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STRUCTURAL,
"Afl The Structural Engineer reviewing the first edition This market leading student text covers the design of structural steelwork to 55 5950 Part 1. the subject In two parts, the first deals with design at an elementary level famlilarislng the reader with BS 5950. Part two then proceeds to cover all aspects of tho design of whole buildings, highlighting the integration of 'elements' to produce economic, safe structures. Tho second edition has been thoroughly and updated to take account of recent research and design developments and a new chapter on plate girders has been added. The revised text retains all the popular features of the original work. In particular, readers v.ill
find that theouthors: • explain concepts clearly • use an extremely practical approach . inciudo nuSieroS vàovkS examples and real scenarios
. cover whole structure design • take the reader step-by.step through the Britkh Standard 11
Structural Stcelwark Desimr to BS 5950 Is a care text for cIvil/structural engineering degree and BTEC HIW/D courses. It will also prove useful to professIonal engineers needing to famillarise themselves with 55 5950 Part land the design of complete buIldings, particularly portal frames.
Li Morris v:as formerly Reader iii Struchiral Enpinecring at the Unlver:;ity rzi flr.nchestor. 31 tht: D R Plum is Lecturer in Structural
Unittrr.ity of
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9 780582 230880 ISBN cawtes, of SuttoN MDCI
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STRUCTURAL STEELWORK DESIGN to BS 1355950 5950 . '.
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Harrow Harrow,England U1gland,London lendon. NewYork Yruk .Pnding A.",nn,},kinsiac Ma ... (hu<~IIJ. ~~n Fflln, Onta,I". ~yd,,~y - New - S.n Francisco Toronsa - Doss MIlls. Onlasla - hydsscy Tokyo Tokyo.Singapore Singap","- .Hong HongKong-Seoul Kong . ~"ouJ Taipei T.ip~l. Cap. Town . Madrid· M",fco CIty. Amu.,tI>m· Mu,,!
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Preface Pearson Education Limited Edinburgh Gate Gnte Harlow OvI20 2JE Essex CM2O EngI and England
and ASSOClnted Associated Companies and Compames throughout throughout the Ihe world VisitliS us 011 on 1he the World Wide Visit Wide lVeb Web at: .com htrp:/Iwww.pearsoneduc.com htrp:l/www.pearsonedtic
All nghts nghts reserved. reserled. No No part part of ofthis this publication publicatIOn may may be be reproduced, reproduced, stored stored in in aa system. or or transmitted transmitted in in any any form fonn or or by byany anymeans, means,electronic, electrontc. retrieval system, mechanical, mechamcal. photocopying, recording or otherwise, without prior pnor wntten the permission pennlsslon of of the the publisher publisher or or aa licence licence permitting pennlttmg restricted restncted copying copymg in lfl the United Kingdom Kingdom tssued Umted Issued by the the Copynght Copynght Licensing Licensmg Agency Agency Ltd, Ltd. 90 Tottenham Tottenham Court Road, WIP OLP 90 Road. London London \VIP Cover designed by by Clu-is Chns Eley Design, DeSign, Reading Reading by The TheRiverside RiversidePrinting PrintmgCo. Co, (Rending) (Reading) Lid Ltd and pnnted by Cover photograph photograph of of Hays Hay'SGallena, Gallena,London, London,courtesy courtesyof ofBritish BritishSteei Steel Typeset by Tradespools Trndespools Lid, Ltd, Frome, Frome, Somerset Somerset Printed PmHed in in Malaysia, f.,blaysla, PP pp
First pnMed pnnted 1996 1996 ISBN 0-582-23088-8 0-582-23088-8
Bntish Bntish Library Library Cataloguing-in-Publication Catalogumg-m-PublicatlOn Data Data A catalogue catalogue record record for for this this book book isIS available available from from the the Bntish BntishLibrary. Library. Wish in 10 thank thank Ward Ward Building Building Components Components for for permission pennlsslOn to to TIle publishers publisherswish The use their their material. matena!.
Structuml steelwork and diploma diploma courses courses Structural steelwork design design ISis usually usually laught taught 10 tn degree and Initial grounding in In the theory of of structures and strength of of matenals matenals_ after an initial design teaching elements and and The design teaching usually usually covers covers both both simple'structurnl simplestructural elements buildings, More often covered covered complete buildings. More complex complex elements elements and and buildings are often In postgraduate postgraduate courses, outlined in in this this text textstill still in courses, but but the the ideas and concepts outlined provide the baSIS for more complicated structures. This book book has has been been the basis prepared pnmarily for the Ihose engineers engmeers in m practice pracllce prepared pnmarily for the student, student, but but also also for those who are not familiar familiar with wilh BS BS 5950: 5950: Structural Stnlctllraillse ofsteel steelwork III buildings. buildillgs. use of work in TIus book book falls falls naturally naturally mlo sets out out in In detail detail the thedesign deSign This into two two pans. Part I seis of elements (beams, columns, columns, etc.) frequently found found in aa structural structural steel steel etc.) frequently framework. Part how these elements are combined combined to 10 form form aa framework. Part Il II shows shows how building frame, building frame, and and should should prove prove cspecllllly especially useful useful to to the the engineer engineer m in the of practtcal praClJcal destgn. deSign. Past Part U 11 also develops other other considerations consideratIOns suck' such as as context of ovemll stability the overall stability of of building building structures. structures. Tliose Those with with some some expenence experience of element deSign usmg the cross-references cross~references to to element design may may prefer prefer to to start start with with Part Part n, II, using re-examine element final chapter considers detailing detailing re-examine element deSign design as as necessary. necessary. A final practice, and pracllcal considerations consideratIOns such as practice, and Ihe the effects effects of a number of practical fabncation and and fire fire protection. protecuon. fabncatioo It is IS assumed that the reader has some knowledge of of structural analysis analYSIS and and Ii baSIC understanding of metallurgy has been gained gained elsewhere. elsewhere. The The that a basic deSign examples examples concentrate concentrate on manual methods methods to ensure ensure aa proper proper design on manual understanding of steelwork compulmg understanding sieelwork behavIOur. behaviour, with with suggestions suggestions where computing could be used. Detailed programs for specific microcomputers are are Increasmgly bemg of complele design deSign packages packages arc are increasingly being Wfmen, written, and and a number of commercially. available commercially. pnncroal documents documents required required by by the the reader reader are: are: The pnncipal BS 5950: 5950; Part. Part It (1985): (1985): Design DeSIgn in 111 szntple SImple cOllstmctlOn; rolled sections. seC/lOllS. constnzctton; hot rolled British Standards Standards Institution. Insiltutlon. British Steelwork deSign. Section properties; member capacities. Steel Steel Steelivork design.Vol Vol i:i: Section properties; ,neniber Construction Institute. Instltule. Construction first of these these documents documents IS form from from The first is also also available available In in abridged extract form British Standards Standards Institution InslltutlOn as: as: British Extracts from from British British Standards Standardsfür forstudents SWdenLS of stmctllral deSIgn. of structural design. Extracts
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vi PREFACE vi PREFACE
Thesecond seconddocument documentISisll\'ailable availableIninextract extractfonn form(dimensIOns (dimensionsand andpropertIes properties The only) minthe the followmg following two two publicatIOns: publications: only) c/seeklist listfor fordesIgners. designers.Steel Steel ConstructIOn Construction lnsbtute. Institute. AA check StructuralSectioJlS SectionstotoBS 1154:4:Po,.t Part1I && BS 1154848: 4848;Part Part 4.4. British BritishSteel Steel St11lcfllrol Corporation. CorporatIOn.
Thelatter tatter two two extracts extracts are are updated updated regularly regularlyand and the the latest latest edition editionshould shouldbe be The used.Ititshould shouldbe benoted notedthat thatsmce since 1995, 1995,the the symbols symbolsused usedininthe the BSC BSC used. publication for for the the mam main dimensIOns dimensions of of rolled rolled sectIOns sections have have been publication been changed changed toto reflectthe theEurocode l3uroeode3 3nomenclature. nomenclature.The Therelevant relevantchanges changesare arenoted notedatatthe the reflect foolof of this this page. page. foot Throughout the the book, book, clause clause references references and and notation given m in ll1roughout notatIOn follow follow those those gIVen BS 5950: Part i I (deSIgn 5950: Part (design 1II in sImple simple alld mid conUnuous continuous construction), construction), except except for for BS iliosechapters chapterswhich whichdeal dealspecifically specificallywith withcompOsite compositeconstruction constructionwhen whenthe the those clause references referencesand andnotahon notationfollow followthose thoseminBS BS 5950: 5950;Part Part3.3.ii (deSign (design of of clause composite beams) beams) and and Part Part 44 (design (design of of floors floors with profiled steel steel sheeting). sheeting). compOSite with profiled The mam main change change m in the the second second edition edition has has been been the the II1trocluctlOn intioduction of of aa Tbe chapter on on the the deSIgn design of of plate plate girders. girders. The The authors authors have have also also taken taken the the chapter opportunity to to update update the the text text III in the the light light of of current current practice practice and and latest design opportunity latest deSign information. mformation. While every every effort effort has has been been made made to While to check check both both calculations calculations and and mterpretatioa of of BS 85 5950 for mterpretatlOn 5950 the tbe authors authors cannot cannot accept accept any any responsibility responsibility for inadvertent errors. errors. Inadvertent LilvI LJM flit? DRP
Acknowledgements Acknowledgements
CONTENTS CONTENTS Preface Preface Foreword Foreword to tofirst first edition edition
PART PART I
I
l.l 1.1 1.2 1.2 1.3 I.] 1.4 1.4 1.5 1.5
i.6 i.6 1.7 i.7 1.8 1.8 1.9 1.9
The authors The authors acknowledge acknowledge the the assistance assistance of ofmany many structural structural engineers engmeers in in industry and whom details details of mdustry and teaching, teaching, 10 10 whom of both both interpretation mterpretalIon and and current current practice have have been been submitted, submllted, and and whose whose helpful helpful comments comments have have been been practice incorporated into into the the text. text, In in particular, particular, the the collaboratIOn collaboration with with P.A. PA. Butter III IUcorporated Rutter in the the initial Initial drafting drafting of ofthe the first first edition edition proved provedinvaluable. invaluable. Extracts Extracts from from 85 BS 5950 5950 are arereproduced reproduced by byperintssion permiSSIOn of ofthe the British British Standards Standards Institution. InstitutIOn. Complete Complete copies COPIes can can be be obtained obtained from from 851 BSI at at Linford Linford Wood, Wood, Milton MiltonKeynes, Keynes,M1C14 l\-iK14 fiLE. 6LE.
2.2 2.2 2.3 2.3 2.4 2.4 2.5 2.5
The The equivalent eqUlvaient Eurocode Eurocode 33 notatton notallon for for the the relevant relevantnotation notatIOn given given in III
2.6 2.6 2.7 2.7
Deli D;;:;:" b '" b/2t bj2r
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INTRODUCTION DESIGN IN IN INTRODUCTION TO TO DESIGN STEELWORK STEEL WORK
3
DeSign Design reqUirements requirements Scope Scopeof ofBS 85 5950 5950 Struclllrai Structural use use of of steelwork steelii,ork m in buildings buildings limit Limit state state deSign design Partial factors Partial safety safety factors Landing Loading Internal forces and and moments moments tniernal forces Stresses deformatIOns Stresses and and deformations Layout calculailons Layout of of calculations Structural theory Structural theory Fonnat of chapters chapters Format of Study references references Study
55 55
LOADING AND AND LOAD LOAD 22 LOADING COMBINATIONS COMBINATIONS 2.1 2.1
Trat T;;:;:t
iid;;:;: AA (T-secdon (T-section oaly) only)
xiii
THE THE DESIGN DEStGN OF OF STRUCTURAL STRUCTURAL STEEL STEEL ELEMENTS ELEMENTS
LlO 1.10
85 BS 5950: 5950: Part Part i.I.
v xi;;
Dead loads loads Dead Imposed loads loads Imposed Wind loads loads Wind combinatIOns Load combinations Example I.I. Loading Loading of ofaasimply Simplysupported supported Example gantry girder girder gantry Example 2. 2. Loading Loading of of continuous continuous spans spans Example Example 3. 3. Loading Loading of of aa portal POrtal frame frame Example Study references references Study
BEAMSIN INBUILDINGS BUILDINGS 33 BEAMS
3.1 3.1 3.2 3.2 3.3 3.3
Beains with with MI full lateral lateral restraint restramt Beams Beams without without full full lateral lateral restraint restramt Beams Simplified design deSign procedures procedures Simplified
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viii viii
CONTENTS CONTENTS
CONTENTS CONTENTS Ix ix
3.4 3.5 3.6 3.7 3.8
Moment capacity capacity of members members (local (local Moment capacity check) check) capacity Buckling reSIStance resistance (member buckling Buckling check) check) Other considerahons considerations Other Example 4. 4. Beam supporting Example supponmg concrete floor slab slab (restrlllned (restrained beam) beam) floor Example 5. Beam supporting Example supponmg plant loads loads (unrestrained beam) (unrestramed Study references
PURLINS AND SIDE SIDE RAILS 44 PURLtNS 4.', 4.1 4.2 4.3 4.4
Desigit reqUirements requirements for for purlins purlins and and side DeSign rails mils Example 6. 6. Purtin Purlin on Example on sloping slopmg roof Example 7. Design DeSign of of side side rail rail Example Example 8. 8. DesIgn Design of of multi-span purlin Example
55 CRANE GIRDERS GIRDERS
30 30
7.7 7.7
31 31
7.8 7.8
31 31 32 32
36 36
47 49 52
99 COMPOSITE COMPOSITE BEAMS BEAMS && SLABS SLABS
H 42
42 42
8.6 8.6
43
8.7 8.7
TRUSSES TRUSSES
65
Types of of truss Types truss and and their their use use Loading and Loading and analysis analYSIS Slenderness of Slenderness of members members Compresston CompreSSion reststance resistance Tension capacity TenSIOn capacity Connections ConnectIOns Example II. Example 11. Roof Rooftruss truss with with sloping slOPing rafter Example 12. 12. Lattice Lattice gtrder gnder Study references references
65 66 67 67 69 69 69 70 71 76 78
77 SiMPLE SIMPLE AND COMPOUND COMPOUND COLUMNS COLUMNS
Column Column bases Design Design of of column bases Brackets DeSign Design of of brackets 'Example',I5. Example.15. DeSign Design of of slab slab base base Example 16. 16. DeSign Example Design of crane girder brncket (face) bracket (face) Example 17. 17. DeSign girder Example Design of crane gtrder 'bracket {lapped} bracket (lapped) Study references
40 40
Crane wheel loads Crane loads Maximum load load effects MaXimum Example 9. Crane girder without Example without lateral lateral restraint along span restramt Crane girder with Example ID. 10, Craoe with lateral lateral restraint restramt Study references
5.1 5.i
9.1I 9. 9.2 9.2
61
9.3 9.3 9.4 9.4
64
9.5 9.5 9.6 9.6 9.7 9.7
7.1 7.1
Types Types of ofcolumn column Axial Axial compression compressIOn Slenderness Slenderness Bending and eccentncity eccentnclty Local Local capacity capacity Overall buckling buckling
80 81 81 82 82 83 83 84 84 85 85
Composite beams Shear of composite composl!e Shear and and moment moment capacity of beams Shear connectors In concrete concrete Local shear in Deflections Deflections slabs Composite slabs Example IS. 18. Composite Composue beam in In building building Example Study references
10 BRACING tO BRACING 10.1 10.1
by bracing bracmg Loading resisted by
10.2 10.2
Sway stability Stvay Multi*storey bracing braCing Multi-storey Single-slorey bracing bracmg Single-storey Be:lIm, truss and column column bracing braCing Beam, wmd girder girder and and side side Example 19. Gable wind braCing bracing l....fulti~storey wind wmd bracing bracmg Example 20. Multi-storey references Study references
10.3 10.3 lOA 10.4 10.5 10.5 10.6 10.6 10.7 10.7
7.2 7.2 7.3 7.3 7.4 7.4 7.5 7.5 7.6 7.6
Example 13. Column for industrial industrial Example 13. building building Example Example 14. 14. Laced Laced column column for for industrial industnal building building Study Study references
13.1 Introduttion 13.1 IntroductIOn 13.2 Design brief 13.2 DeSign brief '3.3 Design information 13.3 DeSign mfonna!lOn 13.4 Design of purlins and sheeting rails 13.4 DeSign of purlins and sheetmg rails 13.5 Spacmg of ofsecondary secondarymembeis members 13.5 Spacing 13.6 DeSign of ofportal portalframe frame 13.6 Design 13.7 Frame stability 13.7 Frame stability 13.8 —- lateral fI,·Iemberstability stability lateral torsional torsIOnal 13.8 Member buckling buckling
226 226
13
14 14
Overall Overallstability stabilityofofbuilding building DeSign of main column base Design of main column base DeSign DesignofoffoundatIOn foundationblock block Other Oilierconsiderations considerations Study Studyreferences references
256 256 265 265 273 273
275 275
277 277 279 279 279 279
DESIGN DESIGN OF OFAN AN OFFICE OFFICE BLOCKBLOCK — COMPOSITE CONSTRUCTION
281
Layout and baSIC chOIces Layout and basic choices Loading Loading Roof Roofbeam beam deSign design Typical floor beam deSign Typical floor beam design Column Column deSign dcsign ConnectiOns Connections Wind Wind bracmg bracing Wind resistance by /Tame action Wind resistance by frame action Study references
DESIGN DESIGN OF OF SINGLE-STOREY SINGLE·STOREY BUILDING BUILDING -- PORTAL PORTALFRAME FRAME CONSTRUCTION CONSTRUCTION
13
13.11 13.11 13.12 13.12 13.13 13.13 13.14 13.14
143 143
DESIGN Of OF SINGLE-STOREY SINGLE-STOREY DESIGN BUILDING -- LATTICE AND BUILDING LATTICE GIRDER GIRDER AND COLUMN CONSTRUCTION CONSTRUCTION COLUMN Design bnef bnef DeSign Preliminary deSign design decisions Preliminary deCISIons Loading Loading Design of of purlins purl ins and and sheeting sheetmg rails rails DeSign Design of of lattice lattice girder girder DeSign Design DeSign of of column coiumn members members Overall stability Overall stability of of building building Design of gable DeSign of gable posts posts Design DeSign of of connections connectIOns Design DeSign of of foundation foundatIOn block block OIlier Other consideratIOns Study Study references references
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First Edition Edition Foreword to First
In 1969, 1969, the Standards InstitutIOn Comml!tee 11/20 B/lO responsible responsible for for the Bntish Untish Standards Institution Committee BS BS 449, 449, a permissible stress stress structural the sn-ucturatsteelwork steetwork design design code. code, lnslIgated instigated the code based based on stale pnnciples pnnclplesand and preparation of preparation of aa new draft code on iimJt limit slate Incorporatmg the the latest latest research research 1010 1010 the structural components components incorporattng the behaviOur behaviour of of structural and complete complete structures. In 1977 1977 and structures.The Thedraft draft was was Issued issuedfor for public public comment in and attracted long and attracted considerable considerable adverse adversecomment commentfrom from an an mdustry industry long aCQuamted with the the simpler Simplerdesign deSignmethods methodsinIn115 BS 449. 449, acquainted It was was realized realized by the the newly constituted constituted ES! BSI Committee, Conumttee, CSB/27, CSBj27,that thataa It of the tbe 11/20 B /20 document document would necessary before be redraft of would be necessary before Itit would' would be acceptable totothe acceptable theconstructIOn constructionmdustry. industry.The Thework work of of redrafting redrafling was funded by by Ibe the European European Coal Coal and and Steel Steel undertaken by undertaken by Constrada, Constrado, partly partly fbnded and the task task was was guided by aa small small steenng steenng group group representing representmg Community, and engmeers, steelwork and the the the Interests of the interests of consulting engineers, steetwork fabncalors and Department of EnVironment. Department of the Environment. Pnor to the the completIon completIOn of the redraft, calibration calibration was was carried carned out out by by the the Poor of the land and and Building Research Research Establishment Establishment to to denve denve SUitable suitable values values for load factors, and and design deSign exercises exercises to compare compure thc the design deSign of ofwhole whole matenal factors, structures 10 code with designs deSigns to US BS 449 449 were were directed directed by by structures to the the draft draft code of these these studies was was to to assess assess whether the Constrado. Constrado. The The object of produce recommendatIOns recommendationstotobe becontained containedwithin within the the new new code code would produce structural be no less safe structural deSigns designswhich which would would be no less safe that} thaq deSIgns designs to to BS BS 449 449 but In overall overall economy. economy. would an improvement In woold give an resulting code code of practice, practice, BS 5950: Part 1, published published in In 1985. 1985, The resulting 1155950: Part I, covenng of hot hot rolled rolledsteel steel covenng the the deSign design In in simple simple and and contmuous continuous construcuon construction of of materials, materials, fabneation fabncauon seclions, and the specihcattott specificatIOn of sections, and Part Part 2, 2, dealing dealing with with the and mdustry while while and erection, erection, achieved achievedthe thegreater greaterSimpliCIty simplicity sought sought by by industry structures lobe to be based based on on the the more more rationai ratIOnal allowiOg the design deSign of allowing the of building structures approach limit state state theory theory than than the the permissible permIssiblestress stressmethod methodofofUS BS4-49, 449. approach of of limit Part 3, 3, which which ISis In in the the Course courseof olpreparation, will give for preparutlOn, will give rccotnmendattons recommendatIOns for Part constructIOn. deSign design in in composite constructIon. Part! is }S explicit explicit inInits ItSdesign deSIgnrecommendations, recommendatIons, the tile code code While US BS 5950: 5950: Part is intended persons who have experience intended to to be be used usedby byappropnately appropnately qualified qualified persons III structural steelwork design deSign and construcilon. IS aa need, need, therefore, for for to constructton. There is
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xiv
XIV
FOREWORD FOREWORD
textfor forunivcrsIty universityand anticolJegc collegestudcnts studentscngaged engagedononcourses coursesIIIinclYH civiland and aatext structuralengmeenng engineenngwhich whichglves givesclear clearguidance guidanceononthe theapplication applicationofofthe the structural code tototypical typicalbuilding building structures structuresby byworked workedexamples exampleswhich whichset setout outthe the code calculationsundertaken undertakenIninthe thedesign designoffice. office. This ThisISisachicved achievedininthis thisbook book calculations throughexplanatory explanatory(ext, text,full full calculatIOns calculations and and reference referencetotoUle theas BS5950: 5950: through Part clauses. Part i clauses. I
Theftrst firstpart partofofthe thebook bookdeals dealswith withthe thedesign designofofvanous vanoustypes typesofof The structuralmembers membersand andthe thesecond secondpart partdeals dealswith withcomplcte completedesigns designsofofthc the structural mostcommonly commonlyencountcred encounteredstructures, structures,namely, namely,smgle-storey single-storeymdustnal industnal most buildingsand andmulti-storey multi-storeyoffice officeblocks. blocks.Apart Apartfrom fromthe therelevant relevantcodes codes and and buildings standards, references referencesare are gtven given 10 to wcll wefl established estabhshed publicatIOns publications commonly commonly standards, foundininthe thedesigner's designers office. office. The The book book should should aiso also be be useful useful to to the found the desIga design engineerTeQumng requinngan anunderstanding understandingofthe of theapplicatIOn applicationof ofthe thelimit limitstate statecode. code. engmeer
PA. Rutter Rotter P.A. Partner, Scot! Scott Wi/son JVI/sonKirkpamck Kirkpatrickand andPartners Partners Partner. Member of of BS 1355950 5950 Committee committee Member
THE DESIGN OF THE DESIGN OF STRUCTURAL STRUCTURAL STEEL ELEMENTS STEEL ELEMENTS
A framework composed of a A simple simple basIs basis for for design design IS is to to consider consider aa structural structural framcwork composed of a sustamed by the element, numbcr number of of elements elements connected connected together. together. Loads Loads are are sustained by the element, Ute connectIOns. In [his and and ItS its reachons reactions transferred transferrcd (0 to other oilier eiements elements via via the connections. In this IS essentml that tbe overall actIOn of the simple simple concept concept for for design design Itit is essenttal that the overall action of the framework an introduction introductIOn to to llle concept of overall framework ISis considered. considered. Therefore Therefore an the concept of overall the structure structure is IS given In terms of bracing systems. stability stability of of the given in terms of bracing systems.
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[] DESIGN INTRODUCTION TO DESIGN IN STEELWORK
Structural Structural steelwork can can be either a single smg!e member member or an assembly assembly of of aa number of of steel steel sections sectIOns connected perfoml aa number connected together together In in such such aa way way that that they perform specified funeuon. specified function.The The function function requIred requiredby byaaclient clientororowner ownerwill will vary van enonnously hut but may may include: mclude: enormously
•
••
• •
• •
•
framesby bywhich whichloads loadsmust mustbe besupported supported safely safely and .and building frames building without undue undue movement, movement, and weatherproof envelope envelope and 10 to which aa weatherproof must be attached; attached; must be chemical supports by bywhich whichloads loads must must be be supported supported hut but chemical plant supports commonly rentnre reqUire no no external external envelope; envelope; which commonly containers which will willretain relamliquids, liquids,granular grnnularmaterials matenalsororgases, gases, and which may may also also be be elevated elevated as structural function; functIOn; and as aa further further stractural masts masts which must must safely safely support support mechanical mechamcal or or electncai electncal and in In which which the Ihe deflecttons, defiecttons, equipment at equipment at specified specified heights, heights and vibrattons fatigue must vibratIOns and and fatigue must be controlled; which will willsupport supportflues fluescarrytog carrymgwaste wastegases gases totosafe safe chimneys which heights; hetghts; bridges which which must must support support traffic traffic and and other other loads loads over over greatly greatly varying of movement movement may may be be varylOg spans spans, and and for for which which degrees degrees of permitted; penmtted; temporary supports supportsused used during dunng the the construction constmctlOn of of some some part ofaa temporary pail of structure, of steelwork. steelwork, concrete, concrete. brickwork, bnckwork,etc., etc., HIit structure. which may be be of which safety safety for forshort shortpenods penodsand andspeed speed of o{assembly assembly are are important. 1I11Portam. which
willbe benoticed nOilcedthat thaiboth bothsafety safetyand andmovement movemeO!ofofthe thestructures structuresdescribed described itIt will are proper functiOn and, together be are Important important for for proper functton and, together with with economy. economy, these these will will be the mam the main consideratIOns considerations when when diSCUSSing discusstng the the deSign destgn method method later. later, It should be noted that the the deslgn design of of only some noted that some of of the the above above structures structures is IS covered by SS 5950 and and hence hence discussed discussed in chapters. BS in the later chapters. Steel vanety of ofcross-sections, cross-sectIOns, aa Steel sectIOns sectionsare arerolled rolled or or fanned formed into aa variety selechon IS shown L I. together together with their their common common selection of of which which ts shown In tn Fig. 1.1, descnpflOns. ofthese these cross-sections cross-sectIOns are are obtatned obtained by the the hot hot descriptions. The The maJonty majority of of steel steel billets 10 mill, while while aa minority, mlnonty,sometimes sDlnettmes involvtng IOvol\'lOg rolling of in aa rolling rolling mill, complex shapes. are steel sheet. sheet Hollow Hollow sections sectIOns are are complex shapes, arecold cold formed formed from steel obtained obtamed by extrusion extrusIOn or or by by bending bendingplates platestotothe therequired reqUiredcross-section, cross-sectIOn,and and
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___________
4
4
STRUCTURAL DESIGN TO TO 8S ES EBSO STRUCTURAL STEEl WORK DESIGN 5950
!NTRODUCTION TO DESIGN IN STEELWORK INTRODUCTION TO DESIGN IN STEEIWORI(
seaming(welding) them themtotofann formtubes. tubes.The TheSectIOns sectionsare areusunlly usuallyproduced produced in in aa seammg varietyof ofgrndcs gradesof ofsteel steelhavmg havingdifferent differentstrengths strengthsnnd andother otherproperties. properties.The The vanety
1.1 1.1
commonestgrade gradeISisknO\vn knownasas43A, rcferred referred 10tosometimes sometimesasasmild mildsteel, steel, commonest 1 In some types of having a yield strength N/mm2. . In some types of havmg a Yield strength in In the the range range 245—275 245-275NJmm structureother othergrades grades(43B, (43B,43C, 43C.etc.). etc.),having havingthe the same same Yield yield strength, strength are are structure more suitable owing to their higher resistance to bnttle fracture. more sUltnble oWing to their higher resistance to bnttle fracture.
Web
Cross-section
.rr1 ]E
Flange Flange
Flange
j
Web
Fiinge_E
i
L 6,1 Dasc,iplion Descnptron Typicai sue
TypIcal sIze
universal beam beam Universal 305 x 127 x 48 U8 305 x 127 x
crass.section
Cross·section
Oescriptinn
Fig 1.1 Steel sections
Oescnption Typical size
TypIcal size
r Equai angle Equal angle 150x 150 18 Angie 150x150xlBAng!e
In to cross-section, the shape In addition addition to cross-section, the shape of of aa steel steel section sectIOn will will include 'mclucte reference to its 115 length, length. curvature curvature if if required, reqUired, euttmg cuttmg and and drilling drilling for for reference to connections. connecttons, etc., etc., all all of of which which are are needed needed to to ensure enSure that that each eacn part part fits fits accurately accurately into mto the the finished fimshed structure. structure.'Fhese TI,ese further furthershaping shapmgprocesses proceSses are are known knmvn as as fabncation fabncatton and and are are earned carned out out in m aa fabncating fabncatlOg works works or or 'shop. 'shop'.ItIt isIS for for this tlus stage stage that that drawings drawmgs giving givingprecise precisedimensions dimenSIOnsof ofthe thesteelwork steelworkwill willbebe required, reqUired, showing shOWing what what the the designer designer intends. mtends. In In many many eases cases these thesedrawings drawings are are oroduced produced by by the the fabneatnrs fabncators rather rather than than by bythe thedesigner designerof ofthe thestructure. structure. The The final final stage stage of of producing produemg aa slnjeture structure in In steeiwnrk steeiwork is 1S the the erection erection or or putting pultlOg together together of of the the vanous vanous elements elements on on site Site to to form fonn the therequired reqUired framework. framework. At At this this final final stage stage the the safety safetyof ofthe thepartly partlyfmsslied fuushed structure structure must must be pnor thought thought given gIVen as as to to how how the the framework framework is IS to to be be erected erected be checked, checked, and and prior in order to define the location of each part with precision. The steelwork In order to define the locahon of each part with precIsion. The steelwork usually usually forms forms only only aaskeleton skeletontotowhich whichother otherbuilding buildingelements ciements(floors, (floors,walls, wails, etc.) frameworks supporting chemical or mechanical plant etc.) are are fixed, fixed, but but in III frameworks supportmg chemical or mechanical plant the thesteelwnrk steelworkmay maybe bethe thesole solestructural structuralmedium. medium.
S
DESIGN DESIGN REQUIREMENTS REQUIREMENTS The The design designof of!lily anystructure structuremust mustbebeJudged judgedby bywhether whether[titfulfils Ililtilsthe thereqlllred required functIon mnmtam an acceptable functionsnfely. safely,can canbebebuilt builtwith witheconomy economy rutd and cnn can maintain an acceptable appearance structural appearancefor forIts itsspecified specifiedlifet[mc. lifetime.hIt foHows follows that that thc the design design of of stilictural steelwork steelwork also also will be beassessed assessedby bythese thesecntena cntena of of safety, safety, economy economy and and appearance. appearance. Safety SafetyISisassessed assessedby byconsidenng ennsidenngthe the strength strength of of the the structure structurerelative relative to to the the ioads loadswhich whichititISisexpected expected toto carry. carry. In In practice, practice, this this asscssment assessment ISis applied applied to not to each eachstructural structuralelement elementIn in turn, turn, but but these these mdividual individual element element checks eltecks are are not suffiCient sufficientwithout 'vithnutconsidering consideringthe theoverall overallsafety safety of of the the fr..tmework. fratnework. The The strength ioads strengthof ofthe thestructural structuralelement elementmust mustalways alwaysexceed exceed the the cffects effectsof of the the loads by of providing by aa margm margin which which1Sisknown known as as aa f..ctor factor of of safety. safety. The The method method of providing the the factor factorof ofsafety safetyISisdiscussed discussedininSectIOn Seehon iA. i.4. In In the the general general sense sense assessment assessmentof ofthe thestrueture structureIncludes includesall allthe thecntena cntena by by which which Its its performance performance will will be be judged, judged, e.g. e.g. strength, strength. deflection, deflection, vibration, vibration, etc. etc. Wh.ile While in in practice practice economy economy of of the the deSign design ISisof ofgreat great Importance importance to to the the owner owner ofthe economic of the fimshed finishedstructure, structure,students students are are rarely rarely reqUired required to to make make aa full full economic assessment. assessment. However, However, two two baSIC basic matters matters should should be be taken taken 1010 into account account. Firstly, Firstly, the the finished finished design design should should match, match, without without exceSSively excessively exceeding, exceeding, as as many many of of the as possible. possible. Clearly, the deSign design cntena entena as Clearly, the the proVISion provision of of excess excess strength strengthmmaa structural m structural element element without without reason reason will will not not be be Judged judged economic. economic. Secondly, Secondly, in structural steelwork steelwork construction constructIOn only only part part of ofthe the cost cost IS in the structural is contamed contained in the rolled rolled steel sectIOns, steel sections, and and aa large large part part ofthe of the cost cost results results from from the the fabncallon fabncahna and and erection process. Consequently, economIC design deSign does does not result from fmding erection process. Consequently, economic not result from finding Ihe smallest and weight weight without without considenng considenngthe the difficulties difficulties of the smallest structural structural Size size and of fabncatlOn. In In many many eases cases repetition repetition of ofa member size and standardizatIOn fabneation. a member size end standardization of of components can lead to to substantial overall savings. savmgs. components can lead substantial overall The appearance of of the the finished fiOlshed structure great importance Importance The appearance structure IS is geneml!y generally of of great owmg to to the the veiy very size Size and and impact impact of offrames frames in m structural structural steel. steel. The Thc owing achievement elegant deSign IS desirable deSirable not not only only in In complete structurcs achievement of of an an elegant design is complete structures but also IS here here that that the the studcnt should try to achieve but also In in small sniall deSign design details. details, It it is student should try to achieve many cases cases these these will will stylish, iteat neat and and balanced balanced solutions solutIOns to to problems problemsset. set.In In many stylish, to be be the the strongest strongest and and most most economic economicsolutions solutIOnsalso. also. prove to prove
Outstand Outsland
TL rH
5
1.2 1.2
SCOPE OF OF BS as5950 5950STRUCTURAL STRUCTURAL USE USE OF OF STEELWORK STEELWORK IN SCOPE IN BUILDINGS BUILDINGS
BS 5950 5950 is is subdivided subdivided into into nine mne parts, parts, each each being bemg published publiShed separately, scparately. Parts 85 Parts to 99 inclusive mciuSlve are are a'vaiting awaitingpublication. publication. and S5 to 33 and
De.flgll En III Simple alld ca,itznuaits COlljlllllOUS COIlSl!1lcilOn: rolled Part I:I: Design Part siniple and co,:su'iictzon: hot liar rolled sectlOlIs (1985) (1985) sections
Part 2:2: Specification SpecijicallOlIfor formaterials, maienals. fabncatlOlI erectIOn: hOI Pail fabrication oar!ami erecrio,i: hot rolledsections sectlOlls(1985) (985) rolled Part3:3: Design DeslglI En III composite composue construcnon constrllCIlOll Pad Part 4:4: Design Deslgll oj'Jloors offloors with withprofiled profiledsteel steelsheeting sheeung (1982) (1982) Part Part5:5: Design DeSIgn of ofcold coldfanned fannedsections sectIOns Part Part 6:6: Design DeSign in l1Jlight fightgauge gaugesheeting, sheetlllg.decking deckingmid alldcladding cladding Part
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6
STRUCTURAL STEELWORK STEELWORK DESIGN DESIGN TO TO ES 8S 5950 5950 STRUCTURAL
7: Part 7: 8: Part 8: 9: Part 9:
INTRODUCTION TO DESIGN DESIGN IN IN STEELWORK STEELWORK INTRODUCTION
SpecificatiolJ cold fanned Specificationfor for matenais materialsand and workmanship: woitnanship: cold sectlOlls sections Deslgll of offire protectio1l sIn/crural steelivork steelwork Design protection for for stntcr,,ral skin design desIgn Stressed skin
I. I.
2. 2.
3. 3. 4.
5. 5.
The purpose of of 135 BS 5950 define comman 5950 IS is to to defiuie commoncntena cntena for for the the design design of structural steelwork in m buildings buildings and and allied allied structures, structures, and and to to give givegUidance guidance to to designers on nn methods methods of assessing compliancewith with those thosecntena. cntena. Part' Part of designers asseSSing compliance this British Bntish Standard deals with with design design in m simple and continuous contmuous construction construction rolled sections. sections. Part Part 22 covers covers the the specification specificahon for for materials, matenals. for hot rolled fabncatlOn and erectIOn. follOWing chapters deSign fabrication erection. The following chapters give exampies examples of the design of buildings principally pnnclpaJly covered covered by by Part Part I1 and and Part Part 22ofof135 BS 5950. 5950. IS also made of of other olher parts Darts for far particular particular design deSign requirements such as Use is composlle these are referred'to appropnate in 10 the the composite canstructJOn,-and consiruction,ind these to where appropriate followmg following chapters. BS 5400 is the appropnate appropriate code for for the deSign design afbridges, of bridges, baSIS for the design deSign of other types types of of and more appropriate appropnate basis and may also be a more for the plated bunkers. plated structure, structuie, e.g. bunkers.
I
I
08
-.J
LIMIT STATE DESIGN
!!!~ i>:tlt
In common 5 -0 adopts ad ~ts a limit common with with most most current UK codes codes of practice, practice, BS 85 5950 state state approach approach to to deSign. design. In In this this approach, approach, the the deSigner designer sele selects a nunt.;~f number of cntena assess the the proper proper functioning functIOning of of the the structure structure and and then then criteria by by which to assess checks whether they have been satisfied. satisfied. The into two two checks whether they have The cntenn criteria are divided into malO groups •main groups based based on on whether whether assessment assessment ISis made made of the the collapse (ultimate) condition, or or normal nonnal worktng working (serviceability) (serviceability) condition. condition. IOcludes: Ultimate limit state includes:
vanatlons; load variations; load combinations; and detailing detailing procedures; procedures; design and fabncahon and and erection erecHon procedures; procedures; fabrication matenal variations. vanatlons. material
faclor can be applied applied at at one one point pomt in In the the design deSign (global (global or or The safety factor pomts t,partial (partial safety overall safety factor), or at several points safety factors). factors), In steelwork deSign aa partial y{ (the (the load load factor) loads design partial safety safety factor factory1 factor) IS is applied applied to to the loads (variations I to 33 above) (variatIOns above) and and another another factor factor y,,, Ym to io the the material matenal strengths strengths (vanahons 5950: Part Part i includes mcludes a value value of of y, y/ for for structural structural (variations 44 and 5). BS 5950: perfonnance within assumes a value of of 1.0 1.0 for for y,,,. i'm' The The performance within the the value value of YI' yj, and assumes use of Y y,, = 1.0 use LO does not imply Imply that Ihat no margin rnargm of safety for material matenal has been been m = Included, SUitable allowance allowance has has been 10 the the design deSign included, but but rather rather tlmt that a suitable been made in strengths given gtven in m e.g. Table Table 1.2 1.2 of ofthis this chapter. chapter. Typical Typ1C[lI values valuesofofYI are given given strengths y1 are 10 with further further values values given m table 2. 2. Application ApplicatIOn in Table Table Lt Li with in BS 85 5950: Part I,I, table the factors factors to different different loads loads in 10 combination combinatIOn is IS given given in m Chapter Chapter 22and and of the throughout the factor ·reflects the throughout the deSign design examples. examples. The The value value of each load factor reflects the be estimated, estimated, and and the the likelihood likelihood of ofthe the accuracy with accuracy with which which a load can be sunultaneous occurrence glVen combination combinatIOn of of loads. loads. simultaneous occurrence of a given
zm
.3 1.3
77
Table 1.1 Partial safety factor for loads
1.5 1.5
Loading
factor Yr Load factor
Dead load load lVd Dead restralnmg uplift uplift Dead load restraining load Wo Imposed load Wind load Ww Combined n~ + +Ii',,,) 1-Fw) Combined loads loads (Wd (lVa + + II',
i.4 1.4 j.O 1.0 1.6 LU
1.4 1.4
1.2 1.2
LOADING
Serviceability limit state state includes: mcludes:
Limiting values for each each cntenon criterion are are given given in inDS Limtting BS 5950: Part t1and and their thelfuse use isIS demonstrated followmg chapters. chapters. The designer deSigner should, should, however, demonstrated in in the following always be aware of always of the the need need for for additional additional or or varied vaned criteria. cnlena.
1.4
PARTIAL SAFETY SAFETY FACTORS Safety are used Safety factors factors are used in all all designs deSIgns to allow for for variabilities van abilities of of load, material, workmanship matenal, workmanship and so on, on, which which cannot cannot be be assessed assessed with with absolute absolute certainty_They suffiCient to certainty. They must be sufficient io cover: cover -
In most cases cases design deSign begins beglOs 'vith with as as accurate as as possible an assessment of of the Inmost loads ID be earned. carned. These may be be given, given, or obtained obtained from from a Bntish Bntish loads to be Standard(1) or other appropriate appropnate source. source. They will be used in in idealized idcalized forms fonns Standardth as either distributed distributed loads loads or point pomt tconcentrated) (conccntrated) loads. loads_ Chapter 22 sets sets out 01lI typIcal loadings loadings and how they are combined combin-ed in 10 design. deSign. typteal and gIves gives examples examples of how IOtal These external external loads, loads, sometimes sometimes called called actions, actions, fonn form only only part of the total forces on a structure, structure, or on on a structural structural elemem. element. The The reactions to forces on 10 the loads on each element element must proceeds_ These must be each must be be obtamed obtained as deSign design proceeds. These reactIOns reactions must camed through through 10 supportmg elements, elements, so that all external external loads, loads, including mcluding earned to supporting weight of members, members, are by the the shortest shortest self weight are transferred through the structure by load load path, path, unlil until the the foundation foundatIOn is IS reached. reached. This process is IS essential essenilUi to to safe safe deSign. A Simple IS shown in In Fig. Fig. 1.2, 1.2, In In which which the the design. simple example example of this this process process is load path path for for the external (snow) on a section sectiOn of of roof roof cladding cladding is IS traced traced load external load (snow) to the the foundation. foundation. The load as a~s well well as as The cladding cladding (sheelmg) (sheeting) cames comes the snow load
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INTRODUCTION TO DESIGN IN STEELWORK INTRODUCTION TO DESIGN IN STEELWORI(
i 'f
9
9
sclfwclght, from a typical Durlin. self weight,and andIhis thiscombined combinedload loadproduces producesaareactIOn reaction from a typical purhn. W on tllc purl in, producing reactIOns the rafter. This Thisconslltules constitutesaaload toad It' on the purlin, producing reactionsPP on on the rafter. Similar rafter self weight conslitute the Similarpurlin purlin reactIOns reactionslogether togetherwith withthe the rafter self weight constitute the rafter Loading raflerload, load,producmg producingreachons reactionsRRfrom fromthe theroof roof truss truss(al (ateach eachnode), node). Loading the the coiumn, and also a theroof rooftruss trussproduces producesvertical verticalreaction reactionTTfrom from the column, and also a honzonlaJ actlOg on honsontalreachon reactionSSifif the the ioading loading ISis non-verticaL non-vertical. These These lOintUn!, turn, acting on and A'!. M. the thecolumn, column produce producefoundatIOn foundationreactIOns reactionsH. H, V V and
1.6 1.6
?uriin
Loads moments Loadswilh with their their reactIOnS reactions may may be be used usedtoto fmd find intema! internal forces forces and and moments within dra\v shear force and within any any structural structural element. element. The The usual usual method method isis to to draw shear force and bending These diagrams graphs showmg how the bendingmoment momentdiagrams{2,J) diagramsi2]) These diagrams are arc graphs showing hon the mternal forces vary along a structural element as a reSUlt a sel of s!atlQnary internal forces vary along a structural element as a resultof of a set of stationary (static) envelopes may be needed in (static) loads: loads:Influence influence lines lines and and moment/force moment/force envelopes may be needed in cases ofmovmg (dynamiC) loads (see Chapter 2), illS also necessary 10 find cases of moving (dynamic) loads (see Chapter 2). It is also necessary to find the members, theaxml axial force force present presentmmaastructural structuralelement, element,particularlY particularly vertlcaj vertical members, and be used to advantage for and in in some some cases casesthe thetorsIOn torsionas aswell. well. Diagrams Diagrams may may be used to advantage for these these forces forces also, also. In many bending In many cases casesdeSign designconcentrates concentrateson onspecific specificvalues valuesof ofmaximum maximum bending moment pOsition, e.g. e,g, mid-span. mid·span, In Inthese these cases moment or or shear shear force force at at aa known known position, cases fonnulae formulae or or coeffiCients coefficients may may be be useful useful and and can can be be obmmed obtained from from standard standard tables or charts(") tables or charlst8t
I neaction
1.7 1.7
nailer
INTERNAL INTERNAL FORCES FORCESAND AND MOMENTS MOMENTS
STRESSES AND STRESSES AND DEFORMATIONS DEFORMATIONS tu the the design deSign of steelwork to !o 55 BS 5950: 5950: Part Part II stresses are used to obtam the In of steelivork stresses are used to obtain the cnpaclties of structural sectIons III bending, shear, aXial force, elC., and any capacities of structural sections in bending, shear. axiat force, etc., and any ofthese these forces. forces. Stresses Stresses used used are are generally generally based based on tltt! Yield combinations of combinations on the yield stresses appropriate thickness required reqUired by by stresses appropnatetotothe thesteel steelquality quality and and maximum maximum thickness the deSigner, BS 5950: 5950; Part Part!, which IS basl!d on the designer,and anddetailed detailedmintable table6,6,85 1, which is based on values sllpulated BS 4360 t !i) The deSign strength Py IS usea, for example, values stipulated in in 55 The design strength p. is used, for example. 10 calculate the moment capacity of ofaasteel steelsection. section. to calculate the moment capacity
Moment capacity capacity M, = P,yS, Moment = where 5, S. is IS the the major major axis aXIs plastic plastiC modulus modulus of ofthe the section. section. where
nasa reactions
roundation Foundation
Fig 1.2 Load tnnscer Fig 1.2 Load tmnsfer
v
Strength isIS used used to to define define an an ultimate ultimate stress stress for particular Situation Strength for aa panicular situation (bending, shear, shear, etc.), include an an adjustnteni adjustment for for partial partialsafety safelY factor (bending, etc.), and and will will include factor (matenal) and buckling (local or overall). (material) and buckling (local or overall). Capacityrefers refers toto aa local local moment momentofofresistance resistance tar (orshear shear or uXlai force) at Capacity or axial force) at sectIOn based based on on the the given given strength strength but but disregarding disregarding overall overall(member) (member) aa section buckling. ReSistance refers to maximum moment with due regard to overall buckling. Resistance refers to maximum moment with due regard to overall {member)buckling. buckling, tmember) someparts pnrts of oflhe deSignitItmay maybebenecessary necessarytoto assess stresses when the InInsome the design assess stresses when ihe steel isis inIIIthe thelinear linearelastic elasticcondition. condition.InInthis thiscase casethe thelinear linearelastic elasticbending bending steel maybebeused16'75 used{6,J} in in which which theorymay theory
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10 STRUCTURALSTEELWORK STEELWORKDESIGN DESIGN TO TO ES 8S 5950 5950 10 STRUCTURAL
INTRODUCTION TO STEELWORK INTRODUCTION TO DESIGN IN STEELWDRK
Al/I = a/j' =
11
1.3Deflection Defieclion limits !imns Table 13
E/J?
Deformations are usually required from Deformations are usually required in in the the design design and and are are denved from elasllc bending theory. theory. The commonest reqUirement elastic The commonest requirement ISis the the calculatIOn calculatton of of deflectIOns. These are found ustng usmg formulae formulae denved from from bending bending theotyttS) theory{8,9) deflections. but in m more complex cases cases may may reqUire require the the use useof ofmoment-area moment-areamethods{lo.lI) methods00"t In general, general, the the deflection deflectIOn due due to to unfactored unfactored imposed imposed or computer programs. In loads only only isIS required. reqUired. Strams normally calculated calculated in in steelwork steelwork design design and and excessive exceSSIve Strains are not normally strams avoided by by limiting limiting stresses stresses and and other otherdesign designparameteRs. parameters. strains are avoided In detail in BS Stresses which should be used in sieelwork steelwork design design are given in vaiues of of design design strengths strengths are are given given 5950; 5950; Part Part I, clause clause 3.1. 3.!. i. Some common values In Table 1.2. 1.2. in
Structural element element Strueturai
DeOection limit to Deflection limit due to unfadored Imposed imposed load load unfactored
Cantilever Beam (bnttle finishes) fimshes) Other beam Purlin Purlin or sheeting rail
Length/ Length/l180 80 Span/36O Span/J60 Span/200 To suit SUIt cladding cladding (but (bUl may be be used) used) span/200 may Span/600 Span/500 Span/500
that the steel grades grades 43A, Note ihat 43A, etc., etc., are are specified specified in in BS 55 4360(5), which defines mechanical and and other properties the mechamcal propertIes of the steel. steeL The most important Important defines the properties for for structural structural use use are are YIeld yteld strength, strength, tensile tensile strength and impact test test properties ~al~es. B, C indicate increasing resistance reSistance to to impact Impact and and values. The deSIgnatIOns destgnattons A, B. no significant Significant change in m the the other other mechanical mechanical properties. properties. bnttle fracture, fracture, with with no The cross-section of aa structural structural member member needs needs to to be be classified classified according according cross-seclton of to 5950: Part Part i, I. clause clause 3.5 3.5 in in order order to to assess assess the the resistance reSIstance to to local local to SS 85 5950: buckling secllon. Cross-sections Cross-sections are classified classified as plastic, plastIC, compact, compact, buckling of the section. semi-compact and and slender slender by by reference reference to to the the breadth/thickness breadth/thickness ratios ratios of semi-compact flange out~tands LI), and also to to the the design design strength. strength. Details Details flange outstands and webs (Fig. 1.1), are gIven classificatIOns of given In in clause 3.5 3.5 and and !able table 77 of BS BS 5950: 5950: Part Part 1. I. The classifications of most grades43 43 and and 50 50 are' are given mosl hot rotted rolled sections sectIOns Rn In grndes given with their their section sectIOn In reference reference (12). (12). properties in 5950: Part 1, Recommended values values of of maximum maximum deflectIOns deflectionsare aregiven givenminBS 855950: I. table 5. Some conunon common values values are reproduced reproduced in in Table 1.3. i.3. Ia In cases cases where where the the IS to (0 support support machinery, machinery. cranes and other other moving moving toads, loads, steelwork structure is more stnngent limitatIOns on deflection deflectton may be necessary. Values of ofmuxlmum strtngent limitations maximum deflectiOns should be checked with the manufacturers manufacturers of of any any machinery to to be be defiections used.
LAYOUT OF OF CALCULATIONS CALCULATIONS
deSign calculations calculatIOns are started, the must first first interpret Interpret the Before design are started, the deSigner designer must client's drawings draWings so so that Ihnt an structural structural arrangement arrangement can be decided to carry the the client's loads foundatIOns. This structural structural arrangement arrangement must musl avoid avoid loads down down 10 to the foundations. intruSIOn into Into space required reqUIred by the client's processes processes or or operations. operations. ItIt isIS intrusion down into mto simple Simple structural structural elements glVen an broken down elements which which are are each given Individual code number by deSIgner (see 15). Calculations CalculatIOns individual by the steel steel designer tsee Chapter IS). Individual element can tItus thus be identified, identified, as deSIgn examples. for an individual as In in the design cssentml in In setting setllng out out calculations, calculatIOns, and and the the designer deSignershould should Clanty isIS essenual to that thal they can can be be checked checked without without constant constant reference reference to make sure that they designer. deslgner. Designers DeSigners develop their own individual Individual styles styles For for setting settmg out their their calculations. calculatIOns. Design DeSIgn offices offices of consulting engineers, local authonties authonties and and contractors often format as house style. The student student contractors ofien use one particular format as a hotise baSIC format such that glVen but adapting it11 usmg a basic should start ustng such as as that given In in the text, hut SUIt the the particular particular structure. structure. to suit
IN/mml) (N/mm2)
275 265 255 245
16 40 40
I.B 1.8
1.8.1 I.B.I
Subdivision
Subdivide the calculatIOns calculations mto into approPfJate.sectlons appropriatesections using Subdivide the usmg subheadings subheadings such such as 'dimensions', 'dimenSIOns', 'loading', 'loading', moment 'moment capacity' capacity' This Thi~makes makeschecking checking of ofthe the as baSIC assumptIOns easter,.and helps the the designer deSignerachieve achieve baste assumptions and the results much easier, and helps neat presentation. presentatIOn. a neat -
1.8.2 1.8.1
-:
-
Sketches Sketches
Engineers Engmeers think pictorially, pictoriully. and and should should develop develop auspatial spatlill awareness. awareness.AAsketch sketch dearly indicate mdicate what the deSigner of numbers will clearly designer Intended, intended, while while 10 in aa stnng stnng of senOus omission omiSSIOn can be be overlooked. overlooked. Sketches In follOWing chapters are an senous in the following the lefi-hand left-hand margin margm where where convenient. convenient. placed in the
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INTRODUCTION TO DESIGN IN STEELWORK 13 INTRODUCTION TO DESIGN ftJ EELWORK fl
References References
the and may be thepropenles propenicsgiven givenIninthese thesepublicatIOns publicationsmust must be be understood understood and may be studied studiedin, in, for for example. example. references references (6, (6. 7). 7). ItItmay the background 10 maybebeuseful usefulatatsome somepOInt pointfor forthe thestudent studenttotoexamine examine the background 10 the therefore_made to the thesteelwork steclworkdeSign designmethod methodand andBS OS5950. 5950. Reference Reference ISms thereforemade to the Steel as explanatory SteelConstructIOn ConstructionInstitute institutepublicatlOn{l)) publication"'t which which ISismtcnded intended as explanatory totoBS OS5950: 5950:Part Part I.I. The have a clear Thedesigner designerof ofsteelwork steelworkelements elementsand and structures structures must must have a clear understanding undersiandingofofthe thetheory theoryof of structures structuresand and strength strength of of matenals. materials. The The student given In studentISisreferred referredtotothe therelevant relevantsections sections of of textbooks textbooks such such asasthose those given In references and further further explanation explanation ISis avoided. references(6) (6)toto(11) (It) and avoided.
Sourcesof of information informationmust must be be quoted quoted t1s: us: Sources
.
•
loading loading
••• •
dimensions dimenSIOns stresses stresses
BritishStandard; Standard; manufacturer's manufacturer'scatalogue; catalogue; British clients brief brief c1ient·s drawingnumber number drawmg OS5950 5950clause clausenumber; number;research researchpaper paper BS
This ensures ensuresthat that future futurequenes quenes about about the the calculatIOns calculationscan canbe be answered answered This quicklyand aridthat thatsubsequent subsequentaiteratlons alterationscan canbe beeasily easilydetected. detected.Ititwill will also also qUickly assist the the deSigner designerwhen when carrytng carrytagout outaasimilar similardeSign designatataaiater laterdate dateand andHtis this aSSISt can be be of of great great value value to to aa stUdent student who who will will one one day day deSign design to tn earnest. earnest. can .BA tt,8.4
Results Results
In deSign, design, results results(or br output), size, load, load, moment, moment, from from one one In output), such such as as member member SIZe, stage of of the calculations may may be be used used as as mput input at at aa further Is stage the calculatIOns further stage. stage. it It IS important therefore therefore that that such such tnformatlon information IS is easily easily obtamed obtained from from the Important the calculations. Such Such results results may may be be highlighted highlighted by by placlOg placing m in an an output output margtn margin caiculations, ton tne the nght~hand nghi-hand side). side), or or by by placing placing In in aa 'box', 'box' or (on ormerely merely by by underliasng underlinmg or or using coloured coloured marker marker pens; pens; in using in this this bnok, book, bold bold type type has has been been used. used. The student studeat should The should attempt attempt to to matntatn mamtam realism realism tn In calculation, calculatIOn, and and avoid avoid quotmg the the eight eight or quotmg or more more digits diglls produced produced by by calculators calcUlators and and micromlcro~ computers. Loading Loading IS is commonly commonly no no more more than accurate, and and computers. than two-figure two~figure accurate. section properties section properties are arc given given to to only only three three figures. figures. No No amount amount of ofcalculation calculatiOn will give give results of higher higher accuracy. accuracy, will results of 1.8.5 I.B.5
RelationshIp Relationship with with drawings drawings Ta most most cases. cases, the the final results of In final results of destgn deSign calculations calculatIons are are member member sizes, SIZes, bolt bolt numbers, so on, Imd so on, all all of of which which information IDfonnatlOn must must be be numbers. cotinection connectIOn layouts layouts bnd conveyed conveyed to to the the fabncator/conlractor fabncator/contractoron ondrawings. drawmgs.ItItis, is.however, howe\'er, common commonin in steelwork design steelwork deSign for for some some of of the the drawings drawings to to be be earned carned out out by by persons oilier other than than the the design design engineer, engineer, such such as as detailing detailing draughtsmen, draughtsmen. techmciSn tcchruciim engineers, is therefore engmeers, or or even even the the fabneaior. fabncator. ItillS therefore essential essential that thatthe the fiaat final output output should should be be clearly clearly marked marked in in the thecalculattons. calculatIOns.Specific Specific requirements reqUirements such such as as connection be clearly clearlysketched. sketched. connecllon details details must must be
1.10 1.10
FORMAT FORMATOF OF CHAPTERS CHAPTERS The The foUowmg followrngchapters chapiers provide providedesign design examples examples of of structural structural steclwork steelwork elements (Purl II). Il). elements (Part (Part J) 1) and and structural structural steelwork steclwork frameworks frameworks (Part The chapter order IS mtended to guide a student with a baSIC knowledge The chapter order is intended to guide a student with a basic knowledge of of the Into steeitvork steciwork deSign. it the theory tlteoty of of structures structures and and strength strength of of materlais maierials Into design. Ii therefore wJth loading, loading, including Including combination combination effccts, and therefore S[arts starts (in (in Part Part I) I) with effects, and proceeds already proceeds toto Simple simple elemenls elements With with which which the the student student ISis probably probably already familiar. elemcnts atsd and those those requmng specml treatment familiar. Later, Later, more more complex complex elements requinng special treatment are the simple Simple elements elements are are combined combined to to form fonn complete complete are mtroduced. introduced. In In Part Part nlithe structures. While this chapter arrangement IS preferred for teaching, Ihe structures. While this chapter arrangement is preferred for teaching, In In die actual elements, the the calculations caicuiatJOlls are usua!Jy actual deSign design of of structural structural steel steel elements, arc usually arranged load order, order, I.e. I.e. as as indicated mdicated in in Section SectiOn 1.5 and Fig. ! .2. This IS arranged in in load t .5 and Fig. .2. This is sometimes reverse construction constructIOn order. order. sometimes known known as as reverse Each chapter chapler (in (in Pan Part 1) I) starts starts with with basic baSIC definitions definitions of ofstructural structural members Each members and the element/frame element/frame follow and how how they they act. act- General General noles notes on on the ihe deSign design of of the follow or more examples and then Ihe deSign calculatIOns are set out for one and then the design calculations are set out for one or more examples demonstratmg the the main malO variations. varmilons. demonstrating The calculations calculatIOns follow follow the layout suggested 1.8.References References 10 The the layout suggested in in SectIOn Section IS. to SS 5950: 5950: Part Part t1 are are given given isierely merely by by quoting quotmg the the appropriate appropnale clause, ciause, e.g. OS e.g. 'clause 2.4.1', 2.4.1', or or table, table, e.g. e.g. 'OS 'SS table table 13'. 13'.References Referencesfor forthe thestructural structural theory 'clause ilieory reqUIred by by the the student, student, or or for forbackground background to(0OS BS5950, 5950, are gIven as study required are given as study toPiCS at at the the end end of ofeach each chapter chapter with with aa numerical numerical reference reference ID the teXi, c.g. topics
in the text, e.g.
(3). (3).
STUDYREFERENCES REFERENCES STUDY 1.9
1.9
STRUCTURAL STRUCTURALTHEORY THEORY
TopiC Topic
Loading I.J. Loading ItIt is IS assumed assumed in in the thefollowing followlDg chapters chapters that that the thereader readerwill willhave haveavailable available aa copy copy of of05 BS5950: 5950:Part PartI I or or extracts extracts from from it. It. Tables Tablesand andcharts chartsfor fordesign deSignwill will not not be be reproduced reproduced in in fill fullinInihe Ihetext textbut butextracts extractswill willbe begiven gIvenwhere where appropnate. appropriate. inInaddition, addition.properties propertiesof ofsteelwork steelworksections secttonswill willbeberequired. requued. These Theseare areavailable avaiiabJefrom fromthe theSteel SteelConstruction Construction InsntuteflSi Inshtute l12 ) The Themeanmg meanmg of of
2. BM BMand andSF SFdiagrams diagrams 2.
Rejerences References BS6399 6399Desmgmm Deslgll LoadiJJg 88 Loadi:igjor forBuildings an dings Pan I:\: Dead Deadand audimposed imposedloods loads(1984) (1984) Part Part 2: 2: Iflnd Windloods loads (1995) (1995) Part Part3:3: imposedroof raoftoads loads(tOSS) (J 988) Pan
Marshal!\Y.T. \V,T. && Nelson H.M. (l990) Examples of Marshall Nelson JIM, (1990) Examplci of bendingmoment momentand and shear force diagrams. SlnJclIIres, bending shear force diagrams. Srrmmc,m,res, pp.23—4. 23-4. Longnian Longman pp.
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Coates R.C RC.,.. CouUe Coates Kong F.K. Shear CauLk M.G. MG. & & Kong F.K. (988) (l988) Shear AnalysIS, forces and bending moments, Structural Analysis. pp, 58-71. Van Remhold pp. 58—71. Van Nostrand Nostrand Reinhold
4. EM Steel DeSigners' Designers Manual, 4. BM and and SF SF coefficients coefficients (1992) (1992) Steel Manllal, pp. 1026-53. 1026—53. Blackwell Blackwcll 5. Steel Steel quality quality 5.
as 4360 (1990) (1990) SpectJkanon Specificotlollfor Weldable Srn/crural 55 4360 for Weldable Structural Steels Steels
of bending bending 6. Theory of
Marshall W.T. W.T. & & Nelson Nelson JIM. ILM.(1990) (1990)Bending Bending stress stress analysIs, pp.134—43. 134-43. Longman analysts, Stn/ctures, Strictures, pp.
7. Theory Theory of ofbending bending 7.
Hellrn EJ.(1985) (1985)Bending, Bending,Afeclsonucs Mechanics of of Matenols, Mater/als, Ilearn EJ. vo!. 62-8. Pergamon vol. i, 1, pp. 62—8. Pergamon -
Deflections 8, DeflecllOns 8.
Marshal! & Nelson Nelson MM. H.M.(1990) (1990) Bending Bending Marshall W.T. & pp.203—IS. 203-18. Longman Longman defonnatlOn, deformation, Stn/cll/res, Strictures. pp.
9. Ocflecttons 9. DeflectIOns
Bearn E.J. (1985)Slope Slopeand anddeflection deflecttonof ofbeams, beams, llearn El (1985) Mechanics AJechanics DJ vo!. I,i.pp. pp.92—107. 92-107. Pergamon Pergamon of Matenals Materials vol.
10. JO. Moment-area Moment-area
Cro:don PL.C. p.L.e.& & Marttn MartinLII. L.II. (990)Asea-momeni Area-moment Crorton (1990) analysIs, So) Solvmg In Structures, Stnlctures, vol. vo!. method meihod of ofanalysis. ring Problems in 2, pp. 25—47. 25-47. Longusan Longman
II Moment-area 11 MomenHlfeamethod method
Conies & Kong Kong F.K. F.K. (1988) (1988) Coales RC, ILC., Coutie Cousie M.G. MG. & Slructura/Analysis, AnalysIS,pp. pp.176—St. 176-81. Moment-area methods, Structural Van Nostrnnd Reinhold RelOhold Van Nostrand
12. Section 12. Section propenies propemes
/1987) SectIOn oroperties properties 11987) Steelwork Steelwork Des/gll Design voL vol. 1, Section member Sleel Construction Construcuon Institute Institute member capacities. capacities. Steel
I I
2
I I
LOADING I ['1 G AND NI D LOAD L0P1,D COMBINATIONS CO NI B I N4OC1T 10 NI S .
-
-
TIle loading loading for for most most structures structures isIS obtained obtamed from from the the appropriate approDnate British Bntish Die Standard{l,21, the loads obtained obtained Standard0'21, Usemanufacturer's manufacturer's data data and and Similar similar sources. The loads IS perceived by the designer deSigner to to must, however, be combined to simulate what is in practice, praCtice, and be be multiplied multiplied by by appropnate appropnate load load factors. factors. The The process process occur in of combiotog combinmg loads loads and including the load factors factors is IS corned carned out out for for simple slmpJe of denvlng the the bending bending moments, moments, shear shear forces, forces, etc., etc., structural elements when deriving which will occur. For For more more complex complex structures structures itIt is IS advantageous advantageous to to include Include Ihe bending bending moments etc., so that that specific specific load factors factors after after deriving denvmg the the load combinatIOns can be be examined examined more more readily. readily. combinations
Dowling 9.3., P.J., Knowles P. & & Owens Owens G.W. G,W, (1988) (988) 13. Background 10 13. Background to BS 55 5950 Dowliog Structural Steel Sfeel Design. DeSign. Steel Steel Construction ConstOlcnon Institute institute
2.1 1.1
LOADS DEAD LOADS
mclude the the following: follOWing: These will include weight of steel of steel steel sectton) section) • own weight steel member member (kg/m (kg/m of permanent parts parts of of building, building, etc., etc., not not normally normallymoveable moveable • other permanent bnck/blocK walls, walls, finishes, finishes, cladding) cladding) (e.g. concrete floor slabs, brick/block They are are calculated calculated either either from from density denSity of ofmatenal matenal (lcg/m3) (kg/m) or or specific specific They lkN/m), or from from manufacturers' manufacturers' data data contained conta,med in in catalogues catalogues or or weight fkN/m3), 2.1 shows shows some some typical typical values; values; these these are are all all permanent permanentloads loads manuals. Table 2.1 "'and combined with the the appropriate appropnate dead load parttal partial safety safety factor factor lsee isee and are combined Secllon i .4). Section 2.1Typical TYPicalvalues valuesofofcommon commonstsucturai structural matenals malenals Table 2J l\Ialerial Matertat
values values have havebeen beencalculated caicuhttedatatevery every pomt. point. Only Only minSimple simple cases cases will will onc one arrangement arrangementof ofma;umum maxixnumloads loadsbebesuffiCIent sufficienttotoproduce produce maxlmwn maxunurn moments moments or or forces forces for for deSign design purposes. purposes.
These wjll will mclude include the the followmg following temporary temporary loads: loads: These
••• • •••
•• •
snow on on roofsH ) snow people people furniture furniture equipment such as as cranes cranes and and other otner machinery machinery equipment such semi-permanent partitions partitions which which are are moveable moveable seml-pennanent
1.5 2.5
EXAMPLE EXAMPLEI.I. LOADING LOADING OF OF AA SIMPLY SIMPLY SUPPORTED SUPPORTED GANTRY GANTRY GIRDER GIRDER
(a) (a)
Dimensions Dimensions Simply 6.0 Simply supported supportedspan span 6.0 m in 3.6m Crane Crane wheels" centres centres 3.6m (See SectIOn 5.1) 5.1) (See further further descnptton descnption In in Section
Imposed loads loads vary vary with with the the functIOn function of of the the room room or or buUding(I), and and some some Imposed typicalvalues values are are shown shown In InTable Table2.2. 2.2.All All imposed imposed loads loads are are based based on on typIcal expenence within within the the construction construction mdustry industry and and the the statlstical statistical analyses analyses of of expenence observed cases, cases.These These are are all all temporary temporary loads loads Dnd and are are combined combined with with the the observed imposed load load partial partial safety safety facior factor (see (see Section 1.4). Imposed SecllOn 1.4),
(b) (b)
Loads Loads specified specsfied Self guder (uniformly (unifonnly distributed) distributed) Self weIght weight of of girder MaXlmum Maximumcrane crane wheel wheel load load (static) (static) WeIght Weight of of crab crab Hook be lified lifted Hook load load to to be
Table 2.2 Table 2.2 Typical Typu::nl values values of ofimposed imposed loads loads Building usage usage Building
l Iniposedload load(kN/m (kN/mi) ) Imposed
Residential (self-contained lieu-contained dwellings) Residentml dwellings) Offices (depending (depending on on room Offices room usage) usage) Educational (classrooms) Educallonal fclassrooms) Theatres (areas (areas with with fixed fixed sealing) seating) Theatres Warehousing (general WarehOUSing 1generni storage) stornge)
1.5 i.5 2.5—5.0 2.5-5.0 3.0 3.0 4.0 4.0 2.4 per per m m height 2.4 height 5.0 5.0
Industrial workshops industTlal workshops
2.3 1.3
Dynamic 6399: Part) 25%. Dynamic effects effects to to be be mduded included ininaccordance accordancewith withBS 586399: Part (I) at at 25%. (For effects and and the the derivation derivatIOn of (For further further details details of of dynamiC dynamic effects of maximum maximum wheel wheel loads Scclion 5.1.) 5. L) loads see Section
(c) (c)
1111 6.0m 6.0 sri
A A
(Cpe (Cp~ — Cp,)q Cpdq
Moments u.d.l.) Moments and and forces forces (due (due to to u.d.l.)
(S,' Fig. Fig. 2.1.) (See 2.i.)
1111 11L
Self weight weIght 'Ttd TVd (faclored bY'!f) = 1.5;.: 12.6kN (factored by yj) = lA x 1.5 x 6.0= 1.4 x 6.0 = 12.6 kN Ultimate midspan == 12.6 x 6/8 6/8 9.0kNm Ultimate midspan BM SM = WdL/B IVdL/8 12.6 x = 9.OkNns RA == R8 Rn = = 12.6/2 12.6/2 = 6.0kN Ultimate reactions RA I VL/2= = WL/2 = 6.OkN
8B
Fig. 2,1 2.]
The wmd wind loads loads used The used in m the the text text are are based based on on CP3, CP3, Cliv, ChV, Part Part2t21 2(2) (as (as 8S6399): Part was issued too late to be incorporated). Basic 8S 6399): Part 2(1) was Issued too late to be mcorporated). BaSIC wind Wind speed, speed, appropnate to and reduced of the the building, building, is IS selected selected and reduced to to aa design deSign appropnate to the the location location of wind speed speed using wmd uSing factors factors which which take take into mto consideration consideralton topography, topography, surrounding surrounding buildings, buildings, height heIght above above ground ground level, level, component componentsize sizeand and'period period of of exposure. exposure. The The design deSign wind wmd speed speed is IS equated equated to to aa dynamic dynamiC pressure pressure qq (kN/m2). roof shape, (kN/m 2 ). Owing Owmg to to building building and and roof shape, openings opemngs in III walls, walls, etc., etc., pressures pressures and msc.Pressure Pressurecoefficients coeffiCients and suctions, suc!Jons, both both external external and and internal, internai, will willarise. external and internal internal (Ce;) (Cp;) may may be be used used as as shown shown in m Section SectIOn 2.7. 2.7. external (Cp... ) and
(d) Id) we 2.1
gc 3.6 c",,3.6m
Cb
L=6,Om L=6.0m
Casn i1 case
xx area area of of element element
and forces vertical wheel wheel load) load) Moments and forces (due (due to to vertical Wheel load load W(' (induding 'If and 'Wheel (including y1 and 1.25 x x 220=440W 220=440kN =1.6 xx 1.25
W, We
m (4)
A
3.0m
15 AA
(t)
"
l=:: 6.0 m L=6.om
8
B
Case22 cate
W
LOAD LOAD COMBINATIONS COMBINATIONS
Wet
lD
a
Loads Loads on on any any structure structuremust mustbe be arranged arranged in in design deSign so so that that the the maximum maximumforce force
AA
or or moment moment tsIS actrieved achieved at at the the point point at inthe thestructure structure betng being considered. considered. Hence Hence all all realistic realistiC load loadcombinattons combinahonsmust must be beconsidered considered to to ensure ensure that thatall allpeak peak
FIg. Fig. 2.2 2.2
-:1
W(h m
___
L=6.Om L=6.0m
1) Ultimate SM BM under under wheel wheel (case (case I) Ultimate =~ 2W,(L/2 -c/4)'/L ~ 22 xx 440(6.0/2 440(6.0/2 - 3.6/4)' /6.0 =~647 647kNm kNmUltimate SM BM under underwheel wheel (case (case2)2) Ultimate 440 xx 6.0/4 6.0/4 660kNm kNm =~ W,L/4 =~ 440 =~660
Wq
c:3.6 m (}) c3.6m
2Y'Io impact) Impact) 25%
11 pOSitions of of moving movmg loads loads to to give give maximum maXimum values values of of moment moment and and 8 The positions B shear force force are given in In Section SectIOn 5.2 and and references references (3, (3, 4), 4). The The maximum maximum each case are are now now given given land (and see see Fig. Fig. 2.2). 2.2). values of each
W'-l
Wind data are given Wind data with With suitable SUitable factors factors and and coeffictents coeffiCients are given in In reference reference (I). (I). The ofobtaining obtammg the the quasi-static quasI-static wind wmd load load used used in in design deSign isisgiven givenin 10 The method method of greater greater detail detail in 10 Sections SectIOns 2.7 2.7 and and 12.4.3. 12.4.3. 2.4 1.4
and forces horizontal wheel wheel load) load) Moments and forces (due (due to horizontal
In addition to the vertical vertlca!loads forces calculated calculated above, honzontal surge loads and forces In
Arrangements of loading loading to produce maximum moments moments and forces forces may may be be Arrangements found by taspection mspectlon of of the the appropriate appropnate influence mfluence lines. The The use use of ofinfluence influence found lines to give give the the required reqUired arrangements arrangements (patterns) (paUerns) of of loading loading isIS described described m lines in mfluence line shows (he the effect, say of ofbending bending moment, moment, references (5, 6). An influence Ihe maxtmum maximum JIM BM is obtained by placing plaCing movmg unit umt load. load. Hence Hence the due to a moving the imposed loads where the mfluence influence IS is of of one one sign sign only, only. e.g. e.g. in Fig. 2.3 the maximum JIM BM due Imposed load of span I1 is IS obtained by by maximum due to imposed load al at the middle of placmg the imposed Imposed load and 3. Load Load on on spans spans 22 and and 44 would would placing load on on spans spans iI and BM of of opposite opposlle sign sign at al the the point pomt considered. considered. produce a JIM deSIgn office, office, standard standard loading loading arrangements used to to speed speed tip up the design In the anangements are used of selection, with with influence Influence lines lines used used for for more more complex complexcases. cases. process of this process
loading hook load honzontal loading due doe to to movement movement of of the the crab crab and and hook load gives gives nse to honzontal 10% of ofthese these loads. loads. moments and forces (see equal to (0% moments and forces (see Sechon Section 5.1) 5.1) equal Honzontal W" .. (inc,. 1.6 x 0.10(200 0.10(200 + +60) 60) Honzontal surge surge load load rv5C (md.'1/) yj) == 1.6 = 41.6kN This is IS divided divided between between 44wheels wheels(assuming (assummgdouble-flanged douhle*ftangedwheels): wheeis): l0,4kN IV".. == 41.6/4 == lO.4kN Honzontal wheel load load Using those for vertical vemcal moments moments and forces: forces: Using calculatIOns calculations similar similar 10 to those Ultimate BM (case 2) Ultimate honzontal honzontal JIM = W"c LI4 = = lOA 6.0/4 = = 15.6 kNm k-Nm = l0.4 x 6.0/4 Ultimate reaction (case 3) 3) Ultimate honzonlal honzontal reaction = - c/L) elL) = W"c{2 — = 14.6 = 10.4f2 14.6k-N kN = 10.4(2 -— 3.6/6.0) =
Cc) (c)
'If value value of of1.6 1.6 maybe may bereduced reduced to to 1.4 lA where where vertical vertIcal and Note that Note that the theyj
MaXimum M. is IS produced produced when when spans spans ,i and and 3 have have the the maximum maXimum Maximum value of Mt loading and have the minimum mmimum loading. loading. Using the the influence mfluence loading and spans spans 2 and 4 have lines shown, shown, the loading patterns producing producmg maximum maXimum effect may may be be lines obtained, and and are are summarized swnmarized below. below. obtained, .i
honzontal crane crune loads loads act together, together, and and this this combination combinationmust must be bechecked checkedinin horizontal SectIOn 5.3). 5.3). practice (see Section 2.. 2.6
EXAMPLE2.2. LOADING LOADING OF OF CONTINUOUS CONTINUOUS SPANS EXAMPLE SPANS
A
Obtam moment, shear shear force force and and reaction reaction for for Obtain the the maximum values of of bending moment, a contmuous at the positions positions noted noted tn in (c), le), Cd), (d), (e) continuous beam beam at and (1). "nd (I). fa) (a)
Load pattern for for mid-span mid-span moment MI Load
" 6
Dimensions
A A
MalO beams, beams, spaced supportmg a coacrete concrete slab slab (spanning (spanning Main spaced at 4.5 m centres, supporting for office office accommodation. accommodatIOn. Steel Steel beams beams are are to to have have four four one way only) for contmuous spans of of 8.0 8.0 m. Assume Assume uniform unifonn section sectlOn properties. propertles. continuous
(b)
FIg. Fig. 2.3 Influence Influence line line for for Al Afti
+
+
ti8~L'l
B.Om tom
Span 1 Spent
B
B.O m ton
C c
Span 2 Span?
8.om B.Om Span J
0
B.Om tom
LI EE
Span
Loading Loading
Self weight of beam beam firush Concrete slab and fimsh (offices) Imposed loading (offices)
MUJomum value of of lvh Maximum M,, ISis produced producedwhen whenspans spans 1,2 1,2 and 4 have the maximum loading has the the minimum mtnimum loading loading (Fig. (Fig. 2.4). 2.4). loading and and span 3 has
dead loading loading (self (selfweight, weight, slab slab and and. finishes) finishes) isISfixed, fixed, but but the the imposed Imposed The dead loading is IS moveable. moveable. The dead loading loading must mu~t be be present present ('1/ = 3.4), 1.4), while while 1.6) or absent. the be present present ()~r the Imposed imposed landing loadingmay maybe (y, == L6)
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Load pattern pattern for Load for shear shear force force 8ab 5 ab Maximum as Ic) (Fig. (Fig. 2.6). 2.6). Maximum as in in Ic)
2.7 2.7
EXAMPLE OF A A PORTAL PORTAL FRAME FRAME EXAMPLE 3.3. LOADING LOADING OF
(a)
Frame Frame
See pitched portal portal with with pinned pmnedfeet; feet; span span 38 38rn, m, spaced spaced at at 66 m See Fig. Fig. 2.7: pitched m centres. Fig. 2.6 Influence line
Fig. 2.6 influence line for 5,b
for
A
A
S~b
1
2
B
B
C
2\
o
3 3
LI 4
E
c
3.4m 3.4 m B
Arrangements of loading Arrangements of loading for for maximum maxImum moments moments and and forces: forces:
(g) (g)
if's 11'1
IF',
IV2
WJ
11'4
57! 571 57! 571
283 283 571 571
57! 571 283 283 283 283
283 283
283 283
571 571
571 571
57! 571 57! 571
57! 571
I.
;t Fig. Fig. 2.7 2.7 Portal Portal frame
Span Spnn loading loading (kN) (kN) For For moment momentMi All For moment .44 For moment Alb For reaction Rb For reaclioll Rb For shear force Sab For shear force
Frames at af 6.0 6.0 m m eaniras cenlres Frames
S.Sm 5.6 m L-_ _ o!> A
Cb) (b)
571 571
Moments Momentsand and forces forces Moments Moments and and forces forces may may be be found found by by any any analytical analytical method, method, but but for forequal equal span spnn cases cases coefficients coeffiCientsare are available(7), where, where, for forexample example All ==(ai1 {all W, ++a1211'2 an W:2 ++ alJ IV}++a141111)L a141V4)L and M1, R,, and 5ab are given by similar expressions. and A/b, Rb and S"b are given by Similar expressIOns. The The coefficients coeffiCientseamay maybebesummanzed: sununanzed:
Windpressures pressures are are based based on for aa location locatIOn inm Wind on aa baSIC basic wmd wind speed speed of of 50m/s 5Dm/s for
Scotland. Scotland. Factors{2} 5,S,and and S2are are both both taken taken as as 1.0 1.0 and S3 as as 0.88 0.88 for for aa height height of of Factorg'1 and factor factors3
IOm. 1Dm.
., "h
The design deSign wind wmd speed speed is is hence hence 44 44 m/so giVing aa dynamic dynamiC pressure pressure ofof The m/s. giving 1.20kN/m'. 1.20 kN/m'.
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22 STRUCTURAL 22 STRUCTURALSTEELWOAK STEELWORKDESIGN DESIGN TO TO 05 8S 5950 5950
LOADING AND AND LOAD LOAD COMBINATIONS LOADING
I
is possible to use a lower this It IS possible 10 lower wind wmdpressure pressure below a height of ofS5 m but ihis makes the analYSIS analysis more more complex complex for very little makes the littlereduction reducttoninmframe framemoments. moments. (c)
It maybe may benoted noled that that case case 4 is similar Similar to to case case 3, but has has lower values and may be discarded. discarded.
Pressure coefficients Pressure
(d)
External pressure coefficients (C,,,) pressure coeffiCIents (Cp~) may be found and and are are summanzed in in the the following table. followmg table. These These values values are are obtained o'btained from from reference reference 12). (2).
Wind on on side side Wind Wind on on end end
BC BC
CD
DE
0.7 0.7 —0.5 -0.5
-1.2 —1.2
-0.4 —0.4 —0.6 -0.6
-0.25 —025
—0.6 -0.6
Dead Dead load on walls (Fig. (Fig. 2.11): 2.11): Self weight weight == 0.9 0.9 xx 5.5 5.5 Cladding = 0.09 0.09 x 5.5 5.5 xx 6.0 6.0 Cladding Total Wm.. Totai
—0.5 -0.5
Internal pressure coefficients (Cpl) (C,,,) should should be be obtamed{l). obtainedtfl, and are taken taken m in pressure coeffiCients and are this 0.2 (maximum) and and — - 0.3 0.3 (minimum) (minimum) which whichare are this example example as as + 0.2 combined algebraically with the of Cp~ above. combined algebraically the values values of Figure 2.8 shows the individual mdividual pressure pressure coefficients, coeffiCients, and and Fig. Fig. 2.9 2.9 shows shows Figure 2.8 shows the the vanous vanous combination combinationcases. cases. 1.2
0.1
0.4
B~0.;"5 " ~
~
0.6
! I!
0.9
0.1
~~5 ~~ Case !a+b) Case 1I In + bI
(
B
B~ I
Ic) le) Internal Internal pressure pressure
Id) lnttrnsl suction Idllnterna! suction
w,
IIIH(l)i)llIIl I I I I I II WI
I!I !I ! 0.6
! ! i I
5.5
I
".Om
Fig. 2.10
Fig. 2.8 2.8
1.4
= 8.OkN = 8.0kN
C
0.6
. (hi uI Wind Wind on end
W" i!!
=
= 5.0 kN 5.0kN = 3.OlcH 3.0kN
~C
~
!i1jWind a) Wind on side
Member loads loads
Dead load load on on roof is calculated on the area (Fig. (Fig. 2.10): 2.10): Dead calculated on the projected projected area = 0.9 kN Self weight \vetght= 0.9 xx 19.3 19.3 = 17.5 l7.5kN Cladding = 009 6.0= l0.5kN = 0.09 x 19.3 19.3 xx 6.0= lO.5kN = 28.OkN Wdr = 28.01eN Total Wd.
C,,,for frame Cp"..for frame member member AD AB
0.8
O.B
0.3
0.3
~~ Case lb + d) dl Cas6 4 fb+
Case 3 )b'+ cI
I
~ I~ I
Fig. 2.11 2.11
Imposed load load on on roof given imposed gIven on on plan plan area area (and (and services) servIces) (Fig. 2.12): 2.12);
Wind loads loads for for case (1)— Wind (1) - wmd on side+ side +internal '"temal pressurc: pressure: Wind load load on on 'vail wall W....... (Fig. 2.13) L20 xx 6.0 k.N 2.13) = = 0.5 0.5 xx 1.20 6.0xx 5.5 5.5== 15 l5kN
".Om
FIg. Fig. 2.12 2.12
Fig. 2.9 2.9
23
Wind pressure on roof roof is is divided into pressure on mto vertical vertical and and horizontal honzontalcomponents components (Fig. 2.14), 2.14),
=
Vertical VertIcal component component W"... -lA 6.0}j; 19.0 = -192kN —1.4xx l.20 1.20 xx 6.0 't 19.0 —l92kN l-lonzontal component Honzonlal component Wwh = —1.4 -1.4 x 1.20 1.20 x 6.0 x 3.4·= -34 kN 3,4 '= —34 -
-
Case Cnse
I. Wind Wind on on side+inIemat side + mlemal L pressure 2. Wind + niemal Wind on on side side + mlernal 2. suction SUCllon 3. Wind Windon onend end ++internal mternal pressure pressure 4. Wind Windon onend end ++ intemal Internal suction
(C,,... — (~p -
AB
C,,,) for tram Cp ;) for framee member BC BC
CD CD
DE DE
0.5
—1.4 -lA
—0.6 -0.6
-0.45 —0.45
1.0 LO
-0.9 —0.9
—0.1 -0.1
0,05 0.05
-0.7 —0.7
—0.8 -0.8
—0.8 -0.8
-0.7 —0.7
-0.2 —0.2
-0.3 —0.3
-0.3 —0.3
-0.2 —0.2
W,"" !! (It I,
I ! ! ! S
I Fig. 2.13 2.13
I!
I
f~~JU3'4m C
I
10Dm 19.0 m
1_..
Fig. 2.14 2.1~
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thesame same manner manner wind wind load load for for each each member member and and each cacti case casemay may be be InInthe calculated.The The loading loading due due to to dead, dead, imposed imposed and and wmd wind loads toads may maybe be calculated. summanzed using using the the posItive positive notation Fig. 2.15. 2.15. summanzed notation La ID Fig.
w,WI
w,W4
I I ! I ! ! ! I BI
w'~ W,
I
g,w,
c B A
Fig. 2.15 Loading
Fig. 2.15 Loading summary summary
! I
I
0 E
H, H,
H.
V.
Each Each force forceorormoment momentmay maybebecalculated calculatedUSing ustngequatIOns equationssImilar similartotothat thatgiven given for forHa H, abovc. above. The TheexpresSIOn expressionof ofcoefficients coefficients(a:) (a) and and loads loads {JV) (IV) as as arrays arrays allows allows ror for combimng combiningby byuse useof ofaacomputer computerprogram. program.This Thiswould wouldbe beof of particular particular advantage advantageififmatnx matnx multiplicatIon multiplicationwas was available. available.
IVV}(aI= Wli"i = IF) By computer or hand the the moments moments and and forces forces are By computer or by by hand are calculated calculated and and tabulated: tabulated:
"C
W,
V. V,
Value of moment or force due to unfactored loads
"4.
Unfactored loads toads (kN) (kN) Unfactored
.
iv, W,
Deadtnadjl'd Dead load Wd
lf'5 W,
00
Imposed load Imposed load H'j Wind case (I) nc..1 Wind case (1) WWl Wind case Wind case (2) (2) W w1 Wind case (3) Wind case (3) Wwl
00 00
00
20 20
—34 -34 —22 -22 —20 -20
40 40 —28 -18
lv, W,
lv, W,
fl'5 W,
W,
28 28 85 86 —192 -192 —123 -123 —109 -109
28 28 86 86 —82 -82 —14 -14 —109 -109
00 00 —IS -15 —2 -2 —20 -20
00 00 —IS -18 —2 -2 —28 -18
IV, IV, Ifj IV,.I rç1 Ww). W IV,,, wJ
Note that that the the wall as it does Note wall dead dead load load Is is not not included included at at this this stage, stage, as does not not produce y1 are are not aot yet of 'If yet included. included. produce aa bending bending moment. moment. Values Values of (e) (e)
(I) (I)
The loads The loads given given in In Fig. Fig. 2.I5 2.l5may may be be used used to to obtain oblam moments moments and and forces forces by by any charts or any analytical analytical method method or orby byuse useof of-charts orcoefficients, coefficients, Each Eachload loadW W has hasan an appropriate appropnate set set of ofcoefllcicntsa coeffiCIents-agtvmg gIVingthe therequired reqUiredmoment momentororforcet63. force(6), For For any any load load case, case, the the effects effects of ofall all six SIX loads loads must must be be summed. summed. For For dead dead load: load: IV, + 721 WI + 73i H0 = j'llWI H" = + 1'21 W2 + j')1 W) ++ 1'41 W4 ++751W5 'Is, Ws ++Vet )-'61W6 W6 ·
i.2W,/ ± + J.2W, l.2W; + I.2Ww.l + 1.2W,,2 J .lIfT" + 1.2W; + I.2W w ;! 1.2W,, + 1.2W1 + 1.2nc, Wind 1.0W,, Group 44 Dead Dead (restraining (rcstramlOg uplift) uplift) ++wind l.DIVd ++1.4W,,, JAJVwl Group l.OIVd ++ J.4W,~.J 1.011',,
oment or Coefficient a!Xfor formmoment or force force
Ii. 11,
V, V. (kN) (kit)
Dead + +imposed Imposed ++wind wmd Group 33 Dead
CnelTicle of
Load
V. V. (kN) (kN)
Group Ii Group 22
and and similarly similarly for for each each force force or ormoment. moment. The calculnt~d and andtabulated: tabulated: The coefficients coefficients for for elastic elastic analyses analyses may maybe becalculatgd
H,, H, (kM (kN)
(kN) (kit)
Combinations Combinations of of load load must must now no'v be be considered considered and and at at this this stage stage the the values values of of }'! may be included. mcluded. Possible Possible combinations combinatIOns are: are:
Forces Forces and and moments moments
Load
R, H,
11. H, (kN) (kit)
:;:
Group 44 isISintended mtcnded for foruse usewhen whenconsidering considenngrestraint restramt against agamst uplift uplift or or Group overturning, i.e. I.C. maximum maximum wind wlOd plus plus minimum minimum dead dead load. load. Some Somecombinacombinaoverturrnng, tlOns may may be be eliminated elimmated by by inspection, mspechon, but but care care must must be be taken takentotoretain retam tions combinations giving glvmg maximum ma;umum values valuesofofopposite opPosite sign. sign. Some Some of ofthe the combinations combinations In Groups 2 2 and and 33have havebeen beendiscardcu discarded inIn the the following folloWlOg combinations in Groups table. table. Forgroup group! combination(1.4W,, (L4W" ++i.6W,): 1.6Wd: For I combination
While these rattles are group 4comhinmllnl conibinomoni(SS tns 5950) 5930) ut sit ma rover 'pu a and • While these Y~lues ate maxima m:ulml topposiie lQPpO~l!e nigs) ~!gn) lame the grQUP" 10 !:OVer uplift :md ovenantung only. Irnwever for thethe effects of a load cornbtnam'nn to be eomnnldeeed ovenummg only. iiUisUunecessaoy, neee5~:1l)'.I"'wever, lorallall effec!$o/ a IOJd comhin1!iOO 10 he eoruidered (nor /scealso :tl~o Chapier l2) Ch2pler 12)_ ."
Maximum and MaXimum and minimum minimum values values may may be be selected selected from the the table. Bending moment diagrams diagrams may may be moment be drawn drawn as as shown shown in In Figs. Figs. 2.16 2.16 and and 2.17. 2.17.
27 27
2.2. Wind Loading Loading
British British Standard Standardinshtute InstituteCP3, Cl'), Chapter Chanter V, V. Part Part 12
3. louds effects 3. MOVing Moving loads effects
Marshal! Nelson n.M. 11.1\'1.(1990) (1990) MovIng MOVing loads loads Marshall W.T. W.T. && Notion and SrnJcfllres. pp. and influence influence lines, lines, Srnscn,res, 79—i06. Longman Longman pp. 79-106.
4.4. Movmg Movmngload load effect.s effects
Wang Influence lines Wang c.l(. C.K. (983) (1983) Influence lines for fur slallcally statically detcmunale lntermedilJle Srnlcrllral determinate beams, beams, interniediname Stn,c:ural Allah'sls, .4nnlvs:s, lcGmw~HiII pp. pp. 4S9-.Q7. 459—67.l...McGraw4liIl
5. influence lines 5. Influence
Coates F.K. (1988) (1988) Coates R.C., ILC.. Coutie Coulie M.G. M.C. && Kong Kong F.K. Mueller-Breslau's Mueller.Brenlau'spnnclple. pnnciple. Model Model analysIs, analysis, StnlcfUral Stn,cnsrul AnalysIS, .9nolvsts, pp. pp. 127-31. 127—31.Van VanNostrand Nostrand Remhold Reinhold
6. influence lines lines 6. Influence
Wang Wang C.K. C.K. (1983) (1983) lnflm:nce Influence lines lines for fur slatlcally statically delennmale determinate beams, beams. lnrermediare )nrernnedio,e Srnlcfllrai S!nsctural AnalysIs. Analysts. pp. 496-503. McGraw-HiIl pp. 496—503. KlcGraw-l-Iill
7. coeffiCients (1992) MOIIIIOI, 7. BM and SF coefficients (1992) DeSign Design theory, theory. Steel Steel DeSIgner's Designer r 3-!aoiaol. pp. 1051—4. 1051-4. Blackwell Blackwell 8. BM and and SF SF coefficients coeffiCients 8. BM
(1992) Sreel Der:griers Designer'sMnnural Mmmal, (1992) DeSign Design theory, theory Stee/ pp. 1080—97. J08[).....97. Blackwe\l Blackwell
C -317
850
A
850
D
B
Fig. 2.16 Fig. 2.16 Bending Bending moments moments for combination for combinatIon !
BMs In kNm
E
252 252
C
FIg. Fig. 2.17 2.17 Bending Bending
moments moments for for combinanon combin!ltlOn 22
D
E
Finally the Finally the effect effect of of wall wall dead dead load load can cnn be be added, added, ififaxial axial force force in in the the columns is columns IS required reqUired (combination (combinatIon 1): I):
Maximum axial force 177 + Ma;umum aXial force at at A A= = 177 + 1.4 lA x x 8.2 8.2 = = 190 190 kN Mimmurn axial force force at at A Mirumum aXlal A= = —209 - 209 + + 1.0 1.0 xx 8.2 8.2 = = —201 -20] kN kN The The alternative alternatIve method method of ofanalysts analysIs isIS by byapplication applicatIOn of ofplasttc plastic theory theory tsee isee Chapter Chapter 13). 13). in in this this elastic elastic analysis analysIs of ofaaportal portal frame, frame, the the loading loading combinations combinations can can be be added, added, but but in inChapter Chapter 13 13 each each load combination combination produces produces its its own urnque umque collapse collapse mechanism, mechamsm, i.e. I.e. each each load load combination combinntlOn must must be be analysed analysed independeatly independentlywhen when applying applymgplastic plastiC theory. theory.
STUDY STUDYREFERENCES REFERENCES Topic Tomc
References References
1. 1. Loading Loading
55 BS6399 6399Design DesJgnLoading Loadingfor forBuildings Buildings Part Part 1: 1: Dead Deadand andimposed imposedloads loads(1984) (1984) Pan 2: Wind loads (1995) Part 2: fnnd foods (1995) Past Part 3: 3: imposed lmposed roof roofloads loads (1988) (1988)
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i
I
BEAMS BEAMSININSUIltllNGS BUlLqiIqo5 29
29
3.1 3.1
131
Many beams Many beamsIninaasteel steelframework frameworkwill willbe berestrained restrainedlalerally laterally by by the thefloors floors which whichtransmit transmitthe theloads loadstotothem. them. Concrete Concrete floor floorsiabs, slabs, and and wall wall or or roof roof cladding, cladding,are aregenerally generallyable abletotogive give this this lateral lateral support supportor or restramt. restraint. Timber Timber floors floorsand andopen opensteel steelfloors floorsare areless lesscertain certain Ininproviding providing restramt. restraint. The The degree deirree ll of be assessed of attachment attachment of of the the flange flange to to the the floor floor muy mayneed needtolobe assessedtt)} AHernaUvely, Alternatively,iatera! lateralrestramt restraintmay maybe be provided provided by by braCing bracing members members at at specific floor slab specific pOints points along along the the beams(IJ. beamstti. If If adequate adequate braCing bracing or or floor slab restramt restraint IS is present present Ihen then laternl lateral torsIOnal torsional instability instability will will be be prevented. prevented. In addition, addition, this this need need not not be be considered considered for for beams beams In in which: which:
BEAMS IN IN BUILDINGS BUILDINGS BEAMS
Most buildings buildings are are Intcnded intended to to provide and to Most provide load load carrying carrymg floors, floors, and to contain contain these within within aa weathertlght weathertight envelope. In some the these envelope. In some structural structural frameworks frameworks the wpathertightfunction function isis not not needed, needed, while while the the load load carrymg carrying funcllon function only only ISis w!!athertight required, e,g. e.g. supporting supporting chemica'l chemicai plunt. plant. The The loads loads to to be be supported reqUired, supported are are often often placed on on floor floor slabs slabs of of concrete, concrete, or or on on steel timber grid grid floors, floors, and and these these steel or or timber placed are In In turn turn supported supported on on the the steelwork steelwork beams. beams. In In some some cases, cases, especially especially in in are ndustnal buildings, on to to the mdustrial buildings, loads loads from from equipment eqUipment may may be be placed placed directly directly on beams without the use use of of aa floor slab. Wind Wind loads loads also also must must be be carned cained to to the beams without the floor slab. the beams by by provision provtslon of of cladding cladding of ofadequate adequate strength, strength. and and by by secondary secondary beams members such such as as purlins purlins and and side members side rails. rails. Beams which which carry carry loads loads from Beams from floors floors or or other other beams beams to to the the columns columns are generally called called main main beams. beams. Secondary beams will will be be provided provided to to transfer transfer generally Secondary beams load to load to the the main mam beams, beams, or or in ID some some cases cases just just to to give give lateral lateral stability stabiiity to to columns, while while themselves in columns, themselves carrying carryIng only only their their self self weight. weight. The The manner manner III which loads are distributed on to which loads are distributed from from the the floors floors on to the the beams beams needs needs careful careful consideration so so that that each each beam of the is designed deSigned for for aa realistic realistjc proportion proportlon of the consideratIOn beam is total load. and two-way two-way spanning spanning total load. Examples Examples of of load load distribution distributIon for for one-way one-way and slabs are are shown shO\\'Jl in ID Fig. Fig. 3.1. 3.1. slabs Beam I
H N CM
E
f
'H'
Beam Beam I1
H f
°.• N
E
m
, ,'j',
Concrete slab or grid tloor spanning in both directions
Basin 11loading loading
Fig. 3.1 Load distribution Fig, 3.1 Load distribution excluding !excluding self weight) self weight)
Beams, will not Beams, In in which which lateral lateral torsIOnal torsional Instability instability will not occur, occur, arc are classed ciassed as as restrained Sechon 3.7. restrained und and are are deSIgned designed as as illustrated illustrated in in Section
3.2
BEAMS BEAMSWITHOUT WITHOUT FULL LATERAL LATERAL RESTRAINT RESTRAINT An understanding m appreciating appreclatmg understanding of of the the behaVIOur behaviour of strutS{2.3) will will be be useful useful in behavIOur of beams beams where thll full lateral latera! restraml proVided. The The the behaviour restraint IS is not not provided. compressIOn budding compression flange flange of of such such members members will will show show aa tendency tendency to to fail fail by by buckling sideways most flexible flexible plane. plane. Design DeSign factors will sideways (laterally) (laterally) In in the most factors which which will Influence be summarized summarized as: as: influence the the lateral lateral stability can be
• • • • •
•• • •• •• ••
L'l
Beam? loading Beam 210adlng
Beam Beam 22 no no loading loading
the mmor axis, aXIS, or the section section ISis bent bent about about a minor or the stiffness, e.g. hollow the section section has has a high torsional siiffiiess, e.g. aa rectangular rectangular hollow sectIOn. section.
the length length of of the member between between adequate adequate lateral lateral restraints; restramts; the of the the cross-section; cross-sechon; the the shape of vanatlOn of of moment moment along along the the beam; beam; the variation the form fonn of of end end restraint reslmml provided; provided; manner in In which which the the load load isis applied, applied, I.e. to tension tension or or the manner i.e. to compressIOn flange. flange. compression
I
i
spinning spanningin1none onedirection direction
AIIIIII'h-. Boom b.
.1
, I
I , 'HH H'Precast units or other Precast units or olher floor floor I
•• ••
and their their effects effects are are discussed discussed in in detail detail in in reference (I), and and are are These factors factors and reference (I), Qut in In clause 4.3.7 BS 5950.The The buckling buckling resistance resistance (Mb) (Mb ) of of aa beam beam set out 4.3.7 of 955950. may be be found found by by use use of of aa number number of ofparameters parameters and and factors: faciors: may
length (Le), (Ld,which whichallows allowsfor forthe theeffects effecls of of end end restraint restTamt Effective length Effective as well well as as type type of ofbeam, beam, and and the the existence eXistence of ofdestabilizing destabilizmgforces. forces. as Minor axis axis slenderness slenderness(A), (1), which which includes Includes lateral lateral stiffuess stiffness in in the the Minor of r,, rv,and andisISdefined definedby by.4;;. == LE/ry. form of form Torsional index index (x), (x), which which isa IS ameasure measureof ofthe the torsional torSIOnal stiffness stiffness of of Torsional cross-sectIOn . aa cross-section. Slenderness factor factorlv), (v),which whichallows allowsfor fortorsional torSIOnal stiffness stiffness and and Slenderness mcludes the the ratio rallo of ofA/x. A/X. includes
Slenderness correction factor ractor (a), (n), which which isIS dependent dependent on 011 the Slenderness the moment vanation variatIOn along along the the beam. beam. moment parameter(a:), (u),which Whichallows allowsfor for section section type typeand aud Buckling parameter Buckling mcludes aa factor factor for forwarping. wnrpmg. includes
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Equivalent slenderness slenderness PH), which which combines combines the the above above paraEquivalent parameters and and from from which which the the bending strength (pa) (Pb) may be derived: meters bending strength
Local Local buckling bucklingcan can be beavoided avoidedby byapplYing applyingaa limitatIOn limitation to to the the width/thickness width/thicknessratios ratiosof of elements elemenisof of the the cross~sectlOn. cross-section.This This leads leads to to the the classificatIon 1.7. classificationof of cross~sectlOns cross-sections discussed discussed in in SectIOn Section 1.7. Where mem'bers are \\'here members are subjected subjected to to bending bending about about both boilsa.xes axes aa combinatlOll combination relationship relationship must must be be satIsfied: satisfied:
== ni,aA nuv}. In addition, addition, an an equ,valcnt used which which allows allows for for the the In equIvalent moment moment factor factor Cm) (m) isis used effect of of moment effect moment vanatson vanahon along along the the beam. beam. }.LT
(a) (a) 3.3 3.3
-(b)
,n== i.O m 1.0 0.94 nn = 0.94 a = 0.9 0.9 (for (for I,I, H Ii and 11 and channel channel sections) sectIOns) or or from from published published tables tables = 1.0 1,0 (for (for other other sectIOns) sections) = A conservative approach approach 15 is allowed allowed in in BS 58 5950 and A conservative and may be used 'in in the design of nflIand deSign and HHsections sectIons only, only, basing baSing the the bending bending strength strength (pa) (Pb) on on the the design strength, strength, the deSign the minor minor axis aXIs slenderness slenderness (A) (1) and and the the torsional torsional index index (x), (x), which for for Ihis this method method may may be be approximated approximated 10 to D/T. Dir. This which Thisapproach approach isis useful useful in In the the preliminary prelimmary sizing SIZing of ofmembers members and and isIS given gn'en inInclause clause4.3.7.7. 4.3.7.7.
Where members members are aXial Where are subjected subjected to to bending bending about about both both axes axes {without (without axial combinatiOn relationship relationship must must be be satisfied: satisfied: load) aa combination
The local local moment The moment capacity capacity (M0) (Mc) nt al any any cntical cntical point pomt along along aa member member must must not not be be less less than than the the applied applied bending bending moment moment at at that that point. pomt. The The moment moment capacity capacity will will depend depend on: on:
m"M,,/Jl-/1> my My/Mc)' 2' ., II mx Air/Ala + + as. where where
the the design deSign strength strength and and the the elastic elastic or or plastic plastic modulus; mOdulus; the the co-existent co-existent shear; shear; the the possibility possibility of of local local buckling buckling of of the the cross-section. cross-sectIOn.
Air !I·Ir- == pyS pyS Mr Afr- i Ij .2pyZ
where where S S is is the the plastic plastic modulus modulus ZZ isis the the elastic elashc modulus modulUS Note .2p1Z Nole that that the the limitation limltalJon I 1.2p is to to prevent prevent the tile onset onset of ofplasticity plasttclty below below y Z is working load. load. For For liD, UB, UC UC and and joist JOIst sections sectIOns the the ratio rallo S/Z S /Zisisless lessthan Lhan1.2 i.2and and the For sections where S/Z S/Z >> 1.2, .2, the plastic plnstic moment moment capacity capacity gnverns governs design. deSign. For sectIOns where the the constant constant 1.2 1.2 isis replaced replacedby bythe theratio mtlO(70) (Yo) of of factored factared load/unfactored load/unfaclored load. load. The The limiintinn limllatlOn-i-j .2p,.z isis therefore purely purely notional notIOnal and and becomes becomes in in practice practice 1
y0p,.Z. 'IoP).z·
If If0.6 0.6 of ofshear shearcapacity capacityisISexceeded, exceeded,some somereduction reductIOnin10Al. Mc will willoccur occuras asset set out out in In clause clause 4.2.6. 4.2.6.
nI", my as1,
eqUivalent uniform unifonn moment moment factors factors are equivalent Mt:y ISis the aXIS but but without without the the Al0, the moment moment capacity capacity about about the theyy axis restnctlOn of l.2p,.Z 1.2py Z (as (ns in In Section SectIOn 3.4). 3.04). restriction
This is IS described described as as aa ·more approach (clause {clause 4.8.3.3.2) 4.8.3.3.2) which which isIS more exact' approach iess conservative conservaUve thlln the 'stmplified' 'sunplified' approach approach (clause (clause 4.8.3.3.1), 4.8.3.3.1), in m which which less than the Mcy is IS defined defined as p,Zyo Also, a ·sllnplified two At,, p,.4.. Also, simplified appr~ach approach for for bending about two axml load) does does not not reduce reduce the the calculations. calculatIOns. axes (without axial
Providing Providing the tbe applied applied shear shearforce force isIS not not more more than than0.6 0.6 of ofthe the shear shearcapacity, capaCity, no no reduction reductIOn in In moment moment capacity capacity isIS needed, needed, and and but but
For senii-cddiAaara,id semi~cdriliJacrand slender sections: (and (and as as aa Simplified simplified method method for for compact compactsections sections In in (a) (a) above) above) ),;fr/Ma + }dv/AIry 2' ., 1I Alt/Ala + Alt/Mn
Members not provided with with full full lateral lateral restraint restraint (Section (Sec non 3.2) 3.2) must be not provided must be (Mb) as as moment moment checked torstonal buckling buckling resistance resistance (Ala) checked for for lateral torsional as well well as capacity. the bending strength strength (Section (Sectton capacity. The The buckling buckling reSistance resistance depends depends on on the the plastic plastic modulus: modulus: 3.2) and the
MOMENT CAPACITY OF OF MEMBERS (LOCALCAPACITY CAPACIfl MOMENT CAPACITY MEMBERS (LOCAL CHECK) CHECK)
•• •• ••
For For plastIC plastic and compact sectiollS: sections: For and. jOist p-r,,/Mat My/(Mc)') ;t ! For un, liD, UC ann joist sections (AIx/Ala]2 + + AlyI(Alcy) For and solid solid sections sections {M,,/kl For RHS, P115,CHS 015 and a )5/ + (AIx/Ala)3' + M y (/lfcy)5!3 ;f.2' II For secltons M,,/Ma+ My/MC) .•"f I For channel, channel, angle angle and and all all other sections where }'-Ix> My axes Al,. are are applied applied moments moments about about xx and yp axes Ala, !I-fey are y axes Ma, Me,. asemoment momentcapacities capacitiesabout aboutxxand and)' axes t
SIMPLIFIED DESIGN DESIGN PROCEDURES PROCEDURES SIMPLIFIED The deSign design of of many many simple beams will The will not require reqUire the Ihe calculation calculatIOn of of all the the above parameters. parameters. In In panlcular, particular, Simply simply supponed supported beams beams carrymg carrying distributed distributed above loads and and not not subjected to destabilizing destabilizmg loads loads will will use: use: loads subjected to
3.4 3.4
a.. 31 Oi
3.b 3,6
OTHER CONSIDERATIONS CONSIDERATIONS OTHER
addition to the above above requirements reqUirements for moment moment capacity and buckling buclding In addition reSistance, a member is IS usually usually reqUired mee;t some deflection deflecllon critena. cntena. resistance, required to to meet These are are outlined outlined in in Section SectIOn 1.5 1.5 and and reference reference (I). (1). These applicatmn of The application of heavy loads or reactions reactions to to aa member may produce high locnl stresses stresses and IS necessary necessary to web local and itit is to check check that that the the web web beanng and web buckling requirements reqUirements are satisfied. satisfied. These reqUirements are generally generally buckling These requirements ns crane girders girders Significant only beams canylng significant only in in beams carrying heavy heavy pomt point loads loads such such as (Chapter 5) 5) or or beams beams supporting supportmg column column members members within within the the span. span. (Chapter
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32 STRUCTURAL STEELWORK DESIGN TO BS 5950
32
STRUCTURAL STEELWORK DESIGN TO BS 5950
SEAMS iN BUiLDINGS
Connectionsmust mustbe beprovided providedatatJunctIOns junctions between between members membersand andmust must Connections safelytransmit transmitthe the calculated calculated loads loads from from one one member member to to the the next. next. A A variety variety safely ofconnectIon connectiondetails details exist exist for for most most common common situations situations and and are are fully fully of described in in the the Se! SC! handbookt41. DesigninfonnatlOn informationfor forconnectIOns connectionsisis described bandbook(4). DeSIgn givenminsection section 6,6. BS BS 5950, 5950, which which Includes includessome someguidance guidanceon onbolt boltspacmg spactag gwen andedge edge distances. distances.Bolt Bolt and and weld weld.sizes and capacities capacitiesare are given given inin and ·sizes and reference (5). (5). reference
Self weight
Dead load slab Dead load screed
!
lOkN 10kN j
I !
A
Dimensions DimenSions
11
7.4m U,
(See Fig. Fig. 3.2.) 3.2.) (See
7474
r-i H
Main beam beam Mam
6.0 m 6.0m 7.4 m m 7.4 2SOmm thick 250 mm thick 4Omm thick 40 mm thick
a
3.Om
\ /
I 1,
B
CE
" 7.4 m ~
7.4 in
2> 7 250 mm slab
250 mm slab EOm 6.0 m
1.4m 1Dm
!ii!g;.:~.~
40 mm screed 40mm licreed
II .1
(c) (c)
..
Imposed Wj : Imposed load load W;: all on rectangles rectangles on triaggles on tuiaogles
1.4 IA xx 77 1.4 x 57 + 1.6 x 42 l.4x57+l.6x42 l.4x 123+1.6x90 IA x 123+ 1.6 x 90
= 10 kN = 10 kN
= = 147 147 kN IN =316kN = 316 kN
BM and and SF SF
Maxunum ultimate moment moment M, M;x (mid-span) (mid~span) Maxanum ultimate 10 x7.4/8+232 x 7.4/8+232 xx 3.0—158 3.0 -158 xx 1.7—74 1.7 -74 xx 10.35=573 10.35=573 kNm leNm = ID
6.83 IN/rn2 Area of siab supported by beam (Fig. 3.3):
FIg. 3.3
3.7 3.7
0.92 x 7A =7 kN 23.7 x 0.250
33
Ftg. 3.5
Maxtmum ultimate 1012 + 232 = 237 kN kN Maximum ultimate shear shear force force F;x F, = = 10)2 + 232 = 237
"
"
Typical bay of Typical bay of larger larger floor floor area area
Fig. 3.2 Slab
and
i-i H
beams
Fig. 3.2 Slnb nnd beams
(d) (d)
.—
Shear capacity capacity Shear
H Using the the design deSign strength strength from from Table Table 1.2 1.2 for for grade grade 43A 43A steel, steel, noting notlng that that Using of section section is IS 15.6 15.6 mm: mm: maxImum thickness thickness of maximum 2
For Forpreliminary prelimmary calculation, calculatIOn, an an estimated estimatedself selfweight weIghtisISincluded. Included. Assume Assume beam sufficiently beamtotobe be533 533xx210 210xx92 92tIE UB(grade (grade43A). 43A).Ins It IS suffiCientlyaccurate accuratetototake take beam beam weight weight of of92 92kgim= kglm =0.92 0.92kN/m. kN/m.Member Membersize sizemust mustbebefinally finally confirmed carnedout. out. confinned after after all all the the design deSIgn cheeks checks have have been beenearned
clause clause 4.2.3 4.2.3
Shear capacity capacity P" == 0.6 0.6 p, Py A" Shear = 0.6xx0.275 0.275 xx 531.1 531.1 xx 10.2=597 10.2 = 897kN leN =0.6 Shearforce force F;x IP" = 0.26 0.26 Shear Therefore, as as F,/P?<0.6, F;xIP" < 0.6,there therewill willbe beno noreduction reduction In m moment moment capacity capacity(see (see Therefore, clause4.2.5). 4.2.5). clause
(e) Moment Momentcapacity capacity (e) Theconcrete concretestab slabprovides provides full full restraint restraint totothe the compression compressIOn flange flange (Fig. (Fig. 3.6), 3.6), The notconsidered. considered.The Thechosen chosenUB VB isIS andlateral lateraltorsional torsionalbuckling buckJingisisnot and
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plastic sectIOn section (blT= (b/T= 67) force is zero. aa plastic 6.7) and and atat mid-span mid-span the the shear shear force
(i)
3
kNm M==pySx=275 x 2370 x 10- =651 kNm
Alas P,4== 1.2*275*2080* /ylez;t 1.2 1.2 PyZ-" 1.2 x 275 x 2080 x
clause 6.3.2
10- 3 = 686 kNm
capacity of ofbolts bolts Pbh=dlpbb: P bb = dtPbb: Beanng capacity
clause 6.3.3.3 dause
Beanng capacity capaCJly of the the angles for bolts, bolts, because because angles fdtpfu) (dip,,,) ISis the the same same as that for Pt, Pbs for for grade43A grade-43A steel steel has has the the same same value as Pbb for bolts, i.e. I.e. for grade 4.6 bolts, 2 460N/mm2. 460 N/mm • In addition, the the beann'g"capaclty tiflhe with Of the angles angles mllst must comply with the the cnterion, &e Pbe i erp,,,/2 erpb112 (e defined defined in Fig. Fig. 3.7).
P,,,=50x Pbs=50 x lOxO.460/2= 10 x 0.460/2 = ll5kN/boli 115 kN/bolt
Section is SectIOn IS satisfactory, satlsfaclOlY.
S:,i :;:tS:
Note mm Note that that the the column columnflange flangewill willalso alsoreqUlre requirechecking checkingififititISisless lessthan than100mm thick. Column thick. Column bolt bolt connection connection ISis satisfactory. satisfactory. Capacities Capacities of bolts bolts and beanng values alternatively be obtamed from values may may alternatively be obtained from reference (5).
39.7
DeflectIon Deflection
~
Deflection (which is a serviceability DeflectIOn (which IS servIceability limit limit state) state) must be be calculated calculated on on the the basis of the unfactored baSIs unfactored imposed imposed loads: loads:
9, -a
~
Forces tN Fo'",kN
(ii)
r-c--- _ Direction Direction
lic=90÷42= 132 kN W.,=90+42=132 kN
ol kN of 89.4 BB.4I:N resultant resultant
Assume the load is IS approximately approximately triangular tnnnguJar and and hence hence formulae fonnulae are are Assume available for deflection available deflectIOn calculationstt) calculalJons(6) = 132w 132 x 7400 /(60)( 205 )( 55 400)( 10 =8.4 mm =8.4 mm 13S table 55 BS table
4
)
01_ a_50 :il 01- 0 BC;J1n ,., 01-
c EE a
,"
0 0 u
"
Jlso
_LL_so Fig. 3.7 3.7
M22 bolts
Assume 99 bolts, 22 mm diameter diameter(grade (grade 4.6) 4.6) as as showa shown in In Fig. Fig. 3.7. Assume
kN (dollble shear) = 242/3 = 80.7 kM (double shear)
=(A/dtn=/ £d2 )due due to10bending bending moment moment =(Aid,,a/Ed2)
88.4 kM kN = 88.4
capacity of ofbolt, bolt,P,,,, P bb == dip,,,, dtPbb Beanng capacity =22* =22 x10.2 10.2 x460=97.6kN x 460=97.6kN
Deflectionlimit limil=7400/360=20.6 DeflectIOn = 74001360 = 20.6 mm mm
The design deSign of of connections connections which which are are both both robust robust and and practicable, practicable, yet yet economic, economiC, is IS developed developed by by expenence. expenence. Typical Typical examples examples may may be be found found inin references (4, 7). 7). The connection ofthe the beam beam must mllst be be able able to to transmit transmi! the the connectIOn at each each end end of ultimate shear force force of of 237 237 kN kN to to the the column column or orother other support. support. The The connection connectIOn forms forms part of of the the beam, beam, i.e. I.e. the the point pOint of ofsupporl support is IS the the column column to to cleat interface. mterface. Design DeSign practice practice assumes assumes that that the the column column bolts bolts support support shear shear force force only, only, while while ihe the beam beam boils bolts early carry shear shear force, force, together together with with aa small small bending moment. moment.
(See Fig. 3.8.)
Maximum honzontal honzonial bolt forces Ma:umum forces are SectIOn 8.4. are discussed discussed in Section
FIg. Fig. 3.8
(g) (g)
BEMIBOLTS BEAM BOLTS
Vertical shear/bolt Vertical
IV,L3/60E1, Wx L JI60Elx 3 7400'/(60
Shear Ps it,, A~, where where A, A~ is IS the the cross-sectional cross~sectlOnai area at at the the root roO! of of Shear capacity capacity p~ P. = ap, bolt thread: thread: the bolt
clause 6.3.3.2 clause
Al, Wa = 573/652 M;./fr/o; 5731652 == 0.88
b x ==
(smgle (single shear)
P. = 0.160 0.160 x*303 303 =48.5 =48.5 kM/bolt kN/bolt
and H sections P,$x Note that for II and sectIons bent about about the the x axis axis the the expression expression PyS~ governs the governs the design. For bending bending about about the the y axis, aXIS, however, however, the the expression expressIOn t.2 p,2, may be be 1.2 p)Zygoverns governs the the design. design. The The factor factor 1.2 1.2 to m this this expression expressIOn may increased to to the ratio Increased ratio factored factored load/unfaciored loadlunfactored load (clause 4.2.5):
Note that because the units (N/mm2) and and S;. S, (cm4) thea the 10- 3 Note umts are Pr (N/mml) (cm)) then must be be included must mcluded for Ala M= in In order order to to obtain obtam the the correct correct unit umt of of kNm. kNm. Alternatively there used for p,.. Alternatively, there is IS no no need need for for 10 10- 3 if 0.275 kM/mm2 kN/mm 1 isis used Pr But BUl
35
ASPbl=Pbb,then thcnbeanng beanngcapacity capacityofofIlic the web web plate plate is IS the the same samC as as for for the the bolt. bolt. ASP,,,p,,,,,
cleats {allOWing mm holes) holes) Shear area area of cleats (allowing for for 24 24mm =0.9(400* \0)( 2 - 3 x 2 x lOx =0.9(400)( 10)( 24)= 24)'=5904 5904mm2 n1ln 2
Shear capacity capacity P" = 0.6 *0.275*5904 x 0.275 x 5904
=974kN 974kN '
check for for bending bending may may also A check also be be carned earned out out but will generally give aa high high to the the applied applied moment(3) capacity relative relnuve to bending capacity
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-
30 STRUO1URAL STEEL WORK DESIGN TO SI $050
36
STRUCTURAL STEELWORK DESIGN TO 8S 5950
8EAMS BEAMSIN INBUILDINGS BUILDINGS 37 37
3.8 3.8
EXAMPLE 5. 5. BEAM BEAMSUPPORTING SUPPORTINGPLANT PLANT LOADS LOADS EXAMPLE (UNRESTRAINED BEAM) BEAM) (UNRESTRAINED
Mum beam beamisissimply simplysupported supportedand andspans spans9.0 9.0mm(Fig. (Fig.3.9). 3.9).Boilers Boilersare are Mam supportedsymrnetncally symmetncally on on secondary secondary beams beams (A (A and and B) B) of ofspan span6.0 6.0 m. m, which which supported areatat 5.0 5.0 m m centres. centres. are
-.--,--1 E o ,.;
It 1.Sm
:'_~'~" .. l.5m
l.5m
•t,J 1'l$ni
I --_~,;-:7_-_ ·1.5m
(c) (c)
-
S.Om
,1=:-1
S.Om
E
m
BM BM and and SF SF Ultimate Ultimate shear shear force force F, =335 kN kN F,== 1912 19/2 xx 422 422 xx 6.0/9.0 6.0/9.0++398 398xxl.0/9.0 1.0/9.0=335 Ultimate Ultimate shear shear force force F, F2 == 19/2 19/2 xx 398 398 xx 8.0/9.0 3.0/9.0+÷ 422 422 xx 3.0/9.0 3.0/9.0 == 504 504 kN kN
E
"
1.Sm
With Fig. 3.12: 3. J 2: With reference reference to to Fig. Ultimate Ultimate moment moment MA =335 = 335 xx 3.0 3.0 == 1005 1005 kNm kNa, =504 x xl.0 = 504 Ultimate Ultimatemoment momentMDM0=504 2.0= 504 kNm kflm Note calculatmg the the moments, moments, the the small small reduction reductlOn due Note that that m in calculating due to to the the self self weight Ignored. weight is ignored.
E o ,,;
Beams B
I,
I!-
~--I~--------I'~------~
I
Fig. 3.9 Plant loads
Fig. 3.9 Plant loads
'.Om
,I,
'.Om
3.12
(d) Cd)
.\
DeSign stTength Py for steel steel 19.7 19.7 mm mm thick thick (grade (grade 43A) 43A)=265 N/mm2 (see Design strength = 265 N/nun2 (see Table 1.2). 1.2). Table clause 4.2.3 d.2.3 clause
This maximum maximum coexistent coexistent shear shearforce force is IS present at pomt B, while the This present at point B, while the maXImum moment moment occurs occurs at at point pomt A. A. maximum
(e) (e)
Momentcapacity capacity Moment
clause clause 3.5.2 3.5.2 The The chosen chosen section sectIOn isIS aaplastic plastiC section section (b/T= (blT=7.7). 7.7).
With With reference reference totoFig. Fig.3.10: 3.10:
flu
10-:; = ll53 k.N kN
FI/P. =335/1153=0.29 =335/1 153 =0.29
The The boilers boilers produce produce reactions reactionsof ofIOU 100 kN kN at at the the end endof ofeach eachsecondary secondarybeam. beam. Allow weight of selfwclght ofeach eachsecondary secondarybeam beam(not Inotdesigned desIgned here). here). Allow 4.0 4.0 kN kNfor forthe theself Assume Assume 610 610 xx 305 305 xx 149 149 1W VB (grade (grade 43) 43) for formain mainbeam. beam.
,I
capacity P" 0.6 pr-i, PvA." Shear capacity P. ==0.6
Loading Loading Boiler Boiler loading loading (each) (each) Open Open steel steel floonng floonng (cariied (earned on on beams beams AA and andB) B) Imposed Imposed load load (outside (outside boiler boiler area) area)
6.0 6.0 m
ShearShear capacity capadty
Flooring =0.3 6.0 == 7.2 Flooring on on beam beamAA = 0.3 xx 4.0 4.0 x x 6.0 7.2 kN kN Imposed Imposed load load on on beam beam AA== 4.5 4.5xx 3.0 3.0 xx 4.0 4.0 == 54.0 54.0 kN kN
Moment capacity capacity Me = pyS;r Moment 265 ,<4570 x 4570 xx 10- 3== 1210 1210 kNm kNm == 265 Moment raho M"dMe = 1005/]210=0.83
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38 TO 8S IS 5950 38 STRUCTURAL STRUCTURALSTEELWORK STEELWORK DESIGN DESIGN TO 5950 (1) (I)
BEAMS IN BEAMS INBUILDINGS BUILDINGS 39 30
Buckling resistance resistance Budding
(g)
The buckling resistance moment of the The resistance moment Ihe beam in In the the part-span part-span AB AB must must be be found, and and in in this part the moment momenl vanes from from 504 kNm to 10 1005 1005 kNm. It It is IS found, assumed thal that the steel steel floonng does not not provide provide lateral lateral restralDt, restraint, but but that that the the assumed floonng does secondary beams beams give give positional and rotatIOnal rotational restraint restraint at at 5.0 m spacing. secondary posltiona! and spacmg. Loading between between the the restTamts restraints ISis of of aa mlllOT minor nature nature (self tself weight weight only) only) and and is is Loading ignored for for use use of OS Ignored SS table table 13 13 as itIt would would affect the moments moments by by less than 10% (see tsee also OS 10% BS table table 16). 16). clause 4.3.5 clallse .;-.3.5
= 5.0 m LE =5.0 Slenderness .1==L5/r, Slenderness..:. LefTy
=5000/69.9=72 (both =5000169.9=72 (both ry and L5 LE given in In mm) mm) Torsional index indexxx ==32.5 TorsIOnal 32.; Aix )Jx
== 2.7 2.2
C
In P3
2!
1
A
L son
I3.OmIZ
4.5m
4.5m
.
AL,. ALT
= nuvA =1JI1vA = J.O .0 x = x 0.886 0.886 xx 0.94 0.94 xx 72 72
Ca kulatlon of the cannot be be Calculation the deflectIOn deflection for for tile the serviceability imposed loading cannot of fonnulae, formulae, which which become become complex complex for for nonnoo* carned easily by the use use of earned easily by the standard cases. Serviceability standard cases. Serviceability point load
W W4=200+54 A =200+54
kN =254kN
~254
1J's=200+40.5 =240 kN kN W8=200+40.5 =240
1045 1045
149~ 588 kN m
355
BM diagram (areas In boidi bold) areas in
found by by the the With reference With reference to to Fig. Fig. 3.l3, 3.13, mid-span deflection may be found moment~area moment-area
method{8.9):
{,6 ~= Jj (Af(x)/E1)dx fM(x)I£1)dx =(149 x 0.667+ 1045 x 2.75 +355 x x 3.333)/ =(l49
Fig. 3.13
N= \' = 0.94 0.94 (for N = 0.5, 0.5, i.e. 1.e. equal equal flanges) a == 1.0 I .0 (for member 11 member not nol loaded loaded between between restraints) restramts) it 11 = = 0.886
Calculation by Macaulay's method(IO.II) gives poml of of maximum maxlmwn Calculation by gives the the point 17.6 mm. deflection deflection 4.65 4.65 m in from from support support 22 with with aa value value of 17.6
= 60 60
DeflectIOn 9000/200 = 45 mm mm Deflection limIt limit == 90001200 = 45
Equivalent needs to be obtatned EqUivalent uniform unifonn moment factor In IrJ needs obtained for a member not loaded between between restraints: restramts:
An treatmg the the load load An approximate approximate estlmate estimate of of deflection deflection ISis often oflen obtamed obtained by by treating as eqUivalent u.d.I. u.d.l. as an equivalent
2 no. 100 100 x 100 x 12 t.s ls 500 long
(ii) (h)
Connection Connection
// P =504/1005=0.50 =504/1005=050 hence
in =0.76 = 0.76
III
Equivalent uniform unifonn moment moment M H = oiMA EqUIvalent mAlA. = 0.76 x 3005=764 lOO; = 764 kNm kNm hence M 1Mb = 764/946 and section is satisfactory. 764/946 = =0.81 0.81 and satisfactory.
Try aa smaller smaller section sectIOn (610 x* 229* 229 x 140 140 liD): UB): cm44 SE=4150 cm
A =99.4 "= 99.4 Pb = 161 161 N/mm2 Nlmml 668 kNm Mb = =668 Ji /Mb = 1.14 I.l4 which is IS not not satisfactory. satisfactory. This companson companson indicates mdicales the the sensitivity senSitiVity of ofthe thebuckling buckling resistance reSistance moment to to small small changes changes tn In section sectIOn properties, properties, particularly particularly to to aareduction reduchon in In flange width. width.
connectIOn at support 2 of the tmnsmlt an ~n ultimate uitimate shear shear The connection the mam beam must transmit force 504 leN kN and 111 Section 3.7(g): 3.7(g): force of 504 and follows follows the the method method given given in
8_50 a—SO 100 100 100 Beam 100 asan 'lOO
., "
E = .5 u
M=504 0.05=25.3 kNm M= 504 x 0.05 = 25.3 kNm
o,~ II
I
M22 bolts M22 bolts
Assume 22 mm bolts (grade 8.8) 8.B) as shown in m Fig. 3.14. ~.14. Assume 22mm
-I WO
'r"-
Fig. Fig. 3.14 3.14 ' (I) (i)
6.2.3 clause 6.2.3 clause dause 6.3.3.3 6.3.3.3
COLUMN BOLTS BOLTS
Vertical shear/bolt F~ F3 = 504/10 Verttcal shearlbolt =;04110 50.4 leN kN 50.4 capacltylbolt P3 P~ = 0.375 0.375 *303 x 303 = 134 114 leN kN Shear capacity/bolt capacity of ofangles/bolt angleslbolt P b:! = 22 22 x 12 12 x 0.460 0.460 = = 121 121 leN kN Beanng capacity but but Pru-/bolt I' SO 50 xx1212*0.460/2 x 0.460/2 = 338 l38 kN the column column flange flange will require checking if if less less than than 32mm 12 mm thick. thick. Note that the connectIOn is satisfactory. satisfactory. Column bolt connection
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=504/5 = 100.8 l0O.8 !eN kN = 504/5 =Md,,,J1d2 =Mdm,.)l.d' =25.3 = 25.3 xO.20/2(0.102+0.202) x 0.20/2(0.10' + 0.20') =50.6kM =50.6 !eN 2 = .J(l00.8 1(100.82+5062) Resultant shear/bolt =113 kN shearfbolt = +50.6 2 ) !eN =0.375 xx 2 x 303 Shear capacity capacity (double (double shear) shear) =227 kN !eN =271 Bearing capacity/bolt i.035 Beanng capacttylbolt P bb =22 = 22 xx 11.9 11.9 x 1.035 =271 leN !eN. Bearing capacity of P5, = 120 120 kM Beanng of web web plate plate P = 22 xx 11.9 x 0.460 !eN bJ =22 243kN hut P" l' 89 x 31.9 11.9 x 0.460/2 = 243 !eN but Vertical shear/bolt shearlbolt Horizontal shear/bolt shearlbolt
clause 6.3.3.3
5~O'6,
50.6
- _~,
Contes M.G. & & Kong Kong F.K. F.K.(1988) (1988) Coates R.c., ILC., Coulie Caulk M.G. Moment-area methods. methods.Structural StructllraiAnalysis, AnalysIs,pp. pp.176—SI. 176-81. Moment.area Van Nostrnnd Nostrand Reinhold Van Retn.hold
10. DeflectIon 10. Deflection
Marshall W.T. & & Nelson Nel!on tiM. H.M.(1990) (1990)Singulanty Singulanty Mnrs-ball W.T. functions, Structures, functIOns, Structures, pp. pp. 233—8. 233-8. Longnsan Longman
II. Deflection 11. DeflectIOn
Ilearn Hearn El. EJ.(1985) (1985)Slope Slopeand anddeflection deflectIOn of ofbeams beams, Mechanics Mer:lramcr of of Materials Malenals vol. vo!. 1, 1, pp. 302—7. l02-7. Pergamon
C
connection is is satisfactory. satisfactory. Beam bolt connection
~,
C'!
41
";'1
;2
(iii) (iii)
,
'll>
i
• Direction of 113 tN resullant resullant
'" \
Shear area of cleats area of deats (24 nun mm holes) = 0.9(500 x 12 x 22 x 32 = 8208 mm mm22 =0.9(500 12 xx 22—5 - 5x 12 x 24) 24)=8208
.
40!
.
a
= 1350 Shear Shear capacity capacity pP.... =0.6 =0.6 xx 0.275 0.275 x 8208 = 1350 leN kN FV/P, =50411350=0.37 =504/1350=0.37 F.jP"
clause 4.2.3 I I
FIg. 3.35 Fig. 3.15
ANGLE CLEATS CLEATS
I
Connection Conne'ctlOll cleat is IS satisfactory. satisfactory.
! STUDY REFERENCES Topic TopIC
References References
f
I. j. Lateral Lateral restraint restraint BS 5950 3950 BS 2. Strut StIllt behaviour behavIOur
Bowling Dowllng P.3., P.J., Knowles Knowle.s P. P. & & Owens Owens G.W. G.W.(3988) (l988) Steel Construction institute Structural Steel Design. DesIgn. Stetd
3. Strut 3. Strut behaviour behavIOur
Coates R.c., R.C., Coutie Coutie M.G. M.G. & & Kong F.K. (3988) Coates (1988) of struts and frameworks, Stntctural Structural Analysis, Instability of Anaiysls, ' Reinhold pp. 58—73. 58-71. Van Nostrand Nostmnd Remhold . pp.
4. Connections 4. ConnectIOns
(3993) JOintS Joints In in Simple Simple Connections Connections val. vol. 1. I. Steel (1993) Steel Construction Institute Canstruchon Insbtute
5. Bolt 5. Bolt details details
(1985) Steelu'ark (985) Steelwork Design DeSIgn vol. I, Section ptoperues, propertles. member capacities. capacities, Sleel Steel Construction ConstructlOn institute Instttute
Joints In (1992) Jomts In Simple Simple Connections vol. voL 2. Steel Steel Construction Institute institute
8. Moment-arcs Moment*llfell method method
r
Marshal! Marshal! W.T. W.T. & & Nelson Nelson H.M. H.M.(3990) fl990)Elastic ElastIcstability analysis. anaiysls. Structure, pp. pp. 420—52. 420-52. Longman
Croxton PP.L.C. Croxton .L.e. && Martin Lii. L.H.(1990) (1990)Area-moment A",,-n,on"" methods of of analysts, analYSIS. Solving SolVing Problems in In Structures Structures vol. VD!. 2. pp. pp.23—47. 2'5-47. Longanan Longman
.
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PURLlNS AND SiOE SIDE RAILS RAILS PURLINS
43
1.0 {fl
~
HI PURLlNS AND SIDE SIDE RAILS RAilS PURLINS
Fig. FIg. 4.2
dead, Imposed by the the dead, imposed and and wmd wind pressure pressure loads loads will will cause the flange restrained by cladding to can, however, how~ver, reverse reverse this this cladding to be be m in compressIOn. compression. Wind Wind suctIOn suction load bad can, In compression. compression. Torsional TorSIOnal arrangement, I.e. arrangement, i.e. the the unrestrained unrestrained flange flange will will be in purl ins restraint to 10aabeam beamInvolves involvesboth both flanges flanges bemg being held held in in pOSition position and for purlins wil! be be tnle only at the the supports. supports. and side rails this will vertical loading (cladding) (cladding) and lioozontal honzontal rails are Side rails are subjected subjected to to both both vertical loading (Wind vertical loading loading is IS loading (wind pressure/suction), pressure/suction), but but In in general the vertical considered to considered to be be taken taken by by the the cladding cladding actmg actmg as as aa deep deep girder. girder. Consequently, to wind) wmd) are are considered considered in in design. deSign. only moments moments In in the the honzontal plane (due to penetrated by holes deSIgn of new In the design new construction construction where where the the cladding cladding is is penetrated conveyors, the design deSign engineer engmeer should should be be satisfied satIsfied access, ductwork ductwork or conveyors, for access, acting in In this manner. manner. that t/le cladding and that the cladding and fixmgs fixings are are capable capable of acting sometimes used the effective effective length length of ofpurlins purl ins and and Sag rods are sometimes used to reduce the In continuous beam design deSIgn (see Section Section 4.4). 4.4). \Vhere Where sag sag side rails, rails, and result in rods· are used, used, prov1Sl0n the end reaction reactIOn on eaves eaves or or apex apex rods provision must must be be made made for the IS no reason why purlins purl ins beams. As IS beams. is shown shown in in the the following following examples, there is biaxial bending m and side side rails and rails should should not not be be deSigned designed as as beams beams subject subject to to biaxial bending in accordance with the normal normal design deSIgn rules. rules. accordance
In and side side rails rails used used in in the the construction ·constructlOn of ofindustrial industrial In the UK purlins and sectlOns. These TIlcse sections sectlOns can can buildings are often fabncated fabncated from from cold cold fonned fonned sections. desIgned in accordance with Part 35 of of BS 5950, 5950, but hut the the load load tables tables for for be designed sectIOns are are frequently frequently based bused on on test test data. data. The sections sectIOns are are marketed marketed by by these sections compames specializing specUllizmg in III this this field field who will normally nonnally give gIVe the the appropnate appropnate companies and allowed allowed loadings laadings in In their their catalogues. catalogues. Sections SectIOns of of this this kind kind are are spans and commonly of channel channel or zed form form as illustrated illustrated in Fig. Fig. 4.1. Although their theIr commonly 4.1. Although IS not covered covered in in this Ihis chapter, chapler, the tbe selection selectIOn of ofcold cold formed formed sections sectIOns isIS design is in Chapter Chapter 12. 12. discussed in
Fig. 4.1 4.1 Cold Cold boned fonned Fig.
4.2
EXAMPLE 6. 6. PURLIN PURLlN ON SLOPING ROOF ROOF
(a)
Dimensions Dimensions
purlins
sectIOns may be be used used as as an an alternative, alternatlve, and and in in sonic some situations situations Hot rolled sections preferred to 10 cold cold formed formed sections. sectIOns. The The design deSign of ofangles angles and and hollow hollow may be preferred sections be camed carned out out by by empirical empirical methods methods which are covered by sections may may be deSign procedure procedure (i.e. non-empincal) non-emplOcal) is IS set set out out m m this this clause 4.12.4. The full design
I,
See Fig. Fig. 4.2: purl ins at at 2.0 m centres; centres; span 6.0 6.0 m m simply Simply supported; supported; rafter rafter See 4.2: purlins
slope 20"
chapter. (b) 4.1 4.!
kN/m'
Dead + insulation insulatIOn panels) 0.15 Dead load load (cladding tcladding+ 0.15 kN/m2 0.75 kNjm 2 (on (on plan) plun) 0.7$ kN/m2 Imposed load load 0.40·kNjm (suctIOn) 0.40kNjm22 (suction) Wind load
DESIGN FOR PURLINS PURLlNS AND AND SIDE SIDE RAILS RAILS DESIGN REQUIREMENTS FOR The bending IS The deSign design of steelwork steelwork In in bending is dependent dependent on on the the degree degree of of lateral lateral of the the restramt the comoression compreSSlOn flange flange and the the' torsional torsIOnal restraint restramt of restraint given to the on the the degree degree of oflateral/torsional lateral/torslOnai restraint restramt given given at at the the beam beam beam, beam, and also on supports. been supports. These These restraints restraints are are given given In in detail detail in in clause clause 4.3 4.3 and and have have been discussed demonstrated in Chapter 3. Side rails and and purlins purlins may may be he discussed and demonstrated have lateral lateral restraint restrmnt of ofthe the compression compressIOn flange flange owing oWing to to the the considered to have of the the cladding, cladding, based based on on adequate adequate fixings flxmgs (clause (clause 4.12.1). Loads Loads presence of will be transferred transferred to the the steel member member via Via the the cladding cladding (see Fig. 4.2), and the
Loading
Wdl W/orW,,.l W,.} or W,or Wdlor
T9
,sIllDIJI::IITi ]iJiIiIij':IUD'::jl hiItJLUii 6.0 m m at 2.0 m centres a12.0
Fig. FIg. 4.3
derivatIOn of of loads, loads, the lhe Reference Reference should should be be made made to to Chapter Chapier 22 for the derivation the area load. directIOn direction In in which which each each will will act, act, and and tIm area appropnate appropriate to to each load. Maximum valuesof of bending bending moment moment and and shear shear force force must must be found at the MaXimum values the ultimate limit state state making making due due allowance allowance for for the the slope angle and including ultimate limit mcluding 'If factors. factors. the y,. Assume purlin }52 xx 76 channel section, sectIOn, grade 43A 43A steel steel (see (see Assume purlin to to be be 152 Fig. 4.3). 4.3).
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44 STRUCTURAL TOES 44 STRUCTURALSTEELWORI( STEELWORI( DESIGN DESIGN TO as 5950 5950
PURLlNS PURLINSAND AND SIDE SIDE RAILS RAiLS 45 45
Cladding 2.0 xx 6.0 6.0 xx 0.15 0.15 == L80 L80 kN kN Cladding 2.0 Self weight weight 6.0 xx O.IB 0.18 ~ = LOB kN Self 6.0 LOB kN Total dead dead load load Wd=2.8B kN kN Tota! Wd~2.BB Imposed load load =2.0 Imposed =2.0cos cos20° 20 xx6.D'x 6.0·x 0.75 0.75 JV,~=8.46kN W, B.46 kN Wind load load = 2.0 xx 6.0 Wind ~2.0 6.0 x x (—0.40) (-0.40) øç= —4.80 Ww~ -4.BO kN kN
Ce) Moment Moment capacity Ce)
0
BS table 77 BS clause 4.2.5 clause
0 results in purlins at the same angle. Components of The rafter rafter slope slope of of 20 20° The results in purl ins at the same angle. Components of load are are used used to to calculate calculate moments momentsabout aboutthe thexxand andy axes, I.e. i.e. nonnal normal and load y axes, tangential to to the the rafter rafter (Fig. (Fig. 4.4). 4.4). As As with with side side rails, rails, it it would tangential would be possible to ignore bending bending in m the the plane plane of of the the cladding, cladding, but but in in practice, practice, biaXial biaxial bending bending is is Ignore usually considered in purlia usually considered in purUn design. deSign.
!
Fig. 4.4 Fig. 4.4
The section The biuxllli bending sectionclassification classificationof of aa c'hannei channel subject subject to to biaxial bending depends depends on on bIT and dlt In this The is b.'Tand d't which in this case are 8.47 and 16.5, respectively. respcciively. The channel channel is therefore therefore aa plastic plastic sectIOn. section. Hence, Hence, the moment moment capacity capacity
M=p551=275 X 130x l03=35.8kNm but must not but Mo; Ma must not exceed exceed 1.2 i.2 PyZz ~ 1.2 xx 275 ~ 37.0 kNm kNm = 1.2 275xx 112 112xx 10-' 11r3=37.0 1 Note that 10 Note 10- must mduded to In kNm, kNm, when when N/mjn2 N/mm2 for must be be included to give give Mo; Ma in for p,y and P and cm3 cm) for Sx- Alternatively, py expressed as as 0.275 0.275 kN/mm2 kN/mm2. p,. may be expressed but aX:Ial forces forces and and stresses used. but this this reqUIres requires care care later when when axial stresses arc are used.
=2.88 cos =2.71 kN Wch =2.88 cos 20° 20 0 =2.71 kN 0 20° =0.99 Wdy ==2.88 2.88 sin sin 20 = 0.99 kN kN 20°0 = = 7.95 Wu; == 8.46 8.46 cos cos 20 7.95 kN kN
x 41.3=11.4 kN m
8.46 sin sin 20 20° = Jt'iy == 8046 = 2.89 2. 89 kN kN 0
jJç1= —4.80 W,,,,"= -4.80 kN kN
The raho S,./Z, Sy/Zy isIS greater greater titan than 1.2 1.2 and and hence hence the the constant constaut 1.2 1.2 is IS replaced replaced by by the ratIO ratio factored factored load/unfactored loadlunfactored load ioad (6.0/[0.99 (6.0/[0.99 + 2.89J == 1.55). 1.55). the ± 2.89]
Note that zero as as wind wind pressure pressure IS is perpendicular perpendicular to Note that W,,.., WilY ISiszcro to the the surface surface on on which it it acts, acts, i.e. I.C. normal nonnal to to the the rafter. rafter. which
Mcy must 1.55 P0y must not not exceed exceed 1.55 p,Z, ~ 1.55 1.55 xx 0.275 0.275 xx 21.0~ 9.0kNm kNm = 21,0= 9.0
Ultimate load load W,.= W1= 1.4 2.71++ 1.6 1.6 xx 7.95 7.95= U1timate \:,4 xx 2.71 = 16.5 16.5 kN kN where 1.4 and 1.6 factors (Section 1.7). lA and 1.6 are are the the appropnate appropnate 4,' Xf 'factors 1.7). where (c) (c) WiN Wk"
The be earned out (SectIOn 3.4): 3.4): The local local capacity capacity check check may may now now be out (Section
(I) (f) 1,
(d) (d)
Shear Shear capacity capacity
Buckling Buckling resistance buckling resisiance resistance moment moment M,, Mb of of the the section section does does not not need need to to be be found found The buckling IS restrained restrained by by the the cladding cladding in In the the x.1' plane plane (Fig. (Fig. 4.6) 4.6) and and because the beam is Instability is IS not considered considered for for aa moment moment about about the the minor mmor axis axiS (Fig. (Fig. 4.7) 4.7) instability (Scction 3.1). 3.1). (Section
Design strengthp,. given in Table forselected the selected is DeSign strength py IS is given In Table 1.2 1.2 and and for the purlin purlin sectionsection IS 275 275 N/mm2 N/mm2 clause clause 4.2.3 4.2.3
Shear ==975mm 975 tom22 Shear area area A,.,. == 152.4 152.4 xx 6.4 6.4 Shear Pa\.'X == 0.6 Shear capacity capacity P 0.6 PyAv,. =0.6x275x975x30'3 =O.6x275 x 975 x 10- 3 =l6lkN =161kN Shear A,, Shear area area A,y==0.9A0 0.9Ao .. =0.9 = 1234mm2 =0.9x2 x2x x76,2 76.2xx9.0 9.0 1234 mm'" Shear Shear capacity capactty Pvy == 0.6 0.6 xx275 275 xx1234 1234xx 10- 3 =204kN =204kN
It It may may be be noted noted that that in in purlin purlin design, deSIgn, shear shear capacity capacity isis usually usually high high relative relative to to shear shear force. force.
Fig. 4.6 4.6 Fig.
Fig. 4.7 4.7 FIg.
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J 7.95 it 600&/(384 12.8mm ox=55 it x 7.95 x6oo0 /(384 itx 205 205 itx 852 852 itx 10 4 )=12.8 mm 28.5 tam b;.=28.5mm Deflection 30mm DeflectIOn limit limit = 6000/200 = 30 mm
value of of M1 MI is much lower ihan than the value value 32.8 12.8kNm The value kNm used earlier, but negattve Sign Indicates that the lower flange 10 the negative sign indicates that ihe flange of the the channel channel is tn ~compresslOn and flange is not restrained. restrained. The and this this flange is not The buckling resistance A·h must therefore he be found. found. effectlve length LE purlin may found from from BS table 9. The effective L5 of the the purlin may be found
(I) (i)
3.0 it LE == 1.0 x 6.0 6.0 = 6.0 6.0 m
4.3.5 clause 4.3.5
2Lr = nuvA
BS ES tabl, table 16 J6
Clause 4.3.7,4 4.3.7.4 ES BS table 11 11
= 14.5 =
A/s =268/14.5 = 268/14.5 = 18 38 A/X V v = 0.49 0.49 it =0.902 = 0.902 11 nn =0.94 (for and y=O) (for P=O fl=0 and y=0) ALT = = 0.94 0.902 it ALT 0.94 xit 0.902 x 0.49 it x 26S 268 = III lIt
BS table 14 14 ES
Connections The connection connectIOn of the the pllrlin cleal purlin 10 to the rafter may be be made made by bolting Ilto it to a cleat as shown m iii Fig. 4.8. 4.8. The design deSign of of these connections connectlons is IS usually nominal nommal due due to the tow the transfer of low reactions reactIOns at the end of the purtins. pUrlins. However, However, the of forces between the should be considered. considered. For ttie the channel section secuon between the purlin purlin and rafter shnuld transfer to to the the rafter through chosen, Wx and IF'. Wy transfer tlrrough a cleat. cleat. Bolts Bolts must must be be chosen, provided but will be nominal provided but nommal due due tn to the the low low reactions reachons invotved mvolved (8.0 (8.0trW kN and 3.0 kN). Chapter 3 gives calculations calculatIOns for a butted bolted connection connectIOn in In more more detail. detail. (continuous) purtins Multi-span (contmuous) purlins may be used and minor,ctianges mlOor.changes in m design deSign considered in Section SectIOn 4.4, 4.4. ' . are considered
Slenderness A Ln'r, = 6000/22.4 = = 268 l == Ldry 268 (which is less than 350 as required requrred lwhich IS by clause 4.7.3.2) where where LE LE and ry are EqUivalent stenderness slenderness lLT are in mm. Equivalent allowing for is given by: allowing for lateral torsional torsional buckling buckling IS
Torsional mdex index xr TorSional
Deflection Deflectton limits for purl purlins specified ill in BS BS table table 55 but but a limit of Defieclton ins are are not specified of span/200 IS is commonly adopted. span/200 adopted.
some situations In combinatton combination with with ioads oath IV,, and nc, sltuallons it could be critical. critical. Tn Wd and Wi, aa lower total load IV W is clearly produced.
= -4.01 —4.01 x 6.0/8 M, =
Purlin Putlin
30811/mm22
Bending strengthpb strength pa may may be be obtained: pa= 108N/mm Bending obtamed: Pb= Buckling resistance resistance Al6 M/)
47
__4__ 7
d,~~:"'=-Cleat CUI tram un equal angta
~"gl'
=puSx = 108 308 it x 330 130 itx 10- 3 = 14.0kNm i4.OkNm
ES BS table J3 13
The overalt may now now be be camed overall buckling buckling check check may carned out out using using an an equivalent eqUIvalent uniform moment factor factor (m) (at) equal equal to to i.0 uniform moment 1.0 (member (member loaded loaded between between
3.Ot/14.0+0.75/(275 itx41.3 3.01/14.0+0.75/(275 41.3 itx 10-'):=0.28 The overall buckling of of the the sectton section isIS therefore therefore satisfactory. satisfactory. diagrams for for bending bending moment moment and and shear shear force force shown shown in m Fig. Fig. 4.5 4.5 The diagrams indicate that that maximum maximum values mdicate values are not coincident COincident and itit is IS not not therefore therefore necessary to to check moment moment capacity in necessary III the presence of of shear shear load. load. Purlin Purlin design does does not not normally normally need need aa check check on on web web beanng and buckling as the deSign applied concentrated concentrated loath loadsare arelow low— - note nole the tow low values of shear force. force. The pariiculariy needed where heavy check for beanng and buckling of of the web is IS particularly concentrated concentrated loads loads occur, occur, and and reference reference may mdybe be made made toto Chapter ChapterS 5 for the the reievasit calculahons. calculations. reievant
4.3
EXAMPLE 7. DESIGN OF OF SIDE SIDE RAIL RAIL EXAMPLE
(a)
Dimensions
(See (See Fig. Fig. 4.9): 4.9): side rails at 2.0 2.0 m m centres: centres; span span 5.0 5.0 m m simply simply supported. supported.
(1) (b)
Loading
Dead Dead load (cladding/insulation panels) panels) Wind load (pressure) (pressure)
0.38 kH/m2 O.ISkN/m' 0.80 kN/m2 O.80kN/m'
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45 STRUCTURAL STEELWORK DESIGN TO ES 5550
48
PURLlNS AND SIDE RAILS 49 PIJRUNS AND SIDE RAILS 49
STRUCTURAL STEELWORK DESIGN TO BS 5950
Side
rail
X
Wind load Wind load
y
N
.
"g-g .0, "E
~
0
,
u
X
E
c E ."
c
" .,
iqi
'0
~~ ~
G
o.
~
~
~
Wind toad Wind load
Maximum values of bending moment and shear force must be found MaxlfllUm values of bending moment and shear force must be found allowing for the thewmd windloading loading(bonzontal) (bonzontal)only only(Fig. (Fig4.9) 49) and and inciuding mciudmgtbe the allowlOg for safety factor y1. safety faclor Yf' Assumethe theside siderail railtotobe be125 125xx75 75xx 10 10 unequal unequalangle, angle,grnde grade43 43steel. steel.An An Assume angle, such as that chosen,chosen1provides providesgreater greaterresistance resistancetotobending bending(higher (higher angle, such as that section properties) about the x axis than they axis, compared to that for an section properties) about the x axis than the y axIS, compared to that for an equalangle angleof of the the same same area area (weIght). (weight). equal Cladding 20 x 50 x 018 I 8OkN ==L80 kN Cladding 2.0 x 5.0 x 0.18 Self weight5.0 5.0xx 0.15 0.15 =0.75 lcN Selfwelgbt =0.75 kN Total dead dead load =2.55kN Total load 'T1d Wd =2.55k:N Windload loadW", rV,,,=2.0 5.0xx0.80 0.80==8.0 Wind = 2.0 xx5.0 8.0 k:N
.
(e) (e) Moment Momentcapacity capacity -
4.11
The loads loads W", and and Wd lVdact actIninplanes planesatatnght nght angles angles producing producing moments moments The about sandy about xrare in axes of ofthe the steel steel section, sectlon, but but only only moments moments about are used used in about x and yaxes dpsign (as discussed in Section 4.!). dS=Slgn (as discussed in SectIOn 4.1).
For cladding, which also Forslflgle singleangles, angles iaternl lateral restraint restraint ISis provided provided by by the the cladding which also ensures bending about the x rous, rather than about a weaker ensures bending about the x axis, rather than about a weaker aXIS axis (Fig. (Fig. 4.11). 4.11). The The moment momentcapaCIty capacityonly onlyof ofthe thesectton section isis therefore therefore Checked. checked. The The section section BS (both BStable table 77 chosen chosenISisdefined definedasassemI-compact semi-compacthavmg havingb!T= bIT=7.5 7.5and anddJT= d/T=12.5 12.5 (both <<15), 15), and 23), hence: and (b + d)IT= 20 1< 1<23), hence: 3 =275 = ID.OkNm =275 xx 36.5 36.5x x10103r10.OkNm M, /Ma = 7.0/10/.0 =0.70 !%4/Ma=7.0/l0/.0=D.70
.
-
SectIOn Section ISis satisfactory. satisfactory. The mclude a check on web beanng The deSIgn design of of side side rails rails does does not not nonnally normally include a check on web bearing and 4.2(g). and buckling, buckling, as as discussed discussed in in Section Section 4.2(g).
pI!! &
80
mz
(c)
£i
A
WkN WkN
(l) (I)
BM and SF
12
With reference Fig. 4.10: With reference to 10 Fig. 4.10:
,Ill! 1 ! 1 I ! i I A
Maximum moment Maximum moment
a
Al, AI., ==lI.2x5.0/8 11.2 x 5.0/8 =7.0kNm =7.0kNm
Mnximumshenr forceFx F, ==11.2/2 Maximum shear force 11.2/2
=5.6kN = 5.6 kN
Deflection CalculatIOn deflectIOn IS on the the serviceability condition, I.e. with Calculation of of deflection is based based on condition, i.e. with unfadored loads. unfactored loads.
SM and SE
(c)
=pyZ;r 4
llIa i%4,
Ww =8.0kN = 5Ww L' /384£1,
ay 4,.
.•
= 5 x 8.0 X 5000' /(384 xx 205 302 x 104) 104) =5x8.Ox 205 xx302x
=21.0mm
Although dause any value, value, use a deflection limit of, Although clause 4.12.2 4.12.2 avoids avoids specifYing specifring any use a deflection limit of, say, 11200=25 U200 = 25 mm mm. say
2 Shear area A, == 1125mm2 Sbear area Av=0.9 =0.9xx125 125xxID ID 1125 mm 3 = 186 Shear capacity F, = 0.6 x 275 x 1225 x 10 Shear capacity P" =0.6 x 275 x 1125 x 10- = 186kN
EXAMPLE 6. B. DESIGN DESIGN OF OF MULTI-SPAN MULTI-SPANPLJRLIN PURLlN EXAMPLE
Contmuity of of aa stnictural structural element element over overtwo two or more spans may be useful In Continuity or more spans maybe useful in order to to reduce reduce the the maximum maximum moments moments to to be be resisted, resisted, and and hence hence the thesection sectIOn order size. and and to to improve unprove the the buckling buckling resistance resistance of ofthe the member. member. size, 1.1.
In general, general, the the bending bending moments moments in In aa continuous contmuous beam beam are iess thun In are less than
those in in simply simply supported supported beams beams of ofthe thesame same span. It should be those span. It should be
The The shear shear capacity capacity isis clearly clearly very very large large relative relatl ve to tothe theshear shearforce. force.
noted, however, however, that that aa two-span two-span beam beam has has the the same same moment moment(WL/8) (WL/8) noted,
the middle middle support support as as the mid-span moment of a Simply atat the the mid-span moment of a simply supportedbeam. beam. supported 2.2.
The resistance resistance of of aa member member ofoflateral lateral torsional torsIOnal buckling The buckling improved by by continuity continUity and and this this isIS reflected reflected in in BS BS table table 16. 16. improved
IS
is
Continuitymaybe may beachieved achievedby byfabricating fabricatingmembers membersofoflength equal to two Continuity length equal to two
;.
be limited limitedby byrequirements reqUirementsfor fordelivery delivery morespans. spans. Length Length will, will, however, however, be orormore -.
NA
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and flexibility flexibility dunng 4.2 aa and dunng site Site erection. erectIOn. For the the purlin designed designed in Section SectIOn 4.2 length of of not (12 m) would length not more more than than two two spans spans (12 would be be acceptable acceptable (they can can be delivered bundled bundled together togetherto to reduce reduceflexibility). flexibility). Conunrnty delivered ConnnUlty can can also also be of site site connectIOns connections capable capableoftransmlltmg of transmitting bending moments. moments. arranged by Use use of Such connections connecl1ons are costly to to fabncate fabncate and and totoassemble assemble and and are are rarely rarelyused used as pUrlins purlins and and side side rails. Using the in small structural structural elements elements such such as the same same example as as m in Section Section 4.2, 4.2, the the design design ISis repeated repeatedfor for aa purlin purlin continuous example contmuous over over of 6.0 two spans spans of 6.0 m. m.
51
Buckling resistance resistance
=
=
= 6000/22.4 6000/22.4 = 268 .lA = LE/ry = n is IS obtained obtamed from ES BS table table 16 16 for for /3 P== 0 and and}'y = M/Ala .0, The factor a Al/Al0 = -.—J j0 as supported moment are equal as the the end end moment moment AI Al and and Simply simply supported moment Mo Al0 are equal (but Opposite hence opposite sign), hence na
7.95 x 600&/(l85 o:r 6000 1/(185 xx205 205 xx 852 852 x 10.\)=5.3mm & ==7.95 = 5.3mm
Maximum MaXimum ultimate ultimate moment moment (at (at central central suppod) support)
10 4 N
---=:::c:] FIg. Fig. 4.12 4.12
(d) (d)
Deflection 6, the imposed Imposed lond state is \s From Example 6. load at at serviceability serviceability limll limit state
DII BM and and SF Sf With reference reference to iD Fig. Fig. 4.12: 4.12:
\:•>/
= 170N/rnm2 170 N/mm2 Mb == 170 170 x 130 130 Xx 10- 3 = 22.1 22.1 kNm
lV,y =2.89kW =2.89kN
oy 5' ==14.Smm 14.5mm
},{, = ~ 16.5 16.5 xx 6.0/8 =~12.4 12.4kNm kNm F, Fi =0.65 = 0.65xx16.5 16.5 = 10.4kW 10.4 kN Si. 4.5 kWm My ==4.5kNm F,, 3.8kW Fy = ~3.SkN
DeflectIOn 6000/200 = 30 mm Deflection limit limit G000/200=3Omm
Shear capacity capacity Shear Shear force force is IS less less than than 0.6 0.6 shear shear capacity, capacity, as as Section SectIOn 4.1(d). 4.Hd).
(e) ES BS
table 77
Moment Moment capacity capacity The 152 is a compact compact sectlon section(bIT= (b/T= 8.45), hence 152 x 76 76 channel channel IS hence /v! =275 '75 x 130 103=35.8kNm Ma. 130 xx 1O-3=35.8kNm Mey =9.OkNm =9.0kNm
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CRANE CRANE GIRDERS GIRDERS
53 53
1st c.
CRANE GIRDERS GJRDERS CRANE Wheel wheel
Si 5.1
+ lift lift 5.2 Overhead crane . 5.2
along lifting. The The along the the crane crane frame, frame, together together with with the the effects effects of non-vertical non-vertical lifting. value IS assessed in in l3S BS 6399t1) 6399(1) at of the the crab crab value of of this this load is at 10% 10% of the sum sum of weight wheels weight and and hook hook load. load. ItIt IS is divided divided equally equally between between the the four crane crane wheels when the the wheels when wheels are double flanged Hanged and and can can act act in in either either direction. direction. hOrizontal load (longitudinal) (longitudinal) is IS the the braking braking load ioad of the whole whole The second horizontal of the crane, and in this case acts along (lie the crane girder at the level of oflhe flange. the top top flange. The value value of of this this load load is is assessed at 5% 5% of each each wheel wheel load, load. and and is is therefore therefore aa maXImum maxunum. As before, the braking braking load load maximum when when the the wheel wheel load load is is a maximum. As before, covers covers acceleration acceleration as as well well as non-vertical lifting. 10 addition, addition, gantry gantry girders girders intended Intended The loads are are summanzed in Fig. 5.3. In carry class class Q3 Q3 and and Q4 Q4 cranes cranes (as (as defined defined in in BS BS 2573: 2573: Part Part I) 1) should should be. be to carry the crabbing crabbing forces forces given given in In clause clause 4.11.2. 4.11.2. designed for the
CRANE CRANE WHEEL WHEEL LOADS LOADS of of aa typical typical overhead overhead crane crane are are shown shown in In Fig. Fig. 5.2. 5.2. The weight weIght or load load associated associated with with each each part part should should be be obtained obtamed from from the the crane crane supplier's supplier's data, data, A and and then be combined to to give give the the crane crane wheel wheel loads, loads. Alieniaitve Altemattve wheel wheel loads loads may may be be given glven directly directly by by the the crane crane manufacturer. manufacturer. Reference Reference may may be be made made toto BS BS 6399: 6399: Part Part Ill) for Ml full details details of ofloading loading effects. effects. The The following follOWing notes notes apply apply to to single smgle crane crane operation operatlOn only. only. The The crab crab with with the the hook hook load load may may occupy occupy any any position posItion on on the the crane crane frame frame up to the minimum mmimum approach approach shown shown in m Fig. 5.2. Hence the vertical load on the nearer nearer pair pair of ofwheels wheels can canbe becalculated, calculated, adding adding an anamount amount for forthe thecrane crane frame, which ~hich isIS usually usually divided divided equally equally between between the the wheels. wheels. Maximum Maximum wheel wheel loads loads are are often often provided provided by bythe thecrane cranemanufacturer. manufacturer. An An allowance allowance for for impact Impact of of25% 25% is15made made for formost most light)medium lightlmedium duty dUty cranes cranes (classes (classes QI Qland andQ2), Q2),and andthis thisisISadded addedtotoeach eachvertical verticalwheel wheelload. load. For For heavy heavy duty duty cranes cranes (classes (classes Q3 Q3 and and Q4) Q4) reference reference should should be be made mllcie to to BS as 2573(2) 2573(2) and and to suppliers supplier"sdata datafor forappropnate appropnate impact Impact values. values. In In addition addition to to the thevertical verticalloads loadstransferred transferred from from the thewheels wheels totothe thecrane crane rail, rail, horizontal hOrIzontal loads loads can can also also develop. develop. The The first first of of these these isIS called ealled surge surge and and -rig acts acts at at nght nghtangles angles(laterally) (laterally) toto the the girder girder and and atat ihe the level level of ofthe the rail. raiL This This ,C surge surge load load covers covers the the acceleration Ilcceleration and and braking braking of ofthe the crab crab when when moving movlOg
carnage carriage
Hook:
industrial buildings house manufactunng processes which involve Industnal buildings corrunonly house rail '--L~_Crane rail heavy items being moved moved from from one one pOint point to to another during assembly, heavy Hems being dunng assembly, fabneation or or plant maintenance. In some cases fabncatlOn cases overhead cranes cranes are are the the best best way of providing providing aa heavy heavy lifting lifting facility facility covenng covering virtually virtually the the whole whole area area of of way of the building. These cranes cranes are are usually usually electrically electrically operated, operated, and ami are are provided provided specialist suppliers. The crane Crane gantry gantry gIrder girder by specialist crane is is usually usually supported on on four four wheels wheels nmnmg running _ _ Crane lUB and and plate plate on {UB on special special crane crane rails. rails. These These rails rails are are not not considered considered to to have have significant significant welded iogeiherl welded logetherj bending strength, and each IS supported supported on on aa crane crane beam beam or orgirder gtrder(Fig. (Fig. 5.1). each is The The design design of of this this girder, girder, but but not not the the rail, rail, is IS part part of of the the sieelwork steelwork designer's designer's Fig. 5.1 brief. bnef. However, However, the position position and attachment of the rail rail on on the the crane crane girder gtrder Fig. 5.1 Crane Crane gantry gantry girder must must be be considered, considered, as as aa bad bad detail detail can can led led to to fatigue fatigue problems, problems, particularly particularly glfder for heavy duty cranes. cranes, The attachment attachment of of the the rail rail should should allow allow future future adjustment to be carried out, out, as as continuous continuous movement movement of of the the crane crane can can cause cause lateral lateral movement movement of of the the raiL raiL
Parts Parts
0
so
-c
'
V a —
a
0
0g 2e a-c
0+
a,
"Pa '0
...g. 5.3 5.3Crane Craneloads loads
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-
34 STRUCTURAL 54 STRUCTURALSTEELWORK STEELWORKDESIGN DESIGN TOSS TO as 5950 5950
CRANE CRANEGIRDERS GIRDERS 55 35
.
The safety safety factor factor '11 7j for for crane crane loads loads(ultimate (ultimate limit limit state) state) isis taken taken as as 1.6, 1.6, Le. i.e. The as for for Imposed imposed loads loads generally generally (SectIOn (Section 1.7). 1.7). Whenever Whenever the the vertical vertical load load and and as the surge' surge load load are are combined combined in in the the deSign design of of aa member, member, the the safety safety factor factor the should however however be be taken taken as as 1.4 table 2). 2). Further detailed detailed should lA for both loads (135 (BS table provisions for forgantry gantrygirders girdersare aregiven givenIninclauses clauses4.11 4.1! and and 2.4.1.2. 2.4.1.2. proVISions
In glrder must In all all cases cases the the effect effect of of self self weight weight (unifomlly (unifomity distributed) of of the girder be added. added. be -
5.3
EXAMPLE CRANE GIRDER LATERAL EXAMPLE 9. 9. CRANE GIRDERWITHOUT WITHOUT LATERAL RESTRAINT RESTRAINT ALONG ALONG SPAN -
(a) 5.2 5.2
Moving loads, loads, such such as as crane crane wheels, wheels, will tvill result result in Movmg In bending moments moments and shear forces forces which which vary vary as as the loads travel along the shear the supporting supportmg girder. gtrder. In simply supported beams beams the the maXimum maximum shear shear force force will will occur immediately slmply.supported munediately support, while while the the maximum bending moment adjacent to aa support, moment will occur occur near, near, - but not not necessarily necessarily at, at, mid~span. mid-span. In In general, general, influence should be influence lineiJ ,4,5} should but used to to find find the the load load positions positioos producmg producing maximum values of of shear force and used bending moment. bending moment. The mnximum effects of of two two moving moving loads loads may maybe maximum effects be found found from from fonnulaetti fonnulae(J) in Section Sectlon 2.5. For aa simply simply supported supported beam beam the the load load as demonstrated demonstrated in positions shown shown In in Fig. Fig. 5.4 5,4 give gwe maximum maXlmum values: values: pOSitions Shear force 1112 -— clL) cIL) Shear force (max) (max) == IT'(2 Bending moment moment (max) (max) == WL/4 WL/4 Bending c/4)2/L or =2W(L/2 =2W(Ll2 — - C/4)2IL
The greater The greater of of the Ihe bending bending moment moment values values should should be be adopted. adopted. The The design deSign of of the the bracket bracket supporting supporting aa crane crane girder girder uses uses the the value value of of maximum from adjacent Simply simply supported beams, as in maximum reaction reaction from in Fig. Fig. 5.4. 504. Where adjacent spans are equal, the the reaction reachon is IS equal equal to to the the shear shearforce, force, i.e. I.e. Reaction (max) 1v12 -— clL) o'L) ReactIOn (max) == IT'(2
C
Wt2— c/LI
(tw
cfr'
ZI L
Shun three Shaar force end and resetian reaction
ZI
~ Bending Banding moment moment
Fig. Fig. 5.4 504 Maximum MaXimum PM, BM. SF SF and and It R
Dimensions Dimensions
MAXIMUM LOAD MAXIMUM LOAD EFFECTS EFFECTS
ikf L/2
Bending Bending moment moment
Span Span of of crane Wheel Wheel centres Minimum Minimum hook hook approach approach Span Span of of crane crane girder
(b)
15.0 m 3.5 3.5 m 0.7 0.7 m 6.5 6.5 m m (Simply (stmply supported)
Loading
Class crabbing forces need be calculated) calculated) Class Q2 (no crabbing Hook load 200 kN 200kN Weight 60 kN Weight of crab Weight of crane (excluding (excluding crab) crab) 270 kN Weight of 270kN
(c)
Wheel Wheel loads
VertJcal wheel load load from: from: Vertical wheel load 200(15.0 200(15.0— - 0.7)1(15.0 hook load 95.3 kN 0.7)/(15.0 x 2) 95.31<14 load 60(15.0 - 0.7)/(15.0 xx 2) crab load 60(15.0 — 28.6 kH kN 2) = 28.6 load 27014 = 67.5 crane load kN 270/4 67.SkN Total vertical vertll:al load load wheel = 191.4kN 191.4 kN per wheel VertIcal Wc (including Impact and yj) 1'/) Vertical load IV. (including allowance allowance for for impact = 1.25 x l.4x lA x191.4=335104 191.4 =335 kN =L25x Where is considered considered acting acting alone alonethen thenVj'V/ is IS 1.6 1.6 and and Where vertical load is Wc becomes becomes 383 383 kN. kN. Lateral(honzonra!) (honzontal) surge surge load load isis10% 10%ofofhook hook±+crab crab load: load: Lateral =0.10(200+60) =26.0104 =26.0kN Total TOlal lateral lateral load load =26/4 26/4 li5 kN per per wheel wheel = = 6.5 (including37) lj) = 1,4 x 6.5 6.5 9.1 kN kN Surge load Whc (including = 1.4 = 9.1 Longitudinal (honzontal) .ISis5% wheel load load and and Longitudinal (honzontal)braking brakingloat;!. load, 5%.of of wheel Including 'VI IS: including is:
-' ,.
1.6 x 191.4 19L4 = 15.3 per wheel wheel 0.05 x 1.6 15.3 kN per
follOWing design deSign itIt will will become become clear clear that that the the cntical cnticalconsiderations considerations In the the following In the support. support. Hence Hence first first sizing slzmg of of the the are lateral lateral buckling, buckling, and web beanng at the are girder would would be be based based on onthese thesecntena. cntena,Assume Assumea 610 a 610 x 305 x 179 VB girder x 305 x 179 1313 IS (grade 43) with extra extra plate (grade (grade 43) top flange. flange. This plate plate ts (grade 43) with 43) welded to top gIve additional udditional srrength Ilange which which is IS assumed to to act act used to give strength to to the the top flange alone to resist reSist the lateral (surge) (surge) loading. loading. A A channel channel section section may be preferred fiat plate; plate; this type of section was commonly commonly used in in the the past. past. Instead of the flat instead
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Dead load load due due to to self weight of of girder (1.79 + 0.36 kN/m) kN/m) and rail Dead (0.25 kN/m) kN/m) Including including YI y, is (0.25 IS
(e) (e)
6.5=2I.8kN tJ'd1.4x2.403< Wd = lA x 2AO x 6.5 =21.8k:N
i:) .1
L
r-
clause 4.2.3 4.2.3 Table 1.2
:
179kg/rn 610 x 305 x 179 kg/m UB UB
I
Fig. 5.5
(f) (0
-
clause 3.5.5 clause (d)
Design strength DeSign flange 22 mm thick: strength for for chosen sectIOn section with flange Py =265 N/mm2 p, =265
=0.6 xO.265 =0.6 x 0.265 xx 617.5 617.5 x 14.1 14.1 = 1380 kN = 041 <<0.60 F,IP,., =0.41 0.60 Shear capacity capacity P", P =0.6 Shear kN =0.6 xx 0.265(307 0.265(307xx23.6+280 23.6+280xx120)=2030 l20)=2030kN F /P, =0.01 -"'yIP,y 0.60 =0.01 <<0.60
:
280 xx 20 plaID plate (36 (35kgimJ kg/nI
Shear capacity Shear
Shear capacity capacitypPs., =0.6p,A, Shear V.T =0.6p'y Av
-
25kg/rn 25 kg/m rail rail
75L-~~
57 57
Moment capacity The chosen chosen section (Fig. 5.6) IS a sechon with 5.6) IS a plastic plastic sechon
bif (internal) bIT (internal) =280/20 = 14 14 bif (external) =75.5110.4 bIT =75.5/10.4 =6.5
BM and SF BM SF Moment due to vertical vertical wheel wheel loads loads is is either W,Ll4 =335 x6.5/4=544kN :r x 6.5/4 =544 kN 3.5/4)2/6.5 or 20ç(L/2 21V,(LI2— - c/4)2/L c14)'IL =2 =2x335(6.5/2 x 335(6.5/2— - 3.5/4)'/6.5 =581 kNm (664 (664kNm = 581 kNm kNm when acting alone) alone) Moment due 10 1' to dead load load =2l.5x6.5/8=IBlcNm =21.5x6.5/8=18kNm Max. ultithate moment M", M, =581 ultimate moment =581++18 18 =599 kNm (682 kNm when acting = 599 kNm (682 kNm acling alone) Although and the wheel Although the dead load load maximum maXImum BM occurs at mtd-span, mid-span, and maximum occurs aa distance maxmlllm distance c/4 cl4away, away,itItisISusual usualtotoassume assumethe thevalue valueofofA-fr Al", to 10 -" the sum of be the of the the maxima maXIma as as shown. shown. Moment due to to surge surge load load =2 =-2xx9.1(6.5/2 9.1(6.512— - 3.5/4)2/6.5 3.5/4)216.5 Moment due
Shear force due due to to vertical vertical wheel wheel loads loads is: IS: (V (2 (2— cIL)==335a IV, - dL) 335(2— - 3.5/6.5) =490kN =490kN (560kN (560kNwhen when acting actmg alone) alone) , Vertical shearforce forcedue duetotodead dead load2I.8/2=1l V~rtlcal shear load=21.8/2= 11 kN kN Max. ultimateshear shearforce forceF",F,=490+ =490+ II Max. ultimate 11 =501 =501 kN kN (571 (571 kN kN when when acting acting alone) alone) LaternI Lateral shear force due to to surge surge load=9.1(2 loarl=9.H2—- 3.5/6.5)=13.3kN 3.5/6.5)= 13.3 kN Max. 13.3 kN Max.. ultimate shear force force FI' = =13.3 kN Max. ultimate reaction R1 =490+21.8 = 512 kN reactIOn R", =490+21.8=512kN
1?. R, ==I3.3kN 13.3kN
'I
I'll
r.o'--L~=J -.~-::las~cNA I
F—
I
,
I
j - 4r.
kNm = 15.8 15.8kNm Max. ultimate moment My = 15.8 Max. 15.8 kNm
,"0 280
I'
'—j__
—
-
FIg. 5.6
ic :
.:,' 4
:'r-:
—-—Plastic NA taxis tax1s ot 01 equal aria) aroa)
the built~uf. built-uy section chosen. chosen, the deSigner designer may need to calculate the For the plastiC the plastic modulus '1), if th.is IS (S:c)' .7}, if properties modulus (5,)t tids is not available in published tables. The properties may be obtained may obtamed from from formulae formulae given in Appendix Appendix A. A.
=280 x 20 Area of of plate pia le Ap =280x20 =5600mm =5600mm22 Total area areaA Total A =228+5600 1O- 2 =284cm2 =284cm 2 =228+5600xx l0_2 Plastic section propeHles properties:: dp =56001(2 =56001(2 xx 14.1)= 198.6 mm(i.e. (i.e.l3Omm 130 mmbelow belowtop top face) face) l4.I)= l98.6nun S, =5520+[14.t =5520+[14.txx198.62+5600(617.5/2+20/2 198.6'+5600(617.512+20/2 — - 198.6)JI0-' =6750 cm3l =6750cm flangeonly) only)=(23.6x3072/4+20 Sy (for top top flange = (23.6 x 3072 /4 + 20 x 280 2/4) I 0-:> =948cm33 =948cm Elastic ElaSllC section sectIOII properties: properties: -
d, =5600(617.5+20)/(284 =5600(617.5+20)1(284 x 102) ID') 4 152 000+228 x 6282 62.8' x 10-' 4I, = 152000+228 +5600(617.5/2+20/2 + 5600(617.5/2 + 20/2 -_62.8)2x 62.8)' x 10-'
TorSIOnal index index may may be be calculated calculated using usmg the th!! appropnate appropnatemethod methodinin Torsional BS appendix appendix 8.2.5.1(c). B.2.5.l(c). 85 h. =617.5+20 ~617.5+20— - 23.6/2 — - 43.6/2=604mm 43.612~604mm )i. rh l +hwtw ==totol area=28400mm2 400mm2 .Eb, total area=28 Eb? + h".?,~ =2 = 2xx307 307 x23.63+280 X 23.6] + 280 x20.03+570.3 x 20.03 + 570J xX 14.1 3 11.91 xlObmm4 x 10b mm 4 ==11.91 ~604r284001(11.91 400/(11.91 xx ID')]'" ~29.5mm xx =604[28 I06)]a =29-5mm
Hence the section section isIS satisfactory. satJsfactory.
8
Fig. 5.7 5.1 FIg.
Wj
.'",
,
.1
n./2
I
(Ii) (h) Rail piatâ plato
clallse clause 4.11.3 BS 13 55 table 13
girder glfder must be be checked for local bucklingtt) buckling(8) (see Fig. Fig. 5.8). 5.8). If Ifnecessary, necessary, load load Introduced to prevent prevent local local buckling buckling of ofthe theweb. web. canymg stiffeners stiffeners must he introduced Dl2', I carrying J-L..L_.-.I-~~t-
rI
-'
b,
45"
;f. 1 599/1790+ = 0.40 599J1790+ 15.8/226 15.8/226=0.40 I
FIg. Fig. 5.8 5.8
M/Ma MiMer ==6831!790~0.38 683/1790 = 0.38
clause4.11.5 clause 4.iI.5
Hence sectIOn chosen is IS satisfaciory. sails factory. Hence section
clause 4.5.2.1 452.1
BS 115table table 9
The buckling resistance Sectlon 3.8(0, The buckling resistance may may be be found found in in the the same same way way as In in Section but allowing a!lowmg for the destabilizing destabilizmg effect of ofthe the surge surge load. load. Hence Hence hut
D/2
=i.0 =1.0 No restraint restramt is IS provided between between the the ends ends of 0 f the the girder. girder. At the supports the diaphragm gives partial restraint restramt against agamst torsion, torsIOn, hut but the the compression compressIOn flange diaphragm IS not not restrained reslramed (Fig. (Fig. 5.7). 5.7). is
0/2
=11
LE ~1.2(L+2D) = l.2(L +20) ~i.2(65+2xO.637)=9.J3 =1.2(6,5+2 x 0.637)=9.33
Dispersion length under wheel DispersiOn lengtlt wheel b1=2x75 =150mm b l =2x75 nt =617+2 x 20 =657mm n!=617+2x20 Web slenderness A slenderness..l
Buckling resistance
m
At load (wheel (wheel loads loads or or reacttons) reactions) the the web web of At points pomts of concentrated concentrated load of die Ihe
,n,/2. I
BS table 27c ~tg) -(g)
Web buckling buckling
D/2
Actmg alone without surge surge Acting
,E
II
599/864 + 15.8/(265 x 609 5991864 609 xx 10 -') ~ 0.97
45"
Combined local capacity check check Combined local
1-
Acting Actmg alone alone without without surge surge
M"", '/1.4 1.4 x265 x 265 xx5320x 5320 x W- J =1970kNm = I97OkNm Ma MC)' =Py Sy where S,,is SylS for for top top flange flange only only =p, 5,, =265 x948 x 10- 3 = 251 kNm 25lkNm usingconstant constant1.4 1.4asasnoted notedabove above and; for i.4Pr ZyZ,.usmg and.t;. for top top flange flange But MC}. ;f.4 1.49,.
J.H~M"x + + M/Al"y
¥LT =nuvA =IIUVA. Ytr
=1.OxO.9x0.80x = LO x 0.9 x 0.80 x 128=102 128= 102
MC)( =PySx Ma 'Py3'x =265 x 6750 x 1O-J=1790kNm =265x6750x plastiCIty at at working working load load (see (see Section Section 3.4), 3.4), M=;f.IA where To prevent plasticity 1.4 py p,.Z~ 4 where factored factored loadlunfactored load/unfactored load= load= 1.4 i.4
Diaphragm Diaphragm
Aix 128/29.5 ~4.3 =4.3 ,vx ~= 128129.5 N N =lcfl(lcf+lif) =9350/(9350 ~93501(9350++5690) 5690)=~0.62 0.62 v =0.80 itu == 1.0 1.0 (conservatively) {conservatively)
=2.5 xx 537/14.1 195 =2.5 537114.1 ==195 Compressivestrength strengthPcP. == 131 N/mm2 Compresslve Buckling reSistance resistance p,~ P., ==(b1 Buckling (b l + "l) tPC t Pc +nt) =1150+657)14.1 xO.131 =(150+657)14.1 xO.131 =i49OlcN =1490kN Max. wheel load Max. load = 383 kN Hence buckling resistance resistance is IS satisfactory. satisfactory.
Minimum beanng length length reqUired required at at support Mimmum stiff beanng b =F~ 1'x ,tUPc) n, mm b! I(tpc) —' - III F, =571 Fx =571 kN kN (support (support reaction) reaction) 309 mm nl=309mm b I=573/(l4.i 0.131) 309= -lOmm — 10mm ~5711(14.1x x 0.131)— - 309~
i.e. no is required at support support for web beanng. I.e. no stiff bearing beanng IS reqUired at
FIg. Fig. 5.9 5.9
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________________
00 STRUCTURAl. STEEL WORK DEBtON TO ES IBID
60
STRUCTURAL STEELWORK DESIGN TO BS 5950
CRANE GIRDERS 61 CRANE GIRDERS 61
Web bearing
(i) (i)
Web bearing
(k) (It) Connection Connection
At thesame samepomts pointsthe theweb webof ofthe thegirder girdermust mustbe bechecked checkedfor forlocai localcrushing(8) At the (see Fig. 5.10). If necessary,beanng beanngstiffeners stiffenersmust mustbe beintroduced introducedto to prevent prevent (see Fig. 5.10). Ifnecessary, localcrushing crushingof of the the web. web. local clause 4.11.5 Load dispersion under wheel = 2(75+43.6+ I6.S)=270mm douse 4.11.5 Load dispersIOn under Wheel = 2(75 +43.6+ 16.5)=270mm
The Thevertical verticalforces forcesare aretransmitted transmittedtotothe thesupporting supportingbracket bracketby bydirect direct beanng beanng from surge load (13.3 and (Fig. 5.11). HonzontaheacllOns are present (Fig. 5.11). Horizontal-reactions are present from surge load (13.3 kN) kIN) and honzontal horizontalbraking braking(I5.3 (15.3kN). kN).The Thesurge surgeload loadisistransmitted transmittedtoto the the column column by by the thediaphragm diaphragm(Fig. (Fig.5.7). 5.7).The Thebraking brakingforce forcewill will be bc transmitted transmitted by by nominal nominal bolts; bolts;provide, provide,say saytwo twolv120 M20 boils bolts (grade (grade4.6). 4.6).
Deanng capacity =270 xx 14.1 14.1x x0.265= 0.265=IOa9kN Beanng capacIty PCrip =270 1009 kN Maximumwheel wheelload load==383kN Maximum 383 kN SA 5.4
The Thedesign designininSection Section5.3 5.3 may may be berepeated repeatedbut butinCluding including aalattice lattice rcstramt restraint 10 to the been the compressIOn compressionflange. flange.InInprachce practicesuch suchaalattice lattice girder girder may may have have been provided purpose, or provided specifically specifically for for this this purpose, or to to support support access accessplatfonns platforms or or walkways walkways(Fig. (Fig.5.12). 5.12).InInmodem modemUK UKpractice practiceItit ISisrarely rarelyeconomic economictotoInclude include such reduce the the beam beam Size, suchaarestramt restraintIII in order order to to reduce size, except except III in heavy heavy mdustnal mdustnal buildings. buildings.
,'\/45'
Rail
Rail flange
15 /cJ45° $75 , 436 •\ . —.-x— 16.5 '..,:,. 16.5
~~~(~~'~~~ 7...L
fl ange
4-plate +pi,t,
~
.
/
Root ol
Rool fillet01 fillel
EXAMPLE EXAMPLE 10. JO.CRANE CRANEGIRDER GIRDERWITH WITHLATERAL LATERAL RESTRAINT RESTRAINT
Under wheel wheel Under
lii
~-L-'6.5 ~=t=23.6
(a) (a)
16.5
Dimensions Dimensions
in 2.5
l1
Fig. 5.10 Fig. 5.10
t At support support At
As Section SectIOn 5.3. 5.3. Fig. 5.12
(b)
Load dispersion at at support: Load disperslOll support:
M Section for the the changed changed self weight). As Section 5.3 5.3 (making (making no no correction correction for self weight).
Minimum stiff — Mimmum stiffbeanng=Fxl(tpyw) - 112 n2=(23.6+ ~(23.6+l6.S)2.S=lOOmm 16.5)2.5~ 100mm F1=571 Ft = 571 kN kN (support (SUpport reaction) reacllon) b ,=571/(14.1 x 0.265) — 100=53mm b! ~571/(l4.1 x 0.265) - 100~53 mm Web beanng of Web beanng beanng at at the the support support requires reqUIres aa minimum muurnum stiff stiffbeanng of53 53 mm; mm; check check that the supports provide this mimmum stiff beanng to prevent web beanng that the supports provide this mlIllmum stiff beanng 10 prevent web beanng capacity of the girder from being exceeded (see Fig 5.10). capacIty of the girder from bemg exceeded (see Fig 5.10).
n,
(c) (c)
(j)
Wheel toads loads Wheel As Sectioo Section 5.3. 5.3. As
Connection As Section 5.3 but the the lattice lattIce transmits transmllS horizontal honzoniai forces forces to 10 the the support: support: hence hence tbe is not not needed. needed. the diaphragm is
Moment Moment capacity capacity
(j)
LattIce Lattice I
The chosen US is The chosen UB IS a plastic section section (bIT ==5.0) 5.0) Moment Momentcapacity capacity,\lex Ala == p,.Sr =0.265 xX 3200= 3200= 849 849 kNm
The lattice girder which provides The provides restraint restramt to to the the crane ·crane girder gIrder top top flange flange isIS . loaded by
(13.06kN kN when when actmg acting alolle) alone) which which is (i) the the restraint restramt force force of of11.45 11 AS 114 k.N 03.06 IS considered to to be be distributed distributed between between the the nodes nodesof of the the lathce; lattice; considered wheel (10.4 (JOAkN when acting actmg alone). alone). (ii) the surge lond ihe surge load of9.1 of 9.1 kN per wheel tiN when
The of the the lateral lateral moment momem M,. M... is IS considered considered later 0) in m The effect effect of later In in part U) combination with with aa restraint restramt force force equal equal 102.5% to 2.5% of ofthe the flange flange force. Note that the value value of of 1% tn the Initial initial version consideredto lobe In the version bf bfclause clause 4.3.2 4.3.2 was considered be too the small.
fu 5 6 7S!9 123456789 <\
ES BS tab/e table 27c 27c
Panel
Minimum beanng length Mimmum stiff stiffbeanng length aia!support support to toresist resist reaction reactIOn of of571 571 tiN. kN. 544.ô/2272mm III =544.612 =272 mm ft A x477/l2.893 A. =2.5 =2.5 x477111.8=93 z Pr Pc ==134 134 N/mm2 Nlmm Mm. beanng length b, b l =571/(12.8 =571/(12.8 xO.134) x 0.134) -—272=61 mm Min. beanng 272=61mm
Bottom chord Bottom
Diagonal
17.1
Post 0 0
I
0 0
21.2
22
21.2 11.2 38.3 51.3 51.J
38.3
51.3
18.4
13,0 13.0
0.3 11.9 11.7
0.3 0 4.3
25.7
51.0 51.0 46,6 46.6 38.2
0
25.7
36.3
46.6 38.1 381
6.1
8,4 8.4 12.5 12.5
0
The applied forces forces may act m either either direction directIOn and and hence hence the the member member forces forces may be either tension tenSlOn or compression. compressIOn. Designing Deslgnmg the chord members members for for aa compression of of 51.3 51.3kN; tiN; use use aa 45 35xx 45 45xx 66 equal equal angle compressIOn
Mm. Min. beanng lengthb, b=l 57 =5711( 12.8 X 0.265) - =8584mm =84 mm beanng length l/( 12.8 x 0.265) — 85
ES iable table 27c BS
L5Ir,,.t.0x813/8.6794 A = L£/r A. rr = J.O x 813/8.67 = 94 Pc = IJ5 N/mmz p =I35N/mm2 Compression resistance reSistance Pc =.A. Pc =Agg Pr =5.09x l35x 135 x 1D- 1 =69kN the diagonals diagonals for for aa compressIOn compression of of 36.2 36.2 kN tiN using single Deslgmng usmg the same same smgle Designing the angle:
Deflection
Wheel load load= Wheel =191.4kN 191.4 kN Calculation as as m in Section Section 5.30) 5.3(j) gives gives o=9.0mm b=9.Omm CalculatIOn Limlt=65oo/600= 10.5 mm mm Limit=6500/600= 10.5
iop chord Top chord
30.0 24.2
33 44 5 5 6 6 7 8 8 9 9
n,=(2l.3 nz =(21.3 + + 12.7)2.5 = 85 mm 85mm
(h)
45' 45.
The truss is analysed analysedby bygrnphics. graphics,calculatiOn, calculation,or orcomputer, computer,glvmg giving member truss IS forces as as tabulated tabulated (kN): (tiN): forces
Web buckling/bearing buckling/bearing
stiff beanng length Therefore, aa stiffbeanng length of ofatatleast least 84 84 ruin mm isISneedcd needed at at the the supports to bearing strength strength for for the give adequatc adequate beanng the girder.
pv0Z1St\t\I'0l bays al at 813.5 813.5mm 8 bays mm
623mm
Fig 5.13 Fig.
9.1 IN
9.1 1N
Fig. 5.14
Flange force may be be estimated estimated as: as: 599/0.523 599/0.523 = 1145 1145 kN (1306 tiN kN when when Flange force tiN (1306 acting actmg alone); alone); see see Fig. 5.13. 5.l3. Restraint force force = = 28.6 28.6kN tiN (32.6 (32.6 kN tiN when actmg acting alone) Restramt alone)
(g)
I
ES table 27c BS
= 1.0 1.0 xX 8l3/cos45°= 813/cos 45° =1150mm 1150 mm Effective lengthLe length L~ = =1150/8.67=133 "A = 1150/8.67 = 133 =83N/mm2z Pc =83N/mm Pc =5.09x83xlO- I =42kN Pr
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64
64
STRUCTURAL STEELWORI< DESIGN TO BS 5950 STRUCTURAL STEELWORK DESIGN TO BS 5950
ThebasIc basiclattlcc latticemember memberISistherefore thereforeaa45 45xx45 45xx66equal'angle equal angleand andmIght mightbebe The fabncatedasasa awelded weldedtruss. truss.For Fora amore moredetailed detailedconsiderahon considerationoroftruss trussdesIgn design rabncated reference should be made to Chapters 6 and 12. reference should be made to Chapters 6 and 12. C
Web buckling and 8. bearing bearing 9. Deflection formulae 9. Deflection fonnulae
TRUSSES TRUSSES
References References 556399 Design Loading Loadingfor forBuildings Buildings as 6399 De.ngn Part Deadand andimposed imposedloads loads(1984) (1984) PIUt I:I:Dead US 2573: Part Rulesfor far the the DesIgn Design0/ ofCranes: Cranes: BS 2573: Part i Rules Specification farclassificatIon. classification,stress stresscalculation calculationand and Specification/or design criteria of structures (1983) 0/ stnlclures (l9B3) desIgn crueria Marshall W.T. J\IarshlllJ W.T. & & Nelson Nelson ELM. H.M. (1990) (1990) Moving Movmg loads loads andinfluence influencelines, lines,Structures, Snctures, pp. 79-106. 79—106. Longman and. CoatesR.c., R,C.,Coutie CoutieM.G. M.G.&& Kong Kong F.K. F.K. (1988) (1988) Coates Mueller-Breslaus ponclpie. pnnciple. Model Structura: Mueller-Breslau;s Model analysis, anaiysls, Structural Analysts, pp. 127—31. Van Nosunnd AnalysIs, pp. 127-31. Van NostTand Remhold Wang c.K. C.K. (1983) (1983) lnfluc;nce Influqncc lines lines for for statically stancally Wang deten-ninatebeams. beams,lntennediate Jntennediate Structural Srrucrural AnalySIS. Analysis, delenmna!e pp. 459-67. McGrnw-HiIl McGraw-Hill pp. Marshall W.T. W.T. & & Nelson Nelson H.M. H.!U.(1990) (990) Plaihc PlastIc bending, bending, l\1arshllIl Structures, pp. Longman pp. 532-6. Longman S(ructflres, Home M.R., MAt., & & Morris Design0/ of Horne Morris L.a. L.J. (198l) (1981) Plastic Plasllc Design Low-Rise Frames, Collins Low-Rise Frames. Collins Dowling PA.. & Owens Owens G.W. G.W. (1988) (1988) Dowling P.J., Knowles Knowles P. P. & Structural Sleel Steel Design. Construction Institute Structural Design. Steel Steel Construction Instltule (1992) Design (1992) DeSign theory, theory, Steel Steel Designers' DesIgners' Manual Manual pp. 1026—50. Black-well pp. 1026-50. Blackwell
" '4' -
-
Trusses steel sectlOns Trussesand andlattice latticegirders girdersare arerabncated fabricatedfrom from the the vanous various steel sections by bolting gusset available, available, Jomed joined together togetherby bywelding weldingororby bolting usually usually Via via gusset (connecUng) the trusses trusses act plane and are usually (connecting) plates. plates. Generally Generally the act in in one one plane and are usually designed pm-Jomted frames, some mam members may be designed designedasaspin-Jointed frames,although althoughsome mammembers may be designed Where members lie in three the truss truss IS known as a as continuous. as continuous. Where members lie in three dimenSIOns dimensions the is icaown as a and lattice gIrders are particularly long spans, as space space frame. Trusses Trusses and lattice girders are particularly SUited suitert to to long spans, as commonly used in bridge they they can canbe bemade madetotoany anyoverall overall depth, depth, and and are are commonly used in bridge construction. construction,InInbuildings buildingsthey theyhave havepartlcuiar particularapplicatIOn applicationfor forroof roofstructures, structures, and members supporting heavy loads loads (columns from floors above) and for and for for members supporting heavy from floors above) and for members longer spans, spans. members having having longer The leads to saving in weIght of steel The use use or of aa greater greater overall overall depth depth leads to aa large large saving in weight of steel a uruversai beam. This saving of matenai cost can offset the compared with compared with a universal beam. This saving of matenal cost can offset the extra costs in In certain certam cases, cases. extra fabncatJOn fabncation costs
'
,
6.1 6.1
.
-x
\
TYPES OF THEIR USE USE TYPES OF TRUSS TRUSS AND AND THEIR selection of or roof rooftrusses trusses is IS shown tu Fig. 6.1, where the roof slopes and A selection shown in Fig. 6. I, where rhe roof slopes and spans the members. members. Hipped Hipped spans dictate dictate the the shape shapeor of the the truss truss and and the the iuyaut layout of of the trusses are spans, economically up to 6m, the lattice girders trusses are used used ror for small small spans, economically up to Gm, the lattice girders and the the mansard mansard for for large large spans. Such trusses are jjghtly for medium medium spans, spans, and for spans. Such trusses are lightly by snow snow and and wmd load, together together with with a small allowance for servIces. loaded by loaded wind load, a small allowance for services. unusual for for lifling liftingfacilities facilitiestotobe besupported supported from from the roof trusses. The It11 isis unusual the roof trusses. The resulting member member and and connection connechon sizes sizes are are therefore therefore relatively relatively small. small. resulting be used used in multi-storeybuildings buildings where column loads Heavy trusses trusses may Heavy may be in multi-storey where column loads from die thefloors Hoorsabove aboveneed need totobe be earned. carned.Examptes Examplesofofthese these are shown In Fig. from are shown in Fig. ofthis this type carry very heavy loads, and are Similar in layout and 6.2. Trusses Trusses of 6.2. very heavy loads, and are similar in layout and member size size totobridge bridgestructures. structureS. member common method method of ofproviding providing stability stabilitytoto a building, whether Single or AAcommon a building, whether single or multi-storey,isistotouse usean anarrangement arrangementofofbracing bracmg members. These are multi-storey, members. These are carry the the horizontal horizontalloads, loads, such as wmd, to essentially formed formed into intoaa truss truss and and cany essentially suds as wind, to by actmg as honzantal and vertical frameworkS. Examples the roundattons the foundations by acting as honzontal and vertical frameworics. Examples are shown m Fig. 6.3. Bracing IS considered in more detail in Chapter 10, and are shown in Fig. 6.3. Bracing is considered in more deiail in Chapter tO, and graphicallyillustrated illustratedininChapters Chapters1212and and1313 In terms of the overall stability graphically in terms of the overall stability of smgle-storey buildings. of single storey buildings
9,'
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sections and graphical means sections means are are all all suiiable0'2' sUllablefl .11— - or computer techniques. arrangetechniques. Several Severalanalyses analysesmay maybe beneeded neededwhere wheredifferent diflèrent arrange. menu of and wind loading must must be considered, of dead, dead. imposed Imposed and wllld loading considered. ments (b) Analysis (b) AnalYSIS of of the the load load beanng beanng member member such as us the rafter rafter as as aa continuous continuous beam ins. In beam supported supportedatat the the nodes nodes and and loaded loaded by by the the purl purlins. In cases where the load positions raftermoment momentmay may be be taken as the pOSitions are uncertain uncertam tIme the rafter WL/6 (clause 4.IOc), 4.1Cc),where whereIVII'isisthe thepurllll purlinload loadand andL1.ISisthe the node node 10 0 IVLl6 (clause node length peqendicular perpendicularintoIV. IV. Ic) Assessment (c) Assessment of ofstresses stresses due due to to eccentricity eccentricity of of ihe IIle conneciions. connecuons. Ideally Ideally the centroidal ceniroidal axes axes of of members members should should meet meet atat the the nodes. nodes. Where Where this the [his is llot not possible the members members and and connectJOns connectionsshould should be be deSigned designed for for IS possible the the moments moments due due to (lie the the eccentncity, eccentnclty, ififsignificant. Significant. Cd) Assessments Assessments of of the the effects effects ofjomt of joint dgidity 'and deflectiOns. deflections. Secondary (d) rigidity -and stresses become Important In trusses having havmg short thick members, members, strcsses become important in some some trusses but may be neglected neglected where where more fdause more slender slender members members are arc used used (clause 4.10).
latimee lalllce
lipped Hipped
Fig. 6.1 6.1 Roof Roof trusses trusses
TRUSSES
1
The overall analysis involve the the summatIOn summation of analYSIS of the truss will ilterefore therefore IIlvo!ve of two two or more effects. Analyses (a) (a) and and (b) (b) must most alwnys always be considered, considered, while (c) effects. Analyses (c) and Cd) may be be avoided avoided by by meeting (d) may meetmg cenain cemun conditions. conditions. 6.3
SLENDERNESS OF OF MEMBERS MEMBERS The slenderness of aa compression member (a (a strut) is slenderness A l of compressIOn member IS given given by by ).=Lclr =
where effective length about (lie the appropnate axis aXIs where LE LE IS is the effective length of Ihe the strut strut about r is \s the the radius radius of of gyration gyratIOn about the appropriaie appropnate axis aXIs Mutii'stnrey Multi.storey building building
Fig. 6.3 6.3 Bracing Bracmg
6.2
Singie-sioray building Single-slOrey building
LOADING AND ANALYSIS ANALYSIS Loading Loading will will consist of of dead, dead, Imposed Imposed and wind wmd loads loads as as described described in in Chapter 2. 2. Combinations CombinatIOns of of loads loads giving glvmg maximum maximum effects in in individual mdividual Chapter members must be considered y,- must be considered (see (see Section Section 2.4) 2.4) and and safety safety factors factors},! mcluded (Section 1.7). 1.7). included to the the truss truss by by other other members members such such as as usually be be transferred transferred to The loads will usually I .2) or or by by beams In in the case of (Fig. 1.2) of a floor floor truss. A wtnd wmd bracing bracmg will will purlins (Fig. be loaded by the the gable gable posts; posts; or or by by side side members members such such as as the the eaves eaves beam. beam_ itItisIS ideal if the ideal the loads can be transferred transferred to the the trust truss at at the the node node points, pomts, but but In Fig. Fig. 1,2) 1.2) this this isIS aol not possible. possible. In In roof rooftruss truss design deSign the the commonly {as (as shown in purlin positions pOSitions may not not be be known known initially, initially. and and allowing allowmg for for the the possibility possibility of purlin purlin changes changes dunng future future re-roofing, re~roofing. aa random random position pOSition for for loads loads isIS olien allowed. often The analysis analYSIS therefore involves mvolves several stages: stages: Analysis of the truss truSS assuming asswrung ptn-joints pm-Jomts (except (except Viereodeel Vierendeel trusses) trusses) and and (a) AnalYSIS loading loading at the nodes. nodes. This proceed using usmg manual manual methods methods —joini - JOInt resolutlOn, This may proceed resolution, method method of of
The requirements clauses 4.7.2 and and 4.7.10 4.7.10 define define the the effective effective lengths lengths of reqUirements of cia uses 4.7.2 the chord and internal inlemal members trusses and are illustraied the chord members of trusses illustrated in Figs. Figs. 6.4 6.4 and and 6.5. These These reqUIrements requirements take take mto into account account the the effect of 6.5. of tlte the nodes nodes (loinis) GOlnls) In she the which divide the top chord into into a number number of in~Dlane in-plane effectl\'e effective lengths. in lateral (out-of-plane) (out-of-plane) directton direction the the purlins purlins reslram restrain the the top top chord. The lateral The radius mdius of appropnaie 10 to aa given strut depends depends on on (he ilic possible possible aXIS axis of of of gyration gyralloo appronnate gIven strut buckling, and these are shown in III Figs. Figs. 6.4 6.4 and and 6.5. 6.5. The e(fecttve effecuve lengths of of discontinuous discontmuous struts are increased Illcreased where wil
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:41
68 STRUCTURAL STEELWORI( DESIGN TO 85 5950
66
STRUcTURAL STEELWORK DESIGN TO BS 5950
TRUSSES TRUSSES69 69
':'c should should not notexceed exceed50. 50.These ThesereqUirements requirementsarearegiven givenIninclauses clauses4.7.13.t(d) 4.7.l3.t(d) and and4.7.9(c). 43.9(c). Slenderness Slendernessfor for any any strut strut should should not not exceed exceed 180 180 for forgeneral general members members resistmg resistingioads toadsother otherthan thanwmd windloads, loads,or or250 250 for for members members reSistIng resistingself self weight weightand andwmd windload loadonly only(clause (clause4.7.3.2). 4,7.3.2). For Foraa member membernomlally normally acting acting as as aa he, tie, but butsubject subjecttotoreversal reversalof ofstress stressdue dueto to Wind, wind, the the slenderness slenderness should should not not exceed exceed350. 350.InIn addition, addition,very verysmall smallsecttons sections should should be be avoided, avoided, so so that that damage damagedunng duringtransport transportand. anderectIOn erectiondoes does not not occur, occur. e.g. e.g.aa llllrumum minimum size size of of angle genemlly. angle would would be be 50_x 50 50 xx 66 generally.
A maximum at l'" maximum 01 0.85s,/rucr si/mar s2lrt'Y s,/m1,. 0.85
The 7. The The The compression compressionreSistance resistanceof ofstruis strutsISis discussed discussed also also III in Chapter Chapter 7. A. and strength compresslve compressive strength strength Pc p0 depends depends on on the the sienderness slenderness A and the the deSign design strength Pr' p,,. Tests Tests on on aXIally axially loaded, loaded, pm~ended pin-ended struts struts show show that that thelt their behavIOur behaviour can can be be represented represented by byaanumber numberof ofcurves curveswhich whichrelate relatetotothe thetype typeof of sectIOn section and and the the axlS axis of of buckling. budding. These These curves curves are are dependent dependenton onmatenal matenal strength strength and and the the the Inelasttc milial initial unperfechons, unperfections. which which affect affect the the Illeiastlc inelastic behaVIOur behaviour and and the inelastic buckling obtamed from from one onc of of four four strut strut buckling load. load. For For deSign design the the value value of of Pc p. isIS obtained by curves curves or or tables tables (BS lBS tables tables 27a 27a toto 27d). 27d). The The appropnate appropnate table table ISis chosen chosen by reference and thickness, thickness, and to to the the iiXJS buckling (US (SS reference to to section section type type and axis of of buckling table 25). 25). The compression compressIOn resistance resistance Pc IS is
liii
RHS RHS
r,~ VI
-\~ u· I V
Fig. 6.4 A for continuous
Fig. 6.4 .-. for conUnuous
chords
chords
COMPRESSION COMPRESSIoN RESISTANCE RESISTANCE
X
U
Single
S'ogl, engte angle
A=meximum at 1", maximum 01 0.7 sllrwor sprw 0.7
either Pc =.4 g Pa for slender slender sections secttons (see (see Section SectIOn 1.7) 1.7) either P0 Pci for or Pc other sections. secltons. P0 =Ag Pc for all other the gross sectional sectIOnal area where Ag is is the Pa IS compresslve strength based on on aa reduced reduced design deSign is the the compressive strength based strength (clause (clause 3.6). 3.6).
L tar
discontinuous strut
L?
63 6.5
CAPACITY TENSION CAPACITY The tension tenSIOn capacity of aa member member is IS The capacity Pr F, of
VI
F, Pr
VI
X X=jiF XjipX —.
Doubie Doublo
'
-— angle angle
A maximum at ,4. ""= maxImum 01 + 30 0.85 0.85Ur,,,a, Urmln or or 0.7 0.7tim,, Ur",,+ 30
vI vi
\, !
V
X
Fig. 6.5 .4 for Fig. 6.5 A for discontinuous
discontinuous struts
struts
-aPt 1:;?
IV
U /,U
.
Single Singla
1",maximum maXImum ot 01
0.85 angle 0.85 Ur..... or 0.7 Urrr + 30 ./ .·x angle U vi V Member connected to gusset lparallelV—Vl Member connected 10 gusset !paralleIV-V)
by bytwo twoor ormore morebolts bolts
=A~py
where where A~ is the the effective effectIve sectional sectIOnal area area as as defined defined in in clause dause 33.3. 3.3.3. Where aa member member is IS connected connected eccentrically eccentncally to to its Its axis aXIs then then allowance allowance Where should be be made made for for the the resulting resulting moment. moment. Alternatively, Altemattvely, such such eccentric eccentnc should effects may may be be neglected negiected by using uSing a lower value of of the the effective efTective area area A~. For For effects single angle angie connected connected through through one one leg: leg: aa single A~ =a, +3al a21(3al +a2) .ir=ai+3ai a2/(3oi+a2)
where where al isIS the the net net sechonal sechonal area area of ofthe theconnected connectedleg leg a2 is IS the tlle sectional sectlOnaj area area of ofthe the unconnected unconnectedleg leg a3 Full details details of ofthese these reduced reduced effective effective areas areas are are given given in In clause 4.6.3. Full clause 4.6.3.
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_____-t •1•
70 STRUCTURALSTE STEELWORK DESIGN TO 5950 70 STRUCTURAl. ELWORK DESIGN TO 8S 555950
b.b 6.6
TRUSSES 71 TRUSSES 71
CONNECTIONS
m Hi ===-1:
ConnectIOns one member Connections are arc reqUired required to to JOin join one member to to another another (internal (internal joints), joints), and to connect jOints). Three main connect the truss truss to to the the rest rest of of the the building (external joints). of connection connectJon are are used: used: types of •• •o ••
,$ F
, I
-0-
I',
bolting bolting to to aagusset gusset plate plate welding to to aa gusset gusset plate to member member welding member member to
c:
these are are shown connectIOn type is is Examples Examples of of these shown 10 an Fig. Fig. 6.6. 6.6. The The chOice choice of connection often made made by the fabncator, fabncator, and and will willdepend depend on on the the available available equipment, eqUipment, oftruss truss and and with weldingbecoming becommgmore moreeconomical economicalthe thelarger largerthe the number numberof with welding member repetitions. repetitions.
at
brF
-0-
-0-
b
3O'fl~r
,I
G:I
/ 'T' '''{''''PI,(~t' ~
6.7 6.1
EXAMPLE 11. II. ROOF WITH SLOPING EXAMPLE ROOF TRUSS TRUSS WITH SLOPING RAFTER RAFTER
(a)
Dimensions (See Fig. Fig. 6.8.) (See Span of truss Span Rise of of truss truss Rise Roof Roof slope Truss spacing spacmg Rafter length
so that that centroidal centroidat axes axes (or (or bolt bolt Ideally, members members should be be connected connected so centre the case case of angles angles or this centre lines lines III in the or tees) tees) meet meetatat aaPOint point (Fig. (Fig. 6.6). 6.6). If if this cannot be be achieved, then both members cannot members and and connection COlmectlon must must be be designed deSigned from the the eccentricity. eccentnclty. inInmany manyeases cases the gusset gusset plate not lie lie ininthe the plane plane platewill 'viii not of the member member centroidal centroidal axes, a.'i:es, but but stresses stresses due are of the due to to this this eccentnclty eccentricity are ignored in in construction constructionusing usmgangles, angles, channels channels and and tees tees (clause lclause 4.7.6c). 4.7.6c). of the the bolts bolts or or welds welds follows followsconventional conventionalmethodst34). methods{l.4l. Bolts DeSign Design of (gi-ade 4.6 4.6 or 8.8) must (grade must be be designed deSigned for forboth bothshear shearstress stressand andbeanng beanngstress stress beusedtS), used('s\ but are are not usually (see Frictiongrip gnpfasteners fasteners may maybe (see Section Sectton 3.7g). Friction unacceptable.Design DeSignstresses stresses mmthe thegusset gusset may may econotnJc unless unless bolt boltslip slipisISunacceptable. economic be checked short beam beam with aa combined axial checked as as aa short axial load. load. Such Such design deSign is IS not not realistic realistic however, however, and and ititisISsufficient sufficienttotocheck checkthe thedirect directstress stress otily onlyatatthe theend end of the the member (Fig. 6.7), 6.7), on on an an area area b xx r tas IlS shown. shown. Minimum bolt usually16mm, 16mm.and andminimum minimumgusset gussetplate platethickness thickness Mimmum boltsize sizeisISusually IS 6mm 6 mm (internal) (internal)oror8 8nun mm(exposed). (exposed). is
mternal and and external external pressure pressure coeffiCients reference (6), the the internal coefficients given given m in reference worst case casefor for wind wind loading loading on on the the roof roof slope slope ISis the the case case'Wind 'wind on end end plus worst internal pressure an outward pressure pressure of mtemai pressure' (Fig. 6.9). This gives gives an of Us'~ng Usng
-(0.7+0.2)0.68=-0.61 kN/m:! —(0.7±0.2)0.68 =—0.6i kN/m2 Note must must be be taken taken of of the the effects effects of of the the slope slope on on the the values values of nf each Note each load as as appropnate SectIOn 2.7(d) 2.7(d) and and Fig. Fig. 6.10). 6.10). appropnate (see scc Section
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_________ _________ 72 STRUCTURAL STEELWORI( DESIGN TO ES 5950
72
TRUSSES
STRUCTURAL STEELWORK DESIGN TO 8S 5950
Deadload toadononeach eachpurlin pudin Dead cladding 4.0 x 2.0 0.12 =0.9GkN ciadding 4.0 x 2.0 xx 0.12 =0.96kN ownweight weightof of purlin purlin (say, (say, 0.11 0.11kN/m) kN/m)and andtruss buss own 0.11xx4.0+8.0/10 4.0±8.0/10 =1.24kN 0.11 =l.24kN Totaldead deadload toad on on purlin purlin Total ==2.2OkN 2.20 kN Deadload toadon onrafter rafter Wd =2.20 xx 8.6212.O=9.48kN Dead =2.20 8.6212.0=9.48kN
CompressIOn Compressionforce force1Sispositive. positive. As and Asthe thewmd windload loadisissuctlOn, suction,the the load load combinatIOns combinations 1.411'" +± lAW", 1.4W,,, and 1.2(TV" 1.21W6+±TV/ +±Ww ) may maybe beIgnored. ignored.
-
(t) SMInInrafter rafter (fl BM
Imposedload toadon on each each purlin purtin =4.0 =4.0 xx 2.0cos21.8° 2.Ocos2l.8° xx O.75=5.56kN 0.75=5.56kN Imposed Imposed loadon onrafter rafter TV/ If1 x 8.6212.0=24.okN Imposed load ==6.46 6.46 x 8.6212.0 = 24.0 kN Windload loadon on each each purlin purtin =4.0 =4.0 xx 2.0 =4.88kN Wind 2.0 xx 0.61 0.61 =4.88 kN (suction) (suction) Windload toadon on rafter rafter Ww Jf1, =4.88 xx 8.62/2.O=21.0k34 (suction) =4.88 8.6212.0=21.0kN (suctIOn) Wind
Fig. 6.10
Fig. 6.10
clause clause 4.JO(c) 4.JO(ç)
Truss forces forces (dead) (dead) Truss
(c) (c)
(g) (g)
y
Nodal forces W6or U5 Nodal forces Wdor WI
l~
Truss analysis analysts isis carned carried out out piacmg placing concentrated concentrated loads toadsatat the the nodes nodesof of the the Truss truss, i.e. dividing thc rafter ioad load proportIOnal proportional to to the the nodal nodal centres centres (Fig. (Fig. 6.11). truss, I.e. dividing the rafter Analysis by manual or computer techniques gives forces as in the table AnalYSIS by manual or computer techruques gIves forces as in the table (owing to to symmetry, symmetry, half half of of the the truss truss only only isis recorded). recorded). (OWIng
Lii I
—x
W6 = 9.48/4 = 2.37 IN
Wd =9.4B/4 "" 2.37 kN W, =24.014=6.OOlcN
Wj
'"
I
24.0/4 = 6.00 kN
V'
Nodal lorces
Nodal forces Ww Fig. 6.11
Fig. 6.11
Truss forces forces (imposed) (imposed) Truss
(d) (d)
6.13
(e)
2
r!t!,-
Ww
V
6.
Member
27,0/4 = 525 kN W..,= 21.0/4 '" 5.25 kN Fig. 6.12
~
Fig. 6.12
b.
Use angles spaced spaced 8 8 mm apart to allow for Use two two -80 —80xx60 60 xx 77 unequal unequal angles apart to allow for gusset gusset plates plates with with spacing spacing washers washers atat quarter quarter pOints. points, I.t!. i.e. 0.54 0.54 m m centres centres (Fig. (Fig. 6.13). IS for for the the single smgie angle angle section and they-y IS 6.13). Note Note thatthey-yaxls that they'—j axis is section and they—y aXIS axis is for the combined combined section. sectIOn.
Slenderness Am Am =2155/26.2=82 =2155/26.2=82 A, 12.7)=42 1, =2155H4 xx 12.7)=42
Wind the truss as shown shown (Fig. Wind forces forces on on the truss are are as (Fig. 6.12) 6.12) and and member member forces forces are are again analysed results given gwen In the table: agam analysed and and the the results tu the table:
W..
~2 \ t I
Rafter Rafter
Truss forces (wind) Truss forces (wind)
6.4 and 4. J0 and clause clause 4.10
(for connections connections at at quarter quarter pOints) IS based based on tlte mmlmwn r of the (for points) where where AC .çi5 on the minimum r of the for single smgle angle) angle) as as the the angle angle can can fail faii as us aacomponent component component (r"" for component between fasteners. fasteners. between
-
w..
Assummg Assuming purlin purlinpositions positionsare arenot not known known Purl in load l.4(0.96+0.44) +1.6 xx 5.56 = 10.9 kN Purlin loadWW= =1.4(0.96±0.44)+I.6 5.56 =lQ.9k34 BM BM=WLI6=1O.9 =WLf6=1D.9xx2.0cos21.8°/6=3.36kNm 2.Ocos2l.8°/63.361thm
The forces are arranged as in (c) but have values of 6.00 kN, instead of The forces are arranged as in (c) but have values of 6.00 kN, instead of 2.37 kN. Member forces are given in 2.37 kN. Member forces are gIven ID the the table. table.
(e)
I
2
2
3
3
4
4
5
5
6
6 7 7 8 8 9 9 10 10 II 11 12 12 13 13 14
I'
Member Member forces fDrces (kN) due due to: tD: Dead
BStable table 25 25 Strut Strut table table for for angles angles is table 27c, fiS is table 27c, and and for for A=92: 2=92: BS table table 27c 27c 115
clause 4.2.5 4.2.5 clause
Compiesslve strength strength p. pc ==123 123N/mm2 N/mm 2 Compresstonresistance resistance Pc =AgPc =Ag pc Compression =21.2 x x 123 x 10-; =261 kN =21.2 123 x lr=261 Moment capacity capaCIty Ma Mcr Moment =PyZ= =p, =275 xx 24.3 24.3 x x I0"36.7kNm 10-3 =6.7 kNm =275 local capacity capacitycheck check(for (forfurther furtherdiscussion diSCUSSIOnsee sceSection SectIOn7.5): 7.5): Simplified local Simplified
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STRUCTURAL STEELWORK STEELWORi< DESIGN DESIGN TO TOES STRUCTURAL as 5950 5950
clause 4.8.3.2 4.8.3.2
F/Ag py p,. + M/M= /tl,J&i,, FlAg
1- I 121.9/(21.2x x275275 105+336/67=071 121.91(21.2 x x 10-')+3.36/6.7=0.71
(I)
!
r",,=lL7mm ryy= 18.2 mm 18.2mm L= L72m L= 1.72 m Fig. 6,5 and 0.85LJr,, = 125 or Fig. A= O.85Urvv = clause 4.7.1O.2a 4.7.iO.2a elm/se A= a.7Ur)'}' +30=96 A Compresslve for A A. == 125 BS 135table table 27c Compressive strength =91 N/mm2 N/mm' for 125 strength Pc P. =91 CompressIOn Compression resistance Pc =Ag =Ag Pc = 6.91 6.91 xx 91 91 x 10- 1 =63 kN Section is satisfactory. SectIOn IS satisfactory.
=Le~"
=2 155/26.2 = 82 =2155/26.2=82 Buckling resistance reSIStance ""h =O.8pb 2" 4 =0.8 xx 275 275 xx 24.3 24.3 xx l0"3=5.35kNm lD- J =5.35kNm
F/Ac pc +mMJAI,, FlAg mAl./klb Pt +
where nt ni= 3.0 l' ! where = 1.0
121.9/261 3.3615.35 121,9/261 ++3.36/5.35
I
= 1.00
m (h) (h)
Connection Connection
Bottom chord Bottom chord design design Check 7, 10 10 and 13. 13. Check the the strength strength of of the the connecilon connectionJommg jotntngmembers members6,6,7. 4.6), see sec Fig 6.14. Assume mm thick gusset Assume an an 88mm gusset plate plate and and 2D'mm 20mm bolts (grade 4,6),
Max =1I3.0kN Max tensIOn lenSlOn = I J3.0kN compression = 39.2 kN Max Max. compressIOn
kN Max. force force (member Max, (member 13)=32.3 13)=3L3 kN
Force change change between between members members 66 and and 77 is is 32.2 32.2 kN. kN. If If members members 66 and and 7 were were 10 to be Jomed joined at at this this connection connection (to (to change change Size size or or reduce reduce fabncallon fabncat,nn lengths) then maximum maxImum force force (member 96.9 kN. then (member 6) 6) would would be 96.9kN.
Use two —75xx 50 50 x 66 unequal Use two -75 unequal angles angles spaced spaced 8mm 8 mm apart apart (semi-compact) (semi-compact) clause 4.6.3.2
== 10,3 10.3 kN
Use 60 x 66 equal equnl angle angle Use 60 60 xx 60
ai =(L—t12)t—Dt al =(L-tl2)/-D/ =175—3)6—22 =(75-3)6-22 xx 6=300mm' calculating aJ: allowing for allowmg for one 22mm 22 mm diameter diameter hole hole in In calculahng UI:
'9
a2 =(50-3) =150—3)6=282mm' ", 6=282mm'
A, A .. =300+5 =300+5x x300 300x x2821(5 2821(5 xx 300+282) 300+282) = 537 mm'/angle =537 mm 2/angle becomes nef net area if A.. becomes jf clause 4,6.3,3 4.6.3.3 is IS satisfied. satisfied. Note that A,
Tension capacityP,F,=A.,py =A, p, ==22 xx 537 TenSIOn capacity 537 xx 0.275=295164 0.275 =195 kN Compression CompressIOn resistance resistance must must be checked assuming lateral lateroi restraint restrmnt to the the bottom chord chordbracing bracmgconnections, conneCllons,maximum maxImumspacing spacing4.64 4.64mm chord at at the the chord (Fig. 6.8). 6.8). 21,4mm r,, r.,~~ =21Amm =4640/21.4=217 A,,, =4640/21.4=217 "'m Ar =580/10.8=54 =5801I0.8=54
A,, =,/(2l7'+54') £5 SS table 27c 27c
20 mm mm boiia boils' 22 mm holes lor br 20
Fig. 6.14 6.14
clause 6.3.2 6.3.2
(for connections connectIOns at at quarter quarter points, pomts, i.e. I.e. 0.58 0.58 m m centre) centre) clause 3.2b clallse 47. 4]3.1b
7
-o(-9¥~t·.o...:·L._.-D-.L?¥f"4-
=224 =224
Maximum =250 Maxlll1um slenderness slenderness = 250 Compression strength P. =34 Pc = 34N/mm2 N/mm 2 Compression resistance resIstance Pc =Ag =A~ Pc pc = 14.4 14.4 x 34 34 x l0" 10- 1=50.4364 =50.4 kN Section is satisfactory. IS satisfactory.
capacity of of bolts bolts (double (double shear) shear) Shear capacity
Pl=p~A~ =160x2x0,245 =160 x 2 x 0.245 = 78kN = Beanng capacity ofbol!s P bb =dtPbb 20 xx 66 xx 0.450 0,450=108kN Hearing capacity of bolts F,,,, dtp,,,,=1 =2 x 20 308 kN Bearing capacity 0.450= 108 kN kN Bearing capacityof ofangles anglesPb~=dIPIn F,,, = dip,,,=2 = 2xx 20 20 xx 6 xx 0.450 = 108 = but F,,, P bs 'f elPbs12 = 40 x 66 xx 0,450/2 0.45012 ==54 54kN kN elp,,,/2 Size of of as as low low as as I166 mm possible. Clearly a bolt size mm would be possible.
Gusset plate Gusset plate stress stress can can be be based based on on an an effective effective width width (Fig. (Fig. 6.7) 6.7) of 60/cos30'—22=47mm. 60/cos30° -22=47mm.
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-i*SM
10 STRUCTURAL STEELWOffi( DESIGN TO ES 5030 STRUCTURAL STEELWORK DESIGN TO SS 5950
76
TRUSSES TRUSSES77 77
1 Plateslress=32.3 stress=32.3 x 103/(8 xx 47)= 47)=86N/min2 Plate 86N/mm 1 (Maximumvalue valueis isp, =275N/mm N/mm2) ) (Ma:umum Py =275
(c) (c)
Theremaining remaining truss h-ussmem"tJe~ membersmay maybe bedesigned designedInin the the same same way, way,with with the the The compressionforce forceusually usuallybemg beingthe themam maindesign designcntenon. cnterjon.ItItshould stiouldbe benoted noted compression thatItit ISis good good practtce prnctice to to limit limit the the number number of of different different member member sizes sizes being being Ihal used, probably probably nol not more more than than four four in in total. total.The The deflecllOn deflection ofa of a roof rooftruss trussisisnot not used, usually cntical critical to to the the desIgn, design, and and the the effects effectsof of sag sag under under load load may maybe be offset offset usually by pre·cambering pre-cambering the the truss truss dunng during fabrication fabncation by. by, say, say, 50 50mm or 1OOmm. lOOmm.IfIf by mm or the deflection deflection isis required required then then hand hand or or the the Vlrtual virtual the or graphical graphical methods(7,B), or work method(9.1Ol, methodt9iO).orornppropnale appropnateprograms programsmay maybe beused. used. ItIt should should be be noted noted work that deflectIon deflection due due to to bolt bolt slip slip can can be be significant significant compared compared with with dead dead load load that elastic deflections. defiections. elastlc
6.8 6.8
EXAMPLE 12. LATTICE GIRDER EXAMPLE 12. LATTICE GIRDER
(a) (a)
Dimensions
t l.1) under the nodal loads gives the member forces (kN) AnalYSIS Analysisof of the the truss trusst121 under the nodal loads gives the member forces (k34) tn in the the table: table:
Top Top chord chord
3
Member Member
Force Force
2I 33
188 188 322 322 402 402 429 429
55 66 77 88
-107 —107 -268 —268 -375 —375 -429 —429
99 10 ID 1I II 12 12 13 13 14 14 15 15 16 to
152 152 -144 -144 1I4 114 -76 —76 76 76
-38 —38 38 38 00
Compression Compression force ISis pOSitive. positive. (d)
Imposed Imposed load load Dead load Dead load on on timber Umber joist: jOist: floor floor finishes finishes
BM BM in in top top chord
(See Fig. Fig, 6.16.) 6.16,)
J~;';~"{/0VV"'/,\A ~O.5m
(b)
Member Member
44
6.16
2
t
Fig. 6.15 Lllttu:e girder
Diagonals Diagonals
Force Force
Joiii ioidi 13.4 IN
limber louis TImber JOlsls at III 0.5 0.5 m m cenirei cenlres
Fig. 6.15 Lattice girder
Bottom Bottom chord chord
Member Member
Dimensions
See Fig. Fig. 6.l5: 6.15: lattice lattice gtrder girder fabncated fabncated from 8.0m; See from tubular tubular sections, sectIOns, span span 8.0m; spacing 3.5 m. spacing 3.5 m.
t
Truss Truss forces forces
0.5 0.5 xx 3.5 3.5 xx0.5 0.5 =0.8SkN =0.88 kN 0.2 x 3.5 x 0.5 0.2 x 3.5 x 0.5=0.35 =0.35kN kN
(e)
BM in continuous contmuot!S top chord=WL/8 chord = WLl8 3M in =13.4 = 13.4 xx LO/8=1.68kNm Top chord
Max. compresSion =429 kN Max. compressionFF=429kN
BM M;.: Max. BMMX
= 1.68 kNm 1.6SkNm
90x-x· 6.3 6.3 I1RS RHS Use 90 xx 90 =34.1 mm rr= 34.lmm 6.4 Slenderness A.A=0.85 l000/34.1 =25 Fig. 6.4 Slenderness =0.85 xx l000/34.i =25 BS table table 25 25 Strut table ES table for for RI-IS RHS is table table 27a 27a 1 BS table 27a 27a ES Compresslve strength Pc =270N/mm Compressive 270N/mm' CompressIon resistance resistance Pc =Ag Pc Pc Compression =20.9 xx 270 270 xx 1O-'=564kN =20.9 564 kN Section chosen chosen isIS plastic plastic (bIT (bIT = 14) 14) Section Moment capacity capacity Mc =Py S Moment =PJ,s =275 xx 65.3 65.3 xx 10- 3 =17.96 17.96kNm =275 kNm clause 4.8.3.2 4.8.3.2 Local clause Local capacity capaCIty check check
weight weIght of of truss truss (estimated)=0.37kN (estimated)=0.37 kN == 1.60 Total load IV,, kN Total load Wd 1.60kN Imposed Imposed london load ontimber umberjoist jotst IT',=4.0 Wj =4.0 xx 3.5 3.5xx0.5=7.OOkN 0.5=7.00kN
Section isissatisfactory sails factory. Section
Load Loadon ongirder girderfrom fromeach eachjoist JOIst== 1.4W,, 1.4WJ ++1.6W, 1.6Wt 1.4 xx 1.60 l3.4kN =1.4 1.60 xx 1.6 1.6 xx7.00= 7.00=13.4kN Concentrated Concentrated load load at at nodes=26.8kN nodes=26.8kN
Lateral torsional 10rsIOnal buckling buckling need need not not be be considered considered ififcompression compressIOnflange flangeisIS Lateral posItively restrained restrmned connected connected to (0 the thetimber tImberjoist Joistfloor, flOOf, or orfor forbox boxsections sectIOns positively lclause8.2.6.1). 8.2.6.1). (clause
FIAgPy+m;.: 1Ma ?S-; i F/A5
429/(20.9 xx 275 275 xx 10- 1)+I.68(17.96=O.84 1.68/17.96=0.84 429/(20.9
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78 78
STRUCTURAL STEELWOAI( STEELWORI( DESIGN STRUCTURAL DESIGN TO TO ES SS 5950 5950
(I) (t)
TRUSSES TRUSSES 79 79
chord Bottom chord Bottom
2. 2. Truss Truss analYSIS analysis
tvlax.tensIon tensionF=429kN F=429kN Max. 3. >.
Use90 90 xx 90 90 xx S Use 5 RIIS rutS Tensioncapacity capacityP,P —A, p, Tension =A t Py 16.9xx275 275xx1010'1 =465 = 16.9 =465 kN Section is SectIOn IS satisfactory. satIsfactory.
BS OS6399 6399DeSign DesignLoading Loadingfor for Buildings Part WilJd loads /oads (J 995) Pan 2: 2: Wind (1995)
7. Truss Truss defleCllon deflection 7.
Mnrshal! Nelson H.M. H.i\L (1990) (1990) Deflection Deflecllon of Marshall W.T. & Nelson of Slructures, 273-98. Longman Slruetures, SIn/Cll/reS, Stn:c:ures, pp. 273—98. Longnian
UseSOx5Ox3.2RHS 50 x 50 x 3.2 RHS Use = 19.1 r~ = 19.1 truss mm Slenderness A =0.7Ur~ =0.7L4'5 =26 Slenderness,l =0.7 xx500/19.1 500!l9,1 eos4S' cos45"=26 Compressive strength strength Pc p. =269N/mm =269 N/mm21 Compresslve Compression resistance =Agg Pr CompressIOn resIstance Pr =A pc = 5.94xx269 269xx10I0"1 ==160kN 160 =5.94 1 163 leN Tension resistance P, = 5.94 x 275 x ID_i TensIOn resistance P, =5.94 275 x 10- = = 163 kN
,. S
Truss dcflectlOn Truss deflection
,
I
(1991) voL 4, 4, pp. pp.1011—10/ 1011-10/10 11991) Steelwork Steelsrork DeSign Design vol. ID Steel Steel Conslfuctton Construction inslltme Institute
4. Welding 4.
I
(See Fig. Fig. 6.17.) (See
Fig. 6.17 Fig, 6.17
deSign Bolt design
Contes Kong F.K. CoatesRC, ftC., eouUI! Coutie !\I.G. M.G. & & Kong F.K. (1988) (1988) AnalYSIS of plane plane trusses, trusses, S/mcrurol Alluly.SIS pp. pp. 441-5 Analysis of Structural A,ia/vsis I—sIJ Van Van Nostrand Nostrand RemJlOld Reinhold
Wang (983) The The gmphical method, Wang C.K. (1983) method, illtermediale Jntennediate 93-99. Perganson Pergamon StntctJ/ral Allalysls,pppp93—99. S;nicniral Analysis,
Virtual work 9. Virtual
Marshall Nelson 11.M. H.M. (1990) l%larsltall W.T. W.T. & Nelson (1990) Virtual work and pnnclplcs. Sfmcfllres, pp SOt—) 501-31.t. Longnsais Longman and energy energy tsnnctples, Structures, pp
tO. Virtual Virtual work 10.
ConIes & Kong Kong F.K. F.K.(1988) (1988) CoatesR.c., ftC., Coulie CoistieM.G. MC. & ApplicatIOns the pnnclple Applications of the pnnetple of of vlnua! vinual work, work, SlmC/Ilroi Stniciurai AIJO(I'SIS, pp.134-75 Van Analysis, pp.134—IS Van Nos!rnnd Nostrand Remhold Reinhold
11. ConnectIOn of ofP145 RHS II. Connection
CIDECT (1985) (1985) Welded Weided joints. jOlnls. Consrnzcrion COlJStnlctJOn u'ith with Hollow Steel Sections, SecllOlJS, pp. pp. 129-42, British Sleet Steel Hollow Steel I 29-42, British Corpor.l!lon Corçiorntton Tubes Tubes DiVISIOn Division
Section is IS satisfactory. Secllon
(h) (h)
Connections Connections All welded All welded joints jOintswith witheontmuous contmuous44mm mm welds. welds.
The maximum maXImum forces forces which may may be be transmitted transmitted between between hollow steel sections are are more more complex complex and and detailed detailed references referencesmay may be beconsulted consultedifif sectIOns requsredtiU reQUlred{ll) As in of section sizes would would be be limIted limited in practice, In Section SectIOn 6.7, 6.7, the number of sectIOn sizes practice, with membersizes sizes. depending depending on the degree degree of with the the allowable allowable vanation vanallOninIIImember repetition repetition expected. expected.
STUDY STUDYREFERENCES REFERENCES Topic TopIC
References References
I. L Truss Trussanalysis analYSIS
Marshall Mnrshnll W.T. \V.T. & & Nelson Nelson H.M. H.M. (1990) Analysis AnalYSIS of of statically slallcally determinate delerrmrlille structures, structures, Structures, Struclllres,pp. pp.10—19. lQ-19. Longman. Longman.
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rrh
ILiC 100 SIMPLE SIMPLE AND AND COMPOUND COLUMNS
Prolerred
171
UC
SIMPLE AND AND COMPOUND COLUMNS
Others Olhars
I
Column cross. 7.1 Column cross· sections 5cclions
•
Columns, sometimes sometimes !mown known as as stanchions. stanchions, are are vertical vertical steel steel members In in both Columns, single and multi-storey 'smgle multj~storey frames. frames. They are pnncipally pnnclpally designed designed to to carry carry axial axiai loads but will 'viii also to moments due to to loads in m compression, compressiOn, but also he he subjected subjected to moments due eccentricities or or lateral lateral loads loads or or as as aa result result of being being part part of a ngid rigid frame. eccentncities frame, i.e continuity moments. moments. In some structures, particularly contmuity some structures, particularly single-storey smgle·storey frames frames and top lengths lengths m in multi~storey multi-storey ftames, frames, the themoments momentsmay mayhave havegreater greatereffect effectm,., in S. top the design design than the than the the axial axtal compression; compressIOn; under under some some loading loading combinations combinations axial tension aXial tensIOn may occur (see Section Section 2.7(d) 2.7(d) and and Table Table 12.7). 12.7). Where columns are are principally principallycompressIOn compressionmembers memberstheir theirbehavIour behaviour and ¶ Where columns design are are similar similar to to those those of struts struts (Section 6.4). The The loads loads carned carried are· are (Sectlon 6.4). deSign usually larger larger than than those usually those in in typical typical truss truss members, members, and and the the simplifications slmplificallOns adopted adopted in truss design (Section (Section 6.2) 6.2) should should not nol be be applied. applied.
RHS
81 81
CHS
1 RSA nsA
VB
RSCs ,"d plate 7.2 Compound sections scctlOns
RSAs "d
-
4
Welded be<
Jalllca
IL
U lDl
C
Laced laced
Battened Baiiened
-'i
.e 7-I 7.1
TYPES TYPES OF OF COLUMN COLUMN
7.3 Column typcs
Typical cross-sections 7.1. While Typical cross~sectlOns used used in In column column design deSign are arc shown shown in in Fig. Fig'c7oonl"o'
St .&
7.2 7.2
AXIAL COMPRESSION COMPRESSION Columns Colwnns with idealized idealized end end connections connectlOns may be be considered considered as failing failing in In an an Euler-type Elller~type buckling buckling mode{1.1) However, However, practical fail by practical columns columns usually usually fail inelastic bending and and do do not not conform conform to melasuc bending io the Euler theory theory asswnptlOns{3,ot). particularly partlculariy with respect to elastic elasuc behaviour. behavIOur. Only Oniy extremely extremely slender slender columns remain linearly elastic up to failure. columns remain linearly elastiC failure. Local Local buckling buckling of of thin thin flanges flanges rarely In practice practIce when when normal normal rolled rolledsections sections are are used, used, as as their theirflange flange rarely occurs occurs in tbicknesses usually of BS BS 5950. 5950. thicknesses usually conform conform to to clause clause 3.5 of The behaviour columns and their ultimate strength are behaVIOur of columns are assessed assessed by by their their slenderness 1i and and the the matennl matenal design deSIgn strengthp,,, strength Py, which which are ,Ire uescribed described bnefly bnefiy in in Section Section 6.4. 6.4. <
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82
STRUCTURAL STEELWORKDESIGN DESIGNTO TOOS 8S 5950 5950 STRuCTURAL STEELWOAK
7.3
SIMPLE SIMPLE AND ANDCOMPOUND COMPOUNDCOLUMNS COLUMNS 83 83
J
SLENDERNESS
The effectIVe smgle~slOrey frame to The effective iength length of of a column column In in a single-siorey frame IS is difficult difficult to as In Fig. 7.4 7.4 and and hence hence 135 SS 5950: assess assessfrom fromconsideratIOns considerationsof offiXity fixity as in Fig. 5950: Part Part II gives should be be gives these thesespeCial specialconsideration. consideration.Clause Clause4.7.2c 4.7.2cand andappendix appendixDl Dl should used these cases. cases. used for these It is IS essential essential to have aa realistic to have realistic assessment assessmentof ofeffectrve effective length, length, but but It it should assessment ISis not open to to debate, debate, should be be realized realized that that this assessment not precIse precise and and is open and IS not not reasonable reasonable to expect expect highly highlyaccurate accumtestrength strength and consequently consequently 11it is prediclJons predictions In in column column deSign. design. interpolatIOn Interpolation between betweenthe thevalues valuesIninSS 85 table table 24 24 might might produce produce more more accurate accurate estimates, estimates,but butsuch suchinterpolatIOn interpolation must nmst reflect the the actual actual restraint restraint conditions.
Slenderness is IS glVen by: iven by: A = Ls/r
where
=
LE = effective effective length length L.g radius of ofgyration gyration rr == radius
columnisISdependent dependent on on the the restraint restramt conditions conditionsatat effective length lengthofofaacolumn The effective each end. pm connections connectIOns exist e:(lst then then LE equals In each end. If If perfect pin equals the the actual length. In practtce, beam beam type members connectcd as well as as practice, type members connectedtoto the the end endof of aa colwnn. column, as gtvmg positional posllional restraint, restraint, provide provide varying varying degrees degrees of of end end restraint restraint from from giving full fixity fixitytotonearly nearlypinned. pinned. virtually full SS table length of ofaa 135 table 24 24 Indicates indicates In in broad broad terms terms the the nommal nominal effective length member, provided the the designer deSigner can can define the amount of of positional positionai column member, define the and rotatlOnal restraint acting actmg on on the the column column ininquestionquestion. The The vanous vanousclasses classes and rotational resiraint of end fixity fixity alluded alluded toto inIn85 BStable table2424represent represent aa crude crude classification, classificatIOn, as as of end Indicated in Fig. Fig. 7.4. 7.4. The The nominal nommal effective effectivelengths lengths tend tend to to be belonger longerthan than indicated those obtamed vaiues as those obtained usmg using the the Euler values as they they take take cogruzance cognizance of of pracl1cal practical there IS (unless a pin is IS conditions: that that IS, conditions: is, there is no no such such condition condition as as 'pinned' pinned' (unless deliberateiy Therefore, deliberately manufactured, manufactured,which which would would be be costly) costly) or or 'fixed'. 'fixed'. Therefore, taking the the case case haVing In reality realitythe theends ends would wouldhave have some some having both both ends ends 'fixed', 'fixed', in flexibility; henceaa value value of of0.7, 0.7, rather rather than than the idealized Euler value of of 0.5, 0.5. is IS flexibility; hence used. these values values in m 135 BS table members used. Although Although these table 24 24 are are adequate adequatefor for column column members mmulti~storey buildings buildingsdesigned deSIgned by the the simple 'sImple design' deSign' method, method, aa more more inmulti-storey accurate assessment sectIOn 5.7 5.7 and and appendix appendix EEofof135 BS 5950: 5950: accurate assessmentISisprovided provided by section columns in m 'ngid' frames (continuous (contmuous construction). constructIOn). Part Ii for columns ngid' frames Part may be different differentabout about the the two two column columnaxes, axes, and and in in The restraint provided maybe practice in steelwork steelwork frames frames this generally bethe be·thecase. case. The effective effective practice this will will generally lengths and so will willbe be lengths In in the the two two planes planes willlherefore will therefore generally be different, and the Ay). For loaded columns columns the the compressive compresslve and A,.). For aXlllily axially loaded the slenderness P . . and strength Pc IS selected selected from 135 BS tables strength P. is tables27a 27a10to27d 27dfor for both both values values of of Slenderness, and the lower lowerstrength strength IS of axis. a.us. slenderness, is used used in in the the deSign, design, mespectlve irrespective of
free In In Free position position
o.n
tOL
01 Pinned
Fig. 7 A End End fixitv/ fixltyl Fig. 7.4 effective length
Fixed Partial Partial In restramt restraint in direcllon direction
lo2L
7.4
BENDING ECCENTRICITY BENDING AND ECCENTRICITY
WA
'4' Wc
4!
'ii
t
'TV Fig. 7.5 lands 7.5 Column toads
In addition to aXial compreSSIOn, columns will usuallybe be subjected subjected to to moments moments axial compression, will usually due due to tohonzontalloading honzontal loading and andeccentnclly eccenincity of of connections connectiotis carrymg carrying vertical loads. loads.The The types typesof ofload loadthat thatcan canoccur occurare aresummanzed sunimanzedininFig. Fig.7.5 7.5in in which: which: WA is the vertical vertical load load from from aaroof rooftruss, truss.taken takenas as applied applied concentncally (clause (clause 4.7.6af2)) 4.7.6a(2)) Ws by side side rails rails (V5 are are honzontal honzonial loads loads due due 10 to wmd, wind, applied by We and Wile are crane gantry loads, applied are the the crane gantry toads, applied through aa bracket at eccentriCity bracket at aa known eccentricity WD is IS the the self self weight of ofthe Ihe column/sheeitng colwnnlsheetmg W£ truss bottom chord chord WE ISis the the resultant resultant force force carned earned through through the tmss
In singie-storey smgle~storey column column and and truss truss structures structures load W£occurs the occurs whenever the columns carry or unequal unequal moments. moments. Values of of the the columns carry unequal unequal honzontal honzontal loads loath or resultant arrangements of resultant force force for for different arrangements of honzontal honzontat loading loading are arc shown shown m in 7.6. Fig. 7.6. In cases cases such not such as as most most beam/column beam/column connectIOns connections whcre_eccentnclty where.eccentnctiy isis not known precisely, preCisely, clause clause 4.7.6 4.7.6 states statesthat thataavalue valueof of100 100mm mm from the column face (flange web as as appropnate) appropnate) should should be be used. used. face (flange or web the design deSign of column under under aa toad load system system such such as as that that shown shown inIII For the of aa column Fig. 7.5, 7.5, load factors factors 'If must he be included IIlciuded inin the. the,calculatlOllS. Each load load may may yj must calculations, Each compnse only. while while compnse one oneor ormore moreload loadtypes, types,e.g. e.g.load loadWowill W5wilt be be wlIla wind load only, W When toads loads are applied applied W4 may conSist consistof of dead deadload, load, Imposed imposed and and wllld wtnd load. \Vhen A may m combinatIOn, different load factors factors may may be be required reqUIred for each each load group group in combination, (see no longer be be clear which which (see Chapter Chapter2). 2).Unlike UnlikeSimple simple load load cases. cases,Ititwill will no forces and mbments. mbments. Complex load load combinatIOn produces the combination produces the highest highest aXial axial forces cases, combinatIOns to to be be cases,as asshown shownmtaFig. Fig. 7.5, 7.5, may may requue requtre all. all possible combinations examined, expenence three three or or four four worst worstarrangements arrangements may maybe be examined, although although with expenence or some some combinations combinatIOns selected. selected.The Thebest bestway wayof of exammmg examining the the effects effects of all or IS tbe use mampulatlOn should should be be arranged arrnnged in 10 the the is the use of of matnces. mainces. Matnx manipulation follOWing manner: following manner
Unfactored load = [IF] [IV] Unfactored load matnx matrix Load factor/combination factor/combinatIOn matrix = [Yf] Load [yj] Factored load matnx matnx rW [W,] Factored load =fW} =[IP] xx [y4 [td y] force and and moment moment coeffiCIent Axial force coefficientmalnx matnx 10:] e] Axtal force force and and moment moment matnx matnx [F4u1=[1V7]T fF,M]=PVy]T x Ia] 10:) Axial This matnx matnx method method isis demonstrated demonstrated in in Section Section 7.7. 7.7.
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84 rmucmmAL S'TEELWORK DESIGN TO ES 6530 STRUCTURAL STEELWORK DESIGN TO SS 5950
84
W;
Loading Loading
(WI + W2)
~ !f;~i1 ',! \it:,\' ,:.:
w1
w2
W,
SIMPLE AND COMPOUND COLUMNS
7.6 7.6
.::.::.!!:
216
FlAg F/AgPcp.++IJ/:rl'vf:/Mb ++11f,Af/p), t?5.My/4~. Z,1- i
where IS the 6.4) wherePr:: Pr 15 the compresslve Compressive strength strength (SectIOn (Section 6.4) m,n",.\" are the eqUivalent unifonn moment factors (Section 3.2) are the equivalent unifoim moment factors (Section 3.2) Mb (Section 3.5) 3.5) M6 ISis the the buck1ing buckling resistance resistance (Section Zy the mmor minor axiS axis elastic elastic sectiOn section modulus modulus (compressIOn tcompression face) face) 4, ISis the
1111 40
11g. 7.6 Resultant force fo'lg. 7.6 Resultant roof truss force botiom {! roof chordtruss bottom
OVERALLBUCKLING BUCKLING OVERALL The an axml The failure failureof ofcolumns, columns,Whether whethercarrymg carryingmoments momentscombined combinedWIth with an axial load of loadorornot, not,will svillcommonly commonlyIOvolve involvemember member buckJing. buckling. The The ilssessment assessment of overall as given overallbuckling bucklingreSistance resistanceInvolves involvesthe thesame samefactors factorsand and procedUres procedures as given toinSectIOn Section3.2, 3.2,but butwith withananadditional additionaltenn termfor foraXial axial load. load. Usmg Usuig aaSimplified simplified 4.8.3.3.1) Ule overall buckling check IS considered approach \clause approach (clause 4.8.3.3.1) the overall buckling check IS Consideredtoto be be satlsfactory satisfactoryifif the the combinatJOn CombinationrelatIOnship relationship isissatisfied. satisfieu.
1 2
16
Jr1
85
A 4.8.3.3.2) may be carned out usmg Mar and A more more exact exact approach approach lciause (clause 4.8.3.3.2) may be carried out using Ma and AI"),, the axes In the the maxlmum buckling bucklingmoments momentsabout about major major and and mmor mmor axes in the Ut.i~ approach: presence presence of of axial axiai load. load. In In this approach:
(Mi + M,l
chord
III:rM:/MC1..t nixA I JA fgg+ + ",).A~\/M"y.,. II
..
" ;'"
,"
7.5
7.5
LOCAL CAPACITY
LOCAL CAPACITY
At any any section sum of of effects effects ofaxml of axial load At sectIOn In in aa column column Use the sum load F F and and moments moments and M). must M..rand must not not exceed exceed the the local local capacity. capacity. This This is IS considered considered to to be be satisfactory if the the combination combination relationship relattonship satisfies sahsfies the the following followmg interaction interaction satisfactory if equation:
:. -~
7.7
EXAMPLE EXAMPLE 13. 13. COLUMN COLUMN FOR FOR INDUSTRIAL INDUSTRIAL BUILDING BUILDING
DimenSions (a) Dimensions
equation:
(See Fig. ·7.7.) (See Fig. Overall height Overall height HeIght of of crane crane rail rail Height bock wall wall Free standing standing brick Free Crane nil rail eccentncity eccentncIty Crane Cladding eccentncity eccentnclty Cladding
i (a) (a) for for plastic plastic and and compact compact sections: sections: 1.113, UC and joist sections ~, UC and jOist sections (M'/M~)' + M/Mry f I
RI-IS, CHS and RHS, CHS and. solid solid sections sectIOns
where of where .Mn;and M0,are Ai,yare the the reduced reduced plastic plastiC moments moments in m die the presence presenceof axial aXial load, load, as as defined defined ininreference reference(5). (5).
I
where sectional areaarea andand A-f,,, and whereAg AgisIStilethegross gross sectIOnal Mo: andA-f,,. Mr::vare are the moment capacities capacitiesdefined definedininSection Section3.4. 3.4.
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86
DESIGN TO TO SS 85 5950 STRUCTURAL STEELWORI< STEELWORK DESIGN 5950
SIMPLE AND AND COMPOUND COMPOUNDCOLUMNS COLUMNS 87 87
Wind on side of ofbuilding building W8. Ws. Wind load land (u.d,l. lu.dJ. above above brickwork) bncbvork) 55 kN kN
W4
Ihese combinations, combinations, three m Of these three cases cases are are selected selected as as shmvn shown illld and lire are given In matrix form h'f]: mntnx fonn[VA:
.<
Self weight (column) Se1fwelght (column) W0 WD Dead load
Wc
Wuc
Cladding CladdingWDC WDC Dead load
0.5
0.
c
'"
;Wvc 'ci a
1r53(5+D)1l6
T E 7!:2 ~ 0 \.SI ;i
WE =3(220+ 150)0.5/(4 150)0.51(4 x 12.5) 12.5)
"
£
= JO.2kN 10.2 kN =
for dead dead load) load) depends depends on on the crane girder dead load The resultant force force (IVE for producing producmg moments In m opposite opposite directions: directions:
TL
Case (ii) (11)
(!il) Case (ill)
1.2 1.2 i.6 1.6
1.2 L2 1.6
1.2 1.2 1.2 !.2 1.2 1.2 1.2 1.2 !.2
IV,. W~
1,4 iA
1",,, JJ'ch
1.4 1,4
1.6 00
IV",
00
0
(d)
Factored loading loading As shown shown in m Section SectIOn 7.4, the the factored faclored loading landing matrix malnx isis [JV,j=[IPJ [W,l=[Jf1 x [rrlkN
= 1.2 i.2 kN
3(20 + 20)0.51(4 xx 12.5) WE=3(20+20)0.51(4 12.5)
Fl9. Fig. 7.7 7.1
Case (I) (I)
=IO.3kN
resultant force force (WE (lVifor on the difference The resultant for crane) depends depends on difference in m the the crane crane (vertical) loading on the near and and fae far' side columns columns (Fig. (Fig. 7.6): 7.6):
E
{
IV,, W, IV,
t6kN 16kN
for wmd) wind) depends depends on on the the difference difference between between the the wmd The resultant force force (JFk (WE for loading landing on the column being bemg designed and that that on on the the similar similar column column on on the the of the the building building (Fig. 7.6). If Iflhe wmd load load on on the the far far side side isIS zero: zero: far side of the wind
£ E
o0
'D
lOkN 10kN
Case (i) (i)
Case (ii) Cnse fiI)
Case (iii) (Hi)
IVA IV.
185
H'B
0 332 332 8,4 8.4
185 185 0
46 66 288 7.2
Note that the crane surge loading (horizontal) resultant force (honzontal) produces produces no resultilllt force m mduces equal loads in III both columns, colwnns, and and in in the the same same direction. directIOn. the truss as itIt induces unfadored loads may 'be be shown In the load load matrix mntnx [if'] {W] kN: kN: All the unfactored shown in
WHe 1V0 W D
lYc IVc
12 19 15.7
W W,oc DC
tV4 IVA
;s'3 IV. Wc H'llC 1V0 IVD
W DC WE
Dead Dead load load
Imposed Imposed load load
W3 !I'd
JPj 78 78
50 00 20 20 00
00 00 00 00 00
ID 10
16 1.2 1.2
00
Crane load vertical øç, JJ'",~
%Vind Crane Wind load load loud horizontal
iv,,, JVch
iv_ IV.
0 0
0 0
—90 -90
220 220
0 66 0
00 00 00
00 00
0 10.3
0 0 0 10.2 10.1
WE
376 376 0
12 19
12 19 19
17.8
26.0
The loads 7.9 lands for each loading case can be shown on diagrams as in In Figs. Figs. 7.8, 7.8,7.9 and 7.10.
Bending moments moments and and axml axial forces forces are are produced produced by by the the factored factored loads. loads. To Bending calculate axml axial force force all all vertical vertical loads loads above above tbe the cross-section bemg being checked checked calculate should be added added together. together. To To calculate calculate bending bending moments momentsthe theproducts productsof of should'be force and and distance distance should should be together. This This is is agam again usefully force be added added together. usefully shown shown 10in aa matrix [a], and bending M. The matnx [Cl1, for for axial axml force force FFund bending moment momentM. The BM BM values values are are to to be be calculated at at three three posltions positions as shown in Fig. Fig. 7.7. 7.7. The The coeffiCIents coefficients for for BM EM are are calculated as shown iii metres. metres. Ul
Local Local capacity Use 305 xx 137 137 UC DC section section (grade (grade 43). This TIus IS plasLlc section sechon Use aa 305 305 xx 305 is aa plastic 2 (bIT dlt == l7.9) =265N/mm (b/T =7.8, =7.8, cUt 17.9) and and has has aa deSign design strengthp), strength p, =265 N/mm2 Moment j\fcx =Py Moment capacity capacity Ala =py S;r =265 =265 xx 2298 2298xx 1O-'=609kNm 103t6O9kNm
—2.5 -2.5
Agpy =174.6 xx 265 A1p, =174.6 265xx ID-"=4630kN 10=4630kN
The coefficients allow for a reduced value The coefficients for for EM BM in in relation relatIOn to to the the load load Wo aUaw value of Wn as as well well as as aa reduced reduced distance distance In in calcuiatIng calculating er.a for 2 and 3. For For of for positions pOSitions 2 and 3. example; example:
CASE (i) clause 4.8.3.2
Local capacJty check check (simplified): iSlmplified): Local capacity
FlAg py +AIJIrIC. +Mfila ;t I I
(7.5/80) x x (7.5/2)=2.81 (7.5/2)=2.81W4 M, = W. (7.5/10) W. hence a=2.81 m. hence Cl = 2.81 m. Similarly the the value value of of tVoc Similarly Woc decreases decreases for for points pOlOts 2 2 aad and 33 aad and this this isIS reflected reflected in the ID the value value of of a: IX:
548/4630+ 125/609 == 0.32 0.32 548/4630+ 125/609
CASE (ii) (ii) capacIty check: check: Local capacity
As As shown shown in In Section Section 7.4, 7 A, the the EM BM and and axial fixml force force matnx matnx is IS
,'
[F, [F, M]=[W,]T M]=[W,]T xx [a] raj (h) (Ii)
Case Case(1) {i} Case Case (ii) (ii) Case Case (iii) (iii)
F5 F,
Mi
548 548 592 592 365 365
M3 AI,
M3 AI,
49 49
87 87
—40 -40 381 381
50 50
125 125 142 142 86 86
167 167
The The EM BMdiagrams diagrams for for each eachload loadcase casemay maybebedrawn drawnasasintoFig.. Fig.7.11. 7.1 !.
Overall buckling buckling Overall For compresslve thethe slenderness A IS based on an an effectivc effective length length For compressivestrength, strength, slenderness A is based on LE_ L5.
Agpcy =174.6 =174.6 xx 86 86 xx 1O-'=150DkN Agpcy For moment moment resistance, resistance, the the buckling buckling strength strength is on the the minor minor For IS always always based based on slenderness: slenderness:
Y ~-381167~-5.7 y= —381/67 = —5.7 nn =0.71 =0.7! A.LT 0.85! xx 0.71 0.7! xx 109=47 =0.71 xx 0.851 ALT =0.71 Buckling 238 N/mm2 Buckling strength sirength Pb ==23814/mm2 Buckling 2298 xx 1O-3=547kNm Suckling reSistance resistancelvh M5=238 =238 x 2298 l0"3=S47kNm
x = 14. (torsIOnal (torsional index) x=14.i mdex) Ak ~7.7 =7.7 ill N ~0.5 =0.5 N 0.71 vv = 0.71
36:5Jl500+ 365/1500+ LO 1.0 xx 38\/547=0.94 381/547=0.94
=
ES table table 14 14 BS
Section 118 UC) Section ISis satisfactory. satisfactory.Using Usinga asmaller smallersection section(305 (305x x305 305xx 118 UC) results results m the Simplified in the simplified buckling bucklingcheck> check> I.
ua =0.851 ~0.851
zm
CASE 0) (I) CASE
ES table 13 BS IQble 13
ES mble rable /8 iS BS
~.&!~XAMPLE EXAMPLE 4. LACED COLUMN FOR INDUSTRIAL BUILDING mO
46
j"
n= in S;1.0, =1,0, 1.0, m LO,as asthe thecolumn columnisISnot notloaded loadedalong alongits Itslength lengthfor forthis this combination. combination.
IJ
~
P =49/125=0.39 P ~491125~0.39
~
Fas
in =0.72 =0.72 m
t __
iiitVA ALT =lIIIV). ALT
ES SS table table J111
=194 = 194 x x 2298 2298 xx 1r3=446ld4m JO- l =446kNm
66
FIACPC±o&AIXIMb FlAg pc + mxM~/A'h -;.
I
U ~}nnclpal cross*s~ctlOn is IS revised revised as os Pnncipal dimensIOns dimensions are are as as SectIOn Section 7.7(a) 7.7(a) bul but the the cross-section shown 7.12). shown (Fig. 7.12).
(b)
Loading As SectIOn 7.7(b). Section 7.7(b).
lSj
"1
I
548/1500+0.72 xx 125/446=0.57 548Jl500+0.72 125/446=0.57
(c) 0.4 n O.4m
rorce and and SM BM Combined force
0.4 m lO.4m I
CombinatIons before (including (including VI 'If factors) Combinations of loads loads may may be considered as before usmg the the same same values values of of W4 w'i to to W5 WE inclusive. mcluslve. Maximum MaXImum values of of SM BM and and using force will vary vary slightly slightly dwmg Qwmg to to the the revised revised cross-section. cross-section. force Consider case J.2 (deau± (dead.+crane + wind± wmd+ imposed): Imposed): case (iii) (iii) only, only, I.e. i.e. 1.2 crane ±
CASE CASE (ii) (ii)
As in ~I .0. As for for case case (i) (i), nn = = 1.0, 1.0, 111 1.0. ES BS table table 18 18
592/l500+0,48 42/446=0.55 592/1500+0.48 xx j42/446=O.S5 FIg. Fig. 7.12 7.12
CASE CASE (iii) (Hi)
ES BS table table 13 JJ
,n= m = tO, 1.0, nn <1.0, ~ 1.0. as as the the column column is IS loaded loaded along along its ItS length length for for ibis Ibis combination, combinatIon.
(d) (d)
Local capacity capacity Local
clause 4.7.8 4.7.8 The The design deSign of of aa laced laced column column may be earned carned out out assuming assummg itIt to to be be aa clause mtegral member, member, and checking the design deSign of of the the lacings. lacmgs. single sntegral -305 x 127 127 xx37 37LID un sections. sections. These These are are semi-compact semi-compactsections sectmns Use two two —305 Use (bIT = 14.2) 14.2) with with aa design deSIgn strength strength p, py=275 N/mm 2 (bIT = 275 N/mm2.
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92 STRUCTURAL STEELWORIC DESIGN TO OS 5950
92
STRUCTURAL STEELWORK DESIGN TO 85 5950
SIMPLE AND COMPOUND COLUMNS
SIMPLE AND COMPOUND COLUMNS93 93
Moment of
inertiaof ofcombined combinedsectIOn section Moment of inertia =2 (337 (337±475 670cm4 I"1. =2 +47.5xx402)c= 40 2)= 152 152 670 cm.j Z == 152 670/55 == 2776cm Z" 152 670155 2776 cm:; clause 4 25 =275 xx 2776 763 kNm clause 4.2.5 Mr;;r =Pyp,Z"Z =275 2776 xx JO- 3 =763kNm Local capacity check Local capacity check F/A
In Thaddition, addition the the iacmgs lacingsshould should carry aatransverse transverse force force equal Lqual toto2.5% 2 5%of ofthe the aXial column force, I.e. 0.025 x 365=9.1 kN. Note that the ongmai value axial column force, i.e. 0.025 x 365 =9.! kN. Note that (he onginal valueofof 1% 1% Inin clause clause4.7.8i 4,7.Siwas wasconsidered considered too toosmalL small.
F/AgPJ'+M~M=11 -
365/(2x 47.5 xx 275 275xx 10-')+326/909=0.50 105±326/9090,50 365/(2 x 47.5
Lacmg 9.lIcos 51.3" = 13.3 kN Lacingforce force= =9.1/cos Sl.3a= 13.3kN Tota! Total force force =75.5+13.3 =755 + 133 =88.8kN =88 8kN
(e) Overall Overall budding buckling
(e)
I.e. I e 44.4 444 kN on on each eachside sideof of column. column
For compressive strength the slendernessAAISis based basedon on an an effective effective For compresslve strength the overall slenderness length
Use Use60 60xx 60 60 xx 66 equaJ equal angle: angle
length:
115 appendix D (fig. 20,1
BS appendix D (fig. 20)
.
L£a LEb
r.
EffectIve fO.80 + =iJ28 .28 m m Effective length length of of lacmg lacuig = -I ± 1.00 002]] = V 10 ;, = 128011 1.7 1280/11.7 == 109 109 Compresslve == IIllJ I N/mnr N/mml Compressive strength strength Pc Pc CompressIOn resistance Pe =A Pc g Compression resistance P. =4gPc 1 =6.91 =6.9!X x[11 Ill xx 10=76.7kN = 76.7 kN 2
clause clause4.7.811 4 78/i
= 1.5 x 10.0= 15.Om, or
= 1.5 x 10.0 = 15.0 m. or =0.85 x IO.O=8.Sm =0.85 x 10.O=8.5m V(JJA) == ';(I.,lA)
:9!-
2:
BS 118table table 27c 27c
=4152 6701(2 xx 47.5)]=40.t =';[152670/(2 47.5)]=40.1 cm cm LgJra Aa =LE.!ra =37 == 1500/40.1 1500/40.1 =37 .4 1b =L£i/rb = 850/12.3 = 69 =850112.3=69 Local slenderness =L/r,y Local slenderness Ac =Ur = 1000/26.7=33 =1000/26.7=38 clause 4.7.Sg Limiting value for 4 =50 ciallse 4.7.8g LimIting vaiue for Ao:- = 50 Overall s(enderness Ab 1.41 Overall siendemess Ab ;t 1.4..1. = lA xx 38 38 53 =53 ES table 27a strength Pe p. based i.e. 69: BS table 27a Compressive Compresslve strength based on on the the highest highest value value of of A, A, I.e. 69: Pc =225 N/mm2 Pe =225 N/mm2
-
Sechon Section is is satisfactory. satisfactory.
STUDY REFERENCES STUDY REFERENCES TopIC Topic
Bending to produce Bending about about the theA—A A-A axis axiS (Fig. (Fig. 7.12) 7.12) can can be be assumed assumed to produce axial axIal forces in the LIE sections. forces m the UB sectIOns. ktial =momentlcentroidaldistance distancebetween between tiEs UBs A.'oal force force =momeniicentroidal =326/0.8=408 =326/0.8 =408kN kN(tension (tensIOnand andcompression) compressIOn) Maximum lviaxlmumcompression compressIOn in m one one UB UB
=408 ± 365/2 = 591 kN =408+365/2=591 Compression P. =Ag Pc CompressIOn resistance resistance Pe =Agpc =47.5 =47.5 xX 225 225 xx 1O-! = 1069 kN Section is satisfactory. Section IS satisfactory.
2
3. 3.
Co!umrl behaviour belmvlour Column
instability of ofsEmis struts and AnalYSIS, and fmmeworks, frameworks, Slrllclura[ Stnictural A,iohsis. pp. 58—71. 58-7l. Van Rcmhold pp. Van Noslrnnd Nostrand Reinhold Dowling P.J., P.J., Knowles Knowles P. P. & & Owens Dwens G.W. (1988) Bowling
4. Column bcliaviour behaVIOur 4.
G.W. (1988) SlruclIIrai Steel Steel Design. Deslga. Steel Steel Construclton 5,rucn,ro/ Construction inslllule. lnsiiiuic l(irby PA. P.A. && Nethercot NethercotB.A. D.A.(1979) (1979)in-plane In·plane ICirby IIlslability of columns, DeSign for Sl11tClurol Stability. instability of columns, Design for Soisceie,'al Srobilio'. Collins Collins
5. Reduced Rcduced plastic plnstlc 5. moment moment
(1987) 51cc/work Steelwork Design, Design. vol. vo!. i, S~clton propertIes (1987) t, Scciion properties membcr capaciiies. capacltics. Sled Stccl Construction ConstructIon Institute J.nStltutc member
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COLUMN COLUMN BASES BASES && BRAC!
fl. lL
(81
Ia
COLUMN BASES BASES & & BRACKETS
LHH Slah Stab hase base
Fig. bases Fig. 8.1 8.1 Column bases
The transfer of force force between onc one element of a structure structure and and the the next next reqUires requires The particular care care by by the the deSigner. designer. A A good good detail detail may may result result from from long long partIcular expenence of of the usc and many examples examples are are expenence use of of structural structural steelwork, steelwork, and available for for students students to to copy .or or adapt(l,2) However, available However, equally good good details details may be developed with less may less expenence expenence providing providing the the following followmg basic baSIC pnnciples are pnnclples are adhered adhered to: tD;
•• ••
••
FL
B.2 8.2
me
GU5Sllled base Guaseted base
DESIGN COLUMN BASES BASES DESIGN OF OF COLUMN Column bases lis Column bases transnut transmtt forces forces and and moments moments from from the the column column to to its about foundatIOn. foundation.The The forces forces will will be be aXial axial loads, loads, shear shear forces forces and and moments moments about either axis aXIs or or any any combination combinatIOn of IS in 10 reality reality probably probably either of them. them. Shear Shear force force is transmitted by frictIOn between foundatlOU concrete, concrete, transmuted by friction between the the base base plate plate and and the the foundation but itIt IS reSisted totally totally by by the the is common common In in deSign design for for this tIns shear shcar force force to to be be resisted holding down bolts. bolts. The common common design deSign case with aXlalload about one one axis. aXIs. The case deals deals with axial load and and moment moment about If the the ratio raho of of moment/axial moment/axial load base lengih, ienglh, If load isis less less than than JJ6 L/6 where where L/. ISis the the base pressure exists eXists over the whole whole base base and and may may be be then aa positive positive beanng then beanng pressure over the calculated from from equilibrium equilibnum alone. alone. NomlOui holding down bolts are are provided provided calculated Nominal holding down bolts In this this case, case, and facl provided locate the the base base plate plate in and at at least least two two are are In in fact provided to to locate accurately. If If the the ratio mllo exceeds exceeds base base length lenglh JJ6, bolts are are accurately. L/6, holding holding down down bolts arrangements are are shown shown in In reqUired to to provide provide aa tensile tensile force. force. Both Both arrangements required
The forces forces to be camed earned must must be be set set out out and and transfened transferred between between different elements elements of of the the connection, i.e. aa realistIc realistic load load path must different connection, I.e. must be assumed assumed within the connection. connechon. Simplicity of detail usually produces produces the the most effective effective and nnd robust robust SimpliCity engtneenng engmeenng solution, solution, provided provided strength strength and and stiffness stiffness requirements reqUirements are satisfied. Any detail detail must be practicable Any practicable and and cost-effective, cost~effect1ve, both from from the the point of of view vie'v of and from of the site pOInt of the the steehvork steelwork fabncator. fabncator, and from that of site erector. erector.
Fig. 8.2. 8.2. Fig.
Column bases and brackets brackets are are connections connectIOns which which carry carry forces forces and and reactions reactIOns to to aa column column element. eiemem. Bases Bases transfer transfer reactions reactIOns from from the the foundation foundation to the the column, column, while brackets brackets may be he used used to to transfer transfer loads loads from from crane gtrders or similar Similar members. In In addition, addition, columns columns may may receive recetve loads loads via via beam beam connections Section 3.7g) 3.7g) or connectIOns 15cc (see SectlOn or cap cap plates plates (see (see Section Section 6.6). 6.6).
F
Al
----)A!
~-J-.~
.>
F
ML
~>k 1>
G
Bearing pfllssure pressure
Blll!
lenSlon tension
8.1 B.I
1..
"
Two Two main mam types types of ofcolumn column base base are are used used and and these these are are shown shown in Ifi Fig. Fig. 8.1. 8.l. Welded Welded or or halted bolted construction construction can can be be used, used, or or aa combination combinahon of ofboth, both, the the decision ornot notthe thebase baseisISattached attachedtotothe thecolumn coluITUl deCISion being beingdependent dependenton onwhether whetheror dunng fabncation, fabnca!1on, or or later later during dunng site site erection. erectIOn. In In genera!, general, the the stmpler slmpler slab slab base IS used used in in small small and andmedium mediumconstruction constructIOn v.'hen when axial axml load load dominates. dommates. base is The The gusseted gussetedbase baseisISused usedininheavy heavyconstruction constructIOnwith withlarger largercolumn columnloads loadsand and where certam amount amount of offixity fiXity isISrequired. reqUIred. where a11 certain Construction ConstructIOn requirements reqUirements and and details details are arc given given in In reference reference (3). (3).
1,1
Fig. 8.2 S.! Beinng Beanngpressure pressure Fig.
COLUMN COLUMN BASES BASES
Ibl
Where tension tensIOn does does occur in In the the holding down bolts, aa number number of ofmethods methods Where ofdesign deSign are arepossible: possible: of ~:
i.
assumed that that the the bearing beanng pressure has a linear linear distribution distributIOn to to aa It11 isIS assumed maximum value of maximum of 0.4 feu, wherej, where j""is ISthe theconcrete concretecube cubestrength. strength. baSIS is IS suggested suggested in in clause 4.13.1 4.13.1 and analysts analYSIS of ofthe the bearing beanng This basis This pressure and and bolt bolt stresses stresses may may follow follow reinforced reinforced concrete concrete theory theory pressure 8.3). (F;g. 8.3). (Fig.
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96 STRUCTURAL 55 5560 96 STRUCTURALSTEELWORK STEELWORK DESIGN DESIGN TO TO BS 5950
COLUMN & BRACKETS COLUMN BASES BASES &
(0'1
'I~~i
M-
"
Breadth B Breadth
bracket bracket ISis necessary necessaty for forthe thechosen chosenstructural structuralarrangement. arrangement.ItItdoes. does, however, however generate generate large large moments momentsminthe thecoiumn column(Section (Section7.7c) 7.7c)and and ISis therefore therefore used used only iayoui. only where where Itit IS is essenliai essential to to the steelwork layout. to the flange Hange of of the column Brackets Brackets may may be be connected connected either either to to tbe the web web or Otto Exampies are shown in 111 Fig. Fig. using using bolts. bolts, welds welds or or aa combinatlon combination of of the the two. two. Examples 8.5. In Fig. Fig. 8.5(a), 8.5(a). the moment moment acts plant! prodUCing the 85. In acts out out of plane producing tension tension In in the bolts, the plane of the the connection connectIOn bolts, while while III in Fig. Fig. 8.5
bearing pressure pressure Ir ""=bearing tensile stress stress In in bolts /1'"= tensile m = modular ratio m '" modular ratio A5 = total iotai bolt bolt cross cross section section As: 0=1.—n d=L-n d\=O.5{d-n}+MlF A, ""=Gmd,A3IB AI 6md 1AJ fB
0=1.— is p,l p mpcfll{mpc + PI) y: M+F(LIJ2—n) C=M+FIU2 nl
d- y/3 Ts C-F C—F T= Ic'" 2ClBy 1r
A rectangular pressure may also be used, pressure distribution distributIOn may used. which which leads leads to aa slightly analYSIS is IS slightly different different plate plate thickness thickness and and bolt bolt sizes. sizes. The The analysts based based on on reinforced rem forced concrete concrete theory theory for for ultimate ultimate limit limitstate. state.
In In general. general. the the first first method method is is used used to to obtain obtam beanng beanng pressures pressures and and bolt bolt stresses. IS obtained obtamed using usmg the the steel steel strength strength stresses. The thickness of the base plate ts N/mm2.2 • P.1'P from from Table Table 1.2 1.2 of of Chapter Chapter I,1,but butnot nolmore morethan than270 270N/mru Maximum MaXimum moment moment in m plate plate 1 L2PJV L2pJp ZZ (clause (clause 4A3.2.3) 4.13.2.3) where where Z Z is is the the elastic elastic modulus modulus of ofthe the plate platesection. section. I For For the the case case of ofconcentnc concentnc forces forces only, only, the the base base plate platethickness thickness may may be be obtained obtamed from from clause clause4.13.2.2. 4.13.2.2.
,I
8.3 8.3
BRACKETS BRACKETS Brackets Brackets are are used used as as an an alternative alternative to to cleated cleatedconnections connectIOns(Section (SectIOn3.7g) 3.7g)only only
where ofthis thissstuatton sItuallon isis the the crane crane where the the latter latterare are unsuitable. unSUitable. AAcommon common case case of girder girder support support on on aacolumn column (Section (Sec lion 5.3k). 5.3k). The The ecceninc eccentnc connection connection by by the the
Bolts Bolts In in tensIOn tension
~j
JPr
~::-n
L3
, 1
W 11-1-1l-~.!'.......j.
L
T
----ID ----
An alternative is sometimes An alternatIVe approximate approximate analysts analysIs IS somettmes used used which which assumes that that the the oenmssible permissible stresses stresses for for steel steel tension and concrete tensIOn and assumes beanng are reached together. This method is is shown shown in m Fig. Fig. BA. 8.4. beanng
p,[~
97 07
~
:;]> la) Face moment (a) Face connected momenl 01 plane) planel tout of
Fig. 83 Brackets Fig. 8.5 Brackets
8.4
"" II II
Bolts III Boils in shear only shear only
{b) Lapped lapped lorsional torsIonal moment moment tb) {in planel plane) (in
OF BRACKETS BRACKETS DESIGN OF are subiected subjectedboth both to to aa vertical vertical shear load and and to to aa moment moment due due io 10 Brackets are shear load tile eccentricity eccentnclty of the vertical vertical load. will vary vary with with the the load. Note that the moment will lie POlllt in III ihe the bracket bracket under consideration. consideratIOn. A bracket may also be be subjected subjected to to point hOflzontal loads, loads, but these are usually of a secondary secondary nature, nature. or or may may be be horizontal but these are usually (Fig. 5.1 5.11). specml detail detail (Fig. covered by aa special I). Vertical loads loads are are supported supported by by welds welds or or by by boIls bolts acting acting inInshear. shear.Ideally. Ideally, Vertical camed by bolts balls in In shear shear (or {or by by welds), welds), but but sonic some moments also should be carried the moments Dse to to bolt bolt tension. tensIOn. bracket arrangements arrangements shown shown 10 in Fig. Fig. 8.5(a) 8.5(a) wil! will gIve give nse Wis divided divided between between the the bolts or weld weld group group unifonnly unifonnly so so Vertlca! load iVis Vertical that: that: WIN kN kN Bolt shenr Bolt shear == (V//I
or or
Weld shear shear Weld
=IT'lL,,, IVIL kNfmm kN/mm w
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where N is the the number number of of bolts where L,. IS is the the Iota! total weld length (mm) Lw
N/mm2 (BS (85 table 215 N/mm2 table 36), weld capacities capacities For a weld design design strength strength of 215 (kN/mm) may (kN/mm) maybe be calculated calculated for foreach each weld weld size. size. Note Note that that the the weld weld size size isIS in 10 fact defined by the leg length fact length,and and the the destgn design dimension is IS the the throat distance, distance, i.e. throat size = leg lengthlV2. Values Values of of weld capacity are given in Le. size = are given In reference (4). Where the the moment bolt shear shear only only (Fig. 85) Where moment produces Droduces bolt 8.5)then thenthe theshear shear on on each bolt is each IS given given approximately approximately by: by:
8.5
EXAMPLE 15. DESIGN DESIGN OF OF SLAB SLAB BASE BASE EXAMPLE IS.
(a)
Dimensions
305 305 x 137 137 UC UC column column 305 x 305 (b)
where where dis the the boll bolt distance distance from from the the bolt bolt group group centroid. centroid. For aa weld group For group the the approximation approximatIOn is: IS:
(c)
Weld shear= Wed"'
(lie moment In some cases cases the moment produces produces bolt tension tensIOn and and in in these these cases cases bolt force is IS considered considered proportional to (0 distance distance from from aaneutral neutral axis. aXIs.The Theneutral neutral axis may be aXIs be taken laken as as dr!7 d r l7 in in deptlit?t. depth(2J, and as as aa result: result:
Loading
All loads loads include mclude appropnate appropnate values vaiues of vi I'r Case MaXimum vertical load load 1400 I
Bolt shear= IVed).,,JEd2
where I~ IS is the 1, 1.~ + + 1. ly for for the the weld group (Fig. (Fig. 8.11). 8.11).
99
Bearing pressure pressure
(See halts (grade 19rade 4.6) 4.6) (See Fig. Fig. 8.6.) 8.6.) Assume Assume base base520 520xx 520 520 plate plate and and four bolts 20 mm mm diameter. TenSion 245 =490mm2 =490mm 2 Tensionbolt boltarea areaA..A,=2 =2 x 245 d =S20—50=470mm =520-50=470mm
where the bolt distance where d is the distance from the the neutral neutral axis. axiS. For a weld weld group: group:
Assuming a concrete cube strengthfcu =30N/mm Assummg =30N/mm22 Permissible prcssure=0.4 Penmssible pressure=OA x 30 30 12.0N/mm2 = 12.0 N/mm2 520)=5.2 N/mm2 Pressure Pressure= = 1400 1400 x 10 3/(520 x 520)= 5.7 N/tnmt I
CASE CASE fii) (ii) LOADING
Weld shear=
Af/F= MIF=60!850=71 mm 60/850 = 7! mm L16 =520/6 =87mm =87 mm 06 =520/6 A/IF C< L/6 L16 M/F Base area ..4=520 520mm Base A =520 xx 520 mm22 =2700cm =2700cm22 Base 520 2/6mm 3 =23 400cm 400 cm: Basemodulus modulusZ=520 Z=520 xx 5202/6mm3=23 Pressure =F/A+,U/Z=6.11 +M/Z= 6.11 N/mm2 Pressure=FIA N/mm 2
J:
where is the eqUlvaleni equivalent second moment of of area IS the second moment area of the weld about about the weld group centroid. cenlroid. The effects of ofvertical verttcalload loadand andmoment momentdue due to to eccentricity eccentncitymust mustbe beadded added either for for individual mdividualbolts, baits,ororfor forpoints POintsinInaaweld weldrun. run.Clearly, Clearly,those those points pomts of of maximum force force or orstress stress need need to to be he checked, checked. which which occur occur at at positions positions furthest from aXIs. from the the group group centroid centroid or orneutral neutral axis. Where shear In in aa bolt, vectonal Where the the moment produces produces shear vectonal addition addition may may be be used. used. In cases cases where the moment produces produces tension tensIOn a combined combined check check may may be be used used (clause 6.3.6.3): 6.3.6.3):
CASE (Hi) LOADING EASE (iii) LOADING Fig. 3.3 8.3
L4 for ordinary F /P~ + F,iP, 1- 1.4 ordinary bolts bolts where where F4 F., isIS the applied applied shear shear P. P., isIS shear capacity capncHy F, Fi isis applied tension tensIOn
Pt ts IS iension tenSIOn capacity capacity
.1',
~ II
50111~ .~ 620
For aa weld weld group group all allcombinations combinations ofofvertical vertICalload loadand amimoment momentproduce produce shear in the shear In the weld, weld, and and vectonal veclonaladdition additionisisused usedasasnecessary. necessary.
FIg. 8.7 Fig, B.7
1
Al/F MIF =85/450=189mm =85/450= l89mm d1 =0.5(470—50)+85 d, =0.5(470-50)+85 xx 10~/450=339mm mm Al=6 =6x x399 399x x1515x x490/520=33.8 490/520=33.8 xxlO3mm2 JO l mm 2 A
distance y(Fig. IS the the solution solutIOn of: The distance y(Fig. 8.7) is .v=' -3(d —dt)v2+Aiv—Aià=0 _dl )y2 +Aly-A,d=·O y}-3(470-399) ),2+33.8 xx ID3 yy -33.8 <170=0 y—3(470—399)y2+33.8 —33.8X x10) l& x 470=0 hence 1'= 288 mm hence t'= 288mm PressureJ~ =6d,FlfBy(3d-y)] =6 x 339 339 x 450 450 x 10'/[520 x 2880 288(3 xx470—288)] 470-288)J =6 =6.<11 N/mml =6.41 N/mm2 Beanng satisfactory (C Beanng pressure pressure sallsfactory « 12N/mm2) l2N/mm2)
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________ 100 STRUCTURAL TO BS 55 5950 5950 100 STRUCTURALSTEELWORK STEELWORK DESIGN DESIGN TO Cd) (d)
=mf,(dly— 1) =mJ;(d/y-l) IS xx 6.41 6.41 (470/288-1)=61 (470/288— t)=6I N/mm' N/mm2 == 15 Force/bolt =6! xx 245 Farcelbolt 14.9 kN =61 245 xx 1O-3=14.9kN Bolt capacity capacity PI F, =PI A, p, At Bolt 195 xx 245 245 xx 10-' =47.8kN =47.8kN == 195
101
101
8.6 8.6
EXAMPLE EXAMPLE 16. 16.DESIGN DESIGNOF OFCRANE cRANE GIRDER GIRDERBRACKET BRACKET (FACE) (FACE)
(a) (a)
Dimensions Dimensions (See (See Fig. Fig. 8.10.) 8.10.) Coiumn Column610 610xx 229 229xx 140 140 UB UB Crane mm Crane girder girder eccentnclty eccentncity 550 550mm
Bolts are are sahsfactory. satisfactory. Bolts
(e) (e)
Plate thickness thickness Plate
(b) (b)
(See Fig. Fig. 8.8.) 8.8.) (See
Loading' Loading MaXimum reaction 462 462 kN Maximum rail rail reaction (including appropnate values of of }j) (including appropnaie Crane diaphragm restraint restramt Crane surge surge load load camed camed by diaphragm
I
Fig. 8.8 Fig. 8.8
Cc)
6.41 N mm2
Maximum bearing bearingpressure pressure from fromcase case(ii) (ii)loading loading=6.41 N/mm'2 M;:l;Iomum = 6.41 Nlmm Maximum BM BM (assummg (assuming constant Maximum constant pressure) pressure) =6.4! xx 520 =6.41 520 xx 1002/2= 100'/2=I6.7kNm 16.7kNm Some reducuon reduction of of BK! may be using the the trapezium Some BM may be found found by by usmg trapezIUm pressure pressure distribution. distribution. Try 25 2Smm mm thick thick plate: plate: Tr)'
Use 191 x 89 (grnde 43A) 43A) offcut of of 457 457 x 191 89 un till (grade Use affcul Maximum BM BM in bracket
221.5
308.5
= 102kNm =t02kNm =0.6py —0.6p, A" A, 3 ==0.6 0.6 x x 275 275 x x 10.6 10.6 x x 463.6 463.6 x x 10=811 kN =8!! kN Shear =462 kN =462kN Shear force force Fv F, F. JP" =0,57 <0.60 < 0.60 F, IF, =0.57 Moment capacity Afr:.T =P S;r; p,y S1 3 =533 kNm =265 2010 xx 10=265 xx 2010 lr3=533 kNm
M.=462 M,=462 x 0.2215 Shear Shear cnpaclty capacity p~ P,
I
·'1
Table Table 1.2 j.2
clause clause 4.13.2.3 4.13.2.3
=
,Pyp =265 265 N/mm2 N/mm2
Plate modulus modulusZZ =520 =520 xx 252/6=54.2 mm3J Plate 25 2/6=54.2 xx lO)mm Moment capacity== 1.2p)1' Z Moment capacity = 1.2 x = 1.2 x 265 265 xx 54.2 54.2 xx IO-J=17.2kNm
FIg. 8.10 Fig.
Bracket is IS satisfactory. satisfactory. Bracket
Plate is Plate IS satisfactory. satisfactory. For For larger larger loads loads and/or and/or moments, moments, aa gusseted gusseted base base may be required, reqwred, particularly particularly if if the the thickness Ihickness of of aa slab slab base base would would otherwise otherwise exceed 50mm. 50 mm. The The design design is IS the tne same same as as given given above, above, but but in m Section SectIOn 8.5(e) 8.5(e) the tbe plate plate modulus Z is is based of plate md gussets. modulus Z based on on the the combined combined effect effect of plate and gussets. At At thicknesses thicknesses greaier greater than tlmn 25 25 mm, mm, steel sleel grades gradesother otherthan than43A 43A may be be needed needed io 10 avoid avoid the the possibility possibility of of bottle bnttle fracture fracture (BS (BS table table 4). 4).
(9 (t)
(d)
Shear force force =462 =462kN !eN Shear Momenl 102kNm kNm ==l02 Moment
Column/base Column/base plate plate weld weld
I'
The The weld weld is, is. commonly commonly designed deSigned to 10carry carrythe the maximum mMtlmUm moment, moment, ignonng Ignonngthe the effect effect of of vertical vertical load. load. All Allcompression compreSSionus IS taken taken m ut direct direct bearing beanng(Fig. (Fig.8.9). 8.9).
_Jt1203kN
Maximum MaXimum tension tensIOn inIDflange flange
Fig. Fig. 8.9 8.9
For =2 For one one flange flange weld we Id length length = 2 xx 308=ól6mm 308 = 616 mm Weld 0.459kN/mm Weld shear=283/616 shear=283/616 =0.459kN/mm Use Use 6mm 6 mmfillet filletweld, weld,capacutyt4) capaclty(4) =0.903 =0.903kN/mm kN/mm
length =4 190+2 420= 1600mm 4 xx 190 + 2 xx 420 = 1600mm Weld length Weld force force (vertical (vertical load) kN/mm load) =462/1600=0.289 =462/1600=0.289kN/mni Weld 2 3 Weld second second moment moment l~ =2 =1Xx 420 112+4 xx 190 190 xx2202 220 Weld l =49xx106mm3 lOomm =49 =AfyJ!~ Weld shear shear(moment) (moment) Weld = 102 xx io3 10-x3 2201(49 x 2201(49x X106) 106 ) =102 = 0.458 kN/mm =0.458kN/mm Note that that In In this this case, case, the the vertical vertical shear and and the the shear shear due due totomoment momentact act Note
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102 STRUCTUnAL JOBS 102 STRUCTURALSTEELWORI< STEELWORK DESIGN DESIGN TO BS 5950 5950 COLUMN COLUMNBASES BASES& &BRACKETS BnACKETS103 103 perpendicular to perpendicular 10 each eachother, other, and andthe the resultant resultant shear shear is is obtatned obtamed by by vectorial vectorial addition. addition.
Resultant Resultant
AIx 0.550=254 kNm A!, =462 =462 x 0.550=254 kNm 2 Shear =0.9(450 4 xx 29)20 x, 2 Sheararea areaA,..4 =0.9(450 -—4 29)20 x, 2== 12 12 020 020 mm mm2 Shear P~ ==O.6p,. Q.6py AA,~ Shear capacity capacity P5 =0.6 x 1O-~=1910kN =0.6 xx 265 265x x12020 12020 x ir'=l9IOkN hence hence FJP~ =0.24 F,/P,024
Second Second moment momentof ofarea area of of plate plate (cm (cm ullltS) units) =2 =2 xx 20 20 xx 450j/12=30380cm.l 4504/l2=30 380 cm4 Minus 'holes: Minus bolt bolt holes: 502 =580 44 xx 20 xx 26 cm.l 26 xX 501 =580cm4 15022 ==5220cm4 4 xx 20 x 26 Xx 150 5220 cm -l Net Net I=24580cm I = 24 580 cm44 Modulus = 24 580/22.5= 580/22.5 = 1090 1090 cm~ Modulus Z Z=24
Connection bolts bolts Connection Shear force force == 462 462 kN kN Shear Out-of-plane moment=462 moment=462 xx 0.2415=112 Out-of-plane 0.2415= 112 kNm kNm
Use eight etght no. 22 mm mm diameter diameter bolts Use no. 22 bolts (grade (grade 8.8). 8.8). Shear/bolt F, F" ==462/8 462/8 Shearlbolt capacity p. P4 Shear capacity A, p, A. Shear =Ps =0.375 xx303 =0.375 303 Bearing capacity of plate plate P Ph,=drpI,, Beanng capacity of b• =drpbl 0.450 =22 x>c2020 xx 0.450 =22 Tensile force force (Fig. (Fig. 8.12) 8.12) Tensile
=57.8kN =57.8kN
=ll4kN =114kN
For provide For brackets brackets of of this lids type type 11it may may be be assumed assumed that that the bolts or welds provide Imeral be lateral restramt restraint to to the the compression compression zones. zones. The The moment capacity should he taken taken as:
i' I
=l98kN = 198kN
Mo=JJ y Zx =265 xx 1090 1090 x 1O-~=289kNm itfc,=PyZ,=265
= 457/7 =65 = 65 mm mm den =45717 F,;::::;Mdm dLd 2
Bracket is IS satisfactory.
:;;.
2 =112 xx 10' 3l5/[(152+1152+2151+3152)?i=IjlkN =112 10) xx 315/[(l5 +11S2+215 2 +3'J5 2)2J=11IkN Tension capacity =p, 4, TenSIOn capacity F, P, =P I AI =0.450 =0.450 x x 303=l36kN 303=136kN
Combined check Combined check
(d)
° k 63.4 0
F,/P4+F,/Pj FsIP.+F/Pd·1.4 1.4
Bolts Bolts are are satisfactory. satisfactory.
olo_
8.7 B.7
EXAMPLE EXAMPLE 17. 17. DESIGN DESIGN OF OF CRANE CRANEGIRDER GIRDER BRACKET BRACKET(LAPPED) (LAPPED)
(See (See Fig. Fig. 8.13.) 8.13.) Column Column 305 305 xx 305 305 xx 158 UC Crane mm Crane girder glfiler eccentncity eccentnclty 550 550 mm
(b) (b)
Loading Loading
sso m
{
0
'I'
28.9~ -----~
o '11 0 - n o !I r 10
° II!M o~:; ! s: o III g ° I~ 0 I Bno. 'i' 800. 0
I
tight right angles angles 10 radius radius to
Fig. Fig. 8.14 8.14
147 147
162 162
Coiumn bolts bolts Column Shear 462 kN Shearforce force= =462kN 0,550=254kNm Moment =462 xx O.550=2S4kNm Use eight no. 27mm 27 mm diameter diameter bolts bolts (grade (grade 8.8) 8.8) on on each each lace. race. eight oo. SheariboJt due =28.9 kN Shear/bolt duetotovertical verticalload=46218 load=462J8xx 22=28.9kN Shear/bolt due to moment = Mdm,,)r.d 2 =254 xx JQ-j x 168/8(902÷ 168/8 (90 2 +1682) 168 2 ) = 147kN =l47kN shear= 162 162 kN/bolt Vector sum sum of shear= capacJtylboit P, P, =Pl ri, Shear capacirwbolt p. A, 459=172kN =375 xx 459=172kN Beanng capacity capacity of ofplate p!aie Pb~ = = dcm, d[Pb~ Beanng =27>cx 20 20 xx 0.450=243kN 0.450=243 kN =27 are satisfactory. satisfactory. Bolts are lllrger Note that the lapped bracket bracket reqUIres requires tWIce twice the tltc number number or of bolts bolts or olaalarger bracket sIze compared compared with with the the face face bracket. size
0-
lltJ Ii
As As Section SectIOn 8.6(b) 8.6(b) Maximum MaXimum reaction reactIOn462 462kN kN (c) (c)
Use Use two two 20mm 20 mm thick thickplates plates(grade (grade 43A) 43A) shaped sbaped as as Fig. Fig. 8.13. 8.13. Maximum Maxlmwn 3M BM in inbracket: bracket:
2.
Jomls in 111 SOap/c Simple Consrn,coo,i, ConstmctlolI. vol. \101. I. i. Steel Slee! ConslructlOn lnstttute lnslitule Construction
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104 STRUCTURAL 104 STRUCTURALSTEELWORI( STEElWORKDESIGN DESIGN TO TO BS BS 5950 5950
3. Column 3. Column bases bases
(1980) Holding Holding Down Do,in Sysrems Systemsfor for Steel Stanchions. (1980) coicrete Society/B C&USteel COrlStroc/ion Construction institute Institute Concrete SocletylBCSAlSteel
4. Weld 4. Weld capacity capacity
(1987) Strength Strength of of fillet fillet welds, Design vol. vol. I, (1987) welds, Steelivork Steelwork DesIgn Section properties, properties, member member capacities, capacities p. 205. Steel SectIOn p. 205. Steel Construction lnstitUle Institute Construction
lot COMPOSITE BEAMS BEAMS & SLABS
The The term ·composlte: can can be be used used of of any any structural structural medium medium III in which which two two or In or more more matenals matenals mteract mteract to to provide provide the the reqUlreCl required strength and stiffness. In steelwork combine steelwork constnll:tlon constrUctionthe the term term refers refers to to cross~sectlOns cross-sections which combine two act togcther, together. steel steel sectlOOS sections with with Concrete concrete In in such such aa way way that that the Iwo Typical Typical cross.sections cross-sectionsof ofbeams beamsand andslabs slabs are areshown shown In in Fig. Fig. 9.1. 9.i.
performance of composite 10 that of of reinforced remforced The performance composite beams beams IS is similar similar to concrete beamsfl >, but there are afe two two main mam differences. differences. Firstly, Firstly, the the steel steel concrete has a significant significant depth moment of of area area may may not not be be section has depth and and its its second second moment Ignored, unlike unlike that that of the the steel bar reinforcement. remforcement. Secondly, the the concrete concrete ignored, reinforcement bond, actIOn, is IS to reinforcement bond, which which ISis essential' essential for for reinforced reinforced concrete action, absent in In composite composite beams be provided by by shear shear absent beams generally generally and and must must be connectton. DeSign therefore follow follow those those connection. Design methods methods for for composite beams therefore rem forced concrete concrete with modifications modificatIOns as indicated. mdicated. Owing OWlIlg to to methods for reinforced methods presence of the the concrete concrete slab, problems of compressIOn flange flange the presence slab, problems of steel compression mstability-and local usually relevant relevant instability-and local buckling buckling of of the the steel steel member member are not usually III simply Simply supported members except dunng dunng erection. erection. in RecommendatIOns for constructIOn are are not notincluded Included Recommendations for deSign design m in composite construction ill Part Part I of of BS BS 5950 5950 but but are are included mcluded in: Ill: m
The advantages advantagesof ofcomposite compositebeams beamscompared comparedwith withnormal normal steelwork steelwork The beams are arethe themcreased increasedmoment momentcapacity capacityand andstiffness, stiffness,or or alternatlvely alternatively the beams the reducedsteel steelsizes stzesfor forthe thesame samemoment momentcapacity. capacity.Apart Apartfrom. from.a saving in in reduced a savmg matenal, the the reduced reducedconstructIOn constructiondepth depthcan canbebeworthwhile worthwhileminmulti-storey multi-storey matenal, frames. The The mam main disadvantage disadvantageof of composlle composite constructIon construciton ISis the the need need to to frames. provide shear shear connectors connectors10 to ensure ensuremteractlOn interaction of of the the parts. parts. provide As inin all all beam beam desIgn, design, shear shearcapacity capacityand andmoment momentcapacIty capacity of of aa As composite section section must must be be shown shown to to be be adequate. adequate.But Butm in addition, addition, the the composite strengthof of the the shear shearconnection connectionmust mustbe besho\'.'1l showntotobe besatisfactory. satisfactory, with with strength regard to to both both connector connector failure failure and and also also local local shear shearfailure failure of of the the regard surrounding concrete concrete(see (seeSectIOn Section9.4). 9.4).For Forfull full mteractlon interaction of of the surrounding the steel steel and and concrete,suffiCIent sufficient shear shear conneclion connection must must be be provided provided to to ensure that the concrete, ensure that the ultimate moment capacity of of the section can can be be reached. reached.Lower Lowerlevels levelsof of the section ultimate moment capacIty connection will will result connectlOn result in m partial partial interaction mteracuon which which isis not not covered covered in in this this chapter(2) Composite beams beamsare areessentially essentiallyTT beams beamswith with wide wide concrete Composite concrete flanges. flanges. The non-uniform non-uniform distribution The distributIOn of oflongitudinal longitudinalbending bendingstress stress must must be be allowed for allowed for and and this this is IS usually usually achieved achieved by by use use of of an an effective effechve breadth breadth for for the concrete concrete flange. flange. For For buildings buildings the may be the the effective effechve breadth breadth B~ may be taken Inken as as one-quarter of of the (simply supported). one-quarter the span span (simply supported). Continuous ContlOuous beams beams and and cantilevers are are treated treated differently differently (see ES 5950: cantilevers (see BS 5950: Part Part 3.1). 3.1).
x0=Ap1Jlo.45 (,.,B,l
Fig. Fig.9.2 9.2 Moment Moment capacity capacity (NA (NA in slab) slab) 8, a,
'I
1 '
Fig. capacity Fig. 9.3 9.3 Moment capacity (NA In steel steel (NA in
Asc= All - 0.225 Aw:A/2_o.225
Stoat
JCIJ8~D~/Pr
beam)
IA -
9,3 9.3
SHEAR SHEAR CONNECTORS CONNECTORS Many of shear shear connector Many forms forms of connector have have been beenused, used,of ofwhich which two two are are shown shown Fig. 9.4, but but the the preferred preferred type type isIS the theheaded headed stud. stud. TIns This combines combines ease ease Fig. 9.4, of fixing fi.XlOg with witheconomy. economy. Shear Shear connectors connectors must of must perform perform the the pnmary pnmaty fUnctIOn of transfernog tmnsfemng shear shear at at the the steel/concrete steel/concrete interface mterface (eqtuvalent (eqmvaient to to function of bond) and between the they have hnve bond) and hence hence control control slip slip between the two two parts. parts. In'·addition, ln'addition, they carrymg tension tensIOn between bel ween the the· parts parts and and the secondnry the secondary function fl.inction of of carrying controllingseparation. separahon. controlling IS The relattonslup relatIOnship between between shear connector is The shearforce forceand andslip slip for for aa given given connector Important in in design deSign where where partmi interactIOn isIS expected. expected. For For the the design deSign in In important partial interaction this section. section, where interaCtion isISassumed, assumed, aa knowledge knowledge of ofonly onlythe the this where full full interaction reqUired. The The max.lmum shear shear force force which which the the connector connector can can sustain sustam isIS required. maximum strengths of strengths of standard standard headed headedstuds studsembedded embeddedinindifferent different normai normal weight concretes are 9.!. concretes are gIVen given In in Table 9.1. The strength strength of ofalternative alternative shear shear connectors connectors can can be be found found by byuse use ofofaa The standard push-out performance of ofall all shear shear standard push-out test test (SS (ES 5400: 5400: Part Part 5). 5). The The performance connectors ISis affected restramt of ofthe thesurrounding surroundingconcrete, concrete, the the connectors affected by by lateral latent restraint In in
9.2 9.1
SHEAR AND AND MOMENT SHEAR MOMENT CAPACITY CAPACITY OF OF COMPOSITE COMPOSITE BEAMS BEAMS The shear capacity of of aa composite The shear capacity Composite beam beam is IS based based on on thc the resistance resistance of of the the web web of of the the steel Sleel section section alone. alone. Calculation CalculatIOn of ofthe the shear shear capacity capacity P. P,. is IS given tn Section 3.7(d): given In SectIOn 3.7(d):
P,=O.flpy .4, Moment Moment capacity capacIty isisbased based on on asstimed assumed ultimate ultimate stress stress conditions conditions shown shown tn Figs. 9.2 and lO Figs. 9.2 and 9,3. 9.3. When When the the neutral neutral axis aXIs lies lies in In the the concrete concrete slab slab (Fig. (Fig.
9.2) 9.2) the the valuc value of of xp may may be be found found by by equilibrium eqUilibriumofofthe thetension tensIOnand and compression compreSSIon forces. forces. The The moment moment capacity capacity lH. is IS given gIven by: by: = Ap,
+ D /2 —x,, /2) Table9,1 9.1—- Shear Shear strength strength of ofheaded headedstuds studs Table
V/ben aXIS lies lies in In the the steel steel section section (Fig. (Fig. 9.3) 9.3) the the value value of As. WIlen the the neutral neutral axis may may be be found found by by equilibnum. equilibnum.The Thecentroid centroidofofthe thecompression compressIOn steel steel As. must must be be located, located, and and moment moment capacity capacity a\i. M .. is IS given given by: by:
presence of of tensIOn tension III in the the concrete, concrete, and and the the type type of of concrete concrete used, used, i.e. i.e. presence normal concrete or lightweight. For design of composite composite beams beams IIIin these these normal concrete or lightweight. For design of cases further further references(2) referencestu) should should be be consulted. consulted. cases The shear shear connection connection in in buildings may be on the the assumption assumption The buildings may be designed designed on that at at the across the the that the ultimate ultimate limit limit state state the the shear shear force force transmitted transmitted across interface ISis distributed distributed evenly evenly beh\'een between the the connectors. connectors. The The shear shear force force IS is mterface based on on the the moment moment capacity based capacity of of the the section sechon and and connector connector force force Qp is shown In In Fig. Fig. 9.5. 9.5. shown
B. xp x, (when Rc =0.45fCIJ B., (when HA NA in m concrete) concrete) BED. NA in in steel) steel) R.,=0.45fCIJ S., Ds (when NA The connector force must he checked: The connector force Qp must be checked:
9.5 9.5
t
Load 1 load
I
r _
't
NJf1 conneciors connectors
IIIT ITT 1 T1 ~ ii T T T T T T T T T T T TAT 1 T 1J 1T1T T A in concrete R, m,on,,,,, IT
0p1 =
°pz = RrINpz
Fig. 9.5 Connector force Fig. 9.5 Connector force
9.4 9.4
An and All! ilfe An and AIi are remforcement areas/unit ler191h
DEFLECTIONS
As calculated at at the the serviceability serllceability As In in steel steel beam beam deSign, design, deflection deflection must must be be calculated limit state, I.e. with the limit state, I.e. with unfactored unfactored loads. loads.The The presence presence of of concrete concrete In in the sec lion means the two two different different elastic dastlc moduli moduii (steel (stec! and nnd concrete) concrete) section means that that the mcluded, which must must be be included, which ISis usually usuallyachieved achievedby by\1se useof of the the transfomled iransfonned (or tor TIleelastic elastiCmodulus moilulusfor forconcrete concrete isIS usually usually eqUIvalent) cross~sechon{J·'1) lie equivalent) modified to creep. Under Under sustained sustamed loadiiig loading the the elastic elastiC modulus modulus isIS modified to allow allow for for creep. about one-third one~third that Er' about that under under short short term term loading. loading. The The mod.ular modular ratio ratio cca ((= if,! Et;) is IS taken taken as as 66 for for short short term term loading, loading, and and 18 term loading. loading. An An Er) I S for for long tong term eqUIvalent ratio ratio a.., may may be used, based based on on the the proportion proportion of of loading loading equivalent be used, considered to long term, term, and is aa linear linear interpolation interpolatIOn between behveen these these considered to be be long and is values. values. The values values of ofneutral neutral axis aXIS depth depth x .. and and equivalent eqUivalent second second moment moment of of The area 19 are arc shown shown in III Fig. Fig. 9.7. 9.7. This This allows allows deflections deflectlOlls to to be be calculated calculated using uSlOg area 2 nonnal elastic elashc formulae fonnulae with with aa value value for for if, E.•for for205 205kN/mm2 kN/rrun normal
Qp;> O.8Q,
N,,, conneciors NPl connectors
4,5
re,ntorcement areas/un,i lerigih
Fig. In concrele concrete Fig. 9.6 9.6 Shear io
where where or or
flI
I
I
B.
LOCAL LOCALSHEAR SHEAR IN INCONCRETE CONCRETE The total shear connection depends not only on the shear connector The total shear connectton depends not only on the shear connector
I
(headed (headed stud, stud, etc.) etc.)hut butalso alsoon onthe tileability abilityofofthe thesurrounding surroundingconcrete concrete toto transmit transmit the the shear sllearstresses. stresses.Longitudinal Longitudinal shear shearfailure failure isISpossible possible on on the the planes Fig.9.6. 9.6.Transverse Transversereinforcement reinforcement combined combined with with the the planes shnwn shown inIIIFig. cuncrete concrete should should give giveaastrength strengthgreater greaterthan thanthe theapplied appliedshear shearper perunit umt length length v,v,such suchthat: that:
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110 STRUCTURAL 110 STRUCTURALSTEELWORK STEELWORK DESIGN DESIGN TO TO 05 BS 5950 5950 96 9.6
COMPOSITE COMPOSITE BEAMS BEAMS & & SLABS Cc) (c)
COMPOSITE SLABS COMPOSITE SLABS Composite slabs slabs are are constructed constnicted from from profiled profiled steel steel sheetlOg sheeting with with two two ComposIte typical sectIOns, sections, as as shown shown In in Fig. Fig. 9.8. 9.8. The sheeting sheetmg alone resists the the typical moments due due to to the the wet wet concrete concrete and and other other construction constnjction loads. loads. When When the the moments concrete has has hardened hardened the the composite section reSists resists moments moments due due to concrete composite sectIOn finishes and and imposed imposed loads. loads. Composite Composite actton actton isis achieved achieved by by bond bond as as well fimshes well as web web tndentations, and in as mdentatlOns, and in some some cases cases by by end end anchorage anchorage where where the the connectors for for composite composite beams beams are are welded welded through through the the sheetmg. sheeting. connectors
111 111
BM BM and SF
Ultimate mOl1lent moment M% = 592 kNm Ultimate Ultimate shear shear force force F;z F, = 242 kN
-
(d) (d)
..
Shear capacity Assume UB. Assume the the beam beamtotobe be406 406xx 140 140 x 46 US.
Shear capacity capacity p~ F, =O.6py Shear A,. O.6p,. A,. =0.6 xx 0.275 0.275 x 402.3 402.3 xx &9=458kN 6.9=458kN Shear capac: force F" IP~ =0.52 =0.52 Fig. 9.8 Fig. 9.8 Profited Profiled sheeting sheeung
(a) te) In most IS controlled controlled by by the the construction construction condition condition ruttier rather In most cases cases design deSign is than by by the as aa composite section. In In general, general, the the failure failure of of composite section. than the performance performance as ihe slab slab lIS as a composite the composue section sectIOn takes takes place place owing oWing to to incomplete Incomplete interaction, I.e. i.e. slip slip on mteractlon, on the the steel/concrete SI eel/concrete interface. interface. For For these these reasons, rellsons, design of of composJle composite stabs sheeting has has evolved slabs with with profiled profiled sheetmg evolved from nom testing. testmg. deSign Details of of the test information mformatlOn are available available from from manufacturers manufacturers and SCl{4) Detllils The effects of of the the sheeting sheetlng profile profile on on connector connector performance performance and and on on beam beam behaviour are are also behaVIOur also given gIVen tn In the the SC! SCI publicatlOn{4)
Use effecllve breadth breadth B, Be as asLL14, 14, i.e. I.e. 1.85 1.85 m. m. Use effective For neutral Fig. 9.2. neutral ru{}S axis In tn the the concrete concrete slab, stab, see see Fig.
-
-
EXAMPLE lB. COMPOSITE EXAMPLE COMPOSITE BEAM BEAM IN IN BUILDING BUILDING The destgn deSign foltows follows that that given given in In Section SectIOn 3.7 3.7 for for aanon-composite non-composite beam. beam. The notation follows follows that of of 55 BS 5950: 5950: Part Part 3.1. 3.i.
(a) {a)
=5900 x 1850 1850 xx 30)=65mm = 5900 xx 2751(0.45 x In mm thick, see sec Fig. Fig. 9.9. 9.9. Th slab slab 250 250mm Moment Moment capaCIty capacity }.Ic=Apy lvi, =Ap, I(D~+DI2-xJ2) /(EJ, +D/2 = 5900 x 2751(250 + 402.312 -6512)1 0- 6 =5900 =619kNm =679kNm
i.asm
M;z IMc =0.84 AI,/Ai,
201
SectIOn satJsfactory. Section is satisfactory. Fig. 9.9 Fig.
(f) U)
-
Loading Loading As As Section Section 3.7b allowing allOWing the the same same self self weight weight of ofbeam. beam.
R, Rc =0.45L. =0.45/"" B, Be Xp J =0.45 xX 30 x 1850 1850 xx 65 =1623kN 55 Xx IO10'=1623kN
(See Fig. 3.2.)
(b) (b)
Shear Shear connectors Force Force
Dimensions Dimensjons
Span Span 7.5m 7.5 m simply Simply supported supported Beams al 6.0 6.0 m m centres centres Beams ai Concrete stab 250 250mm Concrete sillb mm thick thick (feu = 30 N/mm 2 ) soanmag spannmg in m two hvo directions directIOns Finishing screed 40 40 mm mm thick thick
-
x,=Ap, xp=APy I(0.4B,j,,,) 1(0.45Be)~II)
250F4___. 9.7 9.7
Moment capacity
19 mm diameter diameter by by 100 100 mm headed stud stud connectors. 1.':onnectors. Use 19mm mm high beaded
=40+2 xx 100240mm =40+2 100=240mm /(1/2) =R, J(Ll2) 1620)3700=438N/mm = 162013700 = 43 8 Nlmm Longitudinal shear shear capaCity -; 0.8LJ' Jlcu Longitudinal ==0.8 0.8 xx 240 240 xx ]30= J30 =1050 1050N/mm NJmm and and ;>0.03L,f~+0.7A=fy Length of of shear shear path path L~ 4, Lcngth Shear per per unit unit length Shear length v v
hot
S
=0.03 x 240 =0.03 240 xx 30+0.7 xx 0.785 0.785 x 410 410 =441 Nlmm N/mm =441
BRACING
Local shear is satisfactory. Local
(g)
Deflection Deflection Using unfactored unfactoredImposed imposedloads loadsasasininSectIOn Section3.7f, 3.7fW= W= 132kN. l32kN. Using The propertIes properties of of the The the transformed transformed sectionstti sections(4} are: are:
Pig. 9.7 93 Fig.
r=AI(B,1J,) =59001(1850 xx 250)=0.0128 250)=0.0128 =5900J(1850 = 10 10 at! = x,=[250/2+ ID +250)]/(l x,=[25012+ 10 xx 0.0128(20! 0.0128(201 + 250)]J(1 + + lOx 10 x 0.0128) 0.0128) = 176mm = 176mm 4= 79 700 cm' 19=79700cm-i 3 Deflection = WL IVL3I6OEJg = /60Elg DcflectIOn = 132 = 132 xx 7400'1(60 xx 205 205 xx 79700 79700 xx l0')=S.S 10') = 5.5 mm mm Deflection limit limit = =700/36O=2O,6msn DeflectIOn 7001360 = 20,6 mm
Section 9.7 9.7 SecllOll
10.1
Bracmg members" Bracing members, or or braced braced bay bay frames, frames, conslst consist usually usually of simple simple steel sections such such as as flats, fiats. angies, angles, channels channels or hollow sections arranged to to form form aa sections 6.1). Thc members are often arranged, using usmg cross-bracing, cros5~brncmg, so so truss (Section 6.1). The members that tensIOn only only basis. baSIS. that deSign design may may be be on on a tension A loading which which IS honzontal, derived dcnved from a A bracmg bracing will will carry catty loading is usually horizontal, number of of sources:
-
H.
•• ••• ••
Companng the the section section used used (406 (406 x 140 140 xx 46 46 UB) 00) with with that that required reqUired in Companng non-composite (533 x 210 xx 92 non~composlte (533 92 UB) VB) gives gIVes aa clear clear indication mdicatlOn of ofthe the weight weIght saving in composite compOSIte construction. constructIOn. However, as discussed discussed in in savmg achieved achicved in Section taken into 1010 account account in any cost Sectlon 9.1, some other costs must be taken
•
••
companson, companson.
constructIOn. construction.
Topic TopIC
Reference Reference
I. i. Reinforced Remforced concrctc concrete
Kong KR. F.K. & & Evans EvansRH. R.H.(1987) (1987)Reinforced Relliforcedconcrete concrete beams Remforced and beams—- the fhe ultimate ultimate limit state, Reinforced Preslressed Van Nostrand 85-155. Van Nostrand Prestressed Concrete, Concrete, pp. 85—155.
10.2 10.2
Reinhold Remhold
2. 2. Composite Composite cnnstnjction constructIOn
Johnson RI'. R.P.(1982) (1982)Simply Simplysupported supported composite compostte
3. 3. Transformed TransfOlmed
Kong 1,Kong KR. F.K. & & Evans EvansRH. R.H.(1987) (1987)Elashc Elasttctheory, theory, Reinforced Remforced and andFresrressed Prestressed Concrete, Concrete. pp. pp. 157—67. 157-67. Van Van
beams beams and and slab, slab, Composite Composite Structures of ofSteel Steel and and Concrete, 40—100.Granada Granada Publishing· PublishingConcrete. pp. pp. 40-100.
Nostrand Nostrand Reinhold Remhold 4. slabs 4. Composite Composile slabs
wind, wind, crane crane and and machinery machinery loads loath actmg acting honzontally horizontally on a structure; earthquake earthquake loads loads dcnved denved as as an art eqUivalent equivalent static static honzontai honzontai load; notIonal notional loads loads to to ensure sway stability; beam proportiOn of the longitudinal iongltudinal beam or or column column bracmg bracing forces forces as as a proportion force; loads the temporary temporary construction constructIOn stage. stage. loads present present dunng the
In addition, addition, bracmg, bracing, wheUler whether permanent permanent or or temporary, temporary, IS is usually usually necessary for steelwork steelwork erectors level properly framework dunng for erectors to to line line and and level properly the steel framework
STUDY REFERENCES STUDY REFERENCES
cross-section cross~sectlon
LOADING RESISTED LOADING RESISTED BY BY BRACING
U
Lawson Lnwson RISI. R.l\I. (1989) (1989) Design DesIgn of ofComposite Composite Slabs Slabs and and Beans B~am.r wit/i withSteel Steei Decking. Decking. Steel Steel Consiruction Construction institute lnstltute
SWAY STABILITY STABILITY Important that is important that all all structures structures should should have have adequate adequate stiffness stiffness agamst against sway. ItIt IS IS designed deSigned to to resist resist stiffness ISis generally Such stifThess generallypresent present where where the the frame frame is honzontal forces forces due to the wind wmd loading. loading. To To ensure ensure aa minimum mlflllllUm sway sway honzontai prOVISIon, notational notatIOnal forces in clause clause 2.4.2.3 2.4.2.3 applied applied provision, forces are suggested in honzontally: honzontally.
1.0% ofyj I or 1.O%ofYfW", or 0.5% if greater greater 0.5% of 1'1 (JVd+ PV... + IV,) Wf ) if i.4Wd + 1.6W, 1.6Wf vertically. verucalJy. aclmg inmconjunction conjunction with withi.4IVo-lacting This requirement reqUirement IS of the the honzontal honzontal wind wmd or orother otherloads loadsatsd and inIn is In in place of practtce forms fonns aa minimum minimum provision. proVISIOn. practice
-5.
I
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114
STRUCTURAL STEELWORK STEELWORK DESIGN TO SS OS 5950 5950 STRUCTURAL DESIGN TO
lO.3 10.3
BRACING
MLJLTI-STOREY BRACING BRACING MULTI-STOREY
difficuity where door wmdow openings are required. reqUired. TIse The alternatives difficulty where door or window alternatives shown may be used to accommodate openings, but but will will involve involve compression compressIOn of such such members members slenderness slenderness must must be be in the bracing bracmg members, members. in In the the design deSign of kept possible by tubes or hollow hoUow sections, reducmg kept as as low low as possible by lIse tise of tubes sections, and by reducing practicable. lengths tengths as as far as practicable.
in multi-storey multi-slorey frames frames hOflzonlal horizontal forces In forces may be resisted reSisted by: by:
••
ngidty JOlOling jointing die ngid!y the framework framework with with connections connectIOns capable of of resisting the applied moments and analysing reslstmg analysmg the the frame fhune accordingly; accordingiy;
••
providing stiff stiff shear shear concrete concrete walls walls usually usually at stair and lift providing lift wells, wells, and designing these to to absorb all the honzontal loads; and deslgmng these honzontai loads; arranging braced braced bay bay frames arrangmg frames of steel members members forming fomung trusses as as shown In in Fig. 6.3. shown
•o
115
10.4 10.4
SINGLE-STOREY SINGLE-STOREY BRACING The pnnclpal loading which smglepnncspal loading which reqUires requires the the proVISIOn provision of of braclllg bracing 10 in a singlestorey building IS Ihat due due to wind. wmd. In addition addition the longitudinal longitudinal crane forces storey building is that will will reqUITe require braced braced bay bay support. support.The Thehonzontal honzontal (wmd twind and and crane crane surge) loads transverse transverse to to the the building building are are supported supported by by ponai portal name frame actIOn, action, or column cantilever directlOU. cantilever actIOn, action, and and no no further further bracmg bracing ISis needed needed In in this this direction. Longitudinal Longitudinal forces forces do, do, however, however, reqUire require support support by by a braced bay frame as shown In forces anse from from pressures pressures or suciions sllctlOns on on the the slsown in Fig. Fig. 6.3, 5.3. TIle The wmd wind forces frictIOnal drag drag on the the cladding of aa gable end and frictional cladding of both both the roof and sides of SectIOn 12.4.3). 12.4.3). Gable Gable wsnd wmd gtrders girders are are needed needed tlsereforc therefore at at each each building (see Section oflhe and may may be be provided provided at at the the level level of oflhe rafters (low-pitcts) f1ow~Pllch} end of the building, and Ilte raftcrs of tlse the eaves, lO.3. The wmd girders are or at the level of eaves, as as shown shown In in Fig. Fig. 10.3. The gable gable wind vertical side bracing brncmg as as shown, whicls which IS the supported by vertical is also used to support the longitudinal crane forces_ The deSigned to span longitudinal crane forces. The gable gable posts posts themselves themselves are are designed carrymg tlse the wind wmd load ber-ween the gable gable wind wmd girder. girder. vertically carrying between the base and the
in all but In but the the first first case case the the steel steel beams beams and and columns columns may may be be designed deSigned as simply supported. supported. The arrangement of steel bracmg bracing or wind requires wmd towers of of concrete walls requires care to care 10 ensure ensure economy economy and simplicity. simpliCity. Alternattve Alternative arrangements arrangements are are shown shown in In Fig. 10,1. 10. L Symmeincal Symmclncal arrangements arrangements are are preferred preferredas as they Iheyavoid avoidtorsion torsIOnto 10 plan of plan of the the braced braced frames. fTames. The vertical vertical bracing bracmg must be be used used in in conjunction conjunctIOn with with suttable SUitable honzontal honzontal framing. Wind \Vind loads loadsare are transmitted transmittedby bythe thecladding claddingof of the the building buildingon onto the frammg. to the floors, and then to the vertical floors, and then 10 vertical braced braced bays bays or or towers. lowers. Design DeSign should should ensure ensure that that adequate honzontal frames frames exist eXist at at floor floor levels levels to to carry carry these these loads loads to to vertical bracing. bracmg. \'There V/here concrete floors floors are provided no no further further provision provISIon may may be required is also also be reqUired but in In open frame frame industnal Industna! buildings buildings Isorizontal hOrizontal bracing braCing is needed (Fig. 10.1). 10.1). Braced bay frames frames may take take aa number number of ofdifferent different forms fonns as as shown shown so 10 Fig. 10.2. while itit allows Fig. 10.2. Cross-bracing, Cross~bracmg. while allows a tension tensIOn only only design, deSign, creates creates Lack of ol symmetry symmeiry lack requires additional additional reQUlfOS
Symmetry Symmetry Wind w,!nd tower lower
r---l ,, ,, , - ___ J
Fig. Fig. tILl 111.1 Wind Windlowers towers and and bracing braCing -
/ Horizontal HOrIZontal framing or rigid tloor floor
\ Plan outline outline of building building
,
Side bracing
bracing __
,.
,~'~..-? ...... ;' .-'f....
,.
i
...
In truss lower lower chord chord In addition addition some some bracmg bracing may may be be reqUired required by by the· thetniss members. This IS a restraint restraint against cases "tscre wlH:re members. This is against buckling buckling and and isis needed needed in in cases of stress in In the the bottom bottom chord chord can can occur. occur. Lightweight LightweIght roof roofstructures structures reversal of this design deSIgn condition, condition, when wind Wllld suction suctIOn on use Ihe roof roof causes causes often have this compressIOn in In the the lower lower chord chord of orthe the truss. truss. compression
Ponal Irame _LL-,_.l..L_ Mixed bracing braCing
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uoinui.#1UHALSIEI:LWORK0ESIGNT0BsS • ,,,,
.:;0
I nUl" I
UHAl :::iII::ElWORK DESIGN TO BS 5950
10.5
BRACING 117 BRACING 117
BEAM TRUSS AND COLUMN BRACING BEAM TRUSS AND COLUMN BRACING
10.5
Both liexural and compression members mayrequire requireiateral lateralbracmg bracingoror Both flexural and compression members may restraint to improve their buekiing resistance. ThisprovISIon provisionhas hasbeen been restramt to in Improve their buckling reSistance. This discussed the appropriate chapters: discussed in Ule appropnate cbapters: Beams in buildings — Chapter 3 Beams III buildings - ChaPter 3 Crane girders — ChapterS Crane girders Chapter 5 Tnjsses Chapter6 6 Trusses - —Chapter Columns — Chapter 7 Columns - Chapter 7
I~~
L4 'c ;rp .:(e
10.5 10.5Wind Windgirder gIrder
l-jf4P?fSJsJ~f" t t t t 93
Fig. lOA.
]
13
8
Pressure land suction) EN
14 !-11.2)
Wind gmier Hg. 10.6 Wind gader loading toadmg
30 f-24.0)
—112)
1—240)
(c) Ic) truss
—25.6)
30 30 1-2<1.0) 2&w
14 H1.2l (—112)
Member forces may be obtained by any of the methods of analYSIS (SectIOn
Hinge Hinge restraints. resframls,--I-""",,,,,:l (plastic design) (plastic deSIgn)
Member Member
Faclored member force (kN) Factored member force (kN) Pressure Suction Pressure
184 299 299 338 338 -215
I 33
44
Dimensions
r—'.'
Dimensions
(See Fig. 10.5.)
(See Fig. 105.)
Gable end panel widths (6 no.) Gable end panel widths (6 no.) Depth of girder (in plan) Depth of girder (in plan) Side bay width Side bay width Eaves height
f:
SUm S.Omeach each 3.Om 3.Dm 6.Om 6.0m
12.Sm 12.5m
184
22
EXAMPLE (9, GABLE WIND GIRDER AND SIDE BRACING EXAMPLE 19. GABLE WIND GIRDER AND SIDE BRACING
Eaves height
32 32 !_25.6)
—27.2)
Member forces forces Member
Fig. lOA SpeCIal restraints
(a)
34 34 1-27.2J
Member forces may be obtained by any of the methods of analysis (Section 6.2a) and the press~re and suciton cases are shown ID the table; the loads 6.2a) and the pressure and suction cases are shown in the (able; the loads incorporate the the factor factor Yj= lA. incorporate jq= 1.4.
\
Advisable to AdVisable [0 have restraint have restraint at this Joint
Fig. iO.4 Specini restraints
(a)
(25.6I
lattice
at this Jotnt
(0.6 10.6
32 32 {-25.6}
93
;1
Compression member
e~Co\Y\i~
~
Reactions gable stanchions (spannmg vertically) vary as Reactions(excluding (excluding}j) from from gable stanchions (spanning vertically) vary as shown shownIninFig. Fig. 10.6. 10.6. Wind pressure or suctton (in brackets) resuits ill two sets of reactIOns. Wind pressure or suction (in brackets) results In two sets of reactions. These values are denved knowmg Cpt! and Cp, and arc given In These values are derived knowing C,,,, and C,,, and are given in SectIon Section 2.3. 2.3. Longitudinal load from crane (Section 5.1) = 12.0 kN. Longitadinal load from crane (Section 5.l)=s 12.OkN.
-
.1
_I
33 6 baysal 5.0m Obaysats.Om
i—I
In each case, the effective length of the portionof ofthe themember memberm Inmay each the effectiVe length of tile portion m comp""iSie,,j, becase, reduced by providing stnglemembers membersor or frameworks frameworkscapable capableof of may be reduced by providing slOgle resisting the lateral buckiing forces. Thevalues valuesof ofthese theselateral lateralforces forces have reslsUng the lateral The have been assessed from buckling test data forces. andgiven givenIninthe theappropnate appropnateclauses clausesofBS5950.: of BSS9S0. ' been assessed from test data and In some cases, e.g. crane girders, the buckJing force is combmed with other Inlateral someforces cases,soe.g. gIrders, the buckling force IS combined with other'thecrane design of the the bracmg. bracing. of lateral forces m the deSign The designer should always be aware of the need of bracing unusual TIle deSigner should always be aware of the need of bracmg Inin unusual positions, and should examine all compression members, and and compreSSIOn compression positions, and should examine all compreSSIOn members, flanges, to ensure that adequate lateral restraint exists and is satisfactory, restramt eXists and is sahsfactory. flanges, to ensure that adequate lateral Examples of restraints needed in lattice frameworks and portal frames Examples of restramls needed in lattice frameworks and portal frames ,7. (plastic design) are given in Fig. 10.4. ID
,
(b) (b) Loading Loading
-
(plastiC deSIgn) are given
1
<
'. 1
dimensIOns dimensions
55
—2)5 III
66 77 8 8 99 tO10 II11
iii
-133 _ijj 69 69
-46 —46 48 48 -184 —184 -299 —299
Suction -147
—147 -239 —239 -270 —270 172 172
-89
—89 106 106
-55 37 37 -38
—38 147
47 239 239
Compressionisispositive. positive. Compression
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External chord External chord Maximum force MaxImum force (compression) (compressIOn) 338 338kM kN
Use 254 254 xx 146 Use 146 xx 37 37 US un (grade (grade 43) 43) Slenderness A or Slenderness J =O,85Ur.r or = 1.0 J.1r1. =j.O Ury (see Fig. Fig. 6.4 6.4 and (see and Section Sectlon 12.7.2) 12.7.2)
BRACING BRACING
(g)
Side bracing bracing Side ReactIOn gIrder 93 93 kM kN Reaction from from Wind wind girder Crane 12 kN Crane load load 12kM MaXimum Maximum design design ioad load = lA 1.4 xx 93 93
\i"
=
== 130 kN 130kM =L2 93+1.2 xx 12 12 =126kM =126kN =1.2 x 93+1.2
ES table 22 BS I3DtN
-130 —130
=
Compressive sirength Pr =80 Compresslve strength Pc 80N/mm2 N/mml Compression resistance =Ag pc p. CompressIOn reSistance Pc =Ag
12 3 44 5C 6
,
=47.5 xx 80/l0=380kN ~47.5 80/10~380kN
.........4
5
Section is SectIOn IS satisfactory. satisfactory.
" ,,
7
77
It B.Om 6Dm (e) (e)
clause 4.6.3.1 4.6.3.1
clause 3.3.3 clause 3.3.3
Use 203 203 xx 133 Use 133 xx 30 30 UB un (grade (grade 43) 43) Slenderness = 1.0 1.0 xx 5000/31.8 Slenderness A ).= 5000/31.8 == 157 157 2 Compressive strength strength Pc p. ==68N/mm2 Compresslve 68N/mm Compression resistance =32.0 68/I0=2SRkN Compression resistance Pr: = 38.0 x 68/10 = 258 kN
.4,
=860mm2 = 860mm2 TenSion capacity py Tension capacity PI F, =Ac =A, m =860 xx 275 =860 275x x1O-:>=237kN lr=237kN
=
Maximum compression compressioninInstrut strut33= 130 130kM kN
32.3 cm,
Use 133 xx 25 25115 un (grade (grade 43) 43) Use 203 203 x 133
Tension capacity PI F, =A "A,..Py TensIOn capacity Pr
=32.3 =32.3 xx 275/l0=88BltN 275110=888kN
ES table 27b SS
Section SectIOn is IS satisfactory. satisfactory.
(I) (f)
Effecllve area a21(3a!+a2) Effective areaAc A, =a, =a1 +3al +3ai a,/(3ai+a2) al =(100-7/2)7-22 xx 7=522mm2 7=522nm1 2 a, =000—7/2)7—22 Q1 = (65 - 7/2)7 = 43! mm 2 a2 =(65—7/2)7 =431mm2 allOWing mm diameter diameter hole allowing for for one one 22 22mm hole In in connected connected leg leg (lOOmm) (100mm) Ac =522+3 xx 522 522 xx 431/0 4311(3 xx 522+431) 522+431) A,
Maximum Maximum tension tension 299kN 299 kN Effective area area AA,= Effecttve .. = 1.2 1.2 A,,, An.., but but;t Ag Allowing Allowing for for two two 26 26 mm mm diameter diameter holes hoies A,= A .. =1.2 1.2 (32.3—2 (32.3-2 xx 2.6 2.6 xx 0.58)=35.l 0.58)=35.1cm2 cm 2 but
188 188
135 135 -135 —135 270
Use Use 100 100 xx 65 65 x 77 Angle
Maximum compression Maximum compressIOn 239 kN
BS fable table 27b BS 27b
188 188
-130 —130
IVlaximum tension in In diagonals diagona-Is 22 and 4 (assuming {assummg cross-bracmg Maximum tension cross-bracing io to avoid avoid compressIOn) compression) == 188 188 kN. kM.
Fig. 10.7
Internal Internal chord chord
or
With 10.7, the factored member forces forces (kN) (kN) arc: are: With reference reference 10 to Fig. Fig. 10.7, the factored
1
hence mnx. max.A= 2=1.0 hence l.0 xx 5000/34.7= 5000/34.7= )44 ES table table 27b 27b BS
119 119
A ~ 1.0xx6000/31.0=194 6000/31.0 ~ 194 1=1.0 Pc =46 N/mm2 =46N/rnm2 CompreSSiOn resistance resistance Pc = 46/10 = 149 kN kN Compression = 32.3 x 46/10 Forces In Forces in the the eaves eaves girder girder 1,I. and malO main frame frame members members 5, 5, 66 and and 77 should be considered in values of of considered in the the deSign design of these these members members when when appropnate. appropnate. The values these forces forces will in the IIle combination CombinatIOn of offorces forces ihese will need need to to be be adjusted adjusted for for )'fused yjused in member. for each member.
Diagonals)struts Diagonals}struts Maximum Maximum compression compression(diagonal) (diagonal)172kM. 172 kN.
.85 BS table (able 27b 27b
Use Use 203 203 xx 133 133 xx 30 30US UB(grade (grade43) 43) Slenderness 2=1.0 = 1.0xx5830/31.8= 5830/31.8 =183 183 Slenderness A pc=52N/mm:: Compression Compression resistance resistance Pc=38.0 x 52/IC 52/10 =198kN =198kN Maximum compression{strut (strutmember member5)5) == Ill Maximum compressIOn IIIkN kN 2= 1 ~ 1.0 1.0 x x 3000/31.8=94 3000/31.8 ~ 94 Use same section. section. Use same
Dimensions Dimensions (See Fig. (See Fig. 10.8.) 10.8.) storeys atat3.5 7 storeys 3.5mmhigh high width 4.Om 4.0 m Bay width 10 allow door door openings opemngs Cross-bracmg with nlternatJve floors to Cross-bracing with K-bracing K-bracing on on aliernative
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120
STRUCTURAL STEELWORK DESIGN TO OS 5950 STRUCTURAL STEELWORK DESIGN TO as 5950
1 ;.W
(b)
L
Loading Loading Wind loading transmitted to bracing concrete (learslabs slabsatatench eachievel. level. Wind loading transmitted to brncmg byby concrete floor
(b)
(c)
Member forces
Member forces
(c)
Member forces UrN) excluding Yf are given for the lower two storeys Member forces (kN) excluding }'fare glven for the lower two storeys oniy:
PLATE P LATE GIRDERS GIRDERS
993 716 23 El —1304 E91 -1304 34 in —716 ..; 4 -716 ~ 56 472 —197 ~ ~ 67 -197 319 ~ 7 319 -319 8 317 9 -3170 10 0 10 Note that wind loading can act in the reverse direction, which will generally Note thatthe wind loading caninact m the reverse direction, which wiJI generally reverse force direction each member. Member 10, 10, which which has haszero zero load, load, reverse thewill force directIOn In each member. Member however, 317 kN ia this wind reversal case. however, will carry 317 kN in this wmd reversal case. (d) Cross bncing (d) Cross bracing I
'"u 55 x
55—+
NV ZN
~
x
-" l'i
49 42
.910
.
-
~
II
I 1.1
I
Fig. 10.8 Fig. 10.8
Occasionally, of a member cannot be Occasionally the the reqUired required bending bending reSIstance resistance of a member be and provided by the largest available umversaJ beam x 419 caunot x 388 UE) provided by the largest available universal beam (914 (914 419 x 388 TJB) and therefore the designer has 10 resort to uSing a plate guder. Basically, a plate therefore the has to resou to using a plate girder. Basically, a plate fastened girder IS built up from three plates (one web and two flanges) is built up from tiree plates (one web and two flanges) fastened I~shape, see Fig 11. I. There are many examples of plate together together to to fonn an an 1-shape, see Fig 11.1. There are many examples of plate girders, e.g. crane girders m heavy mill buildings, road and rail bridges, roof girders. e.g. crane in millInbuildings, road and rail bridges roof stadia, balcony balcony girders girders concert halls. construction constraction of of stadia, in conced halls.show Ihat the ExammatlOn of older forms of plate gtrders would Examination of older that the the flanges of andplate webgirders plates ISwould madeshow by nvets or boits, via connectlon counecuoa between between the flanges and web plates is made by nvets or bolts, via angle sections, as shown in Fig I U(b). These piate glfoers were relahvely angle sections, as in Fig 11.1(b). These plate werewas relatively more prevalent as the depth of rolled sections, prior togiruers the 19505, limited more prevalent as the depth of rolled sections, prior to the 1950s, was mlimited 0.6 m, whereas today Universal beams are rolled up to 0.92 deep. to about to about 0.6 m, whereas today universal beams are rolled up to 0.92 m decp. Splices for for this this type of plate girder, required because of transportation Splices plate girder, required because of transportabon considerations andlorofmaximum available length of plate, were provided by considerations and/or maxünum available of plate, were provided by means of nveted or bolted cover plates. means of nveted or bolted cover plates. The advent of welding allowed the deSigner the freedom to tai!or~make a The advent of welding allowed the freedom to tailor-make SUIt any any deSign requirement. As thethe chOIce of plate is restricteda member to to suit member
.
Section JO.6e SectIOn JO.6e
tension = 1.4 x 472=661 Maximum lenslOn= lA x 472=661 kN Use 203 x 133 x 30 UB 43) Use 203 x 133 x 30 UB {grade 43) Tension =888 kN TensIOn capacity = 888 kN
(e) -
(e)
K-bradng
compresstonl.4
83 table 27b BS table 27b
MaXlmwn compression=lA x'319=447kN Use 203 x 133 x 37 UB (grade 43) Use 203 x 133 x 37 UB {grade 43) 1 =0.85 x 4030/34.7 = 116 .l =0.85 x 4030/34.7 = 116 = I 14N/mxn2 Po =114N1mm' Compression resistance P, =47.5 CompressIOn resistance Pc =47.5 xx 114/10=541 114/10=541kN kN As in Section lO.6g the forces in colunms 1, 2, 3 and 4 and beams 6, 9 and In Section lO.6g the forces In columns 1,2, 3 and 4 and beams 6, 911nd 10 As must be taken into account in the overall design of these members which 10 must be taken mto imposed account loading in the overall design of these members which will include dead and from floors, etc. wiII includetodead and Imposed etc. The The value valueof ofyr 'If appropnate each combination loading of loadsfrom floors, appropnate to each combination of loads must must be beused llsed(Section (Section2.7). 2.7).
INTRODUCTION
.
c.
requirement. As the choice of plate is restncted
STUDY REFERENCES STUDY REFERENCES Topic TopIC
I. Frame slability 1. Frame slnbility
Reference
Reference
(1988) Stability of Buildings. The histituijon of (1988) Stability o/Buildings. The lnslitulion of Structural Engtneers
Table 11 II.1(11) 1(a) I\.·lmomum Maximum rolled rolled length' lengths Im) im) for selected range of Table of wide wide flnts(l) 'i5i Thski.'.i,i iYidi.i_,i 55 Se u 5l ~\V'."'''''j 11 I! l' U j' lJ ~55 ~jIi n5' J! /4 15 TO
readily accommodate this requirement, reqUirement, see Figs. Figs. IIIl.l(c) accommodate this - lie) and and ll.l(d). II. 1(d).The The deSIgn and design strength strength of nf each each plate plate ISis dependent dependent only oniy on on ItS its own own thickness and grade grade of steel, steel, unlike unlike the the Universal universalsectiOns sections where where the thedeSign design stTength strength IS is based on on the the thickest thickest part part of of the the I-section. 1-section,I.e. i.e.the theflanges. flanges.Also, Also,different differentgrades gradesof of steel sleet can can be be used used for for the the plates plates within within one one girder. girder. e.g. e.g. Ihe the use use of a notch ductile 10 ductile steel, steel, say say SOC, 5CC, for for the the tensIOn tension flange flangeof ofaa road road bridge bridge lfi tn order in elimmate low cycle bnttle brtttle fracture. fracture. eliminate the the possibility possibility of low Plate usually reqUire carrymg stiffeners and intermediate mtermediate Plate gIrders gtrders usually require load carrying stiffeners stiffeners (non-load (non-load carrymg). carrying),see seeFig. Fig. 11.2, 11.2,dividing dividingthe the web web mto into panels panels.. These the following followmg functions: functIOns: These have the
Table 31.1(b) l1.Hb) Maximom Ma;umum rolled rolled lengths lengths dn) (ht) for for selected selected range mnge of of plaies plntes ri_i,)
only only by the Ihe discrete discrete stzes sIzes of of rolled railed plate plate (see (see Table 11.1) 11 I) then then this this form (onn of of construction construction can can be be economic economic in In terms tenns of ofmatenal. matenal.The Theflanges flanges and andweb webare are normally nonnally connected connected together together by by fillet fillet welds, welds. using using semiseml- or or filly fully automatic automatic welding procedures. procedures. The designer deSigner must must assume assume that that the the load load transfer transfer isIS entirely entirely through through the the welds welds as as there there isISno noguarantee guarantee there there IsIS aaperfect perfectbeanng beanng between reqUired, owing owmg to 10 maxtmum ma;umum between flange flange and and web. web. Where a splice splice isis required, length of of plate plate rolled rolled or or because because there there isIS aa change change to In plate plate thickness, thickness, aa full full strength butt strength butt butt weld weld isis required. reqUired. As Asthere there are areaanumber numberof ofdifferent different types types of ofbuU weld, ofthe the appropnate appropnate type typeshould should be be discussed discussed with with the the weld, the the selection selectIOn of fabncator in III order order to toproduce producean aneconomic economicsolution. solullon. Generally, Generally. aa plate plate gtrder girder isIS made madedoobly doublysymmetrical, symmetncal, i.e. I.e.both bothflange flange plates (b) plates are arcidenticaL identical,like likethe theuniversal universalsections, sectIOns,see seeFigs. Figs.11.1(a) l1.1(a)and and11.1 I l.l(b). However, However, should design deSIgn conditions conditions dictate dictate that that aa single smgle axis aXlS of ofsymmetry symmetry isis necessary, necessary, then then aajudicious JudicIOUSchoice chOIceofofdifferent differentplates platesfor forthe theflanges flangescan can
•• ••
clallse 4.4.5.J 4.4.5.1 clause
Load carrying carrying siiffeners stiffenersare are uted usedtotodiffuse diffuse any any concentrated concentrated load load locally, from axial aX131 load load in III locally, mto into the the web. web. Tlus This load load can can result result from columns fTom an columns connected connected toto aa flange flange or or an an end end reactlOn reaction from the mtersectmg Intersecting beam beam member, member,which whichcan canbe be connected connected 10 to either either the flange stiffener ttself. Itself. flange or the stiffener The to control control the the shear shear The sole sole funchon functtonof ofintermediate intermediate sUffeners stiffeners isIS to any web web area/panel the flanges flanges resistance of any buckling resistance area/panel bounded by the and of intermcdiate intermediate stiffeners. stiffeners . and an adjacent pair of IS aa functton function of of elastiC cntical shear buckling The elastic cntical shear buckling of of aa web web panel is aid (known (known as and dh, dll, where a is IS the the distance distance as the the aspect aspect ratio) and between belllg between the the two two stiffeners stiffeners bounding bounding the the web web panel panel being considered, IS the web web considered,dis dis the the actual actual depth depthof of the the web web plaie plate and and Iits thickness. Note umversal sections, the overall depth depth plate thickness. Note that that for for universal sections, the D is allowed allowed for for calculating calculating the the shear shear area. area. Dis
IS relatively relatively thin, thin, the presence of of intcrmediatg i'1termediatt; stiffeners stiffeners isIS When the web web is When useful in in maintaining mamtammg the the I-shape, I-shape, particularly particularly during dunng transportation transportation and and useful erectton. In very deep plate girders, gmlers, additional additional honzontal honzonlal longitudinal longitudinal erection, stiffemng maybe may benecessary necessary in In the the compresston compreSSlOn zone, zone, in In order order to to maintain mamtaman an stiffening economic web thickness. thickness. Tills the economic This particular particular deSign design vanatlOn variation lies outside the scope of of BS 5950: 5950: Part but isIS covered covered in in BS BS 5400 5400 and and therefore therefore will will not not scope Pan I,I, but be discussed discussed here. here. The two two main main forces forces that a plate plate gtrder has to 10 resist resist are are bending bending moment moment The force, though be taken taken into mto and shear force! though axml axial force, force, if present, present, would would need need to be a;ual load) ioad) and and account. Though account. Though tu in reality. reality, the bending bending moment moment tplus Iplus any axtal uSllal assumption assumpuon force would would be resisted reSisted by the whole whole section, the the usual shear force made for for small and medium plate girders isIS that that the the flanges flanges resist resistthc thebending bending made moment and the web cames cames the the shear shear force. force. moment
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••• ~ .... , .... , .........
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iUBSSY 50
! U BS 5950
PLATE GIRDERS PLATE GIRDERS
The further apart the two flanges beammember memberare areposItioned positionedfrom The further apart the two flanges ofofa abeam from the centroidal axis, the better is the memberbending bendingcapacity. capacity.The {he member
• The web is made deliberately thick, removing the necessityfor for • ;The web IS made deliberately thick, removing Ihe necessity
I 11.3 I.3
intermediate stiffeners; the disadvantage
• •
•
that the the weight weight of of Ihe the :inlennediate stiffeners; the disadvantage ISisthat girder would be relatively heavy compared with that obtained by \girder would be relatively heavy compared with that obtained by using the other methods and could increase the cost of the uSing the other methods and could increase the cost of the foundations. However, the overall fabncation costs would be lower costs would be lower ·foundalions. However, Ihe overall fabncatlOn tno stiffeners), which may more than offset the extra material than offset Ihe extra matenal and (no stiffeners), which may more and foundation costs. foundatton costs. The web is made thin enough require intermediate stiffeners The web IS made thin enough toto require intennediate stiffeners 10to control the shear buckiing actidn within any web panel. conrroi the shear buckling action within any web panel. Finally, the web thickness is minimized by taking into account Finally, the web thickness is mmltmzed by taking mto account tension field action, thereby maximizing the effectiveness the tensIon field action, thereby maXimizing the effectiveness of the web and flanges. Tension field action is discussed in Section 11.7. web flanges. Tension field acttOn is discussed in SectIOn 11.7. This and nietliod cannot be applied to gantry girder design. 17tis melhod cannot be applied la gantry girder desIgn.
AA plate of 22 m, IS reqUired 10 carry the plate girder, girder, Simply simply supported supported over over aa span span of 22 in, is required to carry the loads load includes the toadsindicated indicated inin Fig Fig 11.3; 11.3; Ihe the uniformly uniformly distributed distributed dead dead load includes the self uxlili load self weight weight of of the the gIrder. girder. The The concentrated concentrated applied applied load load ISis the the axial load x 46 UC) and the ends of the transferred transferredfrom from aa column column member member (203 (203 xx203 203 x 46 UC) and the ends of the girder supported by 254 x 254 x 73 UC girder and and those those of of adjacent adjacent gIrders girders are are supported by 254 x 254 x 73 UC columns. io 1.5 m owmg to columns.The Thedepth depthof of the the plate plate gIrder girder isis to to be be limited limited to i Sm owing to mlOlmum headroom requirements. reqUirements. minimum headroom IS assumed 10 For For the the purpose purpose ofthis of this example, example, the the top top (compressIOn) tcompression) flange flange is assumed to be prevented from from rotating. The plate gmler IS to be be "restrained 'restrained laterally laterally and and prevented minting. The plate girder is to be steel. fabncnted fabricated from from grade grade 43C 43C steel.
The design of the plate girder must also comply with TIle 4.3, design the (lateral plate girder must also comply with the Ihe guidance guidance given given in clause BSof 5950 torsional buckling buckling of beam members). members). The design of beam clause 4.3, BS 5950 (lateral torsIonal of load canying and intermediate stiffeners is covered by clause 4:5. of load carrymg and intermediate stiffeners IS covered by clause 4~5. The different methods of designing different meUlOds of deslgnmg plate plate girders girders are are illustrated illustrated by by Examples 21—24 inclusive. which include further design information where 21-24 inclusIVe, which include further design informatIOn where appropriate. appropriate.
11.2
DESIGN OF UNSTIFFENED PLATE DESIGN OF UNSTIFFENED PLATE GIRDER GIRDER
itiHiHit III IHI 111111I I h. Flg. FIg. 11.3 Details Delails of ofgirder girder
An unstiffened plate girder is similar to a universal beam section, An unstiffened le girder to a universal beam sectIOn, where where the the web is generallypIa thick enoughIS sImilar not to to necessitate necessJlateshear shearstiffening/intermediate stiffening/intermediate web IS generally thick enough not stiffeners. According to 13S 5950: Part 1, the moment stiffeners. According to BS 5950: Part 1, the moment capacity capacIty of ofan an unstiffened plate girder depends on the value of dli, i.e. unstiffened plale girder depends on the value of dll, I.e. clause 4.4.4.1 clallse 4.4.4.1 clause 4.4.4.2 clause 4.4.4.2
•
•
•
•
If dIr C 63s (thick web), then the If die < 63£ (thick wcb), then Ule moment moment capacity capaCItyof ofthe theplate plate girder, can be determined girder. can be detcrmincd as as for for universal untversal beams, beams,i.e. I.C.according according toto clause 4.2.5 or 4.2.6, B5 5950. or 4.2.6, BS 5950. Ifclause 4.2.5 63s ((lila wet,), then the moment If dll;;:: 63£ (thin web), then the moment eapncity capacity can canbe be calculated by one of two methods or n combination of the two calculated by one of hvo mcthods or n combination of the two methods (clause 4.4.4.2). The rtvo methods are: methods (clause 4.4.4.2). The two methods are: — moment plus any axial load resisted by - moment plus any nxml load reSisted byflanges flanges only, only,and andthe the web is designed for shear only (clause 4.4.5). It is assumed that web IS deSigned for shear only (clause 4.4.5). It is assumed that each flange is subject to a uniform stress Pp each flange IS subject la a uniform stress Pv. — moment and axial load resisted by whole - moment and aXial load resisted by wholesecuon sectIOnwith withthe theweb wcb resisting the combined shear and longitudinal stresses (see resisting tile combined shear and longItudinal slresscs (see clause H.3).
1~50
240 kN 240 kN kN 450 kN
20 kNlm 25 tN/rn kNlm i ! i ! I t 20
j
11.2
125
Similar wllh clauses 3.5 SimilartotouOlversai universalbeams, beams,a aplate plategmler girderhas hastotocomply comply 'vith ciauses 3.5 4.2 (members 10 bending) and 4.3 (laternl torsional (local buckling), (local buckling), 4.2 (members in bending) and 4.3 (lateral torsional reqwre end beanng stiffeners In buckling), buckling),BS BS5950. 5950.Also, Also,a aplate plategIrder girdermay may require end beanng stiffeners in order to transfer the end shear into the supports, order to transfer the end shear into the supports and and load load carrying carrying stiffeners stiffeners within thc span of a where wherelarge largeconcentrated concentratedloads loadshave have10tobe be supported supported within the span of a member. member.
significant contribution. Clause 4.4 of the steel code BS 5950: Part I allows the designer Part 1 allows the designer three Clause 4.4 of the steel code BS 5950: three different ways of proportiomng web piates: plates: different ways of proportIOning web
L6x450 1.6 x450 =720kN =720kN 20 = 28kNlm 1.4 xx 20 28kN/m 25 =40kN/m 1.6 xx 25
1
PLATE PLATE GIRDERS GIRDERS
(d) (d)
I'
85 BS table table 77
Moment and and shear shear force force Moment
Moment Moment capacity capacity In In order order to to maximize maXimize its Its moment moment capacity, capaCity, the the cross-section cross-sectIOn of ofthe the plate plate girder girder should should be be proportioned proporttoned so so as as satisfY the the requirements reqUirements for foraacompact compact section. The moment moment capacity capacityfor for aa compact compactpiate plategirder girder with with aa thick web sectIOn. The web is IS given given by: by:
clause dause 4.2.5 4.2.5
(b) (b)
127 127
The distributions distributIOnsofofbending bendingmoment momentand andshear shear force force m ill the the simply simply supported supported The plate girder girdercan can readily readilybe bedetermined determmed by byconventional conventionalelastic elasticmethods methodsand and plate arc shown tn In Fig. Fig. 11.4. 11.4. are
Mc. =p,.S, Mcx=pyS"
but but
1.2 p72..4 or 1.2 Py or yp,. YPy 2"4 if if 5> S> 1.24 1.22.%
The bIT bIT ratio rallo for for the the outstand outstand of ofthe the compresstnn compreSSIOn flange flange for for aa compact compact built—up section should should not not exceed exceed 8.5£, 8.5e, and and assummg assuming that that the the deSign design strength built-up section strength The plastiC plastic moment capacity of a plate N/mm2,then thenBB::;(17.3T (17.3T -ipy is IS 265 265 N/mm:?, -I- 1). t). The ofa plate girder gmler is: IS: W= Pv [BD Ma =Py [BD 2—(B '-(B -I)(D _2nzv4 -2T)'Jl4 hence
(c) (c)
capacity Shear capacity
7442 C0.265[(t7.3T+25)15002—(l7.3T)(1500 — 2fl2]/(4 xx IQ') 7442 ';0.265[(17.3T+25)1500'-(l7.3T)(1500 - 2T)'J/(4 Solving this equation T =23.4mm Solvmg equatlOn gives gtves T=23.4 mm and hence B =430 =430 mm. Note that that the assumption assumption regarding regardingPy p,,ISiscorrect. correct. Select Select450 450 mm mm xx 25 25 mm from the from the the range of of wide wide Hats flats gIven given IU in Table for the the flanges. flanges, from from which itIt range Table 11.1(a) l!.l(a) for x 25 mm. The actual follows that that the the web web size size must must be 1450mm 1450mmx 25nuu. actual plastic plastiC moment capacity capacity of of the moment the plate girder gIrder isIS
There are arc two design deSign requirements reqUIrements regnrding thickness for regarding tbe the minImum minimum web thickness for ofno no intermediate mtermediate stiffeners, sliffeners, i.e. I.e. the condition of the clame 4,4.2.2 4.4.2.2 clause clause 4.4.2,3 4.4.2.3
servH:eability: for serviceability: flange buckling: buckling: to avoid flange
ut cC dlt :; 250
&tC250 till'; 250 (345Ip;;)
P;f isISthe thedesign deSignstrength strength of ofthe thecompression compressIOn flange. flange. As As the the web web is to where p,1 deliberately made made thick, thick, i.e. I.e.ilkd/tC<63g, 63£,these theserequirements requIrements are are be tieliberately automatically complied complied with. A A quick qUick estimate estunate of the minimum mlmmum web automatically of the thickness can the section seclion (D), I.e. thickness can be be denved denved usmg using the the overall overall depth depth of of the (D), i.e.
and
tf?: lSOO/63=23.8mni 1500/63 =23.8 mm
Therefore, capacity of of the design plate girder (7880 moment capacity deSign plate (7880 kNm) isIS Therefore, the the moment adequate. adequate.
From the relevant table table for for plates piates in m Table 111.1 thenearest nearest appropriate appropriate plate plate l.lb,b,the web isIS 25 25 mm. mm. thickness for the the web thickness the web is IS thick thick,then thenthe the moment moment capacity capacity for for aa plate plate girder girder isIS When the calculllled according according to to clause clause 4.2.5 or or 4.2.6 4.2.6 depending depending on the magnitude magrurude of. calculated ofi the moment, I.e. 1166 kN. The The shear shear the shear shear load load coexistent coexistent with with maximum maximum moment. i.e. 1166kN. capaCity of ofaa web web of ofaabuilt-up built-upsection sectIOnisISdefined definedas: as: capacity
clause 4.2.3
P,,=O.6 p;. A" P?=O.6p, where A,,= Id. A A good good estimate estimate of ofthe the shear shear capacity capacIty of of the the web web can can be where A,.= td. obtamed by d. which which is IS obtained by subshtutmg substituting the the ovemll overall depth depthof of the thegtrder girder(D) (D) for ford, at this this stage: stage: unknown at
P,,=0.6 0.265 xx 25 x t500=5962k}J 1500=5962kN P?=0.6 x 0.265
Use
m
(i)
two 450 45Ommx2Smm two mmx25 mm wide flats 1450 1450 mmx2S mrnx25 mm mm plate plate
UNCTION WELD AT WEB/FLANGE WEB/FLANGE JUNCTION Next, the required for the connection between the flanges and web web Next. the weld size required flanges and is determined from the IS determined from the magnitude magmtude of the honzonial honzontul shear/mm s-hear/mm at at the the web/ web! flange interface, assumtng assumingaafillet fillet weld wefd on on each side of the web, flange tnterface; each side web,
qw
FAfJ'l
=2J;kN/mm = 1540(450 25)737.5/(2 x 18.59 1540(450 x 25)737.5/(2 18.59 x IQ') = 0.35 0.35kN/mm
as and as
mm FW Use 66 mm FW F" $. 0.6 p... F?<0.6J'?,
clause 4.2.5 4.2.5
M =0.265[(450)I500' _(425)l4502)/(4 x I10') .11= =0.265[(450)1500'-(425)1450')/(4 = 0.265 x 29736 =0.265 29736 =7880kNm>>7442 kNm =7880kNm 7442kNm 3 3 t.2p, L2pyZZ= =1.2 1.2xO.265[(450)15003—(425)t4503]/(12 X 0.265«450)1500 -(425)1450 ]1(12 xx 750 750 x 10 3 ] 6 1061/750 **..,=1.2xO.265fI8.59xl0 = t.2 x 0.265f 18.59 x ]1750 , =7882kNm> = 7882 kNm >7880 7880 kNm kNm
i.e. 1l66kN 1t661d4<3577kN I.e. < 3577kN
has aa 'low 'Iow shear shear load'. Note Note that that the the use use of ofDD instead instead of of Then the member has d does not affect affect the the outcome outcome of ofthis thisdesign deSign check. check. ddoes
Table 11.1 reveals that that both both plates plates would would need need to io be An examination exammatlOn of Table Il.l reveals be spliced, as asthe theappropnate appropnatemaximum maximumlengths lengthsavailable availablefrom fromthe therolling rolling mills spliced, (flanges -—.ISISmm; web web -—19 19m) m)are areless lessthan thanthe theoverall overall length length of the Ihe plate (flanges
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w 0 r riLit, I LIHML S I uhLWUIRI< DESIGN TO 85 5950 'LV
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II:;I:;.LWUB!< DESIGN TO 85 5950
PLATE GIRDERS 129 PLATE GIRDERS 129
girder. If tile girder can be transported one unit,then thenmake makea awelded weldedShop shop girder. If the glfder can be transported asas one unit, splice about ISm from nght-hand end, the welds being full strength butts. On ,splice about 18m from nght-hand end, the welds being full strength butts. On the other hand, lithe girder has lobe delivered in two parts owing to transport ! the other hand, irthe girder has to be delivered in hvo parts owing to transport considerations, then make a bolted sitesplice splicenear nearthe thecentre centreofofthe thegirder. girder consideratIOns, then make a bolted site The use of numeneal controlled cutting machines in the modem fabncauon The use of numencal controlled cuttmg mac1lines In the modem fabncatton shops would minimize any wastage, by utilizing theplate plateOffcllts offcutstotoprovide provide shops would mminuze any wastage, by utilizing the stiffeners and other plate componentsforforIhis thisgirder girderand andother otherproJects. projects. stiffeners and other plate components
(iI)1W AT ATTHE THESUPPORTS SUPPORTS
The reachon OCcurs, I.e. Theweb webatatthe thenght-hand nght-handsupport, support,atatwh.ich whichthe thegreater greater reaction occurs, 'c, 1540 kN. aiso to to bebe checked forforbeanng buckling. l540kN, alsoneeds needs checked bearmgand and buckling.ItItIS isassumed assumed that IS 254/2 mm: thata aminimum minimumstiff stiffbeanng bearingprovided providedbybysupport support is 254/2 mm: clause Poop =[254/2 +2.5(25)J x 25 x 0.265=2095 kN clause4.5.3 4.5.3 = [254/2 + 2.5(25)] x 25 x 0.265 = 2095 kN p~ above) Pc=71 =71N/mm2 N/mm' (as (as above) clause Po. = [(254/2 + 750) x = 1555 kN clause4.5.2.1 4.5.2.] F,, =[(254/2 + 750) x25J 25]0.071 0.071 = 1555 kN
(e)
The reqUires no load beanl1g Theweb web isis adequate adequateatat both both supports supports and and therefore therefore requires no load beanng of a thick web has eJimmated the use of stiffeners. stiffeners.InIn this this example, example, the the use use of a thick web has eliminated the use of load loadbearing bearmgstiffeners, stiffeners,thereby therebymlOJmizmg minimizingfabncatlOn. fabrication. See Figs. 11.21 and 11.25 for Ule of this girder. See Figs. 11.21 and 11.25 for theconstruction constructiondetails details of this girder.
Lateral torsional buckling
lateral
~or5ional
,
buckling
; to the design bneL the compression flange ofthe thegm:l:er girderISis According to the deSign brief, the compression flange of restrained laterally and therefore there is no need for a lateral torsional restramed laterally and therefore there IS DO need for a lateral torsIOnal buckling check to be undertaken. the flange had not been restrained, then buckling check to be undertaken. IfIf the flange had not been restrained, then the recommendations of section 4,3, 133 SPSOmust be satisfied. the.recommendatlOns of section 4.3, BS 5950 must be satisfied.
(I) (f)
11.4 11.4
Check bearing capacity and buckling
resistance of of web web Check bearing capacity and buckling resistance
In Ule previous previous example, example, the the bending bending moment was reSisted by the Whole In the moment was resisted by the whole section, while the shear capacity web IS clearly unutilized; this IS section, while the shear capacity of of the the thick thick web is clearly unutilized; Similar to the deSign of wuversal beams. The difference between the this is similar to the design of umversal beams. The difference ber'.veen the uDlversal girders IS that Ihe designer can select the web universal sections sections and and plate plate girders is that the designer can select the web thickness. thickness. The in Example Example 21 21 IS redesigned usmg a thin web plate, In The plate plate guder girder in is redesigned using a thin web plate, in order to make make the the web web work work more effiCIently. This IS achieved by companng order to more efficiently. This is achieved by comparing the design deSign shear shear load with the the web webshear shear reSIstance. baSed on the cnticai the load with resistance, bascu on the critical (qcr) and on the the design design strength (py) as IS the case for thick shear shear strength strength (qer) and not not on strength (p7) as is the case for thick webs. The nonnal design design practice practice for for plate plate gIrders With thin webs will be webs. The normal girders with thin webs will be employed, by which the the bending bending moment and aXial load are assumed 10 be employed, by which moment and axial load are assumed io be by the the flanges flanges and and the the shear shear load load by the web. reSisted wholly resisted wholly by by the web.
At points of concentrated applied load and support reactions the web of At pomts of concentrated applied load and support reactions the web of aa plate girder must be checked for local web bearing and web buckling, If plate girder must be cheCked for local web bennng and web buckling. If necessary, load carrying stiffeners must be introduced to prevent these forms necessary, load carrymg stiffeners must be Introduced to prevent these fonns of local failure. The design checks are similar to those applied to umversal of local failure. The deSign checks are snniiar to those applied to unIversal beams, as outlined in Section 5.3, Example 9. beams, as outlined in SectIOn 5.3, Example 9.
(i)
(0
AT POSITION OF APPLIED COLUMN LOAD AT POSITION OF APPLIED COLUMN LOAD F1= lO5GkN F,= 1056kN acting in the plane of the web of the plate gtrder, i.e. no moment generated in acting m the plane of the web oflbe plate girder, i.e. no moment generated in stiffeners. Assummg that the column base (supported by the compression stiffeners. Assummg that the column base (supported by Ule compression flange) provides a minimum stiff beanng of 203 mm, thea the bearing flange) provides a minimum stiff beanng of 203 mm, then the bennng capacity of the unstiffened web capacity of the unstiffened web at at the the junction junction of ofthe theweb/flange web/flange is: is:
clause 4.5.3 clal1se 4.5.3
=
(a) (a)
clause 4.4.4.2a 4.4.4.2a clause
±fl 2) tP)w
P~rip = (b, +n 2) CPyw =[203 ±2.5(2 x 25)] x 25 x O.2G52170IcN =[203 +2.5(2 x 25)J x 25 x 0.265 =2170 kN
clause 4.5.2.j clause 4.5.2.1
.4 =2.5&t=2.5 x 1450/25
A. =2.5d!t=2.5 x 1450/25 =71 N/mm' p~ =71N/mm2
;
=(b, +1I!l tJJ~ =[(203±2x750)x2s] 0.071 =[(203+2 x 750) x 25J 0.071
The design design assumption assumption that thatthe the moment IS carned only by the flanges The moment is earned only by the flanges means that a good esllmate of flangearea areacan canbe bedetennined. detennmed. means that a good estimate of flange
Ar =7442 1500) = 18 720 mm' 41= 7442 xx 10'/(0.265xx1500) = 13720mm'
==J45 145
BS&b!e table66 The Thelimiting limitingbIT bIT ratIOofof 8.5&(compact (compact sections) IS still applicable, hence BS ratio 8.5e sections) is still applicable, hence theapproximate apprmomateflange flangethickness thickness IS given by: tue is given by:
F,. = (Ii + zi1) P,~
Momentcapacity capacityofofflanges flanges Moment
Within the range 16-40 mm, then AntlclpatlOgthat thatthe theflange flange thickness thickness lies lies within Anticipating the range 16—40 mm. then p,.=265N/mm2 N/mmzand andhence: hence: p,.=265
The associated buckling resistance (F,,) is dependent on the slenderness of of Theunstiffcned associated buckling resistance (Pw ) is dependent on the slenderness the web and a design strength oi2GSNlmm';2 the unstiiTcned web and a deSign strength of 265 N/mm
EXAMPLE GIRDEREXAMPLE 22. 22. DESIGN DESIGN OF OP UNSTIFFENED UNSTIFFENED PLATE PLATE GIRDER.THIN WEBS THIN WEBS
=3020kN =3020kN
T=.J[t8720/l7.3]32.9mm, J[l8720117.3J=32.9mm, say 35 mm T= say 35mm
The web is adequate and therefore requires no load carrying stifTeners under TIle web IS adequate the concentrated load. nnd therefore requires no load carrying stiffeners under
hence hence
the concentrated load.
18720/35 = 535 mm, sny 550 nun BB= =18720/35 '=535 nun. soy SS0 mm
EL
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WELD WELD AT AT WEBIFLANGE WEB/FLANGEJUNCTION JUNCTION
The follows the The determinatIOn determination of of the the weld weld connectmg connecting the the flanges flanges to to the the web web follows the same m Example Example 21. 21. same method method as that outlined in Use Use 6mm 6mniFW FW
of web web resistance of resistance ;ign reqUIrements requirements regarding regarding the the minimum for the ;lgn mmimum web web thickness thickness for the in of )n of no no intermediate mtennediate stiffeners stiffeners are arc the the same same as as in In Example Example 21, 21, i.e. I.e.
As As ill m Exampie Example 21: 21, there there ISis no no need need for for aa lateral lateral torstonai torsional buckling buckling check as as the compression flange IS restramed. the flange is restrained.
te girders which te girders WhICh are are not not designed deSIgned for for tension tenSIOn action, actIOn, the the elastic elastic critical cntIcal xength (qcr) (qcr) for for thm thin webs webs can can be be determmed determined from Tength from table table 21, 21, BS BS 5950, 5950, gg the the d/t d/! and and the the a/d aidvalues. values. In In the thespecial specmlcase caseofofuns4ffened llnstiffened thin thm i/d= 0:', a, therefore therefore the !Id= the appropriate approprIate shear shear strength strength for for aa given gIven dli d/t is is column of i the of table 1 the last lasl column table 21. 21. tluckness (t) (t) is .vcb tiuckness IS unknown unknown at at this fhlS stage, stage, hence hence itIt isis suggested suggested that that aa ss approximately ss approximately half half that that of of the the thick thick web web (Example (Example 21) 21) istS selected, selected, 5j (25/2)mm. (=25/2) mm,which whichinmfact factis15a arofled rolledthickness, thickness,see seeTable Table 11.1(b). I 1.1 (b). ;ults in a dli ratio of 1430/12.5 114 (<250) and 16mm, ;ults in a dll ratIO of ! 4301I2.5 = 114 « 250) and as 1< 16 mm, then then an an on of table 21(b) indicates that, ion of table 21(b) mdicates that,
(d)
Design stiffeners Design of of load load carrying stiffeners ExammatlOn of the detailed detailed calculations calculations for for web web bearing beanng and and buckling buckling in in Examination of the Example 21 would readily 15 mm would would Example 21 would readily mdicate indicate that that aa web web thickness thickness of 15mm reqUIre toad load carrying carrymg stiffeners stiffeners at at both both the the concentrated concentrated load load and and end end reaction reactIOn require
ciause 4.5.1.2 4.5,1.2 clause qcr= 77.4
Lateral Lateral torsional buckling
N/mm2 4.5.1.5 clause 4.11.5
pOSIttons, positions, recommends. for these The code recommends, for fhe the condition condition where where the the outer outer edge edge of these stiffeners is IS not stiffened stiffened (normal stiffeners (normal prachce), practice), that that the the outstand outstand of of the stiffener exceed 191s 6. However, However, where where the the outstand outstand is IS between between I31s 6 and and should not exceed 19ts e, then the the stiffener stiffener design deSIgn must be based on a core core area area of ofthe the stiffeners stiffeners havmg an outstand of I3ls 6. In deriving denving the the compressive compresslve strength strength P. Po of of having for welded plate girders, girders, the the design deSIgn strength strength(p,,) (Py) is IS the the lesser lesser value value stiffeners for the web web or orstiffener, stiffener, less less20 20N/mm2. for the N/mm2.
.veb thickness of of 12.5mm .veb thickness 12.5 mm is IS adequate adequate for for the the part-length part-length from from the the leftleftid lojust right of the applied point load, i.e.> 1012 lcN, but not !d 10 Just nght of the applied pomt load, I.e. > 1012 kN, but not for for the the fer of of the the girder, gIrder, i.e. i.C. C < 1540 1540 kN. kN. Recalculate Recalculate the the shear shear resistance reSIstance of ofthe the etween the )etween the applied applied load load and and the the right-hand nght-hand end, end, using usmg aa 15mm 15mm plate, plate, ias 1as aa shear shear buckling buckling resistance resIstance of: of: ICr =0.llOx 1430x t5.0=2360kN C< 15401cN Va =0.110 x 1430 x 15.0=2360kN 1540kN
(i) (I)
AT POSITION POSITION OF OF APPLIED APPLIED COLUMN COLUMN LOAD LOAD AT
The applied applied load, load, T056kN, 1056 kN, acting actmg in In the plane of of the web web pf ofthe the plate plate girder, girder, The
generated in i.e. no moment generated in stiffeners, stiffeners. Similar Similartotothe thecalc~lations calculations outlined~in outlinedin Example 21, 21, the the bearing beanng capacity capacIty of ofthe the unstiffrned IInsliffened web web is: IS: Example
Pc"p = [(203+2 [(203 + 2xx35) 35)xx15] 15] 0.275 0.275 == I 1126 kN>1056 > I 056kN kN l26kN
clause clallse 4.5.3 4.5.3
The associated assOCIated buckling buckling resistance reSIstance (P w ) is IS dependent dependent on on the the slenderness sienderness of of The 2 the unslijJened web and a deSign strength of 275 N/mm the uns4/jened web and a design strength of 275 N/mm2
d/:=95 N/mm2 dll=95 and and qcr= qcr= 110 110 N/mm2 at at aa 20% 20% increase Increase inmweb webthickness thicknessproduces producesa adisproportionate disproportlOnate ~ (70%) (70%) in In shear shear buckling buckling resistance. resistance. iin from 11.1 (b), the the maximum maxImumavailable availablelength lengthofofplate, pi ate,i.e. I.e. from Table Table 11.1(b), —23 m and and web— web - 19 19 m, rn, means means that that the theweb webplate platehas hastotobe bespliced spliced - 23 m
clause clallse 4.5.2.1 4.5.2.1
A=2.5d!1=2.5 xx 1430115=238 A=2.5d11=2.5 1430/15238 Pc== 30.4 30.4N/mm2 N/mm2 Pc P",=[(203+2x 750) xIS] x 15]0.0304=777 0.0304=777kN 1056kN +2 x 750) kN <<1056kN
tlearly, stiffeners CVISION's are necessary to prevent the local buckling failure of the PDF compression, OCR, web-optimization with PdfCompressor ··':.ClearlYI stiffeners are necessary to prevent the local buckling failure of the .,up}.. <>t
11-. ..
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--_ ••• _-
132 132
STRUCTURAL STEELWORK OESIGN TO 65 5950 STRUCTURAL STEELWORK DESIGN TO SS 5950
PLATE GIRDERS 133 PLATE GIRDERs 133
wliereA is the area of the stiffener in contact with
flangeand and p.,isisthethe where 'A IS the area of the stiffener in contact with thetheflange p.'.~ design strength of the stiffeners. In this example, the stiffeners aresubject subjecttoto desIgn strength of the stifTeners. In this example, the stiffeners are external load and therefore must extend to the flanges,though thoughmay maynot notbebe external load and therefore must extend 10 the flanges, necessarily connected to them, toilers externalload loadillduces inducestensI01I tension111litt/re the necessarily conneeted to them, unless anonexternal Normally the flanges and web would be welded together, using siijJelfer. Normally the flanges and web would be welded together, uSing semi-or fully automatic welding, before the stiffeners are 'fitted', This means seml- or fully automatic welding, before the stiffeners are 'fitted'. This means that the inside corners of the stiffeners need to be coped/chamfered, say that the Inside corners of the stiffeners need to be coped/chamfered, say 15mm, at the Junction of the web and flange, thatthey theydo donot notfoul foulthe thewebl web/ 15 mm, at the Junction of the web and flange, sosothat flange weld. Hence, flange weld. Hence,
c
When load, then the shear load/mm Whenthe thestiffener stiffenerISisalso alsosubject subjecttotoananexternal external load, then the shear load/mm to the above shear load. The ioad by the stiffeners IS must be added must be added to the above shear load. The loadresisted resisted by the stiffeners is the the mmimum load that can be thedifference differencebetween betweenthe theapplied appliedload loadand and the minimum load that can be earned earnedsafely safelyby bythe theunstiffened unstiffenedweb. web.
Note been restralDcd, then L£= L . Notethat thatififthe therotation rotationof ofthe theflange flangehad hadnot not been restrained, then L5= L. AJso, had the column base (compression member) been iateraJ/y Also, had the column base teompression member)not not been laterally restrained, as part of the restrained,then thenthe thestiffeners stiffenerswould wouldneeded neededtotobebedesigned designed as part of the member and the interfacmg connectJQn cheCked any effects compression compression member and the interfacing coonectioa checicedfor for any effects from ' fromstrut strut actIOn. action.
A =2x( 60—15)x 12=3480 mm2 2 A =2 x (160-15) x 12=3480mm p,,IO.8=3480 x 0275/08= il96kN> 2056 leN AAp,,10.8 =3480 x 0.27510.8 = 1196kN > 1056 kN i.e. the stiffeners are adequate in bearing.
I.e. the stiffeners are adequate III beanng. As the outstand of the stiffeners is slightly greater than 13t,e (156), then the As the outsland oflhe stiffeners IS slightly greater than 13ts E (156), then the beat buckling resistance of a stiffened web is based on the stiffener core area area local buckling resIstance of a stiffened web IS based on the stiffener core of 156 x 12 mm2, together with an effective web area limited to 2 x 20r. of 156 x 12 mm!, together with an effective web area limited to 2 x 20t.
Load carrying Fig. 11.5 Load carrying siiffner
AT AT THE THE SUPPORTS SUPPORTS
(Ui) (Iii)
FIg. 11.5
A,
TIle of the member IS greater The reachon reaction (1540 (2540!eN) kN)atat the the nght-hand nght-hand support support of the member is greater than therefore stiffcners are necessary. The thanthe the applied applied point point load. load, and and therefore stiffeners are necessary. The restramt flange apply this location. The restraint conditions conditions with with respect respect to to the the flange apply also also to to this location. The deSign of sliffeners at both supports follows pattern as the prevlOus design of stiffeners at both supports follows aa similar similar pattern as the previous design design caicujattons, calculations, except except that that the the effectJve effective web area 15 limited to only 20t.
15)15=12744mm2 A, = 12(2 x 156)+(2 x 20 x 15)15 = 12 744 mm'
sliffn~r
and the corresponding radius of gyration, about
an axis axis parallel to the the web, web, is and Ihe corresponding rndius of gyratIOn. about an parallel to IS
— /12(2
r—
r=
(2 x 20 x 12(2 x 156 + 15)' ++ (2 x 20 x 15)15) 12 x 12744 12x12744
= S2.7
x 156 + 15)J
web area is limited to only 20t.
Try mm x 15mm wide flat. flat. Try aa 450 4SOmmx l5mm wide
nun
= 52.7 mm It
clause 4.5.1.3 clause 4.5.1.5
is noted that the flange is restrained against lateral IS noted that the flange 15 restTllmiagamst lateral
movement and It movement and rotation. As a result of this flange restraint, it assumed that column rolallon. As a result of Ulis flange rcsW -, It can be e ' assumed that the the column base is restrained laterally. Therefore, the effective the load base IS restrmned laterally. Therefor~ -, re length . ngth CL5) (LE) of oflhe load carrying stiffeners can be taken as O.7L and with a reduced design strength of carrying s!iffeners can be laken as O:1&id I~re uccd deSign strength of where p, is the lesser strength of web or stiffener, hence the (Pv-20), where Py is the less;r stren~-=, ~r 'ffener, hence the reduced strength ts 255 N/mm2, G1 U reduced strength 15 255 N/mm .
The outstand of of the the stiffeners sliffeners IS equivalent to 14.5IJ G, whicil means the core The outstand ts equivalent to 14.5t,n, which means the core area of the sliffeners is reduced reduced to 10 22 xx 195 195 mm x 15 mm. The local buckling area of tlte stiffeaers is turn x 15mm. The local buckling resistance of the stiffened web IS based based on this core area of the stiffeners, plus resistance of the stiffened web is on this core area of the stiffeners, plus of 2Omm 20 mm xx 15mm. 15 mm. effective area of effeehve web web area
,(=0;7 Ur=0.7 x 143015n19.1 252 N/mm2 pc=252N/mm1. Px=A5Pr= xO.252=3210kj4 >l056kN P$=Asp,.= 12744 12744 x O.252=321OkN >I056kN
The buckling resistance of the stiffener is more than satisfactory. The buckling reSlstance of the stiffener IS more than satisfactory.
Use Use two two lfiOmmxl2mm 160 mm x 12 mmflats nats
y
15(390)' + (20 x 15)15' 15)153 12 x 10350 10350 12x
= 84.7mm 84.7 mm Agam, as as the the flange flangeisis restramed agamst lateral movcment and rotation, the Again, restrained against lateral movement and rotation, the (L1J of of the load carrymg stifTeners IS O,7Land with a design effective length iength (L5) effective 2 the load carrying stiffeners is 0.7L and with a design strength of of255 255N/mm the slenderness slcndemeSSlS:. streagth N/mm2,, the Is:.
WELD FOR LOAD CARRYING STIFFENERS WELD FOR LOAD CARRYING STIFFENERS
The minimum weld size required for connecting the stiffeners to the web, TIle mlnlmUm weld size reqUired for connectmg the stiffeners 10 the web, assuming a weld on each side of the stiffener, ts determined as follows: assuming a weld on each side of the stiffener,
clause 4.4,6.7
clause 4.4.6.7
qi =t2/(2 x 5b,)= 152/(2 x 5 x 160)
IS
determmed as follows:
=0.14 kN/mm =0.14kNlmm
Pc =255 =255N/mm Pc N/mm2 P,= 10350xx 0.255=2639 kN> 1540kN P1= 10350 1540kN Thebuckling bucklingresistance resistance of ofthe thestiffener stiffener is satisfactory. Make the load The ts satisfactory. Make the load carrying stiffener for the left-hand endofof girder the same slZe.·Now check the carrying stiffener for the left-hand end girder the same size-Now cheek the bearing capacity of the end stiffener, note that as the stiffener is welded to bearing capacity of the end stiffener, note that as the stiffener is welded to end of girder there IS no copmg, I.e. full stiffener area can be used. end of girder there is no coping,
I.e. full stiffener area can be used.
P
X
15 x 0.275 = 1855 kN > 1470 kN
l470kN
Use 450 mm >< 15 him wide fiut
-
Use
4SOmmxjSa,m wide flat
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134
STnUCTIJRAL STEELWORI< STRUCTURAL STEELWORK DESIGN DESIGN TO SS 5950 5950 To os
PLATE GIRDERS GiRDERS PLATE
However. it It might mIght be be deemed deemed necessary necessary that tbat the the ends ends of of the the plate plme girder girder be be However, transportatIOn and erection. erectIOn. This can can be be lorslOnally restrained restramed dunng transportation torsionally of area of of the the end-beanng end~beanng accomplished by checking the second moment of Sliffeners at the the supports against agamst the the guidance guidance given given in In BS BS 5950: 5950: stiffeners
