, and by current UK practice.

6.3.4. Vertical shear The simplest calculation of shear resistance involves basing the resistance on that of the steel beam alone. Clause 6.3.4(1) indicates that this resistance can be calculated using the plastic shear resistance and so ignoring shear buckling. The clause does permit a contribution from the concrete to be taken. Clauses 6.3.4(2) and 6.3.4(3), respectively, cover a method of determining the shear force that may be carried on the reinforced concrete section and the determination of the resistance of this concrete section. Clause 6.3.4(3) applies also to shear resistance in the transverse direction.


6.3.5. Resistance and stability of steel beams during execution Clause 6.3.5(1) refers to EN 1993-1-1 and EN 1993-2 for the check of the bare steel beams. This covers both cross-section resistance and lateral–torsional buckling. The latter is an important consideration prior to hardening of the concrete.

Clause 6.3.5(1)

6.4. Lateral–torsional buckling of composite beams 6.4.1. General It is assumed in this section that in completed bridges, the steel top flanges of all composite beams will be stabilized laterally by connection to a concrete or composite slab (clause 6.4.1(1)). The rules on maximum spacing of connectors in clause 6.6.5.5(1) and (2) relate to the classification of the top flange, and thus only to local buckling. For lateral– torsional buckling, the relevant rule, given in clause 6.6.5.5(3), is less restrictive. Any steel top flange in compression that is not so stabilized should be checked for lateral buckling (clause 6.4.1(2)) using clause 6.3.2 of EN 1993-1-1 to determine the reduction factor for buckling. For completed bridges, this applies to the bottom flange adjacent to intermediate supports in continuous construction. In a composite beam, the concrete slab provides lateral restraint to the steel member, and also restrains its rotation about a longitudinal axis. Lateral buckling is always associated with distortion (change of shape) of the cross-section (Fig. 6.11(b)). This is not true ‘lateral–torsional’ buckling and is often referred to as ‘distortional lateral’ buckling. This form of buckling is covered by clauses 6.4.2 and 6.4.3. The general method of clause 6.4.2, based on the use of a computed value of the elastic critical moment Mcr , is applicable, but no detailed guidance on the calculation of Mcr is given in either EN 1993-1-1 or EN 1994-2. For completed bridges, the bottom flange may be in compression over most of a span when that span is relatively short and lightly loaded and adjacent spans are fully loaded. Bottom flanges in compression should always be restrained laterally at supports. It should not be assumed that a point of contraflexure is equivalent to a lateral restraint. Design methods for composite beams must take account of the bending of the web, Fig. 6.11(b). They differ in detail from the method of clause 6.3.2 of EN 1993-1-1, but the same imperfection factors and buckling curves are used, in the absence of any betterestablished alternatives. The reference in clause 6.4.1(3) to EN 1993-1-1 provides a general method for use where the method in clause 6.4.2 is inapplicable (e.g. for a Class 4 beam). Clause 6.4.3 makes a similar reference but adds a reference to a further method available in clause 6.3.4.2 of EN 1993-2. During unpropped construction, prior to the presence of a hardened deck slab, the buckling verification can be more complicated and often involves overall buckling

Clause 6.4.1(1)

Clause 6.4.1(2)

Clause 6.4.1(3)

91


h

F

F

(a)

(b)

θ0

Fig. 6.11. (a) U-frame action and (b) distortional lateral buckling

of a braced pair of beams. This situation is discussed further at the end of section 6.4.3.2 of this guide.

6.4.2. Beams in bridges with uniform cross-sections in Class 1, 2 and 3


This general method of design is written with distortional buckling of bottom flanges in mind. It would not apply, for example, to a mid-span cross-section of a beam with the slab at bottom-flange level (Fig. 6.12). The reference to ‘uniform cross-section’ in the title of the clause is not intended to exclude minor changes such as reinforcement details and effects of cracking of concrete. The method cannot be used for Class 4 cross-sections, which is a significant limitation for larger bridges, in which case the methods of clause 6.4.3 should be used. The latter methods are more general. The method is based closely on clause 6.3.2 of EN 1993-1-1. There is correspondence in the definitions of the reduction factor LT , clause 6.4.2(1), and the relative slenderness, LT , clause 6.4.2(4). The reduction factor is applied to the design resistance moment MRd , which is defined in clauses 6.4.2(2) and (3). Expressions for MRd are given by references to clause 6.2. It should be noted that these include the design yield strength fyd which should, in this case, be calculated using M1 rather than M0 because this is a check of instability. If the beam is found not to be susceptible to lateral–torsional buckling (i.e. LT ¼ 1.0), it would be reasonable to replace M1 with M0 . The determination of MRd for a Class 3 section differs from that of Mel;Rd in clause 6.2.1.4(6) only in that the limiting stress fcd for concrete in compression need not be considered. It is necessary to take account of the method of construction. The buckling resistance moment Mb;Rd given by equation (6.6) must exceed the highest applied moment MEd within the unbraced length of compression flange considered.

Fig. 6.12. Example of a composite beam with the slab in tension at mid-span

92


Lateral buckling for a Class 3 cross-section with unpropped construction The influence of method of construction on verification of a Class 3 composite section for lateral buckling is as follows. From equation (6.4), MRd ¼ Mel;Rd ¼ Ma;Ed þ kMc;Ed

(a)

where subscript c is used for the action effect on the composite member. From equation (6.6), the verification is: MEd ¼ Ma;Ed þ Mc;Ed LT Mel;Rd

(b)

which is: LT ðMa;Ed þ Mc;Ed Þ=Mel;Rd ¼ MEd =Mel;Rd

(c)

The total hogging bending moment MEd may be almost independent of the method of construction. However, the stress limit that determines Mel;Rd may be different for propped and unpropped construction. If it is bottom-flange compression in both cases, then Mel;Rd is lower for unpropped construction, and the limit on LT from equation (c) is more severe.

Elastic critical buckling moment Clause 6.4.2(4) requires the determination of the elastic critical buckling moment, taking account of the relevant restraints, so their stiffnesses have to be calculated. The lateral restraint from the slab can usually be assumed to be rigid. Where the structure is such that a pair of steel beams and a concrete flange attached to them can be modelled as an inverted-U frame (clause 6.4.2(5) and Fig. 6.10), continuous along the span, the rotational restraining stiffness at top-flange level, ks , can be found from clause 6.4.2(6). In the definition of stiffness ks , flexibility arises from two sources: . .

Clause 6.4.2(4)

Clause 6.4.2(5) Clause 6.4.2(6)

bending of the slab, which may not be negligible: 1/k1 from equation (6.9) bending of the steel web, which predominates: 1/k2 from equation (6.10).

A third source of flexibility is potentially the shear connection but it has been found71 that this can be neglected providing the requirements of clause 6.4.2(5) are met. There is a similar ‘discrete U-frame’ concept, which appears to be relevant to composite beams where the steel sections have vertical web stiffeners. The shear connectors closest to those stiffeners would then have to transmit almost the whole of the bending moment Fh (Fig. 6.11(a)), where F is now a force on a discrete U-frame. The flexibility of the shear connection may then not be negligible, nor is it certain that the shear connection and the adjacent slab would be sufficiently strong.72 Where stiffeners are present, the resistance of the connection above each stiffener to repeated transverse bending should be established, as there is a risk of local shear failure within the slab. There is at present no simple method of verification. This is the reason for the condition that the web should be unstiffened in clause 6.4.2(5)(b). The restriction need not apply if bracings (flexible or rigid) are attached to the stiffeners, but in this case the model referred to in clause 6.4.3.2 would be used. Clause 6.4.2(7) allows the St Venant torsional stiffness to be included in the calculation. This is often neglected in lateral–torsional buckling models based on buckling of the bottom chord, such as that provided in EN 1993-2 clause 6.3.4.2. No formula is provided for the elastic critical buckling moment for the U-frame model described above. Mcr could be determined from a finite-element model of the beam with a lateral and torsional restraint as set out above. Alternatively, textbook solutions could be used. One such method was given in Annex B of ENV 1994-1-120 and is now in the Designers’ Guide to EN 1994-1-1.5

Clause 6.4.2(7)

6.4.3. General methods for buckling of members and frames 6.4.3.1. General method Reference is made to EN 1993-2 clause 6.3.4 where the method of clause 6.4.2 for beams or the non-linear method of clause 6.7 for columns does not apply.

93


EN 1993-1-1 clause 6.3.4 gives a general method of evaluating the combined effect of axial load and mono-axial bending applied in the plane of the structure, without use of an interaction expression. The method is valid for asymmetric and non-uniform members and also for entire plane frames. In principle, this method is more realistic since the structure or member does, in reality, buckle in a single mode with a single ‘system slenderness’. Interaction formulae assume separate modes under each individual action with different slendernesses that have to subsequently be combined to give an overall verification. The disadvantage is that software capable of both elastic critical buckling analysis and secondorder analysis is required. Additionally, shell elements will be needed to determine elastic critical modes resulting from flexural loading. An alternative method is to use second-order analysis with imperfections to cover both inplane and out-of-plane buckling effects as discussed in sections 5.2 and 5.3 of this guide, but this has the same difficulties as above. The basic verification is performed by determining a single slenderness for out-of-plane buckling, which can include combined lateral and lateral–torsional buckling. This slenderness is a slenderness for the whole system and applies to all members included within it. It takes the usual Eurocode form as follows: rffiffiffiffiffiffiffiffiffiffiffi ult;k op ¼ ð6:64Þ in EN 1993-1-1 cr;op where:

Clause 6.4.3.1(1)

ult;k is the minimum load factor applied to the design loads required to reach the characteristic resistance of the most critical cross-section ignoring out-of-plane buckling but including moments from second-order effects and imperfections inplane, and cr;op is the minimum load factor applied to the design loads required to give elastic critical buckling in an out-of-plane mode, ignoring in-plane buckling.

The first stage of calculation requires an analysis to be performed to determine ult;k . Inplane second-order effects and imperfections must be included in the analysis because they are not otherwise included in the resistance formula used in this method. If the structure is not prone to significant second-order effects as discussed in section 5.2 of this guide, then first-order analysis may be used. The flexural stiffness to be used is important in determining second-order effects and this is recognized by the text of clause 6.4.3.1(1). It will be conservative to use the cracked stiffness Ea I2 throughout if the bridge is modelled with beam elements. If a finite-element shell model is used, the reinforcement can be modelled and the concrete neglected so as to avoid an overestimation of stiffness in cracked zones. Out-ofplane second-order effects may need to be suppressed. Each cross-section is verified using the interaction expression in clause 6.2 of EN 1993-1-1, but using characteristic resistances. Effective cross-sections should be used for Class 4 sections. The loads are all increased by a factor ult;k until the characteristic resistance is reached. The simple and conservative verification given in clause 6.2.1(7) of EN 1993-1-1 becomes: NEd My;Ed þ 1:0 NRk My;Rk

ðD6:4Þ

where NRk and My;Rk include allowance for any reduction necessary due to shear and torsion, if separate checks of cross-section resistance are to be avoided in addition to the buckling check being considered here. NEd and My;Ed are the axial forces and moments at a cross-section resulting from the design loads. If first-order analysis is allowable, the load factor is determined from: NEd My;Ed ult;k þ ¼ 1:0 ðD6:5Þ NRk My;Rk which is given in a Note to clause 6.3.4(4) of EN 1993-1-1. If second-order analysis is necessary, ult;k is found by increasing the imposed loads progressively until one cross-section reaches failure according to expression (D6.4). This is

94


necessary as the system is no longer linear and results from one analysis cannot simply be factored up when the imposed load is increased. The second stage is to determine the lowest load factor cr;op to reach elastic critical buckling in an out-of-plane mode but ignoring in-plane buckling modes. This will typically require a finite-element model with shell elements to predict adequately the lateral–torsional buckling behaviour. The reinforcement can be modelled and the concrete neglected so as to avoid an overestimation of stiffness in cracked zones. If the load factor can only be determined separately for axial loads cr;N and bending moments cr;M , as might be the case if standard textbook solutions are used, the overall load factor could be determined from a simple interaction equation such as: 1 cr;op

¼

1 cr;N

þ

1 cr;M

Next, an overall slenderness is calculated for the entire system according to equation (6.64) of EN 1993-1-1. This slenderness refers only to out-of-plane effects as discussed above because in-plane effects are separately included in the determination of action effects. A reduction factor op for this slenderness is then determined. This reduction factor depends in principle on whether the mode of buckling is predominantly flexural or lateral–torsional as the reduction curves can sometimes differ. The simplest solution is to take the lower of the reduction factors for out-of-plane flexural buckling, , and lateral–torsional buckling, LT , from clauses 6.3.1 and 6.3.2, respectively, of EN 1993-2. For bridges, the recommended reduction factors are the same but the National Annex could alter this. This reduction factor is then applied to the cross-section check performed in stage 1, but this time using design values of the material properties. If the cross-section is verified using the simple interaction expression (D6.4), then the verification taking lateral and lateral–torsional buckling into account becomes: My;Ed NEd þ op NRk =M1 My;Rk =M1 It follows from equation (D6.5) and expression (D6.6) that the verification is: op ult;k 1:0 M1

ðD6:6Þ

ðD6:7Þ

Alternatively, separate reduction factors for axial load and LT for bending moment can be determined for each effect separately using the same slenderness. If the cross-section is verified using the simple interaction expression (D6.4), then the verification taking lateral and lateral–torsional buckling into account becomes: My;Ed NEd þ 1:0 NRk =M1 LT My;Rk =M1

ðD6:8Þ

It should be noted that this procedure can be conservative where the element governing the cross-section check is not itself significantly affected by the out-of-plane deformations. The method is illustrated in a qualitative example for a steel-only member in the Designers’ Guide to EN 1993-2.4

6.4.3.2. Simplified method A simplified method is permitted for compression flanges of composite beams and chords of composite trusses by reference to EN 1993-2 clause 6.3.4.2. Its clause D2.4 provides the stiffness of U-frames in trusses (and plate girders by analogy). The method is based on representing lateral–torsional buckling by lateral buckling of the compression flange. All subsequent discussion refers to beam flanges but is equally applicable to chords of trusses. The method is primarily intended for U-frame-type bridges but can be used for other types of flexible bracing. It also applies to lengths between rigid restraints of a beam compression flange, as is found in hogging zones in steel and concrete composite construction. The use of the method for half-through bridges is discussed in the Designers’ Guide to EN 1993-2.

95


Scope: distortional lateral buckling of a steel flange of a span of an unhaunched composite beam, with varying major-axis bending moment, rigid lateral supports at beam vertical supports, intermediate lateral supports of stiffness Cd (either rigid or flexible) at spacing ℓ and/or continuous lateral restraint of stiffness cc per unit length

No

Is the flange susceptible to lateral buckling, to 6.3.2.4(1) of EC3-1-1, via 6.4.3.2(1) of EC4-2 and 6.3.4.2(1) of EC3-2? Yes or

Elastic critical analysis as a restrained strut to find Ncrit and λLT to EC3-2/6.3.4.2(4)

either

Note 1: all subsequent clause numbers refer to EN 1993-2

Yes

Is there a continuous restraint?

Classify discrete restraints. Is Cd > 4NE/L, to EC3-2/6.3.4.2(6), where L = ℓ? No

END

No

Neglect torsional stiffness of the flange. Simplified analysis to 6.4.3.2(1)

Restraint is flexible, of stiffness Cd. Find cd = Cd/ℓ. L is span length of beam

L, the distance between ‘rigid’ restraints = ℓ

Yes c = cd

Find cc. Then c = cc. Find γ to EC3-2/6.3.4.2(6) c=γ=0 Ncrit = π2EI/L2

Is there a continuous restraint? Yes

Yes

Find Ncrit to eq. (D6.13)

No

Find cc. Then c = cd + cc Find γ to EC3-2/6.3.4.2(6)

Find n to eq. (D6.11). Is n < 1? No

Find Ncrit to eq. (D6.12) – conservative unless n is an integral number. Eqs (6.14) in EC3-2/6.3.4.2(8) are less conservative

This applies for uniform compressive force over length L, to EC3-2/6.3.4.2(6). Is it required to take account of variation of bending moment over this length? No

Yes

Does the bending moment change sign within length L? No

Yes

Replace the bending moment that causes tension in the flange by M2 = 0

Find m from eqs (6.14) in the Note to EC3-2/6.3.4.2(8)

Find Ncrit = mπ2EI/L2 but ≤ π2EI/ℓ2

Find λLT from eq. (6.10) in EC3-2/6.3.4.2(4) (END)

Fig. 6.13. Flow diagram for slenderness for lateral buckling of a compressed flange

The method effectively ignores the torsional stiffness of the beam. This may become significant for rolled steel sections but is generally not significant for deeper fabricated girders. A flow diagram for determining the slenderness LT for a length of beam of uniform depth between rigid lateral supports is given in Fig. 6.13. EN 1993-2 clause 6.3.4.2 allows the slenderness for lateral buckling to be determined from an eigenvalue analysis of the compression chord. The flange (with an attached portion of web

96


Awc = twhwc hwc Af

Fig. 6.14. Definitions for effective compression zone for a Class 3 cross-section

in the compression zone) is modelled as a strut with area Aeff , supported by springs in the lateral direction. These represent restraint from bracings (including discrete U-frames) and from any continuous U-frame action which might be provided by the connection to the deck slab. Buckling in the vertical direction is assumed to be prevented by the web in this model but checks on flange-induced buckling according to Section 8 of EN 1993-1-5 should be made to confirm this assumption. Bracings can be flexible, as is the case of bracing by discrete U-frames, or can be rigid, as is likely to be the case for cross-bracing. Other types of bracing, such as horizontal members at mid-height between beams together with plan bracing or a deck slab, may be rigid or flexible depending on their stiffness as discussed below. Elastic critical buckling analysis may be performed to calculate the critical buckling load, Ncrit . The slenderness is then given by EN 1993-2 equation (6.10): sffiffiffiffiffiffiffiffiffiffiffiffi Aeff fy LT ¼ Ncrit where Aeff ¼ Af þ Awc /3, as shown in Fig. 6.14. This approximate definition of Aeff (greater than the flange area) is necessary to ensure that the critical stress produced for the strut is the same as that required to produce buckling in the beam under bending moment. For Class 4 cross-sections, Aeff is determined making allowance for the reduction in area due to plate buckling. If smeared springs are used to model the stiffness of discrete restraints such as discrete Uframes, the buckling load should not be taken as larger than that corresponding to the Euler load of a strut between discrete bracings. If computer analysis is used, there would be no particular reason to use smeared springs for discrete restraints. This approximation is generally only made when a mathematical approach is used based on the beam-on-elasticfoundation analogy, which was used to derive the equations in EN 1993-2.

Spring stiffnesses for discrete U-frames and other restraints Spring stiffnesses for discrete U-frames may be calculated using Table D.3 from Annex D of EN 1993-2, where values of stiffness, Cd , can be calculated. (It is noted that the notation C rather than Cd is used in Table D.3.) A typical case covering a pair of plate girders with stiffeners and cross-girders is shown in Fig. 6.15 for which the stiffness (under the unit applied forces shown) is: EIv Cd ¼ 3 ðD6:9Þ 2 hv h bq I v þ 3 2Iq Section properties for stiffeners should be derived using an attached width of web plate in accordance with Fig. 9.1 of EN 1993-1-5 (stiffener width plus 30"tw ). If the cross-member is composite, its second moment of area should be based on cracked section properties. Equation (D6.9) also covers steel and concrete composite bridges without stiffeners and cross-girders where the cross-member stiffness is the short-term cracked stiffness of the deck slab and reinforcement, and the vertical-member stiffness is based on the unstiffened web. For continuous U-frames, consideration of this stiffness will have little effect in

97


Neutral axis of cross-girder

hv

h

Iv

Iq

bq

Fig. 6.15. Definitions of properties needed to calculate Cd

raising the buckling resistance, unless the length between rigid restraints is large, and will necessitate an additional check of the web for the U-frame moments induced. For multiple girders, the restraint to internal girders may be derived by replacing 2Iq by 3Iq in the expression for Cd . Equation (D6.9) is then similar to equation (6.8). That differs only by the inclusion of Poisson’s ratio in the stiffness of the web plate and by the assumption that the point of rotation of the compression flange is at the underside of the deck slab, rather than some way within it. The stiffness of other restraints, such as a channel section placed between members at midheight, can be derived from a plane frame model of the bracing system. For braced pairs of beams or multiple beams with a common system, it will generally be necessary to consider unit forces applied to the compression flanges such that the displacement of the flange is maximized. For a paired U-frame, the maximum displacement occurs with forces in opposite directions as in Fig. 6.15 but this will not always be the case. For paired beams braced by a mid-height channel, forces in the same direction will probably give greater flange displacement. A computer model is useful where, for example, the flange section changes or there is a reversal of axial stress in the length of the flange being considered. In other simpler cases the formulae provided in clause 6.3.4.2 of EN 1993-2 are applicable.

Elastic critical buckling load The formula for Ncrit is derived from eigenvalue analysis with continuous springs. From elastic theory (as set out, for example, in Refs 73 and 74), the critical load for buckling of such a strut is: Ncrit ¼ n2 where:

2 EI cL2 þ 2 2 n L2

ðD6:10Þ

I is the transverse second moment of area of the effective flange and web, L is the length between ‘rigid’ braces, c is the stiffness of the restraints smeared per unit length, and n is the number of half waves in the buckled shape.

By differentiation, this is a minimum when: cL4 4 EI which gives: pffiffiffiffiffiffiffiffi Ncrit ¼ 2 cEI n4 ¼

ðD6:11Þ

ðD6:12Þ

Equation (6.12) of EN 1993-2 is: Ncrit ¼ mNE where: NE ¼

98

2 EI ; L2

pffiffiffi cL4 m ¼ 2=2 1:0; ¼ EI

and

c ¼ Cd =l


where Cd is equal to the restraint stiffness and l is the distance between restraints. When these terms are substituted into equation (6.12) of EN 1993-2, equation (D6.12) is produced. When ¼ 4 ¼ 97, then c ¼ 4 EI=L4 , and this model gives the results: Ncrit ¼ 22 EI=L2 , n ¼ 1.0 It is not valid for lower values of c because then n<1, which implies a buckling halfwavelength that exceeds the length L between rigid restraints, and a value of Ncrit lower than that corresponding to a length L. In this case, the buckling load should be taken as: Ncrit ¼

2 EI cL2 þ 2 L2

ðD6:13Þ

Equation (D6.10) assumes that the end restraints that define the length L are ‘rigid’. The definition of ‘rigid’ is discussed below. If intermediate bracings are not rigid, their stiffness can be taken to contribute to ‘c’ but the length L is then defined by the length between rigid bracings. Bracings at supports for typical composite bridges will usually be rigid due to the need for them to provide torsional restraint to the beams.

Short lengths of beam between rigid bracings Equation (D6.12) shows that the critical buckling load from equation (6.12) of EN 1993-2 is independent of the length between rigid restraints. Equation (D6.13) is the basis for the first of equations (6.14) in EN 1993-2 for short lengths between rigid braces. The half wave length of buckling is restricted to the length between braces, and any flexible restraints included in this length increase the buckling load from that for a pin-ended strut of length L. The formulae provided also allow the effects of varying end moments and shears to be taken into account, but they are not valid (and are actually unsafe) where the bending moment reverses within length L. They are as follows: m1 ¼ 1 þ 0:44ð1 þ Þ1:5 þ ð3 þ 2Þ=ð350 50 Þ or m2 ¼ 1 þ 0:44ð1 þ Þ1:5 þ ð0:195 þ ð0:05 þ =100ÞÞ 0:5 if less than above with: ¼ V2 =V1 and ¼ 2ð1 M2 =M1 Þ=ð1 þ Þ for M2 < M1 and V2 < V1 The subscripts in the symbols m1 and m2 correspond to the number of buckling half-waves considered, n. Figure 6.16 enables the equation that gives the lower result to be found, by 1.0

500 450

M2/M1

400

0.8 350

γ

0.6

300 0.4 280 260 0.2 240 0.0 –0.4

–0.2

0.0

0.2

0.4

0.6

0.8

1.0

V2/V1

Fig. 6.16. Values of (other than zero) at which both equations (6.14) in EN 1993-2 give the same value for m. Below each curve, m1 governs

99


V2/V1 = 1 M1 M2

V2/V1 < 1

Fig. 6.17. Effect of shear ratio on the shape of the moment diagram

giving values of at which m1 ¼ m2 . If the actual value of for a buckling length with ratios V2 =V1 and M2 =M1 is lower than that shown in the figure, the equation for m1 governs; if not, m2 governs.

Uniformly compressed flange The beneficial influence of lateral restraint, represented by , is evident for the most adverse case, a uniformly-compressed flange, for which ¼ 1, ¼ 0. Then, m1 ¼ 1:00 þ =100; m2 ¼ 1:00 þ 0:195 0:5 These ratios m are equal when 0:5 ¼ 19.5, or ¼ 380, as shown by the point (1.0, 1.0) in Fig. 6.16. The change from n ¼ 2 to n ¼ 3 can be found from equation (D6.10) which, in terms of , is: Ncrit =NE ¼ n2 þ =ðn2 4 Þ This gives Ncrit for n ¼ 3 equal to that for n ¼ 2 when ¼ 3500 and Ncrit =NE ¼ 13. The equations for m1 and m2 are more complex than equation (D6.10) because their scope includes non-uniform moment. Within the range of from 380 to 3000, the value m2 for uniform moment can be up to 10% higher than from equation (D6.10). This ‘error’ is small and is in part compensated for by the neglect of the torsional stiffness of the beam in this method. At ¼ 3500 it gives Ncrit =NE ¼ 12.5, which is more conservative. It then follows closely the results from equation (D6.10) as increases (and n also increases from 3 to 4). For up to 20 000, the values of m2 differ from the predictions of (D6.10) by only 3.6%. The shear ratio, , in the equations for m1 and m2 , helps to describe the shape of the bending moment diagram between points of restraint. It is linear if ¼ 1.0. If < 1:0, the moments fall quicker than assumed from a linear distribution as shown in Fig. 6.17 and consequently the flange is less susceptible to buckling.

Change of sign of axial force within a length between rigid restraints The lack of validity for moment reversal of equations (6.14) in EN 1993-2 is a problem for a typical composite beam with cross-bracing adjacent to the internal supports. Where the most distant brace from the pier is still in a hogging zone, the moment in the beam will reverse in the span section between braces as shown in Fig. 6.18. In this region, m should not be assumed to be 1.0 as this could lead to over-design of the beam or unnecessary provision

(6.14) not valid

(6.14) valid

= bracing location

Fig. 6.18. Range of validity of equations (6.14) of EN 1993-2

100

(6.14) not valid


Conservative set of moments with V2/V1 = 0, M2/M1 = 0

M1

M2 = –M1 Real moments

Fig. 6.19. Typical calculation of m where bending moment reverses

of additional braces away from the pier, to ensure that the section between innermost braces is entirely sagging and the bottom flange is in tension. A Note to clause 6.3.4.2(7) of EN 19932 provides the option of assuming M2 ¼ 0. If benefit from the restraining stiffness of the deck slab is ignored (i.e. c ¼ 0), and V2 is conservatively taken equal to V1 then this leads to m ¼ 1.88. Where the top flange is braced continuously by a deck, it may be possible to ‘vary’ to produce a less conservative moment diagram. For the case in Fig. 6.19, the use of V2 =V1 ¼ 0, M2 =M1 ¼ 0 achieves the same moment gradient at end 1 as the real set of moments, and a distribution that lies everywhere else above the real moments and so is still conservative. Equations (6.14) of EN 1993-2 then give the value m ¼ 2.24, again ignoring any U-frame restraint. Providing the top flange is continuously braced, the correct m would be greater. It is possible to include continuous U-frame action from an unstiffened web between rigid braces in the calculation of the spring stiffness c. The benefit is usually small for short lengths between braces, and the web plate, slab and shear connection must be checked for the forces implied by such action. Fig. 6.20 shows a graph of m against M2 =M1 with c ¼ 0, for varying V2 =V1 . It is possible to combine equations (6.10) and (6.12) of EN 1993-2 to produce a single formula for slenderness, taking Af ¼ btf for the flange area, as follows: sffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi uð1 þ A =3A Þð f =EmÞ Aeff fy ðAf þ Awc =3Þ fy L2 u wc f y LT ¼ ¼ L ¼ u t Ncrit m2 EI 2 b3 t f 12 btf

µ = V2/V1 = 1.0

0.75

0.50

0.25

0.00

2.2 2.0 1.8 m

1.6 1.4 1.2 1.0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

M2/M1

M1

M2

Fig. 6.20. Values of m ( ¼ Ncrit =NE ) between rigid restraints with ¼ 0

101


so LT

L ¼ 1:103 b

rffiffiffiffiffiffiffisffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi fy A 1 þ wc Em 3Af

ðD6:14Þ

It is still necessary to evaluate Ncrit when checking the strength of the bracings. The formulae in clause 6.3.4.2(7) of EN 1993-2 do not apply directly to haunched girders as they assume that the flange force is distributed in the same way as the bending moment. The general method of using an eigenvalue analysis based on the forces in the compression chord is still applicable. Alternatively, the formulae provided could be applied using the least value of the spring stiffness c within the length considered. The flange force ratio F2 =F1 is used instead of the moment ratio M2 =M1 , with V2 =V1 taken as 1.0, when applying equation (6.14) of EN 1993-2. EN 1993-2 clause p 6.3.4.2(7) allows the buckling verification to be performed at a distance ffiffiffiffi of 0:25Lk ¼ 0:25L= m (i.e. 25% of the effective length) from the end with the larger moment. (The symbols Lk and lk are both used for effective length in 6.3.4.2.) This appears to double-count the benefit from moment shape derived in equations (6.14) of EN 1993-2; but it does not do so. The check at 0:25Lk reflects the fact that the peak stress from transverse buckling of the flange occurs some distance away from the rigid restraint to the flange, whereas the peak stress from overall bending of the beam occurs at the restraint. (In this model, the beam flange is assumed to be pin-ended at the rigid transverse restraints.) Since these two peak stresses do not coexist they are not fully additive, and the buckling verification can be performed at a ‘design’ section somewhere between these two locations. The cross-section resistance must still be verified at the point of maximum moment. There are clearly problems with pffiffiffiffi applying clause 6.3.4.2(7) of EN 1993-2 where the moment reverses as the section 0:25L= m from an end may be a point of contraflexure. In this situation, it is recommended here that the design section be taken as 25% of the distance from the position of maximum moment to the position of zero moment. In addition, if benefit is taken of verification at the 0:25Lk cross-section, the calculated slenderness above must be modified so that it refers to this design section. The critical moment value will be less here and the slenderness is therefore increased. This can be done by defining a new slenderness at the 0:25Lk section such that sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi M1 0:25Lk ¼ LT M0:25Lk where M0:25Lk is the moment at the 0:25Lk section. This procedure is illustrated in Example 6.6 below. It should be noted that the k in M0:25Lk does not imply a characteristic value; this is a design value.

Stiffness of braces The formulae in EN 1993-2 discussed above are only valid where pffiffiffiffiffiffiffiffithe end restraints that define the length L are ‘rigid’. It is possible to equate Ncrit ¼ 2 cEI to 2 EI=L2 to find a limiting stiffness that gives an effective length equal to the distance between rigid restraints, L, but this slightly underestimates the required stiffness. This is because the formulae assume that the restraints are continuously smeared when they are in fact discrete. The former analysis gives a required value for Cd of 4 EI=4L3 , whereas the ‘correct’ stiffness is 42 EI=L3 , which is 62% higher. Interaction with compressive axial load The interaction with axial load is covered in clause 6.4, in clause 6.4.3.1 only, via the general method given in EN 1993-2. Axial load has several effects including: .

102

magnification of the main bending moment about the horizontal axis of the beam (the second-order effect)


.

increase of stress in the compression flange leading to an increased tendency for lateral buckling.

Most bridge cross-sections are either Class 3 or 4 at supports so the stresses from axial load can simply be assumed to be applied to the cracked composite section, and the elastic section resistance can be used. At mid-span, beams are usually Class 1 or 2 and the calculation of a modified plastic moment resistance in the presence of axial load is relatively simple. The plastic neutral axis is so chosen that the total compressive force exceeds the total tensile force by an amount equal to the axial load. Care must however be taken to ensure that the bending resistance is obtained about an axis at the height of the applied axial force assumed in the global analysis. This is important for non-symmetric beams as the elastic and plastic neutral axes for bending alone do not coincide, whereas they do for a symmetric section. Most of clause 6.7 is for doubly-symmetric sections only, but the general method of clause 6.7.2 may be applied to beams provided that compressive stresses do not exceed their relevant limiting values where Class 3 and 4 crosssections are involved. Alternatively, the cross-section can be designed using a conservative interaction expression such as that in clause 6.2.1(7) of EN 1993-1-1: NEd My;Ed þ 1:0 NRd My;Rd where NRd and My;Rd are the design resistances for axial force and moment acting individually but with reductions for shear where the shear force is sufficiently large. A similar interaction expression can be used for the buckling verification with the terms in the denominator replaced by the relevant buckling resistances: NEd MEd þ 1:0 Nb;Rd Mb;Rd

ðD6:15Þ

The value for MEd should include additional moments from in-plane second-order effects (including from in-plane imperfections). Such second-order effects will normally be negligible. The buckling resistance Nb;Rd should be calculated on the basis of the axial stress required for lateral buckling of the compression flange. This method is illustrated in Example 6.6 below.

Beams without plan bracing or decking during construction During construction it is common to stabilize girders in pairs by connecting them with ‘torsional’ bracing. Such bracing reduces or prevents torsion of individual beams but does not restrict lateral deflection. Vertical ‘torsional’ cross-bracing as shown in Fig. 6.21 has been considered in the UK for many years to act as a rigid support to the compression flange, thus restricting the effective length to the distance between braces. Opinion is now somewhat divided on whether such bracing can be considered fully effective and BS

Torsional bracing

Centre of rotation (a)

(b)

Fig. 6.21. Torsional bracing and shape of buckling mode, for paired beams: (a) plan on braced pair of beams showing buckling mode shape; (b) cross-section through braced pair of beams showing buckling mode shape

103


5400:Part 3:200011 introduced a clause to cover this situation which predicts that such bracing is not fully effective. This situation arises because equilibrium of the braced pair under torsion requires opposing vertical forces to be generated in the two girders. Consequently one girder moves up, one moves down and some twist of the girder pair is generated, albeit much less than for an unbraced pair. If the beam span-to-depth ratio is large, the deflections and hence twists can be significant. The Designers’ Guide to EN 1993-24 suggests a method based on BS 5400 Part 3, but in some cases it may lead to the conclusion that plan bracing is necessary. A better estimate of slenderness can be made using a finite-element analysis. A finite-element model of a non-composite beam, using shell elements for the paired main beams and beam elements to represent the bracings, can be set up relatively quickly with modern commercially available software. Elastic critical buckling analysis can then be performed and a value of Mcr determined directly for use in slenderness calculation to clause 6.3.2 of EN 1993-2. This approach usually demonstrates that the cross-bracing is not fully effective in limiting the effective length of the flange to the distance between bracings, but that it is more effective than is predicted by BS 5400. For simplysupported paired girders, a typical lowest buckling mode under dead load is shown in Fig. 6.21.

Example 6.6: bending and shear in a continuous composite beam The locations of the bracings and splices for the bridge used in earlier examples are shown in Fig. 6.22. The two internal beams are Class 3 at each internal support as found in Example 5.4. The cross-sections of the central span of these beams, also shown, are as in Examples 5.4 and 6.1 to 6.4. The neutral axes shown in the cross-section at the piers are for hogging bending of the cracked composite section. B

F

C

A

D

19 000

E

23 400

3800

G

3800

19 000

Location of bracing Location of splice, about 6.0 m from pier

3100

12 990 mm2 6490 mm2

3100 60

250 25

250 400 × 25

209 Plastic NA

1225

1160 × 25 Elastic NA

70

25 haunch NA 400 × 20 1175 × 12.5

1225 998

775 400 × 40

Cross-section for lengths BD and EG

Fig. 6.22. Details for Examples 6.6 and 6.7

104

400 × 30 Cross-section for length DE


The design ultimate hogging moments at internal support B are 2213 kNm on the steel section plus 4814 kNm on the composite section, so MEd ¼ 7027 kNm. The coexisting hogging moment at the braced point C in the central span is 4222 kNm. The vertical shear at point C is 70% of that at the pier B. The hogging bending moment at the splice, where the beam cross-section changes, is 3000 kNm. Lateral–torsional buckling adjacent to the pier and in the main span beyond the brace is checked and the effect of a coexisting axial compression of 1000 kN applied to the composite section is considered. This could arise in a semi-integral bridge with screen walls at its ends.

Elastic resistance to bending at an internal support The elastic section moduli for the cross-section at point B are given in Table 6.2. These are based on the extreme fibres but it would have been permissible to base them on the centroids of the flanges in accordance with EN 1993-1-1 clause 6.2.1(9). To find Mel;Rd , the factor k (clause 6.2.1.4(6)) is found for the top and bottom surfaces of the steel beam, as follows. The result will be used in checks on buckling, so fyd is found using M1 (¼ 1.1). Primary shrinkage stresses are neglected because the deck slab is assumed to be cracked. Table 6.2. Section moduli at an internal support, in 106 mm3 units

Gross steel section Cracked composite section

Top layer of bars

Top of steel section

Bottom of steel section

— 34.05

18.28 50.31

22.20 29.25

For the top flange, 2213=18:28 þ kMc;Ed =50:31 ¼ 345=1:1 so kMc;Ed ¼ 9688 kNm For the bottom flange, 2213=22:20 þ kMc;Ed =29:25 ¼ 345=1:1 so kMc;Ed ¼ 6258 kNm The elastic resistance is governed by the bottom flange, so that: Mel;Rd ¼ 2213 þ 6258 ¼ 8471 kNm The maximum compressive stress in the bottom flange is: a;bot ¼ 2213=22:2 þ 4814=29:25 ¼ 264 N/mm2

Resistance of length BC (Fig. 6.22) to distortional lateral buckling Strictly, the stiffness of the bracing should first be checked (or should later be designed) so that the buckling length is confined to the length between braces. This is done in Example 6.7. The bending-moment distribution is shown in Fig. 6.23. Where no vertical web stiffeners are required, the deck slab provides a small continuous U-frame stiffness. This could be included using Table D.3 of EN 1993-2, case 1a, to calculate a stiffness, c. This contribution has been ignored to avoid the complexities of designing the deck slab and shear connection for the forces implied, and because any need to stiffen the web has not yet been considered. Therefore from clause 6.3.4.2(6) of EN 1993-2, ¼ cL4 =EI ¼ 0. The compression zone of the beam is designed as a pin-ended strut with continuous vertical restraint, and lateral restraint at points 3.80 m apart (Fig. 6.22). Its cross-sectional areas are: .

flange: Af ¼ 400 40 ¼ 16 000 mm2

105


.

web compression zone: Awc ¼ 735 25 ¼ 18 375 mm2 . (Conservatively based on the height of the neutral axis for the composite section. It could be based on the actual depth for the accumulated stress profile, which is 633 mm.)

The second moment of area of the compressed area is calculated for the bottom flange, as the contribution from the web is negligible: I ¼ 40 4003 =12 ¼ 213:3 106 mm4 The applied bending moments at each end of the equivalent strut are: MEd;1 ¼ 7027 kNm; MEd;2 ¼ 4222 kNm From EN 1993-2 clause 6.3.4.2(7): M2 =M1 ¼ 4222=7027 ¼ 0:60 and ¼ V2 =V1 ¼ 0:70 ¼ 2ð1 M2 =M1 Þ=ð1 þ Þ ¼ 2ð1 0:60Þ=ð1 þ 0:70Þ ¼ 0:46 When ¼ 0, the first two of equations (6.14) in clause 6.3.4.2(8) of EN 1993-2 both give: m ¼ 1 þ 0:44ð1 þ Þ1:5 ¼ 1 þ 0:44ð1 þ 0:70Þ0:461:5 ¼ 1:23 If the deck slab is considered to provide U-frame restraint, the value of m for this length of flange is found to be still only 1.26, so there is no real benefit to lateral stability in considering U-frame action over such a short length of beam. From equation (D6.14): rffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi fy L Awc 3800 345 18375 1þ LT ¼ 1:1 ¼ 0:45 1þ ¼ 1:1 3Af b Em 400 210 103 1:23 3 16000 This exceeds 0.2 so from clause 6.3.2.2(4) of EN 1993-1-1 this length of flange is prone to lateral–torsional buckling. The ratio h=b ¼ 1225=400 ¼ 3:1 exceeds 2.0, so from Table 6.4 in clause 6.3.2.2 of EN 1993-1-1 the relevant buckling curve is curve d. Hence, LT ¼ 0.76 from Table 6.3. From equation (6.56) in EN 1993-1-1, clause 6.3.2.2: 2 LT ¼ 0:5½1 þ LT LT 0:2 þ LT ¼ 0:5 1 þ 0:76ð0:45 0:2Þ þ 0:452 ¼ 0:696 LT ¼

1 1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:81 2 2 2 LT þ 2LT LT 0:696 þ 0:696 0:45

Applying this reduction factor gives: Mb;Rd ¼ LT Mel;Rd ¼ 0:81 8471 ¼ 6862 kNm At the internal support, MEd ¼ 7027 kNm (2% higher). However, clause 6.3.4.2(7) of pffiffiffiffi EN 1993-2 provides the option of making this check at a distance of 0:25L= m from the support. This distance is: pffiffiffiffiffiffiffiffiffi 0:25 3800= 1:23 ¼ 857 mm Using linear interpolation, Fig. 6.23, this gives MEd ¼ 6394 kNm. The modified slenderness is: sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffi M1 7027 ¼ 0:45 0:25Lk ¼ LT ¼ 0:47 6394 M0:25Lk This reduces LT from 0.81 to 0.80, so the new resistance is: Mb;Rd ¼ LT Mel;Rd ¼ 0:80 8471 ¼ 6777 kNm This exceeds MEd (6394 kNm), so this check on lateral buckling is satisfied.

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The further condition that the elastic resistance (8471 kNm) is not exceeded at the point of peak moment (7027 kNm) is satisfied. This cross-section should also be checked for combined bending and shear, but in this case the shear is less than 50% of the shear resistance and thus no interaction occurs.

Resistance with a coexisting axial compression of 1000 kN An axial force of 1000 kN is applied at the height of the elastic centroidal axis of the cracked cross-section at the internal support. Susceptibility to in-plane second-order effects is first checked. The second moment of area of the cracked composite section at mid-span, 1:326 1010 mm4 , is conservatively used to determine the elastic critical flexural buckling load for major-axis buckling. Ncr ¼

2 EI 2 210 1:326 1010 ¼ ¼ 28 600 kN 31 0002 L2

From clause 5.2.1(3): cr ¼

28 600 ¼ 29 > 10 1000

so in-plane second-order effects may be neglected. At the internal support, the area of the cracked composite section is 74 480 mm2 , so the axial compressive stress is 1000 103 =74 480 ¼ 13:4 N/mm2 . This changes the total stresses at the extreme fibres of the steel beam to 204 N/mm2 tension and 278 N/mm2 compression. This stress distribution is found to leave the cross-section at the pier still in Class 3. The section at the brace is also in Class 3. The conservative linear interaction between axial and bending resistances of expression (D6.15) will be used: NEd MEd þ 1:00 Nb;Rd Mb;Rd so Nb;Rd must be found, based on buckling of the web and bottom flange under uniform compression. Since the combined stress distribution leads to a Class 3 cross-section throughout the buckling length, the gross section will be used in this analysis, even for the calculation of the compression resistance, below. If the section became Class 4, effective properties should be used. These could be derived either separately for moment and axial force or as a unique effective section under the combined stress field. In either case, the moment of the axial force produced by the shift of the neutral axis should be considered. For axial force alone, the required areas of the cross-sections are: Af ¼ 400 40 ¼ 16 000 mm2 , Awc ¼ 1160 25 ¼ 29 000 mm2 Neglecting U-frame restraint as before, equations (6.14) in clause 6.3.4.2(7) of EN 19932 with M1 ¼ M2 and ¼ 0 give m ¼ 1.00. From equation (D6.14): rffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi fy L Awc 3800 345 29 000 1þ ¼ 1:1 1þ LT ¼ 1:1 ¼ 0:54 3 b Em 400 210 10 1:00 3 16 000 3Af From curve d on Fig. 6.4 of EN 1993-1-1, ¼ 0.74. The axial buckling resistance of the cracked composite cross-section, based on buckling of the bottom flange, is: Nb;Rd ¼ Afyd ¼ 0:74 74 480 345=1:1 ¼ 17 285 kN It should be noted that Nb;Rd does not represent a real resistance to axial load alone as the cross-section would then be in Class 4 and a reduction to the web area to allow for local buckling in accordance with EN 1993-1-5 would be required. It is however valid

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to use the gross cross-section in the interaction here as the cross-section is Class 3 under the actual combination of axial force and bending moment. The use of a gross crosssection also avoids the need to consider any additional moment produced by the shift in centroidal axis that occurs when an effective web area is used. From expression (D6.15), 1000=17 285 þ 6394=6777 ¼ 1:00 which is just satisfactory, with these conservative assumptions. A check of cross-section resistance is also required at the end of the member, but this is satisfied by the check of combined stresses above.

Buckling resistance of the mid-span region of the central span An approximate check is carried out for the 23.4 m length between the two braced points (C and F in Fig. 6.22), at first using the cross-section at the pier throughout to derive the reduction factor. This is slightly unconservative as the cross-section reduces at the splice. An axial force is not considered in this part of the example. 7027

Bending moment (kNm)

6394

4222 M2 = 0,V2 = V1

3000

M2 = 0,V2 = 0

MEd 0 Section: B

0.86

3.80 C

Distance from support (m) D

27.2 F

Fig. 6.23. Bending action effects and resistances for an internal span

It is assumed that maximum imposed load acts on the two side spans and that only a short length near mid-span is in sagging bending, as sketched in Fig. 6.23. Since the bending moment reverses, equations (6.14) in EN 1993-2 are not directly applicable. If the suggestion of clause 6.3.4.2(7) of EN 1993-2 is followed, and M2 is taken as zero at the other brace (cross-section F), the bending-moment distribution depends on the value assumed for V2 , the vertical shear at F. Two possibilities are shown in Fig. 6.23. Their use does not follow directly from the discussion associated with Fig. 6.18, where the moment was assumed to reverse only once in the length between rigid restraints. In Fig. 6.23, the two fictitious sets of moments do not always lie above the real set and are therefore not obviously conservative. However, the interaction of the hogging moment at one end of the beam with the buckling behaviour at the other end is weak when the moment reverses twice in this way. BS 5400:Part 311 included a parameter ‘’ which was used to consider the effect of moment shape on buckling resistance. For no reversal, m is in principle equivalent to 1=2 , although there is not complete numerical equivalence. Figure 6.24 gives comparative values of m0 ¼ 1=2 . This shows that for the worst real moment distribution, where the moment just remains entirely hogging, the value of m0 is greater (less conservative) than for the two possibilities in Fig. 6.23. This shows that the less conservative of these possibilities (V2 ¼ 0) can be used. It gives m ¼ 2.24 from equations (6.14).

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From equation (D6.14) with Awc ¼ 25 735 ¼ 18 375 mm2 : rffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi fy L A 1 þ wc LT ¼ 1:1 b Em 3Af rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 23 400 345 18 375 1þ ¼ 1:1 ¼ 2:05 > 0:2 400 3 16 000 210 103 2:24 Using curve d in Fig. 6.4 of EN 1993-1-1, LT ¼ 0.17. If the slab reinforcement at cross-section C is the same as at the support, Mel;Rd ¼ 8471 kNm, and Mb;Rd ¼ 0:17 8471 ¼ 1440 kNm This is far too low, so according to this method another brace would be required further from the pier to reduce the buckling length. U-frame action is now considered. mʹ = 1.73

mʹ = 2.5

mʹ = 4.0

Fig. 6.24. Variation of m0 with different bending-moment distributions, derived from BS 5400:Part 311

Use of continuous U-frame action – simplified method of EN 1993-2, clause 6.3.4.2 The method of clause 6.4.2 is inapplicable because the cross-section is not uniform along length CF, and a region near the splice is in Class 4 for hogging bending. The method of clause 6.3.4.2 of EN 1993-2 is therefore used. Its spring stiffness c is the lateral force per unit length at bottom-flange level, for unit displacement of the flange. It is related to the corresponding stiffness ks in clause 6.4.2 by: ks ¼ ch2s

ðD6:16Þ

where hs is the distance between the centres of the steel flanges (Fig. 6.10). For most of the length CF the web is 12.5 mm thick so the cross-section for length DE (Fig. 6.22) is used for this check, with hs ¼ 1200 mm. For the bottom flange, Iafz ¼ 30 4003 =12 ¼ 160 106 mm4

ðD6:17Þ

In comparison with this thin web, the slab component of the U-frame is very stiff, so its flexibility is neglected here, as is the torsional stiffness of the bottom flange. In general, the flexibility of the slab should be included. From equations (6.8) and (6.10) with k1 k2 , so that ks k2 , and using equation (D6.16), 3 t c ¼ ks =h2s ¼ Ea w =½4ð1 a2 Þ ¼ 210 106 ð12:5=1200Þ3 =ð4 0:91Þ ¼ 65:2 kN=m2 hs As an alternative, equation (D6.9) could be used for the calculation of c. As discussed in the main text, there are some minor differences in the definition of the height terms, ‘h’. Those in equation (D6.9) seem more appropriate where the slab flexibility becomes important. From EN 1993-2 clause 6.3.4.2 and result (D6.17), ¼ cL4 =EI ¼ 65:2 23:44 =ð210 160Þ ¼ 582 The less conservative assumption, V2 ¼ 0, is used with M2 ¼ 0. This gives ¼ 0, ¼ 2. From Fig. 6.16, the second of equations (6.14) in clause 6.3.4.2 of EN 1993-2 governs.

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With ¼ 0 it gives: m ¼ 1 þ 0:441:5 þ ð0:195 þ 0:05Þ 0:5 ¼ 1 þ 1:245 þ 7:117 ¼ 9:36 U-frame action (the final term) is now a significant contributor to m. (For uniform moment and the same , m ¼ 5.70.) The cross-section reduces at the splice position, approximately 6 m from the pier, so the minimum cross-section is conservatively considered throughout. In equation (D6.14) for LT , the areas in compression are: flange: Af ¼ 400 30 ¼ 12 000 mm2 . web compression zone: Awc ¼ ð683 47Þ 12:5 ¼ 7950 mm2 . (Conservatively based on the height of the neutral axis for the composite section shown in Fig. 6.5, which is 683 mm above the bottom flange.) Hence from equation (D6.14): rffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi fy L A 1 þ wc LT ¼ 1:1 b Em 3Af rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 23 400 345 7950 ¼ 1:1 1þ ¼ 0:942 400 3 12 000 210 103 9:36 Curve d of Fig. 6.4 in EN 1993-1-1 gives LT ¼ 0.49. With Mel;Rd from Example 6.3, the buckling resistance LT Mel;Rd is:

.

Mb;Rd ¼ 0:49 6390 ¼ 3131 kNm which is less than 4222 kNm (the moment at the brace). By inspection, a check at the 0.25Lk design section will also not pass. These checks are however conservative as they assume the minimum cross-section throughout. If reference is made back to the expressions for slenderness given in equations (6.10) of EN 1993-2 and (6.7), it is seen that: sffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffi Aeff fy MRk LT ¼ ¼ Ncrit Mcrit so that Mcrit measured at the brace is effectively Mcrit ¼

MRk 2 LT

¼

6390 1:1 ¼ 7921 kNm 0:9422

For the length of the beam with the larger cross-section, MRk is larger than assumed above. The buckling moment Mcrit is however mainly influenced by the long length of smaller cross-section so that it will be similar to that found above, even if the short lengths of stiffer end section are considered in the calculation.

Use of continuous U-frame action – general method of EN 1993-1-1, clause 6.3.4 A revised slenderness can be determined using the method of EN 1993-1-1 clause 6.3.4. From equation (D6.5) with NEd ¼ 0, within the span between braces, the minimum value of My;Rk ult;k ¼ My;Ed is 8471 1:1 ¼ 2:21 4222 at the brace location, where: My;Rd ¼ 8471 kNm (For the weaker section at the splice location, My;Rk =My;Ed ¼ 6390 1:1=3000 ¼ 2:34Þ:

110


The minimum load factor to cause lateral–torsional buckling is: cr;op ¼

Mcrit 7921 ¼ 1:88 ¼ My;Ed 4222

The system slenderness of the span between the braces, from equation (6.64) of EN 19931-1, is: rffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffi ult;k 2:21 op ¼ ¼ ¼ 1:08 1:88 cr;op and hence the reduction factor, from curve d of Fig. 6.4 in EN 1993-1-1, is op ¼ 0.43. The verification is then performed according to equation (D6.7): op ult;k 0:43 2:21 ¼ 0:86 < 1:0 ¼ 1:1 M1 so the beam is still inadequate. This verification is equivalent to Mb;Rd ¼ 0:43 8471 ¼ 3643 kNm at the brace, which is still less than the applied moment of 4222 kNm. It would be possible to improve this verification further by determining a more accurate value of Mcrit from a finite-element model. However, inclusion of U-frame action has the disadvantage that the web and shear connection would have to be designed for the resulting effects. A better alternative could be the addition of another brace adjacent to the splice location.

Example 6.7: stiffness and required resistance of cross-bracing The bracing of the continuous bridge beam in Example 6.6 comprises cross-bracing made from 150 150 18 angle and attached to 100 20 stiffeners on a 25 mm thick web. The bracings are checked for rigidity and the force in them arising from bracing the flanges is determined. The effects of the 1000 kN axial compressive force are included. Deck slab

Bracing Stiffener effective section

1020

1 kN

3100

1 kN

Fig. 6.25. Cross-bracing for Examples 6.6 and 6.7

The stiffness of the brace was first calculated from the plane-frame model shown in Fig. 6.25. From Fig. 9.1 of EN 1993-1-5, the effective section of each stiffener includes a width of web: 30"tw þ tst ¼ 30 0:81 25 þ 20 ¼ 628 mm Hence, Ast ¼ 628 25 þ 100 20 ¼ 17 700 mm2 This leads to Ist ¼ 9:41 106 mm4 . The deck slab spans 3.1 m. From clause 5.4.1.2 its effective span is 3:1 0:7 ¼ 2:17 m, and its effective width for stiffness is 0:25 2:17 ¼ 0:542 m. Its stiffness is conservatively

111


based on the cracked section, with the concrete modulus taken as Ecm /2 to represent the fact that some of the loading is short term and some is long term. Greater accuracy is not warranted here as the concrete stiffness has little influence on the overall stiffness of the cracked section. From elastic analysis for the forces shown in Fig. 6.25, the stiffness is: Cd ¼ 80 kN/mm From clause 6.3.4.2(6) of EN 1993-2, the condition for a lateral support to a compressed member to be ‘rigid’ is: Cd 42 EI=L3b where, for length BC in Fig. 6.23, Lb ¼ 3.80 m and I ¼ 213:3 106 mm4 Hence, Cd 42 210 213:3=ð3:83 1000Þ ¼ 32:2 kN/mm The support is ‘rigid’.

Cd

Fig. 6.26. Elastic stiffness of bracing to a pin-ended strut

The formula in EN 1993-2 here assumes that supports are present every 3.8 m such that the buckling length is restricted to 3.8 m each side of this brace and that the force in the flange is constant each side of the brace. The limiting spring stiffness for Cd is analogous to that required for equilibrium of a strut with a pin joint in it at the spring position, such that buckling of the lengths each side of the pin joint occurs before buckling of the whole strut into the brace – see Fig. 6.26. A small displacement of the strut at the pin joint produces a kink in the strut and a lateral force on the brace, which must be stiff enough to resist this force at the given displacement. For a given displacement, the kink angle and thus the force on the spring is increased by reducing the length of the strut each side of the spring. This kink force is also increased because the critical buckling load for the lengths each side of the spring is increased by the reduction in length. Here, the flange force is not the same each side of the brace and the length of unbraced flange is greater than 3.8 m on one side of the brace. The kink force is therefore overestimated and the calculated value for Cd is conservative. This confirms the use in Example 6.6 of L as the length between braces. The design lateral force for the bracing is now found, using clause 6.3.4.2(5) of EN 19932. From Example 6.6, LT ¼ 0:45, and the effective area of the compressed flange is: Aeff ¼ 16 000 þ 18 375=3 ¼ 22 120 mm2 From equations in EN 1993-2, clause 6.3.4.2, 210 213:3 1000 1=2 ‘k ¼ ðEI=Ncrit Þ1=2 ¼ ½EI 2LT =ðAeff fy Þ1=2 ¼ 0:45 ¼ 3:43 m 22 120 345

112


The distance between braced points is ‘ ¼ 3:8 m, so ‘k < 1:2 ‘. (Since the brace has been found to be ‘rigid’, and from Example 6.6, m > 1 so that Ncrit > 2 EI=‘2 , ‘k is obviously less than ‘, and this check was unnecessary.) From clause 6.3.4.2(5), the lateral force applied by each bottom flange to the brace is: FEd ¼ NEd /100 From Example 6.6, the greatest compressive stress in the bottom flange at the pier is 278 N/mm2 . Hence: FEd ¼

NEd ¼ 278 22 120=ð100 1000Þ ¼ 61:5 kN 100

The axial force in the bracing is then approximately: 61:5 ¼ 64:7 kN cosðtan 1020=3100Þ 1

There will also be some bending moment in the bracing members due to joint eccentricities.

6.5. Transverse forces on webs The local resistance of an unstiffened and unencased web to forces (typically, vertical forces) applied through a steel flange can be assumed to be the same in a composite member as in a steel member, so clause 6.5 consists mainly of references to EN 1993-1-5. High transverse loads are relatively uncommon in bridge design other than during launching operations or from special vehicles or heavy construction loads, such as from a crane outrigger. Theoretically, wheel loads should be checked but are unlikely ever to be significant. The patch loading rules given in EN 1993-1-5 Section 6 make allowance for failure by either plastic failure of the web, with associated plastic bending deformation of the flange, or by buckling of the web. More detail on the derivation and use of the rules is given in the Designers’ Guide to EN 1993-2.4 The rules for patch loading can only be used if the geometric conditions in EN 1993-1-5 clause 2.3 are met; otherwise EN 1993-1-5 Section 10 should be used. Clause 6.1(1) of EN 1993-1-5 also requires that the compression flange is ‘adequately restrained’ laterally. It is not clear what this means in practice, but the restraint requirement should be satisfied where the flange is continuously braced by, for example, a deck slab or where there are sufficient restraints to prevent lateral–torsional buckling. Clause 6.5.1(1) states that the rules in EN 1993-1-5 Section 6 are applicable to the noncomposite flange of a composite beam. If load is applied to the composite flange, the rules could still be used by ignoring the contribution of the reinforced concrete to the plastic bending resistance of the flange. No testing is available to validate inclusion of any contribution. A spread of load could be taken through the concrete flange to increase the stiff loaded length on the steel flange. There is limited guidance in EN 1992 on what angle of spread to assume; clause 8.10.3 of EN 1992-1-1 recommends a dispersion angle of tan1 2=3, i.e. 348, for concentrated prestressing forces. It would be reasonable to use 458 here, which would be consistent with previous bridge design practice in the UK. Clause 6.5.1(2) makes reference to EN 1993-1-5 clause 7.2 for the interaction of transverse force with axial force and bending. This gives:

Clause 6.5

Clause 6.5.1(1)

Clause 6.5.1(2)

2 þ 0:81 1:4 where: 2 ¼

z;Ed FEd F ¼ ¼ Ed fyw =M1 fyw Leff tw =M1 FRd

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is the usage factor for transverse load acting alone, and 1 ¼

Clause 6.5.2

x;Ed NEd M þ NEd eN ¼ þ Ed fy =M0 fy Aeff =M0 fy Weff =M0

is the usage factor for direct stress alone, calculated elastically. The calculation of 1 should take account of the construction sequence as discussed in section 6.2.1.5 of this guide. It can be seen that this interaction expression does not allow for a plastic distribution of stress for bending and axial force. Even if the cross-section is Class 1 or 2, this will not lead to any discontinuity with the plastic bending resistance at low transverse load as only 80% of the elastic bending stress is considered and the limiting value of the interaction is 1.4. The ratio between the plastic and elastic resistances to bending for typical composite beams is less than 1.4. Clause 6.5.2 covers flange-induced buckling of webs by reference to EN 1993-1-5 Section 8. If the flange is sufficiently large and the web is very slender, it is possible for the whole flange to buckle in the plane of the web by inducing buckling in the web itself. If the compression flange is continuously curved in elevation, whether because of the soffit profile or because the whole beam is cambered, the continuous change in direction of the flange force causes a radial force in the plane of the web. This force increases the likelihood of flange-induced buckling into the web. Discussion on the use of Section 8 of EN 1993-1-5 is provided in the Designers’ Guide to EN 1993-2.4

6.6. Shear connection 6.6.1. General 6.6.1.1. Basis of design Clause 6.6.1.1(1)

Clause 6.6 is applicable to shear connection in composite beams. Clause 6.6.1.1(1) refers also to ‘other types of composite member’. Shear connection in composite columns is addressed in clause 6.7.4, but reference is made to clause 6.6.3.1 for the design resistance of headed stud connectors. Clause 6.6.1.1(2)P Although the uncertain effects of bond are excluded by clause 6.6.1.1(2)P, friction is not excluded. Its essential difference from bond is that there must be compressive force across the relevant surfaces. This usually arises from wedging action. Provisions for shear connection by friction are given in clause 6.7.4.2(4) for columns. Clause 6.6.1.1(3)P ‘Inelastic redistribution of shear’ (clause 6.6.1.1(3)P) is most relevant to building design in the provisions of EN 1994-1-1 for partial shear connection. Inelastic redistribution of shear is allowed in a number of places for bridges including: .

.

.

clause 6.6.1.2 (which allows redistribution over lengths such that the design resistance is not exceeded by more than 10%) clause 6.6.2.2 (which permits assumptions about the distribution of the longitudinal shear force within an inelastic length of a member) clauses 6.6.2.3(3) and 6.6.2.4(3) for the distribution of shear in studs from concentrated loads.

Clause 6.6.1.1(4)P Clause 6.6.1.1(4)P uses the term ‘ductile’ for connectors that have deformation capacity Clause 6.6.1.1(5) sufficient to assume ideal plastic behaviour of the shear connection. Clause 6.6.1.1(5) quantifies this as a characteristic slip capacity of 6 mm.75 Clause 6.6.1.1(6)P The need for compatibility of load/slip properties, clause 6.6.1.1(6)P, is one reason why neither bond nor adhesives can be used to supplement the shear resistance of studs. The combined use of studs and block-and-hoop connectors has been discouraged for the same reason, though there is little doubt that effectively rigid projections into the concrete slab, such as bolt heads and ends of flange plates, contribute to shear connection. This Principle is particularly important in bridges, where the fatigue loading on individual connectors may otherwise be underestimated. This applies also where bridges are to be strengthened by retrofitting additional shear connectors.

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‘Separation’, in clause 6.6.1.1(7)P, means separation sufficient for the curvatures of the two elements to be different at a cross-section, or for there to be a risk of local corrosion. None of the design methods in EN 1994-2 takes account of differences of curvature, which can arise from a very small separation. Even where most of the load is applied by or above the slab, as is usual, tests on beams with unheaded studs show separation, especially after inelastic behaviour begins. This arises from local variations in the flexural stiffnesses of the concrete and steel elements, and from the tendency of the slab to ride up on the weld collars. The standard heads of stud connectors have been found to be large enough to control separation, and the rule in clause 6.6.1.1(8) is intended to ensure that other types of connector, with anchoring devices if necessary, can do so. Resistance to uplift is much influenced by the reinforcement near the bottom of the slab, so if the resistance of an anchor is to be checked by testing, reinforcement in accordance with clause 6.6.6 should be provided in the test specimens. Anchors are inevitably subjected also to shear. Clause 6.6.1.1(9) refers to ‘direct tension’. Load from a maintenance cradle hanging from the steel member is an example of how tension may arise. It can also be caused by the differential deflections of adjacent beams under certain patterns of imposed load, although the resulting tensions are usually small. Greater tension can be produced near bracings as identified by clause 6.6.1.1(13). Where tension is present in studs, its design magnitude should be determined and checked in accordance with clause 6.6.3.2. Clause 6.6.1.1(10)P is a principle that has led to many application rules. The shear forces are inevitably ‘concentrated ’. One research study76 found that 70% of the shear on a stud was resisted by its weld collar, and that the local (triaxial) stress in the concrete was several times its cube strength. Transverse reinforcement performs a dual role. It acts as horizontal shear reinforcement for the concrete flanges, and controls and limits splitting. Its detailing is particularly critical where connectors are close to a free surface of the slab or where they are aligned so as to cause splitting in the direction of the slab thickness. To account for the latter, clause 6.6.1.1(11) should also include a reference to clause 6.6.4 for design of the transverse reinforcement. Larger concentrated forces occur where precast slabs are used, and connectors are placed in groups in holes in the slabs. This influences the detailing of the reinforcement near these holes, and is referred to in Section 8. Clause 6.6.1.1(12) is intended to permit the use of other types of connector. ENV 1994-1-120 included provisions for many types of connector other than studs: block connectors, anchors, hoops, angles, and friction-grip bolts. They have all been omitted because of their limited use and to shorten the code. Clause 6.6.1.1(12) gives scope, for example, to develop ways to improve current detailing practice at the ends of beams in fully integral bridges, where forces need to be transferred abruptly from the composite beams into reinforced concrete piers and abutments. In British practice, the use of ‘bars with hoops’ is often favoured in these regions, and design rules are given in BS 5400-5.11 The word ‘block’ rather than ‘bar’ is used in this Guide to avoid confusion with reinforcing bars. It is shown in Example 6.8 that the shear resistance of a connector of this type can be determined in accordance with EN 1992 and EN 1993. The height of the block should not exceed four times its thickness if the connector is assumed to be rigid as in Example 6.8. Clause 6.6.1.1(13) identifies a problem that occurs adjacent to cross-frames or diaphragms between beams. For multi-beam decks, beams are often braced in pairs such that the bracing is not continuous transversely across the deck. The presence of bracing locally significantly stiffens the bridge transversely. Moments and shears in the deck slab are attracted out of the concrete slab and into the bracing as shown in Fig. 6.27 via the transverse stiffeners. This effect is not modelled in a conventional grillage analysis unless the increased stiffness in the location of bracings is included using a shear flexible member with inertia and shear area chosen to match the deflections obtained from a plane frame analysis of the bracing system. Three-dimensional space-frame or finite-element representations of the bridge can be used to model these local effects more directly.

Clause 6.6.1.1(7)P

Clause 6.6.1.1(8)

Clause 6.6.1.1(9)

Clause 6.6.1.1(10)P

Clause 6.6.1.1(11)

Clause 6.6.1.1(12)

Clause 6.6.1.1(13)

115


Fig. 6.27. Example of bending moments from a deck slab attracted into bracings

The transfer of moment causes tension in the shear connectors on one side of the flange and induces compression between concrete and flange on the other. Welds at tops of stiffeners must also be designed for this moment, which often leads to throat sizes greater than a ‘nominal’ 6 mm. In composite box girders, similar effects arise over the tops of the boxes, particularly at the locations of ring frames, bracings or diaphragms.

Clause 6.6.1.2

6.6.1.2. Ultimate limit states other than fatigue In detailing the size and spacing of shear connectors, clause 6.6.1.2 permits the design longitudinal shear flow to be averaged over lengths such that the peak shear flow within each length does not exceed the design longitudinal shear resistance per unit length by more than 10%, and the total design longitudinal shear does not exceed the total design longitudinal shear within this length. This is consistent with previous practice in the UK and sometimes avoids the need to alter locally the number or spacings of shear connectors adjacent to supports. Clause 6.6.1.2 has little relevance to the inelastic lengths in Class 1 and 2 members covered by clause 6.6.2.2(2) since the longitudinal shear is already averaged over the inelastic zone in this method.

Example 6.8: shear resistance of a block connector with a hoop Blocks of S235 steel, 250 mm long and 40 mm square, are welded to a steel flange as shown in Fig. 6.28, at a longitudinal spacing of 300 mm. The resistance to longitudinal shear, PRd , in a C30/37 concrete is determined. From clause 6.6.1.1(8) the resistance to uplift should be at least 0.1PRd . This is provided by the 16 mm reinforcing bars shown. None of the modes of failure involve interaction between concrete and steel, so their own M factors are used, rather than 1.25, though the National Annex could decide otherwise. The concrete is checked to EN 1992, so its definition of fcd is used, namely fcd ¼ cc fck =C . Assuming that the National Annex gives cc ¼ 1.0 for this situation, the design strengths of the materials are: fyd 225 N/mm2 , fsd ¼ 500=1:15 ¼ 435 N/mm2 , fcd ¼ 1 30=1:5 ¼ 20 N/mm2 The blocks here are so stiff that the longitudinal shear force can be assumed to be resisted by a uniform stress, block say, at the face of each block. Lateral restraint enables this stress to exceed fcd to an extent given in clause 6.7 of EN 1992-1-1, ‘Partially loaded areas’: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi block ¼ FRdu =Ac0 ¼ fcd Ac1 =Ac0 3:0fcd ð6:63Þ in EN 1992-1-1 where Ac0 is the loaded area and Ac1 is the ‘design distribution area’ of similar shape to Ac0 , shown in Fig. 6.29 of EN 1992-1-1.

116


Clause 6.7 requires the line of action of the force to pass through the centres of both areas, but in this application it can be assumed that the force from area b2 d2 /2 is resisted by the face of the block, of area b1 d1 /2, because the blocks are designed also to resist uplift. The dimensions of area Ac1 are fixed by clause 6.7 of EN 1992-1-1 as follows: b2 b1 þ h and

b2 3b1 ;

d2 d1 þ h

and

d2 3d1

Here, from Fig. 6.28, h ¼ 300 40 ¼ 260 mm; b1 ¼ 2 40 ¼ 80 mm; d1 ¼ 250 mm where b2 ¼ 240 mm, d2 ¼ 510 mm. 45

160

45

h = 260

40

16 dia.

120 40

b2

b1

d1 (a)

d2 = 510

(b)

Fig. 6.28. Block shear connector with hoop, for Example 6.8

From equation (6.63), pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ac1 =Ac0 ¼ ð510 240=2Þ=ð250 40Þ ¼ 2:47 ð<3:0Þ and FRdu ¼ P Rd ¼ 0:250 40 2:47 20 ¼ 494 kN The tensile stress in two 16 mm bars from an uplift force of 49.4 kN is 123 N/mm2 . The required anchorage length for a hooped bar is given in clause 8.4.4(2) of EN 1992-1-1. It is proportional to the tensile stress in the bar and depends on its lateral containment, which in this application is good. For sd ¼ 123 N/mm2 the anchorage length is 115 mm, so 120 mm (Fig. 6.28(a)) is sufficient. The welds between the block and the steel flange are designed for the resulting shear, tension, and bending moment in accordance with EN 1993-1-8. A separate check of fatigue would be required using EN 1993-1-9 to determine the detail category and stress range. The comments on clause 6.8.6.2(2) refer. The welds between the bar and the block are designed for the uplift force. The resistance given by this method is significantly less than that from design to BS 5400-5, where the method is based mainly on tests.77 The above method based on clause 6.7 of EN 1992-1-1 would strictly require vertical reinforcement to control splitting from the vertical load dispersal. However, as both push tests and practice have shown this reinforcement to be unnecessary even when using the higher resistances to BS 5400-5, vertical reinforcement need not be provided here. The resistance of a block connector is much higher than that of a shear stud, so where they are used in haunches, the detailing of reinforcement in the haunch needs attention. EN 1992-1-1 appears to give no guidance on the serviceability stress limit in a region where its clause 6.7 is applied. For shear connectors generally, clause 7.2.2(6) refers to clause 6.8.1(3), where the recommended limit is 0.75PRd . This agrees closely with the corresponding ratio given in BS 5400-5.11

117


6.6.2. Longitudinal shear force in beams for bridges 6.6.2.1. Beams in which elastic or non-linear theory is used for resistances of cross-sections Clause 6.6.2.1(1)

Clause 6.6.2.1(1) requires that the design longitudinal shear force per unit length (the ‘shear flow’) at an interface between steel and concrete is determined from the rate of change of force in the concrete or the steel. The second part of the clause states, as a consequence of this, that where elastic bending resistance is used, the shear flow can be determined from the transverse shear at the cross-section considered. To do this, it is implicit that the beam is of uniform cross-section such that the usual expression for longitudinal shear flow, vL;Ed ¼

Vc;Ed Az I

can be used, where: A is the effective transformed area on the side of the plane concerned that does not include the centroid of the section, sometimes named the ‘excluded area’; z is the distance in the plane of bending from the member neutral axis to the centroid of area A; I is the second moment of area of the effective cross-section of the member. The relevant shear Vc;Ed is that acting on the composite section. Where the cross-section varies along its length, the shear flow is no longer directly proportional to the shear on the beam and the following expression should be used: Vc;Ed Az d Mc;Ed Az d Az vL;Ed ¼ þ Mc;Ed ðD6:18Þ ¼ dx dx I I I

Clause 6.6.2.1(2)

Clause 6.6.2.1(3)

Clause 6.6.2.1(4)

Clause 6.6.2.2(1)

118

Equation (D6.18) does not directly cover step changes in the steel cross-section as often occur at splices. In such situations, it would be reasonable to assume that the step change occurs uniformly over a length of twice the effective depth of the cross-section when applying equation (D6.18). Where there is a sudden change from bare steel to a composite section, design for the concentrated longitudinal shear force from development of composite action should follow clause 6.6.2.4. The calculated elastic longitudinal shear flow is strongly dependent on whether or not the concrete slab is considered to be cracked. In reality, the slab will be stiffer than predicted by a fully cracked analysis due to tension stiffening. Clause 6.6.2.1(2) clarifies that the slab should therefore be considered to be fully uncracked unless tension stiffening and over-strength of concrete are considered in both global analysis and section design as discussed under clause 5.4.2.3(7). Clause 6.6.2.1(3) requires account to be taken of longitudinal slip where concentrated longitudinal forces are applied, and refers to clauses 6.6.2.3 and 6.6.2.4. In other cases, clause 6.6.2.1(3) allows slip to be neglected for consistency with clause 5.4.1.1(8).

Composite box girders For box girders with a composite flange, a shear flow across the shear connection can occur due to shear from circulatory torsion, torsional warping and distortional warping. These effects are discussed in the Designers’ Guide to EN 1993-2.4 Clause 6.6.2.1(4) requires them to be included ‘if appropriate’. This influences the design longitudinal shear stress (clause 6.6.6.1(5)), and hence the area of transverse reinforcement, to clause 6.6.6.2. Shear lag and connector slip lead to a non-uniform distribution of force in the shear connectors across the width of the flange. This is discussed in section 9.4 of this Guide. 6.6.2.2. Beams in bridges with cross-sections in Class 1 or 2 Where the bending resistance exceeds the elastic resistance and material behaviour is nonlinear, shear flows can similarly no longer be calculated from linear-elastic section analysis. To do so using equation (D6.18) would underestimate the shear flow where elastic limits are exceeded as the lever arm of the cross-section implicit in the calculation would be overestimated and thus the element forces would be underestimated. Clause 6.6.2.2(1)


A

LA–B

B

LB–C

C x

0

Mel,Rd

Mpl,Rd MEd

M

Fig. 6.29. Definition of inelastic lengths for Class 1 and 2 cross-sections with the slab in compression

therefore requires that account be taken of inelastic behaviour of the member and its component parts in calculation of longitudinal shear. This differs from previous UK practice but is more soundly based in theory. Clause 6.6.2.2(2) gives specific guidance on how to comply with the above requirements where the concrete slab is in compression. For the length of beam where the bending moment exceeds Mel;Rd , the longitudinal shear relationship should be determined from the change in slab force. The relevant length is that between the points A and C in Fig. 6.29. The longitudinal shear force in the length A–B is determined as the difference between slab forces Nc;el at point A and Nc;d at point B. Appropriate shear connection to carry this force is provided within this length. Similar calculation is performed for the length B–C. The spacing of the shear connectors within these lengths is left to the designer. Normally for lengths A–B and B–C, and in the absence of heavy point loads, changes of cross-section, etc., uniform spacing can be used. In a doubtful case, for example within A–B, the slab force Nc at some point within A–B should be determined from the bending moment and appropriate numbers of connectors provided between A and D, say D, and between D and B. The actual relationship between slab force, Nc , and moment, MEd , is shown in Fig. 6.30, together with the approximate expression in Fig. 6.6 of clause 6.2.1.4(6) and the further simplification of Fig. 6.11. It can be seen that both the approximations are safe for the design of shear connection, because for a given bending moment, the predicted slab force exceeds the real slab force. For inelastic lengths where the slab is in tension, clause 6.6.2.2(3) requires the calculation of longitudinal shear force to consider the effects of tension stiffening and possible overstrength of the concrete. Failure to do so could underestimate the force attracted to the shear connection, resulting in excessive slip. As an alternative, clause 6.6.2.2(4) permits the shear flow to be determined using elastic cross-section analysis based on the uncracked cross-section. Elastic analysis can be justified in this instance because the conservative

Clause 6.6.2.2(2)

Clause 6.6.2.2(3)

Clause 6.6.2.2(4)

MEd /Mpl,Rd

1.0

True behaviour Idealisation in Fig. 6.6 Idealisation in Fig. 6.11

Mel,Rd /Mpl,Rd

Ma,Ed /Mpl,Rd

0

Nc,el /Nc,f

Nc /Nc,f 1.0

Fig. 6.30. Variation of longitudinal force in a concrete flange with bending moment

119


assumption of fully uncracked concrete offsets the slightly unconservative neglect of inelasticity.

Clause 6.6.2.3 Clause 6.6.2.3(2)

Clause 6.6.2.3(3)

Clause 6.6.2.3(4)

6.6.2.3. Local effects of concentrated longitudinal shear force due to introduction of longitudinal forces Where concentrated longitudinal forces are applied to a composite section, the stress state can easily be determined some distance away from the point of application from considerations of equilibrium and from the usual assumption that plane sections remain plane. Plane sections do not, however, remain plane in the vicinity of the force, and accurate determination of the length over which longitudinal shear transfers between concrete slab and steel flange together with the magnitude of the peak longitudinal shear flow requires complex analysis. This clause is based on parametric finite-element analyses78 and existing practice. Clause 6.6.2.3 provides simple rules for the determination of the design shear flow between steel and concrete where there is a concentrated longitudinal force, FEd , applied to the concrete slab. Clause 6.6.2.3(2) distinguishes between forces applied within the length of the member and those applied at ends of the members. In the former case, the length over which the force is distributed, Lv , is equal to the effective width for global analysis, beff , plus ed , which is the loaded length plus twice the lateral distance from the point of application of the force to the web centreline: Lv ¼ beff þ ed . For forces applied at an end of a concrete flange, the distribution length is half of the above. The reference to effective width for global analysis means that the simple provisions of clause 5.4.1.2(4) can be used, rather than the effective width appropriate to the crosssection where the force is applied. The force cannot, in general, be transferred uniformly by the shear connection over the above lengths and Fig. 6.12(a) and (b) shows the distribution to be used, leading to equations (6.12) and (6.13). Where stud shear connectors are used, these are sufficiently ductile to permit a uniform distribution of shear flow over the above lengths at the ultimate limit state. This leads to equations (6.14) and (6.15) in clause 6.6.2.3(3). For serviceability or fatigue limit states, the distributions of equations (6.12) and (6.13) should always be used. The shear force VL;Ed transferred to the shear connection is not FEd , as can be seen from Fig. 6.31 for the case of a force applied to the end of the concrete slab. Force VL;Ed is the difference between FEd and the force Nc in the concrete slab where dispersal of FEd into the cross-section is complete. Clause 6.6.2.3(4) allows the dispersal of the force FEd VL;Ed (which for load applied to the concrete as shown in Fig. 6.31 is equal to Nc ) into either the concrete or steel element to be based on an angle of spread of 2 where is tan1 2/3. This is the same spread angle used in EN 1992-1-1 clause 8.10.3 for dispersal of prestressing force into concrete. It is slightly less than the dispersal allowed by clause 3.2.3 of EN 1993-1-5 for the spread through steel elements.

Clause 6.6.2.4

6.6.2.4. Local effects of concentrated longitudinal shear forces at sudden change of cross-sections Clause 6.6.2.4 provides simple rules for the determination of the design shear flow between steel and concrete at ends of slabs where:

Clause 6.6.2.4(1)

.

the primary effects of shrinkage or differential temperature are developed (clause 6.6.2.4(1)) Mc FEd

VL,Ed = FEd – Nc Nc Lv/2 = beff/2 + ed/2 Ma

Ga

Na = VL,Ed

Fig. 6.31. Determination of VL;Ed for a concentrated force applied at an end of the slab

120


Mc

Nc

VL,Ed = Nc T beff

C Ma Ga

Na = Nc T

Fig. 6.32. Determination of VL;Ed for primary shrinkage at an end of a beam .

there is an abrupt change of cross-section (clause 6.6.2.4(2)), such as that shown in Fig. 6.33.

The shear VL;Ed transferred across the concrete and steel interface due to shrinkage or temperature may be assumed to be distributed over a length equal to beff , as discussed in section 6.6.2.3 above. Generally, clause 6.6.2.4(3) requires the distribution of this force to be triangular as shown in Fig. 6.12(c), which leads to equation (6.16). Where stud shear connectors are used, these are again sufficiently ductile to permit a uniform distribution of shear flow over the length beff . This leads to a design shear flow of VL;Ed =beff . A calculation of primary shrinkage stresses is given in Example 5.3 (Fig. 5.11). The force Nc is found from these. It equals the shear force VL;Ed , which is transferred as shown in Fig. 6.32. The determination of VL;Ed caused by bending at a sudden change in cross-section is shown in Fig. 6.33 for the case of propped construction, in which the total moment at cross-section B–B is:

Clause 6.6.2.4(2)

Clause 6.6.2.4(3)

MEd;B ¼ Ma þ Mc þ Nc z For unpropped construction, the stress would not vary linearly across the composite section as shown in Fig. 6.33, but the calculation of longitudinal shear from the force in the slab would follow the same procedure. The length over which the force is distributed and the shape of distribution may be taken according to clause 6.6.2.4(5) to be the same as that given in clause 6.6.2.4(3).

Clause 6.6.2.4(5)

6.6.3. Headed stud connectors in solid slabs and concrete encasement Resistance to longitudinal shear In BS 5400 Part 511 and in earlier UK codes, the characteristic shear resistances of studs are given in a table, applicable only when the stud material has particular properties. There was no theoretical model for the shear resistance. The Eurocodes must be applicable to a wider range of products, so design equations are essential. Those given in clause 6.6.3.1(1) are based on the model that a stud with shank diameter d and ultimate strength fu , set in concrete with characteristic strength fck and mean secant modulus Ecm , fails either in the steel alone or in the concrete alone. The concrete failure is found in tests to be influenced by both the stiffness and the strength of the concrete.

Clause 6.6.3.1(1)

B Mc

VL,Ed = Nc

MEd,B

Nc beff

MEd,A

z Ma Ga Na = Nc = VL,Ed

A

B

Fig. 6.33. Determination of VL;Ed caused by bending moment at a sudden change of cross-section

121


PRd (kN) 100 Normal-density concrete

fu = 500 N/mm2 450

80 400

Density class 1.8

60

20

30

40

50

fck (N/mm2)

25

37

50

60

fcu (N/mm2)

Fig. 6.34. Design shear resistances of 19 mm studs with h=d 4 in solid slabs

This led to equations (6.18) to (6.21), in which the numerical constants and partial safety factor V have been deduced from analyses of test data. In situations where the resistances from equations (6.18) and (6.19) are similar, tests show that interaction occurs between the two assumed modes of failure. An equation based on analyses of test data, but not on a defined model79 PRd ¼ kðd 2 =4Þ fu ðEcm =Ea Þ0:4 ð fck =fu Þ0:35

ðD6:19Þ

gives a curve with a shape that approximates better both to test data and to values tabulated in BS 5400. In the statistical analyses done for EN 1994-1-180;81 both of these methods were studied. Equation (D6.19) gave results with slightly less scatter, but the equations of clause 6.6.3.1(1) were preferred because of their clear basis and experience of their use in some countries. Here, and elsewhere in Section 6, coefficients from such analyses were modified slightly, to enable a single partial factor, denoted V (V for shear), 1.25, to be recommended for all types of shear connection. This value has been used in draft Eurocodes for over 20 years. It was concluded from this study81 that the coefficient in equation (6.19) should be 0.26. This result was based on push tests, where the mean number of studs per specimen was only six, and where lateral restraint from the narrow test slabs was usually less stiff than in the concrete flange of a composite beam. Strength of studs in many beams is also increased by the presence of hogging transverse bending of the slab. For these reasons the coefficient was increased from 0.26 to 0.29, a value that is supported by a subsequent calibration study60 based on beams with partial shear connection. Design resistances of 19 mm stud connectors in solid slabs, given by clause 6.6.3.1, are shown in Fig. 6.34. It is assumed that the penalty for short studs, equation (6.20), does not apply. For any given values of fu and fck , the figure shows which failure mode governs. It can be used for this purpose for studs of other diameters, provided that h=d 4. The reference to the slabs as ‘solid’ means that they are not composite slabs cast on profiled steel sheeting. It does not normally exclude haunched slabs. The ‘overall nominal height’ of a stud, used in equations (6.20) and (6.21), is about 5 mm greater than the ‘length after welding’, a term which is also in use.

Clause 6.6.3.1(2)

122

Weld collars Clause 6.6.3.1(2) on weld collars refers to EN 13918,40 which gives ‘guide values’ for the height and diameter of collars, with the note that these may vary in through-deck stud welding. It is known that for studs with normal weld collars, a high proportion of the shear is transmitted through the collar.76 It should not be assumed that the shear resistances of clause 6.6.3.1 are applicable to studs without collars, as noted by clause 6.6.3.1(4) (e.g.


where friction welding by high-speed spinning is used). A normal collar should be fused to the shank of the stud. Typical collars in the test specimens from which the design formulae were deduced had a diameter not less than 1.25d and a minimum height not less than 0.15d, where d is the diameter of the shank.

Splitting of the slab Clause 6.6.3.1(3) refers to ‘splitting forces’ in the direction of the slab thickness. These occur where the axis of a stud lies in a plane parallel to that of the concrete slab; for example, if studs are welded to the web of a steel T-section that projects into a concrete flange. These are referred to as ‘lying studs’ in published research82 on the local reinforcement needed to prevent or control splitting. Comment on the informative Annex C on this subject is given in Chapter 10. A similar problem occurs in composite L-beams with studs close to a free edge of the slab. This is addressed in clause 6.6.5.3(2). Tension in studs Pressure under the head of a stud connector and friction on the shank normally causes the stud weld to be subjected to vertical tension before shear failure is reached. This is why clause 6.6.1.1(8) requires shear connectors to have a resistance to tension that is at least 10% of the shear resistance. Clause 6.6.3.2(2) therefore permits tensile forces that are less than this to be neglected. (The symbol Ften in this clause means FEd;ten .) Resistance of studs to higher tensile forces has been found to depend on so many variables, especially the layout of local reinforcement, that no simple design rules could be given. Relevant evidence from about 60 tests on 19 mm and 22 mm studs is presented in Ref. 74, which gives a best-fit interaction curve. In design terms, this becomes ðFten =0:85PRd Þ5=3 þ ðPEd =PRd Þ5=3 1

Clause 6.6.3.1(3)

Clause 6.6.3.2(2)

ðD6:20Þ

Where the vertical tensile force Ften ¼ 0:1PRd , this gives PEd 0:93PRd , which is plausible. Expression (D6.20) should be used with caution, because some studs in these tests had ratios h=d as high as 9; but on the conservative side, the concrete blocks were unreinforced.

6.6.4. Headed studs that cause splitting in the direction of the slab thickness There is a risk of splitting of the concrete where the shank of a stud (a ‘lying stud’) is parallel and close to a free surface of the slab, as shown, for example, in Fig. 6.35. Where the conditions of clause 6.6.4(1) to (3) are met, the stud resistances of clause 6.6.3.1 may still be used. The geometric requirements are shown in Fig. 6.35, in which d is the diameter of the stud. A further restriction is that the stud must not also carry shear in a direction transverse to the slab thickness. The example shown in Fig. 6.35 would not comply in this respect unless the steel section were designed to be loaded on its bottom flange. Clause 6.6.4(3) requires that the stirrups shown should be designed for a tensile force equal to 0.3PRd per stud connector. This is analogous with the design of bursting reinforcement at prestressing anchorages. The true tensile force depends on the slab thickness and spacing of the studs and the proposed value is conservative for a single row of studs. No recommendation is given here, or in Annex C, on the design of stirrups where there are several rows of studs.

Clause 6.6.4(1) to (3)

s A

PEd

ev ≥ 6d ev ≥ 6d

≤s, ≤18d v ≥ 14d

A

(a)

View A–A (b)

Fig. 6.35. Examples of details susceptible to longitudinal splitting

123


Some details which do not comply with clause 6.6.4 can be designed using the rules in the informative Annex C, if its use is permitted by the National Annex. In Fig. 6.35, the effects of local loading on the slab and of U-frame action will also cause moment at the shear connection which could cause stud tensions in excess of those allowed by clause 6.6.3.2. This detail is therefore best avoided. Planes of type a–a such as section A–A in Fig. 6.35 should be provided with longitudinal shear reinforcement in accordance with clause 6.6.6.

6.6.5. Detailing of the shear connection and influence of execution It is rarely possible to prove the general validity of application rules for detailing, because they apply to so great a variety of situations. They are based partly on previous practice. An adverse experience causes the relevant rule to be made more restrictive. In research, existing rules are often violated when test specimens are designed, in the hope that extensive good experience may enable existing rules to be relaxed. Rules are often expressed in the form of limiting dimensions, even though most behaviour (excluding corrosion) is more influenced by ratios of dimensions than by a single value. Minimum dimensions that would be appropriate for an unusually large structural member could exceed those given in the code. Similarly, code maxima may be too large for use in a small member. Designers are unwise to follow detailing rules blindly, because no set of rules can be comprehensive.

Clause 6.6.5.1(1)

Resistance to separation The object of clause 6.6.5.1(1) on resistance to separation is to ensure that failure surfaces in the concrete cannot pass above the connectors and below the reinforcement, intersecting neither. Tests have found that these surfaces may not be plane; the problem is threedimensional. A longitudinal section through a possible failure surface ABC is shown in Fig. 6.36. The studs are at the maximum spacing allowed by clause 6.6.5.5(3). Clause 6.6.5.1 defines only the highest level for the bottom reinforcement. Ideally, its longitudinal location relative to the studs should also be defined, because the objective is to prevent failure surfaces where the angle (Fig. 6.36) is small. It is impracticable to link detailing rules for reinforcement with those for connectors, or to specify a minimum for angle . In Fig. 6.36, it is less than 88, which is much too low. The angle obviously depends on the level of the bottom bars, the height of the studs, and the spacing of both the bars and the studs. Studs in a bridge deck usually have a length after welding (LAW) that exceeds the 95 mm shown. Assuming LAW ¼ 120 mm, maximum spacings of both bars and studs of 450 mm, and a bottom cover of 50 mm gives 178, approximately, which is suggested here as a minimum. Studs may need to be longer than 125 mm where permanent formwork is used, as this raises the level of the bottom reinforcement. Other work reached a similar conclusion in 2004.83 Referring to failure surfaces as shown in Fig. 6.36, it was recommended that angle should be at least 158. In this paper the line AB (Fig. 6.36) is tangential to the top of the bar at B, rather than the bottom, slightly reducing its slope.

Concreting Clause 6.6.5.2(1)P Clause 6.6.5.2(1)P requires shear connectors to be detailed so that the concrete can be adequately compacted around the base of the connector. This necessitates the avoidance

55

A

C B

95

30

α

800

Fig. 6.36. Level of bottom transverse reinforcement (dimensions in mm)

124


e(≥ 6d ) a

b d

a

≥ 0.5d

b

Fig. 6.37. Longitudinal shear reinforcement in an L-beam

of excessively close spacings of connectors and the use of connector geometries that might prevent adequate flow of the concrete around the connector. The former could be a consideration at the ends of fully integral bridges where a very high shear flow has to be transferred into the steel beam over a relatively short length. Since the resistances of connectors other than studs are not covered by EN 1994-2, properties of other types of connector could be referred to from a National Annex. The design of a block-and-hoop connector is illustrated in Example 6.8. A novel type of connection could be investigated as part of the testing requirements of clause 6.6.1.1(12).

Loading of shear connection during execution Clause 6.6.5.2(3) particularly concerns the staged casting of concrete flanges for typical unpropped composite bridges. Partly matured concrete around shear connectors in a recently cast length of beam could possibly be damaged by the effects of concreting nearby. The recommended lower limit on concrete strength, 20 N/mm2 , in effect sets a minimum time interval between successive stages of casting. The rule begins ‘Wherever possible’ because there appears to be no evidence of damage from effects of early thermal or shrinkage strains, which also apply longitudinal shear to young concrete. In propped construction, it would be unusual to remove the props until the concrete had achieved a compressive strength of at least 20 N/mm2 , in order to avoid overstressing the beam as a whole. Where the props are removed prior to the concrete attaining the specified strength, verifications at removal of props should be based on an appropriately reduced compressive strength. Local reinforcement in the slab Where shear connectors are close to a longitudinal edge of a concrete flange, use of U-bars is almost the only way of providing the full anchorage required by clause 6.6.5.3(1). The splitting referred to in clause 6.6.5.3(2) is a common mode of failure in push-test specimens with narrow slabs (e.g. 300 mm, which has long been the standard width in British codes). It was also found, in full-scale tests, to be the normal failure mode for composite L-beams constructed with precast slabs.84 Detailing rules are given in clause 6.6.5.3(2) for slabs where the edge distance e in Fig. 6.37 is less than 300 mm. The required area of bottom transverse reinforcement, Ab per unit length of beam, should be found using clause 6.6.6. In the unhaunched slab shown in Fig. 6.37, failure surface b–b will be critical (unless the slab is very thick) because the shear on surface a–a is low in an L-beam with an asymmetrical concrete flange. To ensure that the reinforcement is fully anchored to the left of the line a–a, it is recommended that U-bars be used. These can be in a horizontal plane or, where top reinforcement is needed, in a vertical plane.

Clause 6.6.5.2(3)

Clause 6.6.5.3(1) Clause 6.6.5.3(2)

Reinforcement at the end of a cantilever At the end of a composite cantilever, the force on the concrete from the connectors acts towards the nearest edge of the slab. The effects of shrinkage and temperature can add further stresses74 that tend to cause splitting in region B in Fig. 6.38, so reinforcement in this region needs careful detailing. Clause 6.6.5.3(3)P can be satisfied by providing ‘herring- Clause 6.6.5.3(3)P bone’ bottom reinforcement (ABC in Fig. 6.38) sufficient to anchor the force from the connectors into the slab, and ensuring that the longitudinal bars provided to resist that force are anchored beyond their intersection with ABC.

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A Transverse reinforcement not shown B

Steel beam Studs

Slab

C

Fig. 6.38. Reinforcement at the end of a cantilever

Clause 6.6.5.4

Clause 6.6.5.5(2)

Haunches Haunches are sometimes provided in composite bridges to cater for drainage cross-falls so that the thickness of the slab or deck surfacing need not be varied. The detailing rules of clause 6.6.5.4 are based on limited test evidence, but are long-established.85 In regions of high longitudinal shear, deep haunches should be used with caution because there may be little warning of failure. Maximum spacing of connectors Situations where the stability of a concrete slab is ensured by its connection to a steel beam are unlikely to occur because a concrete slab that is adequate to resist local bridge loading is unlikely to suffer instability from membrane forces. The converse situation, stabilization of the steel flange, is of interest only where the steel compression flange is not already in Class 1 or 2. Where the steel beam is a plate girder, its proportions will often be chosen such that it is in Class 3 for the bare steel condition during construction. This maximises the lateral buckling resistance for a flange of given cross-sectional area. Clause 6.6.5.5(2) is not restrictive in practice. As an example, a plate girder is considered, in steel with fy ¼ 355 N/mm2 , where the top flange has tf ¼ 20 mm, an overall breadth of 350 mm, and an outstand c of 165 mm. The ratio " is 0.81 and the slenderness is: c=tf " ¼ 165=ð20 0:81Þ ¼ 10:2

Clause 6.6.5.5(3)

Clause 6.6.5.5(4)

so from Table 5.2 of EN 1993-1-1, the flange is in Class 3. From clause 6.6.5.5(2), it can be assumed to be in Class 1 if shear connectors are provided within 146 mm of each free edge, at longitudinal spacing not exceeding 356 mm, for a solid slab. The ratio 22 in this clause is based on the assumption that the steel flange cannot buckle towards the slab. Where there are transverse ribs (e.g. due to the use of profiled sheeting), the assumption may not be correct, so the ratio is reduced to 15. The maximum spacing in this example is then 243 mm. Further requirements for composite plates in box girders are given in clause 9.4(7). These also cover limitations on longitudinal and transverse spacings of connectors to ensure Class 3 behaviour. The rule on transverse spacing in Table 9.1 should be applied also to a wide compression flange of a plate girder. The maximum longitudinal spacing in bridges, given in clause 6.6.5.5(3), 4hc but 800 mm, is more liberal than the equivalent rule of BS 5400 Part 5.11 It is based mainly on behaviour observed in tests, and on practice with precast slabs in some countries. Clause 6.6.5.5(4) allows the spacing rules for individual connectors to be relaxed if connectors are placed in groups. This may facilitate the use of precast deck units with discrete pockets for the shear connection (clause 8.4.3(3) refers) but many of the deemed-to-satisfy rules elsewhere in EN 1994-2 then no longer apply. The designer should then explicitly consider the relevant effects, which will make it difficult in practice to depart from the application rules. The effects listed are as follows. .

126

Non-uniform flow of longitudinal shear. If the spacing of the groups of connectors is large compared to the distance between points of zero and maximum moment in the beam, then the normal assumption of plane sections remaining plane will not apply and the calculation of bending resistance to clause 6.2 will not be valid.


.

.

.

Greater risk of slip and vertical separation of concrete and steel. The latter carries a corrosion risk for the steel flange which would be hard to quantify without testing. Buckling of the steel flange. This can be considered by applying clause 4.4(2) of EN 19931-5. The elastic critical buckling stress, cr , can be determined for the discrete supports offered by the particular connection provided, either by finite-element analysis or from standard texts such as Ref. 86. If the latter method is employed, account needs to be taken of the beneficial restraint provided by the concrete against buckling. In the absence of this restraint, the flange would try to buckle in half wavelengths between the studs, alternating towards and away from the concrete. Buckling into the concrete is, in reality, prevented and therefore no rotation of the flange can occur along the line of the studs. Discrete supports which clamp the plate at the stud locations may therefore usually be assumed in determining the critical stress. Local resistance of the slab to the concentrated force from the connectors. Groups of studs apply a force analogous to that from an anchorage of a prestressing cable. The region of transverse tension does not coincide with the location of the group. Both the quantity and the location of the transverse reinforcement required may differ from that given by clause 6.6.6.

Dimensions of the steel flange The rules of clause 6.6.5.6 are intended to prevent local overstress of a steel flange near a shear connector and to avoid problems with stud welding. Application rules for minimum flange thickness are given in clause 6.6.5.7(3) and (5). The minimum edge distance for connectors in clause 6.6.5.6(2) is consistent with the requirements of BS 5400:Part 5.11 The limit is also necessary to avoid a reduction in the fatigue detail category for the flange at the stud location, which from EN 1993-1-9 Table 8.4, is 80 for a compliant stud. Headed stud connectors Clause 6.6.5.7(1) and (2) is concerned with resistance to uplift. Rules for resistance of studs, minimum cover and projection of studs above bottom reinforcement usually lead to the use of studs of height greater than 3d. The limit 1.5 for the ratio d=tf in clause 6.6.5.7(3) could, in principle, influence the design of shear connection for closed-top box girders in bridges, effectively requiring the thickness of the top flange to be a minimum of 13 mm where 19 mm diameter studs are used. Studs of this size are preferred by many UK fabricators as they can be welded manually. It is unlikely that a composite box would have a flange this thin as it would then require considerable longitudinal stiffening to support the plate prior to setting of the concrete. Clause 6.6.5.7(3) would also apply to plate girders adjacent to cross-braces or transverse diaphragms, but practical sizes of flanges for main beams will satisfy this criterion. It could be a consideration for cross-girders. In clause 6.6.5.7(4), the minimum lateral spacing of studs in ‘solid slabs’ has been reduced to 2.5d, compared with the 4d of BS 5950-3-1. This facilitates the use of precast slabs supported on the edges of the steel flanges, with projecting U-bars that loop over pairs of studs. Closely spaced pairs of studs must be well confined laterally. The words ‘solid slabs’ should therefore be understood here to exclude haunches.

Clause 6.6.5.6

Clause 6.6.5.6(2)


Clause 6.6.5.7(4)

6.6.6. Longitudinal shear in concrete slabs The subject of clause 6.6.6 is the avoidance of local failure of a concrete flange near the shear connection, by the provision of appropriate reinforcement. These bars enhance the resistance of a thin concrete slab to in-plane shear in the same way that stirrups strengthen a concrete web in vertical shear. Transverse reinforcement is also needed to control and limit the longitudinal splitting of the slab that can be caused by local forces from individual connectors. In this respect, the detailing problem is more acute than in the flanges of concrete T-beams, where the shear from the web is applied more uniformly. The principal change from earlier codes is that the equations for the required area of transverse reinforcement have been replaced by cross-reference to EN 1992-1-1. Its

Clause 6.6.6

127


provisions are based on a truss analogy, as before, but a more general version of it, in which the angle between members of the truss can be chosen by the designer. It is an application of strut-and-tie modelling, which is widely used in EN 1992. There is, however, a significant difference between the application of EN 1992 and EN 1994. In the latter, the transverse reinforcement may be placed according to the distribution of vertical shear force envelope, or according to the stud forces for sections where the elastic resistance moment is exceeded. In the former, the transverse reinforcement should be placed according to the location of the web compression struts as they intersect the flanges and their subsequent continuation into the flanges. The definitions of shear surfaces in clause 6.6.6.1(2)P and the basic design method are as Clause 6.6.6.1(2)P before. The method of presentation reflects the need to separate the ‘general’ provisions, clauses 6.6.6.1 to 3, from those restricted to ‘buildings’, in EN 1994-1-1 clause 6.6.6.4. Clause 6.6.6.1(4) requires the design longitudinal shear to be ‘consistent with’ that used for Clause 6.6.6.1(4) the design of the shear connectors. This means that the distribution along the beam of resistance to in-plane shear in the slab should be not less than that assumed for the design of the shear connection. For example, uniform resistance to longitudinal shear flow (vL ) should be provided where the connectors are uniformly spaced, even if the vertical shear over the length is not constant. It does not mean, for example, that if, for reasons concerning detailing, vL;Rd ¼ 1:3vL;Ed for the connectors, the transverse reinforcement must provide the same degree of over-strength. The reference to ‘variation of longitudinal shear across the width of the concrete flange’ means that transverse reinforcement could be reduced away from the beam centre-lines, where the longitudinal shear reduces, if flexural requirements permit. In applying clause 6.6.6.1(5), it is sufficiently accurate to assume that longitudinal bending Clause 6.6.6.1(5) stress in the concrete flange is constant across its effective width, and zero outside it. The clause is relevant, for example, to finding the shear on plane a–a in the haunched beam shown in Fig. 6.15, which, for a symmetrical flange, is less than half the shear resisted by the connectors.

Clause 6.6.6.2(1)

Resistance of a concrete flange to longitudinal shear Clause 6.6.6.2(1) refers to clause 6.2.4 of EN 1992-1-1, which is written for a design longitudinal shear stress vEd acting on a cross-section of thickness hf . This must be distinguished from the design longitudinal shear flow vL;Ed used in EN 1994 which is equal to vEd hf . The clause requires the area of transverse reinforcement Asf at spacing sf to satisfy Asf fyd =sf > vEd hf = cot f

ð6:21Þ in EN 1992-1-1

and the longitudinal shear stress to satisfy vEd < fcd sin f cos f

ð6:22Þ in EN 1992-1-1

where ¼ 0:6ð1 fck =250Þ, with fck in N/mm2 units. (The Greek letter (nu) used here in EN 1992-1-1 should not be confused with the Roman letter v (vee), which is used for shear stress.) The angle f between the diagonal strut and the axis of the beam is chosen (within limits) by the designer. It should be noted that the recommended limits depend on whether the flange is in tension or compression, and can be varied by a National Annex. EN 1992-1-1 does not specify the distribution of the required transverse reinforcement between the upper and lower layers in the slab. It was a requirement of early drafts of EN 1992-2 that the transverse reinforcement provided should have the same centre of resistance as the longitudinal force in the slab. This was removed, presumably because it has been common practice to consider the shear resistance to be the sum of the resistances from the two layers. Clause 6.6.6.2(3) refers to Fig. 6.15 which clarifies that the reinforcement to be considered on plane a–a for composite beams is the total of the two layers, Ab þ At . It should be noted that application of Annex MM of EN 1992-2 would necessitate provision of transverse reinforcement with the same centre of resistance as the longitudinal force in the slab.

128


The use of the method is illustrated in Example 6.9. The reference in clause 6.6.6.2(1) is to the whole of clause 6.2.4 of EN 1992-1-1. This includes clause 6.2.4(5) which requires a check of the interaction with transverse bending in the concrete slab on planes of type a–a. This requires the amount of transverse reinforcement to be the greater of that required for longitudinal shear alone and half that required for longitudinal shear plus that required for transverse bending. This rule is illustrated in Fig. 6.39(b), in which Areq;d is the total reinforcement required, and subscripts s and b refer to the reinforcement required for shear and bending, respectively. The rule is more onerous than that in BS 5400 Part 5, where no interaction is needed to be considered on planes of type a–a. This was because a transverse bending moment produces no net axial force in the slab. Reinforcement in tension is then associated with concrete that has a resistance to shear enhanced by the presence of transverse compression. Test evidence supported this model. Consideration of this interaction according to EN 1992-1-1, rather than simple addition of bending and shear requirements, is rational as the truss angle in the upper and lower layers of the slab can be varied to account for the relative demands imposed on them of shear and either direct tension or compression. Annex MM of EN 1992-2 reinforces this, although its use suggests that the ‘50% of longitudinal shear’ rule is optimistic. The interaction on surfaces around the studs differs slightly as discussed below under the heading ‘Shear planes and surfaces’. Clause 6.2.4(105) of EN 1992-2 adds an additional interaction condition for the concrete struts, which was largely due to the strength of belief in its Project Team that the neglect of this check was unsafe. It was less to do with any inherent difference between the behaviour of buildings and bridges. Considerations of equilibrium suggest that EN 1992-2 is correct. The reference in EN 1994-2 to EN 1992-1-1 rather than to EN 1992-2 is therefore significant as it excludes the check. No check of the effect of moment on the concrete struts was required by BS 5400 Part 5. Clause 6.2.4(105) of EN 1992-2 refers to the compression from transverse bending. It is however equally logical to consider the slab longitudinal compression in the check of the concrete struts, although this was not intended. Once again, the use of Annex MM of EN 1992-2 would require this to be considered. It is rare in practice for a concrete flange to be subjected to a severe combination of longitudinal compression and in-plane shear. It could occur near an internal support of a continuous half-through bridge. It is shown in Example 6.11 that the reduction in resistance to bending would usually be negligible, if this additional interaction were to be considered. Neither EN 1992 nor EN 1994 deals with the case of longitudinal shear and coexistent transverse tension in the slab. This can occur in the transverse beams of ladder decks near the intersection with the main beams in hogging zones where the main beam reinforcement is in global tension. In such cases, there is clearly a net tension in the slab and the reinforcement requirements for this tension should be fully combined with that for longitudinal shear on planes a–a.

Shear planes and surfaces Clause 6.6.6.1(3) refers to shear ‘surfaces’ because some of them are not plane. Those of types b–b, c–c and d–d in Fig. 6.15 are different from the type a–a surface because they resist almost the whole longitudinal shear, not (typically) about half of it. The relevant reinforcement intersects them twice, as shown by the factors 2 in the table in Fig. 6.15. For a surface of type c–c in a beam with the steel section near one edge of the concrete flange, it is clearly wrong to assume that half the shear crosses half of the surface c–c. However, in this situation the shear on the adjacent plane of type a–a will govern, so the method is not unsafe. Clause 6.6.6.2(2) refers to EN 1992-1-1 clause 6.2.4(4) for resistance to longitudinal shear for surfaces passing around the shear studs. This does not include a check of the interaction with transverse bending in the slab. This does not mean that such a check may be ignored.

Clause 6.6.6.2(2)

129


Interaction of longitudinal shear and bending should be considered for the reinforcement crossing shear surfaces around the connectors. Slab bottom reinforcement is particularly important since most of the shear transferred by stud connectors is transferred over the bottom part of the stud. The bottom reinforcement crossing surfaces passing around the studs must control the localized splitting stresses generated by the high stud pressures. For surfaces of type b–b and c–c in Fig. 6.15, it is clear that the reinforcement crossing the surface must provide both the resistance to longitudinal shear and any transverse sagging moment present. The reinforcement requirements for coexisting shear and sagging moment should therefore be fully added. For coexisting shear and hogging moment, there is transverse compression at the base of the connectors. BS 5400 Part 5 permitted a corresponding reduction in the bottom reinforcement. This could be the basis of a ‘more accurate calculation’ permitted by clause 6.6.6.2(2). For surfaces of type d–d in Fig. 6.15, the haunch reinforcement crossed by the studs will not be considered in the sagging bending resistance and therefore it need only be designed for longitudinal shear. These recommendations are consistent with those in BS 5400 Part 5.

Clause 6.6.6.3

Minimum transverse reinforcement Clause 6.6.6.3 on this subject is discussed under clause 6.6.5.1 on resistance to separation.

Example 6.9: transverse reinforcement for longitudinal shear Figure 6.39(a) shows a plan of an area ABCD of a concrete flange of thickness hf , assumed to be in longitudinal compression, with shear stress vEd and transverse reinforcement Asf at spacing sf . The shear force per transverse bar is: Fv ¼ vEd hf sf acting on side AB of the rectangle shown. It is transferred to side CD by a concrete strut AC at angle f to AB, and with edges that pass through the mid-points of AB, etc., as shown, so that the width of the strut is sf sin f . Fv A

B θf

Areq’d/As

Ft 2

Fc

1 0.5 D

C

Fv

sf sin θf

sf

0

(a)

0.5

1 Ab/As

(b)

Fig. 6.39. In-plane shear in a concrete flange: (a) truss analogy; and (b) interaction of longitudinal shear with bending

For equilibrium at A, the force in the strut is: Fc ¼ Fv sec f

ðD6:21Þ

For equilibrium at C, the force in the transverse bar BC is: Ft ¼ Fc sin f ¼ Fv tan f

ðD6:22Þ

For minimum area of transverse reinforcement, f should be chosen to be as small as possible. For a flange in compression, the limits to f given in clause 6.2.4(4) of EN 1992-1-1 are: 458 f 26:58

130

ðD6:23Þ


so the initial choice for f is 26.58. Then, from equation (D6.22), Ft ¼ 0:5Fv

ðD6:24Þ

From equation (6.22) in EN 1992-1-1, above, vEd < 0:40 fcd If this inequality is satisfied, then the value chosen for f is satisfactory. However, let us assume that the concrete strut is overstressed, because vEd ¼ 0:48 fcd . To satisfy equation (6.22), sin f cos f 0:48, whence f 378. The designer increases f to 408, which satisfies expression (D6.23). From equation (D6.22), Ft ¼ Fv tan 408 ¼ 0:84Fv . From equation (D6.24), the change in f , made to limit the compressive stress in the concrete strut AC, increases the required area of transverse reinforcement by 68%. The lengths of the sides of the triangle ABC in Fig. 6.39(a) are proportional to the forces Fv , Ft and Fc . For given Fv and sf , increasing f increases Fc , but for f < 458 (the maximum value recommended), the increase in the width st sin f is greater, so the stress in the concrete is reduced.

Example 6.10: longitudinal shear checks The bridge shown in Fig. 6.22 is checked for longitudinal shear, (i) adjacent to an internal support, (ii) at mid-span of the main span and (iii) at an end support. Creep of concrete reduces longitudinal shear, so for elastic theory, the short-term modular ratio is used.

(i) Adjacent to an internal support The 19 150 mm stud connectors are assumed to be 145 mm high after welding. Their arrangement and the transverse reinforcement are shown in Fig. 6.40 and the dimensions of the cross-section in Fig. 6.22. The shear studs are checked first. Although the concrete is assumed to be cracked at the support in both global analysis and cross-section design, the slab is considered to be uncracked for longitudinal shear design in accordance with clause 6.6.2.1(2). The longitudinal shear is determined from the vertical shear using elastic analysis in accordance with clause 6.6.2.1(1) since the bending resistance was based on elastic analysis. 85

280

85

250 145

20 mm bars at 150 mm

Fig. 6.40. Shear studs and transverse reinforcement adjacent to an internal support

From Example 5.2, the modular ratio n0 is 6.36. From Example 5.4, the area of longitudinal reinforcement is As ¼ 19 480 mm2 , which is 2.5% of the effective crosssection of the concrete flange. Relevant elastic properties of the cross-section are given in Table 6.3 for the uncracked unreinforced section (subscript U) and the uncracked reinforced section (subscript UR). The effect on these values of including the reinforcement is not negligible when the long-term modular ratio is used. However, the significant reduction in longitudinal shear caused by cracking is being ignored, so it is accurate enough to use the unreinforced section when calculating longitudinal shear. The height z to the neutral axis is measured from the bottom of the cross-section. The property A z=I, for the whole of the effective concrete flange, is appropriate for checking

131


the shear resistance of the studs, and slightly conservative for the shear surface of type b–b shown in Fig. 6.40. Table 6.3. Elastic properties of cross-section at internal support Modular ratio

zU (mm)

IU (m2 mm2 )

ðA z=IÞU (m1 )

zUR (mm)

IUR (m2 mm2 )

ðA z=IÞUR (m1 )

6.36 23.63

1120 862

38 530 26 390

0.8098 0.6438

1146 959

39 890 31 200

0.8380 0.7284

The calculation using n0 is: 3100 400 250 ð1375 1120Þ þ 25 ð1238 1120Þ A z 6:36 6:36 ¼ ¼ 0:810 m1 I 38 530 103 The design vertical shear forces on the composite section are: long-term, VEd ¼ 827 kN; short-term, VEd ¼ 1076 kN; total, 1903 kN. From equation (D6.18), vL;Ed ¼ 1903 0:810 ¼ 1541 kN/m Groups of four 19 mm diameter studs with fu ¼ 500 N/mm2 are provided at 150 mm centres. Their resistance is governed by equation (6.19): pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi 0:29d 2 fck Ecm 0:29 192 30 33 103 ¼ 83:3 kN=stud PRd ¼ ¼ 1:25 1000 V Thus vL;Rd ¼ 83:3 4=0:15 ¼ 2222 kN/m which exceeds the 1541 kN/m applied. Longitudinal shear in the concrete is checked next. The surface around the studs is critical for the reinforcement check as it has the same reinforcement as the plane through the deck (i.e. two 20 mm diameter bars) but almost twice the longitudinal shear. It is assumed here that there is no significant sagging bending over the top of the main beams adjacent to the internal supports. From equation (6.21) in clause 6.2.4(4) of EN 1992-1-1, assuming a 458 truss angle: Asf fyd =sf > vEd hf =cot f

where

vEd hf ¼ 1541 kN=m

Asf 1541 150=ðcot 458 500=1:15Þ ¼ 532 mm2 < 628 mm2 from two 20 mm bars The planes through the deck slab are critical for the concrete stress check since the length of each concrete shear plane is 250 mm compared to 570 mm for the surface around the studs, which carries twice the shear. The shear stress is: vEd ¼ 1541=ð2 250Þ ¼ 3:08 N=mm2 From equations (6.6N) in clause 6.2.2(6) and (6.22) in clause 6.2.4(4) of EN 1992-1-1: vEd < fcd sin f cos f ¼0:6ð1 30=250Þ ð30=1:5Þ 0:5 ¼ 5:28 N=mm2 so the shear stress is satisfactory. It should be noted that there is another surface around the studs which intersects the top of the haunch with length of 582 mm. It would be possible for such a plane to be more critical.

(ii) Mid-span of the main span The maximum spacing of groups of studs as shown in Fig. 6.41 is now found. The transverse reinforcement is as before and the cross-sectional dimensions of the beam are shown in Fig. 6.22. In the real bridge, the moment at mid-span was less than the elastic resistance

132


to bending and therefore a similar calculation to that in (i) could have been performed. For the purpose of this example, the sagging bending moment at mid-span is considered to be increased to 8.0 MNm, comprising 2.0 MNm on the bare steel section, 2.0 MNm on the ‘long-term’ composite section and 4.0 MNm live load on the ‘short-term’ composite section. a 85

280

85

250 145 a 20 mm bars at 150 mm

Fig. 6.41. Shear studs and transverse reinforcement at mid-span and adjacent to an end support

The distance between the point of maximum moment and the point where the elastic resistance to bending is just exceeded is assumed here to be 4.5 m. The longitudinal shear force between these points is determined by rearranging equation (6.3) and substituting MEd for MRd so that: Nc;f Nc;el MEd Mel;Rd VL;Ed ¼ Nc Nc;el ¼ Mpl;Rd Mel;Rd It is not obvious which modular ratio gives the more adverse result for this region, so calculations are done using both n0 and nL for the long-term loading. The elastic resistance to bending is governed by the tensile stress in the bottom flange. From clause 6.2.1.4(6), Mel;Rd is found by scaling down Mc;Ed by a factor k until a stress limit is reached. The relevant section moduli in m2 mm units are 15.96 for the steel section, and 23.39 short term and 21.55 long term for the composite section. Hence, 2000=15:96 þ ð4000=23:39 þ 2000=21:55Þk ¼ 345 11:7 where 11.7 N/mm2 is the stress due to the primary effects of shrinkage from Example 5.3. (The secondary effects of shrinkage have been ignored here as they are relieving.) This gives k ¼ 0.788, and the elastic resistance moment is: Mel;Rd ¼ 2000 þ 0:788 ð4000 þ 2000Þ ¼ 6728 kNm From Example 6.1, Mpl;Rd ¼ 10 050 kNm. The compressive force Nc;el in the composite slab at Mel;Rd is now required. The primary shrinkage stress, although tensile, is included for consistency with the calculation of Mel;Rd . Its value at mid-depth of the 250 mm flange, from Example 5.3, is 0.68 N/mm2 . The effect of the stresses in the haunch is negligible here. Using the section moduli for mid-depth of the flange, given in Table 6.4, the mean stress is: c;mean ¼ 0:788 ð4000=969:1Þ þ ð2000=1143Þ 0:68 ¼ 3:95 N=mm2 Table 6.4. Section moduli at mid-span for uncracked unreinforced section Modular ratio

zU (mm)

IU (m2 mm2 )

6.36 23.7

1192 951

27 880 20 500

ðA z=IÞU (m1 )

Wc , top of slab (m2 mm)

Wc , mid slab (m2 mm)

Wc , mid haunch (m2 mm)

Wa , bottom flange (m2 mm)

0.8024 0.6843

575.9 882.7

969.1 1143

3890 1692

23.39 21.55

133


Hence, Nc;el ¼ 3:1 0:25 3:95 ¼ 3:06 MN The force for full interaction, Nc;f , is based on rectangular stress blocks: Nc;f ¼ 3:1 0:25 0:85 30=1:5 ¼ 13:18 MN From equation (6.3): VL;Ed ¼ Nc Nc;el ¼

ðNc;f Nc;el ÞðMEd Mel;Rd Þ ð13:18 3:06Þð8:0 6:728Þ ¼ ð10:05 6:728Þ ðMpl;Rd Mel;Rd Þ

¼ 3875 kN The shear flow is therefore vL;Ed ¼ 3875=4:5 ¼ 861 kN/m Similar calculations using n0 ¼ 6.36 for all loading give: Mel;Rd ¼ 6858 kNm, c;mean ¼ 4.33 N/mm2 , and Nc;el ¼ 3.36 MN, from which vL;Ed ¼ 781 kN/m, which does not govern. From (i) above, PRd ¼ 83.3 kN/stud, so the required spacing of groups of three is: sv ¼ 83:3 3=861 ¼ 0:290 m Groups at 250 mm spacing could be used, unless a fatigue or serviceability check is found to govern. Longitudinal shear in the concrete is adequate by inspection, as the shear flow is less than that at the internal support but both the thickness of the slab and the reinforcement are the same.

(iii) End support The shear stud arrangement and transverse reinforcement are as shown in Fig. 6.41 and the cross-sectional dimensions are shown in Fig. 6.22. The studs are checked first. The longitudinal shear is determined from the vertical shear using elastic analysis in accordance with clause 6.6.2.1(1) since the bending moment is less than the elastic resistance. From similar calculations to those for the pier section, the longitudinal shear flow, excluding that from shrinkage and temperature, is found to be vL;Ed ¼ 810 kN/m. The combination factor 0 for effects of temperature recommended in Annex A2 of EN 1990 is 0.6, with a Note that it ‘may in most cases be reduced to 0 for ultimate limit states EQU, STR, and GEO’. Temperature effects are not considered here. The primary and secondary effects of shrinkage are both beneficial, and so are not considered. With PRd ¼ 83.3 kN/stud, the resistance of groups of three at 250 mm spacing is 1000 kN/m, which is sufficient. Longitudinal shear in the concrete is adequate by inspection, as the shear flow is less than that at the internal support but both concrete slab thickness and reinforcement are the same. The studs in all three cases above should also be checked at serviceability according to clause 6.8.1(3). This has not been done here.

Example 6.11: influence of in-plane shear in a compressed flange on bending resistances of a beam The mid-span region of an internal beam in the main span of the bridge is studied. Its cross-section is shown in Fig. 6.22. The imposed sagging bending moments are as in Example 6.10(ii): MEd;a ¼ 2.0 MNm, MEd;c ¼ 2.0 MNm short term plus 4.0 MNm long term. Shear flow is reduced by creep, so the short-term modular ratio, n0 ¼ 6.36, is used for all loading, and shrinkage effects are ignored. The section is in Class 2, but the full plastic resistance moment, Mpl;Rd , is unlikely to be needed. Properties found earlier are as follows: .

134

from Example 6.1, Mpl;Rd ¼ 10 050 kNm


. .

from Fig. 6.2, at Mpl;Rd , depth of slab in compression ¼ 227 mm from Example 6.10, Mel;Rd ¼ 6858 kNm.

Elastic properties of the cross-section are given in Table 6.4. Situations are studied in which a layer of thickness hv at the bottom of the slab is needed for the diagonal struts of the truss model for in-plane shear. The bending resistance is found using only the remaining thickness, hm ¼ 250 hv (mm units). This is ‘the depth of compression considered in the bending assessment’ to which clause 6.2.4(105) of EN 1992-2 could be assumed to refer. As discussed in the main text, it was only intended to refer to transverse bending moment. The haunch is included in the elastic properties, and neglected in calculations for Mpl;Rd . MRd, MNm 10 Mpl,Rd

8

Tension in the steel bottom flange governs

6 Mel,Rd Compression at the top of the slab governs

4

0

50

100

150

200

250

hm (mm)

Fig. 6.42. Influence of thickness of slab in compression on resistance to bending

The effect on the bending resistances of reducing the flange thickness from 250 mm to hm is shown in Fig. 6.42. Reduction in Mpl;Rd does not begin until hm < 227 mm, and is then gradual, until the plastic neutral axis moves into the web, at hm ¼ 122 mm. When hm ¼ 100 mm, the reduction is still less than 10%. The elastic resistance is governed by stress in the bottom flange, and at first increases as hm is reduced. Compressive stress in concrete does not govern until hm is less than 50 mm, as shown. For hm ¼ 100 mm, Mel;Rd is greater than for 250 mm. The vertical shear for which hm ¼ 100 mm, hv ¼ 150 mm is now found and compared with the shear resistance of the steel web, Vbw;Rd ¼ 1834 kN from Example 6.4. The method is explained in comments on clause 6.6.6.2(1). The optimum angle f for the concrete struts, which is 458, is used. It is assumed that the transverse reinforcement does not govern. The property A z=I should continue to be based on the full slab thickness, not on hm , to provide a margin for over-strength materials, inelastic behaviour, etc. Equation (6.22) in EN 1992-1-1 is vEd < fcd sin f cos f . From Example 6.10(i) with fck ¼ 30 N/mm2 , f ¼ 458; this gives vEd < 5.28 N/mm2 . From Fig. 6.41, the width of flange excluded by the shear plane a–a is: ð3100 450Þ=2 ¼ 1325 mm The shear flow on this plane is: vL;Ed ¼ vEd hv ¼ ð1325=3100ÞVEd ðA z=IÞ

ðD6:25Þ

From equation (D6.25) with hv ¼ 150 mm and A z=I from Table 6.4, VEd < 5:28 150 ð3100=1325Þ=0:8024 ¼ 2309 kN This limit is 26% above the shear resistance of the steel web. Evidently, interaction between in-plane shear and bending resistance is negligible for this cross-section.

135


6.7. Composite columns and composite compression members 6.7.1. General Scope

Clause 6.7.1(1)P

Clause 6.7.1(2)P

Clause 6.7.1(3) Clause 6.7.1(4)

Clause 6.7.1(7)

136

A composite column is defined in clause 1.5.2.5 as ‘a composite member subjected mainly to compression or to compression and bending’. The title of clause 6.7 includes ‘compression members’ to make clear that its scope is not limited to vertical members, but includes, for example, composite members in trusses. Composite columns are more widely used in buildings than in bridges, so their treatment here is less detailed than in the Designers’ Guide to EN 1994-1-1.5 Its Example 6.10 on a concrete-encased I-section column is supplemented here by Example 6.12 on a concretefilled steel tube. In this Guide, references to ‘columns’ include other composite compression members, unless noted otherwise, and ‘column’ means a length of member between adjacent lateral restraints. The concept of the ‘effective length’ of a column is not used in clause 6.7. Instead, the ‘relative slenderness’ is defined, in clause 6.7.3.3(2), in terms of Ncr , ‘the elastic critical normal force for the relevant buckling mode’. This use of Ncr is explained in the comments on clause 6.7.3.3. Clause 6.7.1(1)P refers to Fig. 6.17, in which all the cross-sections shown have double symmetry; but clause 6.7.1(6) makes clear that the scope of the general method of clause 6.7.2 includes members of non-symmetrical section.87 Clause 6.7 is not intended for application to members subjected mainly to transverse loading and also resisting longitudinal compression, such as longitudinal beams in an integral bridge. These are treated in this Guide in the comments on beams. The bending moment in a compression member depends on the assumed location of the line of action of the axial force, N. Where the cross-section has double symmetry, as in most columns, this is taken as the intersection of the axes of symmetry. In other cases the choice, made in the modelling for global analysis, should be retained for the analysis of the cross-sections. A small degree of asymmetry (e.g. due to an embedded pipe) can be allowed for by ignoring in calculations concrete areas elsewhere, such that symmetry is restored. No shear connectors are shown in the cross-sections in Fig. 6.17, because within a column length, the longitudinal shear is normally much lower than in a beam, and sufficient interaction may be provided by bond or friction. Shear connectors are normally required for load introduction, following clause 6.7.4. Where the design axial compression is less than Npm;Rd , shown in Fig. 6.19 and Fig. 6.47, it is on the safe side to ignore it in verification of cross-sections. Where there is moderate or high transverse shear, shear connectors may be needed throughout the member. Example 6.11 of Ref. 5 is relevant. The strengths of materials in clause 6.7.1(2)P are as for beams, except that class C60/75 and lightweight-aggregate concretes are excluded. For these, additional provisions (e.g. for creep, shrinkage and strain capacity) would be required.88;89 Clause 6.7.1(3) and clause 5.1.1(2) both concern the scope of EN 1994-2, as discussed above. The steel contribution ratio, clause 6.7.1(4), is the proportion of the squash load of the section that is provided by the structural steel member. If it is outside the limits given, the member should be treated as reinforced concrete or as structural steel, as appropriate.

Independent action effects Clause 6.7.1(7) relates to the N–M interaction curve for a cross-section of a column shown in Fig. 6.19 and as a polygon in Fig. 6.47. It applies where the factored axial compression F NEk is less than Npm;Rd /2, so that reduction in NEd reduces MRd: As this could be unsafe where NEd and MEd result from independent actions, the factor F for NEk is reduced, as illustrated in Fig. 6.34 of Ref. 5.


The ‘bulge’ in the interaction curve is often tiny, as shown in Fig. 6.47. A simpler and more conservative rule, that ignores the bulge, was given in ENV 1994-1-1. It is that if MRd corresponding to F NEk is found to exceed Mpl;Rd , MRd should be taken as Mpl;Rd . It is applicable unless the bending moment MEd is due solely to the eccentricity of the force NEd . It is doubtful if the 20% rule of clause 6.7.1(7) was intended to be combined with the reduction of F from 1.35 to 1.0 for a permanent action with a relieving effect. Where that is done, use of the simpler rule given above is recommended (e.g. in Fig. 6.47, to replace boundary BDC by BC).

Local buckling The principle of clause 6.7.1(8)P is followed by application rules in clause 6.7.1(9). They ensure that the concrete (reinforced in accordance with clause 6.7.5) restrains the steel and prevents it from buckling even when yielded. Columns are, in effect, treated in clause 6.7 as Class 2 sections. Restraint from the concrete enables the slenderness limits for Class 2 to be increased to the values given in Table 6.3. For example, the factor 90 given for a circular hollow section replaces 70 in EN 1993-1-1. Members in Class 3 or 4 are outside the scope of clause 6.7.


Fatigue Verification of columns for fatigue will rarely be needed, but fatigue loading could occur in composite members in a truss or in composite columns in integral bridges. Verification, if required, should be to clause 6.8.

6.7.2. General method of design The ‘general method’ of clause 6.7.2 is provided for members outside the scope of the simplified method of clause 6.7.3, and also to enable advanced software-based methods to be used. It is more a set of principles than a design method. Writing software that satisfies them is a task for specialists. Much of clause 6.7.3 and the comments on it provide further guidance on design of compression members that are outside its scope. Clause 6.7.2(3)P refers to ‘internal forces’. These are the action effects within the compression member, found from global analysis to Section 5 that includes global and local imperfections and second-order effects: Clause 6.7.2(3)P also refers to ‘elasto-plastic analysis’. This is defined in clause 1.5.6.10 of EN 1990 as ‘structural analysis that uses stress/strain or moment/curvature relationships consisting of a linear elastic part followed by a plastic part with or without hardening’. As the three materials in a composite section follow different non-linear relationships, direct analysis of cross-sections is not possible. One has first to assume the dimensions and materials of the member, and then determine the axial force N and bending moment M at a cross-section from assumed values of axial strain and curvature , using the relevant material properties. The M–N– relationship for each section can be found from many such calculations. This becomes more complex where biaxial bending is present.90 Integration along the length of the member then leads to a non-linear member stiffness matrix that relates axial force and end moments to the axial change of length and the end rotations. Clause 6.7.2(8) on stress–strain curves was drafted, as a ‘General’ rule, before clause 5.7 of EN 1992-2 was available and refers only to the Parts 1-1 of Eurocodes 2 and 3. At that time these rules appeared to be incompatible for use for composite structures. Hence, no application rules on non-linear global analysis are given in clause 5.4.3, where further comment is given. In clause 5.7 of EN 1992-2, the intention is that realistic stiffnesses, not design stiffnesses, should be used, on the basis that a small amount of material at the critical section with ‘design’ properties will not alter the overall response. For bridges, the recommended stress– strain curves are based on the characteristic strengths. This is consistent with Informative

Clause 6.7.2

Clause 6.7.2(3)P

Clause 6.7.2(8)

137


Annex C of EN 1993-1-5, for structural steel. Both Eurocodes 2 and 3 refer to their national annexes for this subject. In the absence of references in EN 1994-2 to these Parts of Eurocodes 2 and 3, guidance should be sought from the National Annex. Where characteristic properties are used in non-linear global analysis, further checks on cross-sections are required. An attractive proposition therefore is to use design values of material properties throughout, so that the non-linear analysis itself becomes the verification, provided that the resistance found exceeds the factored loading. This approach is permitted by clause 5.7 of EN 1992-2. However, it may not be conservative for serviceability limit states if significant internal forces arise from indirect actions such that greater stiffness attracts greater internal effects. There is a caveat to this effect in clause 5.7 of EN 1992-2.

6.7.3. Simplified method of design Scope of the simplified method Clause 6.7.3.1 Clause 6.7.3.1(1)

Clause 6.7.3.1(2)

Clause 6.7.3.1(3)

Clause 6.7.3.1(4)

Clause 6.7.3.2(1)

Clause 6.7.3.2(2)

138

The method has been calibrated by comparison with test results.91;92 Its scope, clause 6.7.3.1, is limited mainly by the range of results available, which leads to the restriction 2 in clause 6.7.3.1(1). For most columns, the method requires explicit account to be taken of imperfections and second-order effects. The use of strut curves is limited in clause 6.7.3.5(2) to axially-loaded members. The restriction on unconnected steel sections in paragraph (1) is to prevent loss of stiffness due to slip, that would invalidate the formulae for EI of the column cross-section. The limits to concrete cover in clause 6.7.3.1(2) arise from concern over strain softening of concrete invalidating the interaction diagram (Fig. 6.19), and from the limited test data for columns with thicker covers. These provisions normally ensure that for each axis of bending, the flexural stiffness of the steel section makes a significant contribution to the total stiffness. Greater cover can be used by ignoring in calculation the concrete that exceeds the stated limits. The limit of 6% in clause 6.7.3.1(3) on the reinforcement used in calculation is more liberal than the 4% (except at laps) recommended in EN 1992-1-1. This limit and that on maximum slenderness are unlikely to be restrictive in practice. Clause 6.7.3.1(4) is intended to prevent the use of sections susceptible to lateral–torsional buckling.

Resistance of cross-sections Reference to the partial safety factors for the materials is avoided by specifying resistances in terms of design values for strength, rather than characteristic values; for example in equation (6.30) for plastic resistance to compression in clause 6.7.3.2(1). This resistance, Npl;Rd , is the design ultimate axial load for a short column, assuming that the structural steel and reinforcement are yielding and the concrete is crushing. For concrete-encased sections, the crushing stress is taken as 85% of the design cylinder strength, as explained in the comments on clause 3.1. For concrete-filled sections, the concrete component develops a higher strength because of the confinement from the steel section, and the 15% reduction is not made; see also the comments on clause 6.7.3.2(6). Resistance to combined compression and bending The bending resistance of a column cross-section, Mpl;Rd , is calculated as for a composite beam in Class 1 or 2, clause 6.7.3.2(2). Points on the interaction curve shown in Figs 6.18 and 6.19 represent limiting combinations of compressive axial load N and moment M which correspond to the plastic resistance of the cross-section. The resistance is found using rectangular stress blocks. For simplicity, that for the concrete extends to the neutral axis, as shown in Fig. 6.43 for resistance to bending (point B in Fig. 6.19 and Fig. 6.47). As explained in the comments on clause 3.1(1), this simplification is unconservative in comparison with stress/strain curves for concrete and the rules of EN 1992-1-1. To compensate for this, the plastic resistance moment for the cross-section is reduced by a factor M in clause 6.7.3.6(1).


0.85fcd

fsd

fyd

–

Mpl,Rd

+ fyd Concrete

+ fsd

Steel

Reinforcement

Fig. 6.43. Stress distributions for resistance in bending

As axial compression increases, the neutral axis moves; for example, towards the lower edge of the section shown in Fig. 6.43, and then outside the section. The interaction curve is therefore determined by moving the neutral axis in increments across the section, and finding pairs of values of M and N from the corresponding stress blocks. This requires a computer program, unless the simplification given in clause 6.7.3.2(5) is used. Simplified expressions for the coordinates of points B, C and D on the interaction curve are given in Appendix C of Ref. 5. Further comment is given in Examples 6.10 and C.1 in that Guide and in Example 6.12 here.

Influence of transverse shear Clauses 6.7.3.2(3) and (4), on the influence of transverse shear on the interaction curve, are generally the same as clause 6.2.2.4 on moment–shear interaction in beams. One assumes first that the shear VEd acts on the structural steel section alone. If it is less than 0.5Vpl;a;Rd , it has no effect on the curve. If it is greater, there is an option of sharing it between the steel and reinforced concrete sections, which may reduce that acting on the steel to below 0.5Vpl;a;Rd . If it does not, then a reduced design yield strength is used for the shear area, as for the web of a beam. In a column the shear area depends on the plane of bending considered, and may consist of the flanges of the steel section. It is assumed that shear buckling does not occur. Shear high enough for Vc;Rd to be relied on in design is unlikely to occur in a composite column, so the code does not go into detail here. The reference in clause 6.7.3.2(3) to the use of EN 1992 does not include EN 1992-2, where clause 6.2 was not drafted with columns in mind. Equation (6.2.b) in EN 1992-1-1, which gives a minimum shear strength for concrete regardless of reinforcement content, is not valid for unreinforced concrete as it assumes that minimum reinforcement will be provided according to EN 1992 requirements. It can be inferred that for a concrete-filled tube with no longitudinal reinforcement (permitted by clause 6.7.5.2(1)), the shear resistance Vc;Rd according to EN 1992 should be taken as zero. Simplified interaction curve Clause 6.7.3.2(5) explains the use of the polygonal diagram BDCA in Fig. 6.19 as an approximation to the interaction curve, suitable for hand calculation. The method applies to any cross-section with biaxial symmetry, not just to encased I-sections. First, the location of the neutral axis for pure bending is found, by equating the longitudinal forces from the stress blocks on either side of it. Let this be at distance hn from the centroid of the uncracked section, as shown in Fig. 6.19(B). It is shown in Appendix C of Ref. 5 that the neutral axis for point C on the interaction diagram is at distance hn on the other side of the centroid, and the neutral axis for point D passes through the centroid. The values of M and N at each point are easily found from the stress blocks shown in Fig. 6.19. For concrete-filled steel tubes the factor 0.85 may be omitted. Concrete-filled tubes of circular or rectangular cross-section Clause 6.7.3.2(6) is based on the lateral expansion that occurs in concrete under axial compression. This causes circumferential tension in the steel tube and triaxial compression in the concrete. This increases the crushing strength of the concrete91 to an extent that

Clause 6.7.3.2(3) Clause 6.7.3.2(4)

Clause 6.7.3.2(5)

Clause 6.7.3.2(6)

139


outweighs the reduction in the effective yield strength of the steel in vertical compression. The coefficients a and c given in this clause allow for these effects. This containment effect is not present to the same extent in concrete-filled rectangular tubes because less circumferential tension can be developed. In all tubes the effects of containment reduce as bending moments are applied, because the mean compressive strain in the concrete and the associated lateral expansion are reduced. With increasing slenderness, bowing of the member under load increases the bending moment, and therefore the effectiveness of containment is further reduced. For these reasons, a and c are dependent on the eccentricity of loading and on the slenderness of the member.

Properties of the column or compression member For composite compression members in a frame, some properties of each member are needed before or during global analysis of the frame: Clause Clause Clause Clause

6.7.3.3(1) 6.7.3.3(2) 6.7.3.3(3) 6.7.3.3(4)

. . . .

the steel contribution ratio, clause 6.7.3.3(1) the relative slenderness , clause 6.7.3.3(2) the effective flexural stiffnesses, clauses 6.7.3.3(3) and 6.7.3.4(2), and the creep coefficient and effective modulus for concrete, clause 6.7.3.3(4).

The steel contribution ratio is explained in the comments on clause 6.7.1(4). The relative slenderness is needed to check that the member is within the scope of the simplified method, clause 6.7.3.1(1). Often it will be evident that < 2. The calculation can then be omitted, as is not needed again unless the member resists axial load only. The unfactored quantities E, I and L are used in the calculation of Ncr , so is calculated using in equation (6.39) the characteristic (unfactored) value of the squash load, Npl;Rk , and the characteristic flexural stiffness (EI)eff from clause 6.7.3.3(3). This is the only use of this stiffness in Section 6. The upper limit on is somewhat arbitrary and does not justify great precision in Ncr . Creep of concrete increases the lateral deformation of the member. This is allowed for by replacing the elastic modulus Ecm (in equation (6.40)) by a reduced value Ec;eff from equation (6.41). This depends on the creep coefficient t , which is a function of the age at which concrete is stressed and the duration of the load. The effective modulus depends also on the proportion of the design axial load that is permanent. The design of the member is rarely sensitive to the influence of the creep coefficient on Ec;eff , so conservative assumptions can be made about uncertainties. Normally, a single value of effective modulus can be used for all compression members in a structure. Further discussion is given under clause 5.4.2.2. The correction factor Ke is to allow for loss of stiffness caused by possible cracking of concrete. The condition for ignoring second-order effects within the member is explained in comments on clause 5.2.1(3). Where the ratio cr (¼ Ncr /NEd ) is used, the critical load Ncr is the axial force in the member in the lowest buckling mode of the structure that involves the member. In the rare cases where both ends of a column are detailed so as to behave as pin-ended (as in Example 6.12), Ncr ¼ 2 ðEIÞeff;II /L2 . The flexural stiffness (EI)eff;II is obtained from clause 6.7.3.4(2). In continuous construction, the critical buckling mode involves adjacent members, which must be included in the elastic critical analysis.

Clause 6.7.3.4(1)

140

Analyses for verification of a compression member For the compression members of a frame or truss that are within the scope of clause 6.7.3, a flow chart for calculation routes is given in Fig. 6.44. The relationship between the analysis of a frame and the stability of individual members is discussed both in the comments on clause 5.2.2 and below. If bending is biaxial, the procedure in clause 6.7.3.4 is followed for each axis in turn. Clause 6.7.3.4(1) requires the use of second-order linear-elastic global analysis except where the option of clause 6.7.3.4(5) applies, or route (c) in Fig. 6.44 is chosen and cr > 10 in accordance with clause 6.7.3.4(3). The simplified method of clause 6.7.3.5(2)


Flow chart for global analysis (g.a.) and verification of a compression member in a composite frame, with reference to global and member imperfections (g.imp and m.imp). This is for a member of doubly symmetrical and uniform cross-section (6.7.3.1(1)) and for a particular loading. See Notes 1 and 2. Find αcr to 5.2.1(3), using (EI)eff,II to 6.7.3.4(2) for all compression members by one of methods (a) to (c):

(a) 2nd order g.a. for proportional loading with g.imp and m.imp. Find ratio αcr of asymptotic load to design load. Is αcr ≥ 10? Yes 1st order g.a. with g.imp and m.imp is permitted, but not needed, as results from (a) for the design loading can be used. (Use of 1st order could be more economic)

(b) As box (a) but m.imp not included. Is αcr ≥ 10?

No Use results from (a) for the design loading

Yes

(c) Elastic critical analysis with proportional loading and vertical loads only, with no imp., to find αcr. See Note 3. Is αcr ≥ 10? No

Use 2nd order analysis as above. Allow for 1st and 2nd order m.imp to 6.7.3.4(5)

Yes Do 1st order g.a. with g.imp, and m.imp to 6.7.3.4(4)

No Go to box (a) or (b), above, except that αcr need not be found again. See Note 4

Do 1st order g.a. with g.imp, and m.imp to 6.7.3.4(4)

Analysis complete. Verify resistance of cross-sections of each compression member, to 6.7.3.6 or 6.7.3.7 (END) Note 1. ‘Loading’ means a particular combination of actions, load case and load arrangement. In boxes (a) to (c) the lowest αcr for various loadings is found. The chosen loadings should include that for maximum side-sway, and those that are expected to cause the greatest axial compression in each potentially critical compression member. Note 2. Analysis (a) includes both P–Δ effects from global imperfections and P–δ effects from member imperfections. Note 3. For choice of loadings, see Note 1 and the comments on clause 5.2.1(3) and (4). Note 4. No need to return to (a) or (b) where αcr < 10 only in a local member mode (pin-ended conditions). Then, do first-order g.a. with amplification to 6.7.3.4(5) and verify cross-sections.

Fig. 6.44. Flow chart for analysis and verification for a compression member

is rarely applicable in practice because some first-order bending moment (other than from imperfections) will usually be present; for example, due to friction at bearings. Clause 6.7.3.4(2) gives the design flexural stiffnesses for compression members, for use in all analyses for ultimate limit states. The factor Ke;II allows for cracking, as is required by the reference in clause 5.4.2.3(4) to clause 6.7.3.4. The factor K0 is from research-based calibration studies. Long-term effects are allowed for, as before, by replacing Ecm in equation (6.42) by Ec;eff from equation (6.41). In clause 6.7.3.4(3), ‘the elastic critical load ’ refers to the frame at its lowest buckling mode involving the member concerned; and ‘second-order effects’ means those in the member due to both its own imperfections and global imperfections. When deciding whether second-order effects of member imperfections can be neglected, the effects of global imperfections can be neglected in an elastic critical analysis (route (c) in Fig. 6.44). A second-order analysis for the asymptotic load, route (a), will give the same value for cr whether global imperfections are included or not. They are shown in Fig. 6.44 as included because the same analysis can then give the design bending moments for the member concerned. Clause 6.7.3.4(4) gives the equivalent member imperfection, for use in a global analysis, as an initial bow. It is proportional to the length L of the member between lateral restraints and is defined by e0 , the lateral departure at mid-height of its axis of symmetry from the line joining the centres of symmetry at its ends. The value accounts principally for truly geometric imperfections and for the effects of residual stresses. It is independent of the distribution of bending moment along the member. The curved shape is usually assumed to be sinusoidal, but a circular arc is acceptable. The curve is assumed initially to lie in the plane normal to the axes of the bending moments.

Clause 6.7.3.4(2)

Clause 6.7.3.4(3)

Clause 6.7.3.4(4)

141


M1st,Ed/M20% 0

M1

M20%

M20%/10

NEde0 + Mtr,Ed

0

2.0

1.0 0.6L r (=M2/M1) M2 = rM1

Second-order effect

(a)

0 –1

Second-order effect (b)

–0.5

0

(c)

Fig. 6.45. Bending moments in a column: (a) end moments from global analysis; (b) initial imperfections and transverse loading; and (c) equivalent first-order bending moment

Clause 6.7.3.4(5)

Where member imperfections are not included in the global analysis and cr < 10, clause 6.7.3.4(5) enables these imperfections to be allowed for. It is based on the critical load Ncr;eff for the isolated pin-ended member even where the critical buckling mode for the frame involves sway, such that the effective length of the member exceeds its system length. This is consistent with clause 5.2.2(7)(b) of EN 1993-1-1, which is referred to from clause 5.2.2(1) via clause 5.2.2 of EN 1993-2. (This route also leads to clause 5.2.2(3) of EN 1993-1-1, which sets out these options in detail and is consistent with Fig. 6.44.) The reason for this definition for Ncr;eff is that where necessary (e.g. where cr < 10), the effects of global imperfections and side-sway have been accounted for in the second-order global analysis. This can be seen by using as an example a ‘flagpole’-type column, with both out-of-plumb and initial bow.

Determination of design bending moment for a compression member It is now assumed that for a particular member, with or without lateral load within its length, an analysis in which member imperfections were not included has provided a design axial force NEd and design end moments for one of the principal planes of bending. These are denoted M1 and M2 , with jM1 j jM2 j, as shown in Fig. 6.45(a). Biaxial bending is considered later. The axial force is normally almost constant along the length of the member. Where it varies, its maximum value can conservatively be assumed to be present throughout its length. The factor k in clause 6.7.3.4(5) is proportional to . For the end moments, from Table 6.4, lies between 0.44 and 1.1, but for the member imperfection it is always 1.0. These two values are denoted 1 and 2 . The calculation of 1 is now explained, assuming that the critical bending moment occurs either at the end where MEd ¼ M1 (where no member imperfection or resulting second-order effect is assumed) or within the central 20% of the length of the member. Except where there is lateral loading, the maximum first-order bending moment within this central length is: M20% ¼ M1 ð0:6 þ 0:4rÞ shown in Fig. 6.45(a). This is represented by an ‘equivalent’ first-order design value, given by Table 6.4 as: M1st;Ed ¼ M1 ð0:66 þ 0:44rÞ with a lower limit of 0.44M1 . The ratio M1st;Ed =M20% is shown in Fig. 6.45(c). It is generally 1.1, but increases sharply where r < 0:5, which is where the lower limit of 0.44M1 is reached. This range of r represents significant double-curvature bending. The increase provides protection against snap-through to single-curvature buckling. The moment M1st;Ed is increased by the factor 1 1 NEd =Ncr;eff

142


to allow for second-order effects. This factor is an approximation, as shown in Fig. 2.9 of Ref. 93, which is allowed for by the use of the ratio 1.1 shown in Fig. 6.45(c). Basing Ncr;eff on pin-ended conditions can be conservative where the column is braced with end rotational restraints that produce an effective length less than the column height. Where there is lateral loading within the column length, the bending-moment distribution should be treated as the sum of two distributions, one corresponding to each of the two parts of Table 6.4. In the first half of the table, MEd is then the sum of contributions from member imperfection and lateral load. These are not necessarily in the same direction, because the member imperfection e0 can be in any lateral direction and must be chosen to give the most adverse overall result for the column. Equation (6.43) states that k must be greater than or equal to unity, and this is correct for a single distribution of bending moment. However, for a combination of two distributions, it could be conservative to adjust both values of k in this way when the two sets of moments are treated separately. At mid-length the component due to end moments depends on their ratio, r, and therefore could be small. The appropriate first component is: 1 M1 1 NEd =Ncr;eff without the condition that it should exceed M1 . The imperfection e0 causes a first-order bending moment NEd e0 at mid-length. The resulting contribution to MEd;max is: 2 NEd e0 ; 1 NEd =Ncr;eff

with 2 ¼ 1

This, plus the contribution from any mid-length moment from lateral loading, is added to the first component. The condition k 1 applies to the sum of the two components, and is intended to ensure that the design moment is at least equal to the greatest first-order moment. In biaxial bending, the initial member imperfection may be neglected in the less critical plane (clause 6.7.3.7(1)). The definitions of MEd in clause 6.7.3.4(5) and Table 6.4 may appear contradictory. In the text before equation (6.43), MEd is referred to as a first-order moment. This is because it does not include second-order effects arising within the member. However, Table 6.4 makes clear that MEd is to be determined by either first-order or second-order global analysis, as shown in Fig. 6.44. The simplified method of clause 6.7.3.5 is rarely applicable, as explained in the comment on clause 6.7.3.4(1), so no comment on it is given.

Verification of cross-sections For uniaxial bending, the final step is to check that the cross-section can resist Mmax;Ed with compression NEd . The interaction diagram gives a resistance d Mpl;Rd with axial load NEd , as shown in Fig. 6.18. This is unconservative, being based on rectangular stress-blocks, as explained in the comment on clause 3.1(1), so in clause 6.7.3.6(1) it is reduced, by the use of a factor M that depends on the grade of structural steel. This factor allows for the increase in the compressive strain in the cross-section at yield of the steel (which is adverse for the concrete), when the yield strength of the steel is increased. Where values of Mmax;Ed have been found for both axes, clause 6.7.3.7 on biaxial bending applies, in which they are written as My;Ed and Mz;Ed . If one is much greater than the other, the relevant check for uniaxial bending, equation (6.46), will govern. Otherwise, the linear interaction given by equation (6.47) applies. If the member fails this biaxial condition by a small margin, it may be helpful to recalculate the less critical bending moment omitting the member imperfection, as permitted by clause 6.7.3.7(1).

Clause 6.7.3.5

Clause 6.7.3.6(1)

Clause 6.7.3.7

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6.7.4. Shear connection and load introduction Load introduction The provisions for the resistance of cross-sections of columns assume that no significant slip Clause 6.7.4.1(1)P occurs at the interface between the concrete and structural steel components. Clauses Clause 6.7.4.1(2)P 6.7.4.1(1)P and (2)P give the principles for limiting slip to an insignificant level in the critical regions: those where axial load and/or bending moments are applied to the member. For any assumed ‘clearly defined load path’ it is possible to estimate stresses, including shear at the interface. Shear connection is unlikely to be needed outside regions of load introduction unless the shear strength Rd from Table 6.6 is very low, or the member is also acting as a beam, or has a high degree of double-curvature bending. Clause Clause 6.7.4.1(3) 6.7.4.1(3) refers to the special case of an axially loaded column. Where axial load is applied through a joint attached only to the steel component, the force to be transferred to the concrete can be estimated from the relative axial loads in the two materials given by the resistance model. Accurate calculation is rarely practicable where the cross-section concerned does not govern the design of the column. In this partlyplastic situation, the more adverse of the elastic and fully-plastic models give a safe result Clause 6.7.4.2(1) (clause 6.7.4.2(1), last line). In practice, it may be simpler to provide shear connection based on a conservative (high) estimate of the force to be transferred. Where axial force is applied by a plate bearing on both materials or on concrete only, the proportion of the force resisted by the concrete gradually decreases, due to creep and shrinkage. It could be inferred from clause 6.7.4.2(1) that shear connection should be provided for a high proportion of the force applied. However, models based on elastic theory are over-conservative in this inherently stable situation, where large strains are acceptable. The application rules that follow are based mainly on tests. Few shear connectors reach their design shear strength until the slip is at least 1 mm; but this is not significant slip for a resistance model based on plastic behaviour and rectangular stress blocks. However, a long load path implies greater slip, so the assumed path should not Clause 6.7.4.2(2) extend beyond the introduction length given in clause 6.7.4.2(2). In a concrete-filled tube, shrinkage effects are low, for only the autogenous shrinkage strain occurs, with a long-term value below 104 , from clause 3.1.4(6) of EN 1992-1-1. Radial shrinkage is outweighed by the lateral expansion of concrete in compression, for its inelastic Poisson’s ratio increases at high compressive stress, and eventually exceeds 0.5. Friction then provides significant shear connection. Concrete-filled tubular columns with bearings at both ends have found application in bridge design. Where the whole load is applied to the concrete core through an end plate, Clause 6.7.4.2(3) the conditions of clause 6.7.4.2(3) can be satisfied, and no shear connection is required. This complete reliance on friction for shear transfer is supported by test evidence94;95 and by inelastic theory. For columns of circular cross-section, no plausible failure mechanism has been found for an end region that does not involve yielding of the steel casing in hoop tension and vertical compression. For a large non-circular column, it would be prudent to check behaviour by finite-element analysis. There is further discussion in Example 6.12. Friction is also the basis for the enhanced resistance of stud connectors given in Clause 6.7.4.2(4) clause 6.7.4.2(4). Detailing at points of load introduction or change of cross-section is assisted by the high Clause 6.7.4.2(5) permissible bearing stresses given in clauses 6.7.4.2(5) and (6). An example is given in Ref. 5 Clause 6.7.4.2(6) in which the local design compressive strength of the concrete, c;Rd in equation (6.48), is found to be 260 N/mm2 . Clause 6.7.4.2(7) Clause 6.7.4.2(7) relates to load introduction to reinforcement in a concrete-filled tube. This and some other concessions made at the ends of a column length are based mainly on tests on columns of sizes typical of those used in buildings. Some caution should be exercised in applying them to members with much larger cross-sections. Unless a column is free to sway, a hinge forms in its central region before it fails. End regions that are slightly weaker have little effect on the failure load, because at that stage their bending moments are lower than at mid-length.

144


The absence from clause 6.7.4.2(8) of a reference to EN 1992-2 is deliberate, as its clause 9.5.3 gives a rule that is not required for composite columns. Figure 6.23 illustrates the requirement of clause 6.7.4.2(9) for transverse reinforcement, which must have a resistance to tension equal to the force Nc1 . If longitudinal reinforcement is ignored, this is given by:

Clause 6.7.4.2(8) Clause 6.7.4.2(9)

Nc1 ¼ Ac2 =ð2nAÞ where A is the transformed area of the cross-section 1-1 of the column in Fig. 6.23, given by: A ¼ As þ ðAc1 þ Ac2 Þ=n and Ac1 and Ac2 are the unshaded and shaded areas of concrete, respectively, in section 1-1. The modular ratio, n, should usually be taken as the short-term value.

Transverse shear Clause 6.7.4.3 gives application rules (used in Example 6.11 in Ref. 5) relevant to the principle of clause 6.7.4.1(2), for columns with the longitudinal shear that arises from transverse shear. The design shear strengths Rd in Table 6.6 are far lower than the tensile strength of concrete. They rely on friction, not bond, and are related to the extent to which separation at the interface is prevented. For example, in partially-encased I-sections, lateral expansion of the concrete creates pressure on the flanges, but not on the web, for which Rd ¼ 0; and the highest shear strengths are for concrete within steel tubes. Where small steel I-sections are present within a column that is mainly concrete, clause 6.7.4.3(4) provides a useful increase to Rd , for covers cz up to 115 mm. The enhancement factor is more simply presented as: c ¼ 0:2 þ cz =50 2:5. Concern about the attachment of concrete to steel in partially-encased I-sections appears again in clause 6.7.4.3(5), because under weak-axis bending, separation tends to develop between the encasement and the web.

Clause 6.7.4.3

Clause 6.7.4.3(4)

Clause 6.7.4.3(5)

6.7.5. Detailing provisions If a steel I-section in an environment in class X0 to EN 1992-1-1 has links in contact with its flange (permitted by clause 6.7.5.2(3)), the cover to the steel section could be as low as 25 mm. For a wide steel flange, this thin layer of concrete would have little resistance to buckling outwards, so the minimum thickness is increased to 40 mm in clause 6.7.5.1(2). This is a nominal dimension. Minimum longitudinal reinforcement, clause 6.7.5.2(1), is needed to control the width of cracks, which can be caused by shrinkage even in columns with concrete nominally in compression. Clause 6.7.5.2(2) does not refer to EN 1992-2 because its clause 9.5 introduces a nationallydetermined parameter, which is not needed for composite columns. Clause 6.7.5.2(4) refers to exposure class X0 of EN 1992-1-1. This is a ‘very dry’ environment, with ‘no risk of corrosion or attack’, so this clause is unlikely to be applicable in bridges.

Clause 6.7.5.1(2) Clause 6.7.5.2(1)

Clause 6.7.5.2(2) Clause 6.7.5.2(4)

Example 6.12: concrete-filled tube of circular cross-section This Example is based on columns used for a motorway interchange, as described in Refs 94 and 96. The column shown in Fig. 6.46 is of circular cross-section. It has spherical rocker bearings at each end, of radii 600 mm (internal) and 605 mm (external). Its length between centres of bearings is 11.5 m. Its design will be checked for a design ultimate load NEd ¼ 18.0 MN. The bearing at its lower end is fixed in position. The deformation of the superstructure at ultimate load causes a maximum eccentricity of load of 75 mm. Concerning M for structural steel, clause 5.2.2(7)(a) of EN 1993-1-1 says that ‘no individual stability check for the members according to 6.3 is necessary’ when

145


second-order analysis is used with member imperfections; and clause 6.1(1) of EN 1993-11 then says that M1 is used for ‘instability assessed by member checks’. This appears to permit M0 to be used when second-order analysis and cross-section checks are used, as here, so M0 ¼ 1.0 is used. The properties of the materials, in the usual notation, are as follows. Structural steel: S355; fy ¼ fyd ¼ 355 N/mm2 , based on the use of Table 3.1 of EN 19931-1. (The UK’s National Annex is likely to require the appropriate value to be taken from EN 10025.) From EN 1993, Ea ¼ 210 kN/mm2 . Concrete: C40/50; fck ¼ 40 N/mm2 ; fcd ¼ 40=1:5 ¼ 26:7 N/mm2 . The coarse aggregate is limestone, so from clause 3.1.3(2) of EN 1992-1-1, Ecm ¼ 0:9 35 ¼ 31:5 kN/mm2 ; n0 ¼ 210=31:5 ¼ 6:67.

600

6 mm cement-sand bedding 11 500

35

680

35

Fig. 6.46. Cross-section at each end of concrete-filled tube of Example 6.12

Geometrical properties of the cross-section In the notation of Fig. 6.17(e), d ¼ 750 mm, t ¼ 35 mm, giving d=t ¼ 21:4. The limit, from Table 6.3, is d=t 90ð235=355Þ ¼ 59:6. From clause 6.7.5.2(1), ‘normally no longitudinal reinforcement is necessary’ (unless resistance to fire is required), so As ¼ 0 is assumed. For the concrete: For the steel tube:

area Ac ¼ 3402 ¼ 363 200 mm2 Ic ¼ 3404 =4 ¼ 10 500 106 mm4 area Aa ¼ 3752 363 200 ¼ 78 620 mm2 Ia ¼ 3754 =4 10 500 106 ¼ 5036 106 mm4

Design action effects, ultimate limit state For the most critical load arrangement, global analysis gives these values: NEd ¼ 18:0 MN, of which NG;Ed ¼ 13.0 MN

(D6.26)

MEd;top ¼ 18 0:075 ¼ 1.35 MNm

(D6.27)

As the column has circular symmetry and its imperfection is assumed to lie in the plane of bending, there is no need to consider biaxial bending. For a rocker bearing, the effective length should be taken to the points of contact, and is: L ¼ 11:5 þ 2 0:6 ¼ 12.7 m Depending on the relationship between the rotation and the displacement of the superstructure at the point of support, it can be shown that the points of contact within the two bearings can be such that the ratio of end moments has any value

146


between þ1 and 1. The worst case for the column, assumed here, is: MEd;top ¼ MEd;bot (i.e., r ¼ 1 in Table 6.4). Double-curvature bending (r ¼ 1) causes higher transverse shear force in the column, but it is too low to influence the design.

Creep coefficient From clause 6.7.3.3(4), Ec ¼ Ecm =½1 þ ðNG;Ed =NEd Þ t

(6.41)

The creep coefficient t is ðt; t0 Þ to clause 5.4.2.2. The time t0 is taken as 30 days, and t is taken as ‘infinity’, as creep reduces the stiffness, and hence the stability, of a column. From clause 3.1.4(5) of EN 1992-1-1, the ‘perimeter exposed to drying’ is zero, so that the notional size, h0 ! 1. Assuming ‘inside conditions’ and the use of normal cement, the graphs in Fig. 3.1 of EN 1992-1-1 give:

ð1; 30Þ ¼ 1:4 ¼ t The assumed ‘age at first loading’ has little influence on the result if it exceeds about 20 days. From equation (6.41), Ec;eff ¼ 31:5=½1 þ 1:4ð13=18Þ ¼ 15.7 kN/mm2

(D6.28)

Shrinkage When the upper bearing is placed, the concrete becomes effectively sealed, so its drying shrinkage can be taken as zero. From EN 1992-1-1/3.1.4(6), the autogenous shrinkage is: eca ð1 1Þ ¼ 2:5ð fck 10Þ 106 ¼ 2:5 30 30 106 ¼ 75 106

(D6.29)

Ignoring the restraint from the steel tube, this would cause the 11 m length of column to shorten by only 0.8 mm. Its influence on shear connection is discussed later.

Squash load, elastic critical load, and slenderness From clause 6.7.3.2(1), N pl;Rd ¼ Aa fyd þ Ac fcd ¼ 78:62 0:355 þ 363:2 0:0267 ¼ 27:9 þ 9:70 ¼ 37:6 MN

ðD6:30Þ

From clause 6.7.3.3(1), the steel contribution ratio is: ¼ 27:9=37:6 ¼ 0:74 which is within the limits of clause 6.7.1(4). For to clause 6.7.3.3(2), with C ¼ 1.5, Npl;Rk ¼ 27:9 þ 1:5 9:70 ¼ 42.46 MN

(D6.31)

From clause 6.7.3.3(3), ðEIÞeff ¼ Ea Ia þ Ke Ec;eff Ic ¼ 210 5036 þ 0:6 15:7 10 500 ¼ 1:156 106 kN m2

(6.40)

Hence, Ncr ¼ 2 ðEIÞeff =L2 ¼ 1:156 103 2 =12:72 ¼ 70.74 MN From equation (6.39), ¼ ðNpl;Rk =Ncr Þ0:5 ¼ ð42:46=70:74Þ0:5 ¼ 0.775

147


The non-dimensional slenderness does not exceed 2.0, so clause 6.7.3.1(1) is satisfied. Clause 6.7.3.2(6), on the effect of containment on the strength of concrete, does not apply, as > 0:5. NRd (MN)

Npl,Rd = 37.6 A

30 4.836 20 NEd = 18.0

Mpl,Rd = 6.884

Mmax,Ed = 3149

10

C

7.057

Npm,Rd

9.700 4.85 D

0

2

4

6

B M (MNm) Rd

Fig. 6.47. Interaction polygon for concrete-filled tube

Interaction polygon The interaction polygon corresponding to Fig. 6.19 is determined using formulae that are given and explained in Appendix C of Ref. 5. Clause 6.7.3.2(5) permits it to be used as an approximation to the N–M interaction curve for the cross-section. The data are: h ¼ b ¼ d ¼ 750 mm; r ¼ d=2 t ¼ 375 35 ¼ 340 mm. From result (D6.30), Npl;Rd ¼ 27:9 þ 9:70 ¼ 37.6 MN The equation numbers (C.) refer to those given in Ref. 5. From equation (C.29) with Wps ¼ 0, Wpc ¼

ðb 2tÞðh 2tÞ2 2 3 r ¼ 6803 =4 2 3403 =3 ¼ 52:4 106 mm3 3 4

From equation (C.30): 2 bh 2 ðr þ tÞ3 Wpc ¼ 7503 =4 2 3753 =3 52:4 106 ¼ 17:91 106 mm3 Wpa ¼ 3 4 From equation (C.8), Npm;Rd ¼ Ac fcd ¼ 0:3632 26:7 ¼ 9.70 MN From equation (C.31), hn ¼

Npm;Rd Asn ð2 fsd fcc Þ 9:70 106 ¼ ¼ 71:5 mm 2b fcd þ 4tð2 fyd fcc Þ 2 750 26:7 þ 140ð2 355 26:7Þ

From equation (C.32), Wpc;n ¼ ðb 2tÞh2n ¼ ð750 70Þ 71:52 ¼ 3:474 106 mm3 From equation (C.33), Wpa;n ¼ bh2n Wpc;n ¼ 750 71:52 3:474 106 ¼ 0:358 106 mm3 From equation (C.5), Mmax;Rd ¼ Wpa fyd þ Wpc fcd =2 ¼ 17:91 355 þ 52:4 26:7=2 ¼ 7057 kNm

148


From equation (C.7), Mn;Rd ¼ Wpa;n fyd þ Wpc;n fcd =2 ¼ 0:358 355 þ 3:474 26:7=2 ¼ 173 kNm From equation (C.6), Mpl;Rd ¼ Mmax;Rd Mn;Rd ¼ 7057 173 ¼ 6884 kNm These results define the interaction polygon, shown in Fig. 6.47.

Design maximum bending moment From Table 6.4 with r ¼ 1, ¼ 1, the equivalent first-order bending moment is: M1st;Ed ¼ 1350 1:1 ¼ 1485 kNm From Table 6.5, the equivalent member imperfection is: e0 ¼ L=300 ¼ 12 700=300 ¼ 42.3 mm For NEd ¼ 18 MN, the imperfection moment is 18 42:3 ¼ 761 kNm To check whether second-order moments can be neglected, an effective value of Ncr is required, to clause 6.7.3.4(3). From equation (6.42), ðEIÞeff;II ¼ 0:9ðEa Ia þ 0:5Ec Ic Þ ¼ 0:9 106 ð210 5036 þ 0:5 15:7 10 500Þ ¼ 1:026 1012 kN mm2 Hence, Ncr;eff ¼ 10262 =12:72 ¼ 62.8 MN This is less than 10NEd , so second-order effects must be allowed for. Table 6.4 gives ¼ 1 for the distribution of bending moment due to the initial bow imperfection of the member, so from clause 6.7.3.4(5), the second-order factor is: 1 1 ¼ 1:402 ¼ 1 NEd =Ncr;eff 1 18=62:8 (The factor for the end moments was accounted for in M1st;Ed .) Hence, Mmax;Ed ¼ 1:402ð1485 þ 761Þ ¼ 3149 kNm

Resistance of column From Fig. 6.47 with NEd ¼ 18 MN, Mpl;N;Rd ¼ 4836 kNm. From clause 6.7.3.6(1), the verification for uniaxial bending is: MEd =Mpl;N;Rd 0:9 Here, the ratio is 3149=4836 ¼ 0.65, so the column is strong enough.

Shear connection and load introduction This column is within the scope of clause 6.7.4.2(3), which permits shear connection to be omitted. The significance of this rule is now illustrated, using preceding results. Creep increases shear transfer to the steel. Full-interaction elastic analysis with nL ¼ n0 ð1 þ L t Þ ¼ 17 (clause 5.4.2.2(2)) finds the action effects on the steel to be: Na;Ed ¼ 14.2 MN

and

Ma;Ed ¼ 1.2 MNm

based on MEd ¼ 1.35 MNm near an end of the column. From the rules for shear connection, these transfers would require 90 25 mm studs for the axial force plus 28 for the bending moment, assuming it can act about any horizontal axis. The significance of friction is illustrated by the following elastic analysis assuming that Poisson’s ratio for concrete is 0.5. The bearing stress on the concrete from NEd ¼ 18 MN

149


is 49.6 N/mm2 . Its resulting lateral expansion causes a hoop tensile stress in the steel of 225 N/mm2 and a radial compression in the concrete of 23 N/mm2 – Fig. 6.48. Assuming a coefficient of friction of 0.4, the vertical frictional stress is: Rd ¼ 23 0:4 ¼ 9:2 N/mm2 The shear transfer reduces the compressive stress in the concrete. Using a guessed mean value of Rd ¼ 5 N/mm2 leads to a transfer length for 14.2 MN of 1.33 m. This is less than the introduction length of 2d (¼ 1:50 m here) permitted by clause 6.7.4.2(2). These figures serve only to illustrate the type of behaviour to be expected. In practice, it would be prudent to provide some shear connection; perhaps sufficient for the bending moment. Shrinkage effects are very small. 23 N/mm2 35 23 N/mm2 225 N/mm2

225 N/mm2 680

Fig. 6.48. Radial and hoop stresses near an end of a concrete-filled tube

6.8. Fatigue 6.8.1. General The only complete set of provisions on fatigue in EN 1994-2 is for stud shear connectors. Fatigue in reinforcement, concrete and structural steel is covered mainly by cross-reference to EN 1992 and EN 1993. Commentary will be found in the guides to those codes.3;4 Further cross-reference is necessary to EN 1993-1-9,42 ‘Fatigue’, which gives supplementary guidance and fatigue detail classifications which are not specific to bridges. The fatigue life of steel components subjected to varying levels of repetitive stress can be checked with the use of Miner’s summation. This is a linear cumulative damage calculation for n stress ranges: n X nEi 1:0 N i ¼ 1 Ri

ðD6:32Þ

where nEi is the number of loading cycles of a particular stress range and NRi is the number of loading cycles to cause fatigue failure at that particular stress range. For most bridges, the above is a complex calculation because the stress in each steel component usually varies due to the random passage of vehicles from a spectrum. Details on a road or rail bridge can be assessed using the above procedure if the loading regime is known at design. This includes the weight and number of every type of vehicle that will use each lane or track of the bridge throughout its design life, and the correlation between loading in each lane or track. In general, this will produce a lengthy calculation. As an alternative, clause 9.2 of EN 1993-2 allows the use of simplified Fatigue Load Models 3 and 71, from EN 1991-2, for road and rail bridges respectively. This reduces the complexity of the fatigue assessment calculation. It is assumed that the fictitious vehicle (or train) alone causes the fatigue damage. The calculated stress from the vehicle is then adjusted by factors to give a single stress range which, for N cycles (2 million cycles for structural steel), causes the same damage as the actual traffic during the bridge’s lifetime. This is called the ‘damage equivalent stress range’ and is discussed in section 6.8.4 below. Comments here are limited to the use of the damage equivalent stress method and, hence, a single stress range.

150


The term ‘equivalent constant-amplitude stress range’, defined in clause 1.2.2.11 of EN 1993-1-9, has the same meaning as ‘damage equivalent stress range’, used here and in clause 6.8.5 of EN 1992-1-1 and clause 9.4.1 of EN 1993-2. Fatigue damage is related mainly to the number and amplitude of the stress ranges as seen in expression (D6.32). The peak of the stress range has a secondary influence that can be, and usually is, ignored in practice for peak stresses below about 60% of the characteristic strength. Ultimate loads are higher than peak fatigue loads, and the use of partial safety factors for ultimate-load design normally ensures that peak fatigue stresses are below this limit. This may not be the case for long-span bridges with a high percentage of dead load, so clause 6.8.1(3) specifies a limit to the longitudinal shear force per connector, with a recommended value 0.75PRd , or 0.6PRk for V ¼ 1:25. As fatigue damage to studs may not be evident, some continental countries are understood to be specifying a lower limit, 0.6PRd , in their national annexes. (For welded structural steel, the effect of peak stress is effectively covered in the detail classifications in EN 1993-1-9, where residual stresses from welding, typically reaching yield locally, are catered for in the detail categories.) Most bridges will require a fatigue assessment. Clauses 6.8.1(4) and (5) refer to EN 1993-2 and EN 1992-2 for guidance on the types of bridges and bridge elements where fatigue assessment may not be required. Those relevant to composite bridge superstructures of steel and concrete include: (i) (ii) (iii) (iv) (v)

Clause 6.8.1(3)

Clause 6.8.1(4) Clause 6.8.1(5)

pedestrian footbridges not susceptible to pedestrian-induced vibration bridges carrying canals bridges which are predominantly statically loaded parts of railway or road bridges that are neither stressed by traffic loads nor likely to be excited by wind loads prestressing and reinforcing steel in regions where, under the frequent combination of actions and the characteristic prestress Pk , only compressive stresses occur at the extreme concrete fibres. (The strain and hence the stress range in the steel is typically small while the concrete remains in compression.)

Fatigue assessments are still required in the cases above (with the possible exception of (v)), if bridges are found to be susceptible to wind-induced excitation. The main cause of wind-induced fatigue, vortex shedding, is covered in EN 1991-1-4 and is not considered further here.

6.8.2. Partial factors for fatigue assessment of bridges Resistance factors Mf may be given in National Annexes, so only the recommended values can be discussed here. For fatigue strength of concrete and reinforcement, clause 6.8.2(1) refers to EN 1992-1-1, which recommends the partial factors 1.5 and 1.15, respectively, for both persistent and transient design situations. For structural steel, EN 1993-1-9, Table 3.1 recommends values ranging from 1.0 to 1.35, depending on the design concept and consequence of failure. These apply, as appropriate, for a fatigue failure of a steel flange caused by a stud weld. The choice of design concept and the uncertainties covered by Mf are discussed in Ref. 4. Fatigue failure of a stud shear connector, not involving the flange, is covered by EN 1994-2. The recommended value of Mf;s for fatigue of headed studs is given as 1.0 in a Note to clause 2.4.1.2(6) in the general rules of EN 1994. This is the value in EN 1993-1-9 for the ‘damage tolerant’ assessment method with ‘low consequence of failure’. From clause 3(2) of EN 1993-1-9, the use of the damage tolerant method should be satisfactory, provided that ‘a prescribed inspection and maintenance regime for detecting and correcting fatigue damage is implemented . . .’. A Note to this clause states that the damage tolerant method may be applied where ‘in the event of fatigue damage occurring a load redistribution between components of structural elements can occur’. The first condition does not apply to stud connectors, as lack of access prevents detection of small cracks by any simple method of inspection. For that situation, EN 1993-1-9

Clause 6.8.2(1)

151


recommends use of the ‘safe life’ method, with Mf ¼ 1:15 for ‘low consequence of failure’. The second condition does apply to stud connectors, and the value Mf;s ¼ 1:0 is considered to be appropriate for studs in bridges. Relevant considerations are as follows. Fatigue failure results from a complex interaction between steel and concrete, commencing with powdering of the highly stressed concrete adjacent to the weld collar. This displaces upwards the line of action of the shear force, increasing the bending and shear in the shank just above the weld collar, and probably also altering the tension. Initial fatigue cracking further alters the relative stiffnesses and the local stresses. Research has found that the exponent that relates the cumulative damage to the stress range may be higher than the value, 5, for other welds in shear. The value chosen for EN 1994-2, 8, is controversial, as discussed later. As may be expected from the involvement of a tiny volume of concrete, tests show a wide scatter in fatigue lives, which is allowed for in the design resistances. Studs are provided in large numbers, and are well able to redistribute shear between themselves. One reason for not recommending a partial factor more conservative than 1.0 comes from experience with bridges, where stud connectors have been used for almost 50 years. Whenever occasion has arisen in print or at a conference, the second author has stated that there is no known instance of fatigue failure of a stud in a bridge, other than a few clearly attributable to errors in design. This has not been challenged. Research has identified, but not yet quantified, many reasons for this remarkable experience.97;98 Most of them (e.g. slip, shear lag, permanent set, partial interaction, adventitious connection from bolt heads, friction) lead to predicted stress ranges on studs lower than those assumed in design. With an eighth-power law, a 10% reduction in stress range more than doubles the fatigue life.

6.8.3. Fatigue strength Clause 6.8.3(3)

Clause 6.8.3(4)

The format of clause 6.8.3(3) for shear studs, as in EN 1993-1-9, uses a reference value of range of shear stress at 2 million cycles, C ¼ 90 N/mm2 . It defines the slope m of the line through this point on the log-log plot of range of stress R against number of constant-range stress cycles to failure, NR , Fig. 6.25. Clause 6.8.3(4) modifies the expression slightly for lightweight-aggregate concrete. It is a complex matter to deduce a value for m from the mass of test data, which are often inconsistent. Many types of test specimen have been used, and the resulting scatter of results must be disentangled from that due to inherent variability. Values for m recommended in the literature range from 5 to 12, mostly based on linear-regression analyses. The method of regression used (x on y, or y on x) can alter the value found by up to 3.97 The value 8, which was also used in BS 5400 Part 10, may be too high. In design for a loading spectrum, its practical effect is that the cumulative damage is governed by the highest-range components of the spectrum (e.g. by the small number of very heavy lorries in the traffic spectrum). A lower value, such as 5, would give more weight to the much higher number of average-weight vehicles. This is illustrated in Example 6.13. While fatigue design methods for stud connectors continue to be conservative (for bridges and probably for buildings too) the precise value for m is of academic interest. Any future proposals for more accurate methods for prediction of stress ranges should be associated with re-examination of the value for m. Annex C gives a different design rule for fatigue of lying studs, which is discussed in Chapter 10.

6.8.4. Internal forces and fatigue loadings

Clause 6.8.4(1)

152

For fatigue assessment, it is necessary to find the range or ranges of stress in a given material at a chosen cross-section, caused by the passage of a vehicle along the bridge. Loading other than the vehicle influences the extent of cracking in the concrete, and hence, the stiffnesses of members and the calculated stresses. Cracking depends mainly on the heaviest previous loading, and so tends to increase with time. Clause 6.8.4(1) refers to clause 6.8.3 of EN 1992-1-1. This defines the non-cyclic loading assumed to coexist with the design value


Stress range from cyclic loads (e.g. passage of fatigue load model)

Stress in reinforcing bar

Tension ‘Mean’ stress level from non-cyclic loads

Compression

0

Reduced stress range for the same vehicle

Time

Fig. 6.49. Stress ranges for fatigue verification of reinforcement caused by the same cyclic action at different mean stress levels

of the cyclic load, Qfat : it is the ‘frequent’ combination, represented by X X Gk; j þ P þ 1;1 Qk;1 þ 2;i Qk;i j1

i>1

where the Q’s are non-cyclic variable actions. Traffic will usually be the leading non-cyclic action since the 2 value for traffic recommended in Annex A2 of EN 1990 is zero. With traffic as the leading action, only thermal actions have a non-zero value of 2 and therefore need to be considered. The non-cyclic combination gives a mean stress level upon which the cyclic part of the action effect is superimposed. The importance of mean stress is illustrated in Fig. 6.49 for the calculation of stress range in reinforcement in concrete. It shows that the stress change in the reinforcement for any part of the loading cycle that induces compression in the concrete is much less than the stress change where the slab remains in tension throughout the cycle. Clause 6.8.4(2) defines symbols that are used for bending moments in clause 6.8.5.4. The sign convention is evident from Fig. 6.26, which shows that MEd;max;f is the bending moment that causes the greatest tension in the slab, and is positive. Clause 6.8.4(2) also refers to internal forces, but does not give symbols. Analogous use of calculated tensile forces in a concrete slab (e.g. NEd;max;f ) may sometimes be necessary. Clause 6.8.4(3) refers to Annex A.1 of EN 1993-1-9 for a general treatment of fatigue based on summing the damage from a loading spectrum. As discussed in section 6.8.1 above, this would be a lengthy and complex calculation for most bridges and therefore clauses 6.8.4(4) to (6) provide the option of using simpler load models from EN 1991-2. The damage equivalent stress method for road bridges is based on Fatigue Load Model 3 defined in EN 1991-2 clause 4.6.4, while for rail bridges it is based on Load Model 71. Clause 6.8.4(5) says that the additional factors given in EN 1992-2 clause NN.2.1 ‘should ’ be applied to Load Model 3 where a road bridge is prestressed by tendons or imposed deformations. As Annex NN is Informative, the situation is unclear in a country where the National Annex does not make it available. The load models and their application are discussed in the other guides in this series.24

Clause 6.8.4(2)

Clause 6.8.4(3)

Clauses 6.8.4 to (6)

6.8.5. Stresses Clause 6.8.5.1(1) refers to a list of action effects in clause 7.2.1(1)P to be taken into account ‘where relevant’. They are all relevant, in theory, to the extent of cracking. However, this can usually be represented by the same simplified model, chosen from clause 5.4.2.3, that is used for other global analyses. They also influence the maximum value of the fatigue stress range, which is limited for each material (e.g. the limit for shear connectors in clause 6.8.1(3)). The provisions for fatigue are based on the assumption that the stress range caused by a given fluctuation of loading, such as the passage of a vehicle of known weight, remains approximately constant after an initial shakedown period. ‘Shakedown’ here includes the

Clause 6.8.5.1(1)

153


changes due to cracking, shrinkage, and creep of concrete, that occur mainly within the first year or two. For bridges, most fatigue cycles occur over very short durations as the stress ranges are produced either by the passage of vehicles or by wind-induced oscillations. Cycles of stress from thermal actions also occur but over greater durations. The magnitude and small number of these cycles do not generally cause any significant fatigue damage. The shortterm modular ratio should therefore be used when finding stress ranges from the cyclic action Qfat . Where a peak stress is being checked, creep from permanent loading should be allowed for, if it increases the relevant stress. The effect of tension stiffening on the calculation of stress in reinforcement, clause Clause 6.8.5.1(2)P Clause 6.8.5.1(3) 6.8.5.1(2)P and (3), is illustrated in Example 6.13 and discussed under clause 6.8.5.4 below. It is not conservative to neglect tension stiffening in this calculation for a composite beam as the increased stiffness attracts more stress to the concrete slab and hence to the reinforcement between cracks. For stresses in structural steel, the effects of tension stiffening Clause 6.8.5.1(4) may be included or neglected in accordance with clause 6.8.5.1(4). Tension stiffening here has a beneficial effect in reducing the stresses in the structural steel. Tension stiffening Clause 6.8.5.1(5) should also be considered in deriving stresses for prestressing steel – clause 6.8.5.1(5). For analysis, the linear-elastic method of Section 5 is used, from clause 6.8.4(1). Clause 7.2.1(8) requires consideration of local and global effects in deck slabs. This is also reflected in clause 6.8.6.1(3). When checking fatigue, it is important to bear in mind that the most critical areas for fatigue may not be the same as those for other ultimate limit state calculations. For example, the critical section for shear connection may be near mid-span, since its provision is usually based on the static design, and the contribution to the static shear from dead load is zero there.

Clause 6.8.5.2(1)

Clause 6.8.5.3(1) Clause 6.8.5.3(2)

Concrete For concrete, clause 6.8.5.2(1) refers to clause 6.8 of EN 1992-1-1, where clause 6.8.5(2) refers to EN 1992-2. EN 1992-2 clause 6.8.7(101) provides a damage equivalent stress range method presented as for a spectrum. The method of its Annex NN is not applicable to composite members. As a simpler alternative, EN 1992-1-1 clause 6.8.7(2) gives a conservative verification based on the non-cyclic loading used for the static design. It will usually be sufficient to apply this verification to composite bridges as it is unlikely to govern design other than possibly for short spans where most of the compressive force in concrete is produced by live load. Structural steel Clause 6.8.5.3(1) repeats, in effect, the concession in clause 6.8.5.1(4). Where the words ‘or only MEd;min;f ’ in clause 6.8.5.3(2) apply, MEd;max;f causes tension in the slab. The use of the uncracked section for MEd;max;f would then underestimate the stress ranges in steel flanges, so that cracked section properties should be used for the calculation of this part of the stress range. Reinforcement For reinforcement, clause 6.8.3(2) refers to EN 1992-1-1, where clause 6.8.4 gives the verification procedure. Its recommended value N for straight bars is 106 . This should not be confused with the corresponding value for structural steel in EN 1993-1-9, 2 106 , denoted NC , which is used also for shear connectors, clause 6.8.6.2(1). Using the values recommended in EN 1992-1-1, its expression (6.71) for verification of reinforcement becomes: E;equ ðN Þ Rsk ðN Þ=1:15

(D.6.33)

with Rsk ¼ 162.5 N/mm2 for N ¼ 106 , from Table 6.3N. Where a range E (NE ) has been determined, the resistance Rsk (NE ) can be found from the S–N curve for reinforcement, and the verification is: E ðNE Þ Rsk ðNE Þ=1:15

154

(D6.34)


σs (tension)

σs,max,f

J

G

C

∆σs

H

σs,o

∆σs,f

F

E

K

σs,min,f

A

L

M

D

0 Mcr

B

MEd,min,f

MEd,max,f

MEd,min,f

Fig. 6.50. Stress ranges in reinforcement in cracked regions

Clause 6.8.5.4(1) permits the use of the approximation to the effects of tension stiffening that is used for other limit states. It consists of adding to the maximum tensile stress in the ‘fully cracked’ section, s;o , an amount s that is independent of s;o . The value of s for fatigue verification is modified by replacing the factor of 0.4 in equation (7.5) by 0.2. This is to allow for the reduction in tension stiffening caused by repeated cycles of tensile stress.99 Clauses 6.8.5.4(2) and (3) give simplified rules for calculating stresses, with reference to Fig. 6.26, which is discussed using Fig. 6.50. This has the same axes, and also shows a minimum bending moment that causes compression in the slab. A calculated value for the stress s in reinforcement, that assumes concrete to be effective, would lie on line A0D. On initial cracking, the stress s jumps from B to point E. Lines 0BE are not shown in Fig. 6.26 because clause 7.2.1(5)P requires the tensile strength of concrete to be neglected in calculations for s . This gives line 0E. For moments exceeding Mcr , the stress s follows route EFG on first loading. Calculation of s using section property I2 gives line 0C. At bending moment MEd;max;f the stress s;o thus found is increased by s , from equation (7.5), as shown by line HJ. Clause 6.8.5.4 defines the unloading route from point J as J0A, on which the stress s;min;f lies. Points K and L give two examples, for MEd;min;f causing tension and compression, respectively, in the slab. The fatigue stress ranges s;f for these two cases are shown.

Clause 6.8.5.4(1)

Clause 6.8.5.4(2) Clause 6.8.5.4(3)

Shear connection The interpretation of clause 6.8.5.5(1)P is complex when tension stiffening is allowed for. Clause 6.8.5.5(1)P Spacing of shear connectors near internal supports is unlikely to be governed by fatigue, so it is simplest to use uncracked section properties when calculating range of shear flow from range of vertical shear, clause 6.8.5.5(2). These points are illustrated in Example 6.13. Clause 6.8.5.5(2) Reinforcement and prestressing steel in members prestressed by bonded tendons Where bonded prestress is present, stresses should be determined in a similar manner to the above for reinforcement alone, but account needs to be taken of the difference of bond behaviour between prestressing steel and reinforcement – clause 6.8.5.6(1). Clause 6.8.5.6(2) makes reference to clause 7.4.3(4) which in turn refers to clause 7.3 of EN 1992-1-1 for the calculation of stresses ‘s ’. This is a generic symbol here, and so applies to the stress s;max;f referred to in clause 6.8.5.6(2).

Clause 6.8.5.6(1) Clause 6.8.5.6(2)

6.8.6. Stress ranges Clause 6.8.6.1 is most relevant to the damage equivalent stress method where the complex cyclic loadings from a spectrum of vehicles are condensed into one single stress range which, for N cycles, is intended to give the same damage during the bridge’s lifetime as

Clause 6.8.6.1

155


Clause 6.8.6.1(2)

Clause 6.8.6.1(3)

Clause 6.8.6.2(1) Clause 6.8.6.2(2)

the real traffic. This stress range is determined by applying the relevant fatigue load model discussed in section 6.8.4 and by multiplying it by the damage equivalent factor , according to clause 6.8.6.1(2). The factor is a property of the spectrum and the exponent m, which is the slope of the fatigue curve as noted in clause 6.8.6.1(4). Deck slabs of composite bridge beams are usually subjected to combined global and local fatigue loading events, due to the presence of local wheel loads. The effects of local and global loading are particularly significant in reinforcement design in slabs adjacent to cross-beams supporting the deck slab, in zones where the slab is in global tension. Here, wheel loads cause additional local hogging moments. Clause 6.8.6.1(3) provides a conservative interaction where the damage equivalent stress range is determined separately for the global and local actions and then summed to give an overall damage equivalent stress range. In combining the stress ranges in clause 6.8.6.1(3), it is important to consider the actual transverse location being checked within the slab. The peak local effect usually occurs some distance from the web of a main beam, while the global direct stress reduces away from the web due to shear lag. The reduction may be determined using clause 5.4.1.2(8), even though that clause refers to EN 1993-1-5, which is for steel flanges. A similar damage equivalent factor, v , is used in clause 6.8.6.2(1) to convert the shear stress range in the studs from the fatigue load model into a damage equivalent stress range. For other types of shear connection clause 6.8.6.2(2) refers to Section 6 of EN 1993-1-9. This requires the damage equivalent stress to be determined from its Annex A using the actual traffic spectrum and Miner’s summation. This approach could also be used for shear studs as an alternative, provided that m is taken as 8, rather than 3. For connectors other than studs, the authors recommend that the method of Annex A be used only where the following conditions are satisfied: .

. .

the connectors are attached to the steel flange by welds that are within the scope of EN 1993-1-9 the fatigue stress ranges in the welds can be determined realistically the stresses applied to concrete by the connectors are not high enough for fatigue failure of the concrete to influence the fatigue life.

The exponent m should then have the value given in EN 1993-1-9; m ¼ 8 should not be used. In other situations, fatigue damage to concrete could influence the value of m. The National Annex may refer to guidance, as permitted by the Note to clause 1.1.3(3). Clauses 6.8.6.2(3) Clauses 6.8.6.2(3) to (5) provide a method of calculating the damage equivalent factors to (5) for studs. With the exception of v;1 , those for road bridges are based on those in EN 1993-2 clause 9.5.2, but with the exponents modified to 8 or 18 as discussed in section 6.8.3. In EN 1993-2, an upper limit to is defined in clause 9.5.2, in paragraphs that EN 1994-2 does not refer to. This is because the upper limit is not required for stud shear connectors.

6.8.7. Fatigue assessment based on nominal stress ranges Clause 6.8.7.1

Clause 6.8.7.2(1)

156

Comment on the methods referred to from clause 6.8.7.1 will be found in other guides in this series. The term ‘nominal stress range’ in the heading of clause 6.8.7 is defined in Section 6 of EN 1993-1-9 for structural steel. It is the stress range that can be compared directly with the detail categories in EN 1993-1-9. It is not the stress range before the damage equivalent factors are applied. It is intended to allow for all stress concentration factors implicit within the particular detail category selected. If additional stress concentrating details exist adjacent to the detail to be checked which are not present in the detail category selected (e.g. a hole), these additional effects need to be included via an appropriate stress concentration factor. This factored stress range then becomes a ‘modified nominal stress range’ as defined in clause 6.3 of EN 1993-1-9. For shear connectors, clause 6.8.7.2(1) introduces the partial factors. The recommended value of Mf;s is 1.0 (clause 2.4.1.2(6)). For Ff , EN 1990 refers to the other Eurocodes. The recommended value in EN 1992-1-1, clause 6.8.4(1), is 1.0. Clause 9.3(1) of EN 1993-2 recommends 1.0 for steel bridges.


Clause 6.8.7.2(2) covers interaction between the fatigue failures of a stud and of the steel flange to which it is welded, where the flange is in tension. The first of expressions (6.57) is the verification for the flange, from clause 8(2) of EN 1993-1-9, and the second is for the stud, copied from equation (6.55). The linear interaction condition is given in expression (6.56). It is necessary to calculate the longitudinal stress range in the steel flange that coexists with the stress range for the connectors. The load cycle that gives the maximum value of E;2 in the flange will not, in general, be that which gives the maximum value of E;2 in a shear connector, because the first is caused by flexure and the second by shear. Also, both E;2 and E;2 may be influenced by whether the concrete is cracked, or not. It thus appears that expression (6.56) may have to be checked four times. In practice, it is best to check first the conditions in expression (6.57). It should be obvious, for these, whether the ‘cracked’ or the ‘uncracked’ model is the more adverse. Usually, one or both of the left-hand sides is so far below 1.0 that no check to expression (6.56) is needed.

Clause 6.8.7.2(2)

Example 6.13: fatigue verification of studs and reinforcement The bridge shown in Fig. 6.22 is checked for fatigue of the shear studs at an abutment and of the top slab reinforcement at an internal support. The Client requires a design life of 120 years. Fatigue Load Model 3 of EN 1991-2 is used. The bridge will carry a road in Traffic Category 2 of Table 4.5(n) of EN 1991-2, ‘roads with medium flow rates of lorries’. The table gives the ‘indicative number of heavy vehicles expected per year and per slow lane’ as 500 000, and this value is used. The ‘safe life’ method (defined in clause 3(7)(b) of EN 1993-1-9) is used, as this is likely to be recommended by the UK’s National Annex.

Studs at an abutment The cross-section of an inner beam at the abutments is as shown for length DE in Fig. 6.22. Groups of three 19 mm studs are provided, Fig. 6.41, at 150 mm spacing. The ‘special vehicle’ of Load Model 3 is defined in clause 4.6.4 of EN 1991-2. For this cross-section its passage produces maximum and minimum unfactored vertical shears of þ235 kN and 19 kN. Since the detail is adjacent to an expansion joint, these values should be increased by a factor of 1.3 in accordance with EN 1991-2 Fig. 4.7, so the shear range becomes 1:3 ð235 þ 19Þ ¼ 330 kN. The short-term uncracked properties of the composite beam are used for the calculation of shear flow. From Table 6.3 in Example 6.10, A z=I ¼ 0:810 m1 . The range of shear force per connector is: 0:810 330 0:150=3 ¼ 13:4 kN The shear stress range for the connector is: ¼

13:4 103 ¼ 47:1 N=mm2 192 =4

To determine the damage equivalent stress range, the factor v ¼ v;1 v;2 v;3 v;4 should be calculated in accordance with clause 6.8.6.2(3). From clause 6.8.6.2(4), v;1 ¼ 1:55. The remaining factors are calculated from EN 1993-2 clause 9.5.2 using exponents 8 and 18 in place of those given. For v;2 it would be possible to use the recommended data for Load Model 4 in Tables 4.7 and 4.8 of EN 1991-2. However, the UK’s National Annex to EN 1991-2 is likely to replace these with the BS 5400 Part 10 data, which are given in Table 6.5. From clause 9.5.2(3) of EN 1993-2: !1=5 !1=5 P 8:051 1018 ni Q5i P Qm1 ¼ ¼ ¼ 381:2 kN for checks on structural steel ni 1:000 106

157


Qm1 ¼

!1=8 P ni Q8i P ¼ ni

3:384 1029 1:000 106

!1=8 ¼ 873:3 kN for checks on shear studs

It can be seen from the above that the contribution of the small number of very heavy vehicles is much more significant when the exponent 8 is used. Table 6.5. Vehicle spectrum for fatigue verifications Vehicle ref.

Weight (kN) (Qi )

No. per million vehicles (ni )

(ni Q5i )

(ni Q8i )

18GTH 18GTM 9TT-H 9TT-M 7GT-H 7GT-M 7A-H 5A-H 5A-M 5A-L 4A-H 4A-M 4A-L 4R-H 4R-M 4R-L 3A-H 3A-M 3A-L 3R-H 3R-M 3R-L 2R-H 2R-M 2R-L

3680 1520 1610 750 1310 680 790 630 360 250 335 260 145 280 240 120 215 140 90 240 195 120 135 65 30

10 30 20 40 30 70 20 280 14 500 15 000 90 000 90 000 90 000 15 000 15 000 15 000 30 000 30 000 30 000 15 000 15 000 15 000 170 000 170 000 180 000

6:749 1018 2:434 1017 2:164 1017 9:492 1015 1:157 1017 1:018 1016 6:154 1015 2:779 1016 8:768 1016 1:465 1016 3:797 1017 1:069 1017 5:769 1015 2:582 1016 1:194 1016 3:732 1014 1:378 1016 1:613 1015 1:771 1014 1:194 1016 4:229 1015 3:732 1014 7:623 1015 1:972 1014 4:374 1012

3:363 1029 8:548 1026 9:029 1026 4:005 1024 2:602 1026 3:200 1024 3:034 1024 6:948 1024 4:091 1024 2:289 1023 1:428 1025 1:879 1024 1:759 1022 5:667 1023 1:651 1023 6:450 1020 1:370 1023 4:427 1021 1:291 1020 1:651 1023 3:136 1022 6:450 1020 1:876 1022 5:417 1019 1:181 1017

1:000 106

8:051 1018

3:384 1029

Totals

For a two-lane road, traffic category 2, NObs ¼ 0:5 106 . (The UK’s National Annex to EN 1991-2 may modify this value.) Also from clause 9.5.2(3),

N0 ¼ 0:5 106 Q0 ¼ 480 kN (weight of vehicle for Fatigue Load Model 3)

Therefore v;2

!1=8 Qm1 NObs 1=8 873:3 0:5 106 ¼ ¼ ¼ 1:819 480 0:5 106 Q0 N0

From clause 9.5.2(5) of EN 1993-2: v;3 ¼ ðtLd =100Þ1=8 ¼ 1:023 for the 120-year design life. From clause 9.5.2(6) of EN 1993-2: " #1=8 N2 2 Qm2 8 N3 3 Qm3 8 Nk k Qmk 8 v;4 ¼ 1 þ þ þ... þ N1 1 Qm1 N1 1 Qm1 N1 1 Qm1

158


The influence coefficient from lane 2 is approximately 75% of that from lane 1. As both lanes are slow lanes with N ¼ 0:5 106 vehicles per year, " #1=8 #1=8 " N2 2 Qm2 8 0:5 106 0:75 8 ¼ 1þ ¼ 1:012 v;4 ¼ 1 þ N1 1 Qm1 0:5 106 1:0 v ¼ v;1 v;2 v;3 v;4 ¼ 1:55 1:819 1:023 1:012 ¼ 2:92 From clause 6.8.6.2(1), E;2 ¼ v ¼ 2:92 47:1 ¼ 138 N=mm2 From clause 6.8.7.2(1), Ff E;2 ¼ 1:0 138 ¼ 138 N=mm2 From clause 6.8.3(3), c ¼ 90 N/mm2 , so the fatigue resistance is: c =Mf;s ¼ 90=1:0 ¼ 90 N=mm2 The shear studs are therefore not adequate and would need to be increased. There is no need to check the interaction in clause 6.8.7.2(2) as the stress in the steel flange is small and compressive at an abutment.

Fatigue of reinforcement, global effects Note: throughout this Example, all cross-references commencing ‘NN’ are to Annex NN of EN 1992-2, ‘Damage equivalent stresses for fatigue verification’. The cross-section at an intermediate support is shown in Fig. 6.22. For these crosssections, the axle loads of Fatigue Load Model 3 should be multiplied by 1.75 according to clause NN.2.1(101). The maximum hogging moment from the fatigue vehicle was 1:75 593 ¼ 1038 kNm and the minimum was 1:75 ð47Þ ¼ 82 kNm. In this calculation, the maximum and minimum moments occurred with the vehicle in the same lane. Previous practice in the UK has been to calculate the stress range by allowing the maximum and minimum effects from the vehicle to come from different lanes. Clause 4.6.4(2) of EN 1991-2 however implies that the maximum stress range should be calculated as the greatest stress range produced by the passage of the vehicle along any one lane. The UK’s draft National Annex currently requires the former interpretation (the safer of the two) to be used, but there is no national provision in EN 1991-2 for this to be done. The maximum hogging moment on the composite section from the frequent combination was found to be 3607 kNm. This includes the effects of superimposed dead load, secondary effects of shrinkage, settlement, thermal actions and traffic load (load group 1a). Traffic was taken as the leading variable action and hence the combination factors applied were 1 for traffic loading and 2 for thermal actions. Wind was not considered, as the recommended value of 2 is zero from EN 1990 Table A2.1. From clause 6.8.4(1), the minimum moment from the cyclic þ non-cyclic loading is: MEd;min;f ¼ 3607 82 ¼ 3525 kNm and the maximum is: MEd;max;f ¼ 3607 þ 1038 ¼ 4645 kNm From clause 7.4.3(3) as modified by clause 6.8.5.4(1), the increase in stress in the reinforcement, due to tension stiffening, above that calculated using a fully cracked analysis is: s ¼

0:2 fctm st s

where

st ¼

AI 74 478 22 660 ¼ 2:50 ¼ Aa Ia 55 000 12 280

The reinforcement ratio s ¼ 0:025 and fctm ¼ 2:9 N/mm2 and thus s ¼ 9:3 N/mm2 . From Example 6.6, the section modulus for the top layer of reinforcement is 34:05 106 mm3 . Therefore the stress due to MEd;max;f ignoring tension stiffening is: s;o ¼ 4645=34:05 ¼ 136:4 N/mm2

159


From equation (7.4), the stress including allowance for tension stiffening is: s;max;f ¼ 136:4 þ 9:3 ¼ 146 N/mm2 From clause 6.8.5.4(2), s;min;f ¼ 136:4 3525=4645 þ 9:3 ¼ 113 N/mm2 The damage equivalent parameters are next calculated from Annex NN. Figure NN.1 refers to the ‘length of the influence line’. This length is intended to be the length of the lobe creating the greatest stress range. EN 1993-2 clause 9.5.2 provides definitions of the critical length of the influence line for different situations and these can be referred to. For bending moment at an internal support, the average length of the two adjacent spans may be used, but here the length of the main span has been conservatively used. From Fig. NN.1 for straight reinforcing bars (curve 3) and critical length of the influence line of 31 m, s;1 ¼ 0.98. From equation (NN.103): pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi k2 Nobs =2:0 with Nobs in millions (which is not stated) s;2 ¼ Q From Table 4.5(n) of EN 1991-2, Nobs ¼ 0:5 106 . From EN 1992-1-1/Table 6.3N, k2 ¼ 9 for straight bars. , is given in Table NN.1, but no guidance is given on its selecThe factor for traffic type, Q tion. ‘Traffic type’ is defined in Note 3 of EN 1991-2 clause 4.6.5(1). The definitions given are not particularly helpful: . . .

‘long distance’ means hundreds of kilometres ‘medium distance’ means 50–100 km ‘local traffic’ means distances less than 50 km.

‘Long distance’ will typically apply to motorways and trunk roads. The use of either of the lower categories should be agreed with the Client as the traffic using a road may not be represented by a typical length of journey. ‘Long distance’ traffic is conservatively used ¼ 1:0. Thus, here, so from Table NN.1, Q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p s;2 ¼ 1:0 9 0:5=2:0 ¼ 0:86 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi From equation (NN.104): s;3 ¼ k2 NYears =100 ¼ 9 120=100 ¼ 1:02 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi P Nobs;i k2 . From equation (NN.105): s;4 ¼ Nobs;1 Since both lanes are slow lanes, from Table 4.5 of EN 1991-2, Nobs;1 ¼ Nobs;2 ¼ 0:5 106 and therefore s;4

rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 0:5 þ 0:5 ¼ ¼ 1:08 0:5

From clause NN.2.1(108) and then EN 1991-2, Annex B, the damage equivalent impact factor for surfaces of good roughness (i.e. regularly maintained surfaces) is ’fat ¼ 1.2. From equation (NN.102): s ¼ ’fat s;1 s;2 s;3 s;4 ¼ 1:20 0:98 0:86 1:02 1:08 ¼ 1:11 From clause 6.8.6.1(2) and (7): E ¼ max;f min;f ¼ 1:11 1:0 ð146 113Þ ¼ 37 N/mm2 There is an inconsistency here between EN 1992-2, where s includes the impact factor ’fat and EN 1994-2 where excludes this factor, which is written as .

160


From Table 6.3N of EN 1992-1-1, for straight bars, N ¼ 106 and Rsk ð106 Þ ¼ 162.5 N/mm2 . The verification is carried out using expression (6.71) of EN 1992-1-1, taking s;equ ðN Þ ¼ E above: F;fat s;equ ðN Þ ¼ 1:0 37 ¼ 37 N=mm2

Rsk ðN Þ ¼ 162:5=1:15 ¼ 141 N=mm2 s;fat

The reinforcement has adequate fatigue life under global loading.

Fatigue of reinforcement, local effects Local bending moments are caused by hogging of the deck slab over the pier diaphragm. The worst local effects are here conservatively added to the global effects according to expression (6.53). A stress range of 44 N/mm2 in the reinforcement was determined from the maximum hogging moment caused by the passage of the factored fatigue vehicle of EN 1992-2 Annex NN. (The moments were found using Pucher’s influence surfaces.100 ) Cracked section properties were used for the slab in accordance with clause 6.8.2(1)P of EN 1992-1-1 since the slab remains in tension when the local effect is added under the ‘basic combination plus the cyclic action’ as defined in clause 6.8.3(3) of EN 1992-1-1. The damage equivalent factors from EN 1992-2 Annex NN are the same as above with the exception of s;1 . For local load, the critical length of the influence line is the length causing hogging moment each side of the pier diaphragm. From the Pucher chart used to determine the local moment, the influence of loads applied more than 6 m from the pier diaphragm is approximately zero so the total influence line length to consider is approximately 12 m. From Fig. NN.1, s;1 ¼ 0:91. From equation (NN.102): s ¼ fat s;1 s;2 s;3 s;4 ¼ 1:20 0:91 0:86 1:02 1:08 ¼ 1:03 From clause 6.8.6.1(2) and (7): E ¼ max;f min;f ¼ 1:03 1:0 ð44Þ ¼ 45 N=mm2 Verifying as for the global loading: F;fat s;equ ðN Þ ¼ 1:0 45 ¼ 45 N=mm2

Rsk ðN Þ ¼ 162:5=1:15 ¼ 141 N=mm2 s;fat

The reinforcement has adequate fatigue life under local loading.

Fatigue of reinforcement, combined global and local effects The simple interaction method of clause 6.8.6.1(3) is used. This entails summing the damage equivalent stresses from the local and global loading, so that the total damage equivalent stress E ¼ 37 þ 45 ¼ 82 N=mm2 . In this case, the locations of peak global stress in the reinforcement and peak local stress do coexist because there was no reduction to the slab width for shear lag. The verification is now: 82 N/mm2 141 N/mm2 The reinforcement has adequate fatigue life under the combined loading.

6.9. Tension members in composite bridges The terms ‘concrete tension member’ and ‘composite tension member’ used in this clause are defined in clause 5.4.2.8. Global analysis for action effects in these members and determination of longitudinal shear is discussed in comments on that clause. Clause 6.9(1) concerns members that have tensile force introduced only near their ends. It refers to their design to EN 1992, as does clause 6.9(2), with reference to simplifications

Clause 6.9(1) Clause 6.9(2)

161


A T

F

a

T

C

L

D H

E 2b

2b

T B

G

T

≤1.5b (a)

K

J

≤1.5b (b)

Fig. 6.51. Shear connection to the deck of a bowstring arch bridge

Clause 6.9(3)


Clause 6.9(6)

162

given in clause 5.4.2.8. Clause 6.9(2) applies also to composite tension members, which have shear connectors throughout their length. The subsequent paragraphs concern the distribution of the connectors along the member. A plan of the steel members at deck level near one end of a bowstring arch bridge is shown in Fig. 6.51(a). The arch applies concentrated forces T at points A and B. The force at A is shared between the steel tie AC and the composite deck, shown shaded. The deck has steel edge members such as DE, and spans longitudinally between composite cross-beams FG, etc. The proportion of each force T that is resisted by the deck structure, Td say, depends on the extent to which its stiffness is reduced by cracking of the concrete. The force Td is applied to the deck by diagonal members such as FH. The stiffness of these members also influences the magnitude of Td . Details of bridges of this type are available elsewhere.101 In some bridges, the deck is shear-connected directly to the main tie member, as shown in Fig. 6.51(b). Clause 6.9(3) requires the shear connection for the force Td to be provided within the lengths 1.5b shown. The precise distribution of the connectors along a length such as JK has been studied, using the rules of EN 1994-2 for tension stiffening.102 In this bridge, Newark Dyke, the arch is the top chord of a truss of span 77 m with diagonals that apply longitudinal force also at points such as L in Fig. 6.51(b). Neither paragraph (3) nor (6) defines the length a over which shear connection near point L should be provided, but clause 6.6.2.3 provides guidance. The number of connectors to be provided over a length such as JK can be conservatively found by assuming the deck to be uncracked. In this bridge, the design ultimate force T was about 18 MN, and ‘uncracked’ analysis gave Td 9 MN at mid-span. Fully cracked analysis gave this force as about 5 MN. Accurate analysis found the deck to be in a state of single cracking (explained in comments on clause 5.4.2.8(6)), with a tensile force of about 8 MN. Lower levels of shear connection are, of course, required along the whole length of the deck for other combinations and arrangements of variable actions. Clause 6.9(4)P, a principle that corresponds to clause 6.7.4.1(1)P for compression members, is followed by application rules. For laterally loaded tension members, shear connection within the length is related to the transverse shear in clause 6.9(5), exactly as for composite beams. Where axial tension is applied to the ends of a member through only one material, steel or concrete, the length over which part of the tension should be transferred to the other material (typically by shear connection), is limited by clause 6.9(6). This corresponds to clause 6.7.4.2(2) for compression members. Other provisions of clause 6.7.4.2 may be relevant here.

CHAPTER 7

Serviceability limit states This chapter corresponds to Section 7 of EN 1994-2, which has the following clauses: . . . . .

General Stresses Deformations in bridges Cracking of concrete Filler beam decks


7.1. General Section 7 of EN 1994-2 is limited to provisions on serviceability that are specific to composite structures and are not in Sections 1, 2, 4 or 5 (for global analysis), or in Eurocodes 1990, 1991, 1992 or 1993. Some of these other provisions are briefly referred to here. Further comments on them are in other chapters of this book, or in other guides in this series. The initial concept for a composite bridge is mainly influenced by the intended method of construction, durability, ease of maintenance, and the requirements for ultimate limit states. Serviceability criteria that should be considered at an early stage are stress limits in cross-sections in Class 1 or 2 and susceptibility to excessive vibration. It should not however be assumed that Class 3 and 4 cross-sections require no checks of stress limits at serviceability. For example, if torsional warping or St Venant torsional effects have been neglected at ultimate limit state (ULS), as allowed by a reference in clause 6.2.7.2(1) of EN 1993-2, then the serviceability limit state (SLS) stresses should be checked taking these torsional effects into account. Considerations of shear lag at SLS may also cause unacceptable yielding as the effective widths of steel elements are greater at ULS. Control of crack width can usually be achieved by appropriate detailing of reinforcement. Provision of fire resistance and limiting of deformations have less influence at this stage than in structures for buildings. The important deformations are those caused by imposed load. Limits to these influence the design of railway bridges, but generally, stiffness is governed more by vibration criteria than by limits to deflection. The drafting of the serviceability provisions in the Eurocodes is less prescriptive than for other limit states. It is intended to give designers and clients greater freedom to take account of factors specific to the project. The content of Section 7 was also influenced by the need to minimize calculations. Results already obtained for ultimate limit states are scaled or reused wherever possible. Experienced designers know that many structural elements satisfy serviceability criteria by wide margins. For these, design checks should be simple, and it does not matter if they are conservative. For other elements, a longer but more accurate calculation may be justified. Some application rules therefore include alternative methods.



Clause 7.1(1)P and (2) refer to clause 3.4 of EN 1990. This gives criteria for placing a limit state within the ‘serviceability’ group, with reference to deformations (including vibration), durability, and the functioning of the structure. The relevance of EN 1990 is not limited to the clauses referred to, because clause 2.1(1)P requires design to be in accordance with the general rules of EN 1990. This means all of it except annexes that are either informative or not for bridges.

Serviceability verification and criteria The requirement for a serviceability verification is given in clause 6.5.1(1)P of EN 1990 as: Ed Cd

Clause 7.1(3)

Clause 7.1(4)

Clause 7.1(5)

Clause 7.1(6)

where Ed is the design value of the effects of the specified actions and the ‘relevant’ combination, and Cd is the limiting design value of the ‘relevant’ criterion. From clause 6.5.3 of EN 1990, the relevant combination is ‘normally’ the characteristic, frequent, or quasi-permanent combination, for serviceability limit states that are respectively irreversible, reversible, or a consequence of long-term effects. The quasi-permanent combination is also relevant for the appearance of the structure. For bridges, rules on combinations of actions are given in clause A2.2 of EN 1990. Its clause A2.2.2(1) defines a fourth combination, ‘infrequent’, for use for concrete bridges. It is not used in EN 1994-2, but may be invoked by a reference to EN 1992, or found in a National Annex. Clause 7.1(3) refers to ‘environmental classes’. These are the ‘exposure classes’ of EN 1992, and are discussed in Chapter 4. The exposure class influences the cover to reinforcing bars, and the choice of concrete grade and hence the stress limits. Clause 7.1(4) on serviceability verification gives no detailed guidance on the extent to which construction phases should be checked. The avoidance of excessive stress is one example. Yielding of steel can cause irreversible deformation, and handling of precast components can cause yielding of reinforcement or excessive crack width. Bridges can also be more susceptible to aerodynamic oscillation during erection. In extreme cases, this can lead to achievement of an ultimate limit state. Clause 7.1(5) refers to the eight-page clause A2.4 of EN 1990. It covers partial factors, serviceability criteria, design situations, comfort criteria, deformations of railway bridges and criteria for the safety of rail traffic. Few of its provisions are quantified. Recommended values are given in Notes, as guidance for National Annexes. The meaning of clause 7.1(6) on composite plates is that account should be taken of Section 9 when applying Section 7. There are no serviceability provisions in Section 9. No serviceability limit state of ‘excessive slip of shear connection’ is defined. Generally, it is assumed that clause 6.8.1(3), which limits the shear force per connector under the characteristic combination, and other rules for ultimate limit states, will ensure satisfactory performance in service. No serviceability criteria are specified for composite columns, so from here on, this chapter is referring to composite beams or plates or, in a few places, to composite frames.

7.2. Stresses Excessive stress is not itself a serviceability limit state. Stresses in bridges are limited to ensure that under normal conditions of use, assumptions made in design models (e.g. linear-elastic behaviour) remain valid, and to avoid deterioration such as the spalling of concrete or disruption of the corrosion protection system. The stress ranges in a composite structure caused by a particular level of imposed loading take years to stabilise, mainly because of the cracking, shrinkage and creep of concrete. Stress limits are also intended to ensure that after this initial period, live-load behaviour is reversible.

164


For the calculation of stresses, the principle of clause 7.2.1(1)P says, in effect, ‘take account of everything that is relevant’. It is thus open to interpretation, subject to the guidance in the rest of clause 7.2.1. Four of its paragraphs are worded ‘may’. For persistent design situations, it is usual to check stresses soon after the opening of the bridge to traffic, ignoring creep, and also at a time when further effects of creep and shrinkage have become negligible. Their values are usually found by letting t ! 1 when applying the data on creep and shrinkage in clause 3.1.4 and Annex B of EN 1992-1-1. Assuming that t ¼ 10 years, for example, gives only about 90% of the long-term shrinkage strain and creep coefficient. It may be necessary to include part of the long-term shrinkage effects in the first check, because up to half of the long-term shrinkage can occur in the first three months after the end of curing of the concrete. Clause 7.2.1(4) refers to the primary effects of shrinkage. These are calculated for uncracked cross-sections (Example 5.3). After cracking, these effects remain in the concrete between cracks, but have negligible influence on stresses at the cracked cross-sections, at which stresses are verified. Clauses 7.2.1(6) and (7) refer to tension stiffening. At a cross-section analysed as cracked, its effect is to increase the tensile stress in the reinforcement, as discussed under clause 7.4.3. It has negligible effect on the stress in the steel flange adjacent to the slab, and slightly reduces the compressive stress in the other steel flange. Clause 7.2.1(8) refers to the effects of local actions on the concrete slab, presumably a deck slab. In highway bridges, these effects are mainly the sagging and hogging moments caused by a single wheel, a pair of wheels, or a four-wheel tandem system, whichever is the most adverse. In Load Model 1 these are combined with the effects of distributed loading and the global effects in the plane of the slab. This combination is more adverse where the slab spans longitudinally between cross-beams than for transverse spanning. Longitudinal spanning can also occur at intermediate supports at the face of diaphragms. In combining the stress ranges, it is important to consider the actual transverse location being checked within the slab. The peak local effect usually occurs some distance from the web of a main beam, while the global direct stress reduces away from the web due to shear lag. The global stress distribution allowing for shear lag may be determined using clause 5.4.1.2(8), even though this refers to EN 1993-1-5 which is for steel flanges. For serviceability stress limits, clause 7.2.2 refers to EN 1992 and EN 1993. Both codes allow choice in the National Annex. EN 1992 does so by means of coefficients ki , whereas EN 1993 permits national values for a partial factor M;ser . If any National Annex uses other than the recommended value, 1.0, this could be a source of error in practice, because partial factors for serviceability checks are almost invariably 1.0, and so tend to be forgotten.

Combinations of actions for serviceability checks Clause 7.2 defines these combinations only by cross-reference, so the following summary is given. The limiting stresses can be altered by National Annexes. Clause 7.2.2(2) leads to the following recommendations for concrete. .

.

Clause 7.2.1(1)P

Clause 7.2.1(4)

Clause 7.2.1(6) Clause 7.2.1(7)

Clause 7.2.1(8)

Clause 7.2.2

Clause 7.2.2(2)

Where creep coefficients are based on clause 3.1.4(2) of EN 1992-1-1, as is usual, its clause 7.2(3) gives the stress limit for the quasi-permanent combination as 0.45 fck . In areas where the exposure class is XD, XF or XS, clause 7.2(102) of EN 1992-2 gives the stress limit for the characteristic combination as 0.60 fck . Comment on clause 4.1(1) refers to the XD class in bridges.

From clause 7.2.2(4), the recommended limit for reinforcement is 0.8 fsk under the characteristic combination, increased to 1.0 fsk for imposed deformations. Clause 7.2.2(5) refers to clause 7.3 of EN 1993-2, where the stress limits for structural steel and the force limits for bolts are based on the characteristic combination, with a limit on stress range under the frequent combination. For service loading of shear connectors generally, clause 7.2.2(6) refers to clause 6.8.1(3). That clause refers only to stud connectors under the characteristic combination, and uses a

Clause 7.2.2(4) Clause 7.2.2(5)

Clause 7.2.2(6)

165


factor k that can be chosen nationally. EN 1994-2 envisages the use of other types of connector (for example, in clause 6.6.1.1(6)P). Rules for the use of these, which may be given in a National Annex, from clause 1.1.3(3), should include a service load limit. To sum up, most stress checks are based on characteristic combinations, as are the determination of cracked regions, clause 5.4.2.3(2), and the provision of minimum reinforcement, clause 7.4.2(5). However, limiting crack widths are given, in clause 7.3.1(105) of EN 1992-2, for the quasi-permanent combination.

Clause 7.2.3(1)

Web breathing Clause 7.2.3(1) refers to EN 1993-2 for ‘breathing’ of slender steel web plates. The effect on a slender plate of in-plane shear or compressive stress is to magnify its initial out-of-plane imperfection. This induces cyclic bending moments at its welded edges about axes parallel to the welds. If excessive, it can lead to fatigue failure in these regions. Further comment is given in the Guide to EN 1993-2.4

7.3. Deformations in bridges 7.3.1. Deflections Clause 7.3.1(1)

Clause 7.3.1(2)

Clause 7.3.1(3)

Clause 7.3.1(1) refers to clauses in EN 1993-2 that cover clearances, visual impression, precambering, slip at connections, performance criteria and drainage. For precambering, ‘the effects of shear deformation . . . should be considered’. This applies to vertical shear in steel webs, not to the shear connection. Clause 7.3.1(2) refers to Section 5 for calculation of deflections. Rules for the effects of slip are given in clause 5.4.1.1. They permit deformations caused by slip of shear connection to be neglected, except in non-linear analysis. Clause 5.4.2.1(1) refers to the sequence of construction, which affects deflections. When the sequence is unknown, an estimate on the high side can be obtained by assuming unpropped construction and that the adverse areas of the influence line, with respect to deflection at the point being considered, are concreted first, followed by the relieving areas. Sufficient accuracy should usually be obtained by assuming that the whole of the concrete deck is cast at one time, on unpropped steelwork. The casting of an area of deck slab may increase the curvature of adjacent beams where the shear connectors are surrounded by concrete that is too young for full composite action to occur. It is possible that subsequent performance of these connectors could be impaired by what is, in effect, an imposed slip. Clause 7.3.1(3) refers to this, but not to the detailed guidance given in clause 6.6.5.2(3), which follows. ‘Wherever possible, deformation should not be imposed on a shear connection until the concrete has reached a cylinder strength of at least 20 N/mm2 .’

The words ‘Wherever possible’ are necessary because shrinkage effects apply force to shear connection from a very early age without, so far as is known, any adverse effect.

7.3.2. Vibrations Clause 7.3.2(1)

166

The limit state of vibration is covered in clause 7.3.2(1) by reference to other Eurocodes. Composite bridges are referred to only in clause 6.4.6.3.1(3) of EN 1991-2, which covers resonance under railway loading. This gives ‘lower bound’ values for damping that are the same for composite bridges as for steel bridges, except that those for filler-beam decks are much higher, and the same as for concrete bridges. Alternative values may be given in the National Annex. The specialized literature generally gives damping values for composite floor or deck systems that are between those for steel and for concrete members, as would be expected. In railway bridges, the presence or absence of ballast is a relevant factor. The reference to EN 1993-2 requires consideration of pedestrian discomfort and fatigue under wind-induced motion, usually vortex shedding. The relevant reference is then to EN 1991-1-4.103 Its Annex E provides guidance on the calculation of amplitudes of oscillation while its Annex F provides guidance on the determination of natural frequencies and


damping. The damping values for steel-composite bridges in its Table F.2 this time do lie between the values for steel bridges and reinforced concrete bridges.

7.4. Cracking of concrete 7.4.1. General

In the early 1980s it was found64;104 that for composite beams in hogging bending, the longestablished British methods for control of crack width were unreliable for initial cracks, which were wider than predicted. Before this, it had been found for reinforced concrete that the appropriate theoretical model for cracking caused by restraint of imposed deformation was different from that for cracking caused by applied loading. This has led to the presentation of design rules for control of cracking as two distinct procedures: .

.

for minimum reinforcement, in clause 7.4.2, for all cross-sections that could be subjected to significant tension by imposed deformations (e.g. by effects of shrinkage, which cause higher stresses than in reinforced concrete, because of restraint from the steel beam) for reinforcement to control cracking due to direct loading, clause 7.4.3.

The rules given in EN 1994-2 are based on an extensive and quite complex theory, supported by testing on hogging regions of composite beams.104;105 Much of the original literature is either in German or not widely available106 so a detailed account of the theory has been published in English,107 with comparisons with results of tests on composite beams, additional to those used originally. The paper includes derivations of the equations given in clause 7.4, comments on their scope and underlying assumptions, and procedures for estimating the mean width and spacing of cracks. These are tedious, and so are not in EN 1994-2. Its methods are simple: Tables 7.1 and 7.2 give maximum diameters and spacings of reinforcing bars for three design crack widths: 0.2, 0.3 and 0.4 mm. These tables are for ‘high-bond’ bars only. This means ribbed bars with properties referred to in clause 3.2.2(2)P of EN 1992-1-1. The use of reinforcement other than ribbed is outside the scope of the Eurocodes. The references to EN 1992 in clause 7.4.1(1) give the surface crack-width limits required for design. Typical exposure classes for composite bridge decks are discussed in section 4.1 of this guide. Clause 7.4.1(2) refers to ‘estimation’ of crack width, using EN 1992-1-1. This rather long procedure is rarely needed, and does not take full account of the following differences between the behaviours of composite beams and reinforced concrete T-beams. The steel member in a composite beam does not shrink or creep and has much greater flexural stiffness than the reinforcement in a concrete beam. Also, the steel member is attached to the concrete flange only by discrete connectors that are not effective until there is longitudinal slip, whereas in reinforced concrete there is monolithic connection. There is no need here for a reference to EN 1992-2. Clause 7.4.1(3) refers to the methods developed for composite members, which are easier to apply than the methods for reinforced concrete members. Clause 7.4.1(4) refers to limiting calculated crack widths wk , with a Note on recommended values. Those for all XC, XD and XS exposure classes are given in a Note to clause 7.3.1(105) of EN 1992-2 as 0.3 mm. This is for the quasi-permanent load combination, and excludes prestressed members with bonded tendons. Both the crack width and the load combination may be changed in the National Annex. It is expected that the UK’s National Annex to EN 1992-2 will confirm these recommendations and give further guidance for combinations that include temperature difference. Clause 7.4.1(5) and (6) draws attention to the need to control cracking caused by early thermal shrinkage. The problem is that the heat of hydration causes expansion of the concrete before it is stiff enough for restraint from steel to cause much compressive stress in it. When it cools, it is stiffer, so tension develops. This can occur in regions that

Clause 7.4.1(1)

Clause 7.4.1(2)

Clause 7.4.1(3) Clause 7.4.1(4)

Clause 7.4.1(5) Clause 7.4.1(6)

167


are in permanent compression in the finished bridge. They may require crack-control reinforcement for this phase only. The check is made assuming that the temperatures of the steel and the concrete are both uniform. The concrete is colder, to an extent that may be given in the National Annex. This causes tension, and possibly cracking. Further comment is given in Example 7.1.

7.4.2. Minimum reinforcement

Clause 7.4.2(1)

The only data needed when using Tables 7.1 and 7.2 are the design crack width and the tensile stress in the reinforcement, s . For minimum reinforcement, s is the stress immediately after initial cracking. It is assumed that cracking does not change the curvature of the steel beam, so all of the tensile force in the concrete just before cracking is transferred to the reinforcement, of area As . If the slab were in uniform tension, equation (7.1) in clause 7.4.2(1) would be: As s ¼ Act fct;eff

Clause 7.4.2(2)


where fct;eff is an estimate of the mean tensile strength of the concrete at the time of cracking. The three correction factors in equation (7.1) are based on calibration work.106 These allow for the non-uniform stress distribution in the area Act of concrete assumed to crack. ‘Non-uniform self-equilibrating stresses’ arise from primary shrinkage and temperature effects, which cause curvature of the composite member. Slip of the shear connection also causes curvature and reduces the tensile force in the slab. The magnitude of these effects depends on the geometry of the uncracked composite section, as given by equation (7.2). With experience, calculation of kc can often be omitted, because it is less than 1.0 only where z0 < 1:2hc . (These symbols are shown in Fig. 7.5.) The depth of the ‘uncracked’ neutral axis below the bottom of the slab normally exceeds about 70% of the slab thickness, and then, kc ¼ 1. The method of clause 7.4.2(1) is not intended for the control of early thermal cracking, which can occur in concrete a few days old, if the temperature rise caused by heat of hydration is excessive. The flanges of composite beams are usually too thin for this to occur. It would not be correct, therefore, to assume a very low value for fct;eff . The suggested value of fct;eff , 3 N/mm2 , was probably rounded from the mean 28-day tensile strength of grade C30/37 concrete, given in EN 1992-1-1 as 2.9 N/mm2 – the value used as the basis for the optional correction given in clause 7.4.2(2). The maximum bar diameter may be increased for stronger concrete because the higher bond strength of the concrete compensates for the lower total perimeter of a set of bars with given area per unit width of slab. The difference between 2.9 and 3.0 is obviously negligible. It may be an error in drafting, because in EN 1992, the value 2.9 N/mm2 is used in both places. If there is good reason to assume a value for fct;eff such that the correction is not negligible, a suitable procedure is to assume a standard bar diameter, , calculate , and then find s by interpolation in Table 7.1. The reinforcement in a deck slab will usually be in two layers in each direction, with at least half of it adjacent to the surface of greater tensile strain. The relevant rule, in clause 7.4.2(3), refers not to the actual reinforcement, but to the minimum required. The reference to ‘local depth’ in clause 7.4.2(4) means the depth at the cross-section considered. The rule of clause 7.4.2(5) on placing of minimum reinforcement refers to its horizontal extent, not to its depth within the slab. Analysis of the structure for ultimate load combinations of variable actions will normally find regions in tension that are more extensive than those for the characteristic combination specified here. The regions so found may need to be extended for early thermal effects (clause 7.4.1(5)).

Design of minimum reinforcement for a concrete slab For design, the design crack width and thickness of the slab, hc , will be known. For a chosen bar diameter , Table 7.1 gives s , the maximum permitted stress in the reinforcement, and equation (7.1) allows the bar spacing to be determined. If this is too high or low, is changed.

168


s (mm) 400 hc = 100 mm

150

300 200 200 300 100

0

5 6

8

10

12

16

20

φ (mm)

Fig. 7.1. Bar diameter and spacing for minimum reinforcement in two equal layers, for wk ¼ 0.3 mm and fct;eff ¼ 3.0 N/mm2

A typical relationship between slab thickness hc , bar spacing s and bar diameter is shown in Fig. 7.1. It is for two similar layers of bars, with kc ¼ 1 and fct;eff ¼ 3.0 N/mm2 . Equation (7.1) then gives, for a fully cracked slab of breadth b: (2 =4Þð2b=sÞ ¼ 0:72 3bhc =s Hence, hc s ¼ 0:7272 s (with s in N/mm2 units)

(D7.1) 2

For each bar diameter and a given crack width, Table 7.1 gives s , so the product hc s is known. This is plotted in Fig. 7.1, for wk ¼ 0.3 mm, as curves of bar spacing for four given slab thicknesses, which can of course also be read as maximum slab thickness size and for bar spacing. The shape of the curves results partly from the use of rounded values of s in Table 7.1. The correction to minimum reinforcement given in clause 7.4.2(2) is negligible here, and has not been made. The weight of minimum reinforcement, per unit area of slab, is proportional to 2 =s, which 1 is proportional to 1 s , from equation (D7.1). The value of s increases with bar diameter, from Table 7.1, so the use of smaller bars reduces the weight of minimum reinforcement. This is because their greater surface area provides more bond strength.

7.4.3. Control of cracking due to direct loading Clause 7.4.3(2) specifies elastic global analysis to Section 5, allowing for the effects of cracking. The preceding comments on global analysis for deformations apply also to this analysis for bending moments in regions with concrete in tension. From clause 7.4.1(4), the combination of actions will be given in the National Annex. There is no need to reduce the extent of the cracked regions below that assumed for global analysis for ultimate limit states, so the new bending moments for the composite members can be found by scaling values found previously. At each cross-section, the area of reinforcement will already be known: that required for ultimate loading or the specified minimum, if greater; so the stresses s;o , clause 7.4.3(3), can be found.

Clause 7.4.3(2)

Clause 7.4.3(3)

Tension stiffening A correction for tension stiffening is now required. At one time, these effects were not well understood. It was thought that, for a given tensile strain at the level of the reinforcement, the total extension must be the extension of the concrete plus the width of the cracks, so that allowing for the former reduced the latter. The true behaviour is more complex. The upper part of Fig. 7.2 shows a single crack in a concrete member with a central reinforcing bar. At the crack, the external tensile force N causes strain "s2 ¼ N=As Ea in the bar, and the strain in the concrete is the free shrinkage strain "cs , which is shown as

169


N

N

Tensile strain

εs2 εs(x) εsm εc(x)

εs1 εcm 0

x

εcs Le

Le

Fig. 7.2. Strain distributions near a crack in a reinforced concrete tension member

negative here. There is a transmission length Le each side of the crack, within which there is transfer of shear between the bar and the concrete. Outside this length, the strain in both the steel and the concrete is "s1 , and the stress in the concrete is fractionally below its tensile strength. Within the length 2Le , the curves "s (x) and "c (x) give the strains in the two materials, with mean strains "sm in the bar and "cm in the concrete. It is now supposed that the graph represents the typical behaviour of a reinforcing bar in a cracked concrete flange of a composite beam, in a region of constant bending moment such that the crack spacing is 2Le . The curvature of the steel beam is determined by the mean stiffness of the slab, not the fully cracked stiffness, and is compatible with the mean longitudinal strain in the reinforcement, "sm . Midway between the cracks, the strain is the cracking strain of the concrete, corresponding to a stress less than 30 N/mm2 in the bar. Its peak strain, at the crack, is much greater than "sm , but less than the yield strain of the reinforcement, if crack widths are not to exceed 0.3 mm. The crack width corresponds to this higher strain, not to the strain "sm that is compatible with the curvature, so a correction to the strain is needed. It is presented in clause 7.4.3(3) as a correction to the stress s;o because that is easily calculated, and Tables 7.1 and 7.2 are based on stress. The strain correction cannot be shown in Fig. 7.2 because the stress s;o is calculated using the ‘fully cracked’ stiffness, and so relates to a curvature greater than the true curvature. The derivation of the correction107 takes account of crack spacings less than 2Le , the bond properties of reinforcement, and other factors omitted from this simplified outline. The section properties needed for the calculation of the correction s will usually be known. For the cracked composite cross-section, the transformed area A is needed to find I, which is used in calculating s;o , and Aa and Ia are standard properties of the steel section. The result is independent of the modular ratio. For simplicity, st may conservatively be taken as 1.0, because AI > Aa Ia . When the stress s at a crack has been found, the maximum bar diameter or the maximum spacing are found from Tables 7.1 and 7.2. Only one of these is needed, as the known area of reinforcement then gives the other. The correction of clause 7.4.2(2) does not apply.

General comments on clause 7.4, and flow charts The design actions for checking cracking will always be less than those for the ultimate limit state due to the use of lower load factors. The difference is greatest where unpropped construction is used for a continuous beam with hogging regions in Class 1 or 2 and with lateral–torsional buckling prevented. This is because the entire design hogging moment is carried by the composite section for Class 1 and 2 composite sections at ULS, but at SLS, reinforcement stresses are derived only from actions applied to the composite section in the construction sequence. It is also permissible in such cases to neglect the effects of indirect actions at ULS. The quantity of reinforcement provided for resistance to load should be

170


See the Notes in the section ‘Flow charts for crack-width control’ Exposure classes. For each concrete surface in tension, find the exposure class to clause 4.2 of EN 1992-1-1 and EN 1992-2 (referred to from 7.4.1(1))

Crack widths. Find the limiting crack widths wk and the combination of actions for verification from the National Annex (from the Note to 7.4.1(4))

Do global analyses for this combination to find bending moments in regions where the slab is in tension

For areas of reinforcement found previously, As, determine tensile stresses σs,o in bars adjacent to surfaces to be checked, neglecting primary shrinkage, to 7.2.1(4). Calculate tensile stresses Δσs for tension stiffening, from 7.4.3(3). Determine the tensile stress due to coexisting local actions, σs,loc, and find σs,E = σs,o + Δσs + σs,loc, from 7.2.1(8)

Do you want to find crack widths from EN 1992-1-1, 7.3.4? (7.4.1(2))

Yes

Recommended only for cross-sections with longitudinal prestress by tendons. Outside the scope of this chart (END)

No From 7.4.1(3), use wk and σs,E to find either max. bar spacing ss fromTable 7.2 and calculate bar diameter φ from As or (less convenient) find diameter φ* fromTable 7.1, then φ from 7.4.2(2), and find bar spacing from As

Reduce φ and ss

Do you want to change φ or ss?

Increase ss

No

Reduce φ, at constant As. Effect is to reduce bar spacing

Only possible by increasing As and hence reducing σs,E. This may permit a small increase in φ. It is inefficient and not recommended

Go to flow chart for minimum reinforcement (Fig. 7.4)

Fig. 7.3. Flow chart for control of cracking due to direct loading

sufficient to control cracking. The main use of clause 7.4.3 is then to check that the spacing of the bars is not excessive. Where propped construction is used, the disparity between the design loadings for the two limit states is smaller. A check to clause 7.4.3 is then more likely to influence the reinforcement required.

Flow charts for crack-width control The check to clause 7.4.3 is likely to be done first, so its flow chart, Fig. 7.3, precedes Fig. 7.4 for minimum reinforcement, to clause 7.4.2. The regions where the slab is in tension depend on the load combination, and three may be relevant, as follows. . .

.

Most reinforcement areas are found initially for the ultimate combination. Load-induced cracking is checked for a combination to be specified in the National Annex, to clause 7.4.1(4). It may be the quasi-permanent or frequent combination. Minimum reinforcement is required in regions in tension under the characteristic combination, clause 7.4.2(5).

171


Minimum reinforcement. Find all regions where concrete can be in flexural tension under the characteristic combination of permanent and variable actions, 7.4.2(5), taking account of shrinkage of concrete and effects of temperature and settlement, if any, 7.4.2(1). These regions may be more extensive than those where crack control for effects of loading is required

For each region, propose details of minimum longitudinal reinforcement: bar size φ and spacing ss; usually in two layers, with at least half near the surface with the greater tensile strain (7.4.2(3)). Figure 7.1 is useful where fct,eff ≈ 3 N/mm2

Choose fct,eff, defined in 7.4.2(1). Use N/mm2 units

Calculate φ* = (2.9/fct,eff)φ

For the design crack width wk, find σs,max for φ* from Table 7.1, using interpolation if necessary. (This route is used because φ must be a standard bar diameter, but φ* need not be)

Do you wish to determine minimum reinforcement by a ‘more accurate method’, following clause 7.3.2(1)P of EN 1992-1-1 (from 7.4.2(1))? Yes Appropriate for sections prestressed by tendons, 7.4.2(1). Outside the scope of this flow chart (END)

No Calculate kc from eq. (7.2). Consider whether the recommended values for k and ks are appropriate, and change them if not (unlikely). For unit width of the tensile zone considered, find Act. Find As,min from eq. (7.1) using σs,max

Reduce ss or increase φ (or combination of both)

No Yes

For the proposed reinforcement, find As per unit width of slab. Is As ≥ As,min?

The chosen bar size and spacing are satisfactory as minimum reinforcement, but may not be sufficient to control cracking due to direct loading at the section considered (END)

Fig. 7.4. Flow chart for minimum reinforcement for control of cracking

The following notes apply to these charts. Note 1.

Note 2.

Creep and shrinkage of concrete both increase stress in reinforcement at internal supports of beams, so crack widths are usually verified using the long-term modular ratio for permanent actions. This is assumed here. The flow charts apply to a tension flange of a continuous longitudinal beam. It is assumed (for brevity) that: . .

.

.

. . .

172

areas of reinforcement required for ultimate limit states have been found the flexural stiffnesses Ea I1 are known for the uncracked cross-sections, using relevant modular ratios (inclusion of reinforcement optional) the cracked flexural stiffnesses Ea I2 are known. For these, modular ratios are usually irrelevant, unless double composite action is being used the deck slab is above the steel beam, and at cross-sections considered, maximum tensile strain occurs at the top surface of the slab there is no double composite action early thermal cracking does not govern the symbols As (area of reinforcement) are for a unit width of slab.


Note 3. A second subscript, l for loading, is used in this note to indicate quantities found in the check to clause 7.4.3. Area As;l is usually that required for resistance to ultimate loads. It is assumed, for simplicity, that minimum reinforcement consists of two identical layers of bars, one near each surface of the slab. Area As;l should be compared with the minimum reinforcement area As required when bars of diameter l are used. If As;l < As , minimum reinforcement governs.

7.5. Filler beam decks This clause is applicable to simply-supported or continuous decks of the type shown in Fig. 6.8, spanning longitudinally. The Note to clause 6.3.1(1) permits the use of transverse filler beams according to the National Annex, which should refer also to serviceability requirements, if any. From clause 7.5.1(1), the methods of global analysis for serviceability limit states are the same as for ultimate limit states, clause 5.4.2.9, except that no redistribution of moments is permitted. The word ‘considered ’ is used in clause 7.5.2(1) because some of the clauses may not be applicable. For example, the thickness of the concrete in a filler-beam deck will exceed that in a conventional composite beam, so the effects of heat of hydration will be greater. The temperature difference recommended in the Note to clause 7.4.1(6) may not be appropriate. Tension stiffening is also different (clause 7.5.4(2)). The objective of clause 7.5.3(1) is to ensure that cracking of concrete does not cause its reinforcement to yield. It applies above an internal support of a continuous filler-beam deck, and can be illustrated as follows. It is assumed that all the concrete above the top flanges of the steel beams reaches its mean tensile strength, fctm , and then cracks. This releases a tensile force of Ac;eff fctm per beam, where Ac;eff ¼ sw cst; as in clause 7.5.3(1). The notation is shown in Fig. 6.8. The required condition, with partial factors M taken as 1.0, is: As;min fsk Ac;eff fctm

Clause 7.5.1(1)

Clause 7.5.2(1)

Clause 7.5.3(1)

(D7.2) 2

For concretes permitted by clause 3.1(2), fctm 4:4 N/mm . If, to satisfy expression (7.7), As;min > 0.01Ac;eff , then expression (D7.2) is satisfied if fsk > 440 N/mm2 . In design to EN 1994, normally fsk ¼ 500 N/mm2 , so the objective is met. Clause 7.5.4(1) applies to longitudinal bottom reinforcement in mid-span regions. Widths of cracks in the concrete soffit between the steel beams should be controlled unless the formwork used (shown in Fig. 6.8) provides permanent protection. Otherwise, wide transverse cracks could form under the bottom transverse bars specified in clause 6.3.1(4), putting them at risk of corrosion. Clause 7.5.4(2) says, in effect, that the stress in the reinforcement may be taken as s;o , defined in clause 7.4.3(3). Thus, there is no need to consider tension stiffening in fillerbeam decks.

Clause 7.5.4(1)

Clause 7.5.4(2)

Example 7.1: checks on serviceability stresses, and control of cracking The following serviceability checks are performed for the internal girders of the bridge in Figs 5.6 and 6.22: . . . .

tensile stress in reinforcement and crack control at an intermediate support minimum longitudinal reinforcement for regions in hogging bending stresses in structural steel at an intermediate support compressive stress in concrete at mid-span of the main span.

Tensile stress in reinforcement The cross-section at an intermediate support is shown in Fig. 7.5. All reinforcement has fsk ¼ 500 N/mm2 .

173


3100 60 20 mm bars at 75 mm

20 mm bars at 150 mm

70

125

hc = 250 25 400 × 25

1225

z0

26 Including reinforcement

Elastic neutral axes for n0 = 6.36

Excluding reinforcement

1160 × 25

1120

400 × 40

Fig. 7.5. Elastic neutral axes of uncracked cross-section at internal support

From clause 7.2.1(8), stresses in reinforcement caused by simultaneous global and local actions should be added. Load Model 1 is considered here. The tandem system (TS) and UDL produce a smaller local effect than does Load Model 2, but Load Model 1 produces the greatest combined local plus global effect. Local moments are caused here by hogging of the deck slab over the pier diaphragm. Full local effects do not coexist with full global effects since, for maximum global hogging moment, the axles are further from the pier. The maximum local and global effects are here calculated independently and then combined. A less conservative approach, if permitted by the National Annex, is to use the combination rule in Annex E of EN 1993-2 as noted below clause 5.4.4(1). This enables the maximum global effect to be combined with a reduced local effect and vice versa. The maximum global hogging moment acting on the composite section under the characteristic combination was found to be 4400 kNm. Clause 7.2.1(4) allows the primary effects of shrinkage to be ignored for cracked sections, but not the secondary effects which are unfavourable here. From Table 6.2 in Example 6.6, the section modulus for the top layer of reinforcement in the cracked composite section is 34:05 106 mm3 . The stress due to global effects in this reinforcement, ignoring tension stiffening, is: s;o ¼ 4400=34:05 ¼ 129 N/mm2 From clause 7.2.1(6), the calculation of reinforcement stress should include the effects of tension stiffening. From clause 7.4.3(3), the increase in stress in the reinforcement, due to tension stiffening, from that calculated using a fully cracked analysis is: s ¼

0:4 fctm st s

where

st ¼

AI 74 478 2:266 1010 ¼ ¼ 2:50 Aa Ia 55 000 1:228 1010

The reinforcement ratio s ¼ 0.025 and fctm ¼ 2.9 N/mm2 and thus: s ¼

0:4 2:9 ¼ 19 N=mm2 2:50 0:025

The stress in the reinforcement from global effects is therefore: 129 þ 19 ¼ 148 N/mm2 The stress in the reinforcement due to local load was found by determining the local moment from an analysis using the appropriate Pucher chart.100 This bending moment, 12.2 kNm/m hogging, is not sufficient to cause compression in the slab, so it is resisted solely by the two layers of reinforcement, and causes a tensile stress of 73 N/mm2 in the top layer. The total stress from global plus local effects is therefore: s ¼ 148 þ 73 ¼ 221 N/mm2

174


From clause 7.2.2(4) and from EN 1992-1-1 clause 7.2(5), the tensile stress in the reinforcement should not exceed k3 fsk ¼ 400 N/mm2 (recommended k3 ¼ 0:8, and fsk ¼ 500 N/mm2 ) so this check is satisfied. Where the global tension is such that local effects cause net compression at one face of the slab, iterative calculation can be avoided by taking the lever arm for local bending as the lesser of the distance between the two layers of reinforcement and that derived from considering the cracked section in flexure alone.

Control of cracking due to direct loading – persistent design situation Creep of concrete in sagging regions causes hogging moments to increase, so this check is done at t ! 1. Cracking during construction is considered later. For concrete protected by waterproofing, EN 1992-2 clause 4.2(105) recommends an exposure class of XC3 for which Table 7.101N of EN 1992-2 recommends a limiting crack width of 0.3 mm under the quasi-permanent load combination. Crack widths are checked under this load combination for which the maximum hogging moment acting on the composite section was found to be 2500 kNm. The secondary effects of shrinkage were unfavourable here. The primary effects can be ignored because the derivation of the simplified method (Tables 7.1 and 7.2) takes account of a typical amount of shrinkage.107 Using a section modulus from Table 6.2, the stress due to global effects in the top layer of reinforcement, ignoring tension stiffening, is: s;o ¼ 2500=34:05 ¼ 73 N/mm2 The increase in stress in the reinforcement due to tension stiffening is 19 N/mm2 as found above. From equation (7.4), the reinforcement stress is: s ¼ 73 þ 19 ¼ 92 N/mm2 For this stress, Table 7.1 gives a bar diameter above 32 mm, and Table 7.2 gives a bar spacing exceeding 300 mm, so there is sufficient crack control from the reinforcement provided, 20 mm bars at 75 mm spacing.

Control of cracking during construction Construction is unpropped, so clause 7.4.1(5) is applicable. Effects of heat of hydration of cement should be analysed ‘to define areas where tension is expected ’. These areas require at least minimum reinforcement, from clause 7.4.2. The curvatures for determining secondary effects should be based initially on uncracked cross-sections. Clause 7.4.1(6) requires account to be taken of the effects of heat of hydration for limitation of crack width ‘unless specific measures are taken’ to limit its effects. The National Annex may refer to specific measures. Assuming that no specific measures have been taken, clause 7.4.1(6) applies. Its Note recommends that the concrete slab should be assumed to be 20 K colder than the steel member, and that the short-term modular ratio n0 should be used. It is expected that the National Annex for the UK will replace 20 K by 25 K, the value used here. Referring to Fig. 5.6, it is assumed that the whole of the deck slab has been concreted except for lengths such as CD, which extend for 6 m each side of each internal support. These two 12 m lengths are now cast, over the whole width of the deck. When they harden, it is assumed that the only longitudinal stress in their concrete arises from shrinkage and the temperature difference. The primary effects are uniform over the whole 12 m length. The secondary effects are a hogging bending moment over the whole length of the bridge, which is a maximum at the internal supports, and associated shear forces. It is reasonable to assume that the 12 m lengths are either cracked throughout, or uncracked. In calculating secondary effects, primary effects due to shrinkage can be neglected in cracked regions, from clause 5.4.2.2(8), so only effects of heat of hydration

175


need be considered. A similar conclusion is reached if the method of clause 5.4.2.3(3), ‘15% of the span cracked’, is used. In any case, shrinkage is a minor effect, as shown below. From Example 5.3, the autogenous and long-term shrinkage strains are "ca ¼ 50 106 and "cd ¼ 282 106 respectively. From clause 3.1.4(6) of EN 1992-1-1, drying shrinkage may be assumed to commence at the end of curing. Autogenous shrinkage is a function of the age of the concrete, t in days: "ca ðtÞ ¼ ½1 expð0:2t 0:5 Þ"ca ð1Þ Here, the temperature difference is assumed to reach its peak at age seven days, and curing is conservatively assumed to have ended at age three days. From EN 1992-1-1, less than 1% of the long-term drying shrinkage will have occurred in the next four days, so it is neglected. From the equation above, "ca ð7Þ ¼ ½1 expð0:53Þ 50 106 ¼ 21 106 For temperature, clause 3.1.3(5) of EN 1992-1-1 gives the thermal coefficient for concrete as 10 106 , so a difference of 25 K causes a strain of 250 106 . The resulting sagging curvature of each 12 m length of deck is found by analysis of the uncracked composite section with n0 ¼ 6.36 from Example 5.2. To find whether cracking occurs, the secondary effects are found from these curvatures by global analysis using uncracked stiffnesses. The definition of ‘regions in tension’ is based on the characteristic combination, from clause 5.4.2.3(2). It is suggested here that heat of hydration should be treated as a permanent action, as shrinkage is. The leading variable action is likely to be construction load elsewhere on the deck, in association with either temperature or wind. The permanent actions are dead load and any drying shrinkage in the rest of the deck. The action effects so found from ‘uncracked’ analysis are added to the secondary effects, above, to determine the regions in tension, in which at least minimum reinforcement (to clause 7.4.2) should be provided. If tensile stresses in concrete are such that it cracks, reanalysis using cracked sections and ignoring primary effects gives the tensile stresses in reinforcement needed for crack-width control. Minimum reinforcement will not be sufficient at the internal supports, where reinforcement will have been designed both for resistance to ultimate direct loading and for long-term crack-width control. The seven-day situation studied here is unlikely to be more adverse.

Minimum reinforcement The minimum reinforcement required by clause 7.4.2 is usually far less than that required at an internal support. The rules apply to any region subjected to significant tension and can govern where the main longitudinal reinforcement is curtailed, or near a point of contraflexure. Figure 7.1 gives the minimum reinforcement for this 250 mm slab as 20 mm bars at 260 mm spacing, top and bottom. The top layer has a cross-section only 29% of that of the top layer provided at the pier. This result is now checked. From clause 7.4.2(1): As;min ¼ ks kc k fct;eff Act =s

(7.1)

with ks ¼ 0.9 and k ¼ 0.8. Assuming that the age of the concrete at cracking is likely to exceed 28 days, fct;eff ¼ 3.0 N/mm2 The only term in equation (7.1) that depends on the steel cross-section is kc , given by: kc ¼

176

1 þ 0:3 1:0 1 þ hc =ð2z0 Þ

ð7:2Þ


where hc is the flange thickness (250 mm) and z0 is the distance between the centres of area of the uncracked composite section and the uncracked concrete flange, with modular ratio n0 (6.36 here). It makes little difference whether reinforcement is included or not. Excluding it (for simplicity) and ignoring the haunch, then from Fig. 7.5: z0 ¼ 1500 1120 125 ¼ 255 mm and kc ¼ 1=ð1 þ 250=510Þ þ 0:3 ¼ 0:97 If the area of longitudinal reinforcement at the internal support is included, kc is reduced by only 0.02. Ignoring the difference between 2.9 and 3.0 N/mm2 , Table 7.1 gives the maximum stress for 20 mm bars as 220 N/mm2 for a limiting crack width of 0.3 mm. For a 1 m width of flange, from equation (7.1): As;min ¼ 0:9 0:97 0:8 3:0 1000 250=220 ¼ 2380 mm2 /m Clause 7.4.2(3) requires at least half of this reinforcement to be placed in the upper layer. This gives 20 mm bars at 264 mm spacing; probably 250 mm in practice. A similar calculation for 16 mm bars gives their spacing as 184 mm.

Stress limits for structural steel The hogging moments under the characteristic load combination are found to be 1600 kNm acting on the steel beam and 4400 kNm acting on the composite section. The relevant section moduli are given in Table 6.2 in Example 6.6. The stress in the bottom flange was found to be critical, for which: 1600 4400 þ ¼ 222 N=mm2 Ed;ser ¼ 22:20 29:25 The coexisting shear force acting on the section was found to be 1900 kN, so: 1900 103 ¼ 66 N=mm2 25 1160 if buckling is neglected at SLS. From clause 7.2.2(5) and EC3-2 clause 7.3, stresses should be limited as follows: Ed;ser ¼

Ed;ser ¼ 222

fy ¼ 345 N=mm2 M;ser

fy Ed;ser ¼ 66 pffiffiffi ¼ 199 N=mm2 3M;ser qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi fy 2 2Ed;ser þ 3Ed;ser ¼ 250 ¼ 345 N=mm2 M;ser where M;ser ¼ 1.00. The stresses in structural steel are all satisfactory.

Compressive stress in concrete The cross-section at mid-span is shown in Fig. 6.2. The compressive stress in the slab is checked under simultaneous global and local actions. Both the primary and secondary effects of shrinkage are favourable and are conservatively ignored. The concrete is checked shortly after opening the bridge, ignoring creep, as this produces the greatest concrete compressive stress here, so the short-term modular ratio is used. The sagging moment on the composite section from the characteristic combination was found to be 3500 kNm. The section modulus for the top of the concrete slab is 575:9 106 mm3 from Table 6.4 in Example 6.10. The compressive stress at the top of the slab is: c ¼ 3500=575:9 ¼ 6:1 N/mm2

177


The maximum longitudinal sagging moment from local loading was found to be 20 kNm/m. The short-term section modulus for the top surface of the uncracked reinforced slab is Wc;top ¼ 10:7 106 mm4 /m in ‘concrete’ units, so the compressive stress from local load is c;loc ¼ 20=10:7 ¼ 1:9 N/mm2 . The total concrete stress, fully combining global and local effects, is: c ¼ 6:1 þ 1:9 ¼ 8:0 N/mm2 The global stress is sufficient to keep the whole depth of the slab in compression so this calculation of stresses from local moment using gross properties for the slab is appropriate. If the local moment caused tension in the slab, this would no longer be adequate. Either the stress from local moment could conservatively be calculated from a cracked section analysis considering the local moment acting alone, or an iterative calculation considering the local moment and axial force acting together would be required. In this case, application of the combination rule of EN 1993-2 Annex E would underestimate the combined effect as the global and local loading cases are the same. From clause 7.2.2(2) and from EN 1992-1-1 clause 7.2 with the recommended value for k1 , concrete stresses should be limited to k1 fck ¼ 0:6 30 ¼ 18 N/mm2 so the concrete stress is acceptable.

178

CHAPTER 8

Precast concrete slabs in composite bridges This chapter corresponds to Section 8 in EN 1994-2, which has the following clauses: . . . .

General Actions Design, analysis and detailing of the bridge slab Interface between steel beam and concrete slab


8.1. General Clause 8.1(1) states the scope of Section 8: precast deck slabs of reinforced or prestressed concrete which are either: .

.

Clause 8.1(1)

partial thickness, acting as permanent participating formwork to the in-situ concrete topping, or full thickness, where only a small quantity of concrete needs to be cast in situ to join the precast units together. Figure 8.1 illustrates a typical deck of this type.

Precast slabs within the scope of Section 8 should be fully composite with the steel beam – clause 8.1(2). Non-participating permanent formwork is not covered, for it is difficult both to prevent such formwork from being stressed by imposed loading, and to ensure its durability.

Pocket for shear connectors

Precast or in situ edge beam

Transverse joint

Precast or in situ edge beam Projecting reinforcement not shown

Fig. 8.1. Typical full-thickness precast concrete deck

Precast deck slab unit spanning multiple girders

Clause 8.1(2)


Clause 8.1(3)

Clause 8.1(3) is a reminder that the designer should check the sensitivity of the detailing to tolerances and specify stricter values than those required by EN 1992 (through EN 13670) if necessary. Key issues to consider include: .

.

.

detailing of the precast slabs at pockets to ensure that each pocket is correctly located over the steel beam, that projecting transverse reinforcement will not clash with the shear connection, and that there is sufficient space for concreting (clause 8.4.3(2)) detailing of stitch reinforcement between adjacent precast slabs to ensure that bars do not clash and to satisfy clause 8.3(1) on continuity tolerances on overall geometry of each precast unit so that, where required, abutting units are sufficiently parallel to each other to avoid the need for additional sealing from underneath. The tolerances for steelwork are also important, and are referred to in clause 8.4.1.

8.2. Actions Clause 8.2(1)

Clause 8.2(1) warns that the design of precast deck slabs should consider the actions arising from the proposed construction method as well as the actions given in EN 1991-1-6.30

8.3. Design, analysis and detailing of the bridge slab Clause 8.3(1)


Even where full-thickness slabs are used, some interaction with in-situ concrete occurs at joints, so clause 8.3(1) is relevant to both types of precast concrete slab. Its requirement for the deck to be designed as continuous in both directions applies to the finished structure. It does not mean that the reinforcement in partial-thickness precast slabs or planks must be continuous. That would exclude the use of ‘Omnia’-type planks, shown in Fig. 8.2. Precast planks of this sort span simply-supported between adjacent steel beams and are joined with in-situ concrete over the tops of the beams. The main reinforcement in the planks is not continuous across these joints, but the reinforcement in the in-situ concrete is. In the other direction, the planks abut as shown, so that only a small part of the thickness of the slab is discontinuous in compression. Continuity of reinforcement is again achieved in the slab but not in the planks. The resulting slab (part precast, part in situ) is continuous in both directions. EN 1992-1-1 clause 6.2.5 is relevant for the horizontal interface between the precast and in situ concrete. Examples of bridges of this type are given in Ref. 108. To allow precast slab units to be laid continuously across the steel beams, shear connection usually needs to be concentrated in groups with appropriate positioning of pockets in the precast slab as illustrated in Fig. 8.1. Clause 8.3(2) therefore refers to clause 6.6.5.5(4) for the use of stud connectors in groups. Clause 8.3(3) makes reference to clause 6.6.1.2. This allows some degree of averaging of the shear flow over a length, which facilitates standardisation of the details of the shear connection and the pockets.

In-situ concrete

Precast plank

Fig. 8.2. Typical partial-thickness precast concrete planks

180

Flange of steel beam

CHAPTER 8. PRECAST CONCRETE SLABS IN COMPOSITE BRIDGES

8.4. Interface between steel beam and concrete slab Clause 8.4.1(1) refers to bedding, such as the placing of the slabs on a layer of mortar. Sealing of the interface between steel beam and precast beam is needed both to protect the steel flange from corrosion and to prevent leakage of grout when the pockets are concreted. Where a precast unit is supported by more than two beams, bedding may also be needed to ensure that load is shared between the beams as intended. ‘Bedding’ in clause 8.4.1(1) appears to mean a gap-filling material capable of transferring vertical compression. Where it is intended not to use it, the clause requires special tolerances to be specified for the steelwork to minimise the effects of uneven contact between slab and steel flange. This does not solve the problems of corrosion and grout leakage, for which a compressible sealing strip could be applied to the edges of the flange and around the pocket. There would then still be no direct protection of the top flange by in-situ concrete (other than at a pocket) and so clause 8.4.2(1) requires that a top flange without bedding be given the same corrosion protection as the rest of the beam, apart from the site-applied top coat. If a non-loadbearing anti-corrosion bedding is provided, then the slab should be designed for the transfer of vertical loads only at the positions of the pockets. It would be prudent also to assume that clause 8.4.1(1) on special tolerances still applies. Clause 8.4.3 gives provisions for the shear connection and transverse reinforcement, supplementing Sections 6 and 7. Clause 8.4.3(2) emphasises the need for both suitable concrete mix design and appropriate clearance between shear connectors and precast concrete, allowing for tolerances, in order to enable in-situ concrete to be fully compacted. Clause 8.4.3(3) highlights the need to detail reinforcement appropriately adjacent to groups of connectors. This is discussed with the comments on clause 6.6.5.5(4). EN 1994-2 gives no specific guidance on the detailing of the transverse and longitudinal joints between precast deck units. Transverse joints between full-depth precast slabs at the intermediate supports of continuous bridges are particularly critical. Here, the slab reinforcement must transmit the tension caused by both global hogging moments and the bending moment from local loading. To allow for full laps in the reinforcement, a clear gap between units would need to be large and a problem arises as to how to form the soffit to the joint. One potential solution is to reduce the gap by using interlocking looped bars protruding from each end of adjoining slab units. Such a splicing detail is not covered in EN 1992-2, other than in the strut-and-tie rules. Experience has shown that even if satisfactory ultimate performance can be established by calculation, tests may be required to demonstrate acceptable performance at the serviceability limit state and under fatigue loading. The publication Precast Concrete Decks for Composite Highway Bridges109 gives further guidance on the detailing of longitudinal and transverse joints for a variety of bridge types.

Clause 8.4.1(1)

Clause 8.4.2(1)

Clause 8.4.3 Clause 8.4.3(2) Clause 8.4.3(3)

181

CHAPTER 9

Composite plates in bridges This chapter corresponds to Section 9 of EN 1994-2, which has the following clauses: . . . .

General Design for local effects Design for global effects Design of shear connectors


9.1. General A composite plate comprises a steel plate acting compositely with a concrete slab in both longitudinal and transverse directions. The requirements of Section 9 apply to composite top flanges of box girders, which resist local wheel loads in addition to performing the function of a flange in the global system. Clause 9.1(1) clarifies that this section of EN 1994-2 does not cover composite plates with shear connectors other than headed studs, or sandwich construction where the concrete is enclosed by a top and bottom steel plate. Composite plates can also be used as bottom flanges of box girders in hogging zones. This reduces the amount of stiffening required to prevent buckling. Composite bottom flanges have been used both in new bridges110;111 and for strengthening older structures. Clause 9.1(2) imposes a deflection limit on the steel flange under the weight of wet concrete, unless the additional weight of concrete due to the deflection is included in the calculation. In most bridges where this deflection limit would be approached, the steel top plate would probably require stiffening to resist the global compression during construction. Clause 9.1(3) gives a modified definition for b0 in clause 5.4.1.2 on shear lag. Its effect is that where the composite plate has no projection beyond an outer web, the value of b0 for that web is zero. For global analysis, the effects of staged construction, cracking, creep and shrinkage, and shear lag all apply. Clause 9.1(4) therefore makes reference to clause 5.4, together with clause 5.1 on structural modelling.

9.2. Design for local effects Local effects arise from vertical loading, usually from wheels or ballast, acting on the composite plate. For flanges without longitudinal stiffeners, most of the load is usually carried by transverse spanning between webs, but longitudinal spanning also occurs in the vicinity of any cross-beams and diaphragms. For flanges with longitudinal stiffeners, the direction of spanning depends on the flange geometry and the relative stiffnesses of the various components. It is important to consider local loading for the fatigue check of the studs as the longitudinal shear from wheel loads can be as significant as that from the global loading in low-shear regions of the main member.

Clause 9.1(1)

Clause 9.1(2)

Clause 9.1(3)

Clause 9.1(4)


Clause 9.2(1)

Clause 9.2(2)

Clause 9.2(3)

Clause 9.2(1) permits the local analysis to be carried out using elastic analysis with uncracked concrete properties throughout. This is reasonable because the concrete is likely to be cracked in flexure regardless of the sign of the bending moment. There is therefore no need to distinguish between uncracked and cracked behaviour, although where the steel flange is in tension, the cracked stiffness is likely to be significantly higher for sagging moments than for hogging moments. The same assumption is made in the design of reinforced concrete and is justified at ultimate limit states by the lower-bound theorem of plasticity. Clause 9.2(1) also clarifies that the provisions of Section 9 need not be applied to the composite flange of a discrete steel I-girder, since the flange will not usually be wide enough for significant composite action to develop across its width. A small amount of slip can be expected between the steel plate and concrete slab, as discussed in the comments under clause 9.4(4), but as in beams its effect on composite action is small. Clause 9.2(2) therefore allows slip to be ignored when determining resistances. Excessive slip could however cause premature failure. This needs to be prevented by following the applicable provisions of clause 6.6 on shear connection in conjunction with clause 9.4. Providing the shear studs are designed as above, the steel deck plate may be taken to act fully compositely with the slab. Clause 9.2(3) then permits the section to be designed for flexure as if the steel flange plate were reinforcement. The requirements of EN 1992-2 clause 6.1 should then be followed. The shear resistance may similarly be derived by treating the composite plate as a reinforced concrete section without links according to EN 1992-2 clause 6.2.2 (as modified by clause 6.2.2.5(3)), provided that the spacing of the studs transversely and longitudinally is less than three times the thickness of the composite plate. The studs should also be designed for the longitudinal shear flow from local loading for ultimate limit states, other than fatigue, and for the shear flow from combined global and local effects at serviceability and fatigue limit states. Both punching and flexural shear should be checked. Checks on flexural shear for unstiffened parts of the composite plate should follow the usual procedures for reinforced concrete design. An effective width of slab, similar to that shown below in Fig. 9.1, could be assumed when determining the width of slab resisting flexural shear. Checks on punching shear could consider any support provided by longitudinal stiffeners, although this could conservatively be ignored.

9.3. Design for global effects Clause 9.3(1)

Clause 9.3(2)

184

Clause 9.3(1) requires the composite plate to be designed for the effects induced in it by axial force, bending moment and torsion acting on the main girder. In the longitudinal direction, the composite plate will therefore resist direct compression or tension. Most bridge box girders will be in Class 3 or Class 4 and therefore the elastic stresses derived in the concrete and steel elements should be limited to the values in clause 6.2.1.5 for ultimate limit states. Torsion acting on the box will induce in-plane shear in both steel and concrete elements of the flange. These shear flows can be determined using a transformed section for the concrete as given in clauses 5.4.2.2(11) and 5.4.2.3(6). Checks of the steel flange under combined direct stress and in-plane shear are discussed under the comments on clause 6.2.2.4(3). The concrete flange should be checked for in-plane shear in accordance with EN 1992-2 clause 6.2. Distortion of a box girder will cause warping of the box walls, and thus in-plane bending in the composite plate. The direct stresses from warping will need to be added to those from global bending and axial force. Distortion will also cause transverse bending of the composite plate. Once a steel flange in compression is connected to the concrete slab, it is usually assumed that the steel flange panels are prevented from buckling (providing the shear studs are spaced sufficiently closely – clause 9.4(7) refers). It is still possible, although very unlikely, that the composite plate might buckle as a whole. Clause 9.3(2) acknowledges this possibility and

CHAPTER 9. COMPOSITE PLATES IN BRIDGES

requires reference to be made to clause 5.8 of EN 1992-1-1 for the calculation of the secondorder effects. None of the simple methods of accounting for second-order effects in this clause apply to plates so a general second-order non-linear analysis with imperfections would be required in accordance with EN 1992-1-1 clause 5.8.6. No guidance is given in EN 1992-1-1 on imperfections in plate elements. The imperfection shape could be based on the elastic critical buckling mode shape for the composite plate. The magnitude of imperfection could be estimated as the sum of the plate imperfection given in EN 1090 and the deflection caused by wet concrete, less any specified camber of the plate. EN 1992-1-1 clause 5.8.2(6) provides a criterion for ignoring second-order effects which requires them first be calculated. This is unhelpful. A simpler alternative would be to use the criterion in clause 5.2.1(3) based on an elastic critical buckling analysis of the composite plate. Where account should be taken of significant shear force acting on the studs in both longitudinal and transverse directions simultaneously, clause 9.3(4) requires the force on the connectors to be based on the vector sum. Hence, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi PEd ¼ P2l;Ed þ P2t;Ed

Clause 9.3(4)

where Pl;Ed and Pt;Ed are the shear forces per stud in the longitudinal and transverse directions respectively. This can influence the spacing of the studs nearest to the webs because they are the most heavily stressed from global effects (see section 9.4 below) and also tend to be the most heavily loaded from local effects in the transverse direction.

9.4. Design of shear connectors The effect of clause 9.4(1) and clause 9.3(4) is that local and global effects need only be combined in calculations for serviceability and fatigue limit states, but not for other ultimate limit states. Several justifications can be made for this concession. The main ones are as follows. .

.

.

Clause 9.4(1)

The effects of local loading are usually high only over a relatively small width compared with the total width providing the global resistance. The composite plate will have significant reserves in local bending resistance above that obtained from elastic analysis. Complete failure requires the deformation of a mechanism of yield lines, which is resisted by arching action. This action can be developed almost everywhere in the longitudinal direction and in many areas in the transverse direction also.

There is no similar relaxation in EN 1993-2 for bare steel flanges so a steel flange should, in principle, be checked for any local loading in combination with global loading. It will not normally be difficult to satisfy this check, even using elastic analysis. For serviceability calculations, elastic analysis is appropriate. This greatly simplifies the addition of global and local effects. For the serviceability limit state, the relevant limiting force per connector is that in clause 6.8.1(3), referred to from clause 7.2.2(6). The force per connector should be derived according to clause 9.3(4). When checking the Von Mises equivalent stress in the steel flange plate, the weight of wet concrete carried by it should be included. No guidance is given on the calculation of stud shear flow from local wheel loads, other than that in clause 9.2(1). Longitudinally stiffened parts of composite plates can be designed as beams spanning between transverse members, where present. The rules for effective width of clause 5.4.1.2 would apply in determining the parts of the composite plate acting with each longitudinal stiffener. For unstiffened parts of the composite plate, spanning transversely between webs or longitudinally between cross-beams, a simplified calculation of shear flow could be based on an equivalent simply-supported beam. A reasonable assumption for beam width, as recommended in Ref. 112, would be the width of the load, w, plus four-thirds of the distance to the nearest support, x, as shown in Fig. 9.1.

185


x

2 3 w

4x/3 + w

L

Fig. 9.1. Effective beam width for the determination of shear flow in a composite plate

Clause 9.4(2)

Clause 9.4(3)

Clause 9.4(4)

At cross-sections where there is an abrupt change from composite plate to reinforced concrete section, such as at the web of the box in Fig. 9.2, the method of clause 6.6.2.4 can be used to determine the transverse shear on the studs near the edge of the plate. Although the composite section formed by a steel flange and concrete slab might provide adequate strength against local sagging moments without additional transverse reinforcement, transverse reinforcement is still required in the bottom of the slab to control cracking and prevent splitting of the concrete ahead of the studs. Clause 9.4(2) requires a fairly modest quantity of bottom reinforcement to be provided in two orthogonal directions. It implies that in the absence of such reinforcement, the static design resistances of studs given in clause 6.6.3.1(1) cannot be used as they assume that splitting is prevented. The limiting fatigue stress range for studs provided in clause 6.8.3 is also inappropriate without some transverse reinforcement as splitting will increase the flexural stresses in the stud. Clause 9.4(3) refers to the detailing rules of clause 6.6.5. The minimum steel flange thickness in clause 6.6.5.7(3) is only likely to become relevant where the top flange is heavily stiffened as discussed in the comments on that clause. The force on shear connectors in wide composite flanges is influenced both by shear lag in the concrete and steel flanges and also slip of the shear connection. At the serviceability limit state, these lead to a non-uniform distribution of connector force across the flange width. This distribution can be approximated by equation (9.1) in clause 9.4(4): vL;Ed nw 0:17 x 2 PEd ¼ 3 1 þ 0:15 ð9:1Þ 3:85 b ntot ntot Equation (9.1) was derived from a finite-element study by Moffat and Dowling.113 The study considered only simply-supported beams with ratios of flange half-breadth between webs (b in equation (9.1)) to span in the range 0.05 and 0.20. The stud stiffness was taken as 400 kN/mm. The studs nearest the web can pick up a significantly greater force than that obtained by dividing the total longitudinal shear by the total number of connectors. This is illustrated in Example 9.1 and in Ref. 74. Connectors within a distance of the greater of 10tf and 200 mm are assumed to carry the same shear force. This result is obtained by using x ¼ 0 in equation (9.1) when calculating the stud force and it is necessary to avoid underestimating the force, compared to the finite-element results, in the studs nearest the web. The rule is consistent with practice for flanges of plate girders, where all shear connectors at a cross-section are assumed to be equally loaded. The assumed value of stud stiffness has a significant effect on the transverse distribution of stud force as greater slip leads to a more uniform distribution. Recent studies, such as that in Ref. 98, have concluded that stud stiffnesses are significantly lower than 400 kN/mm. The same value of stiffness is probably not appropriate for both fatigue calculation and serviceability calculations under the characteristic load combination, due to the greater slip, and

186

CHAPTER 9. COMPOSITE PLATES IN BRIDGES

therefore flexibility, possible in the latter case. Nevertheless, the assumed stiffness of 400 kN/ mm is an upper bound and therefore the transverse distribution is conservative. Clause 9.4(5) permits a relaxation of the requirements of clause 9.4(4) for composite bottom flanges of box girders, provided that at least half of the shear connectors required are concentrated near the web–flange junction. ‘Near’ means either on the web or within the defined adjacent width bf of the flange. The rule is based on extensive practice in Germany, and assumes that there is no significant local loading. At the ultimate limit state, plasticity in the flange and increased slip lead to a much more uniform distribution of stud forces across the box, which is allowed for in clause 9.4(6). To prevent buckling of the steel compression flange in half waves between studs, clause 9.4(7) refers to Table 9.1 for limiting stud spacings in both longitudinal and transverse directions. These could, in principle, be relaxed if account is taken of any longitudinal stiffening provided to stabilise the compression flange prior to hardening of the concrete. Most bridge box girders will have webs in Class 3 or Class 4, so it will usually only be necessary to comply with the stud spacings for a Class 3 flange; there is however little difference between the spacing requirements for Class 2 and Class 3.

Clause 9.4(5)


Example 9.1: design of shear connection for global effects at the serviceability limit state The shear connection for the box girder shown in Fig. 9.2 is to be designed using 19 mm stud connectors. For reasons to be explained, it may be governed by serviceability, for which the longitudinal shear per web at SLS (determined from elastic analysis of the cross-section making allowance for shear lag) was found to be 800 kN/m. x 100

3525

250 mm thick

200 580 580 605 1

2

3

4 25 mm thick

Fig. 9.2. Box girder for Example 9.1

From clause 6.8.1(3) the force per connector at SLS is limited to 0.75PRd . From Example 6.10, this limit is 0:75 83:3 ¼ 62:5 kN. It will be found that longitudinal shear forces per stud decrease rapidly with distance from the web. This leaves spare resistance for the transverse shear force and local effects (e.g. from wheel loads), both of which can usually be neglected adjacent to the web, but increase with distance from it.113 The following method is further explained in Ref. 74. It is based on finite-element analyses that included shear lag, and so is applied to the whole width of composite plate associated with the web concerned, not just to the effective width. In this case, both widths are 3525=2 ¼ 1762 mm, denoted b in Fig. 9.1 in EN 1994-2. From clause 9.4(4), nw is the number of connectors within 250 mm of the web, because 10tf > 200 mm. The method is slightly iterative, as the first step is to estimate the ratio nw =ntot , here taken as 0.25. The longitudinal spacing of the studs is assumed to be 0.15 m, so the design shear per transverse row is vL;Rd ¼ 800 0:15 ¼ 120 kN.

187


If all the studs were within 250 mm of the web, their x-coordinate in equation (9.1) would be zero. The number of studs required is found from this equation by putting PEd ¼ 0:75PRd ¼ 62:5 kN and x ¼ 0: 6:25 ¼

120 ½ð3:85ð0:25Þ0:17 3Þ þ 0:15 ntot

whence ntot ¼ 3:88. Design is therefore based on four studs per transverse row, of which one will be within 250 mm of the web. To conform to the assumptions made in deriving equation (9.1), the other studs will be equally spaced over the whole width b, not the width b 250 mm. (This distinction only matters where ntot is low, as it is here.) In a wider box, non-uniform lateral spacing may be more convenient, subject to the condition given in the definition of ntot in clause 9.4(4). For these studs, uniform spacing means locations at x ¼ b=6, b=2 and 5b=6 – that is, at x ¼ 294, 881 and 1468 mm. These are rounded to x ¼ 300, 880 and 1460 mm. The maximum lateral spacing is midway between the webs: 3525 2 1460 ¼ 605 mm. This spacing satisfies the provisions on spacing of clause 9.2(3) (i.e. <3 275 ¼ 825 mm) and also those of Table 9.1 for Class 3 behaviour. It would not be necessary to comply with the latter if the longitudinal stiffeners were close enough to ensure that there is no reduction to the effective width of the plate for local sub-panel buckling; this is not the case here. The shear force PEd per stud is found from equation (9.1), which is: 120 x 2 PEd ¼ þ 0:15 ð3:85ð0:25Þ0:17 3Þ 1 4 1762 Results are given in Table 9.1. Table 9.1. Forces in studs from global effects Stud No.

1

2

3

4

Total

x (mm) PEd (kN)

100 60.7

300 43.2

880 18.6

1460 6.2

— 129

The shear resisted by the four studs, 129 kN, exceeds 120 kN, and no stud is overloaded, so the spacing is satisfactory. The result, 129 kN, differs from 120 kN because equation (9.1) is an approximation. This ratio, 129/120, is a function, given in a graph in Ref. 74, of nw =ntot and of the effective-width ratio beff =b. Where the lateral spacing of the ntot –nw studs is uniform, as here, the ratio exceeds 1.0 provided that nw =ntot < 0:5 and beff =b > 0:7. Its minimum value is about 0.93. If the design needs to be modified, revision of the longitudinal spacing is a simple method, as the ratios between the forces per stud do not then change. From clause 9.4(6), at ULS the studs may be assumed to be equally loaded. Here, their design resistance is: ðntot =0:15ÞðPRd =V Þ ¼ ð4=0:15Þð83:3=1:25Þ ¼ 1777 kN=m This is more than twice the design shear for SLS, so ULS is unlikely to govern. Table 9.1 shows that for SLS, studs 2 to 4 have a reserve of resistance (cf. 62.5 kN) that should be sufficient for transverse shear and local effects.

188

CHAPTER 10

Annex C (Informative). Headed studs that cause splitting forces in the direction of the slab thickness This chapter corresponds to Annex C of EN 1994-2, which has the following clauses: . .

Design resistance and detailing Fatigue strength

Clause C.1 Clause C.2

Annex A of EN 1994-1-1 is for buildings only. Annex B of EN 1994-1-1, ‘Standard tests’, for shear connectors and composite floor slabs, is not repeated in EN 1994-2. Comment on these annexes is given in Ref. 5. Annex C gives a set of design rules for the detailing and resistance of shear studs that are embedded in an edge of a concrete slab, as shown in Figs 6.13 and C.1 of EN 1994-2 and in Fig. 6.35. Details of this type can occur at an edge of a composite deck in a tied arch or half-through bridge, or where double composite action is used in a box girder. The same problem, premature splitting, could occur in a steep-sided narrow haunch. The use of such haunches is now discouraged by the 458 rule in clause 6.6.5.4(1). The rules in Annex C were developed from research at the University of Stuttgart that has been available in English only since 2001.82;114;115 These extensive push tests and finite-element analyses showed that to avoid premature failure by splitting of the slab and to ensure ductile behaviour, special detailing rules are needed. Clause 6.6.3.1(3) therefore warns that the usual rules for resistance of studs do not apply. The new rules, in Annex C, are necessarily of limited scope, because there are so many relevant parameters. The rules are partly based on elaborate strut-and-tie modelling. It was not possible to find rules that are dimensionally consistent, so the units to be used are specified – the only occasion in EN 1994 Parts 1-1 and 2 where this has been necessary. For these reasons, Annex C is Informative, even though its guidance is the best available. The simplified and generally more conservative rules given in clause 6.6.4 do not cover interaction with transverse (e.g. vertical) shear or resistance to fatigue. It will be found that these ‘lying studs’ have to be much longer than usual, and that the minimum slab thickness to avoid a reduction in the shear resistance per stud can exceed 250 mm. The comments that follow are illustrated in Example 10.1, and in Fig. 10.1 where the longitudinal shear acts normal to the plane of the figure and vertical shear acts downwards from the slab to the steel web.


C.1. Design resistance and detailing Clause C.1(1)

Clause C.1(1) gives the static resistance of a stud to longitudinal shear in the absence of vertical shear, which should not be taken as greater than that from clause 6.6.3.1(1). The minimum length h of the stud and the reinforcement details are intended to be such that splitting of the slab is followed by fracture or pulling out of the stud, giving a ductile mode of failure. The important dimensions are a0r;o and v, from the stud to the centre-lines of the stirrup reinforcement, as shown in Fig. 10.1. Equation (C.1), repeated in the Example, uses factor kv to distinguish between two situations. The more favourable, where kv ¼ 1.14, applies where the slab is connected to both sides of the web and resists hogging bending – a ‘middle position’. This requires reinforcement to pass continuously above the web, as shown in Fig. C.1. Some shear is then transferred by friction at the face of the web. Where this does not occur, an ‘edge position’, kv has the lower value 1.0. Details in bridge decks are usually edge positions, so further comment is limited to these. The geometries considered in the Stuttgart tests, however, covered composite girders where the steel top flange was omitted altogether, with the web projecting into the slab. The general symbol for distance from a stud to the nearest free surface is ar , but notation ar;o is used for the upper surface, from the German oben, ‘above’. Its use is relevant where there is vertical shear, acting downwards from the slab to the studs. Allowing for cover and the stirrups, the important dimension is: a0r;o ¼ ar;o cv s =2 If the lower free surface is closer to the stud, its dimension a0r should be used in place of a0r;o . Clause 6.6.4 appears to cover only this ‘edge position’ layout, and uses the symbol ev in place of a0r;o or a0r . Although fck in equation (C.1) is defined as the strength ‘at the age considered’, the specified 28-day value should be used, unless a check is being made at a younger age. The longitudinal spacing of the stirrups, s, should be related to that of the studs, a, and should ideally be uniform. ch = 40

φs = 10

cv = 40

φℓ

a r,o

40 a ʹr,o = 80 33º

β = 10º d = 19

29 Centre-line of slab

45

v = 136 h = 191

Fig. 10.1. Notation and dimensions for lying studs in Example 10.1

190

10

CHAPTER 10. ANNEX C (INFORMATIVE)

For concretes of grade C35/45 and above, the resistance PRd from clause 6.6.3.1 is independent of the concrete grade. Then, equation (C.1) can be used to find a minimum value for a0r such that PRd;L PRd . For example, let a ¼ s, d ¼ 19 mm, fck ¼ 35 N/mm2 , kv ¼ 1, and V ¼ 1.25. Then, PRd ¼ 90.7 kN from equation (6.18) and PRd;L from equation (C.1) does not govern unless a0r < 89 mm, say 90 mm. With 40 mm cover and 10 mm stirrups, the minimum slab thickness is then 2ð90 þ 40 þ 5Þ ¼ 270 mm if the studs are centrally placed, but greater if they are off centre. If a thinner slab is required, the ratio a=s can be increased or more studs provided, to compensate for a value PRd;L < PRd . The limits on v in clause C.1(2) are more convenient for use in practice than the limits on , because the angle is defined by the position of the longitudinal corner bar within the bend of the stirrup (Fig. 10.1), which is difficult to control on site. It follows from clause C.1(3) that the minimum stirrup diameter s is roughly proportional to the stud diameter d. Where a/s ¼ 1.0, s d/2. The expression for interaction between longitudinal and vertical shear, clause C.1(4), is only slightly convex. The vertical shear resistance given in Example 10.1, equation (C.4), is typically less than 40% of the longitudinal shear resistance, being governed mainly by the upper edge distance ar;o . Application of vertical shear to lying studs is best avoided, and can be minimized by spanning the concrete slab longitudinally between cross-beams.

Clause C.1(2)

Clause C.1(3) Clause C.1(4)

C.2. Fatigue strength Equation (C.5) in clause C.2(1), ðPR Þm N ¼ ðPc Þm Nc , differs from equation (6.50) in clause 6.8.3(3) (for the fatigue strength of studs in a non-lying position) only in that symbols P in equation (C.5) appear as in equation (6.50). Both methods use m ¼ 8. In Annex C, Pc is a function of dimension a0r (Fig. C.1, Fig. 10.1), and is 35.6 kN at 2 million cycles, for a0r 100 mm. With typical covers to reinforcement, this corresponds to a lying stud at mid-depth of a slab at least 280 mm thick, where the splitting effect is probably minimal. This value 35.6 kN should therefore correspond to the value given in clause 6.8.3 for at 2 million cycles, 90 N/mm2 . It does so when the stud diameter is 22.4 mm. This is as expected, because in the fatigue tests, only 22 mm diameter studs were used. However, the method of Annex C is provided for studs of diameter 19 mm to 25 mm, for which: ‘numerous FE-calculations show, that the fatigue strength curve should be based on the absolute range of shear force per stud rather than on the range of shear stress’ (p. 9 of Ref. 115). The value of PR given by the rule in clause 6.8.3 is proportional to the square of the shank diameter of the stud. For 19 mm studs, it is only 25.6 kN at 2 106 cycles, which is only 72% of the resistance 35.6 kN given in Table C.1. This is why clause C.2(1) requires the lower of the two values from clause 6.8.3 and Annex C to be used. Clause C.2(2) refers to the recommended upper limit for the longitudinal shear force per connector, which is 0.75PRd . The word ‘longitudinal ’ reveals that the rule for fatigue resistance does not apply where vertical shear is present, which is not clear in clause C.2(1). There were no fatigue tests in combined longitudinal and vertical shear – another reason why application of vertical shear to lying studs is best avoided.

Clause C.2(1)

Clause C.2(2)

Applicability of Annex C The definition of a ‘lying stud’, given in the titles of clause 6.6.4 and Annex C, does not state how far from a free surface a stud must be, for the normal rules for its resistance and the detailing to apply. Clause 6.6.5.3 refers to a ‘longitudinal edge’, not a top surface, but appears to deal with the same problem of splitting parallel to a free surface nearby. Where the edge distance (i.e. ar;o in Fig. 10.1) is less than 300 mm, it specifies ‘U-bars’ (i.e. stirrups) of diameter s 0:5d and an edge distance ar 6d. These limits correspond closely with the results of

191


C

A

B

A

125 50

125

50 50

B C

D

125

125 125 D

(a)

(b)

(c)

Fig. 10.2. Cross-sections at an edge of a slab: (a) edge studs; (b) lying studs in an edge position; and (c) three rows of lying studs

Example 10.1, s ¼ 0:53d and ar ¼ 6:6d, but the stirrups are required to pass around the studs, whereas in Annex C they pass between them. The difference is that the surface parallel to the plane of splitting, AB in Fig. 10.2, is normal to the plane of the slab in one case, not in the other. The minimum height of the stud, about 90 mm to clause 6.6.5.1(1) and 191 mm in Example 10.1, is less significant in the detail in Fig. 10.2(a) than in detail (b). Details may occur where clause 6.6.5.3 is also applicable, but it does not clarify the scope of Annex C. Let us consider the options, as the local thickness hc of the slab in Fig. 10.1 is increased (e.g. by the addition of an upstand haunch) without change to other details. If the top cover is maintained, length a0r;o must increase, so from clause C.1(2) for v, the studs have to be longer. An alternative is to keep a0r;o and v unchanged, by increasing the cover to the legs of the stirrups. When hc exceeds 300 mm (using data from Example 10.1), two rows of studs are possible, Fig. 10.2(b), because the minimum vertical spacing of studs is 2.5d, from clause 6.6.5.7(4). In the absence of vertical shear, the shear resistance is doubled, as is the potential splitting force. For two rows of lying studs, the force Td given by equation (C.2) should be 0.3 times the sum of their resistances. Limits to the applicability of Annex C are further discussed at the end of Example 10.1.

Example 10.1: design of lying studs A bridge deck slab 250 mm thick is to be connected along its sides to steel webs. The final detail is shown to scale in Fig. 10.1. The design ultimate longitudinal shear at each edge is VL;Ed ¼ 600 kN/m. The vertical shear is at first assumed to be negligible. The effect of adding VV;Ed ¼ 60 kN/m is then found. The strengths of the materials are: fyd ¼ 500=1:15 ¼ 435 N/mm2 , fu ¼ 500 N/mm2 and fck ¼ 40 N/mm2 . Resistances in Annex C are given in terms of fck , not fcd , because the partial factor V allows for both the stud material and the concrete. The minimum cover is 40 mm. There is space only for a single row of studs at mid-depth, and 19 mm studs are preferred. Assuming the studs have a height greater than four times their shank diameter, their shear resistance to equation (6.19) is: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðD10:1Þ PRd ¼ 0:29 192 40 35 000=ð1:25 1000Þ ¼ 99:1 kN=stud and to equation (6.18) is: PRd ¼ 0:8 500 192 =ð4 1:25Þ ¼ 90:7 kN=stud

ðD10:2Þ

so equation (6.18) governs. This gives a spacing of 90:7=600 ¼ 0:151 m, so a spacing a ¼ 150 mm is assumed, with stirrups also at 150 mm. Their required diameter s can be estimated from clause C.1(3), equation (C.2): Td ¼ 0:3PRd;L Using PEd;L as an approximation to PRd;L : Td ¼ ð2s =4Þ 500=1:15 0:3 600 1000 0:15

192

CHAPTER 10. ANNEX C (INFORMATIVE)

This gives s 8:9 mm, so 10 mm stirrups are assumed for finding a0r;o . With cover of 40 mm, a0r;o ¼ ½250 2ð40 þ 5Þ=2 ¼ 80 mm From equation (C.1) with kv ¼ 1.0 for an edge position and a ¼ s, PRd;L ¼ 1:4kv ð fck da0r Þ0:4 ða=sÞ0:3 =V ¼ 1:4ð40 19 80Þ0:4 =1:25 ¼ 91:7 kN=stud ðC:1Þ This is greater than result (D10.2), so the shear resistance is governed by equation (6.18). Where this occurs, the term PRd;L in equations (C.2) and (C.3) should be replaced by the governing shear resistance. For studs at 0.15 m spacing, VRd;L ¼ 90:7=0:15 ¼ 605 kN/m which is sufficient. This result requires the diameter of the stirrups to be increased to 8:9 605=600 ¼ 9:0 mm, so 10 mm is still sufficient. The length of the studs is governed by a0r . From clause C.1(2): . .

for uncracked concrete, v maxf110; 1:7 80; 1:7 75g ¼ 136 mm for cracked concrete, v maxf160; 2:4 80; 2:4 75g ¼ 192 mm

To these lengths must be added ch þ s /2 (i.e. 45 mm) plus the thickness of the head of the stud, 10 mm, giving lengths h after welding of 191 mm and 247 mm respectively. In practice, these would be rounded up. Here, h is taken as 191 mm. Figure 10.1 shows that the angle can then be anywhere between 108 and 338. This is roughly consistent with the alternative rule in clause C.1(2), 308. The simpler rules of clause 6.6.4 would require a0r;o to be increased from 80 mm to 6 19 ¼ 114 mm. It would then be necessary for the slab thickness to be increased locally from 250 mm to at least 318 mm. Dimension v would be increased from 136 or 192 mm to 266 mm, requiring much longer studs. In the absence of vertical shear, a check on fatigue to clause C.2(1) is straightforward.

Interaction with vertical shear The design vertical shear is Fd;V ¼ 0:15 60 ¼ 9 kN/stud. Clause C.1(4) requires longitudinal reinforcement with l 16 mm, so this value is used. From equation (C.4) with kv ¼ 1:0 and a ¼ s, PRd;V ¼ 0:012ð fck l Þ0:5 ðda=sÞ0:4 ðs Þ0:3 ða0r;o Þ0:7 kv =V

ðC:4Þ

¼ 0:012ð40 16Þ0:5 ð19Þ0:4 ð10Þ0:3 ð80Þ0:7 1:0=1:25 ¼ 33:8 kN=stud From interaction expression (C.3): ð600 0:15=90:7Þ1:2 þ ð9=33:8Þ1:2 ¼ 1:20 (but 1.0) which is too large. Changing the spacing of the studs and stirrups from 150 mm to 125 mm would reduce result 1.20 to 0.96.

Two or more rows of lying studs The detail shown in Fig. 10(c) is now considered, with materials and studs as before. The deck spans longitudinally, so it can be assumed that no vertical shear is applied to the three rows of lying studs. The preceding calculations apply to the top and bottom rows. For the middle row, a0r increases from 80 mm to 130 mm. From equation (C.1) the resistance of its studs increases from 91.7 kN to 111 kN, so result (D10.2), 90.7 kN, still governs. Assuming uncracked concrete, the minimum height of these studs increases by 1.7 times the change in a0r , from clause C.1(2); that is, from 191 mm to 276 mm. This situation was not researched, and the increase may not be necessary; but the stirrups should be designed for a force Td ¼ 0:3ð3 90:7Þ ¼ 82 kN per 150 mm length of beam, if the full shear resistance of the studs is needed.

193

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Pucher, A. (1977) Influence Surfaces of Elastic Plates. Springer-Verlag Wien, New York. Kuhlmann, U. (1997) Design, calculation and details of tied-arch bridges in composite constructions. In: Buckner, C. D. and Sharooz, B. M. (eds), Composite Construction in Steel and Concrete III, American Society of Civil Engineers, New York, pp. 359ÿ369. Monnickendam, A. (2003) The design, construction and performance of Newark Dyke railway bridge. Proceedings of a Symposium on Structures for High-speed Railway Transportation, Antwerp. IABSE, Zurich. Reports, 87, 42ÿ43. British Standards Institution. Actions on Structures. Part 1-4: General Actions ÿ Wind actions. BSI, London, EN 1991. Randl, E. and Johnson, R. P. (1982) Widths of initial cracks in concrete tension flanges of composite beams. Proceedings of IABSE, Periodica 4/82, P-54/82, Nov., 69ÿ80. Johnson, R. P. and Allison, R. W. (1983) Cracking in concrete tension flanges of composite T-beams. Structural Engineer, 61B, Mar., 9ÿ16. Roik, K., Hanswille, G. and Cunze-O. Lanna, A. (1989) Report on Eurocode 4, Clause 5.3, Cracking of Concrete. University of Bochum, Report EC4/4/88. Johnson, R. P. (2003) Cracking in concrete flanges of composite T-beams ÿ tests and Eurocode 4. Structural Engineer, 81, No. 4, 29ÿ34. Schmitt, V., Seidl, G. and Hever, M. (2005) Composite bridges with VFT-WIB construction method. In: Hoffmeister, B. and Hechler, O. (eds), Eurosteel 2005, vol. B. Druck und Verlagshaus Mainz, Aachen, pp. 4.6-79 to 4.6-83. Yandzio, E. and Iles, D. C. (2004) Precast Concrete Decks for Composite Highway Bridges. Steel Construction Institute, Ascot, Publication 316. Calzon, J. M. (2005). Practice in present-day steel and composite structures. In: Hoffmeister, B. and Hechler, O. (eds), Eurosteel 2005, vol. A. Druck und Verlagshaus Mainz, Aachen, pp. 0-11 to 0-18. Doeinghaus, P., Dudek, M. and Sprinke, P. (2004) Innovative hybrid doublecomposite bridge with prestressing. In: Pre-Conference Proceedings, Composite Construction in Steel and Concrete V, United Engineering Foundation, New York, Session E4, paper 1. Department of Transport (now Highways Agency) DoT (1987) Use of BS 5400:Part 5:1979. London, Departmental Standard BD 16/82. Moffat, K. R. and Dowling, P. J. (1978) The longitudinal bending behaviour of composite box girder bridges having incomplete interaction. Structural Engineer, 56B, No. 3, 53ÿ60. Kuhlmann, U. and Ku¨rschner, K. (2001) Behavior of lying shear studs in reinforced concrete slabs. In: Eligehausen, R. (ed.), Connections between Steel and Concrete. RILEM Publications S.A.R.L., Bagneux, France, pp. 1076ÿ1085. Kuhlmann, U. and Ku¨rschner, K. (2006) Structural behavior of horizontally lying shear studs. In: Leon, R. T. and Lange, J. (eds), Composite Construction in Steel and Concrete V. American Society of Civil Engineers, New York, pp. 534ÿ543.

Index Notes: references to ‘beams’ and to ‘columns’ are to composite members; cross-references to EN 1992 and EN 1993 are too numerous to be indexed action effect see actions, effects of actions 6, 8 accidental 56 arrangement of 62ÿ4 combinations of 3, 6, 11, 15, 31, 62ÿ4 characteristic 46 for serviceability 164ÿ6, 171 frequent 48, 64, 153 infrequent 164 quasi-permanent 48 effects of 8 de-composition of 51ÿ2 envelopes of 63 global with local and fatigue 156, 161 and serviceability 165, 174ÿ5, 177ÿ8 at failure 56ÿ7, 72 in composite plates 184ÿ5 independent 136ÿ7 local 161, 165, 183ÿ4 primary 12, 48, 168 secondary 12, 48 second-order 31, 33ÿ4, 103, 107, 141ÿ2, 149, 185 indirect 12, 44, 60, 138, 170ÿ1 permanent 15ÿ16 temperature 48, 120 see also fatigue load models; forces, concentrated; loading analysis, elastic, of cross-sections see beams; columns; etc. analysis, global 8, 29ÿ66 cracked 50ÿ1 elastic 30, 36ÿ7, 40, 42ÿ53 elasto-plastic 137 finite-element 31, 34, 39, 93, 111 first-order 31ÿ3, 38 grillage 52ÿ3 non-linear 8, 36, 56, 72, 138, 185 of filler-beam decks 52ÿ3

of frames 29 rigid-plastic 8 second-order 8, 31ÿ4, 38, 64ÿ5, 94, 140ÿ1 uncracked 50ÿ1 see also cracking of concrete; loading, elastic critical analysis, local 183ÿ4 analysis, rigorous 39 Annex, National see National Annex annexes, informative 2, 4ÿ5, 191ÿ2 application rules 7 arches see bridges, tied-arch assumptions in Eurocodes 7 axes 8ÿ9 beams axial force in 81, 83ÿ4, 86ÿ9, 105, 111, 161ÿ2 bending resistance of 67ÿ84 hogging 73ÿ4, 77ÿ9, 83 sagging 72ÿ3, 83 cantilever 68, 125 Class of 12, 57ÿ60 concrete-encased 4, 52, 59, 68, 89 concrete flange of 13, 71, 109 cross-sections of 67ÿ89 Class 1 or 2 118ÿ20 and axial force 83 and filler beams 53 and global analysis 36 and indirect actions 44ÿ5 and reinforcement 20 and resistance to bending 69 and serviceability 163, 170 and vertical shear 80 Class 3 37, 108 Class 3 and 4 20, 80ÿ2, 103, 163 Class 4 77ÿ9, 85ÿ9, 94, 97, 107 classification of 29, 37, 57ÿ61, 71, 82 elastic analysis of 58ÿ9, 69, 75ÿ7


beams – cross-sections of (continued ) plastic analysis of 36 sudden change in 120ÿ1, 186 curved in plan 4, 68ÿ9, 89 curved in elevation 69, 114 flexural stiffness of 46 haunched 102, 117, 122, 126ÿ7, 189 of non-uniform section 4 shear connection for see shear connection shear resistance of 67, 79ÿ83 see also analysis; buckling; cantilevers; cracking of concrete; deflections; filler beams; flange, effective width of; imperfections; interaction; shear . . . ; slabs, concrete; vibration; webs bearings 11, 15, 62, 64, 141, 145 bedding see slabs, precast concrete bending, bi-axial 71 bending moments accumulation of 12 and axial force 59, 68, 102ÿ3 elastic critical 93 in columns 142ÿ3 redistribution of 18, 37, 53 bolts, holes for 68 bolts, stiffness of 30, 37ÿ8 bond see shear connection box girders 82 distortion of 68ÿ9 shear connection for 116, 118 torsion in 45ÿ6, 72 see also composite plates bracing, lateral 35ÿ6, 69, 91, 111ÿ3, 115 and buckling 97 and slip of bolts 38ÿ9 stiffness of 96ÿ9, 102ÿ4 breadth of flange, effective see flange, effective width of Bridge Code (BS 5400) 1, 39, 56, 59, 70, 89, 104 bridges cable-supported 4, 23, 40 durability of see corrosion; durability for pedestrians 151 integral 31, 33, 36, 68, 72, 105, 115 railway 151 strengthening of 114 tied-arch 35, 161ÿ2, 189 U-frame 30 see also box girders; filler beams British Standards BS 5400, see Bridge Code BS 5950 39, 59 BS 8110 70 buckling distortional lateral 35, 91, 105ÿ7 flange-induced 68, 114 flexural 77, 89, 184ÿ5 in columns 34, 136, 140ÿ2 lateral 95

202

lateral-torsional 31, 34, 45, 67, 76ÿ7, 90ÿ104, 111, 138 local 37ÿ8, 57ÿ9, 76, 127, 137 see also beams, Class of of plates 29, 37ÿ8, 184ÿ5 of webs in shear 68, 80, 82 see also bending moments, elastic critical; filler beams cables 4, 23 camber 166 cement, hydration of 175ÿ6 see also cracking of concrete CEN (Comite´ Europeén Normalisation) 1ÿ2 Class of section see beams, cross-sections of class, structural 26ÿ7 Codes of Practice, see British Standards; EN . . . columns 64ÿ6, 136ÿ50, 164 analysis of 29, 140ÿ3 axially loaded 144 bending resistance of 71 bi-axial bending in 140, 143, 146 concrete-encased 4, 138, 145 concrete-filled 4, 144ÿ50 cross-sections of interaction diagram for 138ÿ9 non-symmetrical 33, 47, 136 design methods for 31, 137ÿ43 effective stiffness of 33, 137, 140, 149 moment-shear interaction in 139, 148 out-of-plumb 65ÿ6 second-order effects in 138 shear in 139, 145 squash load of 138, 140, 147 steel contribution ratio for 136, 140, 147 transverse loading on 143 see also buckling; bending moments; cracking of concrete; creep of concrete; length, effective; imperfections; loading, elastic critical; load introduction; reinforcement; shear connection; slenderness, relative; stresses, residual composite action, double 4, 183, 189 composite bridges, see bridges; Bridge Code composite plates 4, 183ÿ8 compression members 136ÿ50 see also columns concrete compaction of 124ÿ5 lightweight-aggregate 17, 19, 22, 136, 152 over-strength of 47, 50, 118ÿ9 partial factors for 13ÿ14 precast 27ÿ8, 62 properties of 17ÿ19 spalling of 4 strength classes for 17ÿ18, 26 strength of 13, 17ÿ18, 70 stress block for 18, 70ÿ1, 138 thermal expansion of 22

INDEX

see also cracking of concrete; creep of concrete; elasticity, modulus of; prestress; shrinkage of concrete; slabs connecting devices 20ÿ1 connections, see joints connector modulus, see shear connectors, stiffness of construction 3, 47ÿ8, 103ÿ4, 164, 166 loads 6, 12 methods of 12, 180 propped 121, 171 unpropped 12, 76, 91ÿ3, 121 see also erection of steelwork Construction Products Directive 14 contraflexure, points of 32, 35 corrosion 25, 27 at steel-concrete interface 27ÿ8, 127, 181 of reinforcement 25ÿ7 cover 25ÿ7, 89ÿ90, 138, 145 cracking of concrete 46, 152ÿ4, 167ÿ73 and global analysis 29, 32, 36, 46ÿ7, 50, 52ÿ3 and longitudinal shear 47, 118 control of 163, 173 load-induced 169ÿ70, 175 restraint-induced 168ÿ9, 175ÿ7 early thermal 167ÿ8, 172ÿ3, 175ÿ6 in columns 47, 140, 141, 145 creep coefficient 19, 42ÿ3, 53ÿ4, 140 creep multiplier 42ÿ3 creep of concrete 12, 17, 19, 32, 42ÿ5, 53 in columns 45, 140, 147 secondary effects of 44 see also modular ratio; elasticity, modulus of cross-sections see beams, cross-sections of; columns, cross-sections of curves, buckling resistance 34 damage, cumulative 153, 155ÿ6 see also factors, damage equivalent damping factor 166ÿ7 definitions 8 deflections see deformations deformations 166 limits to 163 deformation, imposed 12, 44, 49, 90 design, basis of 11ÿ16 design, methods of see beams; columns; slabs; etc. design, mixed-class 36ÿ7 Designers’ guides v, 2 to EN 1990 14 to EN 1993ÿ2 35, 58, 68ÿ9, 72, 77, 79, 82, 95, 104, 113, 114, 118 to EN 1994-1-1 60, 73, 93, 136 diaphragms 115ÿ6 dimensions 14 dispersion, angle of 113, 120 distortion of cross sections 68, 72, 82, 184 ductility see reinforcement, fracture of; structural steels

durability 25ÿ8, 179 effective length see length, effective effective width see beams; flanges; slabs, composite effect of action see actions, effects of eigenvalue see loading, elastic critical elasticity, modulus of for concrete 18, 33, 140 for shear 45 EN 1090 7, 185 EN 10025 60, 72, 146 EN 13670 5, 7, 180 EN 13918 23, 122 EN 1990 v, 2, 6, 14ÿ15, 25, 29, 48, 56, 164 EN 1991 v, 2, 6, 12, 48, 151, 166 EN 1992 v, 2, 5 EN 1993 v, 2, 6 EN 1994-1-1 v, 2 EN 1994-2 v, 2 EN 1998 3, 7 ENV 1994-1-1 5, 137 environmental class see exposure class equilibrium, static 15ÿ16 erection of steelwork 7, 39 European Standard see EN . . . examples bending and vertical shear 104ÿ11 block connector with hoop 116ÿ8 composite beam, continuous 60ÿ2, 104ÿ11 composite column 136, 145ÿ50 concrete-filled tube 145ÿ50 control of crack width 175ÿ7 cross bracing 111ÿ3 distortional lateral buckling 105ÿ8 effective width 41ÿ2 elastic resistance to bending 77ÿ9 fatigue 157ÿ61 in-plane shear in a concrete flange 130ÿ1 longitudinal shear 131ÿ4 lying studs 192ÿ4 modular ratios 53ÿ4 plastic resistance to bending 72ÿ3 resistance to bending and shear 85ÿ6, 104ÿ11 with axial compression 86ÿ9, 107ÿ8 serviceability stresses 173ÿ5, 177ÿ8 shear connection for box girder 187ÿ8 shrinkage effects 54ÿ6 transverse reinforcement 130ÿ1 execution see construction exposure classes 26ÿ7, 164ÿ5, 167, 175 factors, combination 6, 48ÿ9, 56ÿ7 factors, damage equivalent 156, 161 factors, partial see partial factors factors, reduction 92ÿ3, 95 fatigue 15ÿ16, 137, 150ÿ61 analysis for 37 load models for 150, 153, 157ÿ61 of joints 30

203


fatigue (continued ) of reinforcement 19ÿ20, 154ÿ5, 159ÿ61 of shear connectors 47, 150, 155ÿ6, 183, 191 of structural steel 27, 90, 127, 150ÿ1, 154, 157 partial factors for 14, 151ÿ2 filler beams 29, 52ÿ3, 60, 89ÿ91, 166, 173 finite-element methods 68, 93ÿ4, 104, 115 see also analysis, global fire, resistance to 25, 163 flanges concrete see beams; slabs effective width of 39ÿ41, 68, 111 plastic bending resistance of 82 steel 104, 186 flow charts v for classification of sections 57 for compression members 141 for control of cracking 171ÿ2 for global analysis 46, 62ÿ6 for lateral buckling 96 forces, concentrated 114, 120ÿ1 forces, internal 137 formwork, permanent 62, 179ÿ81 formwork, re-usable 76 foundations 7, 16 fracture toughness 12 frame, inverted-U 35, 93, 95ÿ8, 101, 106, 109ÿ11 frames, composite 8, 35, 64ÿ6, 141 braced 47 see also analysis, global; buckling; imperfections geometrical data 15 see also imperfections girders see beams; box girders ground-structure interaction 30 see also bridges, integral haunches see beams, haunched Highways Agency 1 hole-in-web method 58, 59ÿ60, 73, 80 impact factor 160 imperfections 7, 14, 29, 31ÿ2, 33ÿ6 and lateral buckling 91 in columns 65, 138, 141, 143 in plates 185 interaction, partial and full 69 ISO standards 5, 8 italic type, use of vÿvi jacking, see prestress joints 22, 30 between precast slabs 181 stiffness of 38 length, effective 34, 64, 136, 143 see also slenderness, relative

204

limit states serviceability 6, 37, 56, 163ÿ78 STR (structural failure) 67 ultimate 56, 67ÿ162 loading 12 arrangement of 31 construction 12 elastic critical 31ÿ3 for beams 97ÿ102 for columns 140, 147ÿ8 for composite plates 185 wheel 56ÿ7, 84, 165, 183, 185 see also actions load introduction in columns 136, 144ÿ5, 149ÿ50 in tension members 161ÿ2 lying studs see studs, lying materials, properties of 17ÿ23 see also concrete; steel; etc. mesh, welded see reinforcement, welded mesh modular ratio 42ÿ3, 53ÿ4 modulus of elasticity see elasticity, modulus of moment of area, torsional second 46 moments see bending moments; torsion nationally determined parameter 1ÿ2, 62, 84, 91 National Annexes 1, 3, 11, 56 and actions 48, 56, 153, 158ÿ9, 164, 169 and analysis, global 56, 58 and beams 38 and columns 138 and combination factors 48 and materials 13, 20, 22, 59, 61, 70, 146 and partial factors 116, 151 and resistances 95, 124, 128, 156ÿ7 and serviceability 27, 164, 166ÿ7, 173ÿ5 and shear connectors 125, 166 national choice 2 national standards 1 NDPs 1ÿ2 normative rules 2ÿ3 notation see symbols notes, in Eurocodes 1, 12 partial factors 2, 3, 9ÿ16 for fatigue 14, 151ÿ2, 156

F , for actions 6, 15, 19, 45, 51, 55

M , for materials and resistances 13ÿ15, 92, 165, 192 plastic theory see analysis, global, rigid-plastic; beams, cross-sections of, Class 1 and 2 plate girders see beams plates, buckling of see buckling plates, composite 41, 72, 126, 164, 183ÿ8 plates, orthotropic 52 Poisson’s ratio 46, 144

INDEX

prestress 4, 8 by jacking at supports 4, 8, 12, 49 by tendons 49, 77, 155, 167 transverse 4 principles 4, 7, 12 propping see construction, methods of provisions, general 2 push tests see shear connectors, tests on quality, control of 26 redistribution see bending moments; shear, longitudinal references, normative 5ÿ7 reference standards 3, 5ÿ7 regulatory bodies 12 reinforcement 9, 19ÿ21, 22 and lying studs 190ÿ1 ductility of 20, 59, 71 fracture of 21ÿ2, 59 in beams for crack control 167ÿ73 for shrinkage 19 minimum area of 59, 168ÿ9, 176ÿ7 transverse 115, 124, 127ÿ30 in columns 138, 145ÿ6 in composite plates 186 in compression 70 in filler-beam decks 90 in haunches see beams, haunched yielding of 39 welded mesh (fabric) 19ÿ21, 59, 71 see also cover; fatigue resistances 14 see also beams, bending resistance of; etc. restraints, lateral see bracing, lateral rotation capacity 12, 22, 58 see also joints safety factors see partial factors scope of EN 1994-2 4ÿ5, 36, 136, 138 section modulus 8 sections see beams; columns; etc. separation 8, 115, 124 serviceability see limit states settlement 12, 30 shakedown 153ÿ4 shear see columns, shear in; shear, longitudinal; shear, vertical; etc. shear connection 2, 8, 68, 114ÿ35 and execution 124ÿ5 and U-frame action 93, 111 by adhesives 23, 114 by bond or friction 23, 114, 136, 144, 149ÿ50, 190 design of 82, 155 detailing of 124ÿ7, 180, 189ÿ93 for box girders 185ÿ8 full or partial 69 in columns 144ÿ5, 149ÿ50

see also fatigue; load introduction; reinforcement, in beams, transverse; shear connectors; slip, longitudinal shear connectors 115, 156 and splitting see studs, lying angle 5, 115 bi-axial loading of 185, 188 block with hoop 5, 114, 116ÿ7 channel 5 ductility of 114 fatigue strength of 151, 191 flexibility of see stiffness of force limits for 151, 165ÿ6 in young concrete 166 partial factors for 14 perforated plate 22 spacing of 59, 91, 119, 124ÿ5, 162, 180, 184ÿ5, 187ÿ8 stiffness of 18, 69, 164, 186ÿ7 tension in 115 tests on 5, 189 types of 5, 22 see also studs, welded shear flow 116, 118 shear lag see width, effective shear, longitudinal 47, 68, 114, 118ÿ21, 127ÿ30 see also columns, shear in; composite plates; shear connection; shear flow shear, punching 84, 184 shear ratio 100 shear, vertical 29 and bending moment 80ÿ3, 87 and lying studs 191, 193 in deck slabs 84, 184 in filler-beam decks 91 see also buckling shrinkage of concrete 19, 53 and cracking 169ÿ70, 174ÿ5 autogenous 19, 55, 144, 147 effects of 35, 45, 165ÿ6, 172 in tension members 50 modified by creep 43ÿ4, 54ÿ6 primary 12, 54ÿ6, 76, 120, 133ÿ4 secondary 12, 134 see also cracking of concrete situations, design 165 skew 89 slabs, concrete 53 reinforcement in 125 splitting in 115, 123ÿ4, 189ÿ93 see also plates, composite slabs, precast concrete 115, 125ÿ7, 179ÿ81 slenderness, relative for beams 92, 94, 97, 101ÿ2 for columns 136, 140, 147ÿ8 slip capacity 114 slip, longitudinal 8, 38ÿ9, 69, 166 in columns 138, 144 in composite plates 184, 186ÿ7 software for EN 1994 32ÿ3, 75, 94

205


splices 38, 118 squash load see columns, squash load of stability see equilibrium, static standards see British Standards; EN . . . standards, harmonised 14 steel see reinforcing steel; structural steel; yielding of steel steel contribution ratio see columns steelwork, protection of see durability stiffeners, longitudinal 59, 82, 185 stiffeners, transverse web 35, 68ÿ9, 93, 97, 116 stiffness, effective, of reinforcement 51 stiffness, flexural see beams; columns; etc. stiffness, torsional 45ÿ6 strength of a material 13ÿ14 characteristic 13, 15 see also resistance stress block for concrete 13 stresses accumulation of 12, 81, 84, 86 bearing 144 design, at serviceability limit state 164ÿ6 in concrete 26, 165, 177ÿ8 in reinforcement 165 in steel 165, 177 equivalent, in steel 72, 185 excessive 164 fatigue 150ÿ1 mid-plane, in steel 76ÿ7 residual, in steel 31, 34ÿ5, 141 shrinkage see shrinkage of concrete temperature see temperature, effects of stress range, damage equivalent 155ÿ7 stress resultant see actions, effects of structural steels 20ÿ2 partial factors for 9, 14ÿ15 thermal expansion of 22 strut, pin-ended 32 studs, lying 115, 123ÿ4, 189ÿ93 studs, welded 3, 22ÿ3, 121ÿ3 detailing of 127, 189ÿ93 ductility of 120 length after welding 122 resistance of 121ÿ2, 189, 191 tension in 123 weld collar of 23, 115, 122ÿ3 see also fatigue; shear connection, detailing of; shear connectors

206

subscripts 8ÿ9 superposition, principle of 31 support, lateral see bracing, lateral sway 142 symbols 8ÿ9, 13ÿ14, 20 temperature, effects of 48ÿ9, 53, 153 temporary structures 11 tendons see prestress tension field 79ÿ80 tension members 23, 49ÿ52, 161ÿ2 tension stiffening and cracking 47, 169ÿ70 and longitudinal shear 118ÿ9, 162 and stresses 154ÿ5, 160, 165, 173ÿ5 and tension members 50ÿ1 testing see shear connectors tolerances 14, 49, 180ÿ1 torsion 45ÿ6, 68, 72, 91, 163, 184 traffic, road, type of 160 truss analogy 128 trusses, members in 49, 136ÿ7, 140 and buckling 95 and effective widths 41 tubes, steel see columns, concrete-filled U-frame see frame, inverted-U units 189 uplift see separation variables, basic 12 vibration 163, 166ÿ7 warping, resistance to 72, 118, 163, 184 webs breathing of 166 effective area of 37ÿ8 holes in 68, 90 transverse forces on 113ÿ4 see also hole-in-web method; shear, vertical web stiffeners see stiffeners, transverse web width, effective 29, 183ÿ4 see also beams; flanges, effective width of worked examples see examples yielding of steel 37

Designers' Guide to en 1994-2_Eurocode 4_Design of Composite Steel and Concrete Structures

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