Guidebook for the Design of ASME Section VIII Pressure Vessels 3th Edition - James R. Farr-Maan H. Jawad (L) 'I'

GUIDEBOOK FOR THE DESIGN OF ASME SECTION VIII PRESSURE VESSELS Third Edition

by

James R. Farr Wadsworth, Ohio

Maan H. Jawad Camas, Washington

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© 2006 by ASME, Three Park Avenue, New York, NY 10016, USA (www.asme.org) All rights reserved. Printed in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. INFORMATION CONTAINED IN THIS WORK HAS BEEN OBTAINED BY THE AMERICAN SOCIETY OF MECHANICAL ENGINEERS FROM SOURCES BELIEVED TO BE RELIABLE. HOWEVER, NEITHER ASME NOR ITS AUTHORS OR EDITORS GUARANTEE THE ACCURACY OR COMPLETENESS OF ANY INFORMATION PUBLISHED IN THIS WORK. NEITHER ASME NOR ITS AUTHORS AND EDITORS SHALL BE RESPONSIBLE FOR ANY ERRORS, OMISSIONS, OR DAMAGES ARISING OUT OF THE USE OF THIS INFORMATION. THE WORK IS PUBLISHED WITH THE UNDERSTANDING THAT ASME AND ITS AUTHORS AND EDITORS ARE SUPPLYING INFORMATION BUT ARE NOT ATTEMPTING TO RENDER ENGINEERING OR OTHER PROFESSIONAL SERVICES. IF SUCH ENGINEERING OR PROFESSIONAL SERVICES ARE REQUIRED, THE ASSISTANCE OF AN APPROPRIATE PROFESSIONAL SHOULD BE SOUGHT. ASME shall not be responsible for statements or opinions advanced in papers or . . . printed in its publications (B7.1.3). Statement from the Bylaws. For authorization to photocopy material for internal or personal use under those circumstances not falling within the fair use provisions of the Copyright Act, contact the Copyright Clearance Center (CCC), 222 Rosewood Drive, Danvers, MA 01923, tel: 978-750-8400, www.copyright.com. Library of Congress Cataloging-in-Publication Data Farr, James R. Guidebook for the design of ASME Section VIII pressure vessels/by James R. Farr, Maan H. Jawad.—2nd ed. p. cm. Includes bibliographical references and index. ISBN 0-7918-0172-1 1. Pressure vessels—Design and construction. 2. Structural engineering. I. Jawad, Maan H. II. Title. TA660. T34 F36 2001 681’.76041—dc21 2001046096 British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library. Woodhead Publishing ISBN-13: 978-1-84569-119-6 Woodhead Publishing ISBN-10: 1-84569-119-9


Dedicated to the memory of Barbara R. Farr and dedicated to her friend and pal, Dixie Jawad.



PREFACE

TO

THIRD EDITION

The ASME Boiler and Pressure Vessel Code, Section VIII, is a continuously changing document. It contains the latest, safe, and economical rules for the design and construction of pressure vessels and pressure vessel components. The contents of many chapters have been updated to incorporate improvements made in design methods since the book was originally issued. Some of the changes are extensive while others are minor. The brittle fracture rules have been extensively updated and expanded to reflect the latest technology. The heat exchanger chapter has been completely re-written to reflect extensive changes made recently in VIII-1. A new Quick Reference Guide is included in Appendix A to help the reader locate various sections of the latest edition of the VIII-1 code. Metric units were added to some of the figures to reflect recent metrication of the ASME code. James R. Farr Wadsworth, Ohio Maan H. Jawad Camas, Washington December 2005

v



PREFACE

TO

SECOND EDITION

The ASME Boiler and Pressure Vessel Code, Section VIII, is a live and progressive document. It strives to contain the latest, safe and economical rules for the design and construction of pressure vessels, pressure vessel components, and heat exchangers. A major improvement was made within the last year by changing the design margin on tensile strength from 4.0 to 3.5. This reduction in the margin permits an increase in the allowable stress for many materials with a resulting decrease in minimum required thickness. This was the first reduction in this design margin in 50 years and was based upon the many improvements in material properties, design methods, and inspection procedures during that time. Chapters and parts of chapters have been updated to incorporate the new allowable stresses and improvements which have been made in design methods since this book was originally issued. Some of these changes are extensive and some are minor. Some of the examples in this book have changed completely and some remain unchanged. This book continues to be an easy reference for the latest methods of problem solving in Section VIII. James R. Farr Wadsworth, Ohio Maan H. Jawad St. Louis, Missouri July 2001

vii



PREFACE

TO

FIRST EDITION

The ASME Boiler and Pressure Vessel Code, Section VIII, gives rules that pertain to the design, materials selection, fabrication, inspection, and testing of pressure vessels and their components. With few exceptions, the rules that cover the design of components tend to be complicated to implement. This book was written to demonstrate the application of the design rules to various components. Other rules, such as those pertaining to fabrication, inspection, testing, and materials are not covered here. This book is intended as a reference for designers of pressure vessels and heat exchangers. The theoretical background of the equations used here was kept to a minimum, since such background can be obtained from other references. Note also that while the design requirements of such components as shells and heads are interspersed throughout the ASME Code, design requirements pertaining to some specific components are given here in one chapter for easy reference. The emphasis in this book is on solved examples, which illustrate the application of the various equations given in the ASME Section VIII Code. Chapter 1 of this book covers background information and general topics—such as allowable stresses and joint efficiencies—applicable to all components. Chapter 2 is for the design of cylindrical shells under internal and external loads. Chapter 3 covers the design of dished heads and transition sections that are under internal and external loads. Chapter 4 considers flat plates, covers, and flanges. Openings are reviewed in Chapter 5, and Chapter 6 covers special components of VIII-1, such as stayed construction, jacketed components, half-pipe jackets, and noncircular vessels. Chapter 7 covers heat exchangers, and Chapter 8 covers stress categories, fatigue, and other special analysis of components. James R. Farr Wadsworth, Ohio Maan H. Jawad St. Louis, Missouri January 1998

ix



ACKNOWLEDGMENTS We are indebted to many people and organizations for their help in preparing this book and revisions and for permitting us to use material supplied by them. Thanks is given to our former co-workers, fellow ASME Committee Members, and many other persons who have given us suggestions and corrections. Special thanks is given to Mr. Mike Bytnar of the Nooter Corporation for providing us with the front cover photograph, to Mr. Thomas P. Pastor of HSB Global Standards for permitting us to use their new Quick Reference Guide for VIII-1, and to our editors Mary Grace Stefanchik and Tara Smith at ASME Press for their continued guidance.

xi



TABLE

OF

CONTENTS

Preface to Third Edition ......................................................................................................................................v Preface to Second Edition .................................................................................................................................vii Preface to First Edition .......................................................................................................................................ix Acknowledgments ..............................................................................................................................................xi List of Figures..................................................................................................................................................xvii List of Tables ....................................................................................................................................................xxi Chapter 1 Background Information...................................................................................................................................1 1.1 1.2 1.3 1.4 1.5 1.6

Introduction ...............................................................................................................................................1 Allowable Stresses.....................................................................................................................................2 Joint Efficiency Factors .............................................................................................................................4 Brittle Fracture Considerations..................................................................................................................9 Fatigue Requirements ..............................................................................................................................19 Pressure Testing of Vessels and Components..........................................................................................23 1.6.1 ASME Code Requirements .......................................................................................................23 1.6.2 What Does a Hydrostatic or Pneumatic Pressure Test Do?......................................................23 1.6.3 Pressure Test Requirements for VIII-1......................................................................................23 1.6.4 Pressure Test Requirements for VIII-2......................................................................................25

Chapter 2 Cylindrical Shells .............................................................................................................................................27 2.1 2.2

2.3 2.4

Introduction .............................................................................................................................................27 Tensile Forces, VIII-1 ..............................................................................................................................27 2.2.1 Thin Cylindrical Shells .............................................................................................................27 2.2.2 Thick Cylindrical Shells............................................................................................................33 Axial Compression ..................................................................................................................................35 External Pressure .....................................................................................................................................41 2.4.1 External Pressure for Cylinders with DO/t ≥ 10 .......................................................................41 2.4.2

External Pressure for Cylinders with DO/t < 10 .......................................................................44

2.4.3

Empirical Equations ..................................................................................................................45

xiii


xiv

2.5 2.6

Table of Contents

2.4.4 Stiffening Rings.........................................................................................................................46 2.4.5 Attachment of Stiffening Rings ................................................................................................49 Cylindrical Shell Equations, VIII-2.........................................................................................................51 Miscellaneous Shells ...............................................................................................................................52 2.6.1 Mitered Cylinders......................................................................................................................52 2.6.2 Elliptical Shells .........................................................................................................................53

Chapter 3 Spherical Shells, Heads, and Transition Sections .........................................................................................55 3.1 Introduction .............................................................................................................................................55 3.2 Spherical Shells and Hemispherical Heads, VIII-1.................................................................................55 3.2.1 Internal Pressure in Spherical Shells and Pressure on Concave Side of Hemispherical Heads ................................................................................................................55 3.2.2 External Pressure in Spherical Shells and Pressure on Convex Side of Hemispherical Heads ................................................................................................................59 3.3 Spherical Shells and Hemispherical Heads, VIII-2.................................................................................61 3.4 Ellipsoidal Heads, VIII-1.........................................................................................................................62 3.4.1 Pressure on the Concave Side ...................................................................................................62 3.4.2 Pressure on the Convex Side.....................................................................................................63 3.5 Torispherical Heads, VIII-1 .....................................................................................................................65 3.5.1 Pressure on the Concave Side ...................................................................................................65 3.5.2 Pressure on the Convex Side.....................................................................................................67 3.6 Ellipsoidal and Torispherical Heads, VIII-2............................................................................................68 3.7 Conical Sections, VIII-1 ..........................................................................................................................70 3.7.1 Internal Pressure........................................................................................................................71 3.7.2 External Pressure.......................................................................................................................80 3.8 Conical Sections, VIII-2 ..........................................................................................................................89 3.9 Miscellaneous Transition Sections ..........................................................................................................93 Chapter 4 Flat Plates, Covers, and Flanges.....................................................................................................................97 4.1 Introduction .............................................................................................................................................97 4.2 Integral Flat Plates and Covers................................................................................................................97

4.3 4.4

4.5

4.6

4.2.1 Circular Flat Plates and Covers.................................................................................................97 4.2.2 Noncircular Flat Plates and Covers...........................................................................................99 Bolted Flat Plates, Covers, and Flanges................................................................................................101 4.3.1 Gasket Requirements, Bolt Sizing, and Bolt Loadings ..........................................................101 Flat Plates and Covers With Bolting .....................................................................................................102 4.4.1 Blind Flanges & Circular Flat Plates and Covers...................................................................102 4.4.2 Noncircular Flat Plates and Covers.........................................................................................102 Openings in Flat Plates and Covers.......................................................................................................102 4.5.1 Opening Diameter Does Not Exceed Half the Plate Diameter...............................................103 4.5.2 Opening Diameter Exceeds Half the Plate Diameter .............................................................104 Bolted Flange Connections With Ring Type Gaskets ...........................................................................104 4.6.1 Standard Flanges .....................................................................................................................105 4.6.2 Special Flanges........................................................................................................................113


Table of Contents

4.7

xv

Spherically Dished Covers ....................................................................................................................118 4.7.1 Definitions and Terminology...................................................................................................118 4.7.2 Types of Dished Covers ..........................................................................................................119

Chapter 5 Openings .........................................................................................................................................................125 5.1 Introduction ...........................................................................................................................................125 5.2 Code Bases for Acceptability of Opening.............................................................................................125 5.3 Terms and Definitions............................................................................................................................126 5.4 Reinforced Openings—General Requirements .....................................................................................126 5.4.1 Replacement Area ...................................................................................................................126 5.4.2 Reinforcement Limits..............................................................................................................126 5.5

Reinforced Opening Rules, VIII-1 ........................................................................................................128 5.5.1 Openings With Inherent Compensation ..................................................................................128 5.5.2 Shape and Size of Openings ...................................................................................................128 5.5.3 Area of Reinforcement Required ............................................................................................129 5.5.4 Limits of Reinforcement .........................................................................................................131 5.5.5 Area of Reinforcement Available............................................................................................132 5.5.6 Openings Exceeding Size Limits of Section 5.5.2.2 ..............................................................142

5.6

Reinforced Opening Rules, VIII-2 ........................................................................................................144 5.6.1 Definitions ...............................................................................................................................144 5.6.2 Openings Not Requiring Reinforcement Calculations ...........................................................144 5.6.3 Shape and Size of Openings ...................................................................................................146 5.6.4 Area of Reinforcement Required ............................................................................................146 5.6.5 Limits of Reinforcement .........................................................................................................146 5.6.6 Available Reinforcement.........................................................................................................147 5.6.7 Strength of Reinforcement Metal............................................................................................148 5.6.8 Alternative Rules for Nozzle Design ......................................................................................148 Ligament Efficiency Rules, VIII-1 ........................................................................................................154

5.7

Chapter 6 Special Components, VIII-1..........................................................................................................................159 6.1 Introduction ...........................................................................................................................................159 6.2 Braced and Stayed Construction .............................................................................................159 6.2.1 Braced and Stayed Surfaces....................................................................................................159 6.2.2 Stays and Staybolts .................................................................................................................162 6.3 Jacketed Vessels.....................................................................................................................................163 6.3.1 Types of Jacketed Vessels .......................................................................................................163 6.3.2 Design of Closure Member for Jacket to Vessel.....................................................................164 6.3.3 Design of Openings in Jacketed Vessels .................................................................................165 6.4 Half-Pipe Jackets ...................................................................................................................................168 6.4.1 Maximum Allowable Internal Pressure in Half-Pipe Jacket...................................................169 6.4.2 Minimum Thickness of Half-Pipe Jacket................................................................................171 6.5 Vessels of Noncircular Cross Section....................................................................................................175 6.5.1 Types of Vessels ......................................................................................................................175 6.5.2 Basis for Allowable Stresses ...................................................................................................176


xvi

Table of Contents

6.5.3 6.5.4

Openings in Vessels of Noncircular Cross Section.................................................................180 Vessels of Rectangular Cross Section .....................................................................................185

Chapter 7 Design of Heat Exchangers ...........................................................................................................................189 7.1 Introduction ...........................................................................................................................................189 7.2 Design of Tubesheets in U-Tube Exchangers .......................................................................................189 7.2.1 Nomenclature ..........................................................................................................................189 7.2.2 Preliminary Calculations .........................................................................................................194 7.2.3 Design Equations ....................................................................................................................195 7.3 Fixed Tubesheets ...................................................................................................................................208 7.3.1 Nomenclature ..........................................................................................................................208 7.3.2 Preliminary Calculations .........................................................................................................209 7.3.3 Design Equations ....................................................................................................................212 7.4 Additional Rules ....................................................................................................................................223 7.5 Methods of Attaching Tubes-to-Tubesheet............................................................................................223 7.5.1 Push Out Tests.........................................................................................................................227 7.5.2 Tube-to-Tubesheet Locking Mechanism.................................................................................228 7.6 Expansion Joints ....................................................................................................................................229 Chapter 8 Analysis of Components in VIII-2................................................................................................................233 8.1 Introduction ...........................................................................................................................................233 8.2 Stress Categories....................................................................................................................................233 8.3 Stress Concentration ..............................................................................................................................239 8.4 Combinations of Stresses ......................................................................................................................239 8.5 Fatigue Evaluation .................................................................................................................................244 References .......................................................................................................................................................249 Appendices......................................................................................................................................................251 Appendix A—Guide to VIII-1 Requirements..................................................................................................251 Appendix B—Material Designation ................................................................................................................255 Appendix C—Joint Efficiency Factors ............................................................................................................257 Appendix D—Flange Calculation Sheets........................................................................................................279 Appendix E—Conversion Factors ...................................................................................................................285 Index ...............................................................................................................................................................287


LIST

OF

FIGURES

Figure Number 1.1 1.2 E1.1 1.3 1.4 1.5 1.6 1.7 2.1 2.2 2.3 2.4 2.5 E2.8 2.6 2.7 E2.13 2.8 2.9 2.10 3.1 E3.4 3.2 3.3 3.4 3.5 3.6 3.7 E3.11 E3.12 E3.13 3.8 3.9

Welded Joint Categories (ASME VII-1) ...............................................................................................5 Category C Weld ...................................................................................................................................9 ...............................................................................................................................................................9 Some Governing Thickness Details Used for Toughness (ASME VIII-1) .........................................12 Impact-Test Exemption Curves (ASME VIII-1) .................................................................................15 Charpy Impact-Test Requirements for Full Size Specimens for Carbon and Low Alloy Steels With Tensile Strength of Less Than 95 ksi (ASME VIII-1) ...............................................................16 Reduction of MDMT Without Impact Testing (ASME VIII-1)..........................................................17 Fatigue Curves for Carbon, Low Alloy, Series 4XX, High Alloy Steels, and High Tensile Steels for Temperatures Not Exceeding 700∞F (ASME VIII-2)........................................................18 .............................................................................................................................................................28 Comparison of Equations for Hoop Stress in Cylindrical Shells .......................................................29 .............................................................................................................................................................30 Chart for Carbon and Low Alloy Steels With Yield Stress of 30 ksi and Over, and Types 405 & 410 Stainless Steels ................................................................................................37 C Factor as a Function of R/T (Jawad, 2004).....................................................................................38 .............................................................................................................................................................39 Geomatic Chart for Cylindrical Vessels Under Extrenal Pressure (Jawad and Farr, 1989) ...............42 Some Lines of Support of Cylindrical Shells Under External Pressure (ASME VIII-1)...................43 .............................................................................................................................................................48 Some Details for Attaching Stiffener Rings (ASME VIII-1)..............................................................50 MiteredBend........................................................................................................................................53 Elliptical Cylinder ...............................................................................................................................54 .............................................................................................................................................................57 .............................................................................................................................................................59 .............................................................................................................................................................63 .............................................................................................................................................................66 .............................................................................................................................................................67 .............................................................................................................................................................69 .............................................................................................................................................................71 .............................................................................................................................................................74 .............................................................................................................................................................75 .............................................................................................................................................................78 .............................................................................................................................................................86 .............................................................................................................................................................89 Inherent Reinforcement for Large End of Cone-to-Cylinder Junction (ASME VIII-2).....................90 xvii


xviii List of Figures

3.10 3.11 3.12 3.13 4.1 4.2 E4.5 E4.6 E4.7 4.3 E4.8 5.1 5.2 5.3 E5.1 E5.2 E5.3.1 E5.3.2 E5.4 5.4.1 5.4.2 5.5 5.6 E5.5 E5.6 E5.7 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13 6.14 E6.8 7.1 7.2a 7.2b 7.2c 7.2d 7.2e 7.2f

Values of Q for Large End of Cone-to-Cylinder Junction (ASME VIII-2)........................................91 Inherent Reinforcement for Small End of Cone-to-Cylinder Junction (ASME VIII-2).....................92 Values of Q for Small End of Cone-to-Cylinder Junction (ASME VIII-2)........................................93 .............................................................................................................................................................94 Some Acceptable Types of Unstayed Flat Heads and Covers (ASME VIII-1) ..................................99 Multiple Openings in the Rim of a Flat Head or Cover With a Large Central Opening (ASME VIII-1) ..................................................................................................................................104 Ring Flange Sample Calculation Sheet.............................................................................................106 Welding Neck Flange Sample Calculation Sheet .............................................................................109 Reverse Welding Neck Flange Sample Calculation Sheet................................................................114 Spherically Dished Covers With Bolting Flanges (ASME VIII-1) ..................................................119 Example Problem of Spherically Dished Cover, Div. 1 ...................................................................122 Reinforcement Limits Parallel to Shell Surface................................................................................127 Chart for Determining Value of F for Angle q (ASME VIII-1 and VIII-2) .....................................130 Determination of Special Limits for Setting t,r for Use in Reinforcement Calculations .................131 Example Problem of Nozzle Reinforcement in Ellipsoidal Head, Div. 1 ........................................132 Example Problem of Nozzle Reinforcement of 12 in. ¥ 16 in. Manway Opening, Div. 1 ..............135 Example Problem of Nozzle Reinforcement of Hillside Nozzle, Div. 1..........................................137 Example Problem of Nozzle Reinforcement of Hillside Nozzle, Div. 1..........................................138 Example Problem of Nozzle Reinforcement With Corrosion Allowance, Div. 1 ............................140 (ASME VIII-1) ..................................................................................................................................143 (ASME VIII-1) ..................................................................................................................................143 Nozzle Nomenclature and Dimensions (Depicts General Configurations Only) (ASME VIII-2) ...145 Limits of Reinforcing Zone for Alternative Nozzle Design (ASME VIII-2) ...................................149 Example Problem of Nozzle Reinforcement in Ellipsoidal Head, Div. 2 ........................................149 Example Problem of Nozzle Reinforcement of 12 in. ¥ 16 in. Manway Opening, Div. 2 ..............152 Example Problem of Nozzle Reiforcerment of Series of Openings Div. 1 ......................................156 Typical Forms of Welded Staybolts (ASME VIII-1) ........................................................................161 Typical Welded Stay for Jacketed Vessel (ASME VIII-1) ................................................................161 Some Acceptable Types of Jacketed Vessels (ASME VIII-1)...........................................................164 Some Acceptable Types of Closure Details (ASME VIII-1) ............................................................166 Some Acceptable Types of Penetration Details (ASME VIII-1) ......................................................170 Spiral Jackets, Half-Pipe and Other Shapes......................................................................................171 Factor K for NFS 2 Pipe Jacket ........................................................................................................172 Factor K for NFS 3 Pipe Jacket ........................................................................................................173 Factor K for NFS 4 Pipe Jacket ........................................................................................................174 Vessels of Rectangular Cross Section ...............................................................................................176 Vessels of Rectangular Cross Section With Stay Plates ...................................................................178 Vessels of Obround Cross Section With and Without Stay Plates and Vessels of Circular Cross Section With a Stay Plate..................................................................................................................179 Plate With Constant-Diameter Openings of Same or Different Diameters ......................................180 Plate With Multidiameter Openings..................................................................................................181 Example Problem of Noncircular Vessels (ASME VIII-1)...............................................................186 Various Heat-Exchanger Configurations (TEMA, 1999)..................................................................190 Tubesheet Integral with Shell and channel (ASME. VIII-1) ............................................................191 Tubesheet Integral with Shell and Gasketed with channel, Extended as Flange (ASME. VIII-1) .................................................................................................................................192 Tubesheet Integral with Shell and Gasketed with Channel, not Extended as a Flange (ASME. VIII-1) .................................................................................................................................193 Tubesheet Gasketed with Shell and Channel (ASME. VIII-1).........................................................194 Tubesheet Gasketed with Shell and Integral with Channel, Extended as a Flange (ASME. VII-1) ..................................................................................................................................195 Tubesheet Gasketed with Shell and Integral with channel, not Extended as a Flange (ASME. VIII-1) .................................................................................................................................196


List of Figures xix

7.3 7.4 7.5 7.6a 7.6b

Tubesheet Geometry (ASME. VIII-1)...............................................................................................197 Typical Untubed Lane Configurations (ASME VIII-1) ....................................................................198 Curves for the Determination of E*/E and v* (ASME. VIII-1) .......................................................199 Tubesheet Integral with Shell and Channel (ASME. VIII-1) ...........................................................210 Tubesheet integral with Shell and Gasketed with Channel, Extended as a Flange (ASME. VIII-1) .................................................................................................................................211 7.6c Tubesheet Integral with Shell and Gasketed with Channel, not Extended as a Flange (ASME. VIII-1) .................................................................................................................................212 7.6d Tubesheet Gasketed with Shell and Channel (ASME. VIII-1).........................................................213 7.7 Xa vs.Zd, Zv, or Zm(ASME.VIII-1) ...................................................................................................214 7.8 Xa vs. Fm with negative Q3 (ASME. VIII-1) ....................................................................................215 7.9 Xa vs.Fm with positive Q3 (ASME. VIII-1) ......................................................................................222 7.10 Some Acceptable Types of Tube-to-Tubesheet Attachments (ASME VIII-1)..................................225 7.11 Some Acceptable Types of Tube-to-Tubesheet Strength Welds (ASME VIII-1) .............................226 7.12 Schematic Detail of a Hydraulic Tube Expander .............................................................................000 7.13 Typical Detail of Grooves in a Clad Tubesheet ................................................................................227 7.14 Some Bellows-Type Expansion Joints ( ASME VIII-1) ...................................................................230 7.15 Some Flanged and Flued Expansion Joints ( ASME VIII-l) ............................................................231 E8.1 ...........................................................................................................................................................238 8.1 Linearizing Stress Distribution .........................................................................................................240 E8.4 Model of a Finite Element Layout in a Flat Head-to-Shell Junction ...............................................243 8.2 Fatigue Curves for Carbon, Low Alloy, 4XX High Alloy, and High Strength Steels for Temperatures Not Exceeding 700∞F (ASME VIII-2) ................................................................245 8.3 Cyclic Curves ....................................................................................................................................246 A.1 ...........................................................................................................................................................000 C.1 ...........................................................................................................................................................257 C.2 ...........................................................................................................................................................258 C.3 ...........................................................................................................................................................259 C.4 ...........................................................................................................................................................260 C.5 ...........................................................................................................................................................261 C.6 ...........................................................................................................................................................262 C.7 ...........................................................................................................................................................263 C.8 ...........................................................................................................................................................264 C.9 ...........................................................................................................................................................265 C.10 ...........................................................................................................................................................266 C.11 ...........................................................................................................................................................267 C.12 ...........................................................................................................................................................268 C.13 ...........................................................................................................................................................269 C.14 ...........................................................................................................................................................270 C.15 ...........................................................................................................................................................271 C.16 ...........................................................................................................................................................272 C.17 ...........................................................................................................................................................273 C.18 ...........................................................................................................................................................274 C.19 ...........................................................................................................................................................275 C.20.E ...........................................................................................................................................................276 D.1 Fig. D.1—Ring Flange With Ring-Type Gasket...............................................................................279 D.2 Fig. D.2—Slip-On or Lap-joint Flange With Ring-Type Gasket .....................................................280 D.3 Fig. D.3—Welding Neck Flange With Ring-Type Gasket ...............................................................281 D.4 Fig. D.4—Reverse Welding Neck Flange With Ring-Type Gasket..................................................282 D.5 Fig. D.5—Slip-On Flange With Full-Face Gasket............................................................................283 D.6 Fig. D.6—Welding Neck Flange With Full-Face Gasket .................................................................284



LIST

OF

TABLES

Table Number 1.1.1

Criteria for Establishing allowable Stress Values for VIII-1 Except for UCI, UCD, and ULT Material and for Bolting (ASME II-D)...................................................................................2 1.1.2 Criteria for Establishing allowable Stress Values for VIE-1 for UCI, UCD, and ULT Material (ASME II-D)...........................................................................................................................................3 1.2 Criteria for Establishing Design Stress Intensity Values for VIII-2 Except for bolting (ASME II-D)...........................................................................................................................................3 1.3 Stress Values for SA-515 and SA-516 Materials ...................................................................................4 1.4 Allowable Stress Values for Welded Connections..................................................................................5 1.5 Maximum Allowable Efficiencies for Arc- and Gas-Welded Joints (ASME VIII-1) ............................6 El.1 Stress Categories.....................................................................................................................................8 1.6 Assignment of Materials to Curves (ASME VIII-1) ............................................................................10 1.7 Minimum Design Metal Temperatures in High Alloy Steels Without Impact Testing ........................18 2.1 Tabular Values for Fig. 2.4 ...................................................................................................................36 3.1 Factor KOfor an Ellipsoidal Head With Pressure on the Convex Side.................................................64 3.2 Values of D for Junctions at the Large Cylinder Due to Internal Pressure..........................................72 3.3 Values of D for Junctions at the Small Cylinder Due to Internal Pressure..........................................73 3.4 Values of D for Junctions at the Large Cylinder Due to External Pressure.........................................80 E3.14 Allowable Stress and Pressure Data .....................................................................................................90 6.1 Example of Pressure Used for Design of Components......................................................................163 6.2 Closure Detail Requirements for Various Types of Jacket Closures..................................................169 6.3 Penetration Detail Requirements ........................................................................................................171 7.1 Joint Efficiencies fr for various tube-to-tubesheet attachments (ASME VIII-1)................................228 8.1 Primary Stress Category .....................................................................................................................234 8.2 Structural Discontinuity......................................................................................................................235 8.3 Thermal Stress ....................................................................................................................................235 8.4 Stress Categories and Their Limits (ASME VIII-2)...........................................................................236 8.5 Classification of Stresses (ASME VIII-2) ..........................................................................................237 8.6 Some Stress Concentration Factors Used in Fatigue .........................................................................239 E8.4 Summary of Finite Element Output ...................................................................................................244 B.1 Carbon Steel Plate ..............................................................................................................................235 B.2 Chrome-Moly Steel Plate Specifications, SA-387 .............................................................................235 B.3 Chrome-Moly Steel Forging Specifications, SA-182.........................................................................235 B.4 Chrome-Moly Steel Forging Specifications, SA-336.........................................................................256 B.5 Quench & Tempered Carbon and Alloy Steel Forgings, SA-508 ......................................................256

xxi



CHAPTER

1 BACKGROUND INFORMATION 1.1 INTRODUCTION In this chapter some general concepts and criteria pertaining to Section VIII are discussed. These include allowable stress, factors of safety, joint efficiency factors, brittle fracture, fatigue, and pressure testing. Detailed design and analysis rules for individual components are discussed in subsequent chapters. Since frequent reference will be made to ASME Section VIII Divisions 1 and 2, the following designation will be used from here on to facilitate such references. ASME Section VIII, Division 1 Code will be designated by VIII-1. Similarly, VIII-2 will designate the ASME Section VIII, Division 2 Code. Other ASME code sections such as Section II Part D will be referred to as II-D. Equations and paragraphs referenced in each of these divisions will be called out as they appear in their respective Code Divisions. Many designs rules in VIII-1 and VIII-2 are identical. These include flange design and external pressure requirements. In such cases, the rules of VIII-1 will be discussed with a statement indicating that the rules of VIII-2 are the same. Appendix A at the end of this book lists the paragraph numbers in VIII-1 that pertain to various components of pressure vessels. Section VIII requires the fabricator of the equipment to be responsible for its design. Paragraphs UG-22 in VIII-1 and AD-110 in VIII-2 are given to assist the designer in considering the most commonly encountered loads. They include pressure, wind forces, equipment loads, and thermal considerations. When the designer takes exceptions to these loads either because they are not applicable or they are unknown, then such exceptions must be stated in the calculations. Similarly, any additional loading conditions considered by the designer that are not mentioned in the Code must be documented in the design calculations. Paragraphs U-2(a) and U-2(b) of VIII-1 give guidance for some design requirements. VIII-2, paragraph AD110 and the User’s Design Specifications mentioned in AG-301 provide the loading conditions to be used by the manufacturer. Many design rules in VIII-1 and VIII-2 are included in the Appendices of these codes. These rules are for specific products or configurations. Rules that have been substantiated by experience and used by industry over a long period of time are in the Mandatory Appendices. New rules or rules that have limited applications are placed in the Non-Mandatory Appendices. Non-Mandatory rules may eventually be transferred to the Mandatory section of the Code after a period of use and verification of their safety and practicality. However, guidance-type appendices will remain in the Non-Mandatory section of the Code. The rules in VIII-1 do not cover all applications and configurations. When rules are not available, Paragraphs U-2(d), U-2(g), and UG-101 must be used. Paragraph U-2(g) permits the engineer to design components in the absence of rules in VIII-1. Paragraph UG-101 is for allowing proof testing to establish maximum allowable working pressure for components. In VIII-2 there are no rules similar to those in UG-101, since VIII-2 permits design by analysis as part of its requirements. This is detailed in Paragraphs AD-100(b), AD-140, AD-150, and AD-160 of VIII-2. Many pressure vessel designers and vessel users are of the opinion that a vessel or component will be thinner, and subsequently lower cost, using the VIII-2 rules compared with using the VIII-1 rules. This opinion 1


2

Chapter 1

is usually based on the design factor on tensile strength of 3.5 for VIII-1 while the factor for VIII-2 is lower at 3.0. The design factor on yield strength for both VIII-1 and VIII-2 is 1.5. However, as can be seen from a comparison of some design calculations for the same geometry and loadings, the thickness required for a VIII-2 component may be greater than that required for a VIII-1 component. (For example, compare the calculations of an opening in a 2:1 ellipsoidal head given in Example Design Problems 5.1 and 5.5). Often, this is caused by differences in design formulas and analysis methods as well as the design criteria and allowable stresses. For many materials, the allowable stress is set by the yield strength criteria rather than by the tensile strength criteria and, consequently, the allowable stress value at a temperature may be the same.

1.2 ALLOWABLE STRESSES Except for UCI, UCD, and ULT material and for bolting, the criteria for establishing allowable stress values for VIII-1 is detailed in Appendix 1 of II-D and summarized in Table 1.1.1. For UCI, UCD, and ULT material, the criteria for establishing allowable stress values for VIII-1 is detailed in Appendix P of VIII-1 and summarized in Table 1.1.2. The criteria for establishing the design stress intensity values for VIII-2 is detailed in Appendix 2 of II-D and summarized in Table 1.2. The criteria for establishing allowable stress values for bolting for both VIII-1 and VIII-2 is detailed in Appendix 2 of II-D. The allowable stress at design temperature for most materials is the lessor of 1/3.5 the minimum effective tensile strength or 2/3 the minimum yield stress of the material for temperatures below the creep and rupture values. The controlling allowable stress for most bolts is 1/5 the tensile strength. The minimum effective tensile stress at elevated temperatures is obtained from the actual tensile stress curve with some adjustments. The tensile stress value obtained from the actual curve at a given temperature is multiplied by the lessor of 1.0 or the ratio of the minimum tensile stress at room temperature obtained from ASTM Specification for the given material to the actual tensile stress at room temperature obtained from the tensile strength curve. This quantity is then multiplied by the factor 1.1. The effective tensile stress is then equal to the lessor of this quantity or the mini-

TABLE 1.1.1 CRITERIA FOR ESTABLISHING ALLOWABLE STRESS VALUES FOR VIII-1 EXCEPT FOR UCI, UCD, AND ULT MATERIAL AND FOR BOLTING (ASME II-D)

Nomenclature RT = ratio of the average temperature dependent trend curve value of tensile strength to the room temperature tensile strength RY = ratio of the average temperature dependent trend curve value of yield strength to the room temperature yield strength SRavg = average stress to cause rupture at the end of 100,000 hr SRmin = minimum stress to cause rupture at the end of 100,000 hr SC = average stress to produce a creep rate of 0.01 %/1000 hr ST = specified minimum tensile strength at room temperature, ksi SY = specified minimum yield strength at room temperature


Background Information

3

TABLE 1.1.2 CRITERIA FOR ESTABUSHING ALLOWABLE STRESS VALUES FOR VIII-1 FOR UCI, UCD, AND ULT MATERIAL (ASME VIII-1)

TABLE 1.2 CRITERIA FOR ESTABLISHING DESIGN STRESS INTENSITY VALUES FOR VIII-2 EXCEPT FOR BOLTING (ASME II-D)

mum tensile stress at room temperature given in ASTM. This procedure is illustrated in example 4.1 of Jawad and Farr (reference 14, found at back of book). The 1.1 factor discussed above is a constant established by the ASME Code Committee. It is based on engineering judgment that takes into consideration many factors. Some of these include increase in tensile strength for most carbon and low alloy steels between room and elevated temperature; the desire to maintain a constant allowable stress level between room temperature and 500°F or higher for carbon steels; and the adjustment of minimum strength data to average data. Above approximately 500°f or higher the allowable stress for carbon steels is controlled by creep-rupture rather than tensile-yield criteria. Some materials may not exhibit such an increase in tensile stress, but the criterion for 1.1 is still applicable to practically all materials in VIII-1. Table 1.1.1 also gives additional criteria for creep and rupture at elevated temperatures. The criteria are based on creep at a specified strain and rupture at 100,000 hours. The 100,000 hours criterion for rupture corresponds to about eleven years of continual use. However, VIII-1 does not limit the operating life of the equipment to any specific number of hours. The allowable stress criteria in VIII-2 are given in II-D of the ASME Code. The allowable stress at the design temperature for most materials is the smaller of 1/3 the tensile strength or 2/3 the yield stress. The design temperature for all materials in VIII-2 is kept below the creep and rupture values. Table 1.2 summarizes the allowable stress criteria in VIII-2. A sample of the allowable stress Tables listed in Section II-D of the ASME Code is shown in Table 1.3. It lists the chemical composition of the material, its product form, specification number, grade, Unified


4

Chapter 1

TABLE 1.3 STRESS VALUES FOR SA-515 AND SA-516 MATERIALS Line Nominal Product No. Composition Form 19 20

Line No.

CS CS

Plate Plate

Min. Tensile Strength ksi

Min. Yield Stress ksi

70 70

38 38

19 20

Spec No.

Type/ Grade

SA-515 SA-516

70 70

Alloy Desig./ Class Cond./ Size/ UNS No. Temper Thick, in. K03101 K02700

… …

P-No.

… …

1 1

I

Applic. & max. Temp. Limits (NP = Not Permitted) (SPT = Supports only) III

VIII-1

External Pressure Chart No.

1000 850

700 700

1000 1000

CS-2 CS-2

Group No. 2 2

Notes

G10, S1.T2 G10, S1.T2

Maximum Allowable Stress, ksi, for Metal Temperature, °F, Not Exceeding Line –20 to No. 100 150

200

300

400

500

600

650

700

19 20

20.0 20.0

20.0 20.0

20.0 20.0

20.0 20.0

19.4 19.4

18.8 18.8

18.1 18.1

20.0 20.0

20.0 20.0

750 14.8 14.8

800

850

900

12.0 12.0

9.3 9.3

6.7 6.7

Note: G10, S1, T2 are described in II-D and pertain to metallurgical information

Numbering System (UNS), size, and temper. This information, with very few exceptions, is identical to that given in ASTM for the material. The Table also lists the P and Group numbers of the material. The P numbers are used to cross reference the material to corresponding welding processes and procedures listed in Section IX, “Welding and Brazing Qualifications,” of the ASME Code. The Table also lists the minimum yield and tensile strengths of the material at room temperature, maximum applicable temperature limit, External Pressure Chart reference, any applicable notes, and the stress values at various temperatures. The designer may interpolate between listed stress values, but is not permitted to extrapolate beyond the published values. Stress values for components in shear and bearing are given in various parts of VIII-1, VIII-2, as well as II-D. Paragraph UW-15 of VIII-1 and AD-132 of VIII-2 lists the majority of these values. A summary of the allowable stress values for connections is shown in Table 1.4. Some material designations in ASTM as well as the ASME Code have been changed in the last 20 years. The change is necessitated by the introduction of subclasses of the same material or improved properties. Appendix B shows a cross reference between older and newer designations of some common materials. The maximum design temperatures allowed in VIII cannot exceed those published in Section II-D. VIII-1 defines design temperature as the mean temperature through the cross section of a component. VIII-2 defines design temperature as the mean temperature in the cross section of a component, but the surface temperature cannot exceed the highest temperature listed in II-D for the material. This difference in the definition of temperature in VIII-1 and VIII-2 can be substantial in thick cross sections subjected to elevated temperatures.

1.3 JOINT EFFICIENCY FACTORS In the ASME Boiler Code, Section I, as well as in VIII-2, all major longitudinal and circumferential butt joints must be examined by full radiography, with few exceptions. VIII-1, on the other hand, permits various levels of examination of these major joints. The examination varies from full radiographic to visual, depending on various factors specified in VIII-1 and by the user. The degree of examination influences the required



5

TABLE 1.4 ALLOWABLE STRESS VALUES FOR WELDED CONNECTIONS VIII-1 Component Fillet weld Fillet weld Groove weld Groove weld Nozzle neck Dowel bolts Any location

Type of Stress

Stress Value

Reference

tension shear tension shear shear shear bearing

0.55S* 0.49S 0.74S 0.60S 0.70S 0.80S 1.60S

UW-18(d) UW-15(c) UW-15(c) UW-15(c) UG-45(c) II-D II-D

Type of Stress

Stress Value

Reference

tension shear tension shear shear bearing

0.5Sm* 0.5Sm* 0.75Sm 0.75Sm 0.6Sm Sy

AD-920 AD-920 AD-920 AD-920 AD-132.2 AD-132.1

* S = allowable stress for VIII-1 construction

VIII-2 Component Fillet weld Fillet weld Groove weld Groove weld Nozzle neck Any location

* Sm = stress intensity values for VIII-2 construction

FIG. 1.1 WELDED JOINT CATEGORIES (ASME VIII-1)

thickness through the use of Joint Efficiency Factors, E. The Joint Efficiency Factors, which are sometimes referred to as Quality Factors or weld efficiencies, serve as stress multipliers applied to vessel components when some of the joints are not fully radiographed. These multipliers result in an increase in the factor of safety as well as the thickness of these components. In essence, VIII-1 vessels have variable factors of safety, depending on the degree of radiographic examination of the main vessel joints. As an example, fully radiographed longitudinal butt-welded joints in cylindrical shells have a Joint Efficiency Factor, E, of 1.0. This factor corresponds to a safety factor of 3.5 in the parent material. Nonradiographed longitudinal butt-welded joints have an E value of 0.70. This reduction in Joint Efficiency Factor corresponds to a factor of safety of 5.0 in the plates. This higher factor of safety due to a nonradiographed joint results in a 43% increase in the required thickness over that of a fully radiographed joint.


Chapter 1

TABLE 1.5 MAXIMUM ALLOWABLE JOINT EFFICIENCIES1,5 FOR ARC- AND GAS-WELDED JOINTS (ASME VIII-1)

(continues)

6


TABLE 1.5 (CONTINUED)



7

8

Chapter 1

ASME VIII-1 identifies four joint categories that require E factors. They are Categories A, B, C, and D as shown in Fig. 1.1. Category A joints consist mainly of longitudinal joints as well as circumferential joints between hemispherical heads and shells. Category B joints are the circumferential joints between various components as shown in Fig. 1.1, with the exception of circumferential joints between hemispherical heads and shells. The attachment of flanges to shells or heads is a Category C joint. The attachment of nozzle necks to heads, shells, and transition sections is categorized as a Category D joint. The four joint categories in VIII-1 do not apply to items such as jacket closure bars, tubesheet attachments, and ring girders. The degree of examination of the welds attaching these components to the shell or head is not covered in VIII-1. Most designers assign an E value of 1.0 when calculating the shell or head thickness at such junctions. This is justified since in most cases the strain in the hoop direction, and hence hoop stress, is close to zero at the junction due to the restraint of tubesheet or bars. The type of construction and joint efficiency associated with each of joints A, B, C, and D is given in Table 1.5. The categories refer to a location within a vessel rather than detail of construction. Thus, a Category C weld, which identifies the attachment of a flange to a shell, can be either fillet, corner, or butt welded, as illustrated in Fig. 1.2. The Joint Efficiency Factors apply only to the butt-welded joint in sketch (c). The factors do not apply to sketches (a) and (b) since they are not butt welded. The Joint Efficiency Factors used to design a given component are dependent on the type of examination performed at the welds of the component. As an example, the Joint Efficiency Factor in a fully radiographed longitudinal seam of a shell course is E = 1.0. However, this number may have to be reduced, depending on the degree of examination of the circumferential welds at either end of the longitudinal seam. Appendix B shows some typical components and their corresponding Joint Efficiency Factors. Example 1.1 Problem Determine the category and Joint Efficiency Factor of the joints in the heat exchanger shown in Fig. E1.1. The channel side is spot radiographed. The longitudinal seam, b, is a single-welded butt joint with a backup bar. The shell side is not radiographed. The longitudinal and circumferential seams m and 1 are singlewelded butt joints with backup bars. The jacket longitudinal seam, n, is a single-welded butt joint, without backup bar. Solution The joint categories of the various joints can be tabulated as given in Table E1.1: TABLE E1.1 STRESS CATEGORIES Joint

Location

Category

(a) (b) (c) (d) (e) (f)

Channel-to-flange connection Longitudinal channel seam Channel-to-tubesheet weld Nozzle-to-channel weld Flange-to-nozzle neck Pass partition-to-tubesheet weld

C A C D C None

(g) (h) (i) (j) (k) (l) (m) (n) (o)

Tube-to-tubesheet weld Shell-to-tubesheet weld Jacket bar-to-inner-shell weld Jacket to bar-to-outer-shell weld Nozzle-to-jacket weld Longitudinal shell seam (m)Head-to-shell seam Longitudinal jacket seam Skirt-lo-head seam

None G None None D A B A None

Joint Efficiency Does not apply 0.80 Does not apply Does not apply Does not apply Does not apply. See also UW-15(c) and UW-18(d)of VIII-1 Does not apply. See also UW-20 of VIII-1 Does not apply Does not apply Does not apply Does not apply 0.65 0-65 0.60 Does not apply



9

FIG. 1.2 CATEGORY C WELD

FIG. E1.1

1.4 BRITTLE FRACTURE CONSIDERATIONS Both VIII-1 and VIII-2 require the designer to consider brittle fracture rules as part of the material and design selection. The rules for carbon steels are extensive and are discussed first. VIII-1 has two options regarding toughness requirements for carbon steels. The first is given in Paragraph UG-20(f) and allows the designer to exempt the material of construction from impact testing when all of the following criteria are met: 1. The material is limited to P-No.1, Gr. No. 1 and 2, and thickness shall not exceed that given in (a) or (b) below:


10

Chapter 1

TABLE 1.6 ASSIGNMENT OF MATERIALS TO CURVES (ASME VIII-1)

2. 3. 4. 5.

(a) 1/2 in. for materials listed in Curve A in Table 1.6. (b) 1 in. for materials listed in Curve B, C, or D in Table 1.6. The completed vessel shall be hydrostatically tested per UG-99(b) or (c) or to 27-4. Design temperature between –20°F and 650°f, incl. Occasional operating temperatures colder than –20°f lare acceptable when due to lower seasonal atmospheric temperatures. The thermal and mechanical shock loadings are not a controlling design requirement. Cyclical loading is not a controlling design requirement.

The above requirements are intended for, but not limited to, relatively thin carbon steel vessels operating in a service that is neither severe in thermal and pressure cycling nor in extreme cold temperatures. Vessels



TABLE 1.6 (CONTINUED)


11

12

Chapter 1

of low alloy steel or those with carbon steel operating beyond the scope of Paragraph UG-20(f) require an evaluation for brittle fracture in accordance with the rules of UCS-66. The procedure consists of 1. Determining the governing thickness in accordance with Fig. 1.3. 2. Using Fig. 1.4 to obtain the temperature that exempts the material from impact testing. If the specified Minimum Design Metal Temperature, MDMT, is colder than that obtained from the figure, then impact testing in accordance with Fig. 1.5 is required. The specified MDMT is usually given by the user, while the calculated MDMT is obtained from VIII-1. The calculated MDMT is kept equal to or colder than the specified MDMT. 3. The temperature obtained from Fig. 1.4 may be reduced in accordance with Fig 1.6 if the component operates at a reduced stress. This is detailed in Paragraph UCS-66(b) of VIII-1.

FIG. 1.3 SOME GOVERNING THICKNESS DETAILS USED FOR TOUGHNESS (ASME VIII-1)



At a ratio of 0.35 in Fig. 1.6, the permitted temperature reduction drops abruptly. At this ratio, the stress in a component is about 6000 psi. At this stress level, experience has shown that brittle fracture does not occur regardless of temperature level. 4. The rules in VIII-1 also allow a 30°F reduction in temperature below that obtained from Fig. 1.4 when the component is post-weld heat treated but is not otherwise required to be post-weld heat treated by VIII-1 rules.

FIG. 1.3 (CONT’D)


13

14

Chapter 1

NOTE: Variations of tg= governing thickness of the welded joint

FIG. 1.3 (CONT’D) The toughness rules for ferritic steels with tensile properties enhanced by heat treatment are given in Paragraph UHT-6 of VIII-1. The rules require such steels to be impact tested regardless of temperature. The measured lateral expansion as defined by ASTM E-23 shall be above 0.015 in. The toughness rules for high alloy steels are given in Paragraph UHA-51 of VIII-1. The permissible Minimum Design Metal Temperature for base material is summarized in Table 1.7. Similar data are given in VIII-1 for the weld material and weld qualifications. Thermally heated stainless steels may require impact testing per the requirements of UHA-51(c). The rules for toughness in VIII-2 are different than those in VIII-1. However, the concepts of exemption curves and Charpy impact levels are similar in VIII-2 and VIII-1. The toughness requirements for carbon and low alloy steels are given in Paragraph AM-218 of VIII-2. High alloy steels are covered in Paragraph AM-213. Example 1.2 Problem Determine the Minimum Design Metal Temperature, MDMT, for the reactor shown in Fig. E1.2. Let the shell, head, pad, and ring material be SA-516 Gr. 70 material. Flange and cover material is SA-105. Pipe



15

FIG. 1.4 IMPACT-TEST EXEMPTION CURVES (ASME VIII-1)

material is SA-106. The required shell thickness is 1.75 in., and the required head thickness is 0.86 in. The required nozzle neck thickness is 0.08 in. Assume a joint efficiency of 1.0 and no corrosion allowance. Solution Shell SA-516 specifications require the material to be normalized when the thickness exceeds 1.5 in. Thus, from Table 1.6, Curve D is to be used for normalized SA-516 Gr. 70 material. Using Fig. 1.4 and a governing


16

Chapter 1

FIG. 1.5 CHARPY IMPACT-TEST REQUIREMENTS FOR FULL SIZE SPECIMENS FOR CARBON AND LOW ALLOY STEELS WITH TENSILE STRENGTH OF LESS THAN 95 KSI (ASME VIII-1)

thickness of 2.0 in., we get a minimum temperature of –5°F. The ratio of required thickness to actual thickness is 1.75/2.0 = 0.88. Using Fig. 1.6 for this ratio, we obtain 12°F. Hence, MDMT = –5–12 = –17°F. Head For a 1 in. thick head, SA-516 specifications permit a non-normalized material. Thus, from Table 1.6, Curve B is used. Using Fig. 1.4 and a governing thickness of 1.0 in., we get a minimum temperature of 30°F. The



17

FIG. 1.6 REDUCTION OF MDMT WITHOUT IMPACT TESTING (ASME VIII-1)

ratio of re required thickness to actual thickness is 0.86/1.0 = 0.86. Using Fig. 1.6 for this ratio, we obtain 14°F. Hence, MDMT = 30 –14 = 16°F. Stiffener For a 0.75-in. stiffener, Curve B of Table 1.6 is to be used. Using Fig. 1.4 and a governing thickness of 0-75 in., we obtain a minimum temperature of 15°F. Since stresses cannot be established from VIII-1 rules, the MDMT = 15°F. Pad The material will be normalized since it is 2.00 in. thick. Curve D of Table 1.6 is used. From Fig. 1.4 and a governing thickness of 2.0 in., we obtain a minimum temperature of –5°F. Since stresses cannot be established from VIII-1 rules, the MDMT = –5°F.


18

Chapter 1

TABLE 1.7 MINIMUM DESIGN METAL TEMPERATURES IN HIGH ALLOY STEELS WITHOUT IMPACT TESTING

Nozzle Neck From Table 1.6, Curve B is to be used for a nozzle neck of 0.258-in. thickness. From Fig. 1.4, minimum temperature is –20°F. The ratio of required thickness to actual thickness is 0.08/0.258 × 0.875 = 0.36. Using Fig. 1.6 and this ratio, we get 130°F. Hence, MDMT = –20 – –130 = –150°F. Flange Since the flange is ANSI B16.5, it is good to –20°F. Cover From Fig. 1.3(c), the controlling cover thickness is 2.5/4 = 0.625 in. Curve B applies for this material, and the MDMT = 5°F. Therefore, the MDMT for this reactor is governed by the head with a value of 16°F. A colder value can be obtained by impact testing the various components. Thus, assuming a specified MDMT of –15°F is required, then the head, stiffener, pad, and cover need impact testing.



19

FIG. E1.2

1.5 FATIGUE REQUIREMENTS Presently, VIII-1 does not list any rules for fatigue evaluation of components. When fatigue evaluation of a component is required in accordance with UG-22 or U-2(g) of VIII-1, the general practice is to use the VIII2 fatigue criteria as a guidance up to the temperature limits of VIII-2. At temperatures higher than those given in VIII-2, the rules of III-H are followed for general guidance. Other fatigue criteria, such as those given in other international codes and ASME B31.3, may also be considered as long as the requirements of U-2(g) of VIII-1 are met. VIII-2 contains detailed rules regarding fatigue. Paragraph AD-160 gives criteria regarding the need for fatigue analysis. The first criterion is listed in Paragraph AD-160.1 and is based on experience. Vessels that have operated satisfactorily in a certain environment may be cited as the basis for constructing similar vessels operating under similar conditions without the need for fatigue analysis. The second criterion for vessel components is based on the rule that fatigue analysis is not required if all of Condition A or all of Condition B is satisfied, as noted below. Condition A Fatigue analysis is not required for materials with a tensile strength of less than 80 ksi when the total number of cycles in (a) through (d) below is less than 1000. a. The design number of full range pressure cycles including startup and shutdown. b. The number of pressure cycles in which the pressure fluctuation exceeds 20% of the design pressure, c. Number of changes in metal temperature between two adjacent points. These changes are multiplied by a factor obtained from the following chart in order to transform them to equivalent cycle number.


20

Chapter 1

Metal Temperature Differential, °F 50 or less 51 to 100 101 to 150 151 to 250 251 to 350 351 to 450 Higher than 450

Factor 0 1 2 4 8 12 20

d. Number of temperature cycles in components that have two different materials where a difference in the value (α1 – α2)∆T exceeds 0.00034. Where, α is the coefficient of thermal expansion and ∆T is the difference in temperature. Condition B Fatigue analysis is not required when the following items (a) through (f) are met: a. The number of full range pressure cycles, including startup and shutdown, is less than the number of cycles determined from the appropriate fatigue chart, Fig. 1.7, with an Sa value equal to 3 times the allowable design stress value, Sm. b. The range of pressure fluctuation cycles during operation does not exceed P(1/3)(Sa/Sm), where P is the design pressure, Sa is the stress obtained from the fatigue curve for the number of significant pressure cycles, and Sm is the allowable stress. Significant pressure cycles are defined as those that exceed the quantity P(1/3)(S/Sm). S is defined as S = Sa taken at 106 cycles when the pressure cycles are ≤ 106. S = Sa taken at actual number of cycles when the pressure cycles are > 106. c. The temperature difference between adjacent points during startup and shutdown does not exceed Sa/(2Eα), where Sa is the value obtained from the applicable design fatigue curve for the total specified number of startup and shutdown cycles. d. The temperature difference between adjacent points during operation does not exceed Sa/(2Eα), where Sa is the value obtained from the applicable design fatigue curve for the total number of significant fluctuations. Significant fluctuations is defined as those exceeding the quantity S/(2Eα), where S is as defined in (b) above. Adjacent points are defined in AD-160.2, Condition A, Paragraph (c) of VIII-2. e. Range of significant temperature fluctuation in components that have materials with different coefficient of expansion or modulus of elasticity and that do not exceed the quantity Sa/[2(E1 α1 – E2α2)], where α is the coefficient of thermal expansion and E is the modulus of elasticity. Significant temperature fluctuation is that which exceeds the value S/ [2(E1 α1 – E2 α2)], where S is as defined in (b) above. f. Range of mechanical loads does not result in stress intensities whose range exceeds the Sa value obtained from the fatigue chart. The third criterion for nozzles with nonintegral reinforcement is given in Paragraph AD-160.3 of VIII-2 and is very similar to Conditions A and B detailed above. Example 1.3 Problem A pressure vessel consisting of a shell and two hemispherical heads is constructed from SA 516–70 carbon steel material. The self-reinforced nozzles in the vessel are made from type SA 240–304 stainless steel material. The vessel is shut down six times a year for maintenance. At start-up, the full pressure of 300 psi and full temperature of 400°F are reached in two hours. The maximum ∆T between any two points during start-up is 250°F. At normal operation, the ∆T is negligible. At shutdown, the maximum ∆ T is 100°F.


FIG. 1.7 FATIGUE CURVES FOR CARBON, LOW ALLOY, SERIES 4XX, HIGH ALLOY STEELS, AND HIGH TENSILE STEELS FOR TEMPERATURES NOT EXCEEDING 700°F (ASME VIII-2)



21

22

Chapter 1

Determine the maximum number of years that this vessel can be operated if a fatigue evaluation is not performed. Let the coefficient of expansion for carbon steel be 6.5 × 10–6 in./in./°F and that for stainless steel be 9.5 × 10–6 in./in./°F. Solution From Condition A, determine the number of cycles in one year. a. Number of full pressure cycles for one year is 6. b. This condition does not apply for this case. c. From the chart, the 250°F difference in temperature during start-up corresponds to 4 cycles. The 100°F difference in temperature during shutdown corresponds to 1 cycle. Thus total equivalent cycles due to temperature in one year is (4 + 1) 6 = 30 cycles. d. At nozzle attachments, the quantity (9.5 × 10–6 – 6.5 × 10–6) 400 is equal to 0.0012. Since this value is greater than 0.00034, the equivalent cycles per year = 6. Total cycles per year due to (a), (c), and (d) = 6 + 30 + 6 = 42. Number of years to operate vessel if fatigue analysis is not performed = 1000/42 = 23.8 years. Example 1.4 Problem A pressure vessel has an inside diameter of 60 in., internal pressure of 300 psi, and design temperature of 500°F. The shell thickness is 1/2 in. at an allowable stress level of 18,000 psi (material tensile stress = 70 ksi). The thickness of the hemispherical heads is 1/4 in. at an allowable stress level of 18,000 psi. Integrally reinforced nozzles are welded to the shell and are also constructed of carbon steel with an allowable stress of 18,000 psi. At start-up, the full pressure of 300 psi and full temperature of 500°F are reached in eight hours. The maximum ∆T between any two points during start-up is 60°F. At normal operation, the ∆T is negligible. At shutdown, the maximum ∆T is 50°F. Determine if the shell and heads are adequate for 100,000 cycles without the need for fatigue analysis. From II-D, the coefficient of expansion for carbon steel is 7.25 × 10–6 in./in./°F and the modulus of elasticity is 27.3 × 106 psi. Use Fig. 1.7 for a fatigue chart. Solution Condition B is to be used. a. Three times allowable stress at the nozzle location is = 3(18,000) = 54,800 psi. Using Fig. 1.7 for this value gives a fatigue life of 4200 cycles. b. This condition does not apply. c. From Fig. 1.7, with 100,000 cycles, the value of Sa = 20,000 psi. The value of Sa/(2Eα) = 20,000/(2 × 37.3 × 106 × 7.25 × 10–6) = 51°F. Since this value is less than 60°F, the specified cycles are inadequate. The designer has two options in this situation. The first is to perform fatigue analysis, which is costly. The second option, if it is feasible, is to reduce the ∆T at startup to 51°F. d. This condition does not apply. e. This condition does not apply. f. This condition does not apply. Example 1.5 Problem In Example 1.4, determine the required thickness of the shell and heads for 1,000,000 cycles without the need to perform fatigue analysis.



23

Solution From Fig. 1.7, with a cycle life of 1,000,000, the value of Sa = 12,000 psi. From Condition B, subparagraph (a), the maximum stress value for the shell is (12,000/3) = 4,000 psi. The needed shell thickness = 0.5 × 18,000/4000 = 2.25 in. The required head thickness = 0.25 × 18,000/4000 = 1.13 in. The maximum ∆T at start-up or shutdown cannot exceed Sa/(2Eα) = 12,000/(2 × 27.3 × 106 × 7.25 × 10–6) = 31°F, otherwise a fatigue analysis is necessary.

1.6 PRESSURE TESTING OF VESSELS AND COMPONENTS 1.6.1 ASME Code Requirements Pressure vessels that are designed and constructed to VIII-1 rules, except those tested in accordance with the requirements of UG-101, are required to pass either a hydrostatic test (UG-99) or a pneumatic test (UG-100) of the completed vessel before the vessel is U-stamped. Pressure vessels that are designed and constructed to VIII-2 rules also are required to pass either a hydrostatic test (Article T-3) or a pneumatic test (Article T4) before the U2-stamp is applied. Each component section of the ASME Boiler and Pressure Vessel Code has a pressure test requirement that calls for a pressure test at or above the maximum allowable working pressure indicated on the nameplate or stamping and in the Manufacturer’s Data Report before the appropriate Code stamp mark may be applied. Under certain conditions, a pneumatic test may be combined with or substituted for a hydrostatic test. When testing conditions require a combination of a pneumatic test with a hydrostatic test, the requirements for the pneumatic test shall be followed. In all cases, the term hydrostatic refers not only to water being an acceptable test medium, but also to oil and other fluids that are not dangerous or flammable; likewise, pneumatic refers not only to air, but also to other nondangerous gases that may be desirable for “sniffer” detection.

1.6.2 What Does a Hydrostatic or Pneumatic Pressure Test Do? There is always a difference of opinion as to what is desired and what is accomplished with a pressure test. Some persons believe that the pressure test is meant to detect major leaks, while others feel that there should be no leaks, large or small. Some feel that the test is necessary to invoke loadings and stresses that are equivalent to or exceed those loadings and stresses at operating conditions. Others feel that a pressure test is needed to indicate whether a gross error has been made in calculations or fabrication. In some cases, it appears that the pressure testing may help round out corners or other undesirable wrinkles or may offer some sort of a stress relief to some components.

1.6.3 Pressure Test Requirements for VIII-1 1.6.3.1 Hydrostatic Test Requirements. A hydrostatic pressure test is the preferred test method. A pneumatic test or a combination of pneumatic/hydrostatic test is conducted only when a hydrostatic test cannot be done. Except for certain types of vessels that are discussed later, the hydrostatic test pressure at every point in the vessel shall be at least 1.3 times the maximum allowable working pressure multiplied by the ratio of the allowable tensile stress value at test temperature divided by the maximum allowable tensile stress value at design temperature. As an alternative, a hydrostatic test pressure may be determined by calculations agreed upon by the user and the manufacturer. In this case, the MAWP (maximum allowable working pressure) of each element is determined and multiplied by 1.3 and then adjusted for the hydrostatic head. The lowest value is used for the test pressure, which is adjusted by the test temperature to design temperature ratio. In any case, the test pressure is limited to that pressure which will not cause any visible permanent distortion (yielding) of any element. The metal temperature of the vessel or component to be tested is recommended to be at least 30°F above the MDMT to be marked on the vessel but need not exceed 120°F, to minimize the risk of brittle fracture. The test pressure shall not be applied until the vessel and its contents are at about the same temperature. If the test temperature is above 120°F. it is recommended that inspection be delayed until the temperature of the vessel is


24

Chapter 1

120°F or less. Also, it is recommended that a liquid relief valve set to 11/3time the test pressure be installed in the pressure test system in those cases where the vessel under test may warm up while test personnel are absent. After the test pressure is reached and the other items noted above are satisfied, the pressure is reduced to the test pressure divided by 1.3. At that time welded joints, connections, and other areas are visually examined for leaks and cracks. The visual examination may be waived if a gas leak test is to be applied, if hidden welds have been examined ahead of time, and if the vessel will not contain a lethal substance. Venting shall be provided at all high locations where there is a possibility of air pockets forming during the filling of the vessel for testing. The general rules for hydrostatic testing do not call for a specific time for holding the vessel at test pressure. The length of this time may be set by the Authorized Inspector or by a contract specification. 1.6.3.2 Pneumatic Test Requirements. For some vessels, it is necessary to apply a pneumatic test in lieu of a hydrostatic test. This may be due to any number of reasons, including vessels designed and supported in such a manner that they cannot be safely filled with liquid and vessels that cannot tolerate any trace of water or other liquids. If a vessel is to be pneumatically tested, it shall first be examined according to the requirements of UW-50. This paragraph requires that welds around openings and attachments be examined by MT or PT before testing. Except for certain vessels, the pneumatic test pressure at every point in the vessel shall be 1.1 times the maximum allowable working pressure multiplied by the ratio of the allowable tensile stress value at test temperature divided by the allowable tensile stress value at design temperature. For pneumatic testing, the metal temperature of the vessel or component shall be at least 30°F above the MDMT to be marked on the vessel. The test pressure shall be gradually increased to no more than half of the full test pressure and then increased in steps of one-tenth of the test pressure until the full test pressure is reached. After that, the pressure shall be reduced to the test pressure divided by 1.1 and all areas are to be examined. All other requirements for hydrostatic testing shall be observed, including the waiving of the visual examination, provided the same limits are met. 1.6.3.3 Test Requirements for Enameled or Glass-lined Vessels. The maximum test pressure for enameled and glass-lined vessels does not have to be any greater than 1.0 MAWP unless required by the Authorized Inspector or by a contract specification. Higher test pressure may damage the enameled or glass coating. All other rules for hydrostatic testing apply. 1.6.3.4 Test Requirements for Vessels Built to the Rules of Parts UCI or UCD. For those vessels designed and constructed to the rules of Part UCI for Cast Iron and Part UCD for Cast Ductile Iron, where the factor of safety on tensile strength to set the allowable tensile stress values is 10 and 5, respectively, the multiplier for the hydrostatic test pressure is set differently. For Part UCI, the test pressure shall be 2.5 MAWP, but is not to exceed 60 psi for a design pressure less than 30 psi and 2.0 MAWP for a design pressure equal to or greater than 30 psi. For Part UCD, the test pressure shall be 2.0 MAWP. With these changes, the remaining rules of UG-99 are followed.

1.6.3.5 Test Requirements for Vessels Built to the Rules of Part ULT. Alternative rules for the design and construction of vessels to operate at cold temperatures as low as –320°F are given in Part ULT. These rules permit the use of increased allowable tensile stress values at temperatures colder than ambient temperature to as low as –320°F for 5%, 8%, and 9% nickel steels, 5083 aluminum alloy, and Type 304 stainless steels. Other materials listed in both Section II and Subsection C may be used for vessels and parts for design at cold temperature with the allowable tensile stress values set by the value at 100°F. When the vessel is designed and constructed to Part ULT rules, special hydrostatic testing requirements are necessary due to the fact that the material is stronger at design temperature than at ambient test temperature. The vessel shall be hydrostatically tested at ambient temperature with the test pressure held for 15 minutes and either of the following criteria may be applied:



25

a. A standard hydrostatic test as described in 1.6.3.1 is used, but with the ratio of allowable stresses not applied and the test pressure shall be 1.4 MAWP, if possible, instead of 1.3 MAWP. b. In applying (a), the membrane stress in the vessel shall not exceed 0.95 of the specified minimum yield strength nor 0.5 of the specified minimum tensile strength. In complying with these stress limits, the ratio of hydrostatic test pressure divided by the MAWP may be reduced below 1.4, but it shall be not less than 1.1 MAWP. If the value comes out less than 1.1 MAWP, a pneumatic test shall be conducted using the rules of UG-100, but omitting the adjustment for the allowable tensile stress ratio. A vessel to be installed vertically may be tested in the horizontal position, provided the test pressure is applied for 15 minutes at not less than 1.4 MAWP, including a pressure equivalent to the liquid head in operating position. 1.6.3.6 Proof Testing to Establish MAWP. In addition to the hydrostatic or pneumatic pressure test of the completed vessel, a pressure proof test is permitted to establish the MAWP of vessels and vessel parts for which the strength cannot be calculated with assured accuracy. The rules for such a pressure proof test are given in UG-101 of VIII-1 and may be based on yielding or on bursting of the vessel or vessel part. Proof tests must be witnessed by the Authorized Inspector, who indicates acceptance by signing the Manufacturer’s Data Report Form. Duplicate or similar parts to that part which has had its MAWP established by a proof test according to the requirements of UG-101(d) of VIII-1 may be used without a proof test of their own, but shall be given a hydrostatic or pneumatic pressure test as part of the completed vessel pressure test.

1.6.4 Pressure Test Requirements for VIII-2 1.6.4.1 Hydrostatic Test Requirements. Except for glass-lined and enameled vessels, the hydrostatic test pressure at every point in the vessel shall be 1.25 times the design pressure (or MAWP) to be marked on the vessel multiplied by the ratio of the design stress intensity value at test temperature divided by the design stress intensity value at design temperature. For glass-lined or enameled vessels, the hydrostatic test pressure shall be at least equal to, but need not exceed, the design pressure (or MAWP). Similar to the alternative in VIII-1, the hydrostatic test pressure may be determined by calculations agreed upon between the User and the Manufacturer and shall be described in the Design Report. The hydrostatic test pressure shall not exceed a value that results in the following: a. A calculated primary membrane stress intensity Pm of 90% of the tabulated yield strength Sy at test temperature; b. A calculated primary membrane plus primary bending stress intensity Pm + Pb not to exceed the limits given below: [Pm + Pb] ≤ 1.35 Sy, when Pm ≤ 0.67 Sy

(1.1)

[Pm + Pb] ≤ 2.35 Sy – 1.5 Pm, when 0.67 Sy < Pm ≤ 0.90 Sy

(1.2)

1.6.4.2 Pneumatic Test Requirements. A pneumatic test is permitted only when one of the following prevails: a. Vessels cannot be safely filled with water due to their design and support system; b. Vessels in which traces of testing liquid cannot be tolerated. When a pneumatic test is permitted in lieu of a hydrostatic test, except for glass-lined and enameled vessels, the pneumatic test pressure at every point in the vessel shall be 1.15 times the design pressure (or


26

Chapter 1

MAWP) to be marked on the vessel multiplied by the ratio of the design stress intensity value at test temperature divided by the design stress intensity value at design temperature. For glass-lined or enameled vessels, the pneumatic test pressure shall be at least equal to, but need not exceed, the design pressure (or MAWP). The pneumatic test pressure shall not exceed a value that results in the following: a. A calculated primary membrane stress intensity Pm of 80% of the tabulated yield strength Sy at test temperature; b. A calculated primary membrane plus primary bending stress intensity Pm + Pb not to exceed the following limits: [Pm + Pb] ≤ 1.20 Sy, when Pm ≤ 0.67 Sy

(1.3)

[Pm + Pb] ≤ 2.20 Sy – 1.5 Pm, when 0.67 Sy < Pm ≤ 0.80 Sy

(1.4)


CHAPTER

2 CYLINDRICAL SHELLS 2.1 INTRODUCTION The rules for cylindrical shells in VIII-1 and VIII-2 take into consideration internal pressure, external pressure, and axial loads. The rules assume a circular cross section with uniform thickness in the circumferential and longitudinal directions. Design requirements are not available for elliptic cylinders or cylinders with variable thicknesses and material properties. However, such construction is not prohibited in VIII in accordance with Paragraphs U-2(g) of VIII-1 and, AG-100(b) and (d) of VIII-2. The design and loading conditions given in VIII-1 are discussed first in this chapter, followed by the rules in VIII-2.

2.2 TENSILE FORCES, VIII-1 The governing equations and criteria for the design of cylindrical shells under tensile forces are given in several paragraphs of VIII-1. The tensile forces arise from various loads such as those listed in Paragraph UG22 and include internal pressure, wind loads, and earthquake forces.

2.2.1 Thin Cylindrical Shells The required thickness of a cylindrical shell due to internal pressure is determined from one of two equations listed in Paragraph UG-27. The equation for the required thickness in the circumferential direction, Fig. 2.1 (a), due to internal pressure is given as t = PR/(SE – 0.6P),

when

t < 0.5R

or

P < 0.385SE

(2.1)

where E = Joint Efficiency Factor P = internal pressure R = internal radius S = allowable stress in the material t = thickness of the cylinder This equation can be rewritten to calculate the maximum pressure when the thickness is known. It takes the form P = SEt/(R + 0.6t)

(2.2)

It is of interest to note the similarity between Eq. (2.1) and the classical equation for circumferential membrane stress in a thin cylinder (Beer, et al, 2001), given by t = PR/SE 27


(2.3)

28

Chapter 2

FIG. 2.1

The difference is in the additional term of 0.6P in the denominator. This term was added by the ASME to take into consideration the nonlinearity in stress that develops in thick cylinders, i.e., when the thickness of a cylinder exceeds 0.1R. This is demonstrated in Fig. 2.2 for circumferential stress calculated by three different methods. The first is from Eq. (2.3), the theoretical equation for thin cylinders; the second is from Eq. (2.1); and the third is from Lamé’s theoretical equation for thick cylinders and is discussed later as Eq. (2.12). Similarly, the equation for the required thickness in the longitudinal direction, Fig. 2.1(b), due to internal pressure is given as t = PR/(2SE + 0.4P),

with

t < 0.5R

or

P < 1.25SE

(2.4)

or in terms of pressure, P = 2SEt/(R – 0.4 t)

(2.5)

Notice again the similarity between Eq. (2.4) and the classical equation for longitudinal stress in a thin cylinder given by


Cylindrical Shells 29

FIG. 2.2 COMPARISON OF EQUATIONS FOR HOOP STRESS IN CYLINDRICAL SHELLS

t = PR/2SE

(2.6)

Equations (2.1) and (2.4) are in terms of the inside radii of cylinders. In some instances, the outside radius of a shell is known instead. In this case, the governing equation for circumferential stress is expressed in terms of the outside radius RO. This equation, which is obtained from Eq. (2.1) by substituting (RO – t) for R, is given in VIII-1, Appendix 1, Article 1-1, as t = PRO/(SE + 0.4P),

with

t < 0.5RO

or

P < 0.385SE

P = SEt/(RO – 0.4t)

(2.7) (2.8)

VIII-1 does not give an equation for the thickness in the longitudinal direction in terms of outside radius RO. Such an expression can be obtained from Eq. (2.4) as


30

Chapter 2

FIG. 2.3

t = PRO/(2SE + 1.4P)

(2.9)

P = 2SEt/(RO – 1.4t)

(2.10)

or in terms of P,

Equations (2.1) through (2.10) are applicable to solid wall as well as layered wall construction. Layered vessels consist of thin cylinders wrapped around each other to form a thick cylinder, Fig. 2.3. At any given cross section, a–a, the total thickness consists of individual plate material as well as weld seams. The Joint Efficiency Factor for the overall thickness of a layered vessel is calculated from the ratio E = (Σ Eiti)/t

(2.11)

where E = overall Joint efficiency Factor for the layered cylinder Ei = Joint Efficiency Factor in a given layer t = overall thickness of a layered cylinder ti = thickness of one layer The rules in VIII-1 assume that the longitudinal welds in various layers are staggered in such a way that E in Eq. (2.11) is essentially equal to 1.0. Example 2.1 Problem A pressure vessel is constructed of SA 516-70 material and has an inside diameter of 8 ft. The internal design pressure is 100 psi at 450°F. The corrosion allowance is 0.125 in., and the joint efficiency is 0.85. What is the required shell thickness if the allowable stress is 20,000 psi? Solution Refer to Paragraph UG-27 of VIII-1. The quantity 0.385SE = 6545 psi is greater than the design pressure of 100 psi. Thus, Eq. (2.1) applies. The inside radius in the corroded condition is equal to



R= 48 + 0.125 = 48.125 in. t = [PR/(SE – 0.6P)] + corrosion = [100 × (48.125)/(20,000 × 0.85 – 0.6 × 100)] + 0.125 = 0.41 in.

The calculated thickness is less than 0.5R. Thus, Eq. (2.1) is applicable. A check of Eq. (2.4) for the required thickness in the longitudinal direction will result in a t = 0.27 in., including corrosion allowance. This is about 60% of the thickness obtained in the circumferential direction. Example 2.2 Problem A pressure vessel with an internal diameter of 120 in. has a shell thickness of 2.0 in. Determine the maximum permissible pressure if the allowable stress is 20 ksi. Assume E = 0.85. Solution For the circumferential direction, the maximum pressure is obtained from Eq. (2.2) as P = 20,000 × 0.85 × 2.0/(60 + 0.6 × 2.0) = 556 psi

For the longitudinal direction, the maximum pressure is obtained from Eq. (2.5) as P = 2 × 20,000 × 0.85 × 2.0/(60 – 0.4 × 2.0) = 1149 psi

Thus, the maximum pressure permissible in the vessel is 556 psi. Example 2.3 Problem A vertical boiler is constructed of SA 516-70 material and built in accordance with the requirements of VIII1. It has an outside diameter of 8 ft and an internal design pressure of 450 psi at 709°F. The corrosion allowance is 0.125 in., and the joint efficiency is 1.0. Calculate the required thickness of the shell if the allowable stress is 17,500 psi. Also, calculate the maximum allowable additional tensile force in the axial direction that the shell can withstand at the design pressure. Solution From Eq. (2.7), the required thickness is t = 450 × 48/(17,500 × 1.0 + 0.4 × 450) + 0.125 = 1.222 + 0.125 = 1.35 in.


32

Chapter 2

From Eq. (2.10), the maximum allowable axial pressure is P = 2 × 17,500 × 1.0 × 1.222/(48 – 1.4 × 1.222) = 924.0 psi

Subtracting from this value the internal pressure of 450 psi results in the additional equivalent pressure P′, that can be applied to the cylinder during operation. P′ = 924.0 – 450 = 474.0 psi

Total corroded metal area of cylinder = π(R2O – R2) = π(482 – 46.7782) = 363.9 in.2

Hence, total allowable force in cylinder during operation is F = 474.0 × 363.9 = 172,500 lb

Example 2.4 Problem What is the required thickness of a layered cylinder subjected to an internal pressure of 1400 psi? Let R = 72 in., S = 18 ksi, ti = 0.25 in. The longitudinal seams of the layers are staggered circumferentially so that any cross section will have only one longitudinal joint with an efficiency of 0.65. Solution This problem must be solved by trial and error. Let E = 1.0. Then from Eq. (2.1), t = 1400 × 72/(18000 × 1.0 – 0.6 × 1400) = 5.87 in.

Try 24 – 1/4 in. layers with a total thickness of 6.0 in. The joint efficiency from Eq. (2.11) for the total cross section is E = (23 × 1.00 + 1 × 0.65)/24 = 0.985

Using this Joint Efficiency Factor, recalculate the required thickness: t = 1400 × 72/(18000 × 0.985 – 0.6 × 1400) = 5.97 in.

Since this thickness is less than the assumed thickness of 6.0 in., the solution is complete. Hence, 24 – 1/4 in. layers are adequate.



2.2.2 Thick Cylindrical Shells The VIII-1 code is routinely referenced in constructing vessels with internal pressures higher than 3000 psi. Special consideration must be given to details of construction, as specified in Paragraph U-1(d) of VIII-1. As the ratio of t/R increases beyond 0.5, the thickness given by Eq. (2.1) becomes nonconservative, as illustrated in Fig. 2.2. A more accurate equation that determines the thickness in a thick cylinder, called Lamé’s equation, is given by SE = P(R2O + R2)/(R2O – R2)

(2.12)

where RO and R are outside and inside radii, respectively. By substituting the relationship RO = R + t into this expression, Eq. (2.12) becomes t = R(Z1/2 – 1)

(2.13)

where Z = (SE + P)/(SE – P)

Equation (2.13) is used in Appendix 1-2 of VIII-1 to determine the required thickness in thick cylinders for the conditions t > 0.5R or P > 0.385SE. This equation can also be written in terms of pressure as P = SE[(Z – 1)/(Z + 1)]

(2.14)

where Z = [(R + t)/R]2

For longitudinal stress, t = R(Z1/2 – 1),

with

t > 0.5R

or

P > 1.25SE

(2.15)

where Z = (P/SE) + 1

Equation (2.15) can be written in terms of pressure, P, as P = SE(Z – 1)

(2.16)

where Z = [(R + t)/R]2

The thick cylinder expressions given by Eqs. (2.12) through (2.16) can be expressed in terms of outside radii as follows. For circumferential stress, t = RO(Z1/2 – 1)/Z1/2,

with

t > 0.5R

or

P > 0.385SE

where Z = (SE + P)/(SE – P)


(2.17)

34

Chapter 2

or in terms of pressure, P = SE [(Z – 1)/(Z + 1)]

(2.18)

where Z = (RO/R)2 = [RO/(RO – t)]2

For longitudinal stress with t > 0.5R or P > 1.25SE, t = RO(Z1/2 – 1)/Z1/2

(2.19)

where Z = (P/SE) + 1

or in terms of pressure, P, P = SE(Z – 1)

(2.20)

where Z = (RO/R)2 = [RO/(RO – t)]2

All of the equations given so far are in terms of internal pressure only. VIII-1 does not give any equations for calculating stresses in cylinders resulting from wind and earthquake loads. One method of calculating these stresses is given in Section 2.3. Example 2.5 Problem Calculate the required shell thickness of an accumulator with P = 10,000 psi, R = 18 in., S = 20,000 psi, and E = 1.0. Assume a corrosion allowance of 0.25 in. Solution The quantity 0.385SE = 7700 psi is less than the design pressure of 10,000 psi. Thus, Eq. (2.13) is applicable. Z = (SE + P)/(SE – P) = (20,000 × 1.0 + 10,000)/(20,000 × 1.0 – 10,000) = 3.0 t = R(Z1/2 – 1) = (18.25)(3.00.5 – 1.0) = 13.36 in.

Total t = 13.36 + 0.25 = 13.61 in.



Example 2.6 Problem What is the required thickness in Example 2.5 if the design pressure is 7650 psi and the corrosion allowance is zero? Solution The quantity 0.385SE = 7700 psi is greater than the design pressure of 7650 psi. Thus, Eq. (2.1) is applicable. t = PR/(SE – 0.6P) = 7650 × 18/(20,000 × 1.0 – 0.6 × 7650) = 8.94 in.

It is of interest to determine the accuracy of Eq. (2.1) by comparing it with the theoretical Eq. (2.13), which gives Z = (SE + P)/(SE – P) = (20,000 × 1.0 + 7650)/(20,000 × 1.0 – 7650) = 2.239 t = R(Z1/2 – 1) = 18(2.2390.5 – 1.0) = 8.93 in.

This comparison demonstrates the accuracy of the “simple-to-use” Eq. (2.1) over a wide range of R/t ratios. Example 2.7 Problem What is the maximum stress in a layered vessel subjected to an internal pressure of 15,000 psi? The outside diameter is 24 in., and the inside diameter is 11 in. Solution The thickness of 6.50 in. is greater than 0.5R. Thus, either Eq. (2.17) or Eq. (2.13) may be used, since both the outside and inside diameters are given. Both of these equations are in terms of the quantity Z, which is a function of stress S. Solving for S in these equations is not easy. However, since both of these equations were derived from Eq. (2.12), we can use it directly to solve for S. Thus, SE = 15,000 (122 + 5.52)/(122 – 5.52) = 22,980 psi

2.3 AXIAL COMPRESSION Vessel components are frequently subjected to axial compressive stresses caused by such items as wind, dead loads, earthquake, and nozzle loads. The maximum compressive stress is limited by either the


36

Chapter 2

TABLE 2.1 TABULAR VALUES FOR FIG. 2.4

allowable tensile stress, using a Joint Efficiency Factor of 1.0, or the allowable compressive stress, whichever is less. The allowable tensile stress controls thick cylinders, while the allowable compressive stress controls thin cylinders. The procedure for calculating the allowable axial compressive stress in a cylinder is given in Paragraph UG-23 of VIII-1 and is based on a theoretical equation with a large L/D ratio (Jawad, 2004). It consists of calculating the quantity A = 0.125/(RO/t)

(2.21)

where A = strain RO = outside radius of the cylinder t = thickness and then using a modified stress-strain diagram furnished by the ASME to determine the permissible axial compressive stress, B. The ASME plots modified stress-strain diagrams, called External Pressure Charts, for various materials at various temperatures on a log-log scale. One such chart for carbon steel is shown in Fig. 2.4. The strain, A, is plotted along the horizontal axis, and a stress, B, along the vertical axis. The majority of the materials listed in the stress tables of II-D or VIII-1 construction have a corresponding External



FIG. 2.4 CHART FOR CARBON AND LOW ALLOY STEELS WITH YIELD STRESS OF 30 KSI AND OVER, AND TYPES 405 & 410 STAINLESS STEELS Pressure Chart (EPC). Tabular values of the curves in these charts are also given in II-D, for example those shown in Table 2.1 for Fig. 2.4. If the calculated value of A falls to the left of the stress-strain line in a given External Pressure Chart, then B must be calculated from the equation B = AE/2

(2.22)

where E = modulus of elasticity of the material at design temperature The modulus of elasticity, E, in Eq. (2.22) is obtained from the actual modified stress-strain diagrams furnished by the ASME, such as those shown in Fig. 2.4. It is of interest to note that the stress B in the External Pressure Chart, Fig. 2.4, has a value of half the stress obtained from the actual stress-strain curve of the given material. This was done by the ASME in order to utilize these charts for other loading conditions, such as external pressure on cylindrical shells as well as axial compression and vacuum on heads with various shapes. Thus, the stress, B, from Fig. 2.4 for carbon steel at room temperature corresponding to a strain, A, of 0.1 in./in. is 17.6 ksi. This is half the actual yield stress of 35.2 ksi for this material, as obtained from the actual given stress-strain curve. Also, the value of the modulus of elasticity obtained from the elastic portion of the curve by finding the slope between any two points along the curve is half the actual indicated value. If we substitute Eq. (2.21) into Eq. (2.22), we find that in the elastic range, the buckling equation for design becomes B = E/16(RO/t)

This can also be written as B = 0.0625E/(RO/t)


(2.23)

38

Chapter 2

FIG. 2.5 C FACTOR AS A FUNCTION OF R/T (JAWAD, 2004)

The theoretical equation for the critical axial buckling stress used by the ASME is given by σcr = 0.6E/(RO/t)

(2.24)

A comparison of the design Eq. (2.23) and the critical axial buckling stress Eq. (2.24) indicates that a factor of safety of about ten was used by the ASME. However, experiments performed subsequently to the publication of Eq. (2.24) have shown that a more realistic critical axial buckling stress equation is of the form σcr = 0.6CE/(RO/t)

(2.25)

where C is obtained from Fig. 2.5. A comparison of Eq. (2.23) with Eq. (2.25) indicates that the factor of safety varies from a conservative value of 10.0 for small RO/t ratios to an unconservative value of 1.0 for large RO/t ratios. This fact should be considered when designing cylinders with large diameter to thickness ratios. VIII-1 allows an increase of 20% in the value of B obtained from Fig. 2.4 or calculated from Eq. (2.22) when live loads, such as wind and earthquake, are considered. Wind and earthquake loads are usually obtained from various standards, such as ASCE-7 and the Uniform Building Code. Example 2.8 Problem The tower shown in Fig. E2.8 has an empty weight of 60 kips. The contents weigh 251 kips. Determine the required thickness of the supporting skirt. Allowable tensile stress is 16 ksi. Use Fig. 2.4 for axial compression calculations. The temperature of the skirt is 200°F at the base and 800°F at the top. Solution Assume t = 3/8 in.



FIG. E2.8 Axial force = 60 + 251 = 311 kips Axial compressive stress = force/area of material in skirt = 311,000/π × 96 × 0.375 = 2750 psi

The bending moment due to wind forces at the bottom of skirt using a vessel projected area of 8 ft is M = 32 × 8 × 36 × (36/2 + 34 + 26) + 24 × 8 × 34 × (34/2 + 26) + 20 × 8 × 26 × (26/2) = 718,848 + 280,704 + 54,080 = 1,053,632 ft-lb

Notice that in many applications, the projected area must be increased beyond 8 ft to take into consideration such items as insulation, ladders, and platforms. Also, the moment may have to be modified for shape and drag factors. The bending stress is obtained from the classical equation for the bending of beams: Stress = Mc/I


40

Chapter 2

where c = maximum depth of the cross section from the neutral axis I = moment of inertia M = applied moment and for thin circular cross sections, this equation reduces to Stress = M/(πR2Ot) = 1,053,632 × 12/(π × 482 × 0.375) = 4660 psi Total compressive stress = 2750 + 4660 = 7410 psi

The allowable compressive stress is calculated from Eq. (2.21). A = 0.125/(48/0.375) = 0.00098

From Fig. 2.4 with A = 0.00098 and temperature of 200°F, we get B = 12,000 psi, which is the allowable compressive stress. Thus, the selected thickness is adequate at the bottom of the skirt. Note that the thickness would have been inadequate if the temperature at the bottom was 800°F. Now let us check the thickness at the top of the skirt. The axial stress due to dead load stays the same. The bending moment becomes M = 32 × 8 × 36 × (36/2 + 34 + 26 – 16) + 24 × 8 × 34 × (34/2 + 26 – 16) + 20 × 8 × 10 × (10/2) = 571,392 + 176,256 + 8000 = 755,648 ft-lb Bending stress = 755,648 × 12/(π × 482 × 0.375) = 3340 psi Total compressive stress = 2750 + 3340 = 6090 psi

From Fig. 2.4 with A = 0.00098 and temperature of 800°F, we get B = 7,000 psi. Thus, the selected thickness is adequate at the top of the skirt. Maximum tensile stress at bottom of skirt = 4660 – 2750 = 1910 psi Maximum tensile stress at top of skirt = 3340 – 2750 = 590 psi



Both of these values are less than 16,000 psi, which is the allowable tensile stress for the skirt. These calculations show that t = 3/8 in. for the skirt is satisfactory. This thickness may need to be increased in actual construction to take into account such items as opening reinforcements, corrosion, out-of-roundness considerations, and handling factors. Example 2.9 Problem What is the allowable compressive stress in an internal cylinder with Do = 24 ft, t = 3/16 in., and design temperature = 900°F? Use Fig. 2.4 for the External Pressure Chart. Solution From Eq. (2.21), A = 0.125/(120/0.1875) = 0.0002

From Fig. 2.4, this A value falls to the left of the curve for 900°F. Therefore, Eq. (2.22) must be used. The value of E is obtained from Fig. 2.4 as 20.8 × 106 psi for 900°F. Hence, allowable compressive stress B is B = 0.0002 × 20,800,000/2 = 2080 psi

2.4 EXTERNAL PRESSURE External pressure on cylindrical shells causes compressive forces that could lead to buckling. The equations for the buckling of cylindrical shells under external pressure are extremely cumbersome to use directly in design (Jawad, 2004). However, these equations can be simplified for design purposes by plotting them so that the minimum buckling strain is expressed in terms of length, diameter, and thickness of the cylinder. These plots are utilized by the ASME as discussed next. The rules and factors of safety in VIII-1 and VIII-2 are identical for external pressure. Accordingly, references in this section are made to paragraphs in VIII-1 only.

2.4.1 External Pressure for Cylinders with DO/t ≥ 10 The ASME uses plots to express the lowest critical strain, A, in terms of the ratios L/DO and DO/t of the cylinder, as shown in Fig. 2.6. The designer calculates the known quantities L/DO and Do/t and then uses the figure to determine buckling strain, A. To correlate buckling strain to allowable external pressure, the designer uses the modified stress-strain diagram of Fig. 2.4 to obtain a B value. The allowable external pressure can then be determined from this B value, as explained below. Accordingly, the procedure in ASME VIII-1 Paragraph UG-28 for determining the allowable external pressure for cylinders with DO/t ratios equal to or greater than ten consists of the following steps: 1. 2. 3. 4.

Assume a value of t for the cylinder. Calculate the quantities L/DO and DO/t. Use Fig. 2.6 with the calculated values of L/DO and DO/t and establish an A value. Use an External Pressure Chart such as Fig. 2.4 with an A value and determine a B value from the appropriate temperature chart. 5. Calculate the allowable external pressure from the equation P = (4/3)(B)/(DO/t)


(2.26)

42

Chapter 2

FIG. 2.6 GEOMETRIC CHART FOR CYLINDRICAL VESSELS UNDER EXTERNAL PRESSURE (JAWAD AND FARR, 1989)

6. When A falls to the left of the curves, the value of P is determined from P = 2AE/3(DO/t)

(2.27)

where E = modulus of elasticity Note that the curves in Fig. 2.6 are based on a thin cylinder simply supported at the ends with external pressure acting laterally and on the ends. These curves can also be used, conservatively, for cases where the pressure is on the sides only, as is the case with jacketed vessels. These curves can also be used, conservatively, for cylinders with fixed rather than simply supported ends. The effective length of a cylinder,L, needed to use Fig. 2.6, can sometimes be difficult to establish when the cylinder is attached to other components, such as heads and transition sections. Figure 2.7 is provided by VIII1 to define the effective length of some commonly encountered cylinders. The effective length of cylinders with spiral stiffeners is not addressed by VIII-1. Some designers double the value of the effective length between spiral stiffeners to take into consideration their reduced effective support strength. The effective length of cylinders with variable thickness between supports is not addressed by VIII-1. In such cases, careful consideration must be given for using an effective thickness since buckling may occur in the thinner sections. The factor of safety for the allowable external pressure obtained by using Eq. (2.26) or (2.27) is three against buckling and also against yield.



FIG. 2.7 SOME LINES OF SUPPORT OF CYLINDRICAL SHELLS UNDER EXTERNAL PRESSURE (ASME VIII-1)

Example 2.10 Problem What is the required thickness of a cylindrical shell with length equal to 20 ft and outside diameter equal to 5 ft? The cylinder is subjected to an external pressure of 15 psi at 500°? Use Fig. 2.4 for the External Pressure Chart.


44

Chapter 2

Solution Try t = 0.50 in. Then L/DO = 4.0 and DO/t = 120.00. From Fig. 2.6, we obtain a value of A = 0.00022. Then from Fig. 2.4, B = 3000 psi. The allowable pressure is obtained from Eq. (2.26) as p = (4/3) (3000)/120.00 = 33.3 psi

Since this value is higher than 15 psi, try a new thickness of 0.375 in. Then L/DO = 4.0 and DO/t = 160.00. From Fig. 2.6, we get A = 0.00017, and from Fig. 2.4, we see that A falls to the left of the curves. Thus, from Eq. (2.27) P = 2 × 0.00017 × 27.0 × 106/(3 × 160.00) = 19.1 psi

A new trial of t = 5/16 in. results in an unacceptable allowable pressure of 12.2 psi. Thus, the required thickness to be used is 3/8 in. The allowable compressive hoop stress is then S = P RO/t = 19.1 × 30/0.375 = 1530 psi

2.4.2 External Pressure for Cylinders with DO/t < 10 When DO/t is less than 10, the allowable external pressure is taken as the smaller of the values determined from the following two equations: Pa1 = [2.167/(DO/t) – 0.0833] B

Pa2 =

2S [1 – 1/(DO/t)] DO/t

(2.28)

(2.29)

where B is obtained as discussed above. For values of (DO/t) of less than or equal to 4, the A value is calculated from A = 1.1/(DO/t)2

(2.30)

For values of A greater than 0.10, use a value of 0.10. The value of S is taken as the smaller of two times the allowable tensile stress, or 0.9 times the yield stress of the material at the design temperature. The yield stress is obtained from the External Pressure Chart of the material by using twice the B value obtained from the extreme right-hand side of the termination point of the appropriate temperature curve. The factor of safety in Eqs. (2.28) and (2.29) varies from 3.0 for DO/t = 10 to about 1.67 for DO/t = 2. This gradual reduction in the factor of safety as the cylinder gets thicker is justified since buckling ceases to be a consideration and the factor of safety for external pressure is kept the same as that for internal pressure, which is 2/3 Sy.



Example 2.11 Problem The inside cylinder of a jacketed vessel has an outside diameter of 20 in., a length of 72 in., and a thickness of 5 in. What is the maximum allowable jacket pressure? Use Fig. 2.4 for an External Pressure Chart. Let the design temperature be 300°. The allowable stress from tension is 17,500 psi. Solution Calculations give L/DO = 3.60 and DO/t = 4.0. And since DO/t = 4.0, Eq. (2.30) must be used. Hence, A = 1.1/(4.0)2 = 0.0688

From Fig. 2.4, B = 17,800 psi. From Eq. (2.28), Pa1 = [(2.167/4.0) – 0.0833] 17,800 = 8160 psi

The yield stress of the material is (0.9)(2B) or 32,040 psi. Twice the allowable stress is 35,000 psi. Hence S = 32,040 psi is to be used. From Eq. (2.29), Pa2 = (2 × 32,040/4.0)(1 – 1/4.0) = 12,020 psi

Therefore, the allowable jacket pressure in accordance with VIII-1 is 8160 psi. Notice, however, that this pressure is greater than 0.385S, an indication that thick-shell equations may have to be used. Such equations for external pressure are not in VIII-1 yet.

2.4.3 Empirical Equations It is of interest to note that Fig. 2.6 can only be used for (DO/t) of up to 1000. Larger values are not permitted presently by the ASME. One approximate equation (Jawad. 2004) that is frequently used by designers for large (DO/t) ratios was developed by the U.S. Navy and is given by P = 0.866E/(L/DO)(DO/t)2.5

(2.31)

where E = modulus of elasticity, psi P = allowable external pressure, psi This equation incorporates a factor of safety of 3 and a Poisson’s ratio of 0.30. Many pressure vessels are subjected routinely to vacuum as well as axial loads from wind and dead load. Section VIII does not give any method for calculating the allowable compressive stress due to combined effect of vacuum and axial loads. One such method uses an interactive equation similar to the one used for calculating the buckling of beam columns. Booton (Booton, et. al, 1977) suggested a conservative equation of the form F/Fal + p/pal ≤ 1.0


46

Chapter 2

Where, F Fal P pal

= actual axial pressure = allowable axial pressure = actual external pressure = allowable external pressure

Example 2.12 Problem Solve Example 2.10 using Eq. (2.31). Solution From Example 2.10, t = 0.375 in., L/DO = 4.0, DO/t = 160.00, and E = 27,000 ksi. Then from Eq. (2.31), P = 0.866 × 27,000,000/(4.0)(160)2.5 = 18.1 psi

This approximate value differs from the answer in Example 2.10 by about 6%.

2.4.4 Stiffening Rings The required thickness of a shell with a given diameter under a specified external pressure can normally be reduced by shortening the shell’s effective length. The length can be reduced by providing stiffening rings at various intervals, as shown in Fig. 2.7. The required moment of inertia of such rings is determined from Is = [D2OLs(t + As/Ls)A]/14

(2.32)

I′s = [D2OLs(t + As/Ls)A]/10.9

(2.33)

or

where Is = required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in.4 I′s = required moment of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about their combined neutral axis, in.4 The effective length of the shell is taken as 1.10(DOts)1/2. Ls = half the distance from the center line of the stiffening ring to the next line of support on one side, plus half the distance from the center line of the ring to the next line of support on the other side. A line of support is (1) a stiffening ring, (2) jacket bar, (3) circumferential line on a head at one-third the depth of the head, (4) cone-to-cylinder junction. As = area of the stiffening ring, in.2 t = minimum thickness of the shell, in. ts = nominal thickness of the shell, in. To design the stiffening ring 1. Assume first an area, As, of the stiffening ring and calculate the available moment of inertia. I, or Is. 2. Calculate B from the equation B = 0.75 [PDO/(t + As/Ls)]


(2.34)


3. Use the appropriate External Pressure Chart and determine an A value. 4. If B falls below the left end of the temperature line, calculate A from A = 2B/E

(2.35)

5. Solve Eq. (2.32) or (2.33) for the required moment of inertia. 6. The furnished moment of inertia must be greater than the required one.

Example 2.13 Problem Calculate the required thickness of the shell and the required moment of inertia of the stiffening ring shown in Fig. E2.13(a). The shell and ring material are SA 285-C. External pressure is 12 psi, and the design temperature is 100°F. Solution Try t = 1/4 in. Then DO/t = 240.0 and L/DO = 12(10 + 2.5/3)/60 = 2.17. From Fig. 2.6, A = 0.00016, and from Eq. (2.27), P = 2 × 0.00016 × 29,000,000/3(240) = 12.9 psi

Thus, a shell thickness of 1/4 in. is adequate. For the stiffening ring, try a 3 × 1/4 in. hard way bar, as shown in Fig. E2.13. For ease of calculations, assume that the stiffening ring is not integral with the shell. Hence, Eq. (2.32) can be used. The moment of inertia of the bar is bd3/12. Thus, I = 0.25 × 3.03/12 = 0.56 in.4

From Eq. (2.34), B = 0.75[12 × 60.0/(0.25 + 0.25 × 3.0/140.0)] = 2110 psi

Since this value falls below the left end of the material line in Fig. 2.4, we use Eq. (2.35): A = 2 × 2110/29,000,000 = 0.000146

From Eq. (2.32), Is = [60.02 × 140.0(0.25 + 0.25 × 3.0/140.0) 0.000146]/14 = 1.34 in.4


48

Chapter 2

FIG. E2.13

Since this number is larger than the actual moment of inertia of the ring (0.56 in.4), the assumed ring is inadequate and a larger ring is required. However, before such a new ring is chosen, let us use the effective moment of inertia of the existing ring and shell and compare that to Eq. (2.33). From Eq. (2.33), I′s = [60.02 × 140.0(0.25 + 0.25 × 3.0/140.0) 0.000146]/10.9 = 1.72 in.4

The effective centroid of the shell-ring section, Fig. E2.13(b), is h = [4.26 × 0.25 × 0.125 + 0.25 × 3.0 (1.5 + 0.25)]/(4.26 × 0.25 + 0.25 × 3.0) = 0.796 in.

The actual moment of inertia is I = 4.26 × 0.253/12 + 4.26 × 0.25 × 0.6712 + 0.25 × 3.03/12 + 0.25 × 3.0 × 0.9542 = 0.006 + 0.480 + 0.563 + 0.683 = 1.73 in.4



Thus, using the composite section results in a 1/4 in. × 3 in. stiffener that is adequate.

2.4.5 Attachment of Stiffening Rings Details of the attachment of stiffening rings to the shell are given in Fig. UG-30 of VIII-1, which is reproduced’ in Fig. 2.8. The welds must be able to support a radial pressure load from the shell of PLs. This is based on the code assumption that the stiffening rings must support the total lateral load if the shell segments between the rings collapse. Also, the code requires that the welds support a shear load of 0.01 PLsDO. This shear load is arbitrary and is based on the assumption that if the rings buckle, bending moments occur and generate shear forces. VIII-1 also has other requirements pertaining to stitch welding and gaps between the rings and the shell. These requirements are given in Paragraphs UG-29 and UG-30 of VIII-1. Example 2.14 Problem Calculate the required size of the double fillet welds attaching the stiffening ring shown in Fig. E2.13(b) of Example 2.13 to the shell. Let the allowable tensile stress of SA 285-C at 100°F be 15,700 psi. Solution The radial load, F1, on the rings is equal to PLs. PLs = 12.0 × 140 = 1680 lb/in. of circumference

Allowable tensile stress in the fillet weld from Table 1.4 is 0.55S = 8635 psi. The total load carried by weld is Total load = number of welds attaching ring × size of weld × allowable stress = 2 × W × 8635

Hence, the required weld size, W, is W = 1680/(2 × 8635) = 0.10 in.

Use 2 1/4 in. continuous fillet welds, in accordance with the minimum requirements of UG-30(f). Shearing force, V, on the weld is V = 0.01 PLsDO = 0.01 × 12 × 140 × 60 = 1008 lb

Allowable shearing stress in fillet weld from Table 1.4 is 0.55S = 8635 psi. From strength of materials, the equation for shear stress is given by t = VQ/It

where Q is at the location of the weld, as shown in Fig. E2.13(b), and is given by Q = 4.26 × 0.25(0.796 – 0.125) = 0.71 in.2


50

Chapter 2

FIG. 2.8 SOME DETAILS FOR ATTACHING STIFFENER RINGS (ASME VIII-1)



Hence, τ = 1008 × 0.71/1.73(2 × 0.25) = 827 psi < 8635 psi

2.5 CYLINDRICAL SHELL EQUATIONS, VIII-2 The equation for the design of thin cylindrical shells is given in Section AD-201 of VIII-2 and is expressed as t = PR/(S – 0.5P),

for

P < 0.4S

(2.36)

where P = internal pressure R = inside radius S = allowable stress at design temperature t = required thickness This equation is derived from the basic equation for thin cylindrical shells, t = PR/S, by substituting for the inside radius, R, the quantity R + t/2. Equation (2.36) is applicable as long as the axial tensile force, F, is not larger than 0.5PR. When F exceeds 0.5PR, then the required thickness of the cylinder is governed by the equation for longitudinal tensile stress as follows: t = (0.5PR + F)/(S – 0.5P)

(2.37)

For values of P > 0.4S, the ratio of t/R increases beyond the scope of Eq. (2.36) and the design equation in VIII-2 is based on the plastic theory of thick cylindrical shells (Prager and Hodge, 1965). This equation is given by t = R(eP/S – 1)

for

P > 0.4S

(2.38)

Example 2.15 Problem Calculate the required thickness of a cylindrical shell with an inside diameter of 60 in. and an allowable stress of 20 ksi. Let the pressure be (a) 1000 psi, (b) 8500 psi. Solution (a) Since P < OAS, Eq. (2.36) applies. Then, t = 1000 × 30/(20,000 – 0.5 × 1000) = 1.54 in.

(b) Since P > 0.4S, Eq. (2.38) applies. Then, t = 30(e8500/20000 – 1) = 15.89 in.


52

Chapter 2

2.6 MISCELLANEOUS SHELLS 2.6.1 Mitered Cylinders Mitered cylinders, Fig. 2.9, are used in nozzle connections, transition sections, and reducers. Neither VIII-1 nor VIII-2 give design rules. The piping code (ASME B31.3, 2002) gives design equations for various miters. The basic equation for the allowable pressure in a shell with a single miter is P = (SEt/R){1/[1 + 0.643(R/t)1/2 tan θ]}

for

θ < 22.5°

(2.39)

P = (SEt/R){1/[1 + 1.25(R/t)1/2 tan θ]}

for

θ > 22.5°

(2.40)

where E = Joint Efficiency Factor P = allowable pressur, psi R = radius of the shell in accordance with Fig. 2.9, in. S = allowable stress t = shell thickness, in. α = angle of change in the direction of the miter joint, Fig. 2.9 θ = α/2 ASME B31.3 gives further information, such as equations for multiple miters, curved miters, and length of tapers. Example 2.16 Problem A mitered cylinder has an inside radius of 24 in., α of 40°, a design pressure of 500 psi, an allowable stress of 15,000 psi, and a joint efficiency of 0.85. Determine the required thickness. Solution From Eq. (2.1) for a straight cylinder, t = 500 × 24/(15,000 × 0.85 – 0.6 × 500) = 0.96 in.

Try t = 1 in. θ = 40/2 = 20°. Thus, Eq. (2.39) governs. P = (15,000 × 0.85 × 1.0/24){1/[1 + 0.643(24/1.0)1/2 tan 20]} = 248 psi

This is inadequate. Try t = 1.8125 in. P = (15,000 × 0.85 × 1.8125/24){1/[1 + 0.643(24/1.8125)1/2 tan 20]} = 520 psi > 500 psi

Use t = 113/16 in.



FIG. 2.9 MITERED BEND

2.6.2 Elliptical Shells Elliptical shells, Fig. 2.10, are encountered occasionally by the pressure vessel designer. The stresses, away from discontinuities, in the shell due to internal pressure can be approximated by using the membrane theory of elliptical cylinders (Flugge, 1967). The basic equation for hoop stress is expressed as t = Pa2b2/SE(a2 sin2 φ + b2 cos2 φ)3/2

where a = major radius of the ellipse, in. b = minor radius of the ellipse, in. E = Joint Efficiency Factor P = design pressure, psi S = allowable stress, psi t = thickness, in. φ = angle as defined in Fig. 2.10 Example 2.17 Problem Solve Example 2.1 if the cylinder is elliptical in cross section with the major diameter equal to 100 in. and the minor diameter equal to 92 in. Solution a = 50.125 in., b = 46.125 in., P = 100 psi, S = 20,000 psi, and E = 0.85


54

Chapter 2

FIG. 2.10 ELLIPTICAL CYLINDER For φ = 0°, Eq. (2.41) gives t = 100 × 50.1252 × 46.1252/20,000 × 0.85(50.1252 sin2 0 + 46.1252 cos2 0)3/2 + 0.125 = 0.320 + 0.125 = 0.45 in.

For φ = 90°, Eq. (2.41) gives t = 100 × 50.1252 × 46.1252/20,000 × 0.85(50.1252 sin2 90 + 46.1252 cos2 90)3/2 + 0.125 = 0.250 + 0.125 = 0.375 in.

Use t = 0.45 in. This thickness is about 10% higher than that for a cylinder with a circular cross section having an average diameter of 96 in.


CHAPTER

3 SPHERICAL SHELLS, HEADS, AND TRANSITION SECTIONS 3.1 INTRODUCTION Sections VIII-1 and VIII-2 contain rules for the design of spherical shells, heads and transition sections. Head configurations include spherical, hemispherical, torispherical, and ellipsoidal shapes. Transition sections include conical and toriconical shapes. The design rules for most of these shapes differ significantly in VIII1 and VII-2. This difference is due to the design approach used in developing the equations for VIII-1 and VIII-2. In this chapter a brief description of the various kinds of heads is given.

3.2 SPHERICAL SHELLS AND HEMISPHERICAL HEADS, VIII-1 3.2.1 Internal Pressure in Spherical Shells and Pressure on Concave Side of Hemispherical Heads The required thickness of a thin spherical shell due to internal pressure is listed in Paragraph UG-27 and is given by t = PR/(2SE – 0.2P),

when

t < 0.356R

or

P < 0.665SE

(3.1)

where E = Joint Efficiency Factor P = internal pressure R = internal radius S = stress in the material t = thickness of the head This equation can be rewritten to calculate the maximum pressure when the thickness is known. It then takes the form P = 2SEt/(R + 0.2t)

(3.2)

Notice the similarity between Eq. (3.1) and the classical equation for the membrane stress in a spherical shell (Beer et al, 2001), given by t = PR/2SE 55


(3.3)

56

Chapter 3

The difference is in the additional term of 0.2P in the denominator. This term was added by the ASME to take into consideration the nonlinearity in stress that develops in thick spherical shells. The designer should be aware that Eq. (3.1) determines the thickness based on pressure only. Large spherical shells for liquid storage usually have low internal pressure. Thus, the governing thickness is controlled by the liquid weight rather than Eq. (3.1). One method for determining the thickness in such spheres is given in (API 620, 2001) In some instances, the outside radius of a shell is known rather than the inside radius. In this case the governing equation is obtained from Eq. (3.1) by substituting (Ro – t) for R. The resulting equation is given in VIII-1, Appendix 1, Article 1-1, as t = PRo/(2SE + 0.8P),

with

t < 0.356Ro

or

P < 0.665SE

(3.4)

or P = 2SEt/(Ro – 0.8t)

(3.5)

As the ratio of t/R increases beyond 0.356, the thickness given by Eq. (3.1) becomes nonconservative. This is similar to the case for cylindrical shells discussed in 2.2. The ASME VIII-1 equation for thick spherical shells is given by t = R(Y1/3 – 1)

(3.6)

where Y = 2(SE + P)/(2SE – P)

Equation (3.6) is used in Appendix 1–3 of VIII-1 to determine the required thickness in thick spherical shells for the conditions where t > 0356R or P > 0.665SE. This equation can also be written in terms of pressure as P = 2SE[(Y – 1)/(Y + 2)]

(3.7)

where Y = [(R + t)/R]3

The thick shell expressions given by Eqs. (3.6) and (3.7) can be stated in terms of outside radii as t = Ro(Y1/3 – 1)/Y1/3,

with

t > 0.356R

or

P > 0.665SE

(3.8)

where Y = 2(SE + P)/(2SE – P)

or in terms of pressure, P = 2SE[(Y – 1)/(Y + 2)]

(3.9)

where Y = (Ro/R)3 = [Ro/(Ro – t)]3

Equations (3.1) through (3.9) are also applicable to hemispherical heads with pressure on the concave side. This is illustrated in Fig. 3.1. For an applied internal pressure in compartment A, the hemispherical heads


Spherical Shells, Heads, and Transition Sections

57

FIG. 3.1 abc and def are subjected to concave pressure and Eqs. (3.1) through (3.9) may be used. Paragraph UG-32(f) of VIII-1 gives the rules for the design of hemispherical heads due to pressure on the concave side. Example 3.1 Problem A pressure vessel is constructed of SA 516-70 material and has an inside diameter of 8 ft. The internal design pressure is 100 psi at 450°F. The corrosion allowance is 0.125 in. and the joint efficiency is 0.85. What is the required thickness of the hemispherical heads if the allowable stress is 20,000 psi? Solution The quantity 0.665SE = 11,300 psi is greater than the design pressure of 100 psi. Thus, Eq. (3.1) applies. The inside radius in the corroded condition is equal to R = 48 + 0.125 = 48.125 in.

The total head thickness is t = PR/(2SE – 0.2P) + corrosion = 100 × (48.125)/(2 × 20,000 × 0.85 – 0.2 × 100) + 0.125 = 0.142 + 0.125 = 0.27 in.


58

Chapter 3

The calculated thickness is less than 0.356R. Thus, Eq. (3.1) is applicable. Example 3.2 Problem A pressure vessel with an internal diameter of 120 in. has a head thickness of 1.0 in. Determine the maximum pressure if the allowable stress is 20 ksi. Assume E = 0.85. Solution The maximum pressure is obtained from Eq. (3.2) as P = 2 × 20,000 × 0.85 × 1.0/(60 + 0.2 × 1.0) = 565 psi

Example 3.3 Problem A vertical unfired boiler is constructed of SA 516-70 material and built in accordance with the requirements of VIII-1. It has an outside diameter of 8 ft and an internal design pressure of 450 psi at 550 °F. The corrosion allowance is 0.125 in. and the joint efficiency is 1.0. Calculate the required thickness of the hemispherical head if the allowable stress is 19,700 psi. Solution From Eq. (3.4), the required head thickness is t = 450 × 48/(2 × 19,700 × 1.0 + 0.8 × 450) + 0.125 = 0.543 + 0.125 = 0.67 in.

Example 3.4 Problem Calculate the required hemispherical head thickness of an accumulator with P = 10,000 psi, R = 18 in., S = 15,000 psi, and E = 1.0. Assume a corrosion allowance of 0.25 in. Solution The quantity 0.665SE = 9975 psi is less than the design pressure of 10,000 psi. Thus, Eq. (3.6) applies. Y = 2(SE + P)/(2SE – P) = 2(15,000 × 1.0 + 10,000)/(2 × 15,000 × 1.0 – 10,000) = 2.5 t = R(Y1/3 – 1) = (18.25)(2.51/3 – 1.0) = 6.52 in.

Total head thickness = 6.52 + 0.25 = 6.77 in.



59

FIG. E3.4

The required thickness of the shell for this vessel is calculated in Example 2.5. Attaching the head to the shell requires a transition with a 3:1 taper, as shown in Fig. UW-13.1 of VIII-1. This taper, however, is impractical to make in this case since the thickness of the head is about two-thirds the radius. One method of attaching the head to the shell is shown in Fig. E3.4.

3.2.2 External Pressure in Spherical Shells and Pressure on Convex Side of Hemispherical Heads The procedure for calculating the external pressure on spherical shells is given in Paragraph UG-28(d) of VIII-1 and consists of calculating the quantity A = 0.125/(RO/t)

(3.10)

where A = strain Ro = outside radius of the spherical shell t = thickness and then using a stress-strain diagram similar to Fig. 2.4 to determine a B value. The allowable external pressure is calculated from Pa = B/(Ro/t)

(3.11)

If the calculated value of A falls to the left of the stress-strain line in a given External Pressure Chart, then Pa must be calculated from the equation


60

Chapter 3

Pa = 0.0625E/(Ro/t)2

(3.12)

where E = modulus of elasticity of the material at design temperature The modulus of elasticity, E, in Eq. (3.12) is obtained from the actual stress-strain diagrams furnished by the ASME, such as those shown in Fig. 2.4. Equations (3.10) and (3.11) are also applicable to hemispherical heads with pressure on the convex side, as mentioned in Paragraph UG-33(c) of VIII-1. This is illustrated in Fig. 3.1. For an applied internal pressure in compartment B, the hemispherical head def is subjected to convex pressure and Eqs. (3.10) and (3.11) may be used. Example 3.5 Problem Determine the required thickness of the head in Example 3.1 due to an external pressure of 10 psi. Solution From Example 3.1, the required thickness for internal pressure is 0.14 in. We will use this thickness as our assumed t. Then from Eq. (3.10), A = 0.125/[(48 + 0.125 + 0.14)/0.14] = 0.00036

From Fig. 2.4, B = 4,700 psi. And from Eq. (3.11), Pa = 4,700/(48.265/0.14) = 13.6 psi

Since this pressure is larger than the design pressure of 10 psi, the minimum calculated thickness of 0.14 in. is adequate. Example 3.6 Problem What is the required thickness of a hemispherical head subjected to external pressure of 15 psi? Let Ro = 150 in. and design temperature = 900°F. The material is SA 516-70. Solution Assume t = 0.25 in. Then from Eq. (3.10), A = 0.125/(150/0.25) = 0.00021

Since the A value is to the left of the 900°F material line in Fig. 2.4, we have to use Eq. (3.12). Pa = 0.0625 × 20,800,000/(150/0.25)2 = 3.6 psi



61

Since this value is less than 15 psi, a larger thickness is needed. Try t = 0.50 in. A = 0.125/(150/0.50) = 0.00042

From Fig. 2.4, B = 4500 psi. And from Eq. (3.11), Pa = 4500/(150/0.50) = 15 psi

The selected thickness of 0.50 in. is adequate for the 15 psi external pressure. The thickness may have to be increased due to handling and fabrication requirements.

3.3 SPHERICAL SHELLS AND HEMISPHERICAL HEADS, VIII-2 The required thickness of a spherical shell due to internal pressure is given in Paragraph AD-202 of VIII-2 as t = 0.5PR/(S – 0.25P),

when

P < 0.4S

(3.13)

where P = internal pressure R = internal radius S = stress in the material t = thickness of the hemisphere As the ratio of P/S increases beyond 0.4, the thickness given by Eq. (3.13) becomes nonconservative. This is similar to the case for cylindrical shells discussed in 2.2. The VIII-2 equation for thick hemispherical heads is given by ln [(R + t)/R] = 0.5P/S,

when

P > 0.4S

(3.14)

This can be written also as t = R(e0.5P/S – 1),

when

P > 0.4S

(0.315)

When meridional forces, F (for instance, wind and earthquake loads) are present on the head, then Eq. (3.13) is modified as follows: t = (0.5PR + F)/(S – 0.25P),

when P < 0.4S

(3.16)

where F = Meridional force, lb/in. of circumference. F is taken as positive when it is in tension and negative when it is in compression. When F is larger than 0.5PR, then buckling could occur and the rules for external pressure must be considered. The rules for calculating the required thickness of hemispherical heads subjected to pressure on the concave side are given in Paragraph AD-204.1 of VIII-2. The rules are identical to those for spherical shells given by Eqs. (3.13) through (3.16). The procedure and the factors of safety for calculating the allowable external pressure on spherical shells in VIII-2 are given in Paragraph AD-320. The rules and factors of safety are identical to those given in VIII-1.


62

Chapter 3

Similarly, the rules in VIII-2 for calculating the allowable pressure on the convex side of hemispherical heads are given in Paragraph AD-350.1. They are identical to those given in VIII-1. Example 3.7 Problem Determine the required thickness for a hemispherical head subjected to an internal pressure of 10,000 psi. Let S = 20 ksi, R = 20 in. Solution P/S = 0.5. Since this ratio is larger than 0.4, Eq. (3.15) must be used. t = 20 (e0.5 × 10.000/20.000 – 1) = 5.68 in.

3.4 ELLIPSOIDAL HEADS, VIII-1 3.4.1 Pressure on the Concave Side A commonly used ellipsoidal head has a ratio of base radius to depth of 2:1 (Fig. 3.2a). The shape can be approximated by a spherical radius of 0.9D and a knuckle radius of 0.17D, as shown in Fig. 3.2(b). The required thickness of 2:1 heads due to pressure on the concave side is given in Paragraph UG-32(d) of VIII1. The thickness is obtained from the following equation: t = PD/(2SE – 0.2P)

(3.17)

P = 2SEt/(D + 0.2t

(3.18)

or in terms of required pressure,

where D = inside base diameter E = Joint Efficiency Factor P = pressure on the concave side of the head S = allowable stress for the material t = thickness of the head Ellipsoidal heads with a radius-to-depth ratio other than 2:1 may also be designed to the requirements of VIII-1. The governing equations are given in Appendix 1-4 of VIII-1 as t = PDK/(2SE – 0.2P)

(3.19)

where K = (1/6)[2 + (D/2h)2]

and D/2h varies between 1.0 and 3.0. The 1.0 factor corresponds to a hemispherical head. The K equation is given in Article 1-4(c) of Appendix 1 of VIII-1. Equation (3.19) can be expressed in terms of the required pressure as P = 2SEt/(KD + 0.2t)


(3.20)


63

FIG. 3.2

These equations can also be written in terms of the outside diameter, Do. Thus, t = PDoK/[2SE + 2P(K – 0.1)]

(3.21)

P = 2SEt/[KDo – 2t(K – 0.1)]

(3.22)

or in terms of required pressure

It is of interest to note that VIII-1 does not give any P/S limitations for the above equations. Nor does it have any rules for ellipsoidal heads when the ratio of P/S is large.

3.4.2 Pressure on the Convex Side The thickness needed to resist pressure on the convex side of an ellipsoidal head is given in Paragraph UG33 of VIII–1. The required thickness is the greater of the two thicknesses determined from the steps below.


64

Chapter 3

1. Multiply the design pressure on the convex side by the factor 1.67. Then use this new pressure and a joint efficiency of E = 1.0 in the appropriate equations listed in Eqs. (3.17) through (3.22) to determine the required thickness. 2. Determine first the crown radius of the ellipsoidal head. Then use this value as an equivalent spherical radius to calculate a permissible external pressure in a manner similar to the procedure given for spherical shells in Section 3.2.2. The procedure consists of calculating the quantity A = 0.125/(KoDo/t)

(3.23)

where A = strain Ko = function of the ratio Do/2ho and is obtained from Table 3.1 Do = outside base diameter of the ellipsoidal head t = thickness Then, using a stress-strain diagram similar to Fig. 2.4, determine the B value. The allowable pressure is calculated from Pa = B/(KoDo/t)

(3.24)

If the calculated value of A falls to the left of the stress-strain line in a given External Pressure Chart, then Pa must be calculated from the equation Pa = 0.0625E/(KoDo/t)2

(3.25)

where E = modulus of elasticity of material at design temperature The modulus of elasticity, E, in Eq. (3.25) is obtained from the actual stress-strain diagrams, such as those shown in Fig. 2.4, furnished by the ASME. Example 3.8 Problem Calculate the required thickness of a 2.2:1 head with an inside base diameter of 18 ft, design temperature of 100°F, concave pressure of 200 psi, convex pressure of 15 psi, allowable stress is 17,500 psi, and joint efficiency of 0.85. The head is made of low-carbon steel. Solution For Concave Pressure From Eq. (3.19), with K = (1/6)[2 + (2.2)2] = 1.14,

TABLE 3.1 FACTOR KO FOR AN ELLIPSOIDAL HEAD WITH PRESSURE ON THE CONVEX SIDE Do/2ho Ko Do/2ho Ko

3.0 1.36 1.8 0.81

2.8 1.27 1.6 0.73

2.6 1.18 1.4 0.65

2.4 1.08 1.2 0.57

2.2 0.99 1.0 0.50


2.0 0.90


65

t = (200 × 216.0 × 1.14)/(2 × 17,500 × 0.85 – 0.2 × 200) = 1.66 in.

For Convex Pressure 1. First calculate the pressure and thickness. P = 1.67 × 15 = 25.1 psi t = (25.1 × 216.0 × 1.14)/(2 × 17,500 × 1.0 – 0.2 × 25.1) = 0.18 in.

2. For external pressure, we determine Ko from Table 3.1 as 0.99. Let minimum t = 1.66 in. Do = 216 + (2 × 1.66) = 219.32

From Eq. (3.23), A = 0.125/(0.99 × 219.32/1.66) = 0.00096

From Fig. 2.4, B = 12,000 psi. Pa = 12,000/(0.99 × 219.32/1.66) = 91.7 psi

Thus, minimum t = 1.66 in.

3.5 TORISPHERICAL HEADS, VIII-1 3.5.1 Pressure on the Concave Side Shallow heads, which are commonly referred to as Flanged and Dished heads, or F&D, can also be built to VIII-1 rules, in accordance with Paragraph UG-32(e). The most commonly used F&D heads can be approximated by a spherical radius, L, of 1.0D and a knuckle radius, r, of 0.06D, as shown in Fig. 3.3. The required thickness of such heads due to pressure on the concave side is obtained from t = 0.885PL/(SE – 0.1P)

(3.26)

P = SEt/(0.885L + 0.1t)

(3.27)


where E = Joint Efficiency Factor L = inside spherical radius P = pressure on the concave side of the head


66

Chapter 3

FIG. 3.3

S = allowable stress for the material t = thickness of the head Torispherical heads with various spherical and knuckle radii may also be designed to the requirements of VIII-1. The governing equations are given in Appendix 1-4 as t = PLM/(2SE – 0.2P)

(3.28)

where M = (1/4)[3 + (L/r)1/2]

and L/r varies between 1.0 and 16.67. The 1.0 ratio corresponds to a hemispherical shell. The M equation is given in Article 1-4(d) of Appendix 1 of VIII-1. Equation (3.28) can be expressed in terms of the required pressure as P = 2SEt/(LM + 0.2t)

(3.29)

These equations can also be written in terms of the outside radius, Lo, as t = PLoM/[2SE + P(M – 0.2)]

(3.30)

P = 2SEt/[MLo – t(M – 0.2)]

(3.31)




67

FIG. 3.4 The theoretical membrane stress distribution in the circumferential, Nθ, and meridional, Nφ, directions in shallow heads due to internal pressure are shown in Fig. 3.4. Both the circumferential and meridional stresses at the crown of the head are tensile with a magnitude of S = Pa2/2bt. However, at the base of the head, the meridional stress is tensile with magnitude S = Pa/2, while the circumferential stress is compressive with a value of S = (Pa/2t)[2 – (a/b)2]. This compressive stress, which is not considered by Eq. (3.28), could cause buckling of the shallow head as the ratio of D/t increases. One way to avoid such failure is to calculate the thickness based on an equation (Shield and Drucker, 1961) that takes buckling into consideration and is expressed as nP/Sy = (0.33 + 5.5r/D)(t/L) + 28(1 – 2.2r/D)(t/L)2 – 0.0006

where D = base diameter of head, in. L = spherical cap radius, in. n = factor of safety P = design pressure, psi r = knuckle radius, in. Sy = yield stress of the material, psi t = thickness, in. This equation normally results in a thickness that is greater than that calculated from Eqs. (3.26), (3.28), or (3.30) for shallow heads with large D/t ratios. Paragraph UG-32(e) of VIII-1 states that the maximum allowable stress used to calculate the required thickness of torispherical heads cannot exceed 20 ksi, regardless of the strength of the material. This requirement was added in the code to prevent the possibility of buckling of the heads as the thickness is reduced due to the use of materials with higher strength.

3.5.2 Pressure on the Convex Side For pressure on the convex design, the buckling rules for calculating F&D head thicknesses are the same as those for ellipsoidal heads, with the exception that the outside crown radius of the F&D head is used in lieu of the quantity KoDo.


68

Chapter 3

Example 3.9 Problem Calculate the required thickness of an F&D head with an inside base diameter of 18 ft, design temperature of 100°F, internal (concave) pressure of 200 psi, external (convex) pressure of 15 psi, allowable stress is 17,500 psi, and joint efficiency of 0.85. The head is made of low-carbon steel. Solution For Concave Pressure Using L = 216.0 in., r = 0.06 × 216 = 13.0 in., and M = (1/4)[3 + (216/13)1/2] = 1.77, we get from Eq. (3.28) t = (200 × 216.0 × 1.77)/(2 × 17,500 × 0.85 – 0.2 × 200) = 2.58 in.

For Convex Pressure 1. Find the pressure and the thickness. P = 1.67 × 15 = 25.1 psi t = (25.1 × 216.0 × 1.77)/(2 × 17,500 × 1.0 – 0.2 × 25.1) = 0.27 in.

2. Let t = 2.58 in. Then Outside radius = 216 + (2 × 2.58) = 221.16 in.

From Eq. (3.23), A = 0.125/(221.16/2.58) = 0.0015

From Fig. 2.4, B = 14,000 psi. Pa = 14,000/(221.16/2.58) = 163 psi > 15 psi

Thus, t = 2.58 in.

3.6 ELLIPSOIDAL AND TORISPHERICAL HEADS, VIII-2 The required thickness for ellipsoidal as well as torispherical heads is obtained from Paragraph AD-204 and Article 4-4 of VIII-2. The procedure utilizes a chart, Fig. 3.5, which takes into consideration the possibility of buckling of thin shallow heads, as discussed in the previous section. The design consists of calculating the quantities P/S and r/D first and then using Fig. 3.5 to obtain the quantity t/L, and thus t. The thickness for



FIG. 3.5


69

70

Chapter 3

2:1 ellipsoidal heads is obtained by using the r/D = 0.17 curve, while the thickness for a standard F&D head is obtained by using the r/D = 0.06 curve. Figure 3.5 is plotted from the following equation: t = LeA

(3.32)

where A = A1 + A2 + A3 A1 = –1.26176643 – 4.5524592 (r/D) + 28.933179 (r/D)2 A2 = [0.66298796 – 2.2470836 (r/D) + 15.682985 (r/D)2 ][In(P/S)] A3 = [0.26878909 × 10–4 – 0.42262179 (r/D) + 1.8878333 (r/D)2][In(P/S)]2 D = base diameter, in. L = crown radius, in. P = design pressure, psi r = crown radius, in. S = allowable stress, psi t = thickness Example 3.10 Problem An F&D head with a 6% knuckle is subjected to 40 psi of pressure. What is the required thickness if D = 168 in.? Use VIII-2 and then VIII-1 rules. S = 20,000 psi for VIII-2, and 15,000 psi for VIII-1. Solution Using VIII-2 rules, P/S = 0.002

and

r/D = 0.06

From Fig. 3.4, t/L = 0.003

and

t = 0.003 × 168 = 0.50 in.

Using VIII-1 rules and Eq. (3.26), t = 0.885 × 40 × 168/(15,000 × 1.0 – 0.1 × 40) = 0.40 in.

Note that normally the thickness obtained for a given component is 33% higher in VIII-1 than that obtained from VIII-2, since the allowable stress in VIII-1 is 4.0 while that in VIII-2 is 3.0. However, as is illustrated above, in the case of F&D heads, this may not be so. The reason is in the safety factors imbedded in the equations of VIII-1 and VIII-2.

3.7 CONICAL SECTIONS, VIII-1 Conical shells and transition sections have a variety of configurations, as shown in Fig. 3.6. The required thicknesses of the conical and knuckle regions are calculated in a different manner. In addition, conical sections without a knuckle that are attached to shells result in an unbalanced force at the junction that must be considered by the designer. VIII-1 provides rules for the design of the junctions. These rules differ for internal and external pressure.



71

FIG. 3.6

3.7.1 Internal Pressure For internal pressure, the design equation for a conical section is given by t = PD/[2 cos α (SE – 0.6P)],

where

α < 30°

(3.33)

where t = required thickness, in. P = internal pressure, psi D = inside diameter of conical section under consideration, in. S = allowable tensile stress, psi E = Joint Efficiency Factor Equation (3.33) can be expressed in terms of internal pressure as P = 2SEt cos α/(D + 1.2t cos α)

Equations (3.33) and (3.34) can also be expressed in terms of outside diameter as


(3.34)

72

Chapter 3

t = PDo/[2SE cos α + P(2 – 1.2 cos α)]

(3.35)

P = 2SEt cos α/[Do – t(2 – 1.2 cos α)]

(3.36)

Equations (3.33) to (3.36), which are applicable at any angle α, are limited by VIII-1 to α ≤ 30°. When the angle α exceeds 30°, then VIII-1 requires a knuckle at the large end, as shown in Fig. 3.6(c) and (e). This type of construction will be discussed later in this section. After determining the thickness of the cone for internal pressure, the designer must evaluate the cone-toshell junction. The cone-to-shell junction at the large end of the cone is in compression due to internal pressure, in most cases. The designer must check the junction for required reinforcement needed to contain the unbalanced forces in accordance with Paragraph 1-5 of Appendix 1 of VIII-1. The required area is obtained from ArL = (k QL RL/Ss E1) (1 – ∆/α) tan α

(3.37)

where, ArL = required area at the large end of the cone, in. E1 = Joint Efficiency Factor of the longitudinal joint in the cylinder Ec = modulus of elasticity of the cone, psi Er = modulus of elasticity of the reinforcing ring, psi Es = modulus of elasticity of the cylinder, psi k = 1 when additional area of reinforcement is not required y/SrEr but not less than 1.0 when a stiffening ring is required QL = axial load at the large end, lb/in., including pressure end-load RL = large radius of the cone, in. Sc = allowable stress in the cone, psi Sr = allowable stress in the reinforcing ring, psi Ss = allowable stress in the cylinder, psi y = SsEs for the reinforcing ring on the shell ScEc for the reinforcing ring on the cone ∆ = angle obtained from Table 3.2 The area calculated from Eq. (3.37) must be furnished at the junction. Part of this area may be available at the junction as excess area. This excess area can be calculated from the equation AeL = (ts – t)(RLts)1/2 + tc – tr)(RLtc/cos α)1/2

(3.38)

where AeL = available area at the junction, in.2 t = minimum required thickness of the shell, in. tc = nominal cone thickness, in.

TABLE 3.2 VALUES OF ∆ FOR JUNCTIONS AT THE LARGE CYLINDER DUE TO INTERNAL PRESSURE P/SsE1 ∆, deg. P/SsE1 ∆, deg.

0.001 11 0.006 25

0.002 15 0.007 27

0.003 18 0.008 28.5

0.004 21 0.0091 30

0.005 23

NOTE: (1) ∆ = 30° for greater values of P/SsE1.



73

tr = minimum required thickness of the cone, in. ts = nominal shell thickness, in. If this excess area is less than that calculated from Eq. (3.37), then additional area in the form of stiffening rings must be added. The cone-to-shell junction at the small end of the cone is in tension due to internal pressure, in most cases. The designer must check the junction for required reinforcement in accordance with Paragraph 1-5 of Appendix 1 of VIII-1. The required area at the small end of the cone is obtained from Ars = (k Qs, Rs/Ss E1)(1 – ∆/α) tan α

(3.39)

where Ars = required area at the small end of the cone, in.2 Qs = axial load (including pressure end load) at small end, lb/in. Rs = small radius of the cone, in. ∆ = angle obtained from Table 3.3 The area calculated from Eq. (3.39) must be furnished at the junction. Part of this area may be available at the junction as excess area. This excess area can be calculated from the equation Aes = 0.78(Rsts)1/2 [(ts – t) + (tc – tr)/cos α]

(3.40)

If this excess area is less than that calculated from Eq. (3.39), then additional area in the form of stiffening rings must be added. When the angle α exceeds 30°, VIII-1 requires a knuckle at the large end, as shown in Fig. 3.6(c) and (e). The required thickness for the knuckle (called a flange) at the large end of the cone is obtained from the equation t = PLM/(2SE – 0.2P)

(3.41)

where M = (1/4)[3 + (L/r)1/2] L = Di/2 cos α Di = inside diameter at the knuckle-to-cone junction = D – 2r (1 – cos α) r = inside knuckle radius, in. Equation (3.41) can be expressed in terms of the required pressure P = 2SEt/(LM + 0.2t)

(3.42)

Equations (3.41) and (3.42) can also be written in terms of the outside diameter, Do, as t = PLoM/[2SE + P(M – 0.2)]

(3.43)

TABLE 3.3 VALUES OF ∆ FOR JUNCTIONS AT THE SMALL CYLINDER DUE TO INTERNAL PRESSURE P/SsE1 ∆, deg. P/SsE1 ∆, deg.

0.002 4 0.08 24

0.005 6 0.10 27

0.010 9 0.1251 30

0.02 12.5

0.04 17.5

NOTE: (1) ∆ = 30° for greater values of P/SsE1


74

Chapter 3

FIG. 3.7

or in terms of required pressure, P = 2SEt/[MLo – t(M – 0.2)]

(3.44)

When a knuckle is used at the cone-to-shell junction, the diameter at the large end of the cone is slightly less than the diameter of the cone without a knuckle, as shown in Fig. 3.6. Thus, the design of the cone as given by Eq. 3.33 is based on diameter Di rather than on the shell diameter. ASME VIII-1 does not give rules for the design of knuckles (flues) at the small end of cones. One design method uses the pressure-area procedure (Zick and Germain, 1963) to obtain the required thickness. Referring to Fig. 3.7 for terminology, we can determine the required thickness based on membrane forces in the flue and adjacent cone and shell areas from tf = (180/απr)[P(K1 + K2 + K3)/1.5SE – K4 – K5]

where E = Joint Efficiency Factor K1 = 0.125 (2r + D1)2 tan α – απr2/360 K2 = 0.28D1(D1ts) 1/2 K3 = 0.78K6(K6tc) 1/2 K4 = 0.78tc(K6tc) 1/2


(3.45)


75

K5 = 0.55ts(D1ts) 1/2 K6 = [D1 + 2r(1 – cos α)]/2 cos α P = internal pressure, psi S = allowable stress, psi tc = thickness of the cone, in. tf = thickness of the flue, in. ts = thickness of the shell, in. α = flue angle, deg. The flue angle is normally the same as the cone angle. Example 3.11 Problem Determine the required thickness of the cone, the two cylinders, and the area at the cone-to-cylinder junctions shown in Fig. E3.11. Let axial compressive load at cone vicinity from mounted equipment = 50 kips. Small Cylinder Allowable stress, psi Joint Efficiency Factor Modulus of elasticity, ksi Pressure, psi

15,000 0.85 27,000 100

Cone 16,000 1.0 29,000 100

Large Cylinder 17,500 0.85 25,000 100

Reinforcing Ring 13,000 — 30,000 —

FIG. E3.11


76

Chapter 3

Solution Small Shell The required thickness from Eq. (2.1) is t = 100 × 60/(15,000 × 0.85 – 0.6 × 100) = 0.47 in.

Use t = 1/2 in. Cone From Eq. (3.33), the cone thickness is calculated as t = 100 × 2 × 7 × 12/[2 cos 28(16,000 × 1.0 – 0.6 × 100)] = 0.60 in.

Use t = 5/8 in. Large Shell Again, using Eq. (2.1), we get t = 100 × 7 × 12/(17,500 × 0.85 – 0.6 × 100) = 0.57 in.

Use t = 5/8 in. Large Cone-to-Shell Junction Assume that a reinforcing ring, if needed, is to be added to the shell. Then from Eq. (3.37), we calculate the stiffness ratio, k, as k = 17,500 × 25,000,000/(13,000 × 30,000,000) = 1.12

The axial loads are given by QL = PRL/2 – axial equipment load = 100 × 84/2 – [50,000/(2π84)] = 4105 lb/in.

Next, we need to calculate the need for reinforcement in accordance with Table 3.2. P/SsE1 = 100/17,500 × 0.85 = 0.0067

From Table 3.2, ∆ = 26.4°. Reinforcement is needed since α = 28°. The amount of reinforcement is calculated from Eq. (3.37):



77

ArL = (1.12 × 4105 × 84/17,500 × 0.85)(1 – 26.4/28) tan 28 = 0.79 in.2

The available area in the shell and cone due to excess thickness is calculated from Eq. (3.38): AeL = (ts – t)(RLts)1/2 + (tc – tr)(RLtc/cos α)1/2 = (0.625 – 0.57)(84 × 0.625)1/2 + (0.625 – 0.60)(84 × 0.625/cos 28)1/2 = 0.040 + 0.193 = 0.59 in.2

The additional area needed at the large junction = 0.79 – 0.59 = 0.20 in.2 Use a 2 in. × 1/4 in. bar rolled the hard way. Small Cone-to-Shell Junction Assume that a reinforcing ring, if needed, is to be added to shell. Then, the stiffness ratio is obtained from k = 15,000 × 27,000,000/(13,000 × 30,000,000) = 1.04

The axial loads are equal to Qs = PRs/2 – axial equipment load = 100 × 60/2 – [50,000/(2π60)] = 2867 lb/in.

The need for reinforcement is obtained from Table 3.3. P/SsE1 = 100/15,000 × 0.85 = 0.0078

From Table 3.3, ∆ = 7.68°. Since this is less than 28°, reinforcement is required in accordance with Eq. (3.39), Ars = (1.04 × 2867 × 60/15,000 × 0.85)(1 – 7.68/28) tan 28 = 5.41 in.2

In order to determine what excess area, if any, is available at the cone-to-shell junction, we must calculate the required thickness of the cone at the small junction. This information is needed because the cone thickness used so far is based on the large diameter rather than on the small one. From Eq. (3.33), the minimum cone thickness at the small end is t = 100 × 2 × 5 × 12/[2 cos 28(16,000 × 1.0 – 0.6 × 100)] = 0.43 in.

From Eq. (3.40), the available area is


78

Chapter 3

Aes = 0.78(60 × 0.50)1/2 [(0.50 – 0.47) + (0.625 – 0.43)/cos 28] = 1.07 in.2

The additional area needed at the small junction = 5.41 – 1.07 = 4.34 in.2 Therefore, use a 4.5 in. × 1 in. bar rolled the hard way. Example 3.12 Problem Determine the required thickness of the cone, knuckle, flue, and the two cylinders shown in Fig. E3.12. Let P = 100 psi, S = 16,000 psi, and joint efficiency = 0.85. Solution Small Shell From Eq. (2.1), t = 100 × 60/(16,000 × 0.85 – 0.6 × 100) = 0.44 in.

Use t = 1/2 in.

FIG. E3.12



Large Shell From Eq. (2.1), t = 100 × 7 × 12/(16,000 × 0.85 – 0.6 × 100) = 0.62 in.

Use t = 5/8 in. Knuckle at Large End The thickness of the knuckle is obtained from Eq. (3.41). Di = 7 × 2 × 12 – 2 × 10(1 – cos 25) = 166.13 in. L = 166.13/(2 × cos 25) = 91.65 in. M = (1/4)[3 + (91.65/10)1/2] = 1.51 t = 100 × 91.65 × 1.51/(2 × 16,000 × 0.85 – 0.2 × 100) = 0.51 in.

Use t = 9/16 in. Cone From Eq. (3.33), with D = Di = 166.13 in., t = 100 × 166.13/[2 cos 25(16,000 × 0.85 – 0.6 × 100)] = 0.67 in.

Use t = 11/16 in. Flue From Eq. (3.45), the required thickness of the flue is K1 = 0.125(2 × 4 + 120)2 tan 25 – 25 π42/360 = 951.51 K2 = 0.28 × 120(120 × 0.50)1/2 = 260.26 K6 = [120 + 2 × 4(1 – cos 25)]/2 cos 25 = 66.62


79

80

Chapter 3

K3 = 0.78 × 66.62(66.62 × 0.6875)1/2 = 351.67 K4 = 0.78 × 0.6875(66.62 × 0.6875)1/2 = 3.62 K5 = 0.55 × 0.50(120 × 0.50)1/2 = 2.13 t = (180/25π4)[100(951.51 + 260.26 + 351.67)/1.5 × 16,000 × 0.85 – 3.62 – 2.13] = 1.10 in.

3.7.2 External Pressure The design of conical shells for external pressure follows the same procedure as that for cylindrical shells given in sections 2.4.1 and 2.4.2, with the following exceptions: Item Thickness Diameter Length

Cylinder t of cylinder Do of cylinder L of cylinder

Cone te = (t of cone)(cos α) DL = outside large diameter of the cone Le = (L/2)(1 + Ds/DL), where L is obtained from Fig. 3.6.

(3.46) (3.47) (3.48)

After designing the cone for external pressure, the cone-to-shell junctions must be evaluated. Due to external pressure, the cone-to-shell junction at the large end of the cone is tension, in most cases. The designer must check the junction for required reinforcement in accordance with Paragraph 1-8 of Appendix 1 of VIII1. The required area is obtained from ArL = (kQLRL/SsE1){1 – (0.25)[(PRL – QL)/QL](∆/α)} tan α

(3.49)

where all terms are the same as those in Eq. (3.37) and ∆ is obtained from Table 3.4. The area calculated from Eq. (3.49) must be furnished at the junction. Some of this area may be available as excess area at the junction. This excess area can be calculated from the equation AeL = 0.55(DLts)1/2 (ts + tc/cos α)

(3.50)

TABLE 3.4 VALUES OF ∆ FOR JUNCTIONS AT THE LARGE CYLINDER DUE TO EXTERNAL PRESSURE P/SsE1 ∆, deg. P/SsE1 ∆, deg. P/SsE1 ∆, deg.

0 0 0.04 21 0.20 47

0.002 5 0.08 29 0.25 52

0.005 7 0.10 33 0.30 57

0.010 10 0.125 37 0.351 60

0.02 15 0.15 40

NOTE: (1) ∆ = 60° for greater values of P/SsE1



81

If this excess area is less than that calculated from Eq. (3.49), then additional area in the form of stiffening rings must be added. In addition to having a sufficient reinforcement area, the cone-to-shell junction must have an adequate moment of inertia to resist external pressure forces when the junction is considered as a line of support. The required moment of inertia is calculated as follows: 1. Determine the quantity ATL from the equation ATL = LLts/2 + Lctc/2 + As

(3.51)

where As = area of the stiffening ring LL = effective length of the shell Lc = effective length of the cone = [L2 + (RL – Rs)2]1/2 L = axial length of the cone 2. Calculate the quantities M = (R2L – R2S/(3RL tan α) + LL/2 – (RL tan α)/2

(3.52)

FL = PM + (axial forces other than pressure)(tan α)

(3.53)

and

3. Calculate B from the equation B = 0.75(FLDL/ATL)

(3.54)

4. Enter the appropriate EPC and determine and A value. 5. If B falls below the left end of the temperature line, calculate A from the equation A = 2B/Ex

(3.55)

where Ex is the smaller of Ec, Er, or Es. 6. Calculate the moment of inertia from one of the following equations: Is = AD2LATL/14

(3.56)

I′s = AD2LATL/10.9

(3.57)

or

where Is = required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in.4 I′s = required moment of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about their combined neutral axis, in.4 7. The required moment of inertia must be greater than the furnished one. The cone-to-shell junction at the small end of the cone due to external pressure is in compression, in most cases. The designer must check the junction for required reinforcement in accordance with Paragraph 1–8 of Appendix 1 of VIII-1. The required area is obtained from Ars = (kQsRs/SsE1) tan α


(3.58)

82

Chapter 3

The area calculated from Eq. (3.58) must be furnished at the junction. Some of this area may be available at the junction as excess area. This excess area can be calculated from the equation Aes = 0.55(Dsts)1/2[(ts – t) + (tc – tr)/cos α]

(3.59)

If this excess area is less than that calculated from Eq. (3.58), then additional area in the form of stiffening rings must be added. In addition to having a sufficient area, the cone-to-shell junction must have an adequate moment of inertia to resist external pressure forces when the junction is considered as line of support. The required moment of inertia is calculated as follows: 1. Determine the quantity ATS from the equation ATs = LSts/2 + Lctc/2 + As

(3.60)

N = (R2L – R2S)/(6Rs tan α) + Ls/2 + (Rs tan α)/2

(3.61)

Fs = PN + (axial forces other than pressure)(tan α)

(3.62)

where As = area of the stiffening ring LS = effective length of the shell Lc = effective length of the cone = [L2 + (RL – Rs)2]1/2 L = axial length of the cone 2. Calculate the quantities

and

3. Calculate B from the equation B = 0.75(FsDS/ATS)

(3.63)

4. Enter the appropriate EPC and determine an A value. 5. If B falls below the left end of the temperature line, calculate A from the equation A = 2B/Ex

(3.64)

where Ex is the smaller of Ec, Er, or Es. 6. Calculate the moment of inertia from one of the following equations: Is = AD2SATS/14

(3.65)

I′s = AD2sATS/10.9

(3.66)

or

where Is = required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in.4 I′s = required moment of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about their combined neutral axis, in.4 7. The required moment of inertia must be greater than the furnished one.



83

When the cone is flanged and flued, then the required thickness of the cone is determined as before, except that Eq. (3.48) is replaced by Le = r1 sin α + (Lc/2)[(Ds + DL)/DLs]

for sketch (c) in Fig. 3.6

Le = r2(Dss/DL) sin α + (Lc/2)[(Ds + DL)/DL]

(3.67)

for sketch (d) in Fig. 3.6

Le = [r1 + r2(Dss/DLs)] sin α + (Lc/2)[(Ds + DL)/DLs]

(3.68)

for sketch (d) in Fig. 3.6

(3.69)

Example 3.13 Problem Check the calculated thicknesses in Example 3.11 due to an external pressure of 15 psi. The axial compressive load at cone vicinity from mounted equipment = 50 kips. Figure 2.4 applies for external pressure. Notice that the modulus of elasticity, shown in Fig. 2.4, for external pressure calculations must be used and is different from the values listed below for junction reinforcement. The design temperature is 100°F. Small Cylinder Allowable stress, psi Joint Efficiency Factor Modulus of elasticity, ksi External P, psi Effective L

15,000 0.85 27,000 15 10 ft

Cone

Large Cylinder

16,000 1.0 29,000 15 —

17,500 0.85 25,000 15 20 ft

Reinforcing Ring 13,000 — 30,000 — —

Solution Small Shell From Example 3.11, use t = 1/2 in. Do = 2(60 + 0.5) = 121 in.

Then L/Do = 0.99 and Do/t = 242. From Fig. 2.6, A = 0.00038. From Fig. 2.4, B = 5500 psi. From Eq. (2.26), P = (4/3)(5500)/(242) = 30.3 psi > 15 psi

By trial and error, it can be shown that for t = 0.40 in., P = 15 psi. Large Shell From Example 3.11, try t = 5/8 in. Do = 2(84 + 0.625) = 169.3 in.

Then L/Do = 1.42 and Do/t = 271. From Fig. 2.6, A = 0.00021. From Fig. 2.4, B = 3050 psi. From Eq. (2.26),


84

Chapter 3

P = (4/3)(3050)/(271) = 15.0 psi

Cone From Example 3.11, use t = 5/8 in. Do = 2(84 + 0.625) = 169.3 in.

From Eq. (3.46), te = 0.625 cos 28 = 0.552 in. From Eq. (3.47), DL = 169.3 in. From Fig. E3.11, L = (7 – 5)/tan 28 = 3.76 ft. From Eq. (3.48), Le = (3.76 × 12/2)(1 + 121/169.3) = 38.68 in.

Then Le/DL = 0.23 and DL/te = 307. From Fig. 2.6, A = 0.00121. From Fig. 2.4, B = 13,000 psi. From Eq. (2.26), P = (4/3)(13,000)/(307) = 56.5 psi > 15 psi

By trial and error, it can be shown that for t = 0.33 in., P = 15 psi. Large Cone-to-Shell Junction Assume that a reinforcing ring, if needed, is to be added to the shell. Then from Eq. (3.49), we calculate the stiffness ratio, k, as k = 17,500 × 25,000,000/(13,000 × 30,000,000) = 1.12

The axial loads are given by QL = PRL/2 – axial equipment load = –15 × 84.625/2 – [50,000/(2π84.625)] = – 728.7 lb/in.

Next, we must calculate the need for reinforcement in accordance with Table 3.4. P/SsE1 = 15/(17,500 × 0.85) = 0.001

From Table 3.4, ∆ = 2.5°. Hence, reinforcement is needed. The amount of reinforcement is calculated from Eq. (3.49): ArL = (1.12 × 728.7 × 84.625/17,500 × 0.85) {1 – (0.25)[15 × 84.625 + 728.7)/728.7](2.5/28)} tan 28 = 2.32 in.2

The available area in the shell and cone due to excess thickness is calculated from Eq. (3.50).



85

AeL = 0.55(84.625 × 0.625)1/2 (0.625 + 0.625/cos 28) = 5.33 in.2

Therefore, the junction is inherently reinforced and no additional rings are required. Next, determine the required moment of inertia at the large junction needed for external pressure. From Eq. (3.51), and assuming a ring area of 1.0 in2, ATL = (240)(0.625)/2 + (3.67 × 12)(0.625)/2 + 1.0 = 89.8 in.

From Eq. (3.52), M = (84.6252 – 60.52)/(3 × 84.625 tan 28) + (240/2) – (84.625 tan 28)/2 = 123.44

From Eq. (3.53), FL = 15 × 123.44 + (50,000/2π84.625)(tan α) = 1901.6

From Eq. (3.54), B = 0.75(1901.6 × 84.625/89.8) = 1344

From Eq. (3.55), A = 2 × 1344/25,000,000 = 1.08 × 10–4

From Eq. (3.57), I′s = 1.08 × 10–4 × 84.6252 × 89.8/10.9 = 6.37 in.4

A trial run indicates that the available moment of inertia is inadequate without a stiffening ring. Assume that a 2 in. × 1/2 in. ring rolled the hard way will be used. The available moment of inertia is obtained from Fig. E3.13(a) and (b). The neutral axis is at x1 = [(5.64 × 0.6252/2) + (5.64 × 0.625 × 2.65/2) + [2 × 0.5(– 1.0)]/[(5.64 × 0.625) + (5.64 × 0.625) + (2 × 0.5)] = (1.10 + 4.67 – 1.00)/(3.53 + 3.53 + 1.0) = 0.59 in.


86

Chapter 3

FIG. E3.13

x2 = 2.65/2 – 0.59 = 0.74 in. x3 = 1 + 0.59 = 1.59 in.

The moments of inertia of the cone about its two major axes are Ix_ = 5.643 × 0.625/12 = 9.34 in.4 I_y = 0.6253 × 5.64/12 = 0.12 in.4 I_x _y = 0.0

The moment of inertia of the cone around an axis y through its centroid is Iy = Ix_ sin2 α + I_y cos2 α + 2I_x _y sin αcos α



87

= 9.34 sin2 28 + 0.12 cos2 28 + 0.0 = 2.15 in.4

The total moment of inertia of the composite section = (I of shell about its neutral axis) + (area of shell) (distance to composite section neutral axis)2 + (I of cone about the y axis through its centroid) + (area of cone) (distance to composite section neutral axis)2 + (I of the stiffener about its neutral axis) + (area of stiffener) (distance to composite section neutral axis).2 I = 0.6253 × 5.64/12 + (5.64 × 0.625)(0.59 – 0.625/2)2 + 2.15 + (5.64 × 0.625)(0.74)2 + 2.03 × 0.5/12 + (2.0 × 0.5)(1.59)2 = 0.12 + 0.27 + 2.15 + 1.93 + 0.33 + 2.53 = 7.33 in.4

Hence, use 2 in. × 1/2 in. ring at the junction. Small Cone-to-Shell Junction Assume that a reinforcing ring, if needed, is to be added to shell. Then, the stiffness ratio is obtained from k = 15,000 × 27,000,000/(13,000 × 30,000,000) = 1.04

The axial loads are equal to Qs = PRs/2 – axial equipment load = –15 × 60.5/2 – [50,000/(2π60.5)] = –585.3 lb/in.

Reinforcement is required in accordance with Eq. (3.58). Ars = [1.04 × 585.3 × 60.5/(15,000 × 0.85)] tan 28 = 1.54 in.2

From Eq. (3.59), the available area is Aes = 0.55(2 × 60.5 × 0.5)1/2[(0.5 – 0.4) + (0.625 – 0.33)/cos28] = 1.86 in.2

Hence, the small junction is inherently reinforced and no additional rings are needed. Next, determine the required moment of inertia at the small junction needed for external pressure. From Eq. (3.60), ATs = (120)(0.5)/2 + (3.76 × 12)(0.625)/2 + 0.0 = 44.1 in.


88

Chapter 3

From Eq. (3.61), N = (84.6252 – 60.52)/(6 × 60.5 tan 28) + 120/2 – (60.5 tan 28)/2 = 62.06 in.

From Eq. (3.62), Fs = 15 × 62.06 + (50,000/2π60.5)(tan 28) = 1000.8 lbs/in.

From Eq. (3.63), B = 0.75(1000.8 × 60.5/44.1) = 1030 psi

From Eq. (3.64), A = 2 × 1030/27,000,000 = 7.63 × 10–5

From Eq. (3.66), I′s = 7.63 × 10–5 × 60.52 × 44.1/10.9 = 1.13 in.4

The available moment of inertia is obtained from Fig. E3.13(c) and (d). The neutral axis is at x1 = [(4.26 × 0.52/2) + (4.76 × 0.625 × 2.23/2)]/[(4.26 × 0.5) + (4.76 × 0.625)] = 0.75 in. x2 = 0.365 in.

The moments of inertia of the cone about its two major axes are Ix_ = 4.763 × 0.625/12 = 5.62 in.4 I_y = 0.6253 × 4.76/12 = 0.10 in.4 I_x _y = 0.0

The moment of inertia of the cone around an axis y through its centroid is Iy = Ix_ sin2 α + I_y cos2 α + 2I_x _y sin α cos α = 5.62 sin2 28 + 0.10 cos2 28 + 0.0 = 1.32 in.4



89

FIG. 3.8

I = 0.53 × 4.26/12 + (4.26 × 0.5)(0.75 – 0.5/2)2 + 1.32 + (4.76 × 0.625)(0.365)2 = 0.04 + 0.53 + 1.32 + 0.4 = 2.29 in.4

Hence, an adequate moment of inertia is available at the junction.

3.8 CONICAL SECTIONS, VIII-2 The required thickness of a conical shell in VIII-2 subjected to internal pressure is obtained from Eqs. (2.36) through (2.38). These equations, which are for cylindrical shells, are used with the radius R perpendicular to the surface of the cone, as shown in Fig. 3.8. VIII-2 does not list any rules for toriconical heads. The need for reinforcement at the large end of a cone-to-cylinder junction subjected to internal pressure is obtained from Fig. 3.9. The figure is used to determine a maximum angle α when the P/S value is known. If the actual angle α is less than that obtained from the figure, then no additional reinforcement is needed and the original thickness is adequate. On the other hand, if the actual angle α is larger than that obtained


90

Chapter 3

FIG. 3.9 INHERENT REINFORCEMENT FOR LARGE END OF CONE-TO-CYLINDER JUNCTION (ASME VIII-2)

from the figure, then additional reinforcement is needed in accordance with Fig. 3.10. The Q factor obtained from Fig. 3.10 is multiplied by the value of the original shell thickness at the large end to establish a new thickness at the junction. The same procedure is utilized at the small end of the cone, but with Figs. 3.11 and 3.12. The Q factor in this case is multiplied by the required thickness of the shell at the small end of the cone. Figures 3.9 through 3.12 apply to internal pressure only. For external pressure, the rules and factors of safety in VIII-2 are identical to those in VIII-1. Example 3.14 Problem Determine the required thickness of the cone, the two cylinders, and the area at the cone-to-cylinder junctions shown in Fig. E3.11. Allowable stress and pressure data is given in Table E3.14.

TABLE E3.14 ALLOWABLE STRESS AND PRESSURE DATA

Allowable stress, psi Pressure, psi

Small Cylinder

Cone

15,000 100

16,000 100

Large Cylinder 17,500 100

Reinforcing Material 13,000 100



FIG. 3.10 VALUES OF Q FOR LARGE END OF CONE-TO-CYLINDER JUNCTION (ASME VIII-2)

Solution Small Shell The required thickness from Eq. (2.36) is t = 100 × 60.0/(15,000 – 0.5 × 100) = 0.40 in.

Use t = 7/16 in. Cone From Eq. (2.36), with R = 7 × 12/cos 28 = 95.14 in., the cone thickness is calculated as t = 100 × 95.147(16,000 – 0.5 × 100) = 0.60 in.

Use t = 5/8 in.


91

92

Chapter 3

FIG. 3.11 INHERENT REINFORCEMENT FOR SMALL END OF CONE-TO-CYLINDER JUNCTION (ASME VIII-2) Large Shell Again, using Eq. (2.36), we get t = 100 × 7 × 12/(17,500 – 0.5 × 100) = 0.48 in.

Use t = 1/2 in. Large End of Cone-to-Shell Junction The P/S ratio is P/S = 100/17,500 = 0.0057

From Fig. 3.9, maximum α = 19°. Hence, reinforcement is required. From Fig. 3.10, Q = 1.30. The required thickness at the large end of the cone-to-shell junction is tr = Q × t ratio of shell stress to reinforcement stress (must be equal to or greater than 1.0) = 1.30 × 0.48 × 17,500/13,000 = 0.84 in.

Use t = 7/8 in., distributed as detailed in Fig. 3.10. Small End of Cone-to-Shell Junction Here the P/S ratio is P/S = 100/15,000 = 0.0067



93

FIG. 3.12 VALUES OF Q FOR SMALL END OF CONE-TO-CYLINDER JUNCTION (ASME VIII-2)

From Fig. 3.11, maximum α = 1.5°. Hence, reinforcement is required. From Fig. 3.12, Q = 2.15. The required thickness at the small end of the cone-to-shell junction is tr = 2.15 × 0.40 × 15,000/13,000 = 0.99 in.

Use t = 1 in., distributed as detailed in Fig. 3.12.

3.9 Miscellaneous Transition Sections Rules for stiffening rings needed in conical-to-cylindrical shell attachments are well covered in VIII-1. However, rules for stiffening rings needed in attaching other than conical shells are not presently covered by the VIII-1


94

Chapter 3

rules. The designer can usually obtain the required stiffening ring sizes by using the pressure-area method (Jawad, 2004). The use of this method for the design of a pressure vessel is shown in the following example. Example 3.15 Figure 3.13 shows a vessel consisting of two partial spherical components supported by a cylindrical skirt. Design the spherical components and check their intersection for the requirement of a stiffening ring. Let P = 250 psi, R = 30 inches, E = 1.0, and S = 20,000 psi in tension.

FIG. 3.13



95

Solution The pressure-area method is based on the premise that the vessel area occupied by the applied pressure must be resisted by the metal surrounding such area. Thus, for the shaded triangle “ade”, 1/ 2

(R) (R dθ) P = (R dθ)(t)(SE)

or, t = PR/(2 SE).

This equation is the same as Eq.(3.3) derived from the classical strength of materials theory. Accordingly, components “cf” and “fg” can be designed by Eq.(3.1) as t = PR/(2SE – 0.2P) = 250 × 30/(2 × 20000 × 1.0 – 0.2 × 250) = 0.188”.

From Fig.3.13, we notice that the horizontally shaded area “acf” is contained by the metal along the arc “cf”. Similarly, the horizontally shaded area “bfg” is contained by the metal along the arc “fg”. The vertically shaded area “afb”, on the other hand, is not contained by any metal in the pressure boundary of the spherical components. Accordingly, a stiffening ring is needed. The cross sectional area of the stiffening ring needed to contain the pressure in triangle “afb” is obtained from the equation (1/2)(0.707R)(0.707R)(2) P = A(S)

or A = 0.5R2 P/S. A = 0.5 × 302 × 250/20000 A = 5.625 in2.

A stiffening ring with a cross sectional area of 5.625 in2 must be attached at point “f” to satisfy the equations of equilibrium.



CHAPTER

4 FLAT PLATES, COVERS, AND FLANGES 4.1 INTRODUCTION Flat plates, covers, and flanges are used extensively in pressure vessels. Circular plates are used for most applications; however, there are some applications where the flat plate is obround, square, rectangular, or some other shape. When a flat plate or cover is used as the end closure or head of a pressure vessel, it may be an integral part of the vessel by virtue of having been formed with the cylindrical shell or welded to it or it may be a separate component that is attached by bolts or some quick-opening mechanism utilizing a gasketed joint attached to a companion flange on the end of the shell. Flat plates and covers may contain no openings, a single opening, or multiple openings. To satisfy the loadings and allowable stresses, the plate may need to be of an increased thickness or it may require reinforcement from attachments. The equations for the design of unstayed plates and covers based on a uniform thickness with a uniform pressure loading over the entire surface are described in UG-34 of VIII-1 and Article D-7 of VIII-2. For flat plates and covers with either single or multiple openings, design requirements are given in UG-39 of VIII-1 and in AD-530 of VIII-2. See Chapter 7 and Part UHX of VIII-1 for design requirements of tubesheets with multiple openings as used in shell-and-tube heat exchangers. For the design of flat plates and covers which are attached by bolting that causes an edge moment due to the gasket and bolt loading action, both gasket seating loads and operating loads shall be considered in a similar manner to that required for determining the acceptability of a bolted, flanged joint. Since the loadings and dimensions required for analysis of bolted, flat plates and flanges in both VIII-1 and VIII-2 are very similar, if not the same, they will be treated together. Spherically-dished covers are considered in section 4.7 and in Appendix 1-6 of VIII-1.

4.2 INTEGRAL FLAT PLATES AND COVERS Since the design rules of circular flat plates and covers is the same in VII-1 and VIII-2, only the references in VIII-1 will be used. Design rules for noncircular flat plates and covers are given only in VIII-1.

4.2.1 Circular Flat Plates and Covers In UG-34, the minimum required thickness of a circular, flat plate which is integrally formed with or attached to a cylindrical shell by welding or a special clamped connection is calculated by using the following equation: t = d(CP/SE)1/2 97


(4.1)

98

Chapter 4

where d = effective diameter of the flat plate, in. C = coefficient between 0.10 and 0.33, depending on corner details P = design pressure, psi S = allowable stress at design temperature, psi E = butt-welded joint efficiency of the joint within the flat plate t = minimum required thickness of the flat plate, in. For further description of the terms given above, refer to Fig. 4.1 (a) through (i), or Fig. UG-34 of VIII-1. Depending on the shape and welding details of the corner, a value of C is selected. A value of E, the buttweld joint efficiency within the flat plate, is required if the diameter of the head is sufficiently large that the head needs to be made of more than one piece. The value of E depends on the degree of NDE1 performed. It is not a weld efficiency of the head-to-shell corner joint! Example 4.1 Problem Using the rules of UG-34 of VIII-1, determine the minimum required thickness of an integral flat plate with an internal pressure of P = 2,500 psi, an allowable stress of S = 17,500 psi, and a plate diameter of d = 24 in. There are no butt welded joints within the head. There is a corrosion allowance, c.a. = 3/16 in. The corner details conform to Fig. 4.1 sketch (b-2) assuming that m = 1. Solution From Fig. 4.1, sketch (b-2), C = 0.33 × m = 0.33(1) = 0.33 d = 24, dc = 24 + (2)(c.a.) = 24.375 in.

From Eq. (4.1), tc = (24.375)[(0.33)(2500)/(17,500)(1)]1/2 = 5.29 in. tr = tc + c.a. = 5.48 in.

Example 4.2 Problem Determine the minimum corner radius to make Example 4.1 valid. Solution First, the minimum required thickness of the cylindrical shell, tr, must be calculated using Eq. (2.1). tc = PRc/(SE – 0.6P) = (2500)(12.1875)/(17,500 × 1 – 0.6 × 2500) = 1.904 in.

For t > 11/2 in., rmin = 0.25tc = 0.25(1.904) = 0.476 in. 1

NDE is the nondestructive examination of the butt joint.


Flat Plates, Covers, and Flanges

99

FIG. 4.1 SOME ACCEPTABLE TYPES OF UNSTAYED FLAT HEADS AND COVERS (ASME VIII-1)

4.2.2 Noncircular Flat Plates and Covers When the flat plate or cover is square, rectangular, elliptical, obround, or any shape other than circular, the minimum required thickness is calculated in the same manner as for a circular plate, except for the addition


100

Chapter 4

of a factor to compensate for the lack of uniform membrane support obtained in a circular plate. This factor, Z, is related to the ratio of the length of the short dimension, d, to the length of the long dimension, D, and is determined by Z = 3.4 – (2.4d/D) ≤ 2.5

(4.2)

Using this value of Z, the minimum required thickness of an integrally attached noncircular flat plate is calculated from the following equation: t = d(ZCP/SE)1/2

(4.3)

where all terms are the same as in Eq. 4.1. Example 4.3 Problem Using the rules in UG-34 of VIII-1, determine the minimum required thickness of the flat end plate of a rectangular box header which is 8 in. × 16 in. and has an internal pressure of P = 350 psi and S = 15,000 psi. The plate is integrally welded into place. There is no corrosion allowance and no butt-welded joints in the plate. Solution (1) Using Eq. (4.2), determine the stress multiplier, Z, and use it to solve Eq. (4.3). (2) From Eq. (4.2), Z = 3.4 – [2.4(8/16)] = 2.2

(3) Using Eq. (4.3), t = 8[(2.2)(0.33)(350)/(15,000)(1)]1/2 = 1.04 in.

Example 4.4 Problem Determine the maximum permissible length of a rectangular plate which is 8 in. wide by 1/2 in. thick and has P = 100 psi, and S = 15,000 psi. There is no corrosion allowance, and no butt-welded joints in the plate. Solution (1) Rearrange Eq. (4.3), above, to solve for Z as follows: Z = SEt2/CPd2 = (15,000)(1)(0.5)2/(0.33)(100)(8)2 = 1.78

(2) Rearrange Eq. (4.2), above, to solve for D as follows: D = 2.4d/(3.4 – Z) = [(2.4)(8)]/[3.4 – 1.78] = 11.9 in.

A plate that is 8 in. wide by 1/2 in. thick can have a maximum length of 11.9 in.



101

4.3 BOLTED FLAT PLATES, COVERS, AND FLANGES Bolted connections are used on pressure vessels because they permit easy disassembly of components. The bolted connection may consist of a flat plate or so-called blind flange, a loose-type flange, or an integral-type flange. An early method to analyze bolted flange connections with gaskets entirely within the circle enclosed by the bolt holes was developed by Taylor Forge in 1937 (Waters, 1937). This method was further developed by the Code Committee for use in various sections of the ASME Code. In their latest form, flange rules are in Appendix 2 of VIII-1 and Appendix 3 of VIII-2. Rules for pairs of flanges with metal-to-metal contact outside of the bolt circle are given in Appendix Y of VIII-1.

4.3.1 Gasket Requirements, Bolt Sizing, and Bolt Loadings Appendix 2 of VIII-1 provides design rules for flanges under internal and external pressure. Determination of gasket requirements, bolt sizing, and bolt loading is the same for a bolted flate plate or blind flange, loosetype flange, and integral-type flange. Loadings are developed for gasket seating or bolt-up condition and for hydrostatic end load or operating condition. Guidance is given for the selection of the gasket and design factors, m, the gasket factor, and y, the gasket unit seating load, psi. Once the gasket material and sizing is determined, the bolt-up and operating loads are determined, bolting is selected, and the design bolt loading is calculated. 4.3.1.1 Gasket Design Requirements. The selection of gasket type and material is set by the designer after considering the design specifications. Once the gasket is chosen, the m and y factors may be selected from Appendix 2 of VIII-1. The values of m and y in the table are nonmandatory, and different values of m and y, which are either higher or lower, may be used if data are available to indicate acceptability. After the gasket material and type are selected, the effective gasket width, b, is determined by the following procedure: 1. The basic gasket seating width, bO, is selected from Table 2-5.2 of VIII-1. 2. When bO ≤ 1/4 in., b = bO, and when bO bO > 1/4 in., b = 0.5(bO)1/2. From b, the values of G and hG can be determined. 4.3.1.2 Bolt Sizing and Bolt Loadings. mined as follows:

The required bolt load for operating condition, Wm1, is deter-

Wm1 = H + Hp = 0.785G2P + (2b × 3.14GmP)

(4.4)

The required bolt load for gasket seating condition, Wm2, is determined as follows: Wm2 = 3.14bGy

(4.5)

Once the required bolt loads are determined, the required bolt area for each loading condition can be calculated as follows: Am1 = Wm1/Sb

(4.6)

Am2 = Wm2/Sa

(4.7)

and

The total required cross-sectional area of the bolts, Am, is the greater of Am1 or Am2. The bolts are selected so that Ab, the actual bolt area, is equal to or greater than Am. The bolt load used for the design of flanges, W, is then determined from the following.


102

Chapter 4

For operating condition, W = Wm1

(4.8)

W = 0.5(Am + Ab)Sa

(4.9)

For gasket seating condition,

where Sa shall be not less than the allowable tensile stress value in II-D, psi. In addition to safety, Eq. (4.9) provides some protection from overbolting during gasket seating at atmospheric temperature before the internal pressure is applied. Where additional protection is desired or required by the design specifications, the following equation is used: W = AbSa

(4.10)

4.3.1.3 Check for Gasket Crushout. Although not considered in VIII-1, it is prudent to design against crushout of the gasket by determining the minimum gasket width using the following: Nmin = AbSa/2yπG

(4.11)

with changes of m and/or y being permitted, as described in 4.3.1.1.

4.4 FLAT PLATES AND COVERS WITH BOLTING 4.4.1 Blind Flanges & Circular Flat Plates and Covers Before calculating the minimum required thickness of a blind flange, flat plate, or cover, determine if there is a suitable blind flange available from any of the flange standards listed in Table U-3 of VIII-1 and Table AG-150.1 of VIII-2. Standard flanges are acceptable without further calculations for diameters and pressure/temperature ratings in the respective standards when of the types shown in Fig. 4.1, sketches (j) and (k). When there is no standard flange available, the minimum required thickness of the circular flat plate is calculated by using the following equation: t = d[(CP/SE) + (1.9WhG/SEd3)]1/2

(4.12)

where the definitions of terms are as given in 4.2.1 and the determination of W and hG is as given in 4.3.1.1.

4.4.2 Noncircular Flat Plates and Covers When the flat head is square, rectangular, elliptical, obround, or some other noncircular shape and utilizes bolting, the minimum required thickness of the noncircular flat head is calculated by using the following equation: t = d[(ZCP/SE) + (6WhG/SELd2)]1/2

(4.13)

where the terms are as explained in earlier sections.

4.5 OPENINGS IN FLAT PLATES AND COVERS Rules for compensation or reinforcement required for openings in flat plates and covers are given in UG-39 of VIII-1. Single, small openings which do not exceed the size limits UG-36(c)(3)(a) & (b) and are not



103

greater than one-fourth the plate or cover diameter are integrally reinforced and do not require reinforcement calculations. Note: Particular care should be taken when standard blind flanges are used so as to not exceed the size permitted by the Standard. When the opening size exceeds these limits, the pressure/temperature ratings are no longer valid and calculations are required.

4.5.1 Opening Diameter Does Not Exceed Half the Plate Diameter For a single opening when the opening diameter does not exceed half the plate diameter, standard reinforcement calculations may be made, keeping in mind that the total reinforcement required is: Ar = 0.5dtr

(4.14)

where Ar = required area of reinforcement, in.2 d = opening diameter in the flat plate, in. t = minimum required thickness of the flat plate, in. The reason for the factor of 0.5 instead of 1.0 is that, unlike the situation in cylindrical shells and formed heads, the stress distribution through the thickness of a flat plate is primary bending stress instead of primary membrane stress. For multiple openings in which no diameter is greater than half the plate diameter, no pair of openings has an average diameter greater than one-fourth the plate diameter, and the spacing between pairs of openings is no less than twice the average diameter, Eq. 4.14 for the minimum required thickness of a plate with a single opening may be used. If the spacing between pairs of openings is less than twice the average diameter, but not less than 11/4 the average diameter, the amount of reinforcement is calculated by adding the required reinforcement of the pair to no less than 50% of the required reinforcement between the pair. If the spacing is less than 11/4 the average diameter, U-2(g) applies. In all cases, the width of the ligament between the pair of openings shall be no less than one-fourth of the diameter of the smaller of the pair, and the edge ligament between an opening and the edge of the plate shall be no less than one-fourth of the diameter of that opening. In VIII-1, as an alternative to the rules for reinforcement of a single opening, the following procedure may be used to determine the minimum requried thickness of a flat plate. 1. In Eqs. (4.1) and (4.3), a value of 2C or 0.75, whichever is less, may be used, except for sketches (b-1), (b-2), (e), (f), (g), and (i) of Fig. 4.1, for which 2C or 0.50, whichever is less, must be used. 2. In Eqs. (4.12) and (4.13), the quantity under the square root sign must be doubled before solving for t. As an alternative to the rules for multiple openings, when the spacing for all pairs of openings is equal to or greater than twice the average diameter of that pair, the alternative rules for single openings may be used. When the spacing is less than twice but equal to or more than 11/4 the average diameter of the pair, the required plate thickness shall be determined by the alternative rule for single openings multiplied by a factor, h, that is defined as h = (0.5/e)1/2

(4.15)

e = [(p – davg)/p]smallest

(4.16)

where e = smallest ligament efficiency of all pairs p = center-to-center spacing of a pair to get e davg = average diameter of the same pair to get e


104

Chapter 4

FIG. 4.2 MULTIPLE OPENINGS IN THE RIM OF A FLAT HEAD OR COVER WITH A LARGE CENTRAL OPENING (ASME VIII-1)

Again, in all cases, the width of the ligament between the pair of openings shall be no less than one-fourth of the diameter of the smaller of the pair, and the edge ligament between an opening and the edge of the plate shall be no less than one-fourth of the diameter of that opening.

4.5.2 Opening Diameter Exceeds Half the Plate Diameter When the opening is a single, circular, centrally-located opening in a circular flat plate, the plate shall be designed according to Appendix 14 of VIII-1. For small openings which are located in the rim of the flat plate surrounding a large opening, as shown in Fig. 4.2, and the plate is to be analyzed according to Appendix 14, the rules in section 4.5.1 (above) for single openings and for multiple openings shall be followed, using as the minimum requried thickness a plate thickness which satisfies the rules of Appendix 14. As an alternative, the thickness determined according to Appendix 14 may be multiplied by √2 = 1.414 for a single opening in the rim or for multiple openings which satisfy the ligament spacing for a flat plate which had its thickness set by the √2 rule.

4.6 BOLTED FLANGE CONNECTIONS WITH RING TYPE GASKETS As described earlier in section 4.3, Appendix 2 of VIII-1 and Appendix 3 of VIII-2 contain rules for the design of bolted flange connections with gaskets entirely within a circle enclosed by the bolt holes. Selection of the gaskets and determination of the bolt sizes, bolt loading, and loading moment arms are obtained in the



105

same manner for flat heads or blind flanges as they are for integral and loose flanges. Using these loadings and moment arms, gasket seating moments and operating moments are determined and stresses are calculated and compared with allowable stress values. The procedure for calculating stresses and acceptability of stresses is essentially the same for welding neck flanges as for loose, slip-on, or ring-type flanges. For Code consideration, there are three types of flanges: loose, integral, and optional (see Fig. 4.3). Loose means that, for calculations, the flange ring provides the entire strength of the flange, even though the ring is attached to the vessel or pipe by threads or welds. Integral means that, for calculations, the ring-and-shell or ring-andpipe combination provides the strength of the flange, and the assumption is made that the connection between these parts has enough strength that they act together. Optional means that the connection is basically integral; however, it may be calculated as loose, which requires only one stress to be calculated. The difference in these various flanges is the line of load application and the magnitude of the loads. However, the applied moments are determined in a similar manner.

4.6.1 Standard Flanges Typical of the loose-type flanges are the ring flange, which usually is made from a flat plate formed into a ring, and the slip-on or lap-joint flange. Either of these may have a hub, but the weld size and strength are not great enough to have the flange and vessel or pipe act together. Loadings are determined in the same manner as described in sections 4.3.1 and 4.4.1. Example 4.5 presents an example problem and filled-in Sample Calculation Sheet for a ring flange with a ring-type gasket. See Appendix D for blank fill-in Sheet D.I (Ring Flange with Ring-Type Gasket) and blank fill-in Sheet D.2 (Slip-on or Lap-Joint Flange with Ring-Type Gasket). Integral-type flanges usually contain a tapered hub and flange ring which may be integrally formed or the hub and neck are welded together and act integrally so that each carries part of the loadings. Example 4.6 gives an example problem and filled-in sheet for a welding neck flange with a ring-type gasket. See Appendix D for blank fill-in Sheet D.3 (Welding Neck Flange with ring-Type Gasket). Example 4.5 Problem Using the rules in Appendix 2 of VIII-1, determine the minimum required thickness of a ring flange, shown in Fig. E4.5, with the following design data: Design pressure = 2000 psi; Design temperature = 650°F; Flange material is SA-105; Bolting material is SA-325 Gr. 1; Gasket is spiral-wound, fiber-filled, stainless steel, 13.75 in. I.D. × 1.0 in. wide; No corrosion allowance. Solution (1) The allowable tensile stress of the bolts from II-D at gasket seating and operating conditions (design temperature) is Sa = Sb = 20.2 ksi. (2) The allowable tensile stress of the flange from II-D at gasket seating is Sfa = 20.0 ksi and operating conditions is Sfo = 17.8 ksi. (3) The diameter of the gasket’s line-of-action, G, is determined as follows: bO = N/2 = 0.5 in.

and

b = 0.5(bO)1/2 = 0.3535 in.

G = 13.75 + (2 × 1) – (2 × 0.3535) = 15.043 in.

(4) With N = 1, b = 0.3535, m = 3.0, and y = 10,000, the bolt loadings and the number and diameter of bolts are


106

Chapter 4

FIG. E4.5 RING FLANGE SAMPLE CALCULATION SHEET

H = (π/4)G2p = (π/4)(15.043)2(2000) = 355,500 lb Hp = 2bπGmp = 2(0.3535)π(15.043)(3.0)(2000) = 200,500 lb Wm1 = H + Hp = 355,500 + 200,500 = 556,000 lb Wm2 = πbGy = π(0.3535)(15.043)(10,000) = 167,100 lb Am = the larger of Wm1/Sb

or

Wm2/Sa = 556,000/20.2 = 27.5 in.2



Using 20 bolts of 11/2-in. diameter, Ab = actual bolt area = 20(1.41) = 28.2 in.2 Wa = 0.5(Am + Ab)Sb = 0.5(27.5 + 28.2)(20,200) = 562,600 lb WO = Wm1 = 556,000 lb

(5) Using Eq. (4.10), the gasket crushout width is Nmin = AbSa/2yπG = (28.2)(20,200)/2(10,000)(π)(15.043) = 0.6 in. < 1.0 in. actual

(6) The total flange moment for gasket seating condition is: Flange Load HG = Wa = 562,600 lb

Lever Arm hG = 0.5(C – G) = 3.729 in.

Flange Moment MGS = HG × hG = (562,600)(3.729) = 2,098,000 in.-lb

(7) The total flange moment for operating condition is: Flange Loads HD = (π/4)B2p = (π/4)(12.75)2(2000) = 255,400 lb HG = Hp = 200,500 lb HT = H – HD = 355,500 – 255,400 = 100,100 lb


107

108

Chapter 4

Lever Arms hD = 0.5(C – B) = 0.5(22.5 – 12.75) = 4.875 in. hG = 0.5(C – G) = 0.5(22.5 – 15.043) = 3.729 in. hT = 0.5(hD + hG) = 0.5(4.875 + 3.729) = 4.302 in.

Flange Moments MD = HD × hD = (255,400)(4.875) = 1,245,000 in.-lb MG = HG × hG = (200,500)(3.729) = 747,700 in.-lb MT = HT × hT = (100,100 × 4.308) = 431,200 in.-lb Mo = MD + MG + MT = 2,424,000 in.-lb

(8) Shape factor from Appendix 2 of VIII-1 for K = A/B = 26.5/12.75 = 2.078

From Fig. 2–7.1 of VIII-1, Y = 2.812. (9) The minimum required thickness of the flange is the larger tmin of: For gasket seating condition: tmin = [(MGSY)/(SfaB)]1/2 = [(2,098,000)(2.812)/(20,000)(12.75)]1/2 = 4.81 in.

For operating condition: tmin = [(MOY)/(SfB)]1/2 = [(2,424,000)(2.812)/(17,800)(12.75)]1/2 = 5.48 in.



109

Example 4.6 Problem Using the rules of Appendix 2 of VIII-1, determine the minimum required thickness of a welding neck flange, of the type shown in Fig. E4.6 with the following design data:

FIG. E4.6 WELDING NECK FLANGE SAMPLE CALCULATION SHEET


110

Chapter 4

Design pressure = 2000 psi; Design temperature = 650°F; Flange material is SA-105; Bolting material is SA-325 Gr. 1; Gasket is spiral-wound, fiber-filled, stainless steel, 13.75 in. I.D. × 1.0 in. wide; No corrosion allowance. Note: This flange has facing details, gasket size, and bolting that are the same as those given in Example 4.5, except that this is a welding neck flange instead of a ring flange. Solution (1) The allowable tensile stress of the bolts from II-D at gasket seating and operating conditions (design temperature) is Sa = Sb = 20.2 ksi. (2) The allowable tensile stress of the flange from II-D at gasket seating is Sfa = 20.0 ksi and operating conditions is Sfo = 17.8 ksi. (3) The diameter of the gasket’s line-of-action, bolt loadings, bolt number and diameter, and crushout width are the same as in Steps 3–5 of Example 4.5. (4) The total flange moment for the gasket seating condition is the same as in Step 6 of Example 4.5. MGS = 2,098,000 in.-lb

(5) The total flange moment for the operating condition is: Flange Loads HD = (π/4)B2p = (π/4)(10.75)2(2000) = 181,500 lb HG = Hp = 200,500 lb HT = H – HD = 355,500 – 181,500 = 174,000 lb

Lever Arms hD = R + 0.5g1 = 2.5 + 0.5(3.375) = 4.188 in. hG = 0.5(C – G) = 0.5(22.5 – 15.043) = 3.729 in. hT = 0.5(R + g1 + hG) = 0.5(2.5 + 3.375 + 3.7285) = 4.802 in.



111

Flange Moments MD = HD × hD = (181,500)(4.188) = 760,100 in.-lb MG = HG × hG = (200,500)(3.729) = 747,700 in.-lb MT = HT × hT = (174,000)(4.802) = 835,500 in.-lb MO = MD + MG + MT = 2,343,000 in.-lb

(6) Shape factors from Appendix 2 of VIII-1 for K are K = A/B = 26.5/10.75 = 2.465

From Fig. 2-7.1 of VIII-1, T = 1.35

Z = 1.39

Y = 2.29

U = 2.51

g1/gO = 3.375/1.0 = 3.375 hO = (BgO)1/2 = [(10.75)(1.0)]1/2 = 3.279 h/hO = 6.25/3.279 = 1.906

From Appendix 2 of VIII-1, F = 0.57

V = 0.04

f = 1.0

e = F/hO = 0.57/3.279 = 0.174 d = (U/V)hOg2O = (2.51/0.04)(3.279)(1)2 = 205.76

(7) MGS = 2,098,000 in.-lb and Sfa = 20.0 ksi and MO = 2,343,000 in.-lb and Sfo = 17.8 ksi. Since at gasket seating condition, the moment is smaller and the allowable stress is larger, only the operating condition is calculated.


112

Chapter 4

Assume a flange thickness of t = 4.0 in. L = [(te + 1)/(T) + (t)3/d] = 1.256 + 0.311 = 1.567

Longitudinal hub stress: SH = fMO/Lg21B = (1)(2,343,200)/(1.567)(3.375)2(10.75) = 12,210 psi

Radial flange stress: SR = [(4/3)te + 1]MO/Lt2B = (1.928)(2,343,200)/(1.567)(4)2(10.75) = 16,760 psi

Tangential flange stress: ST = [(YMo/t2B) – ZSR] = {[(2.29)(2,343,200)/(4)2(10.75)] – (1.39)(16,760)} = 7900 psi

Combined stresses: 0.5(SH + SR) = 0.5(12,210 + 16,760) = 14,490 psi 0.5(SH + ST) = 0.5(12,210 + 7900) = 10,060 psi

(8) Allowable stresses: SH ≤ 1.5Sf : 12,210 psi < 26,700 psi SR ≤ Sf : 16,760 psi < 17,800 psi ST ≤ Sf :

7900 psi < 17,800 psi

0.5(SH + SR) ≤ Sf : 14,490 psi < 17,800 psi 0.5(SH + ST) ≤ Sf : 10,060 psi < 17,800 psi

Since all actual stresses are less than the allowable stresses, the selection of t = 4.0 in. is adequate. If an optimum minimum thickness of the flange is desired, calculations must be repeated with a smaller value of t until one of the calculated stresses or stress combinations is approximately equal to the allowable stress, even though other calculated stresses are less than the allowable stress for that calculated stress.



113

4.6.2 Special Flanges Rules for special flanges with different geometry and/or loading are given in Appendix 2 of VIII-1. Included are: 2–9 for split loose flanges, 2–10 for noncircular shaped flanges with a circular bore, 2–11 for flanges subject to external pressure, 2–12 for flanges with nut-stops, and 2–13 for reverse flanges. Flanges with other geometry and loading shall follow U-2(g). 4.6.2.1 Reverse Flanges. Reverse flanges are described in Appendix 2-13 of VIII-1. They are similar to standard flanges, except some of the loads on the flange ring cross section may be applied at different locations and in a reverse direction, possibly causing a reverse moment. VIII-1 has chosen to use the term αB to convert a standard flange to a reverse flange. Example 4.7 gives an example problem and a filled-in sheet for a reverse welding neck flange with a ring-type gasket. See Appendix D for a blank fill-in Sheet D.4 (Reverse Welding Neck Flange with Ring-Type Gasket). The method of analysis for a reverse flange is similar to that used for an integral flat head with a large, single, circular, centrally-located opening, as given in Appendix 14 of VIII-1. For both analyses, a special limitation of the geometry is given. When K ≤ 2, calculated stresses are acceptable; however, when K > 2, calculated stresses become increasingly conservative. For this reason, use of the analysis procedure should be limited to K ≤ 2. Example 4.7 Problem Using the rules of Appendix 2 of VIII-1, determine the minimum required thickness of a reverse welding neck flange, shown in Fig. E4.7, with the following design data: Design pressure = 2000 psi; Design temperature = 650°F; Flange material is SA-105; Bolting material is SA-325 Gr. 1; Gasket is spiral-wound, fiber-filled, stainless steel, 13.75 in. I.D. × 1.0 in. wide; No corrosion allowance. Note: This flange has facing details, gasket size, and bolting that are the same as those given in Example 4.5; however, this is a reverse welding neck flange with different flange dimensions. Solution (1) The allowable tensile stress of the bolts from II-D at the gasket seating and operating conditions (design temperature) is Sa = Sb = 20.2 ksi. (2) The allowable tensile stress of the flange from II-D at the gasket seating is Sfa = 20.0 ksi and operating conditions is Sfo = 17.8 ksi. (3) The diameter of the gasket line-of-action, bolt loadings, bolt number and diameter, and the crushout width are the same as in Steps 3–5 of Example 4.5. (4) The total flange moment for the gasket seating condition is the same as in Step 6 of Example 4.5. MGS = 2,098,000 in.-lb

(5) The total flange moment for operating condition is: Flange Loads HD = (π/4)B2p = (π/4)(22.875)2(2,000) = 821,900 lb


114

Chapter 4

FIG. E4.7 REVERSE WELDING NECK FLANGE SAMPLE CALCULATION SHEET



HG = Wm1 – H = 556,000 – 355,500 = 200,500 lb HT = H – HD = 355,500 – 821,900 = –466,400 lb

Lever Arms hD = 0.5(C + g1 – 2go – B) = 0.5(22.5 + 1.8125 – 2 × 1.8125 – 22.875) = – 1.094 in. hG = 0.5(C – G) = 0.5(22.5 – 15.043) = 3.729 in. hT = 0.5[C – 0.5(B + G)] = 0.5[22.5 – 0.5(22.875 + 15.043)] = 1.771 in.

Flange Moments MD = HD × hD = (821,900)(– 1.094) = – 899,200 in.-lb MG = HG × hG = (200,500)(3.729) = 747,700 in.-lb MT = HT × HT = (–466,400)(1.771) = – 826,000 in.-lb Mo = MD + MG + MT = – 977,500 in.-lb

Use the absolute value in the calculations. (6) Shape factors from Appendix 2 of VIII-1 for K are K = A/B′ = 26.5/13.25 = 2.0

From Fig. 2-7.1 of VIII-1; assuming µ = 0.3:


115

116

Chapter 4

T = 1.51

Z = 1.67

Y = 2.96

U = 3.26

αR = (1/K2) {1 + [3(K + 1)(1 – µ)]/πY} = 0.419 TR = [(Z + µ)(Z – µ)] αRT = 0.857 YR = αRY = 1.241 UR = αRU = 1.366 g1/go = 1.0 ho = (Ago)1/2 = [(26.5)(1.8125)]1/2 = 6.930 h/ho = ∞

From Appendix 2 of VIII-1: F = 0.909

V = 0.550

f = 1.0

e = F/ho = 0.909/6.930 = 0.131 d = (UR/V)hog2o = 56.543

(7) MGS = 2,098,000 in.-lb and Sfa = 20.0 ksi and Mo = 977,500 in.-lb and Sfo = 17.8 ksi. Since the moment at operating condition is less than 0.5 times the moment at gasket seating condition with a slightly less allowable stress, only the gasket seating condition is calculated. Assume a flange thickness of t = 4.0 in. L = {[(te + 1)/TR] + (t3/d)} = 1.778 + 1.131 = 2.909

Longitudinal hub stress: SH = fMGS/Lg21B′ = [(1)(2,098,000)]/[2.909)(1.8125)2(13.25)] = 16,570 psi



117

Radial flange stress: SR = {[(4/3)te + 1]MGS}/Lt2B′ = [(1.699)(2,098,000)]/[(2.909)(4)2(13.25)] = 5780 psi

Tangential flange stress: ST = [(YRMGS/t2B′) – ZSR(0.67te + 1)]/β = {[(1.241)(2,098,000)/(4)2(13.25)] – [(1.67)(5,780)(1.351)]}/(1.699) = 4610 psi

Combined stresses: 0.5(SH + SR) = 0.5(16,570 + 5,780) = 11,180 psi

0.5(SH + ST) = 0.5(16,570 + 4610) = 10,590 psi

Tangential flange stress at B′: S′T = (MGS/t2B′){Y – [2K2(0.67te + 1)/(K2 – 1)L]} = [(2,098,000)/(4)2(13.25)]{2.96 – [2(2)2(1.351)/(3)(2.909)]} = 17,040 psi

(8) Allowable stresses: SH ≤ 1.5Sf : 16,570 psi < 26,700 psi SR ≤ Sf : 5780 psi < 17,800 psi ST ≤ Sf : 4610 psi < 17,800 psi 0.5(SH + SR) ≤ Sf : 11,180 psi < 17,800 psi 0.5(SH + ST) ≤ Sf : 10,590 psi < 17,800 psi S′T ≤ Sf : 17,040 psi < 17,800 psi

4.6.2.2 Full-Face Gasket Flanges. Although Fig. 4.1 (p), Shows a flange with a full-face gasket which permits part of the gasket to lie outside the bolt circle, no design procedure exists for such a flange. This type of gasket may be used with either a loose or an integral flange. See Appendix D for blank fillin Sheets D.5 (Slip-on Flange with Full-Face Gasket) and D.6 (Welding Neck Flange with Full-Face Gasket). One of the basic differences with a full-face gasket is that a reverse moment is generated from that part of the gasket loading outside the bolt circle.


118

Chapter 4

Most often, a full-face gasket is used where the m and y factors are relatively low, so that the bolt loading is kept within acceptable limits. A full-face gasket design generally results in the total moments from gasket seating and from operation to be fairly low, and consequently, only a nominal flange thickness is required. However, bolt loads are usually higher. 4.6.2.3 Flat-Face Flange with Metal-to-Metal Contact Across the Face or at the Outer Edge. Appendix Y of VIII-1 contains rules for the design of a flat-face flange with metal-to-metal contact across the whole face or with a metal spacer added to the outer edge between pairs of flanges. Gasket loadings usually are small, as most gaskets are of the self-sealing type. In order to make an analysis easier, assemblies are classified and individual flanges are categorized. Once this is established, the rules for analysis are given in VIII-1. Classification of Assemblies Class 1: A pair of flanges which are identical except for the gasket groove Class 2: A pair of nonidentical flanges in which the inside diameter of the reducing flange exceeds half the bolt circle diameter Class 3: A flange combined with a flat head or a reducing flange with an inside diameter that is small and does not exceed half the bolt circle diameter Categories of Flanges Category 1: An integral flange or an optional flange calculated as an integral flange Category 2: A loose-type flange with a hub that is considered to add strength Category 3: A loose-type flange with or without a hub—or an optional type calculated as a loose type— where no credit is taken for the hub The analysis of an Appendix Y flange is similar to that made for an Appendix 2 flange, except for the additional load and moment caused by the contact or prying effect. The contact force, HC, and its moment arm, hc involve an interaction between the bolt elongation and the flange deflection and the moments from the bolt loading and pressure loading. The bolt loading for the operating condition is Wm1 = H + HG + HC

(4.17)

4.7 SPHERICALLY DISHED COVERS Rules are given in Appendix 1-6 of VIII-1 for designing spherically dished covers with a bolting ring, acting in the same manner as a flange ring, attached integrally to a segment of a sphere. The equations given in VIII-1 are approximate and may be conservative because they do not take into account the discontinuity condition which exists at the intersection of the ring and head and that would distribute forces and moments between the two parts relative to their resistance (stiffness). The Code procedures and equations assume the entire loadings at the intersection are taken by the bolting ring alone. The rules of VIII-1 permit the use of a discontinuity analysis, should the designer so choose.

4.7.1 Definitions and Terminology Symbols and terms used for spherically dished covers are t = minimum required thickness of the spherical head segment, in. L = inside spherical or crown radius, in.



119

P = internal design pressure or MAWP, psi S = maximum allowable tensile stress value, psi T = flange thickness, in. Mo = total moment applied to the ring, in.-lb β1 = angle formed by the tangent to the center line of the dished cover thickness at its point of intersection with the flange ring β1 = arc sin [B/(2L + t)] A = outside diameter of the flange, in. B = inside diameter of the flange, in. C = bolt circle diameter, in.

4.7.2 Types of Dished Covers Spherically dished covers may be either one piece, where the ring and head are one continuous thickness of plate, or two pieces, where the ring and head are separate pieces which are welded together (no joint efficiency is required) as shown in Fig. 4.3. 4.7.2.1 Ring and Head of Uniform Thickness.

This is the type of cover shown in Fig. 4.3(b).

(a) Head thickness is t = (5PL)/(6S)

FIG. 4.3 SPHERICALLY DISHED COVERS WITH BOLTING FLANGES (ASME VIII-1)


(4.18)

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Chapter 4

(b) Flange ring thickness using a ring gasket is T = {[Mo/SB]/[(A + B)/(A – B)]}1/2

(4.19)

(c) Flange ring thickness using a full face gasket is T = 0.6{[P/S]/[B(A + B)(C – B)/(A – B)]}1/2

4.7.2.2 One Piece With Uniform Thickness.

(4.20)

This is the type of cover shown in Fig. 4.3(c).

(a) Head thickness is t = (5PL)/6S

(4.21)

(b) Flange ring thickness using a ring gasket and round bolt holes is T = Q + {[1.875Mo(C + B)]/[SB(7C – 5B)]}1/2

(4.22)

Q = (PL/4S)[(C + B)/(7C – 5B)]

(4.23)

where

(c) Flange ring thickness using a ring gasket and bolt holes slotted through the edge is T = Q + {[1.875Mo(C + B)]/[SB(3C – B)]}1/2

(4.24)

Q = (PL/4S)[(C + B)/(3C – B)]

(4.25)

where

(d) Flange ring thickness using a full-face gasket and round bolt holes: T = Q + {Q2 + [3BQ(C – B)/L[}1/2

(4.26)

Q = (PL/4S)[(C + B)/(7C – 5B)]

(4.27)

where

(e) Flange ring thickness using a full-face gasket and bolt holes slotted through the edge T = Q + {Q2 + [3BQ(C – B)/L]}1/2

(4.28)

Q = (PL/4S)[(C + B)/(3C – B)]

(4.29)

where

4.7.2.3 Ring and Head.

This is the type of cover shown in Fig. 4.3(d).

(a) Head thickness: t = (5PL)/(6S)


(4.30)


121

(b) Flange ring thickness: The flange ring thickness is determined by combining the circumferential ring stress and the tangential bending stress, as follows: (1) Circumferential ring stress = Sc = PD/2T

(4.31)

(2) Tangential ring stress = St = YMo/T2B

(4.32)

(3) Total ring stress = ST = Sc + St = (2S/T)(F) + (S/T2)(J)

(4.33)

(4) Rearranging terms and solving for the flange-ring thickness, T,

T = F + (F2 + J)1/2

(4.34)

F = [PB(4L2 – B2)1/2]/[8S(A – B)]

(4.35)

J = [Mo(A + B)]/[SB(A – B)]

(4.36)

where

and

These rules may be used for either internal pressure or external pressure. The term P is an absolute value for either internal or external pressure. The value of Mo is determined by combining the moments from bolt loading and gasket loading with the moment caused by the internal pressure loading at the head-to-ring intersection. When Mo is used in the equations, the absolute value is used. Example 4.8 shows the analysis of a spherically-dished cover of the ring-and-head type that matches the flange in Example 4.6. Example 4.8 Problem A spherically-dished cover of the type shown in Fig. E4.8 is to be attached to the flange described in Example 4.6. Determine the minimum required thickness of the head and flange ring. There is no corrosion allowance, and no joint efficiency is required. The flange material is SA-105, and the head material is SA-516 Gr. 70. The dish radius is 0.9 I.D. Solution (1) The allowable tensile stress for SA-516 Gr. 70 at 650°F is Sa = 18.8 ksi and for SA-105 at 100°F is Sfa = 20.0 ksi and at 650°F is Sfo = 17.8 ksi. (2) The dish radius, L, of the spherical head segment is L = 0.9B = 0.9(10.75) = 9.675 in.

(3) The minimum required thickness of the head segment, using Eq. (4.17), is th = 5PL/6Sa = (5)(2,000)(9.675)/(6)(18,800) = 0.858 in.

Assume that the thickness of the head is Th = 1.0 in.


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FIG. E4.8 EXAMPLE PROBLEM OF SPHERICALLY DISHED COVER, DIV. 1

(4) The head-to-ring angle, β1, using the equation given in VIII-1, 1-6(b) is β1 = arc sin[B/(2L + t)] = arc sin{10.75/[2(9.675) + 1]} = 31.89°

(5) The total flange moment for the gasket seating condition is the same as in Step 4 of Example 4.6. MGS = 2,098,000 in.-lb

(6) The total flange moment for the operating condition is: Flange Loads Same flange loads as in Example 4.6, plus an additional load occurs from the horizontal component, Hr, due to the internal pressure load on the spherical head. HD = 181,500 in.-lb HG = 200,500 in.-lb HT = 174,000 in.-lb

Hr = HD cot β1 = (181,500)(1.607) = 291,700 in.-lb



123

Lever Arms hD = 0.5(C – B) = (22.5 – 10.75) = 5.875 in. hG = 3.729 in. hT = 4.802 in.

hr is obtained by trial, using an assumed flange thickness and the perpendicular head thickness at the headto-ring intersection. The thickness of the head parallel to the flange ring face is: tp = th sec β1 = (1)(1.1778) = 1.178 in. hr = 0.5(T – tp)

Assume that the thickness of the flange ring is T = 5.375 in. hr = 0.5(5.375 – 1.178) = 2.099 in.

Flange Moments MD = HD × hD = (181,500)(5.875) = 1,066,000 in.-lb MG = HG × hG = (200,500)(3.729) = 747,700 in.-lb MT = HT × hT = (174,000)(4.802) = 835,500 in.-lb Mr = Hr × hr = (291,700)(2.099) = 612,300 in.-lb

Mo = MD + MG + MT – Mr = 2,037,000 in.-lb

(7) The minimum required thickness is the larger of the thicknesses determined for the gasket seating condition and for the operating condition by using the equations in Appendix 1-6(g)(2) as follows: For gasket seating condition:


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F = [PB(4L2 – B2)1/2]/[8Sa (A – B)] = {(2,000)(10.75)[4(9.675)2 – (10.75)2]1/2}/[8(20,000)(26.5 – 10.75)] = 0.137 J = [MGS/(Sa)(B)][(A + B)/(A – B)] = [2,098,000/(20,000)(10.75)][(26.5 + 10.75)/(26.5 – 10.75)] = 23.079 T = F + (F2 + J)1/2 = (.137) + [(.137)2 + 23.079]1/2 = 4.943 in.

For operating condition: F = [PB(4L2 – B2)1/2]/[8Sfo (A – B)] = {(2,000)(10.75)[4(9.675)2 – (10.75)2]1/2}/[8(17,800)(26.5 – 10.75)] = 0.154 J = [Mo/(Sfo)(B)][(A + B)/(A – B)] = [2,037,000/(17,800)(10.75)][(26.5 + 10.75)/(26.5 – 10.75)] = 25.177 T = F + (F2 + J)1/2 = (.154) + [(.154)2 + 25.177]1/2 = 5.174 in.

Since this is less than the 5.375 in. assumed, the thickness is acceptable. As with all calculated stress where a value of thickness, T, is assumed, a lesser value of required thickness may be determined by repeated assumptions of thickness and further calculations until the assumed thickness and the calculated thickness are the same.


CHAPTER

5 OPENINGS 5.1 INTRODUCTION Openings through the pressure boundary of a vessel require extra care to keep loadings and stresses at an acceptable level. Loads may be generated from both internal and external pressure and from applied external loadings. An examination of the pressure boundary may indicate that extra material is needed near the opening to keep stresses from loadings at an acceptable level. This may be provided by increasing the wall thickness of the shell or nozzle or by adding a reinforcement plate around the opening. At some openings, there may be a nozzle to which is attached external piping generating external forces and moments from dead loads or thermal expansion. At other openings only a blind flange or flat cover with little or no available reinforcement may exist. In designing openings, two types of stresses are important: primary stresses, including both primary membrane stress and primary bending stress; and peak stresses for fatigue evaluation. Although UG-22 of VIII-1 and AD-110 of VIII-2 require that both types be considered when evaluating loadings, in VII-1 rules are given only for calculating the primary membrane stresses.

5.2 CODE BASES FOR ACCEPTABILITY OF OPENING VIII-1 gives two methods for examining the acceptability of openings in the pressure boundary for pressure loading only. Other loadings shall be considered separately. The first method, the reinforced opening or area replacement method, is used when that area which was to carry the primary membrane stress is missing due to the opening. To replace this area, close-in substitute areas are called upon to carry the stress. The second method is called the ligament efficiency method. This method examines the area of metal remaining between adjacent openings compared with the area of metal that was there before the openings existed. The primary membrane stress and shear stress are then examined for acceptability. Curves have been developed to simplify this examination. For single openings, only the reinforced opening method is used, while for multiple openings, either the reinforced opening method or the ligament efficiency method may be used. Although the reinforced opening method and the ligament efficiency method are not developed on the same basis, VIII-1 permits either one to be used. It is appropriate to use whichever method is more liberal, that is, the method giving the lower value for the increase in thickness. Consequently, both methods may require examination. Article D-5 of VIII-2 contains reinforced opening rules for a satisfactory design for pressure loading only when a fatigue analysis is not required. It does not contain provisions for added loadings from piping or other loadings which may be imposed. In lieu of meeting the reinforced opening rules of Article D-5, an opening may be considered subject to the requirements of Appendices 4 and 5 of VIII-2.

125


126

Chapter 5

5.3 TERMS AND DEFINITIONS Many terms and definitions used for openings, reinforcements, and ligaments are the same for VIII-1 and VIII-2. Some of the most common terms and definitions are given below, while others are given where they are used. d = diameter or chord of the opening in the plane being examined d1, d2, dn = diameters or chords of various or adjacent openings D = inside diameter of the cylindrical shell Do = outside diameter of the cylindrical shell tn = nominal wall thickness of the nozzle t = nominal wall thickness of the shell te = height of the reinforcement base (see Fig. UG-40 of VIII-1)

5.4 REINFORCED OPENINGS—GENERAL REQUIREMENTS 5.4.1 Replacement Area 5.4.1.1 Design for Internal Pressure. When there is an opening through the shell, except for flat heads, primary membrane stresses which develop from the pressure loading over the area formed by the opening diameter and the minimum required thickness are interrupted. A substitute pathway is required. For flat heads, the situation is similar, except primary bending stresses are interrupted. The assumption is made that since primary bending stresses are maximum at the surfaces and zero at the centerline of the thickness while primary membrane stresses are uniform across the wall thickness, the replacement area for flat heads needs to be only half the area required for cylindrical shells and formed heads. The method presented for determining any needed reinforcement examines the region around the opening for available areas to carry the primary stress around the opening. Since stress is related to the load and crosssectional area, areas are substituted when making calculations. Placement and location of the replacement area is important. The replacement area should be close to the opening; but care should be taken, if temperature is a consideration, not to generate an area of high thermal stresses. If it is not too difficult to place some of the replacement area inside as well as outside of the vessel wall, try to place about two-thirds of the replacement area on the outside and one-third on the inside. 5.4.1.2 Design for External Pressure. Although the procedure for evaluating stresses for external pressure is based on a buckling and stability analysis, the method for determining the reinforcement requirements for openings in shells under external pressure is very similar to that for shells under internal pressure, but with the following changes: (a) The minimum required thickness of the shell is based on the external pressure requirements and may be called tre instead of tr. (b) The replacement area required is 50% of that required for internal pressure.

5.4.2 Reinforcement Limits The stress analysis basis used in the ASME Code to analyze the nozzle reinforcement is called Beams on Elastic Foundation (Hetenyi, 1946). This method determines the effectiveness of the material close to the opening for carrying loads. Reinforcement limits are developed parallel and perpendicular to the shell surface near the opening. Although the method is a simplified application of the elastic foundation theory, experience has shown that it does a good job. Values from two equations are used to set the reinforcement limits measured along the vessel wall surface. The greater value sets the horizontal limit for that opening. The first value is equal to d, and the second value


Openings

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FIG. 5.1 REINFORCEMENT LIMITS PARALLEL TO SHELL SURFACE

is equal to 0.5d + t + tn as shown in Fig. 5.1. The relationship of the nozzle wall thickness compared to the opening diameter or chord dimension, as appropriate, usually decides which of the two values controls. For Section VIII-1, the reinforcement limits measured perpendicular to the shell surface are also set by two limits; however, in this case, the smaller value is used. Using the beam on elastic foundation theory for a cylindrical shell, the damping wavelength is a function of 1/β, where β = 1.285/(rt)1/2 for µ = 0.3. When this vertical limit was set by the ASME Code committee years ago, the assumption was made that r/t of 10 was appropriate. This gave


128

Chapter 5

L = 1/β = (rt)1/2/1.285 = (0.1r2)1/2/1.285 = 0.246r = 2.46t

(5.1)

where r and t are basic terms of nominal dimensions related to either the shell or nozzle. For application in various Code sections, the value was rounded to 2.5t. For VIII-1, it remains the smaller of 2.5T or 2.5Tn where T is the nominal shell or head thickness and Tn is the nominal nozzle wall thickness. For VIII-2, instead of assuming a fixed value of r/t, the actual values of r and t are used to set the vertical limits which depend on the nozzle details, as shown in Fig. AD-540.1 of VIII-2.

5.5 REINFORCED OPENING RULES, VIII-1 In calculating the nozzle reinforcement requirements for VIII-1, it should be recognized that many of the requirements were developed years ago, based on the information available at that time. However, engineering experience and additional data have shown that these rules are satisfactory for most designs and so are still used. As mentioned previously, reinforced opening rules are for pressure loading only. Other loadings that need to be evaluated shall be considered separately using such methods as those given in Welding Research Council Bulletin No. 107, among others.

5.5.1 Openings with Inherent Compensation Openings in vessels which are not subjected to rapid fluctuations in pressure do not require reinforcement calculations [UG-36(c)(3)] if the following dimensional requirements are met: (1) When using welded and brazed nozzles with a diameter not larger than: (a) 3-1/2 in. diameter in a plate with a thickness ≤ 3/8 in. (b) 2-3/8 in. diameter in a plate with a thickness > 3/8 in. (2) When using threaded, studded, or expanded nozzles with a diameter not larger than (a) 2-3/8 in. diameter in all plate thicknesses. (3) When two openings are used, their centers shall be no closer than (d1 + d2). (4) When two openings in a cluster of three or more are used, their centers shall be no closer than: (a) For cylinders and cones: (1 + 1.5 cos j) (d1 + d2) (5.2) (5.3) (b) For double-curved shells and heads: 2.5(d1 + d2) where d1, d2 = diameter of adjacent openings φ = angle between the line connecting the centerlines of the two openings being considered and the longitudinal axis.

5.5.2 Shape and Size of Openings 5.5.2.1 Shape of Opening. Openings in cylindrical shells and formed heads are usually circular, elliptical, or obround. The latter shape is often developed for a nonradial nozzle opening. However, any other shape is also permitted, but there may be no method of analysis given in the Code. 5.5.2.2 Size of Opening. For openings in a cylindrical shell, the rules given in UG-36 through UG-42 of VIII-1 are limited to the following sizes:


Openings

129

1. In shells 60 in. and less in diameter, the opening shall not exceed 0.5D or 20 in. 2. In shells over 60 in. in diameter, the opening shall not exceed 0.33D or 40 in. When the size of the opening meets these limits, the rules given in UG-36 through UG-42 and in sections 5.5.3 through 5.5.5 (below) shall be used. When the size of the opening exceeds these limits, the rules given in Appendix 1–7 of VIII-1 and in section 5.5.6 shall be met as well.

5.5.3 Area of Reinforcement Required 5.5.3.1 Opening in a Cylindrical Shell (Except Nonradial Hillside). The total cross-sectional area of reinforcement required for any plane through the center of the opening is determined by: A = dtrF

(5.4)

where tr = minimum required thickness of the seamless shell, based on the circumferential stress calculated by Eq. (2.1) F = correction factor to obtain minimum required thickness of the shell on the plane being examined f = 1.0, except for integrally reinforced openings listed in Fig. UW-16.1, where F = 0.5(cos2θ + 1) is permitted θ = angle of the plane being examined from the longitudinal plane The value of the F-factor vs. θ is plotted in Fig. 5.2. The F-factor corrects the minimum required thickness for all planes between 0° (the longitudinal plane) and 90° (the circumferential plane). This correction is necessary to adjust for a minimum required thickness when the plane being examined is somewhere between the longitudinal plane and the circumferential plane. 5.5.3.2 Opening in a Cylindrical Shell (Nonradial Hillside). The total cross-sectional area of reinforcement required for a plane through the center of the opening at a nonradial hillside nozzle is determined by: A = dtr

(5.5)

where d = chord length at the midsurface of the thickness required, excluding the excess thickness available for reinforcement tr = minimum required thickness of a seamless shell on the plane being examined If the longitudinal plane is being examined, the value of tr is determined by Eq. (2.1). If the circumferential plane is being examined, tr is determined by 0.5tr or by Eq. (2.4). 5.5.3.3 Opening in a Spherical Shell or Formed Head. The total cross-sectional area of reinforcement required for a plane through the center of an opening in a formed head is determined by A = dtr

where d = diameter or chord dimension of the opening tr = minimum required thickness of the spherical shell or formed head (1) When the opening and its reinforcement are entirely within the spherical part of a torispherical head [see Fig. 5.3(a)], tr is the minimum required thickness for a torispherical head using M = 1.


(5.6)

130

Chapter 5

FIG. 5.2 CHART FOR DETERMINING VALUE OF F FOR ANGLE Θ (ASME VIII-1 AND VIII-2)

(2) When the opening is in a cone or conical shell, tr is the minimum required thickness of a seamless cone of diameter D, measured where the nozzle centerline pierces the inside wall of the cone. (3) When the opening and its reinforcement are in an ellipsoidal head and are located within a circle at the center of the head and the circle has a diameter equal to 0.8 shell diameter


Openings

131

FIG. 5.3 DETERMINATION OF SPECIAL LIMITS FOR SETTING TR FOR USE IN REINFORCEMENT CALCULATIONS

[see Fig. 5.3(b)], tr is the minimum required thickness of a seamless spherical shell of radius K1D, where D is the shell diameter and K1 is given in Table UG-37 of VIII-1.

5.5.4 Limits of Reinforcement As described in 5.4.2, limits of reinforcement are determined in both the vertical and the horizontal direction. Excess cross-sectional area of material within these limits is available for reinforcement. 5.5.4.1 Parallel to Shell Surface. When the opening dimensions are within the limits given in section 5.4.2, the horizontal limits are the greater of: (1) d or (2) 0.5d + t + tn. 5.5.4.2 Perpendicular to Shell Surface. tical limits are the smaller of:

When the opening is within the limits in section 5.4.2, the ver-


132

Chapter 5

(1) 2.5t or (2) 2.5tn + te.

5.5.5 Area of Reinforcement Available When the reinforcing limits do not extend outside of an area where the required thickness and limits are available equally on each side of the opening centerline, the following equations may be used to determine the area of reinforcement available: (1) Area available in vessel wall, A1, is the larger of: A1 = (2d – d)(Et – Ftr)

(5.7)

or A1 = [2(0.5d + t + tn) – d](Et – Ftr)

(5.8)

(2) Area available in nozzle wall, A2, is the smaller of: A2 = (5t)(tn – trn)

(5.9)

or A2 = (5tn + 2te)(tn – trn)

(5.10)

Example 5.1 Problem Using the rules of VIII-1, determine the reinforcement requirements for an 8 in. I.D. nozzle which is centrally located in a 2:1 ellipsoidal head, as shown in Fig. E5.1. The nozzle is inserted through the head and attached by a full penetration weld. The inside diameter of the head skirt is 41.75 in. The head material is

FIG. E5.1 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT IN ELLIPSOIDAL HEAD, DIV. 1


Openings

133

SA-516 Gr. 70, and the nozzle material is SA-106 Gr. C. The design pressure is 700 psi, and the design temperature is 400°F. There is no corrosion allowance, and the weld joint/quality factor efficiency is 1.0. Solution (1) The allowable tensile stress for both SA-516 Gr. 70 and SA-106 Gr. C at 400°F is 20.0 ksi. Therefore, fr = 1.0. (2) Using UG-32(d), the minimum required thickness of a 2:1 ellipsoidal head without an opening is: tr = (PD)/(2SE – 0.2P) = (700 × 41.75)/(2 × 20000 × 1.0 – 0.2 × 700) = 0.733 in. Nominal thickness used is 0.75 in.

(3) According to Rule (3) of tr in UG-37(a), when an opening and its reinforcement are in an ellipsoidal head and are located entirely within a circle the center of which coincides with the head and the diameter is equal to 80% of the shell diameter, tr is the thickness required for a seamless sphere of radius K1D, where D is the shell I.D. and K1 is 0.9 from Table UG-37 of VIII-1. For this head, the opening and its reinforcement shall be within a circle with a diameter of 0.8D = (0.8)(41.75) = 33.4 in. (4) The radius is R = K1D = 0.9(41.75) = 37.575 in.

This radius is used in UG-32(f) to determine the tr for reinforcement calculations as: tr = (PR)/(2SE – 0.2P) = [(700)(37.575)]/[2(20,000)(1.0) – 0.2(700)] = 0.625 in.

(5) Using UG-27(c)(1), the minimum required nozzle thickness is: trn = (PRn)/(SE – 0.6P) = [(700)(4)]/[(20,000)(1.0) – 0.6(700)] = 0.143 in. Nominal thickness used is 1.125 in.

(6) Limit parallel to head surface: X = d or (0.5d + t + tn), whichever is larger. = 8 in. or (4 + .75 + 1.125 = 5.875 in.)

Use X = 8 in. (7) Limit perpendicular to head surface: Y = 2.5t or 2.5tn, whichever is smaller. = 2.5(.75) = 1.875 in. or 2.5(1.125) = 2.81 in.

Use Y = 1.875 in.


134

Chapter 5

(8) Size limit of the opening is 2X = 2(8) = 16 in.

This is less than the limit of 33.4 in. determined in (3). Therefore, the provision to use the spherical head rule is valid. (9) Reinforcement area required by UG-37(c) of VIII-1 is Ar = dtrF + 2tntrF(1 – fr) = (8)(.625)(1) + 0 = 5.00 in.2

(10) Reinforcement area available in the head is A1 = d(Et – Ftr) – 2tn(Et – Ftr)(1 – fr1)

When fr1 = 1.0, the second term becomes zero. Therefore, A1 = d(t – tr) = (8)(.75 – .625) = 1.00 in.2

(11) Reinforcement area available in nozzle is A2 = 2Y(tn – tm) = (2)(1.875)(1.125 – .143) = 3.68 in.2

(12) Reinforcement area available in fillet welds is: A4 = 2(.5) tw2 = 2(.5)(.75)2 = 0.56 in.2

(13) Total reinforcement area available in head, nozzle, and welds is: AT = A1 + A2 + A4 = 1.00 + 3.68 + .56 = 5.24 in.2

Area available of 5.24 in.2 is larger than the area required of 5.00 in.2 (14) Determination of weld strength and load paths: According to UW-15(b), weld strength and load path calculations for pressure loading are not required for nozzles which are like the one shown in Fig. UW-16.1(c) of VIII-1. Since this nozzle is similar to that detail, no calculations are required. Example 5.2 Problem Using the rules of VIII-1, determine the reinforcement requirements for a 12 in. × 16 in. opening for a manway as shown in Fig. E5.2. The manway forging is inserted through the vessel wall and attached by a full penetration weld. The 12 in. dimension lies along the longitudinal axis of the vessel. The manway cover seals


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135

FIG. E5.2 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF 12 IN. × 16 IN. MANWAY OPENING, DIV. 1 against the outside surface of the manway forging. The I.D. of the shell is 41.875 in. The shell material is SA-516 Gr. 70 and the manway forging is SA-105. The design pressure is 700 psi, and the design temperature is 400°F. There is no corrosion allowance, and all joint efficiencies/quality factors are E = 1.0. Solution (1) The allowable tensile stress for both SA-516 Gr. 70 and SA-105 at 400°F is 20.0 ksi. Therefore, fr = 1.0. (2) Using UG-27(c)(1), the minimum required thickness of the shell is: tr = (PR)/(SE – 0.6P) = [(700)(20.9375)]/[(20,000)(1.0) – 0.6(700)] = 0.749 in. Nominal thickness used is 0.75 in.

(3) The manway forging is elliptical. Since there are no equations for determining the minimum required thickness of an elliptical shell in VIII-1, the rules of U-2(g) are followed. For an elliptical shell, equation (2.41) for minimum required thickness is given in section 2.6.2 of this book. The maximum value of minimum required thickness is used for all planes as follows: trn = Pa2b2/SE(a2 sin2 φ + b2 cos2 φ)3/2 = [(700)(8)2(6)2]/{(20,000)(1.0)[(8)2(1)2 + (6)2(0)2]3/2 = 0.373 in. Nominal thickness used is 1.375 in.


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(4) Examination of the longitudinal plane. (a)Limit parallel to shell surface whichever is larger. X = d or (0.5d + t + tn) = 12 in. or (6 + .75 + 1.375 = 8.125 in.)

Use X = 12 in. (b)Limit perpendicular to shell surface Y = 2.5t or 2.5tn, whichever is smaller. = 2.5(.75) = 1.875 in. or 2.5(1.375) = 3.437 in.

Use Y = 1.875 in. (c) Reinforcement area required by UG-37(c) of VIII-1 is Ar = dtrF + 2tntrF(1 – fr) = (12)(.749)(1.0) + 0 = 8.988 sq. in. when fr1 = 1.0.

(d) Reinforcement area available in the shell is: A1 = d(E1t – Ftr) – 2tn(E1t – Ftr)(1 – fr1)

When fr1 = 1.0, the second term becomes zero; therefore, A1 = (12)(.75 – .749) = 0.012 in.2

(e) Reinforcement area available in nozzle is: Outward:

A2 = 2Y(tn – trn) = 2(1.875)(1.375 – .373) = 3.758 in.2

Inward:

A3 = 2Y(tn) = 2(1.875)(1.375) = 5.156 in.2

(f) Reinforcement area available in fillet welds is: A4 = 2(.5)tw2 = 2(.5)(.75)2 = 0.562 in.2

(g) Total reinforcement area available from shell, nozzle, and welds is: AT = A1 + A2 + A3 + A4

= 0.012 + 3.758 + 5.156 + 0.562 = 9.488 in.2


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137

Area available of 9.488 in.2 is larger than required area of 8.988 in.2 (5) Examination of the circumferential plane. The opening has a 16 in. dimension on this plane, but F = 0.5 and fr = 1.0. (a) Reinforcement area required by UG-37(c) of VIII-1 is: Ar = dtrF = (16)(.749)(0.5) = 5.992 in.2

(b) Total reinforcement area available from shell and nozzle is: AT = > 5.992 in.2

Area available of >9.488 in.2 is larger than area required of 5.992 in.2 Note that the increase in diameter from 12 in. to 16 in. would increase the limit parallel to the shell surface, X. However, since AT is larger than Ar, the design is satisfactory and there is no need to consider the increased limit in our evaluation of this design. (6) Determination of weld strength and load paths. According to UW-15(b), weld strength and load path calculations for pressure loading are not required for nozzles like the one shown in Fig. UW-16.1(c) of VIII-1. Since this nozzle is similar to that one, no calculations are required. Example 5.3 Problem Using the rules of VIII-1, determine the reinforcement requirements for a 5.625 in. I.D. nozzle which is located on a hillside or non-radial position on the circumferential plane as shown in Figs. E5.3.1 and E5.3.2. The nozzle wall abuts the vessel wall and is attached by a full penetration weld. The I.D. of the shell is 41.875 in. The shell material is SA-516 Gr. 70, and the nozzle material is SA-106 Gr. B. The design pressure is 700

FIG. E5.3.1 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF HILLSIDE NOZZLE, DIV. 1


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Chapter 5

FIG. E5.3.2 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF HILLSIDE NOZZLE, DIV. 1

psi, and the design temperature is 400°F. There is no corrosion allowance, and all joint efficiency/quality factors are E = 1.0. Solution (1) At 400°F, the allowable tensile stress for SA-516 Gr. 70 is 20.0 ksi and for SA-106 Gr. B, it is 17.1 ksi. Therefore, fr = 17.1/20.0 = 0.855 (2) Using UG-27(c)(1), the minimum required thickness of the shell is: tr = (PR)/(SE – 0.6P) = [(700)(20.9375)]/[(20,000)(1.0) – 0.6(700)] = 0.749 in. Nominal thickness used is 0.75 in.

(3) Using UG-27(c)(1), the minimum required thickness of the nozzle is: trn = (PRn)/(SE – 0.6P) = [(700)(2.8125)]/[(17,100)(1.0) – 0.6(700)] = 0.118 in. Nominal thickness used is 1.5 in.

(4) Examination of the longitudinal plane: (a) Although allowable stresses of the shell and nozzle are different, since the nozzle weld is a full penetration weld abutting the shell wall, fr = 1.0 for both Ar and A1 and fr = 0.855 for A2. (b) Limit parallel to shell surface: X = d or (0.5d + t + tn), whichever is larger. = 5.625 in. or (2.81 + .75 + 1.5 = 5.062 in.)

Use X = 5.625 in.


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139

(c) Limit perpendicular to shell surface: Y = 2.5t or 2.5tn, whichever is smaller. = 2.5(.75) = 1.875 in. or 2.5(1.5) = 3.75 in.

Use Y = 1.875 in. (d) Reinforcement area required according to VIII-1, UG-37(c) is: Ar = dtrF = (5.625)(.749)(1.0) = 4.213 in.2

(e) Reinforcement area available in the shell is: A1 = (2X – d)(t – tr) = (11.25 – 5.625)(.75 – .749) = 0.005 in.2

(f) Reinforcement area available in the nozzle is: A2 = 2Y(tn – trn)(fr) = 2(1.875)(1.5 – .118)(.855) = 4.431 in.2

(g) Reinforcement area available in fillet welds is: A4 = 2(.5)tw2 = 2(.5)(.75)2 = 0.562 in.2

(h) Total reinforcement area available from shell, nozzle, and welds is: AT = A1 + A2 + A4 = .005 + 4.431 + .562 = 4.998 in.2

Area available of 4.998 in.2 is larger than area required of 4.213 in.2 (5) Examination of the circumferential plane: (a)Since this is a nonradial plane, it is necessary to determine the chord length measured diagonally across the opening (chord length 1–2 in Fig. E5.3.2) based on the midpoint of the minimum required thickness of the shell, tr = 0.749 in. Based on the geometry, chord length 1–2 = 12.217 in. Therefore, d′ = 12.217 in. and F = 0.5 on the circumferential plane. (b) Reinforcement area required using the chord length 1–2 is: A′r = d′ trF = (12.217)(.749)(0.5) = 4.575 in.2

(c) Based on the longitudinal plane, total reinforcement area available from shell and head is: AT = 4.998 in.2


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Area available of 4.998 in.2 is larger than the required area of 4.575 in.2 Since AT based on the longitudinal plane is larger than A′r even without any consideration of an increase in the limit parallel to the shell surface due to the increased d′, the design is satisfactory. (6) Determination of weld strength and load paths. According to UW-15(b), weld strength and load path calculations for pressure loading are not required for nozzles of the type shown in Fig. UW-16.1(a) of VIII-1. Since this nozzle is similar to that one in detail, no calculations are required. Example 5.4 Problem Using the rules of VIII-1, determine the reinforcement requirements for a 6.0 in. I.D. nozzle which is located in a cylindrical shell, as shown in Fig. E5.4. The nozzle abuts the vessel wall and is attached by a full penetration weld. The I.D. of the shell is 30 in. The shell material is SA-516 Gr. 60, and the nozzle is SA-106 Gr. B. The design pressure is 1000 psi, and the design temperature is 100°F. The corrosion allowance is 0.125 in., and all joint efficiency/quality factors are E = 1.0. When a corrosion allowance is considered, all calculations are based on the corrosion allowance being fully corroded away. Solution (1) The allowable tensile stress for SA-516 Gr. 60 and SA-106 Gr. B is 17.1 ksi. Therefore, fr = 17.1/17.1 = 1.0. (2) Using UG-27(c)(1), the minimum required thickness, tr, of the shell is: R = r + c.a. = 15.0 + 0.125 = 15.125 in.

FIG. E5.4 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT WITH CORROSION ALLOWANCE, DIV. 1


Openings

tr = (PR)/(SE – 0.6P) = [(1000)(15.125)]/[(17,100)(1.0) – .6(1000)] = 0.917 in. Tr = 0.917 + 0.125 = 1.042 in. Nominal thickness used is 1.125 in.

(3) Using UG-27(c)(1), the minimum thickness, tm, of the nozzle is: Rn = rn + c.a. = 3.0 + 0.125 = 3.125 in. tm = (PRn)/(SE – 0.6P) = [(1000)(3.125)]/[(17,100)(1.0) – 0.6(1.000)] = 0.189 in. Tm = 0.189 + 0.125 = 0.314 in. Nominal thickness used is 1.375 in.

(4) Limit parallel to the shell surface: X = d or (0.5d + t + tn), whichever is larger. = 6.25 in. or [3.125 + (1.125 – 0.125) + (1.375 – 0.125)] = 5.375 in.

Use X = 6.25 in. (5) Limit perpendicular to the shell surface: Y = 2.5t or 2.5tn, whichever is smaller. = 2.5(1.125 – 0.125) = 2.50 in. or 2.5(1.375 – 0.125) = 3.125 in.

Use Y = 2.50 in. (6) Reinforcement area required according to VIII-1, UG-37(c) is: Ar = dtrF = (6.25)(0.917)(1.0) = 5.731 in.2

(7) Reinforcement area available in the shell is: A1 = (2X – d)(t – tr) = (6.25)[(1.125 – 0.125) – 0.917] = 0.519 in.2

(8) Reinforcement area available in the nozzle is: A2 = 2Y(tn – tm) = (2)(2.5)[(1.375 – 0.125) – 0.189] = 5.305 in.2


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(9) Total reinforcement area available from shell and nozzle is: AT = A1 = A2 = 0.519 + 5.305 = 5.824 in.2

This is larger than area required of 5.731 in.2 (10) Determination of weld strength and load paths. According to UW-15(b), weld strength and load path calculations for pressure loading are not required for nozzles of the type shown in Fig. UW-16.1(a) of VIII-1. Since this nozzle is similar to that one in detail, no calculations are required.

5.5.6 Openings Exceeding Size Limits of Section 5.5.2.2 When the opening diameter is large compared to the diameter of the shell or head in which this opening is located, experience shows that some of the reinforcing area should be placed close to the edge of the opening. Therefore, for such openings, the following special requirements shall be met in addition to those listed in sections 5.5.3, 5.5.4, and 5.5.5. 5.5.6.1 Area of Reinforcement Required. The total cross-sectional area of reinforcement required for any plane through the center of the opening is determined by: A = 0.67 dtrF

(5.11)

5.5.6.2 Limits of Reinforcement Parallel to Shell. The horizontal limits are the larger of: (1) 0.75d or (2) 0.5d + t + tn. 5.5.6.3 Limits of Reinforcement Perpendicular to Shell. The vertical limits are exactly the same as those given in section 5.5.4.2. 5.5.6.4 Stresses When Nozzle Radius/Shell Radius ≤ 0.7. In VIII-1, a method is provided for examining membrane and bending stresses in radial nozzle openings when the nozzle radius/shell radius ≤ 0.7. Other alternatives given in Code Case 2236 may also be used. This method is as follows: Membrane stress, Sm, and bending stress, Sb, for either Case A for a nozzle with a reinforcing pad or Case B for a nozzle with integral reinforcement are calculated and compared to the allowable stress value. (1) The membrane stress using the limits given in Fig. 5.4.1 is calculated for Eq. (5.12) for Case A or by Eq. (5.13) for Case B as follows: Case A:

Sm = P{R[Rn + tn + (Rmt)1/2] + Rn [t + te + (Rnmtn)1/2]}/As

(5.12)

Case B:

Sm = P{R[Rn + tn + (Rmt)1/2] + Rn [t + (Rnmtn)1/2]}/As

(5.13)

(2) The bending stress using the greater of the limits given in Fig. 5.4.1 or Fig. 5.4.2 is calculated by using Eqs. (5.14), (5.15), and (5.16) for both Cases A and B as follows: M = P{[(Rn)3/6] + RRne}

(5.14)

a = e + t/2

(5.15)

Sb = Ma/I

(5.16)


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FIG. 5.4.1 (ASME VIII-1)

FIG. 5.4.2 (ASME VIII-1)


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where in addition to the definitions in section 5.3, As = shaded area in Fig. 5.4.1 or Fig. 5.4.2, Case A or Case B, in.2 I = moment of inertia of the shaded area about the neutral axis, in.4 a = distance between the neutral axis of the shaded area and the inside surface of the vessel wall, in. Rm = mean radius of the shell, in. Rnm = mean radius of the nozzle neck, in. e = distance between the neutral axis of the shaded area and the midwall of the shell, in. Sm = membrane stress calculated by Eq. (5.12) or Eq. (5.13), psi Sb = bending stress calculated by Eq. (5.16), psi. (3) Calculated stresses are compared with allowable stress, S, as follows: Sm ≤ S Sm + Sb ≤ 1.5S

5.6 REINFORCED OPENING RULES, VIII-2 Although there is a close similarity between the reinforcement rules in VIII-1 and VIII-2, there are some differences. Reinforcement limits and spacing given in Article D-5 of VIII-2 are based upon the damping length of a beam on an elastic foundation using the actual dimensions. In lieu of using these rules, the rules of Appendix 4 and Appendix 5 of VIII-2 may be used.

5.6.1 Definitions Symbols and terms used in opening calculations in VIII-2 are d = diameter or chord dimension of the opening in the given plane, in. tr = minimum required thickness of the shell or head without the opening, in. F = 1.00 when the plane under consideration is in the spherical portion of a head or when the plane contains the longitudinal axis of a shell. For other planes through the shell, the value of F from Fig. 5.2 is used, except when a reinforcing pad is used. Rm = mean radius of the shell or head at the opening, in. t = nominal thickness of the shell or head at the opening, in. trn = minimum required thickness of the nozzle, in. tn = nominal thickness of the nozzle, in. rm = r + 0.5tn = mean radius of the nozzle, in. r2 = transition radius between the nozzle and the vessel, in. tp = nominal thickness of the connecting pipe, in. K = 0.73r2 when a transition radius r2 is used and the smaller of the two legs when a fillet weld transition is used, in. For h, L, x, and other dimensions, see Fig. 5.5 in this text and Fig. AD-540.1 of VIII-2.

5.6.2 Openings Not Requiring Reinforcement Calculations No reinforcement calculations are required when the following criteria are met for circular openings: (1) Single openings with dmax = 0.2(Rmt)1/2 and two or more openings within a circle with a diameter ≤ 2.5 (Rmt)1/2. The sum of the diameters is ≤ 0.25 (Rmt)1/2. (2) Center-to-center spacing ≥ 1.5(d1 + d2).


Openings

FIG. 5.5 NOZZLE NOMENCLATURE AND DIMENSIONS (DEPICTS GENERAL CONFIGURATIONS ONLY) (ASME VIII-2)


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Chapter 5

(3) Center-to-edge distance of another local stressed area, where PL is greater than 1.1 Sm, is equal to 2.5 (Rmt)1/2.

5.6.3 Shape and Size of Openings Openings are usually circular or elliptical or of a shape formed by the interaction of a circular or elliptical cross section with another surface. The limits given below also apply: (1) (2) (3) (a)

The ratio of the large to small dimension of the opening is limited to 1.5. The ratio d/D ≤ 0.50. The arc length between centerlines of openings is limited to no less than Three times the sum of the radii for formed heads and longitudinal axis of a cylindrical shell; (b) Two times the sum of the radii for the circumferential direction. (4) Rules shall be satisfied for all planes.

For an opening with a shape and size not within these limits, design-by-analysis shall be used.

5.6.4 Area of Reinforcement Required In determining the area of reinforcement required for an opening in an VIII-2 vessel, each opening shall be examined by two criteria: the entire area provided within limits and two-thirds of the area provided within more restrictive horizontal limits. The total cross sectional area of reinforcement required for any plane through the center of the opening is determined by: A = dtrF

5.6.5 Limits of Reinforcement 5.6.5.1 Parallel to the Shell Surface For 100% of the Required Reinforcement The horizontal limits are the greater of: (1) d or (2) 0.5d + t + tn. For 2/3 of the Required Reinforcement Area The horizontal limits are the greater of: (1) 0.5d + 0.5(Rmt)1/2 or (2) 0.5d + t + tn. 5.6.5.2 Perpendicular to the Shell Surface For the Nozzles Shown in Fig. 5.5(a) and (b): When h < 2.5tn + K, the perpendicular limits are the greater of


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(1) 0.5(rmtn)1/2 + K or (2) 1.73x + 2.5tp + K In either case, the limit shall not exceed either 2.5t or L + 2.5tp. When h ≥ 2.5tn + K, the perpendicular limits are the greater of (1) 0.5(rmtn)1/2 + K or (2) 2.5tn In either case, the limit shall not exceed 2.5t. For the Nozzle Shown in Fig. 5.5(c): When 45 deg. > θ ≥ 30 deg., the perpendicular limits are the greater of (1) 0.5(rmt′n)1/2 or (2) L′ + 2.5tp In either case, the limit shall not exceed 2.5t. When θ < 30 deg., the perpendicular limits are the greater of (1) 0.5(rmt′n)1/2 or (2) 1.73x + 2.5tp In either case, the limit shall not exceed 2.5t. For the Nozzle Shown in Fig. 5.5(d): The perpendicular limits are the greater of (1) 0.5(rmtn)1/2 + te + K or (2) 2.5tn + te + K In either case, the limit shall not exceed 2.5t. In all cases, te ≤ 1.5t and ≤ 1.73W, where W = width of the reinforcing element, in. te = thickness of the reinforcing element, in.

5.6.6 Available Reinforcement Metal contributing to the required area of reinforcement shall lie within the reinforcement limits given in section 5.6.5 and shall be limited to a material which meets the following criteria: (a) Area in excess of that required to carry the primary membrane stress, (b) Area of nozzle wall in excess of that required to carry the primary membrane stress, (c) Weld metal within the shell and nozzle wall which required PQR,


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(d) Full-penetration weld joining weld pad to nozzle neck, (e) Other weld areas meeting requirements of AD-570, (f) Metal meeting the following requirement: [(αr – αv) ∆T] ≤ 0.0008

(5.17)

where, αr = mean coefficient of expansion of the reinforcing metal, in./in.°F αv = mean coefficient of expansion of the vessel metal, in./in.°F ∆T = operating temperature range from 70°F to operating temperature, °F

5.6.7 Strength of Reinforcement Metal (a) Sn/Sv ≥ 0.8, (b) For Sn/Sv > 1.0, use Sn/Sv = 1.0 maximum where Sn = design stress intensity value of the nozzle material, ksi Sv = design stress intensity value of the vessel material, ksi

5.6.8 Alternative Rules for Nozzle Design An acceptable alternative to the regular reinforcement requirements may be used subject to special limitations and other reinforcement requirements. 5.6.8.1 Limitations (a) The reinforcement will have a circular cross section and be perpendicular to shell. (b) The reinforcement will have all integral construction using corner fillet radii.

5.6.8.2 Required Reinforcement Area, Ar. d/(Rtr)1/2 is: d/(Rtr)1/2 <0.20 ≥0.20 & <0.40 ≥0.40

The required minimum reinforcement area related to

In Cylinders None except r2 required {4.05[d/(Rtr)1/2]1/2 – 1.81}dtr 0.75dtr

In Spheres & Heads None except r2 required {5.40[d/(Rtr)1/2]1/2 – 2.41}dtr dtr cos φ where φ = sin–1(d/D)

5.6.8.3 Limits of Reinforcing Zone. Metal included in meeting the reinforcing area requirements shall be located in the zone boundary shown in Fig. 5.6. Example 5.5 Problem Using the rules of Article D-5 of VIII-2, determine the reinforcement requirements of an 8 in. I.D. nozzle which is centrally located in a 2:1 ellipsoidal head as shown in Fig. E5.5. The nozzle is inserted through the head and attached by a full penetration weld. The inside diameter of the head skirt is 41.75 in. The head material is SA-516 Gr. 70, and the nozzle material is SA-106 Gr. C. The design pressure is 700 psi, and the design temperature is 500°F. There is no corrosion allowance.


FIG. 5.6. LIMITS OF REINFORCING ZONE FOR ALTERNATIVE NOZZLE DESIGN (ASME VIII-2)

FIG. E5.5 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT IN ELLIPSOIDAL HEAD, DIV. 2


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Solution (1) The allowable stress intensity for SA-516 Gr. 70 is 20.5 ksi, and for SA-106 Gr. C it is 21.6 ksi. Since the nozzle material is stronger than the head material, no adjustment is required. (2) Using Fig. AD-204.1, the minimum required thickness of a 2:1 ellipsoidal head is: P/S = 700/20,500 = 0.034

which gives t/L = 0.021 L = 0.9D = 0.9(41.75) = 37.575 in. t = 0.021(37.575) = 0.789 in. Nominal thickness used is 1.0 in.

(3) Using AD-201(a), the minimum required nozzle thickness is: tm = PR/(S – 0.5P) = (700)(4)/[21,600 – 0.5(700)] = 0.132 in. Nominal thickness used is 1.125 in.

(4) Limits parallel to head surface are (a) For 100% of the required reinforcement: X = d or (0.5d + t + tn), whichever is larger = 8 in. or (4 + 1 + 1.25 = 6.125 in.)

Use X = 8 in. (b) For 2/3 of required reinforcement: X′ = r + 0.5(Rmt)1/2

or

(0.5d + t + tn), whichever is larger

= [4 + 0.5(38.075 × 1)1/2 = 7.085 in.] or 6.125 in.

Use X′ = 7.085 in. (5) Limits perpendicular to the head surface are calculated as shown below. (a) Determine which limits given in section 5.6.5.2 for Fig. 5.4(a) apply by: (i)

h < 2.5tn + K

or (ii)

h ≥ 2.5tn + K

(b) h = 2.50 in. 2.5tn = (2.5)(1.125) = 2.81 in.

K = 0.25 in. 2.5tn + K = 2.81 + 0.25 = 3.06 in.

Limit (5)(a)(i) applies where h < 2.5tn + K


Openings

(c) Determine the perpendicular limit as follows: Y = 0.5(rmtn)1/2 + k or 1.73x + 2.5tp + K, whichever is larger, but not more than 2.5t nor (L + 2.5tp) = 0.5[(4.5625)(1.125)]1/2 + 0.25 = 1.383 in. = 0 + 2.5(1.125) + 0.25 = 3.063 in. = 2.5(1) = 2.5 in. = (4) + 2.5(1.125) = 6.813 in. = 2.5 in.

(6) Reinforcement area required according to AD-520 of VIII-2 is: Ar = dtrF = (8)(0.789)(1.0) = 6.312 in.2 2/3Ar = 2/3(6.312) = 4.208 in.2

(7) Using the 100% limit, reinforcement area available in the head is A1 = (t – tr)(2X – d) = (1.0 – 0.789)(2 × 8 – 8) = 1.688 in.2

(8) Reinforcement area available in the nozzle is A2 = 2Y(tn – trn) = 2(2.5)(1.125 – 0.132) = 4.965 in.2

(9) With the 100% limit, total reinforcement area available in the head and nozzle is: AT = A1 + A2 = 1.688 + 4.965 = 6.653 in.2

Area available of 6.653 in.2 is larger than area required of 6.312 in.2 (10) With the 2/3 limit, reinforcement area available in the head is: A1 = (t – tr)(2X′ – d) = (1.0 – 0.789)(2 × 7.085 – 8) = 1.302 in.2

(11) With the 2/3 limit, total reinforcement area available in the head and nozzle is AT = A1 + A2 = 1.302 + 4.965 = 6.267 in.2

Area available of 6.267 in.2 is larger than the required area of 4.208 in.2


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FIG. E5.6 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF 12 IN. × 16 IN. MANWAY OPENING, DIV. 2

Example 5.6 Problem Using the rules of VIII-2, determine the reinforcement requirements for a 12 in. × 16 in. opening for the manway shown in Fig. E5.6. The manway forging is inserted through the vessel wall and attached by a full penetration weld. The 12 in. dimension lies along the longitudinal axis of the vessel. The manway cover seals against the outside surface of the manway forging. The I.D. of the shell is 41.875 in. The shell material is SA-516 Gr. 70, and the manway forging is SA-105. The design pressure is 700 psi, and the design temperature is 500°F. There is no corrosion allowance. Solution (1) The allowable stress intensity for SA-516 Gr. 70 is 20.5 ksi, and for SA-105 it is 19.4 ksi. An adjustment of fr = 19.4/20.5 = 0.946 is required for A2. (2) Using AD-201(a), the minimum required thickness of the shell is: tr = (PR)/(SE – 0.5P) = [(700)(20.9375)]/[(20500)(1.0) – 0.5(700)] = 0.728 in. Nominal thickness used is 1.0 in.

(3) The manway forging is elliptical. Since there are no equations for determining the minimum required thickness of an elliptical shell in VIII-2, one needs to be located. For an elliptical shell, Eq. (2.41) for minimum required thickness is given in section 2.6.2. The maximum value of the minimum required thickness is used for all planes being examined as follows:


Openings

trn = Pa2b2/SE(a2 sin2Φ + b2 cos2Φ)3/2 = [(700)(8)2(6)2]/{(19,400)(1.0)[(8)2(1)2 + (6)2(0)2]3/2} = 0.385 in. Nominal thickness used is 1.0 in.

(4) Examination of the longitudinal plane: (a) Determine the limit parallel to the shell surface. (i) For 100% of reinforcement area required: X = d or (0.5d + t + tn), whichever is larger = 12 in. or (6 + 1 + 1 = 8.0 in.)

Use X = 12 in. (ii) For 2/3 of reinforcement area required: X′ = r + 0.5(Rmt)1/2

or

(0.5d + t + tn), whichever is larger

= [6 + 0.5(21.4375 × 1)1/2] = 8.315 in. or 8.0 in.

Use 8.315 in. (b) Determine the limit perpendicular to the shell surface. (i) Determine which limits given in section 5.6.5.2 for Fig. 5.4(a) apply by (a)

h < 2.5tn + K

or h ≥ 2.5tn + K

(b)

(ii) h = 2.50 in. 2.5tn = (2.5)(1.0) = 2.50 in. K = 0.25 in. 2.5tn + K = 2.75 in.

Limit (5)(a)(i)(a) applies where h < 2.5tn + K (iii) Determine the perpendicular limits Y = 0.5(rmtn)1/2 + K

or

1.73x + 2.5tp + K, whichever is larger,

but not more than 2.5t nor (L + 2.5tp) = 0.5(6.625 × 1.0)1/2 + 0.25 = 1.537 in. = 0 + 2.5 × 1.0 + 0.25 = 2.75 in. = 2.5(1) = 2.50 in. = 3 + 2.5(1.0) = 5.50 in. = 2.50 in.


153

154

Chapter 5

(c) Reinforcement area required is For 100% reinforcement: Ar = dtrF + 2tntr(1 – fr) = (12)(0.728)(1) + 2(1.0)(0.728)(1 – 0.946) = 8.814 in.2

For 2/3 A: 2/3Ar = 5.876 in. (d) Using the 100% limit, reinforcement area available in the shell is: At = (t – tr)(2X – d) = (1 – 0.728)(2 × 12 – 12) = 3.264 in.2

(e) Reinforcement area available in the nozzle wall is: Outward:

A21 = 2(2.5)(1.0 – 0.385)(0.946) = 3.075 in.2

Inward:

A22 = 2(2.5)(1.0)(0.946) = 4.730 in.2

(f) Total reinforcement area available from the shell and nozzle is: AT = A1 + A21 + A22 = 3.264 + 3.075 + 4.730 = 11.069 in.2

Area available of 11.069 in.2 is larger than area required of 8.814 in.2 (g) Using the 2/3 limit, reinforcement area available in the shell is A1 = (1 – 0.728)(2 × 8.315 – 12) = 1.259 in.2

(h) With the 2/3 limit, total reinforcement area available from the shell and nozzle is AT = 1.259 + 3.075 + 4.730 = 9.064 in.2

Area available of 9.064 in.2 is larger than area required of 5.876 in.2 (5) Examination of circumferential plane: The opening has a 16 in. dimension on this plane, but F = 0.5. (a) Reinforcement area required according to AD-520 of VIII-2 is Ar = dtrF = (16)(0.728)(0.5) = 5.824 in.2

(b) Total reinforcement area available from the shell and nozzle is AT = >9.064 in.2

Area available of 9.064 in.2 is larger than area required of 5.824 in.2 Since area is acceptable without any adjustment of limits, the design is satisfactory.

5.7 LIGAMENT EFFICIENCY RULES, VIII-1 Section VIII-1 permits two methods for calculating the replacement metal removed at openings. They are the reinforced opening method and the ligament efficiency method. The ligament efficiency method considers


Openings

155

the metal area removed from the pressure boundary and the metal remaining between the two or more openings in the pressure boundary. No metal is considered from any nozzle attached at the opening. The ligament efficiency curves apply only to the cylindrical shell of a pressure vessel where the circumferential stress has twice the intensity of the longitudinal stress. Once this was established, Rankine’s Ellipse of Stress was used to determine the tension stress and the shear stress on any diagonal ligament plane. Using these data for tension and shear, curves were developed with respect to the longitudinal plane (circumferential stress) for various values of θ, p′ d, and t. Equations were developed to make the determination of the ligament efficiency an easy task. These equations and a plot of the curves are given in Figs. UG-53.5 and UG-53.6 of VIII-1. Calculations should be made using both the reinforced opening method and the ligament efficiency method; the thinner required plate thickness should be selected. Example 5.7 Problem Using the reinforcement requirements give in UG-37 through UG-42 of VIII-1, determine the minimum required thickness of a 36 in. I.D. cylindrical shell that has a series of openings in the pattern shown in Fig. E5.7. The openings are 2.50 in. diameter on a staggered pattern of three longitudinal rows on 3.0 in. circumferential spacing and 4.50 in. longitudinal spacing. The design pressure is 600 psi at the design temperature of 500°F. Shell material is SA-516 Gr. 70, and nozzle material is SA-210 Gr. C. There is no corrosion allowance. The openings are not located in or near any butt welded joint. Solution (1) The allowable tensile stress for both SA-516 Gr. 70 and SA-210 Gr. C at 500°F is 20.0 ksi. Therefore, fr = 1.0. (2) Using UG-27(c)(1) and assuming a seamless shell with E = 1.0, the minimum required thickness of the shell for reinforcement calculations is tr = (PR/(SE – 0.6P) = [(600)(18)]/[(20,000)(1.0) – 0.6(600)] = 0.550 in.

(3) For comparison with the ligament efficiency method, determine the reinforcement requirements based only on the shell thickness (without consideration of the nozzle thickness). Since the reinforcement area available comes only from the shell, the shell thickness will have to be increased. A trial thickness will be assumed and verified. Try t = 2tr = 2(0.550) = 1.100 in. Assume a nominal thickness of t = 1.25 in. (4) Limit parallel to the shell surface is X = d or (0.5d + t), whichever is larger. X = 2.50 in. or (1.25 + 1.25 = 2.50 in.). Use X = 2.50 in. (5) Examine the longitudinal plane, 1–2. (a) With an actual center-to-center spacing of 4.5 in., the reinforcing limits of 2X = 2(2.5) = 5.0 in. exceeds the actual spacing of 4.5 in. Therefore, the reinforcement limits overlap, and the rules of UG-42 apply. Those limits state that no reinforcement area shall be used more than once. (b) Reinforcement area required is: Ar = dtrF = (2.5)(0.550)(1.0) = 1.375 in.2

(c) Reinforcement area available in the shell is: A1 = (spacing – d)(t – tr) = (4.5 – 2.5)(1.25 – 0.550) = 1.400 in.2


156

Chapter 5

FIG. E5.7 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF SERIES OF OPENINGS, DIV. 1

(6) Examine the diagonal plane, 2–3. (a) With a circumferential spacing of 3 in. and a longitudinal spacing of 4.5 in., the diagonal center-tocenter spacing is p′ = [(3)2 + (2.25)2]1/2 = 3.75 in. And, Θ = tan–1 (3/2.25); Θ = 53.13°. With a diagonal spacing of 3.75 in., the reinforcement limits of 5.0 in. exceed the spacing. Therefore, the limits overlap, and the rules of UG-42 apply. From Fig. UG-37 of VIII-1, F = 0.68 for Θ = 53.13°. (b) Reinforcement area required is Ar = dtrF =(2.5)(0.550)(0.68) = 0.935 sq. in.

(c) Reinforcement area available in the shell is A1 = (spacing – d)(t – Ftr) = (3.75 – 2.5)(1.25 – 0.68 × 0.550) = 1.095 in.2

(7) Both the longitudinal and diagonal planes are satisfactory with t = 1.25 in., and the nozzle thickness is not considered. (8) As an alternative, the reinforcement requirements may be based on a combination of shell area and nozzle area. (9) As determined from VIII-1, UG-37(c)(l), the minimum required thickness of the nozzle with E = 1.0 is: trn = (PR)/(SE – 0.6P) = [(600)(1.25)]/[(20,000)(1.0) – 0.6(600)] = 0.038 in.

(10) With the spacing that close, it is doubtful that very much thickness over the minimum required thickness could be added and be able to weld the tubes. Therefore, most, if not all, of the reinforcement area will come from the shell.


Openings

157

Example 5.8 Problem Using the rules of VIII-1, determine the minimum required thickness of the same shell given in Example 5.7 using the ligament efficiency rules of section 5.7 instead of the reinforcement rules of section 5.5. Solution (1) Determine the minimum longitudinal ligament efficiency or equivalent longitudinal efficiency and compare it with the longitudinal butt joint efficiency. The lesser efficiency is used to calculate the minimum required thickness of the shell. (2) Determine the longitudinal ligament efficiency based on the longitudinal spacing of 4.5 in. as follows: E = (p – d)/p = (4.5 – 2.5)/(4.5) = 0.44

(3) Determine the equivalent longitudinal ligament efficiency from the diagonal efficiency using Fig. UG53.6 of VIII-1 as follows: p′ = p = 3.75 in.; Θ = 53.13°; d = 2.5 in.; p/d = 1.5, which gives E = 0.38.

(4) Using the minimum efficiency of E = 0.38 (which assumes that ligament efficiency is less than butt joint efficiency), calculate tr from UG-27(c)(l) as follows: tr = (PR)/(SE – 0.6P) = [(600)(18)]/[(20,000)(0.38) – 0.6(600)] = 1.492 in.



CHAPTER

6 SPECIAL COMPONENTS, VIII-1 6.1 INTRODUCTION To meet the design and loading requirements of UG-22 of VIII-1, many design equations, charts, tables, and curves for most standard components such as shells and heads are provided. However, there are some design rules for components with special geometries and configurations which require additional consideration. For VIII-2, procedures are given for design by analysis of components with special geometries. Consequently, very few rules for design of special geometries are given. This chapter contains guidance for the design of some of those special geometries in VIII-1 as follows: (1) (2) (3) (4)

Braced and Stayed Construction (UG-47 through UG-50, and UW-19) Jacketed Vessels (Appendix 9) Half-pipe Jackets (Appendix EE) Vessels of Noncircular Cross Section (Appendix 13)

6.2 BRACED AND STAYED CONSTRUCTION Rules for braced and stayed construction are in UG-47 through UG-50 and in UW-19 of VIII-1. Stays are used in pressure vessels to carry part or all of the pressure loading when it is desirable and possible to reduce the span and/or thickness of a tubesheet, sideplate, or other pressure component. Opposite stayed surfaces are “tied” together by staybolts, tubes, or baffles, which carry pressure loading as tension members. Depending on the number of ties, the thickness of braced and stayed surfaces may be less than when the surfaces are not stayed, because the loading is now resisted by both bending moments and bending strength and by tensile strength of the stays.

6.2.1 Braced and Stayed Surfaces For braced and stayed surfaces tied together with threaded-end or welded-in staybolts of uniform diameter and symmetrical spacing, the following formulas apply for determining the minimum required thickness or internal design pressure: t = p(P/SC)1/2

(6.1)

P = t2SC/p2

(6.2)

where t = minimum required thickness of the stayed plate, in. P = internal design pressure (or MAWP), psi S = maximum allowable stress value in tension at design temperature, psi 159


160

Chapter 6

p = maximum pitch between staybolts, in. C = constant, the value of which depends on details of the staybolt end design as follows: C = 2.1 for welded-in stays or threaded-end stays screwed through plates ≤ 7/16 in. thickness with the threaded ends riveted over; C = 2.2 for welded-in stays or threaded-end stays screwed into or through plates > 7/16 in. thickness with threaded-ends riveted over; C = 2.5 for threaded-end stays screwed through plates and fitted with single nuts outside the plate, or with inside and outside nuts without washers, and for stays screwed into plates not less than 1.5 times the diameter of the staybolt measured on the outside of the staybolt diameter. If washers are used, they shall be at least half as thick as the plate being stayed. C = 2.8 for threaded-end stays with heads not less than 1.3 times the diameter of the stays screwed through plates or made with a tapered fit and having the heads formed on the stays before installing them and with threaded-ends not riveted over; C = 3.2 for threaded-end stays fitted with inside and outside nuts and outside washer, where the diameter of washer is not less than 0.4p and the thickness of washer is not less than the thickness, t, of the surface being stayed. 6.2.1.1 Special Limitations for Threaded-End Stay Construction (a) Minimum thickness of plate to which stays can be attached, other than outer cylindrical or spherical plates, is 5/16 in. (b) When two plates are stayed together and only one requires staying, the C value is set by the plate requiring staying. (c) Maximum pitch for threaded-end staybolts is 8-1/2 in. (d) When the spacing is unsymmetrical due to interference, half of the spacing on each side of the stay being considered measured to the adjacent stay shall be used for loading.

6.2.1.2 Special Limitations for Welded-In Stay Construction (a) Required thickness of the plate shall not exceed 1 1/2 in. When plate thickness is greater than 3/4 in., the pitch shall be 20 in. or less. (b) The maximum pitch for welded-in stays is 15ds, where ds, is the diameter of the staybolt. (c) Welded-in stay details shall conform to one of those shown in Fig. 6.1. (d) Welds do not require radiography. (e) Welds may require postweld heat treatment; see US-40(f) of VIII-1.

6.2.1.3 Welded Stays for Jacketed Vessels. vessel meets the following criteria:

Welded stays, shown in Fig. 6.2, are permitted when the

Vessel design pressure is ≤ 300 psi. Required thickness of the plate does not exceed 1/2 in. Minimum fillet weld size is not less than the plate thickness. Allowable fillet weld load is calculated according to UW-18(d), and inside welds are visually examined before assembly. (e) Maximum diameter or width of the hole in the plate is 1 1/4 in. (a) (b) (c) (d)

6.2.1.4 Welded Stays for Dimpled and Embossed Assemblies. Welded stays may be used in construction of a dimpled or embossed assembly where a dimpled or embossed plate is welded to another dimpled or embossed plate or to a plain plate and the following rules from Appendix 17 of VIII-1 are met: (a) A welded attachment is made by fillet weld around the edge of the opening, or when the plate thickness with the opening is ≤ 3/16 in. and the hole diameter is ≤ 1 in., the hole may be


Special Components, VIII-1 161

FIG. 6.1 TYPICAL FORMS OF WELDED STAYBOLTS (ASME VIII-1)

FIG. 6.2 TYPICAL WELDED STAY FOR JACKETED VESSEL(ASME VIII-1)

filled with weld metal. The allowable load for the weld shall be equal to the product of the thickness of the plate containing the opening, the perimeter of the opening, the allowable stress of the weaker of the plates being joined, and a fillet weld joint efficiency of 0.55. (b) When MAWP is determined by a UG-101 proof test of the dimpled or embossed assembly, a representative panel may be used that is rectangular with at least 5 pitches in each direction and not less than 24 in. in either direction.


162

Chapter 6

(c) For a plain plate welded with one of the methods listed below, the minimum required thickness or the maximum allowable working pressure shall be determined by Eq. (6.1) or (6.2) using a value of C = 3.0. The welding procedures are: (1) Resistance seam welding (2) gas tungsten arc seam welding without filler metal (3) plasma arc seam welding without filler metal (4) submerged-arc seam welding with filler metal (d) For a plain plate with other methods of welding than those listed above, the minimum required thickness or maximum allowable working pressure is calculated by using Eqs. (6.1) and (6.2) with the appropriate C value.

6.2.2 Stays and Staybolts 6.2.2.1 Load Carried by a Stay (a) The area supported by a stay is based on the full pitch dimensions with the area of the stay subtracted. The load carried by that stay is the product of the area supported by the stay times the internal design pressure (MAWP).

6.2.2.2 Minimum Required Area of a Stay (a) The minimum required area of a stay at its least cross section, the smaller of the area at the root of the threads, or at any lesser cross section, is obtained by dividing the load carried by a stay (from the calculation in section 6.2.2.1) by the allowable stress of the stay material at design temperature and multiplying this result by 1.10. (b) Stays made from two or more parts joined by welding shall have the minimum required area of cross section of a stay determined the same way as in (a), above, but using a butt-weld joint efficiency of 0.60.

6.2.2.3 Special Requirements for Threaded-End Stays (a) Stays screwed through a plate shall extend two threads minimum and shall be riveted over or upset, or they shall extend through with enough threads to be fitted with a threaded nut. (b) If the stay end is upset for threading, it shall be fully annealed.

Example 6.1 Problem Using the rules in UG-47 through UG-50 and Appendix 17 of VIII-1, determine the maximum allowable working pressure (MAWP) of a dimpled plate/plain plate assembly which is resistance seam welded on 5 in. centers. Both plates are 1/4 in. thick SA-285 Gr.A with a design temperature of 150°F and no corrosion allowance. Solution The allowable tensile stress for II-D for SA-285 Gr.A at 150°F is 12.9 ksi. Since the joint is resistance seam welded, according to section 6.2.1.4(c), C = 3.0. Using Eq. (6.2), P = [(0.25)2(12,900)(3)/(5)2] = 96.8 psi



Example 6.2 Problem A flat plate is stayed by welded-in stays. The design pressure is 115 psi at a design temperature of 100°F using SA-516 Gr. 60. Stays are located on 8 in. centers. There is no corrosion allowance. What is the minimum required thickness of the stayed plate? Solution (1) The allowable tensile stress from II-D for SA-516 Gr. 60 at 100°F is 17.1 ksi. (2) If the plate thickness is > 7/16 in., C = 2.2. If the plate thickness is ≤ 7/16 in., C = 2.1. Using C = 2.2 in Eq. (6.1), the minimum required thickness of the plate is t = (8)[(115)/(17,100)(2.2)]1/2 = 0.442 in.

The minimum required thickness of 0.442 in. is greater than 0.438 in. (7/16 in.) and C = 2.2 is the correct factor to use.

6.3 JACKETED VESSELS Jacketed vessels, as considered in Appendix 9 of VIII-1, applies also to the jacketed portion of the vessel, including the wall of the inner vessel and the wall of the jacket, the closure between the inner vessel and jacket, and other components, such as stiffeners, which carry pressure stresses. Jacketed vessels usually are chosen to provide a chamber or annulus region in which a liquid or gas under pressure or vacuum is used to heat and/or cool the inner vessel contents and to provide an insulation chamber. Half-pipe jackets attached around the outside of the vessel are considered separately, in section 6.5. Although pressure within the inner vessel or in the annulus may be equal to or less than 15 psi, where that pressure or vacuum combines with a pressure or vacuum within the inner vessel or annulus to produce a combined loading on the inner vessel wall or the jacket wall which is greater than the individual loading, the combined loading is considered within the scope of VIII-1. Table 6.1 shows various combinations of design pressure for the inner vessel and for the jacket with the actual pressure to be used for design of components.

6.3.1 Types of Jacketed Vessels Jacketed vessels are categorized into Types, which provides a way to assign closures and other design requirements. These Types are shown in Fig. 6.3 and are defined as follows: Type 1—Jacket of any length confined entirely to the cylindrical shell Type 2—Jacket covering a portion of the cylindrical shell and one head

TABLE 6.1 EXAMPLE OF PRESSURE USED FOR DESIGN OF COMPONENTS Design Pressure in Inner Vessel –15 +15 +200 +100

Design Pressure in Annulus +200 +100 –15 +15

Pressure Used for Design of Inner Vessel –215 +15 & –100 +215 +100 & –15

Pressure Used for Design of Jacket +200 +100 –15 +15


164

Chapter 6

FIG. 6.3 SOME ACCEPTABLE TYPES OF JACKETED VESSELS (ASME VIII-1)

Type 3—Jacket covering a portion of the head Type 4—Jacket with added stay or equalizer rings to the cylindrical shell portion to reduce effective length Type 5—Jacket covering the cylindrical shell and any portion of either head Any combination is permitted as long as the most stringent requirements are met.

6.3.2 Design of Closure Member for Jacket to Vessel 6.3.2.1 Nomenclature. ts trj trc tc tj

The symbols and terms used to design jacket closures are

= nominal thickness of the inner vessel wall, in. = minimum required thickness of the jacket wall, in. = minimum required thickness of the closure member, in. = nominal thickness of the closure member, in. = nominal thickness of the jacket wall, in.



tn = nominal thickness of the nozzle wall, in. r = corner radius of the torus closure, in. Rs = outside radius of the inner vessel, in. Rj = inside radius of the jacket, in. Rp = radius of the opening in the jacket at the penetration, in. P = design pressure in the jacket annulus, psi S = maximum allowable tensile stress value, psi j = jacket space, in. This is equal to the inside radius of the jacket minus the outside radius of the inner vessel. L = design length of the jacket section as follows: 1. Distance between the inner vessel tangent line plus one-third of the head depth if no stiffeners exist 2. Center-to-center distance between adjacent stiffening rings or jacket closure 3. Distance from first stiffening ring or closure to tangent line plus one-third of the head depth a, b, c, Y, and Z = minimum weld dimensions for the attachment of closure members, in. 6.3.2.2 Closure Design Details. Closure members between the jacket and inner vessel are designed as various combinations of simple cantilevers or guided cantilevers, depending on the rigidity of the attachment details. Specific design requirements shall be met depending upon the type of jacket and type of closure used. Some acceptable types of closure details are shown in Fig. 6.4. Table 6.2 gives the required closure member size and weld size details for the various types of jacket closures permitted. Closure Detail Dimension Requirements for Various Types of Jacket Closures: trc ≥ trj; r ≥ 3tc; trc ≤ 5/8 in.; Y ≥ 0.7tc trc ≥ trj; r ≥ 3tc; trc ≤ 5/8 in.; Y ≥ 0.83tc trc ≥ trj trc ≥ 0.707j(P/S)1/2; permits fillet weld with throat ≥ 0.7tc trc ≥ 0.707j(P/S)1/2; requires groove weld with throat of tc trc [from VIII-1, UG-32(g), Eq. (4)] ≥ trj; Θ ≤ 30° trc ≥ larger of 2trj or 0.707j(P/S)1/2; trj ≤ 5/8 in.; Y ≥ smaller of 0.75tc or 0.75ts; Z ≥ tj trc ≥ larger of 2trj or 0.707j(P/S)1/2; Y ≥ smaller of 1.5tc or 1.5ts trc ≥ 1.414[(PRsj)/S]1/2; j = {[2St2s)/(PRj)] – 0.5(ts + tj)}; Y ≥ smaller of 1.5tc or 1.5ts Meet the details and dimensions shown in the sketch trj ≤ 5/8 in. trj ≤ 5/8 in.; welds shall meet the details shown in Fig. 6.4, sketches (i-1) or (i-2), which is the same as Fig. 9-5 in VIII-1 [13] For conical and toriconical jackets shown in Fig. 6.4, sketches (f-1) through (f-3) and (g-1) through (g-6).

[1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

(6.3) (6.4) (6.5) (6.6) (6.7) (6.8) (6.9) (6.10) (6.11) (6.12) (6.13)

6.3.3 Design of Openings in Jacketed Vessels The design of openings in jacketed vessels includes reinforcement for the opening and nozzle detail for the inner vessel and construction details for openings in the jacket. For openings (penetrations) in the jacket of the type shown in Fig. 6.5, the jacket is considered as stayed and needs no reinforcement calculations. Only pressure membrane loading is considered for the openings and penetration in these rules. Other loadings given in UG-22 of VIII-1 shall be considered. 6.3.3.1 Openings in Inner Vessel. The design of openings in the inner vessel shall be according to the rules given in Chapter 5 for combinations of loadings due to internal pressure, external pressure (vacuum), or both. No consideration shall be made for cross-sectional area from the jacket and closure.


166

Chapter 6

FIG. 6.4 SOME ACCEPTABLE TYPES OF CLOSURE DETAILS (ASME VIII-1)

6.3.3.2 Jacket Openings and Penetrations. No reinforcement is required for openings in the jacket when the penetrations are as shown in Fig. 6.5. Jacket penetrations shall conform to that shown in Fig. 6.5 and Table 6.3. Special Penetration Detail Requirements [1] [2] [3] [4]

Jacket welded to nozzle wall, with details as shown in Fig. 6.5(a) trc determined for the shell under external pressure trc ≥ ttj trc1 determined for shell under external pressure trc2 determined from the following formulas: When no tubular section exists between jacket and torus,


(6.14)


FIG. 6.4 (CONT’D)

trc2 = Pr/(SE – 0.6P)

(6.15)

When tubular section exists between jacket and torus, trc2 = PRp/(SE – 0.6P)

(6.16)

where E = weld efficiency from Table UW-12 of VIII-1 for either the circumferential weld in the torus for the equation using r or for any weld in the opening closure for the equation using Rp, radius of penetration.


168

Chapter 6

FIG. 6.4 (CONT’D)

6.4 HALF-PIPE JACKETS The rules in Appendix EE of VIII-1 for half-pipe jackets are given only for the design condition where there is positive pressure inside the head or shell and positive pressure inside the half-pipe jacket and for NPS 2, NPS 3, and NPS 4 pipe size jackets with vessel diameters between 30 in. and 170 in. Obviously, there are



TABLE 6.2 CLOSURE DETAIL REQUIREMENTS FOR VARIOUS TYPES OF JACKET CLOSURES Closure Detail Fig. 6.4(a) Fig. 6.4(b-1) Fig. 6.4(b-2) Fig. 6.4(b-3) Fig. 6.4(c) Fig. 6.4(d-1) Fig. 6.4(d-2) Fig. 6.4(e-1) Fig. 6.4(e-2) Fig. 6.4(f-1) Fig. 6.4(f-2) Fig. 6.4(f-3) Fig. 6.4(g-1) Fig. 6.4(g-2) Fig. 6.4(g-3) Fig. 6.4(g-4) Fig. 6.4(g-5) Fig. 6.4(g-6) Fig. 6.4(h) Fig. 6.4(k) Fig. 6.4(l)

Type 1 [1] [3] [3] [4] [6] [7] [7] [7] [7] [8] [8] [8] [10] [10] [10] [11] [11] [11] — — —

Type 2

Type 3

Type 4

Type 5

[2] [3] [3] [5] — — — — — [9] [9] [9] [10] [10] [10] [11] [11] [11] — [13] [13]

— [3] [3] [5] — — — — — [9] [9] [9] [10] [10] [10] [11] [11] [11] [12] — —

[2] [3] [3] [5] — — — — — [9] [9] [9] [10] [10] [10] [11] [11] [11] — — —

— [3] [3] [5] — — — — — [9] [9] [9] [10] [10] [10] [11] [11] [11] — — —

NOTES: [ ] Indicates the dimensional requirements listed below for that combination. — Indicates that the combination is not permitted.

other combinations of pressure loading that need to be considered separately from these rules. Some combinations are (1) (2) (3) (4) (5)

Positive pressure inside the shell and negative pressure inside the jacket Negative pressure inside the shell and positive pressure inside the jacket Negative pressure inside the shell and inside the jacket Wind or earthquake loading with various combinations Addition of cyclic loading to any combination

Some of these combinations may be more severe than the conditions considered here. In addition to the half-pipe jacket configuration, there may be jackets of other geometries such as angles, channels, and circular segments, as shown in Fig. 6.6.

6.4.1 Maximum Allowable Internal Pressure in Half-Pipe Jacket The maximum allowable internal pressure, P′, in a half-pipe jacket attached to a cylindrical shell is determined (Jawad, 1994) as follows: P′ = F/K

(6.17)

where P′ = maximum allowable internal pressure in the jacket, psi F = 1.5S – S′ (but F shall not exceed 1.5S), psi (6.18) S = maximum allowable tensile stress at design temperature of the vessel material, psi S′ = actual longitudinal stress in the shell or membrane stress in the head due to internal pressure


170

Chapter 6

FIG. 6.5 SOME ACCEPTABLE TYPES OF PENETRATION DETAILS (ASME VIII-1) and other axial forces, psi. When axial forces are negligible, S′ shall be taken as PR/2t. When the combination of axial force and pressure stress (PR/2t) is such that S′ would be a negative number, then S′ shall be taken as zero. K = factor from Fig. 6.7 for NPS 2, from Fig. 6.8 for NPS 3, and from Fig. 6.9 for NPS 4



TABLE 6.3 PENETRATION DETAIL REQUIREMENTS Penetration Detail

Special Requirements

Fig. 6.5(a) Fig. 6.5(b) Fig. 6.5(c) Fig. 6.5(d) Fig. 6.5(e-1) Fig. 6.5(e-2) Fig. 6.5(f)

[1] [2] [3] [2] [4] [4] [5]

NOTE: [] Indicates special penetration detail requirements.

FIG.6.6 SPIRAL JACKETS, HALF-PIPE AND OTHER SHAPES P = internal design pressure of the vessel, psi R = inside radius of the shell or head, in. D = 2R

6.4.2 Minimum Thickness of Half-Pipe Jacket The minimum thickness of a half-pipe jacket is determined from T = (P1r)/(0.85S1 – 0.6P1)


(6.19)

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Chapter 6

FIG. 6.7 FACTOR K FOR NPS 2 PIPE JACKET where T = minimum thickness of the half-pipe jacket, in. r = inside radius of the jacket, in. (see Fig. 6.6) S1 = allowable stress of the jacket material at design temperature, psi P1 = internal design pressure in the jacket, psi The weld thickness attaching the half-pipe jacket to the vessel shall have a throat thickness not less than the smaller of the jacket or shell thickness. Special consideration of fillet welds may be required for vessels in cyclic service.



FIG. 6.8 FACTOR K FOR NPS 3 PIPE JACKET

Example 6.3 Problem Using the rules of Appendix EE of VIII-1, determine the minimum required thickness of a cylindrical shell with an internal design pressure of 350 psi and an externally attached NPS 4 schedule IOS half-pipe jacket at an internal design pressure of 500 psi in noncyclic service. The inside diameter of the shell is 36 in. and E = 1.0. The design temperature is 100°F. The shell material is SA-516 Gr. 70. The jacket material is SA-53 Gr.S/A. There is no radiography of any jacket joints and no corrosion allowance.


174

Chapter 6

FIG. 6.9 FACTOR K FOR NPS 4 PIPE JACKET

Solution (1) From II-D, allowable stresses are: SA-516-70 = 20.0 ksi and SA-53-S/A = 13.7 ksi (2) Minimum required thickness of the cylindrical shell using UG-27(c)(1) of VIII-1 is tr = (PR)/(SE – 0.6P) = (350)(18)]/[20,000 × 1.0 – 0.6 × 350] = 0.318 in. Nominal thickness used is t = 3/8 in.



(3) From Fig. 6.9 with D = 36 in., and t = 3/8 in., K = 40. S′ = PR/2t = (350 × 18)/(2 × 0.375) = 8400 psi P′ = F/K = (1.5 × 20,000 – 8400)/40 = 540 psi > 500 psi

(4) Assuming NPS 4 Schedule 10S, tj = 0.120 × 0.875 = 0.105 in. rj = [(4.5/2) – 0.105] = 2.145 in.

(5) Minimum thickness of the half-pipe jacket using Eq. (6.19) is trj = [(500)(2.145)]/[0.85 × 13,700 – 0.6 × 500] = 0.095 in. < 0.105 in. actual

(6) Minimum fillet weld size is 0.120 × 1.414 = 0.170 in. Use 3/16 in. (7) Summary: Use shell thickness of 3/8 in.; half-pipe jacket of NPS 4 Schedule 10S; and 3/16 in. fillet weld size to attach half-pipe jacket to shell.

6.5 VESSELS OF NONCIRCULAR CROSS SECTION The rules in Appendix 13 of VIII-1 for vessels of noncircular cross section are limited to vessels having rectangular or obround cross sections. As such, the walls are subject to both tension and bending due to internal pressure. Stresses are determined in the walls and end plates from pressure loadings, including effects of stiffening, reinforcing, and staying members. The rules in this section are limited to vessels of noncircular cross section with a straight longitudinal axis. Cross sections which do not have a straight longitudinal axis, such as a torus, are not contained in this section. Often, vessels of noncircular cross section contain openings of various diameters. In addition, the opening may be of uniform diameter through the wall thickness of the vessel or may be of several different diameters through the wall thickness. Consideration of the openings is made by ligament efficiency procedures. In addition to considering many different opening sizes and the effects of stiffeners, stayplates, and reinforcement, this analysis considers vessels with different thicknesses of plate on various sides of the vessel cross section. Furthermore, using this procedure can enable an engineer to consider not only the effects of opening efficiency, but also the joint efficiency of butt welded joints. There are no rules given for calculating openings by the reinforced opening method. If that method is chosen for noncircular cross section vessels, U-2(g) shall be followed. A structural frame analysis is used where moments of inertia and stiffness of various members are determined and equations are developed by equating the end rotations and deflections. From this analysis, shears and moments are obtained, which then are used to calculate the membrane stress and bending stress, comparing them with an allowable stress. For vessels of noncircular cross section that do not have equations given in Appendix 13 of VIII-1, U-2(g) shall be followed.

6.5.1 Types of Vessels Although there are many vessels with noncircular cross section, the rules of Appendix 13 are limited to single wall vessels with essentially rectangular and obround cross sections and one circular vessel with a single diametrical stay plate. Vessels of rectangular cross section are shown in Fig. 6.10; vessels of rectangular cross section with stay plates are shown in Fig. 6.11; and vessels of obround cross section with and without stay plates and a circular cross section with a single diametrical stay plate are shown in Fig. 6.12. Detailed equations for each of these cross sections plus example problems are given in Appendix 13 of VIII-1.


176

Chapter 6

FIG. 6.10 VESSELS OF RECTANGULAR CROSS SECTION

6.5.2 Basis for Allowable Stresses The calculated primary membrane stress shall not exceed the allowable tensile stress at design temperature, (S) given in II-D, for the material multiplied by the butt-welded joint efficiency (E) when applicable. When the calculated primary membrane stress and the calculated primary bending stress are combined, the following limits shall be met:



FIG. 6.10 (CONT’D)


178

Chapter 6

FIG. 6.11 VESSELS OF RECTANGULAR CROSS SECTION WITH STAY PLATES



FIG. 6.12 VESSELS OF OBROUND CROSS SECTION WITH AND WITHOUT STAY PLATES AND VESSELS OF CIRCULAR CROSS SECTION WITH A STAY PLATE


180

Chapter 6

(1) For a rectangular cross section, 1.5 times the allowable tensile stress at design temperature multiplied by the butt weld joint efficiency when applicable, 1.5 SE; (2) For other cross sections (such as structural shapes), the lesser of (a) 1.5 times the allowable tensile stress at design temperature multiplied by the butt weld joint efficiency when applicable, 1.5 SE, or (b) 0.67 times the yield stress at design temperature, 0.67 Sy, except when greater deformation is acceptable where 0.90 times the yield strength at design temperature—but not to exceed 0.67 specified minimum yield strength—may be used.

6.5.3 Openings in Vessels of Noncircular Cross Section As stated previously, the only method for considering openings in vessels of noncircular cross section is the ligament efficiency method. The reinforced opening method is not described nor considered, except by U-2(g). When the ligament efficiency method is used, it appears in the equations for both the membrane stress and the bending stress. If the vessel or header is welded, the butt weld joint efficiency, E, also must be determined. The ligament efficiencies, em and eb, are applied only to the plates in which the openings are located. When both em and eb are less than E, the membrane stress and bending stress are calculated on the gross area of the section using E = 1.0. Those membrane and bending stresses are then divided by em and eb, respectively, to get the stresses based on the net area.

FIG. 6.13 PLATE WITH CONSTANT-DIAMETER OPENINGS OF SAME OR DIFFERENT DIAMETERS



When both em and eb are equal to or greater than E, the membrane stress and bending stress are calculated on the gross area of the section using the appropriate E, which depends on the butt weld joint examination chosen. 6.5.3.1 Ligament Efficiency for Constant-Diameter Openings. For plates with constant-diameter openings, shown in Fig. 6.13, the ligament efficiency for both membrane stress and bending stress is the same. When the diameters of the openings are different, it is necessary to determine the equivalent diameter by averaging the two diameter as follows: DE = 0.5(d1 + d2)

(6.20)

em = eb = (p – DE)/p

(6.21)

The ligament efficiencies then are

6.5.3.2 Ligament Efficiency for Multi-Diameter Openings for Membrane Stresses. For many applications of openings in plates, the opening may have more than one diameter through the plate thickness, as shown in Fig. 6.14. For example, in air-cooled heat exchangers, the diameter increases through the thickness, and in rolled-in tube arrangements, lands of slightly larger diameter are used for holding the tube in place. The equations for determining the ligament efficiency for multidiameter openings for membrane stress are similar to those for a constant diameter opening, with the major difference being the need to determine the equivalent diameter of each multi-diameter opening. For each multi-diameter opening, an equivalent diameter, DE1, DE2, etc., is determined as follows: DE = (d0T0 + d1T1 + d2T2 + … + dnTn)/t

(6.22)

This calculation is repeated for an adjacent tube, and the results are then combined to obtain an equivalent diameter for the combination of the two multidiameter openings as follows: DE = 0.5(de1 + de2)

(6.23)

If de1 and de2 are equal, DE = de. The ligament efficiency for the membrane stress is: em = (p – DE)/p

FIG. 6.14 PLATE WITH MULTIDIAMETER OPENINGS


(6.24)

182

Chapter 6

6.5.3.3 Ligament Efficiency for Multi-Diameter Openings for Bending Stress. Determining the ligament efficiency for multi-diameter openings in order to calculate the bending stress requires locating the neutral axis of the various sets of diameters and thicknesses and determining the effective moment of inertia. From this, the ligament efficiency for bending stress is then determined. The basic structural mechanics equations for doing this are X = Σ AX/Σ A

(6.25)

I = Σ AX2

(6.26)

where Σ AX = b0T0(0.5T0 + T1 + T2 + . . + Tn) + b1T1(0.5T1 + T2 + . . + Tn) + b2T2(0.5T2 + . . + Tn) + bnTn(0.5Tn)

(6.27)

Σ A = b0T0 + b1T1 + b2T2 + . . + bnTn

(6.28)

X = ΣAX/Σ A

(6.29)

I = ΣI0 + ΣAX2

(6.30)

Also,

I = (b0T30)/12 + (b1T31)/12 + (b2T32)/12 + . . + (bnT3n)/12 + b0T0(0.5T0 + T1 + T2 + . . + Tn – X)2 + b1T1(0.5T1, + T2 + . . + Tn – X)2 + b2T2(0.5T2 + . . + Tn – X)2 + bnTn(X – 0.5Tn)2

(6.31)

c = larger of X or t – X

(6.32)

bE = p – DE

(6.33)

c = 0.5t

(6.34)

I = (bEf3)/1/2

(6.35)

The width of a ligament is:

Other properties are:



And solving: c/I = (bEt2)

(6.36)

DE = p – [(6I)/(t2c)]

(6.37)

With

The ligament efficiency for bending stress is: eb = (p – DE)/p

(6.38)

Example 6.4 Problem Using the rules of Appendix 13 of VIII-1, determine the membrane and bending ligament efficiencies in a pressure vessel of square cross section, in which H = h = 6 in. and t1 = t2 = 0.75 in. and there is a single row of 1.5 in. diameter holes on 4 in. center-to-center spacing. Solution Using Eq. (6.21), calculate the ligament efficiencies as follows: em = eb = (4 – 1.5)/(4) = 0.625

Example 6.5 Problem A pressure vessel contains a single row of openings that are alternately spaced on 4 in. and 3 in. center-tocenter spacings. The opening diameters also alternate, with the first one being 1.5 in. diameter and the next one being 1.25 in. diameter. Assuming the butt-joint efficiency is higher than the ligament efficiency, determine the minimum ligament efficiency for setting the thickness of the vessel. Solution Using Eq. (6.20), determine the equivalent diameter, DE, as follows: DE = 0.5(1.5 + 1.25) = 1.375 in.

The ligament efficiency is based on the minimum spacing of p = 3 in. by using the equivalent diameter of DE = 1.375 in. as follows: em = eb = (3 – 1.375)/(3) = 0.542

Example 6.6 Problem Using the rules in Appendix 13 of VIII-1, determine the membrane ligament efficiency of a pressure vessel which is 1.50 in. thick and contains a row of multi-diameter openings on 3.50 in. center-to-center spacing, as shown in Fig. 6.14. The dimensions of all of the multidiameter openings are d0 = 1.625 in.

T0 = 0.125 in.

d1 = 1.5 in.

T1 = 1.125 in.

d2 = 1.375 in.

T2 = 0.25 in.


184

Chapter 6

Solution Using Eq. (6.22), determine the equivalent diameter, de, of the opening as follows: de = [(1.625)(0.125) + (1.5)(1.125) + (1.375)(0.250)]/(1.5) = 1.490 in.

Since the equivalent diameter of each opening is the same, de = DE, and ligament efficiency is determined by using Eq. (6.21) as follows: em = (3.5 – 1.490)/(3.5) = 0.574

Example 6.7 Problem Determine the bending ligament efficiency of the pressure vessel in Example 6.6, above. Solution Using the dimensions given in Example 6.6, determine the values for b as follows: p = 3.5 in. T0 = 0.125 in.

d0 = 1.625 in.

b0 = 3.5 – 1.625 = 1.875 in.

T1 = 11.125 in.

d1 = 1.5 in.

b1 = 3.5 – 1.5 = 2.0 in.

T2 = 0.25 in.

d2 = 1.375 in.

b2 = 3.5 – 1.375 = 2.125 in.

Then Σ AX = 1.875 × 0.125(0.0625 + 1.125 + 0.25) + 2.0 × 1.125(0.5625 + 0.25) + 2.125 × 0.25(0.125) = 2.2314

ΣA = (1.875 × 0.125) + (2.0 × 1.125) + (2.125 × 0.25) = 3.0156

X = (2.2314)/(3.0156) = 0.7400

I = (1/12)[(1.875)(0.125)3 + (2.0)(1.125)3 + (2.125)(0.25)3] + 1.875 × 0.125(0.0625 + 1.125 + 0.25 – 0.74)2 + 2.0 × 1.125(0.5625 + 0.25 – 0.74)2 + 2.125 × 0.25(0.74 – 0.125)2 = 0.5672



c = larger of 0.74 or (1.50 – 0.74) = 0.76 in.

DE = 3.5 – [(6)(0.5672)]/(1.5)2 (0.76) = 1.526 in.

eb = (p – DE)/p = (3.5 – 1.526)/3.5 = 0.564

6.5.4 Vessels of Rectangular Cross Section One of the least complex vessels of noncircular cross section is one with a rectangular cross section. Basic equations are given here for that cross section, which is shown in Fig. 6.10, sketch (1). Equations for other cross sections are given in Appendix 13-7 of VIII-1. (1) Membrane stress: (a) Short-side plate: Sm = Ph/2t1

(6.39)

Sm = PH/2t2

(6.40)

(Sb)N = (Pc/12I1) {– 1.5H2 + h2[(1 + α2K)/(1 + K)]}

(6.41)

(Sb)Q = (Ph2c/12I1)[(1 + α2K)/(1 + K)]

(6.42)

(Sb)m = (Ph2c/12I2) {–1.5 + [(1 + α2K)/(1 + K)]}

(6.43)

(Sb)Q = (Ph2C/12I2)[(1 + α2K)/(1 + K)]

(6.44)

(ST)N = Eq. (6.39) + Eq. (6.41)

(6.45)

(ST)Q = Eq. (6.39) + Eq. (6.42)

(6.46)

(ST)M = Eq. (6.40) + Eq. (6.43)

(6.47)

(ST)Q = Eq. (6.40) + Eq. (6.44)

(6.48)

(b) Long-side plate:

(2) Bending stress (a) Short-side plate:

(b) Long-side plate:

(3) Total stress (a) Short-side plates:

(b) Long-side plates:


186

Chapter 6

FIG. E6.8 EXAMPLE PROBLEM OF NONCIRCULAR VESSEL (ASME VIII-1)

Example 6.8 Problem A noncircular cross section vessel has outside dimensions of 7.25 in. × 7.25 in. and is 0.625 in. thick, as shown in Fig. E6.8. Material is SA-516 Gr. 70. Design temperature is 650°F, and design pressure is 150 psi. There is no corrosion allowance. All but weld joints are radiographed, so that E = 1.0. One side of the vessel contains a single row of openings which are 2.53 in. diameter on a center-to-center spacing of 3.5 in. Is the assumed thickness of t = 0.625 in. adequate to satisfy the rules of Appendix 13 of VIII-1? If not, what thickness is required? Solution Following the rules in section 6.5.4, t = t1 = t2 = 0.625 in. h = H = 7.25 – 2(0.625) = 6.00 in. I = bd3/12 = (1)(0.625)3/12 = 0.0203 in.4

(1) The membrane and bending ligament efficiencies according to Eq. (6.21) are as follows: E = em = eb = (3.5 – 2.53)/(3.5) = 0.277

(2) Determine the stresses according to section 6.5.4 as follows: (a) membrane stress, using Eq. (6.39) or Eq. (6.40), is Sm = (150)(6)/(2)(0.625) = 720 psi



(b) Bending stress at midpoint of the side, using Eq. (6.41) or Eq. (6.43) and α = 1 and K = 1, is (Sb)N = [(150)(6)2(0.3125)/12(0.0203)][–1.5 + (1 + 1)/(1 + 1)] = 3460 psi

(c) Bending stress at the corner, using Eq. (6.42) or Eq. (6.44) and a = 1 and K = 1, is (Sb)Q = (150)(6)2(0.3125)/12(0.0203)][(1 + 1)/(1 + 1)] = 6930 psi

(d) Total stress at the midpoint is Sm + (Sb)M = 720 + 3460 = 4180 psi

(e) Total stress at the corner is: Sm + (Sb)Q = 720 + 6930 = 7650 psi

(f) From II-D, allowable stress for SA-516 Gr.70 at 650∞F is S = 18.8 ksi. Allowable design stresses are: SE = (18,800)(0.277) = 5200 psi 1.5SE = (1.5)(18,800)(0.277) = 7810 psi

(g) Calculated stresses vs. allowable stresses: Sm ≤ SE: 720 psi < 5200 psi Sm + Sb ≤ 1.5SE: 7650 psi < 7810 psi

All calculated stresses are less than allowable design stresses; therefore, t = 0.625 in. is OK.



CHAPTER

7 DESIGN

OF

HEAT EXCHANGERS

7.1 INTRODUCTION Heat exchangers are considered the workhorse in chemical plants and refineries. They come in all shapes, sizes, and configurations and are essential in extracting or adding heat to various process fluids. Figure 7.1 (TEMA, 1999), shows various shapes of commonly used heat exchangers. Design rules for heat exchanger components are covered in various parts of VIII-1. Details of tube-totubesheet welds are given in non mandatory Appendix A. Mandatory rules for tubesheet design are given in part UHX for U-tube and fixed tubesheets. Three types of tubesheets are covered by the design. They are simply supported, integral, and extended as a flange. Design rules for flanged and flued as well as bellowstype expansion joints are given in mandatory appendix 26. Other rules that govern the construction of heat exchangers are given in the Tubular Exchanger Manufacturers Association Standard (TEMA, 1999). These rules govern design, tolerances, baffle construction, and other details of heat exchangers not listed in VIII-1. Similar construction rules for bellowstype expansion joints are given in the Expansion Joint Manufacturers Association Standard (EJMA, 1998). Design of tubesheet in a U-tube heat exchanger is based on the classical theory of the bending of a circular plate subjected to pressure. The perforation of the plate is taken into consideration in VIII-1 and so is the stiffening effect of the attached tubes and attached shells. Design of the tubesheet in a fixed-tube heat exchanger is based on the classical theory of the bending of a circular plate on an elastic foundation. Theoretical design equations in this case are extremely complicated. VIII-1 rules include numerous assumptions and simplifications in order to provide practical design rules. Design equations for tubesheets of U-tube and fixed heat exchangers are given in the remainder of this chapter. A brief discussion of various expansion joints is also given.

7.2 DESIGN OF TUBESHEETS IN U-TUBE EXCHANGERS Section VIII-1 gives the design equations for six various tubesheet configurations. These are shown in Fig.7.2. The design of tubesheets in U-tube exchangers takes into consideration the ligament between the tubeholes as well as the stiffness of the attached tubes and shell components. The design of the tubesheet is lengthy but straightforward.

7.2.1 Nomenclature The following nomenclature is used in the design equations of tubesheets with U-tubes. The nomenclature is based on Fig.7.2 for tubesheet configuration, Fig.7.3 for tubesheet geometry, and Fig.7.4 for tube layout. 189


190

Chapter 7

FIG. 7.1 VARIOUS HEAT-EXCHANGER CONFIGURATIONS (TEMA, 1999)


Design of Heat Exchangers

191

PARAMETERS ρs = Ds/Doρc = Dc/Do βs = [12(1 – νs2)]1/4/[(Ds + ts) ts]1/2, ks = (βs Es ts3)/[6(1 – νs2)] λs = [6 Ds ks/h3][1 + h βs + (h2 βs2/2)] δs = [Ds2/4 Es ts][1 – νs/2], ωs = ρs ks βs δs(1 + hβs) βc = [12(1 – νc2)]1/4/[(Dc + tc) tc]1/2, kc = (βc Ec tc3)/[6(1 – νc2)] λc = [6 Dc kc/h3][1 + h βc + (h2 βc2/2)] δc = [Dc2/4 Ec tc] [1 – νc/2], ωc = ρc kc βc δc (1 + h βc) K = A/Do F = [(1 – ν*)/E*] [λs + λc + E (ln K)] M* = MTS – ωs Ps + ωc Pt

FIG. 7.2A TUBESHEET INTEGRAL WITH SHELL AND CHANNEL (ASME. VIII-1)

The tube pattern layout is normally on a triangular or square pattern. Subsequent equations will refer to the following symbols and definitions: A = outside diameter of tubesheet AL = total area of untubed lanes = UL1 LL1 + UL2 LL2 + … (limited to 4Do p) C = bolt circle diameter ct = tubesheet corrosion allowance on the tube side = 0 in the uncorroded condition Dc = inside channel diameter Do = equivalent diameter of outer tube limit circle, Fig.7.3a. Ds = inside shell diameter d = diameter of tube hole dt = nominal outside diameter of tubes d* = effective tube hole diameter E = modulus of elasticity of tubesheet material at tubesheet design temperature Ec = modulus of elasticity for channel material at design temperature Es = modulus of elasticity for shell material at design temperature


192

Chapter 7

PARAMETERS ρs = Ds/Doρc = Dc/Do βs = [12(1 – νs2)]1/4/[(Ds + ts) ts]1/2, ks = (βs Es ts3)/[6(1 – νs2)] λs = [6 Ds ks/h3][1 + h βs + (h2 βs2/2)] δs = [Ds2/4 Es ts][1 – νs/2], ωs = ρs ks βs δs (1 + h βs) K = A/Do F = [(1 – ν*)/E*][λs + E(ln K)] M* = MTS – ωs Ps – [Wc (C – Gc)/2πDo] FIG. 7.2B TUBESHEET INTEGRAL WITH SHELL AND GASKETED WITH CHANNEL, EXTENDED AS A FLANGE (ASME. VIII-1)

Et = modulus of elasticity of tube material at tubesheet design temperature E* = effective modulus of elasticity of tubesheet in perforated region Gl = midpoint of contact between flange and tubesheet Gc = diameter of channel gasket load reaction Gs = diameter of shell gasket load reaction h = tubesheet thickness hg = tube side pass partition groove depth, Fig.7.3c h’g = effective tube side pass partition groove depth. LL1, LL2, … = length(s) of untubed lane(s), Fig.7.4 τx = expanded length of tube in tubesheet (0< tx < h), Fig.7.3b MAX [(a),(b),(c),…] = greatest of a, b, c,… MIN [(a),(b),(c),…] = smallest of a,b,c,… Ps = Shell side internal design pressure Pt = tube side internal design pressure p = tube pitch p* = effective tube pitch ro = radius of outermost tube hole center, Fig.7.3a



PARAMETERS ρs = Ds/Doρc = Gc/Do βs = [12(1 – νs2)]1/4/[(Ds + ts) ts]1/2, ks = (βs Es ts3)/[6(1 – νs2)] λs = [6Ds ks/h3][1 + h βs + (h2 βs2/2)] δs = [Ds2/4 Es ts][1 – νs/2], ωs = ρs ks βs δs (1 + h βs) K = A/Do F = [(1 – ν*)/E*] [λs + E(ln K)] M* = MTS – ωs Ps – [Wc(G1 – Gc)/2π Do] FIG. 7.2C TUBESHEET INTEGRAL WITH SHELL AND GASKETED WITH CHANNEL, NOT EXTENDED AS A FLANGE (ASME. VIII-1)

S = allowable stress for tubesheet material at tubesheet design temperature Sc = allowable stress for channel material at design temperature Ss = allowable stress for shell material at design temperature St = allowable stress for tube material at tubesheet design temperature Sy,c = yield strength for channel material at design temperature Sy,s = yield strength for shell material at design temperature SPS,c = allowable primary plus secondary stress for channel material at design temperature SPS,s = allowable primary plus secondary stress for shell material at design temperature tc = channel thickness ts = shell thickness tt = nominal tube wall thickness UL1, UL2, … = center-to-center distance(s) between adjacent tube rows of untubed lane(s), but not to exceed 4p, Fig.7.4 Wc = channel flange design bolt load for the gasket seating condition Ws = shell flange design bolt load for the gasket seating condition Wmax = maximum flange design bolt load = MAX [(Wc), (Ws)] µ = basic ligament efficiency for shear


193

194

Chapter 7

PARAMETERS ρs = Gs/Doρc = Gc/Do K = A/Do F = [(1 – ν*)/E*] [E(ln K)] M* = MTS + [Wmax(Gc – Gs)/2π Do] FIG.7.2D TUBESHEET GASKETED WITH SHELL AND CHANNEL (ASME. VIII-1)

µ* = effective ligament efficiency for bending νc = Poisson’s ratio of channel material νs = Poisson’s ratio of shell material ν* = effective Poisson’s ratio in perforated region of tubesheet ρ = tube expansion depth ratio = tx/h (0< ρ <1)

7.2.2 Preliminary Calculations The following quantities are determined first in order to calculate the moments and stresses in the tubesheets. Figure 7.3b. For tubes extended through the tubesheet as shown in Fig.7.3b, Do = 2 ro + dt m = (p – dt)/p d* = MAX {[dt – 2 tt (Et/E)(St/S) ρ], [dt – 2 tt]} Ck = 4 MIN [(AL), (4 Do p)]/p Do2 p* = p/(1 – Ck)1/2 µ* = (p* – d*)/p* h’g = MAX [(hg – ct), (0)]


(7.1)


195

PARAMETERS ρs = Gs/Doρc = Dc/Do βc = [12(1 – νc2)]1/4/[(Dc + tc) tc]1/2, kc = (βc Ec tc3)/[6(1 – νc2)] λc = [6 Dc kc/h3][1 + h βc + (h2 βc2/2)] δc = [Dc2/4 Ec tc][1 – νc/2], ωc = ρc kc βcδc (1 + h βc) K = A/Do F = [(1 – ν*)/E*][λc + E (ln K)] M* = MTS + ωc Pt + [Ws (C – Gs)/2πDo] FIG. 7.2E TUBESHEET GASKETED WITH SHELL AND INTEGRAL WITH CHANNEL, EXTENDED AS A FLANGE (ASME. VIII-1)

Figure 7.3d. For tubes welded to the backside of the tubesheet as shown in Fig.7.3d, Do = 2 ro + d µ = (p – d)/p Ck = 4 MIN [(AL), (4 Do p)]/π Do2

(7.2)

p* = p/(1 – Ck)1/2 µ* = (p* – d)/p* h’g = MAX [(hg – ct), (0)]

7.2.3 Design Equations As shown in paragraph U-2(a) of VIII-1, the user or his agent establishes the general geometry of the heat exchanger including the inside diameter of the shell and the number of required tubes to handle the


196

Chapter 7

PARAMETERS ρs = Gs/Doρc = Dc/Do βc = [12(1 – νc2)]1/4/[(Dc + tc) tc]1/2, kc = (βc Ec tc3)/[6(1 – νc2)] λs = [6Dc kc/h3][1 + h βc + (h2 βc2/2)] δc = [Dc2/4 Ec tc] [1 – νc/2], ωc = ρc kc βc δc (1 + h βc) K = A/Do F = [(1 – ν*)/E*] [λc + E(ln K)] M* = MTS + ωc Pt + [Ws (G1 – Gs)/2π Do] FIG. 7.2F TUBESHEET GASKETED WITH SHELL AND INTEGRAL WITH CHANNEL, NOT EXTENDED AS A FLANGE (ASME. VIII-1)

heat-transfer requirements. Design of the shell, channel and tubes is based on cylindrical equations as outlined in Chapter 2. Tubesheet design consist of calculating the thickness based on bending moments or shearing loads due to applied shell side pressure, tube side pressure, or flange moments as the case may be. Different loading conditions might have to be investigated for a final design. The magnitude of the bending moment in the tubesheet is influence by the stiffness of the attached shell or channel. Accordingly, the stress in the shell or channel needs to be checked when the tubesheet is attached to either of these components. The design procedure is iterative and starts by assuming a tubesheet thickness. The stress in the tubesheet is then determined. A new thickness is chosen when the calculated stress is greater than the allowable stress. The design procedure is in general accordance with the following steps. Step 1. Calculate the dimensional parameters from Eqs.(7.1) or (7.2) as applicable Step 2. Estimate an approximate tubesheet thickness using the shear stress equation h = [1/(4µ)][Do/(0.8S)]Ps – Pt|

Step 3. From Fig.7.5 calculate the quantities E* and ν* relative to h/p. Step 4. Calculate the quantity MTS = (Do2/16) [(ρs – 1)(ρs2 + 1) Ps – (ρc – 1)(ρc2 + 1) Pt

And then calculate the “parameters” from Fig.7.2 for any specific configuration.


(7.3)


197

FIG. 7.3 TUBESHEET GEOMETRY (ASME, VIII-1)

Step 5. For each loading condition calculate the maximum moment in the tubesheet Mp = [M* – (Do2/32)(F)(Ps – Pt)]/(1 + F) Mo = Mp + (Do2/64)(3 + ν*)(Ps – Pt)

(7.4)

M = MAX [|Mp|, |Mo|]

Step 6. For each loading condition calculate the bending stress in the tubesheet σ = 6M/[(µ*)(h – h’g)2]


(7.5)

198

Chapter 7

FIG. 7.4 TYPICAL UNTUBED LANE CONFIGURATIONS (ASME, VIII-1) If σ < 2 S then the assumed thickness h is adequate. Otherwise, increase h and recalculate σ. Step 7. For each loading condition calculate the shearing stress in the tubesheet at the outer edge of the perforated region τ = [1/(4µ)](Do/h) |Ps – Pt|

If τ ≤ 0.8 S then the assumed thickness h is adequate. Otherwise, increase h and recalculate τ.


(7.6)


199

FIG. 7.5 CURVES FOR THE DETERMINATION OF E*/E AND ν* (ASME VIII-1)

Step 8. For each loading condition calculate the stress in the attached shell and/or channel, as the case might be, as follows Shell Stress σs,m = (Ds2)(Ps)/[4 ts (Ds + ts)] σs,b = (6 ks/ts2) {βs δs Ps + 6 [(1 – ν*)/E*)](Do/h3)(1 + h βs/2) × [Mp + (Do2/32)(Ps – Pt)]} σs = σs,m| + |σs,b| ≤ 1.5 Ss


(7.7)

200

Chapter 7

FIG. 7.5 (CONT’D) Channel Stress σc,m = (Dc2)(Pt)/[4 tc (Dc + tc)] σc,b = (6 kc/tc2) {βc δc Pt + 6 [(1 – ν*)/E*)](Do/h3)(1 + h βc/2) × [Mp + (Do2/32)(Ps – Pt)]}

(7.8)

σc = σc,m| + |σ c,b| ≤ 1.5 Sc

Example 7.1 Problem A U-tube heat exchanger with a fixed tubesheet has details as shown in Fig.7.2a. Check the thicknesses of the shell, channel, tubes, and tubesheet for the design data shown in the table below.



201

FIG. 7.5 (CONT’D)

Design Pressure Temperature Shell and channel material Joint efficiency Tube material Tubesheet material S for shells and tubesheet material S for tube material

Shell Side

Tube Side

50 psi 300 °F SA 516-70 0.70

700 psi 300 °F SA 516-70 1.0 SA 688–304 SA 266 – Cl 2

20,000 psi

20,000 psi 16,100 psi


202

Chapter 7

FIG. 7.5 (CONT’D)

Additional Design Data: Ds = 59.0 in. tt = 0.035 in.

Dc = 52.0 in. ro = 23.0 in.

E = Ec = Es = 28,000 ksi νs=νc = 0.3 S = Sc = Ss = 20,000 psi

ts = 0.4375 in.

tc = 1.5 in.

p = 1.25 in.

dt = 1.0 in.

Et = 27,000 ksi ρ = 0.9

hg = 0.1875 inch

tubes are on an equilateral triangular pitch St = 16,100 psi



203

Solution Channel Shell From Eq.(2.1), t = 700 (52.0/2)/(20,000 × 1.0 – 0.6 × 700) = 0.93 inch

This thickness is less than the actual furnished thickness of 1.5 inch. Use 1.5 inch. The 1.5 inch thickness will be needed to satisfy the bending requirements of the tubesheet design as discussed later. Shell-Side Shell From Eq.(2.1), t = 50 (59.0/2)/(20,000 × 0.7 – 0.6 × 50) = 0.106 inch

This thickness is less than the actual furnished thickness of 0.4375 inch. Use 0.4375 inch. The 0.4375 inch thickness will be needed to satisfy the bending requirements of the tubesheet design as discussed later. Tubes For internal pressure, Eq.(2.7) gives t = 700 (1.00/2)/(16,100 × 1.0 + 0.4 × 700) = 0.0214 inch

This thickness is less than the actual furnished thickness of 0.035 inch. Use 0.035 inch. External pressure calculations show that the furnished tube thickness is adequate to resist the external pressure of 50 psi. Tube Sheet Since the tube-side pressure is much larger than the shell-side pressure, the tubesheet will be designed based on tube-side pressure acting by itself with zero pressure on the shell-side. Step 1. From Fig.7.3a, Do = 2 (ro) + dt = 2 × 23.0 + 1.0 = 47.0 µ = (p – dt)/p = (1.25 – 1.00)/1.25 = 0.2 d* = MAX {[dt – 2 tt (Et/E)(St/S) ρ], [dt – 2 tt]} = MAX {[1.0 – 2 × 0.035 (27/28)(16100/20000)0.9], [1.0 – 2× 0.035]} = MAX (0.9511, 0.9300) = 0.9511 AL = UL1 LL1


204

Chapter 7

= 6.0 × 47.0 = 282.0 Ck = 4 MIN [(AL), (4 Do p)]/π Do2 = 4 MIN [282.0, (4 × 47.0 × 1.25)]/π × 47.02 = 4 × 235/6939.7782 = 0.1355 p* = p/(1 – Ck)1/2 = 1.25/(1 – 0.1355)1/2 = 1.3444 µ* = (p* – d*)/p* = (1.3444 – 0.9511)/1.3444 = 0.2925 h’g = MAX [(hg – ct), (0)] = MAX [(0.1875 – 0), 0] = 0.1875 inch

Step 2. An approximate thickness, h, is calculated from Eq.(7.3) as h = [1/(4µ)][Do/(0.8S)] |Ps – Pt | =[1/(4 × 0.2)][47.0/(0.8 × 20,000)][700 – 0] = 2.57 inch.

For most cases this thickness is too small as an approximation because it is based on shear rather than bending stress. Assume a trial h = 5.25 inch. This thickness is actually based on a number of previous trial runs made for this tubesheet. Step 3. From Fig.7.5, h/p = 5.25/1.25 = 4.2 E*/E = 0.26, or E*= 0.26 × 28,000,000 = 7,280,000. and ν*=0.36.

Step 4. From Fig.7.2A, the following PARAMETERS are obtained A = 59.0 + 0.4375 = 59.4375 ρs = Ds/Do = 59.0/47.0 = 1.2553 ρc = Dc/Do = 52.0/47.0 = 1.1064



The equation for MTS becomes MTS = (Do2/16) [(ρs – 1)(ρs2 + 1) Ps – (ρc – 1)(ρc2 + 1) Pt = (47.02/16) [(1.2553 – 1)(1.25532 + 1) 0 – (1.1064 – 1)(1.10642 + 1)(700) =–22,866 βs = [12(1 – νs2)]1/4/[(Ds + ts) ts]1/2 = [12(1 – 0.32)]1/4/[(59.0 + 0.4375) 0.4375]1/2 = 0.3565 ks = (βs Es ts3)/[6(1 – νs2)] = (0.3565 × 28,000,000 × 0.43753)/[6(1 – 0.32)] = 153,086 λs = [6 Ds ks/h3][1 + h βs + (h2 βs2/2)] = [6 × 59.0 × 153086/5.253][1 + 5.25 × 0.3565 + (5.252 × 0.35652/2)] = 374,508 × 4.6228 = 1,731,287 δs = [Ds2/(4 Es ts)][1 – νs/2] = [59.02/(4 × 28,000,000 × 0.4375)][1 – 0.3/2] = 0.0000604 ωs = ρs ks βs δs (1 + h βs) = 1.2553 × 153086 × 0.3565 × 0.0000604 (1 + 5.25 × 0.3565) = 11.8786 βc = [12(1 – νc2)]1/4/[(Dc + tc) tc]1/2 = [12(1 – 0.32)]1/4/[(52.0 + 1.5) 1.5]1/2 = 0.2029 kc = (βc Ec tc3)/[6(1 – νc2)] = (0.2029 × 28,000,000 × 1.53)/[6(1 – 0.32)] = 3,512,144


205

206

Chapter 7

λc = [6 Dc kc/h3][1 + h βc + (h2 βc2/2)] = [6 × 52.0 × 3512144/5.253][1 + 5.25 × 0.2029 + (5.252 × 0.20292/2)] = 19,937,596 δc = [Dc2/4 Ec tc][1 – νc/2] = [52.02/4 × 28,000,000 × 1.5][1 – 0.3/2] = 0.00001368 ωc = ρc kc βc δc (1 + h βc) = 1.1065 × 3512144 × 0.2029 × 000001368(1 + 5.25 × 0.2029) = 22.2803 K = A/Do = 59.4375/47.0 = 1.2646 F = [(1 – ν*)/E*][λs + λc + E (ln K)] = [(1 – 0.36)/7,280,000][1,731,287 + 19,937,596 + 28,000,000(ln 1.2646)] = 2.4829 M* = MTS – ωs Ps + ωc Pt = –22866 – 0 + 22.2803 × 700 = –7270

Step 5. The moments are calculated from Eqs.(7.4) as Mp = [M* – (Do2/32)(F)(Ps – Pt)]/(1 + F) = [–7270 – (47.02/32)(2.4829)(0 – 700)]/(1 + 2.4829) = 32,360 Mo = Mp + (Do2/64)(3 + ν*)(Ps – Pt) = 32,360 + (47.02/64)(3 + 0.36)(0 – 700) = –48820 M = MAX [Mp, Mo] = MAX [32360, 48820] M = 48820



207

Step 6. Calculate the bending stress in the tubesheet from Eq.(7.5) σ = 6 M/[(µ*)(h – h’g)2] = 6 × 48820/[(0.2925)(5.25 – 0.1875)2] = 39070 psi < 2 S = 40,000 psi

Tubesheet thickness of 5.25 inches is adequate.

Step 7. Calculate the shearing stress in the tubesheet at the outer edge of the perforated region from Eq.(7.6) τ = [1/(4µ)](Do/h) Ps – Pt = [1/(4 × 0.2)](47.0/5.25) 0 – 700 = 7833 psi < 0.8 S = 16,000 psi

Step 8. Calculate the stress in the attached shell and channel in accordance with Eqs.(7.7) and (7.8) Shell Stress σs,m = (Ds2)(Ps)/[4 ts (Ds + ts)] =0 σs,b = (6 ks/ts2) {βs δs Ps + 6 [(1 –ν*)/E*)](Do/h3)(1 + h βs/2) × [Mp + (Do2/32)(Ps – Pt)]} = (6 × 153086/0.43752) {0.3565 × 0.00006038 × 0 + 6 [(1 – 0.36)/7,280,000)](47.0/5.253) × (1 + 5.25 × 0.3565/2) × [32360 + (47.02/32)(0 – 700)]} = (4,798,777) [0 + 6(8.7912 × 10–8)(0.3248)(1.9358)(32360 – 48322)] = –25,400 psi σs = σs,m + σs,b = 25, 400 psi < 1.5S = 30,000 psi

Channel Stress σc,m = (Dc2)(Pt)/[4 tc (Dc + tc)] = (52.02)(700)/[(4× 1.5) (52.0 + 1.5)] = 5896 psi σc,b = (6 kc/tc2) {βc δc Pt + 6 [(1 – ν*)/E*)](Do/h3)(1 + h βc/2) × [Mp + (Do2/32)(Ps – Pt)]} = (6 × 3,512,144/1.52) {0.2029 × 0.00001368 × 700 + 6 [(1 – 0.36)/7,280,000)](47.0/7.253) × (1 + 5.25× 0.2029/2) × [32360 + (47.02/32)(0 – 700)]} = (9,365,717)[0.001943 + 6(8.791 × 10–8)(0.3248)(1.5326)(–15962)]


208

Chapter 7

= –21,054 psi σc = σc,m + σc,b = 5896 + 21054 = 26,950 psi < 1.5S = 30,000 psi

7.3 FIXED TUBESHEETS The term “fixed tubesheets” apply to those heat exchangers in which two tubesheets are used with the tubes acting as stays and shell acting as a support at the outside circumference. The rules in VIII-1 give design equations for various types of edge support for the tubesheets. They are simply supported, fixed, and flanged as shown in Fig. 7.6. The design of such tubesheets is based on the theory of plates on elastic foundation and is more complicated than that for tubesheets in U-tube construction.

7.3.1 Nomenclature Symbols and terms used for fixed tubesheets include the following A = outside diameter of tubesheet ac = radial channel dimension, in. ao = equivalent radius of outer tube limit circle as = radial shell dimension, in. C = bolt circle diameter Dc = inside channel diameter DJ = inside diameter of the expansion joint at its convolution height Ds = inside shell diameter dt = nominal outside diameter of tubes E = modulus of elasticity for tubesheet material at T Ec = modulus of elasticity for channel material at Tc Es = modulus of elasticity for shell material at Ts Et = modulus of elasticity for tube material at Tt Gl = midpoint of contact between flange and tubesheet Gc = diameter of channel gasket load reaction Gs = diameter of shell gasket load reaction h = tubesheet thickness J = ratio of expansion bellows to shell axial rigidity (J = 1.0 if no bellows) KJ = axial rigidity of expansion bellows, total force/elongation k = constant accounting for the method of support for the unsupported tube span under consideration. = 0.6 for unsupported spans between two tubesheets = 0.8 for unsupported spans between a tubesheet and a tube support = 1.0 for unsupported spans between two tube supports L = tube length between inner tubesheet faces Lt = tube length between outer tubesheet faces = unsupported tube span under consideration MAX [(a),(b),(c),…] = greatest of a, b, c,… Nt = number of tubes Pe = effective pressure acting on tubesheet Ps = shell side internal design pressure. For shell side vacuum use a negative value of Ps.



209

Pt = tube side internal design pressure. For tube side vacuum use a negative value of Pt. S = allowable stress for tubesheet material at T Sc = allowable stress for channel material at Tc Ss = allowable stress for shell material at Ts St = allowable stress for tube material at Tt Sy = yield strength for tubesheet material at T Sy,c = yield strength for channel material at Tc Sy,s = yield strength for shell material at Ts Sy,t = yield strength for tube material at Tt SPS = allowable primary plus secondary stress for tubesheet material at T SPS,c = allowable primary plus secondary stress for channel material at Tc SPS,s = allowable primary plus secondary stress for shell material at Ts T = tubesheet design temperature Ta = ambient temperature Tc = channel design temperature Ts = shell design temperature Tt = tube design temperature Ts,m = mean shell metal temperature along shell length Tt,m = mean tube metal temperature along tube length tc = channel thickness ts = shell thickness tt = tube thickness W = channel flange design bolt load for the gasket seating condition. αs,m = mean coefficient of thermal expansion of shell material at Ts,m αt,m = mean coefficient of thermal expansion of tube material at Tt,m γ = axial differential thermal expansion between tubes and shell ν = Poisson’s ratio of tubesheet material νc = Poisson’s ratio of channel material νs = Poisson’s ratio of shell material νt = Poisson’s ratio of tube material

7.3.2 Preliminary Calculations The following quantities are determined first in order to calculate the moments and stresses in the tubesheets. Step 1. Calculate the dimensional parameters from Eqs.(7.1) and (7.2) as applicable. Step 2. Calculate the quantities ao = Do/2

ρs = as/ao

χs = 1 – Nt (dt/2 ao)2

ρc = ac/ao χt = 1 – Nt [(dt – 2 tt)/2 ao]2

where, as and ac are obtained from Fig.7.6. Step 3. Calculate the shell axial stiffness Ks, tube axial stiffness Kt and stiffness factors Ks,t and J. Ks = π ts (Ds + ts) Es/L Kt = π tt (dt – tt) Et/L Ks,t = Ks/(Nt Kt) J = 1/[1 + (Ks/KJ)]


210

Chapter 7

PARAMETERS as = Ds/2 βs = [12(1 – νs2)]1/4/[(Ds + ts) ts]1/2 ks = (βs Es ts3)/[6(1 – νs2)] λs = [6Ds ks/h3] [1 + h βs + (h2 βs2/2)] δs = [Ds2/4 Es ts][1 – (νs/2)] ac = Dc/2 βc = [12(1 – νc2)]1/4/[(Dc + tc) tc]1/2 kc = (βc Ec tc3)/[6(1 – νc2)] λc = [6 Dc kc/h3][1 + h βc + (h2 βc2/2)] δc = [Dc2/4 Ec tc][1 – (νc/2)] γb = 0 FIG. 7.6A TUBESHEET INTEGRAL WITH SHELL AND CHANNEL (ASME. VIII-1)

PARAMETERS as = Ds/2 βs = [12(1 – νs2)]1/4/[(Ds + ts) ts]1/2 ks = (βs Es ts3)/[6(1 – νs2] λs = [6 Ds ks/h3][1 + h βs + (h2 βs2/2)] δs = [Ds2/4 Es ts][1 – (νs/2)] ac = Gc/2 βc = 0 kc = 0 λc = 0 δc = 0 FIG. 7.6B TUBESHEET INTEGRAL WITH SHELL AND GASKETED WITH CHANNEL, EXTENDED AS A FLANGE (ASME. VIII-1)



PARAMETERS as = Ds/2 βs = [12(1 – νs2)]1/4/[(Ds + ts) ts]1/2 ks = (βs Es ts3)/[6(1 – νs2)] λs = [6 Ds ks/h3][1 + h βs + (h2βs2/2)] δs = [Ds2/4 Es ts][1 – (νs/2)] ac = Gc/2 βc = 0 kc = 0 λc = 0 δc = 0 γb = (Gc – G1)/Do

FIG. 7.6C TUBESHEET INTEGRAL WITH SHELL AND GASKETED WITH CHANNEL, NOT EXTENDED AS A FLANGE (ASME. VIII-1)

PARAMETERS as = Gs/2 βs = 0 ks = 0 λs = 0 δs = 0 ac = Gc/2 βc = 0 kc = 0 λc = 0 δc = 0 γb = (Gc – Gs)/Do

FIG. 7.6D TUBESHEET GASKETED WITH SHELL AND CHANNEL (ASME. VIII-1)


211

212

Chapter 7

7.3.3 Design Equations The design of fixed tubesheets is an iterative process. A tubesheet thickness is first assumed and its stress is calculated. If the stress of the assumed tubesheet is too high then a new thickness is assumed and the process repeated. The design process is complicated by the fact that differential thermal expansion normally occurs between the tubes, tubesheet, and the shell. Hence different loading conditions must be investigated as well. The design procedure generally proceeds as follows, Step 1. Some of the loading conditions to be evaluated are Loading Case 1: Pt is acting, Ps = 0, differential thermal expansion = 0 Loading Case 2: Ps is acting, Pt = 0, differential thermal expansion = 0 Loading Case 3: Pt is acting, Ps is acting, differential thermal expansion = 0 Loading Case 4: Pt =0, Ps = 0, differential thermal expansion is acting Loading Case 5: Pt is acting, Ps = 0, differential thermal expansion is acting Loading Case 6: Ps is acting, Pt = 0, differential thermal expansion is acting Loading Case 7: Pt is acting, Ps is acting, differential thermal expansion is acting The value of h’g is set to zero for thermal stress load cases 4, 5, 6, and 7. Step 2. Determine the preliminary values from the equations in Section 7.3.2. Step 3. Estimate an approximate tubesheet thickness. A minimum value is that obtained from the shear equation due to the larger of the shell side or tube side pressure and is given by h = ao P/(1.6 S µ)

Step 4. From Fig.7.5 calculate the quantities E* and ν* relative to h/p. Step 5. Calculate the quantity Xa = [24(1 – ν*2) Nt Et tt (dt – tt) ao2/(E* L h3)]1/4 and obtain the quantities Zd, Zv, and Zm from Fig. 7.7.

Step 6. Calculate the “parameters” from Fig.7.6 for any specific configuration. Step 7. Calculate the constants K = A/Do F = [(1 – ν*)/E*] [λs + λc + E (ln K)] Φ = (1 + ν*) F Q1 = (ρs – 1 – Φ Zv)/(1 + Φ Zm) QZ1 = (Zd + Q1 Zv) Xa4/2 QZ2 = (Zv + Q1 Zm) Xa4/2 U = [Zv + (ρs – 1)Zm] Xa4/(1 + Φ Zm)

Step 8. Calculate γ as follows For loading cases 1, 2, and 3: γ = 0 For loading cases 4, 5, 6, and 7: γ = [αt,m (Tt,m – Ta) – αs,m (Ts,m – Ta)] L



FIG. 7.7 XA VS. ZD, ZV, OR ZM (ASME. VIII-1)

Step 9. Calculate the quantities ωs = ρs ks βs δs (1 + h βs) ωs* = [ao2 (ρs2 – 1)(ρs – 1)/4] – ωs ωc = ρc kc βc δc (1 + h βc) ωc* = ao2 {[(ρc2 + 1)(ρc – 1)/4] – (ρs – 1)/2} – ωc

Step 10. Calculate equivalent pressures P’s, P’t, Pγ, Pw, Prim and effective pressure Pe P’s = {χs + 2(1 – χs) νt + (2/Ks,t)(Ds/Do)2 νs – (ρs2 – 1)/(J Ks,t) – (1 – J)[DJ2 – (2 as)2]/(2 J Ks,t Do2)} Ps P’t = [χt + 2(1 – χt) νt + (1/J Ks,t)] Pt


213

214

Chapter 7

Pγ = [(Nt Kt)/(π ao2)] γ Pw = –[(Uγb)/(2 π ao2)] W Prim = – (U/ao2) (ωs* Ps – ωc* Pt) Pe = {J Ks,t/{1 + J Ks,t [QZ1 + (ρs – 1) QZ2]}}(P’s – P’t + Pγ + Pw + Prim)

Step 11. Calculate the quantities Q2 = [(ωs* Ps – ωc* Pt) + (γbW/2 π)]/(1 + Φ Zm) and Q3 = Q1 + [(2 Q2)/(Pe ao2)] then using Xa and Q3 find the value of Fm from Figs. 7.8 or 7.9.

FIG. 7.8 XA VS. FM WITH NEGATIVE Q3 (ASME. VIII-1)



FIG. 7.9 XA VS. FM WITH POSITIVE Q3 (ASME. VIII-1)

Step 12. Calculate the bending stress in the tubesheet from the equation σ = (1.5 Fm/µ*)[2 ao/(h – hg’)]2 Pe

For loading cases 1, 2, and 3:

|σ| < 1.5 S

For loading cases 4, 5, 6, and 7:

|σ| < SPS

Step 13. Calculate the shearing stress in the tubesheet at the outer edge of the perforated region τ = (1/(2 µ))(ao/h) Pe

For all loading cases:

|τ| < 0.8 S

Step 14. Calculate the stress at the outermost tube row. Fq = (Zd + Q3 Zv) Xa4/2 σt,o = [(Ps χs – Pt χt) – Pe Fq]/(χt – χs)


215

216

Chapter 7

When the tube stress is in tension For loading cases 1, 2, and 3:

|στ,o| < St

For loading cases 4, 5, 6, and 7:

|στ,o| < 2 St

When the tube stress is in compression Calculate the allowable buckling stress as follows t = K rt = {[dt2 + (dt – 2 tt)2]1/2}/4 Ft = t/rt Ct = [(2 π2 Et)/Sy,t]1/2 Fs = MAX [(3.25 – 0.5 Fq), (1.25)]

Fs need not be greater than 2.0

When Ct < Ft then Stb = MIN {[(π2 Et)/(Fs Ft2)], [St]} When Ct > Ft then Stb = MIN {[(Sy,t/Fs)(1 – (Ft/2 Ct))], [St]} Step 15. Calculate the stress in the shell integral with the tubesheet. σs,m = [ao2/2(as + ts) ts][Pe + (ρs2 – 1)(Ps – Pt)] + [as2/2(as + ts) ts] Pt σs,b = (6 ks/ts2) {βs[δs Ps –(νs as σs,m)/Es] + [6(1 – ν*2)/E*][ao3/h3] (1 + hβs/2) × [Pe(Zv + ZmQ1) + 2 ZmQ2/ao2]} σs = σs,m + σs,b

< 1.5 Ss for loading cases 1, 2, and 3

σs = σs,m + σs,b

< SPS,s for loading cases 4, 5, 6, and 7.

Step 16. Calculate the stress in the channel integral with the tubesheet. σc,m = [ac2/2(ac + tc) tc] Pt σc,b = (6kc/tc2){βc δc Pt – [6(1 – ν*2)/E*][ao3/h3](1 + hβc/2) × [Pe (Zv + Zm Q1) + 2 Zm Q2/ao2]} σc = σc,m + σc,b

< 1.5 Sc for loading cases 1, 2, and 3

σc = σc,m + σc,b

< SPS,c for loading cases 4, 5, 6, and 7.

Example 7.2 Problem A fixed-tubesheet heat exchanger, with the tubesheet extended as a flange, has details as shown in Fig. 7.6C. The shell has no expansion joint. The tubes are on a triangular pitch. Check the thickness of the shells, tubes, and tubesheet. The design data are as shown in the table below.



Design

Shell Side

Tube Side

Pressure Temperature Shell and channel material Joint efficiency Tube material Tubesheet material S for shells and tubesheet material S for tube material Sy for tube material

370 psi 300°F SA 516-70 1.0

110 psi 300°F SA 516-70 0.85

217

SA 249-316L SA 516-70 20,000 psi

16,700 psi 14,200 psi 18,000 psi

Additional Design Data: Ds = 46.0 in.

Dc = 46.0 in.

tt = 0.083 in.

ro = 21.5 in.

E = Es = 28,000 ksi

tc = 0.4375 in.

p = 1.875 in.

dt = 1.5 in.

Et = Ec = 27,000 ksi t,x = 3.0 in.

UL = 0 αt = αc = 9.0 ×

ts = 1.0 in.

= 36.0 in. αs = 6.26 ×

10–6in./in./°F

Nt = 1835

J = 1.0

h′g = 0

ρ = 0.9

A = 49.0

Gl = 47.875

L = 264.0 in.

10–6in./in./°F

ν = νs = νc = νt = 0.3 AL = 0

Gc = 47.875

k = 1.0

Solution Channel Shell From Eq.(2.1), t = 110 (46.0/2)/(16,700 × 0.85 – 0.6 × 110) = 0.178 inch

This thickness is less than the actual furnished thickness of 0.4375 inch. Use 0.4375 inch. The 0.4375 inch thickness will be needed to satisfy the bending requirements of the tubesheet design as discussed later. Shell-Side Shell From Eq.(2.1), t = 370 (46.0/2)/(20,000 × 1.0 – 0.6 × 370) = 0.426 inch

This thickness is less than the actual furnished thickness of 1.0 inch. Use 1.0 inch. The 1.0 inch thickness will be needed to satisfy the bending requirements of the tubesheet design as discussed later. Tubes For internal pressure, Eq.(2.7) gives t = 110 (1.5/2)/(14,200 × 1.0 + 0.4 × 110) = 0.0058 inch


218

Chapter 7

This thickness is less than the actual furnished thickness of 0.083 inch. Use 0.085 inch. External pressure calculations show that the furnished tube thickness of 0.083 inch is adequate to resist the external pressure of 370 psi. Tube Sheet Step 1. Loading case 2 will be used to illustrate the calculation sequence. In this case, Ps = 370 psi and Pt = 0 psi and the differential temperature = 0. In actual design all seven loading must be investigated. Step 2. Preliminary values are obtained from Section 7.3.2 From Eq.(7.1), Do = 2 × 21.5 + 1.5 = 44.5 µ = (1.875 – 1.5)/1.875 = 0.2 d* = MAX {[1.5 – 2 × 0.083(27000/28000)(14200/20000)0.9], [ 1.5 – 2 × 0.083]} d* = MAX [1.3977, 1.3340] = 1.3977 Ck = 0 p* = 1.875 µ* = (1.875 – 1.3977)/1.875 = 0.2546

From Section 7.3.2, ao = 44.5/2 = 22.25 ρs = (46/2)/22.25 = 1.0337 ρc = (47.875/2)/22.25 = 1.0758 χs = 1 – 1835(1.5/(2 × 22.25))2 = –1.085 χt = 1 – 1835((1.5 – 2 × 0.083)/(2 × 22.25))2 = –0.649

and the stiffnesses are given by Ks = π(1.0)(46 + 1.0)(28000000)/264 = 15,660,400 Kt = π(0.083)(1.5 – 0.083)(27000000)/264 = 37,788 Ks,t = 15660400/(1835 × 37788) = 0.2258 J = 1.0

Step 3. An approximate minimum tubesheet thickness is calculated as h = (22.25)(370)/[1.6(20000)(0.2546)] = 1.01 inch



After a number of trials, a thickness of h = 6.75 inches will be checked. Step 4. From Fig. 7.5 with a triangular pitch, h/p = 6.75/1.875 = 3.6

ν* = 0.4

E*/E = 0.21

E* = 5,880,000

Step 5. Xa = [24(1–0.42)(1835)(27000000)(0.083)(1.5–0.083)(22.25)2/(5880000 × 264 × 6.753)]1/4 = (5.8157 × 1013/47.7411 × 1010)1/4 = 3.3222

and from Fig.7.7, Zd = 0.05

Zv = 0.10

Zm = 0.48

Step 6. From Fig.7.6C, as = 46.0/2 = 23.0 βs = [12(1– 0.32)]1/4/[(46 + 1.0) 1.0]1/2 = 0.2652 ks = (0.2652 × 28000000 × 1.03)/[6(1– 0.32)] = 1,360,000 λs = [6 × 46 × 1360000/6.753][1 + 6.75 × 0.2652 + (6.752 × 0.26522/2)] = 5,360,824 δs = [462/4 × 28000000 × 1.0][1 – (0.3/2) ] = 1.6059 × 10–5 ac = 47.875/2 = 23.9875 βc = 0 kc = 0 λc = 0 δc = 0 γb = 0


219

220

Chapter 7

Step 7. K = 49/44.5 = 1.10 F = [(1 – 0.4)/5880000] [ 5360824 + 0 + 28000000 (ln 1.10)] = 0.8193 Φ = (1 + 0.4)(0.8193) = 1.1471 Q1 = (1.0337 – 1 – 1.1471 × 0.10)/(1 + 1.1471 × 0.48) = –0.0522 Qz1 = (0.05 + (–0.0522)(0.10)) 3.32224/2 = 2.7272

QZ2 = (0.10 + (–0.0522)(0.48)) 3.32224/2 = 4.5634

U = [0.10 + (1.0337 – 1)(0.48)] 3.32224/(1 + 1.1471 × 0.48)

= 9.1268

Step 8. γ=0

Step 9. ωs = 1.0337 × 1360000 × 0.2652 × 1.6059 × 10–5 (1 + 6.75 × 0.2652) = 16.705 ωs* = [22.252 (1.03372 – 1)(1.0337 – 1)/4 ] – 16.705 = –16.4191 ωc = 1.0758 × 0 × 0 × 0 (1 + 6.75 × 0) = 0 ωc* = 22.252 {[(1.07582 + 1)(1.0758 – 1)/4] – (1.0337 – 1)/2} – 0 = 11.8972

Step 10. Calculate equivalent pressures P’s = {–1.085 + 2(1 – (–1.085)) 0.3 + (2/0.2258)(46/44.5)2 0.3 – (1.03372 – 1)/(1.0 × 0.2258) – 0}(370) = {–1.085 + 1.2510 + 2.8394 – 0.3035 – 0}(370) = 1000 psi



P’t = [–0.649 + 2(1 – (–0.649)) 0.3 + (1/(1.0)(0.2258))] (0) =0 Pγ = [(1835 × 37788)/(π × 22.252)] 0 =0 Pw = –[(9.1268 × 0)/(2 π 22.252)]W =0 Prim = –(9.1268/22.252) (–16.4191 × 0 – 0 × 370) =0 Pe = {(1.0 × 0.2258)/{1 + (1.0 × 0.2258)[2.7272 + (1.0337 – 1) 4.5634]}} × (1000 – 0 + 0 + 0 + 0) = {0.2258/1.6501}(1000) = 136.8 psi

Step 11. Q2 = [(–16.4191 × 370 – 11.8972 × 0) + (0W/2 π) ]/(1 + 1.1471 × 0.48) = –3917.8

and Q3 = –0.0522 + [(2)(–3917.8))/(136.8 × 22.252)] = –0.17

From Fig. 7.9, Fm = 0.09 Step 12. The stress in the tubesheet due to bending is given by σ = (1.5 × 0.09/0.2546)(2 × 22.25/6.75)2 (136.8) = (0.5302)(43.4326)(136.8) = 3150 psi < 1.5 S

Step 13. The shearing stress in the tubesheet is given by τ = (1/(2 × 0.2))(22.25/6.75)(241.5) = 2010 psi < 16000 psi


221

222

Chapter 7

Step 14. Stress in the outermost tube row Fq = (0.05 + 0.0119 × 0.10)(3.3222)4/2 = 3.1179 σt,o = [(370 (–1.085) – 0 (–0.649)) – 136.8 × 3.1179]/((–0.649) – (–1.085)) = (–828)/0.44 = –1882 psi

Check allowable buckling stress t = 1.0 × 36 = 36.0 rt = {[1.52 + (1.5 – 2×0.083)2 ]1/2}/4 = 0.5018 Ft = 36.0/0.5018 = 71.74 Ct = [(2 π2 × 27000000)/18000]1/2 = 172 Fs = MAX [(3.25 – 0.5 × 3.1179), (1.25)] = 1.69 When Ct > Ft then Stb = MIN {[(18000/1.69)(1 – (71.74/2×172))], [14200]} = 8430 psi. Hence the tube stress of 1882 is adequate.

Step 15. Membrane stress in the shell is σs,m = [22.52/2(23+1.0)1.0][136.8 + (1.03372–1)(370–0)] + [232/2(23 + 1.0) 1.0] 0 =0

Bending stress in the shell is σs,b = (6 × 1360000/1.02){0.2652[1.6059 × 10–5 × 370 –(0.3 × 23 × 0)/Es] +[6(1 – 0.42)/5880000][22.253/6.753](1 + 6.75 × 0.2652/2) × [136.8(0.10 + (0.48)(–0.0522)) + 2 × 0.48 × 3917.8/22.252]} = (8160000){0.2652[0.00594] + [8.5714 × 10–7][35.8164](1.8951) × [10.2523 + 7.5972]} = (8160000){0.001575 + (0.0000582)(17.8495)} = 21,300 psi < 30,000 psi



223

Thus, for this loading case, a tubesheet thickness of 6.75 inch is adequate. A reduction of this thickness will result in an overstress of the shell junction.

7.4 ADDITIONAL RULES The ASME VIII-1 code contains additional tubesheet rules for cases where the thickness of the shell or channel attached to the tubesheet is different from the remaining corresponding shell or channel. Rules are also available for calculating the effect of plasticity at the tubesheet-to-shell or tubesheet-to-channel junction as well as the effect of radial differential thermal expansion between the tubesheet and adjacent shell or channel. The ASME VIII-1 code also contains detailed rules for designing floating tubesheets. The equations for floating tubesheet design are similar to those for designing fixed tubesheets.

7.5 METHODS OF ATTACHING TUBES-TO-TUBESHEET Tubes are attached to tubesheets by welding, expansion, or a combination of these two methods. 1. Welding: Welding of tubes-to-tubesheets is made by full strength welds, partial strength welds, full penetration back sided welds, or seal welds. a) Full strength welds: Figure 7.10 sketches (1), (2), (3), and (4) show details of some full strength welds. The thickness of fillet welds through the throat must not be less than the tube thickness. Larger weld thicknesses might be required due to various combinations of tube, tubesheet, and weld materials. Figure 7.11 shows some permissible details of fillet and bevel weld attaching tubes to tubesheets. VIII-1 also gives equations for calculating the minimum length size of weld details in Fig.7.11 due to various parameters and material combinations. b) Partial strength welds: Details of some partial penetration welds are given in Fig.7.10 sketches (5), (6), (7), (8), and (9). The strength of such welds is determined by push out tests described briefly in section 7.5.1. c) Back sided welds: This method of attachment is more expensive than the other welded methods and consists of butt welding of the tube to the far side of the tubesheet using Tig welding, Fig.7.10 sketch (10). This method of attachment is used where the process design does not permit any fluid to accumulate between the tube outside wall and the tube hole in the tubesheet. d) Seal welds: These are non-strength welds applied to prevent leakage of fluids between the tubeside and shellside of heat exchangers. The strength of the tube-to-tubesheet joint is achieved by expansion rather than welding. 2. Expansion: Expansion of tubes into tubesheets is done by rolling, hydraulic expansion, or explosive expansion. a) Rolling: This is the most common and economic method of expanding the tubes into tubesheets. Rolling of tubes into tubesheets is accomplished by utilizing tapered rollers. These rollers, when inserted into the tubes, extrude the tubes to a larger diameter until the tube outside surface makes contact with the tubesheet hole surface. Further rolling of the tubes creates an extruding action that results in a reduction in the original tube thickness and elongation of the tubes. This reduction and elongation action may be detrimental to the tube performance during operation in some chemical processes due to stress corrosion cracking. The strength of such attachments is determined by push out tests conducted in accordance with the VIII-1 code. b) Hydraulic Expansion: This method consists of inserting a pressure devise, Fig. 7.12, into the tube in order to apply the required expansion force. Hydraulic pressures of upwards of 75,000 psi may be applied in order to expand the tube and lock it in the tubesheet hole.


224

Chapter 7

FIG. 7.10. SOME ACCEPTABLE TYPES OF TUBE-TO-TUBESHEET ATTACHMENTS (ASME, VIII-1)



225

FIG. 7.11. SOME ASSEPTABLE TYPES OF TUBE-TO-TUBESHEET STRENGTH WELDS (ASME, VIII-1)

Hydraulic expansion causes minimal reduction in the tube thickness and shortening of the tube length. This shortening of the tubes may be detrimental to the tube performance during operation in some chemical processes due to axial tensile forces that develop in the tube. The strength of tube to tubesheet joints due to hydraulic expansion is determined by push out tests. c) Explosive Expansion: Explosives are inserted into the tube junction and detonated to expand the diameter of the tube. The expansion mechanism is similar to hydraulic expansion. The expansion causes minimal reduction in the tube thickness and shortening of the tube length. The strength of such attachments is determined by push out tests.


226

Chapter 7

FIG. 7.12. SCHEMATIC DETAIL OF A HYDRAULIC EXPANDER.

FIG. 7.13. TYPICAL DETAIL OF GROOVES IN A CLAD TUBESHEET.



227

3. Grooves. Grooves in tube holes of tubesheets, Fig. 7.13, are an essential part of securing the tubes to the tubesheet. This is especially true when the tubes have a lower modulus of elasticity than the tubesheet such as titanium and zirconium. During the rolling process the tube walls, if they are thin enough, will flow into the grooves to form a locking mechanism. VIII-1 does not give rules for groove dimensions. TEMA, however, specifies double grooves for tubes 5/8 inch and larger. The groove dimensions are 1/8 inch in width and 1/64 inch in depth. For tubesheets less than one inch in thickness, the rules require only one groove. For hydraulic expansion the grooves are to be 1/4 inch wide. Extensive tests done by Jawad (Jawad, Clarkin, and Schuessler 1987) indicate that these dimensions may not be adequate for thick tubes. Jawad’s suggestion is that the width of the groove should not be less than the quantity 1.56(ro t)1/2 where, ro is the tube outside radius and t is the tube thickness. Subsequent publications by Yokell and various Japanese authors seem to support this criteria.

7.5.1 Push Out Tests Push out tests are performed in order to determine the ability of the joint to withstand axial loads in the tubes. These axial loads are due to pressure end loads in u-tube heat exchangers or axial mechanical and thermal loads in fixed heat exchangers. The joints are either partially welded, expanded, or a combination of these two methods. The tests are usually performed by the fabricator of the equipment in accordance with Appendix A of VIII-1. The test consists of inserting tubes into a tubesheet, securing them by partial welding or expanding, and then pushing them out. The force required to push the tubes out is compared to the axial tensile force required to break the tubes. The ratio, fr, of these two forces determines the strength ratio of the joint. Table 7.1 lists the various combinations of tube joints and their corresponding fr factors. Usually a push out test is required for each new piece of equipment due to the infinite variation is design parameters between equipment. These design parameters include such items as spacing between the tubes, tube layout patterns (triangular or rectangular), and diameter of the tubes. There are numerous variables that effect the strength of tube to tubesheet joints. Some of these variables include weld details, rolling amount, yield strength differential between tube and tubesheet materials, thickness of tubes and tubesheet, modulus of elasticity differential between tubes and tubesheet, differential thermal expansion between tube and tubesheet, surface finish of the outside surface of the tubes, surface finish of tube holes, ovality of tubes and tube holes, and initial clearance between the outside diameter of the tube and the diameter of the hole. Each of these variables could have a profound influence on the strength of the joint.

7.5.2 Tube-to-Tubesheet Locking Mechanism Expansion of the tubes into the tubesheet results in a locking mechanism whereby the tubesheet holds the tube into place. The theoretical derivation of this locking mechanism involves elastic-plastic equations of the tubes and tubesheets (Jawad, Clarkin, and Schuessler 1987). Jawad’s paper includes correlation of theoretical equations with numerous experimental values. The locking mechanism takes place as the tubes are slowly expanded into the tube holes. When the tubes are initially expanded, the tubes wall stress increases until it reaches the yield stress. Upon further expansion of the tubes the stress remains at yield (assuming elastic-perfectly plastic material) while the strain goes past the elastic range into the plastic range. Further expansion causes the tube to deform plastically until the tube wall touches the tubesheet hole wall. With further expansion, the tubesheet region around the expanding tube is stressed first elastically and then plastically. The rest of the ligament in the tubesheet between the tube holes remains elastic. When the


228

Chapter 7

TABLE 7.1

JOINT EFFICIENCIES fr FOR VARIOUS TUBE-TO-TUBESHEET ATTACHMENTS (ASME VIII-1)

expansion of the tube finally ceases, all tube and tubesheet components that are stressed elastically as well as plastically spring back elastically. The tubesheet, which is stressed partly elastic and partly plastic, springs back substantially more than the tube due to the elastic-plastic interaction within the ligaments. The spring back action creates high pressures between the tube and the tubesheet causing compressive hoop stress in the tubes and tensile stress in the tubesheet. This pressure causes the tube to be locked in place. The locking forces, when done properly, result in a joint that is as strong as the tube strength through its cross section. The mechanism explained above works well as long as the modulus of elasticity and yield strength of the tubes and tubesheet are the same and, as long as the ligament between the tubes is large. Unfortunately this condition rarely occurs in practice. Extensive tests have shown that when the modulus of elasticity of the tubes is substantially lower than the tubesheet, as is the case with titanium and zirconium tubes



229

attached to carbon steel tubesheets, the tube spring back is twice that of the tubesheet and the joint tends to be loose with a small locking pressure, if any. In such situations, grooves and or full strength welds are the only option. Tests have shown that tubes with yield stress substantially higher than the tubesheet require much higher rolling forces to achieve the same locking forces as tubes having the same yield stress as the tubesheet. Tests have also shown that expansion of tubes in tubesheets where the thickness of the tubesheet is less than one fourth the diameter of the tube does not result in significant locking forces. Tests have verified theoretical analysis in that joints of tubes with higher coefficient of thermal expansion than tubesheets, such as copper tubes in steel tubesheets, become tighter at elevated temperature operations. Conversely joints of tubes with lower coefficient of thermal expansion than tubesheets become looser at elevated temperature operations. Tests have shown that tube hole roughness has a great influence on joint strength. Items such as burnishing, polishing, tool marks, and rust films have significant effect. The effect of such factors have not been fully evaluated at the present time. Also, tube hole roughness has a great influence on joint leakage. Items such as burnishing, polishing, tool marks, and rust films have significant effect. The effect of such factors on leakage have not been sufficiently evaluated at present. Tests have also shown that a joint with high locking mechanism does not necessarily have a high resistance to leakage, and vice versa. Tubes, such as stainless steel, titanium, and high strength coppers, that do not have classical elastic-perfectly plastic stress-strain diagrams require higher forces to lock in place than materials with classical elastic-perfectly plastic stress-strain diagrams of similar dimensions. This is due the variable stress-strain diagram that required a much greater force than that calculated by the above mentioned simplified theoretical analysis. Thick tubes with BWG gage of 10 and smaller are harder to seal than thin tubes. This is especially true when the yield strength of the tube is substantially higher than the tubesheet. Grooves and additional welding might be needed to secure such tubes to the tubesheet.

7.6 EXPANSION JOINTS Two types of expansion joints are covered in VIII-1. The first is bellows-type and is given in Mandatory Appendix 26 of VIII-1. Some typical bellows-type expansion joints are shown in Fig. 7.14. They include unreinforced as well as reinforced construction. The approximate required thickness of an unreinforced bellows, Fig. 7.14a, due to pressure is obtained from the pressure-area method as

T = P(Db + W)/S[(π – 2) + 4W/q]

where T is the total bellows thickness (T = nt) and Db, q, n, t, and W are defined in Fig. 7.14. P is the internal pressure, and S is the allowable stress. For the omega expansion bellows, Fig. 7.14c, the approximate required thickness due to pressure is obtained from the pressure-area method as T = (P/2Sr){[(Db + (tc/2)) ro/(π – φ)] + r2}

Where T is the total bellows thickness (T = nt) and Db, n, r, ro, t, tc, and φ are defined in Fig. 7.14. P is the internal pressure, and S is the allowable stress. φ, in radians, is obtained from the quantity φ = sin–1 ro/r. The designer may use these two equations to determine a trial thickness and then use the equations given in VIII-1 to obtain stresses due to various loading conditions at various locations in the bellows and collar rings. VIII-1 also gives fatigue, and other pertinent equations for designing the bellows.


230

Chapter 7

FIG. 7.14. SOME BELLOWS-TYPE EXPANSION JOINTS



231

FIG. 7.15. SOME FLANGED AND FLUED EXPANSION JOINTS (ASME VIII-1)

The second type of expansion joints is the flanged and flued expansion joints, shown in Fig. 7.15. They are discussed in Mandatory Appendix 5 of VIII-1. The method of analysis for pressure and expansion is not explicitly given in VIII-1, but is left to the discretion of the designer. A preliminary design can be made by assuming the configuration to be a structural frame. Subsequently, the stresses can be established at various parts of the expansion joint by analyzing it as a combination of plate and shell components.



CHAPTER

8 ANALYSIS

OF

COMPONENTS IN VIII-2

8.1 INTRODUCTION Section VIII-2 requires stress analysis of vessel components when explicit design formulas are not given. This includes flued-in heads, head-to-shell junctions, expansion joints, thermal stresses, and stresses in components due to loads other than pressure. In performing the stress evaluation, the designer must determine the maximum stress at a given point or location. When computer programs such as ANSYS and NASTRAN are used to determine the stress, the output usually consists of the total combined stress at a given point. This stress must then be separated into its components of membrane, bending, and peak stresses. This is necessary in order to compare each of these components to a corresponding allowable stress given in VIII-2 or to properly establish an allowable fatigue life. In this chapter only stress categories, stress concentrations, combinations of stresses, and fatigue evaluation are discussed in accordance with the definitions and requirements of VIII-2.

8.2 STRESS CATEGORIES Stress in any component and location is classified by VIII-2 as one of three categories—primary, secondary, and peak stresses. Primary stress, such as hoop stress in a cylinder due to internal pressure, is developed by the imposed loading and is necessary to satisfy the laws of equilibrium. It is not self-limiting in that gross distortion or failure of the structure will occur if its value substantially exceeds the yield stress. This primary stress is divided into two subcategories in VIII-2. They are primary membrane and primary bending stresses. The longitudinal and circumferential stresses in a cylinder due to internal pressure are classified as primary membrane stress. The primary membrane stress is again subdivided into two categories in VIII-2. They are referred to as general primary membrane and local primary membrane stresses. Examples of these primary stresses are given in Table 8.1. Primary bending stress in VIII-2 refers to such items as the bending of a flat cover or a dished head due to internal pressure. Secondary stress is developed when the deformation of a component due to applied loads is restrained by other components. Secondary stress is self-limiting in that local yielding can redistribute the stress to a tolerable magnitude without causing failure. An example of secondary stress is the bending stress that develops at the attachment of a body flange to the shell. This attachment is referred to as a gross structural discontinuity. Other examples of gross structural discontinuity are given in Table 8.2. Another example of secondary stress is certain thermal stresses. These are referred to as general thermal stress. A typical example of this stress is the longitudinal bending stress that occurs along a vessel skirt due to temperature gradients along the length of the skirt. Other examples of general thermal stress are given in Table 8.3. 233


234

Chapter 8

TABLE 8.1 PRIMARY STRESS CATEGORY

The third category of stress defined in VIII-2 is peak stress. Peak stress is so local that it does not cause any noticeable distortion in a component, but it may cause fatigue cracks or brittle fracture. Examples of peak stress are notch concentrations; local hot spots; local structural discontinuity, as defined in Table 8.2; and local thermal stress, as defined in Table 8.3 VIII-2 establishes limits for the three stress categories discussed so far. These limits are given in Table 8.4. The rationale for these limits are given in various publications (see such references as ASME, 1968; ASME, 1969; and Jawad and Farr, 1989). VIII-2 also lists the stress categories for some commonly encountered loading conditions and vessel components. These are given in Table 8.5. Example 8.1 Problem A cylindrical shell with a flat cover, see Fig. E8.1, is subjected to an internal design pressure of 800 psi and an internal operating pressure of 700 psi. The allowable stress intensity value for the material from II-D is 20 ksi. What stress intensity values should be calculated at sections a–a and b–b, and what are the allowable stress intensities at these locations? Solution Section a–a From Table 8.5, flat heads develop general primary membrane stress, Pm, and primary bending stress, Pb, at the central region due to the internal design pressure of 800 psi. From Table 8.4 the allowable general primary membrane stress intensity, Pm, is equal to Sm (20 ksi). The allowable primary bending stress intensity, Pb, is equal to 1.5Sm (30 ksi).


Analysis of Components in VIII-2

TABLE 8.2 STRUCTURAL DISCONTINUITY

TABLE 8.3 THERMAL STRESS


235

236

Chapter 8

TABLE 8.4 STRESS CATEGORIES AND THEIR LIMITS (ASME VIII-2)

Section b–b From Table 8.5, flat heads develop local primary membrane stress, PL, and secondary stress, Q, at the junction with the shell due to internal pressure. From Table 8.4 the allowable local primary membrane stress, PL, due to the design pressure of 800 psi is equal to 1.5Sm (30 ksi). The total allowable stress due to local pri-



237

TABLE 8.5 CLASSIFICATION OF STRESSES (ASME VIII-2)

NOTES: (1) Consideration must also be given to the possibility of wrinkling and excessive deformation in vessels with large diameter-to-thickness ratio. (2) If the bending moment at edge is required to maintain the bending stress in the center region within acceptable limits, the edge bending is classified as Pb; otherwise, it is classified as Q. (3) Consider possibility of thermal stress ratchet. (4) Equivalent linear stress is defined as the linear stress distribution which has the same net bending moment as the actual stress distribution.

mary membrane plus secondary stresses (PL + Q) is equal to 3Sm (60 ksi). It should be noted that the two stress values, PL + Q, must be calculated at the operating pressure of 700 psi rather than at the design pressure when comparing them to 3Sm, as shown in Table 8.4.


238

Chapter 8

TABLE 8.5 (CONT’D)

FIG. E8.1



239

TABLE 8.6 SOME STRESS CONCENTRATION FACTORS USED IN FATIGUE Location

Stress Concentration

Fillet welds Nozzle in spherical segment Nozzle in cylindrical segment Tangential nozzle in cylinder Backing strips Bolts Cracklike defect

4.0 2.2 3.3 5.5 2.0 membrane 2.5 bending 4.0 5.0

VIII-2 Paragraph 5–112 4–612 4–612 4–614 AD–412.1 5–122 5–111

8.3 STRESS CONCENTRATION The stress concentration at a given location must be included in the stress analysis in order to establish the fatigue life of that component. VIII-2 lists a few stress concentration factors for fillet welds and nozzle penetrations due to internal pressure, as shown in Table 8.6. All other stress concentration factors are usually obtained from handbooks such as Peterson (Peterson, 1974), experimental data, or a detailed stress evaluation using finite element analysis. It must be remembered that substantial inaccuracies could occur in using the finite element analysis if a large mesh is used near a stress concentration. The designer must exercise great judgment in establishing the correct mesh size near such concentrations in order to obtain accurate results. Example 8.2 Problem Categorize the stresses at section b-b in Example 8.1 if a stress concentration factor (SCF) of 4.0 is used at that location due to weld details. Solution Section b–b From Example 8.1, flat heads develop PL and Q at the junction with the shell due to internal pressure. Similarly, PL due to design pressure is equal to 1.5Sm (30 ksi). The quantity PL + Q due to operating pressure is equal to 3Sm (60 ksi). Also, from Table 8.4, the quantity (SCF)(PL + Q) at the operating pressure of 700 psi must be used to find the quantity Sa in determining the fatigue life of this section.

8.4 COMBINATIONS OF STRESSES In order to compare the actual stress at a given location to the allowable stress in VIII-2, the designer must categorize the calculated stress as primary, secondary, or peak. The designer must then combine the categories in the appropriate fashion in order to compare them to the allowable stresses given in Table 8.4. Identifying the stress category may be difficult sometimes, since the output of many finite element calculations are programmed to display only the principal, or effective, stress at a location. Thus, the designer has to either instruct the program to itemize the stresses or manually separate them into various categories. Separating the stress output from a finite element (FE) program into its components is called “linearization.” A typical computer output may look like the solid line shown in Fig. 8.1. The FE line must be divided into a membrane stress and bending stress as shown. These values can then be compared to the allowable stress given in Table 8.4. A typical output of a detailed stress analysis consists of three normal stresses, σr, σl, σh, and three shearing stresses τrl, τrh, τlh. From these six stresses, the designer can obtain three principal stresses σ1, σ2, σ3, by using the classical equation


240

Chapter 8

FIG. 8.1 LINEARIZING STRESS DISTRIBUTION

σmaxmin = (σi + σj)/2 ± [(σi – σj)2/4 + τ2ij]1/2

(8.1)

The maximum stress intensity defined in VIII-2 is the absolute value of the larger of the following values S12 = σ1 – σ2 S13 = σ1 – σ3 S23 = σ2 – σ3

The maximum stress intensity is compared with allowable values in Table 8.4. Example 8.3 Problem The forces and bending moments in sections a–a and b–b due to design pressure in Example 8.1 were calculated from the classical theory of plates and shells as Section a–a Membrane force in the radial direction = 2602.3 Ib Membrane force in the hoop direction = 0.0 Bending moment in the radial direction = 89,052.0 in.-lb Bending moment in the hoop direction = 89,052.0 in.-lb



241

Section b–b Bending moment in the axial direction = 5988.0 in.-lb Bending moment in the tangential direction = 1796.4 in.-lb Shearing force in the radial direction = 2602.3 lb Membrane force in the axial direction = 9600 lb Membrane force in the hoop direction = 0 (assuming the shell cannot grow radially at this location. A more accurate solution of this problem can be obtained by taking into consideration the inward deflection of point b-b due to the edge rotation. The value of this deflection can be taken as the edge rotation times half the flat cover thickness.) Determine the stress values at sections a-a and b-b in accordance with the VIII-2 procedures and compare them with the allowable stresses. Solution Section a–a The membrane stress is Pm = force/t = 2602.3/6.375 = 410 psi

The bending stress is Pb = 6M/t2 = 6 × 89052/6.3752 = 13,150 psi

From Table 8.4, Allowable Pm = 20,000 psi > 410 psi Allowable Pm + Pb = 30,000 psi > 13,560 psi (13150 + 410)

Section b–b The axial membrane stress at design pressure is Pm = force/t = 9600/1.0 = 9600 psi

The axial membrane stress at operating pressure is Pm = (operating P/design P)(force/t) = (700/800)(9600/1.0) = 8400 psi

The axial bending stress at operating pressure is Q = (operating P/design P)(6M/t2) = (700/800)(6 × 5988.0/1.02) = 31,440 psi

From Table 8.4, Allowable Pm = 20,000 psi > 9600 psi Allowable Pm + Q = 60,000 psi > 39,840 psi (8400 + 31,440)


242

Chapter 8

The hoop membrane stress at design pressure is Pm = force/t = 0/1.0 = 0 psi

The hoop membrane stress at operating pressure is Pm = (operating P/design P)(force/t) = (700/800)(0/1.0) = 0 psi

The hoop bending stress at operating pressure is Q = (operating P/design P)(6M/t2) = (700/800)(6 × 1796.4/1.02) = 9430 psi

From Table 8.4, Allowable Pm = 20,000 psi > 0 psi Allowable Pm + Q = 60,000 psi > 9430 psi (9430 + 0)

Example 8.4 Problem A finite element (FE) analysis was performed on a flat head-to-shell junction, shown in Fig. E8.4. Three different loading conditions were calculated. They were pressure, mechanical, and thermal loading. The results of the FE stress output are shown in Table E8.4. Assume the operating and design pressures are the same and all initial stress values are equal to zero. Assume the allowable stress value to be 14 ksi. Calculate the primary membrane stress and the secondary stress at the junction. Solution Primary Membrane Stress Table 8.4 indicates that primary membrane stress is produced by mechanical loads only. Thus, in Table E8.4 under the Membrane Stress part, only the pressure, mechanical, or a combination of pressure and mechanical are to be used. Thermal stresses are ignored in this case. The FE results indicate that there is a shearing stress in the r,l plane. Thus the two principal stresses, σ1 and σ2, in this plane are calculated from Eq. (8.1), while the third principal stress is σh. The three principal stresses become

Pressure, psi Mechanical, psi Pressure plus mechanical, psi

σ1

σ2

σh

6200 1370 7230

–900 30 –530

11,400 500 11,900

And the maximum stress intensity values are given by Maximum Stress Intensity, psi Pressure Mechanical Pressure plus mechanical

12,300 1340 12,430

Allowable Pm = 14,000 psi > 12,430 psi



243

FIG. E8.4 MODEL OF A FINITE ELEMENT LAYOUT IN A FLAT HEAD-TO-SHELL JUNCTION

Secondary Stress Table 8.4 indicates that secondary stress is produced by mechanical and thermal loads. The FE results indicate that there is a shearing stress in the r,l plane. Thus the two principal stresses, σ1 and σ2, in this plane are calculated from Eq. (8.1), while the third principal stress is σh. The three principal stresses become σ1 Pressure, psi Mechanical, psi Pressure plus mechanical, psi Pressure plus mechanical plus thermal, psi

2100 1760 3650 12,920

σ2

σh

–540 140 –1650 –1520

11,000 –700 10,300 14,000

And the maximum stress intensity values are given by Maximum Stress Intensity, psi Pressure Mechanical Pressure plus mechanical Pressure plus mechanical plus thermal

11,540 2460 11,950 15,520

Allowable Pm + Q = 42,000 psi > 15,520


244

Chapter 8

TABLE E8.4 SUMMARY OF FINITE ELEMENT OUTPUT

8.5 FATIGUE EVALUATION When a fatigue evaluation is required in accordance with AD-160 of VIII-2 or by the user or a qualified engineer, it shall be performed in accordance with the requirements of Appendix 5 of VIII-2. The number of cycles are evaluated from fatigue charts such as the one shown in Fig. 8.2 for carbon steel. In most applications the process cycle is easily determined. Each cycle consists of a start-up condition in pressure and temperature, a steady state, and then shutdown of pressure and temperature. This is illustrated in Fig. 8.3a. More complex cycles often occur where there is reversal of stress, as shown in Fig. 8.3b. On occasion, complicated cycles occur, such as the one shown in Fig. 8.3c. For each cycle the designer determines the maximum stress range, which is the algebraic difference between the maximum and minimum stress intensities in a cycle. The alternating stress, which is half the maximum stress range, is then obtained. With this value, the fatigue chart is then used to obtain the number of permissible cycles for each stress range. A Cumulative Usage Factor is then determined for each type of cycle considered by the designer.


FIG. 8.2 FATIGUE CURVES FOR CARBON, LOW ALLOY, 4XX, HIGH ALLOY, AND HIGH STRENGTH STEELS FOR TEMPERATURES NOT EXCEEDING 700°F (ASME VIII-2)



245

246

Chapter 8

FIG. 8.3 CYCLIC CURVES

Example 8.5 Problem Use the Peak stress values given in Example 8.4 to determine the fatigue life at the location indicated in Fig. E8.4. Use Fig. 8.2 for a fatigue chart. Solution Table 8.4 indicates that the peak stress must be combined with the membrane and bending stresses to determine fatigue life.



247

Peak Plus Secondary Stress From Table E8.4, we combine the peak stresses for pressure, mechanical, and thermal conditions with those of membrane plus bending stresses. This gives peak plus membrane plus bending stress, as shown below: σr Pressure, psi Mechanical, psi Pressure plus mechanical, psi Pressure plus mechanical plus thermal, psi

–2000 1250 –750 –950

σl

σh

σrl

2000 3900 5900 7900

11,000 –1950 9050 10950

200 –1950 –1750 –1050

The three principal stresses become

Pressure, psi Mechanical, psi Pressure plus mechanical, psi Pressure plus mechanical plus thermal, psi

σ1

+σ2

σh

2010 4930 6400 8020

–2010 220 –280 –1070

11,000 –1950 9050 10,950

And the maximum stress intensity values are given by Maximum Stress Intensity, psi Pressure Mechanical Pressure plus mechanical Pressure plus mechanical plus thermal

13,010 6880 9330 12,020

The maximum alternating stress is Sa = 13010/2 = 6500 psi

From Fig. 8.2, with Sa equal to 6500 psi, the maximum number of cycles is > 1,000,000.



REFERENCES API, 2001, American Petroleum Institute, Recommended Rules for Design and Construction of Large, Welded, Low-Pressure Storage Tanks, API 620, Washington, D.C., API. ASCE, 2002, American Society of Civil Engineers, Minimum Design Loads for Buildings and Other Structures, ASCE 7-02, New York, ASCE. ASME, 2004a, American Society of Mechanical Engineers, Boiler and Pressure Vessel Code, Section VIII, Division 1, Pressure Vessels, New York, ASME. ASME, 2004a, American Society of Mechanical Engineers, Boiler and Pressure Vessel Code, Section VIII, Division 2, Alternative Rules for Pressure Vessels, New York, ASME. ASME, 2002, American Society of Mechanical Engineers, B31.3, Process Piping, New York, ASME. ASME, 1969, American Society of Mechanical Engineers, Criteria of the ASME Boiler and Pressure Vessel Code for Design by Analysis in Sections III and VIII, Division 2, New York, ASME. ASME, 1968, American Society of Mechanical Engineers, Section VIII-Division 2 of the ASME Boiler and Pressure Vessel Code-Guide to Alternative Rules for Pressure Vessels, New York, ASME. Beer, F.P., Johnston, E.R., Eisenberg, E.R., and Staab, G., 2001, Mechanics of Materials, New York, McGraw Hill. Booten, M., and Tennyson, T.C., 1977, “Design Criteria for Elastic Buckling of Circular Cylinders Under Combined Loading”, Proc. Int. Coll. On Stability of Structures Under Static and Dynamic Loads, Washington, D.C., ASCE. ICBO, 1997, International Conference of Building Officials, Uniform Building Code, Whittier, CA, ICBO. Jawad, M.H., Clarkin, E.J., and Schussler, R.E., 1987, Evaluation of Tube-to-Tubesheet Junctions, Journal of Pressure Vessel Technology, New York, ASME. Jawad, M.H., and Farr, J.R., 1989, Structural Analysis and Design of Process Equipment, New York, ASME Press. John Wiley & Sons. Jawad, M.H., 1994, Background of the Half-Pipe Jacket Rules in Section VIII, Division 1, Journal of Pressure Vessel Technology, New York, ASME. Jawad, M.H., 2004, Design of Plate and Shell Structures, New York, ASME Press. G + W Taylor-Bonney, Bulletin 502: Modern Flange Design, 7th Edition, Southfield, MI, G + W. Peterson, R.E., 1974, Stress Concentration Factors, New York, John Wiley & Sons. 249


250

References

Prager, W., and Hodge, P.G., 1965, Theory of Perfectly Plastic Solids, New York, John Wiley & Sons. Shield, R.T., and Drucker, D.C., June 1961, “Design of Thin-Walled Torispherical and Toriconical Pressure Vessel Heads,” Journal of Applied Mechanics, New York, ASME. TEMA, 1999, Tubular Exchanger Manufacturers Association, Inc., Standards of Tubular Exchanger Manufacturers Association, 8th ed., Tarrytown, NY. TEMA. Waters, E. O., Wesstrom, D. B., and Williams, F. S. G., 1937, “Formulas for Stresses in Bolted Flanged Connections,” Transactions of the ASME, New York, ASME. Zick, L. P., and Germain, A. R., May 1963. “Circumferential Stresses in Pressure Vessel Shells of Revolution,” Journal of Engineering for Industry, New York, ASME.


APPENDIX A GUIDE TO VIII-1 REQUIREMENTS

251


Downloaded From: http://ebooks.asmedigitalcollection.asme.org/ on 01/05/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use Inside Edges UG-76

Flanged and Flued Expansion Joints Appendix 5

Bellows-Type Expansion Joints Appendix 26

Structural Attachments U-1(e); UG-5, 22, 54, 82; UCS-66(a); UHT-28, 85; ULW-22; ULT-30; Appendix G

Stiffener Plate UG-5, 22, 54, 82

Plug Welds UW-l 7, 19; UW-37

Corrosion UG-16; UG-25; UF-25; NF-13; UHA-6; UCL-25; UHT-25; Appendix E

Applied Linings or Integrally Clad Plate UG-26; Part UCL; Appendix F; Appendix 27

Studs and Bolts UG-12; UCS-10; UNF-12; UHA-12; UCI-12; UCD-12; Appendix 2-2; Appendix 24

Nuts and Washers UG-13; UCS-11; UNF-13; UHA-13; Appendix 2-2

Head Skirt UG-32, 33; UW-13; UHT-19

To

nce lera

F-2

;U -81

UG

9

Category A Longitudinal Joint UW-2, 3, 9, 11, 12, 33, 35; UCS-19; UNF-19; UHA-21; UHT-17, 20; ULW-17; ULT-17

Junction Weld UW-9

Backing Strip UW-2; Table UW-12, UCS-66(a); UCS-66(h)

Openings in or Adjacent to Welds UG-36; UW-14; ULT-18

Bars and Shapes UG-14; UCS-12; UNF-14

Nuts and Washers UG-13, UCS-11, UNF-13; UHA-13; UCI-12, UCD-12

Bolts UG-12; UCS-10; UNF-12, UHA-12; UCI-12; UCD-12

Plate UG-5,10,16, 76, 79; UW-5; UCS-6; UNF-6

Pipes and Tubes UG-8, 10, 16, 31; UCS-9, 27

Charpy Impact Tests (Toughness) UG-20(f), UG-84, UCS-66, 67, 68, UNF-65, UHA-51, UHT-5, 6, 82; ULT-5

Heat Treatment UG-11, 85; UW-10, 40, 49; UCS-56, 85; UNF-56, 79; UHA-32, 44, 105; UCL-34; UF-31, 52; UHT-56, 80, 81; ULW-26; ULT-56; Appendix 31, R

Forgings UG-6, 11; Part UF; UCS-7; UNF-7; Appendix 22

Castings UG-7, 11, 24; UCS-8; UNF-8; UHA-8; Part UCI; Part UCD; Appendix 7

Material and Heat Treatment Table

Attachment Details UW-9,13, 33; UHT-20; ULW-17; ULT-17

Forming UG-79, 96; UF-28; UCS-79; UHA-44; UNF-77; UHT-79; ULT-79

Saddle-Type Nozzle, Category D UW-2, 3, 11,16; UHT-17,18

Ellipsoidal Heads UG-16, 79, 81; UCS-79 Internal Pressure UG-32; UHT-32; Appendix 1; Appendix L-1 External Pressure UG-33; UHT-33; Appendix L-6 Spin Holes; UW-34

Category B Welded Joint UW-2, 3, 11, 33, 35; UCS-19; UNF-19; UHA-21; UHT-17; ULT-17

Openings UG-36-42; ULW-16,18; Appendix 1-7, L-7

Category C Welded Joint UW-2, 3, 11; UNF-19; UHT-17; ULW-17; ULT-17; Appendix 2-4

Full-Faced Gasket Appendix 1-6

Spherically-Dished Covers UG-11, 35.1; UCI-35; UCD-35; Appendix 1-6

Nozzle Neck UG-16, 27, 43, 45; UW-13, 16; UHT-18; ULW-18; ULT-18

Multiple Openings UG-42, 53

Bolted Flange Connections UG-44, App. 2, S, Y

Slip-On Flange UG-11, 44, Appendix 2

2004 EDITION

Lap Joint and Loose-Type Flange UG-11, 44; UW-2; Appendix 2; Appendix S, Y

Protective Devices UG-125-137, Appendix 11; Appendix M

Half Pipe Coils Appendix EE

Category B Circumferential Joints UW-2, 3, 9,12, 33, 35; UHT-17, 20; UCS-19; ULW-17; ULT-17; UNF-19; UHA-21

Corner Joints UG-93; UW-9,13, 18 Appendix 28

Welded Flat Heads UG-34; UW-13; ULW-17; Appendix 14

Opening in Flat Heads UG-39; Appendix 14

Side Plates, Rectangular Vessels UW-13; Appendix 13

Welded Connection UW-15, 16; UHT-17-18; ULW-18; ULT-18

Stiffening Rings UG-29, 30; UCS-29; UNF-30; UHA-29, 30; UHT-28-30; ULT-29; Appendix L

Shell Thickness UG-16; UCS-27; Appendix 32 Internal Pressure UG-27; Appendix 1, L External Pressure UG-28; UCS-28; UNF-28, UHA-28; UCI-28, 29; UCL-26; UCD-28, UHT-27; ULW-116; ULT-28; Appendix L

Unequal Thickness UW-9,13, 33, 42; UHT-34

Hemispherical Head UG-16, 79, 81; UCS-79; UHT-34; UNF-79 Internal Pressure UG-32; UHT-32; Appendix 1, L External Pressure UG-33; UHA-31; UCL-26; UHT-33; Appendix L

Nonpressure Parts U-1(e)(2), UG-4, 22, 54, 55, 82; UW-5, 27, 28, 29; UHT-85; ULW-22; ULT-30

This guide illustrates some types of construction provided for under ASME Section VIII, Division 1 Code, and is subject to the rules in the current edition of the Code. This guide should be used only as a quick reference.

Quick Reference Guide ASME BOILER AND PRESSURE VESSEL CODE SECTION VIII, DIVISION 1

THE HARTFORD STEAM BOILER INSPECTION AND INSURANCE CO. OF CONNECTICUT

Downloaded From: http://ebooks.asmedigitalcollection.asme.org/ on 01/05/2016 Terms of Use: http://www.asme.org/about-asme/terms-of-use Weld Neck and Integral-Type Flange UG-11, 44; Appendix 2, S

Threaded Openings UG-36, 43, 46

Fillet Welds UW-9, 12, 13, 18, 36; UCL-46

Scope and Applicability Part UG-General requirements for all construction and all materials Subsection B Requirements for methods of fabrication Part UW Welding Part UF Forging Part UB Brazing Subsection C Requirements for classes of material Part UCS Carbon and low alloy steels Part UNF Nonferrous materials Part UHA High alloy steels Part UCI Cast iron Part UCL Clad plate and corrosion resistant liners Part UCD Cast ductile iron Part UHT Ferritic steels with tensile properties enhanced by heat treatment Part ULW Layered Construction Part ULT Low Temperature Materials Part UHX Rules for Shell and Tube Heat Exchangers Mandatory Appendices 1-33 Nonmandatory Appendices A-Y, DD, EE, FF, GG

Code Jurisdiction for Piping..............................U-1 Design Pressure ................UG-21; UG-98; UCD-3 Design Temperature......UG-20; UCL-24; UCD-3 Dimpled or Embossed Assemblies ...Appendix 17 Inspector’s Responsibility....................U-2; UG-90 Loadings ............................UG-22; Appendix G, H Low Temperature Service .......UW-2; UCS-66-68; UHA-51; Part ULT Manufacturer’s Responsibility.............U-2; UG-90 Material, General ..........................UG-4, 10, 11, 15 a. Bolts and Studs ...................................UG-12 b. Castings .................................................UG-7 c. Forgings.................................................UG-6 d. Nuts and Washers ...............................UG-13 e. Pipes and Tubes ...................................UG-8 f. Plates .....................................................UG-5 g. Rod and Bar ........................................UG-14 h. Standard Parts ...............................UG-11, 44 i. Welding..................................................UG-9 Material Identification, Marking and Certification...........................UG-77, 93, 94 Material Tolerances.......................................UG-16

Nameplates, Stamping and Reports ....UG-115-120; UHT-115; ULW-115; ULT-115; App. W & 18 Nondestructive Examination a. Liquid Penetrant ........................Appendix 8 b. Magnetic Particle........................Appendix 6 c. Radiography..................................UW-51, 52 d. Ultrasonic ..................................Appendix 12 Porosity Charts ......................................Appendix 4 Pressure Tests.................UG-99, 100, 101; UW-50; UCI-99; UCD-99; ULT-99 Quality Control System .............U-2; Appendix 10 Quick Actuating Closures ...U-1; UG-35; App. FF Service Restrictions ............................UW-2; UB-3; UCL-3; UCD-2 Stress, Maximum Allowable ..........UG-23; App. P; UCS-23; UNF-23; UHA-23; UCD-23; UHT-23; ULT-23; UCL-23; UCI-23; User’s Responsibility .....................UG-125; U-2 Welding Preheat Recommendations....................... Appendix R Welding Qualifications...................UW-26 thru 29; ULT-82; UHT-82; UHA-52; UNF-95

Torispherical Head UG-16, 79, 81; UCS-79; UNF-79 Internal Pressure UG-32; UHT-32; Appendix 1-4, L External Pressure UG-33; UHT-33; Appendix L Spin Holes; UW-34

Knuckle Radius UG-32; UCS-79; Appendix 1-4

GENERAL NOTES

Inspection Openings UG-46

Tubes UG-8, 16, 27, 28, 31; UCS-9; Appendix 23

Baffle UG-5

Small Welded Fittings UG-11, 43; UW-15, 16

Conical Heads UHT-19 Internal Pressure UG-32; UHT-32; Appendix 1-4, 5 External Pressure UG-33; UHT-33; Appendix 1-8

One-Half Apex Angle UG-32, 33; UW-3; Appendix 1-5, 1-8

Reinforcing Ring for Conical Reducers Appendix 1-5, 8

Telltale Hole UW-15(d)

Head Attachment UW-12, 13; UHT-34; ULW-17

Introduction Subsection A

Yoke UG-11

Flued Openings UG-32, 38, 46

Manhole Cover Plate U-1(e)(3) UG-11, 34, 46

Flange Attachment UW-2,15,16; ULW-18, Appendix 2, Fig. 2-4

Tolerance UG-80; UF-27; Appendix 27 Appendix L-4

Category D Welded Joint UW-2, 3, 11, 16, 18; UNF-19; UHT-17, 18

O R G A N I Z AT I O N

©2005. The Hartford Steam Boiler Inspection and Insurance Co. of Connecticut. All rights reserved.

Bolted Flange or Ring Gasket Appendix 2, S, Y

Optional-Type Flange UG-14, 44; UW-13; Appendix 2, S, Y

Studded Connections Appendix 30; UG-12; UG-43, 44; UW-16

Conical Shell Reducer UG-32, 33, 36; UHT-19; Appendix 1-5, 1-8

Support Skirt U-1(e); UG-5, 22, 54; UCS-66(a); UHT-28, 85; ULW-22; ULT-30; Appendix G

Telltale Holes for Corrosion UG-25; UCL-25

Staybolts UG-14, 47-50; UG-83; UW-19

Stayed Surfaces UG-27, 47

Tube to Tube Sheet Joints UW-20; Appendix A; UHX-15

Bars, Structural Shapes and Stays UG-14; UW-19

Blind Flange and Flat Head Bolted UG-11, 34, 44 Tube Sheet Design Part UHX U-2(g); TEMA; BS 5500

Internal Structures UG-5; UG-82; Appendix D

Category C Welded Joint UW-2, 3, 11

Welded Stayed Construction UG-47; UW-19, 37

Jacketed Vessels UG-27, 28, 47; ULW-22; Appendix 9, 21

Jacketed Vessel Closure Ring Appendix 9

Reinforcement of Openings with Pad UG-36, 37, 40, 41, 42, 82; UW-14, 15, 16; UHT-18; ULW-18; Appendix 1-7, L-7


APPENDIX B MATERIAL DESIGNATION TABLE B.1 CARBON STEEL PLATE

TABLE B.2 CHROME-MOLY STEEL PLATE SPECIFICATIONS, SA-387

TABLE B.3 CHROME-MOLY STEEL FORGING SPECIFICATIONS, SA-182

255


256

Appendix B

TABLE B.4 CHROME-MOLY STEEL FORGING SPECIFICATIONS, SA-336

TABLE B.5 QUENCH & TEMPERED CARBON AND ALLOY STEEL FORGINGS, SA-508


JOINT

APPENDIX C EFFICIENCY FACTORS

JOINT EFFICIENCY FACTORS SECTION VIII, DIVISION 1

FIG. C.1

257


258

Appendix C


FIG. C.2


Joint Efficiency Factors 259


FIG. C.3


260

Appendix C


FIG. C.4




FIG. C.5


262

Appendix C


FIG. C.6




FIG. C.7


264

Appendix C


FIG. C.8




FIG. C.9


266

Appendix C


FIG. C.10




FIG. C.11


268

Appendix C


FIG. C.12




FIG. C.13


270

Appendix C


FIG. C.14




FIG. C.15


272

Appendix C


FIG. C.16




FIG. C.17


274

Appendix C


FIG. C.18




FIG. C.19


276

Appendix C

EXAMPLE CALCULATIONS FOR JOINT EFFICIENCY FACTORS SECTION VIII, DIVISION 1

FIG. C.20.E



FIG. C.20.E (CON’D)



APPENDIX D FLANGE CALCULATION SHEETS FLANGE CALCULATION SHEETS Blank fill-in calculation sheets are given for the following types of flanges: Sheet D.1—Ring flange with ring-type gasket Sheet D.2—Slip-on or lap-joint flange with ring-type gasket Sheet D.3—Welding neck flange with ring-type gasket Sheet D.4—Reverse welding neck flange with ring-type gasket Sheet D.5—Slip-on flange with full-face gasket Sheet D.6—Welding neck flange with full-face gasket

FIG. D.1 RING FLANGE WITH RING-TYPE GASKET

279


280

Appendix D

FIG. D.2 SLIP-ON OR LAP-JOINT FLANGE WITH RING-TYPE GASKET


Flange Calculation Sheets

FIG. D.3 WELDING NECK FLANGE WITH RING-TYPE GASKET


281

282

Appendix D

FIG. D.4 REVERSE WELDING NECK FLANGE WITH RING-TYPE GASKET


Flange Calculation Sheets

FIG. D.5 SLIP-ON FLANGE WITH FULL-FACE GASKET


283

284

Appendix D

FIG. D.6 WELDING NECK FLANGE WITH FULL-FACE GASKET


APPENDIX E CONVERSION FACTORS CONVERSION OF U.S. CUSTOMARY UNITS TO SI UNITS1 Quantity

Multiply U.S. Units

By Factor

Linear Dimension

inches (in.) feet (ft.) square inches (in.2) square feet (ft.2) U.S. gallons cubic feet (ft3) pounds (lbm) pounds (lbf) inch-pounds (in.-lb) pounds/sq.in. (psi) bars Btu degrees Fahrenheit (°F) ksi sq. root inches (ksi✓in.)

0.0254 0.3048 0.0006452 0.092903 0.003785 0.02832 0.4536 4.448 8.851 6,894.8 100,000.0 1,005.056 tc = (tf – 32)/1.8 1.099 × 106

Area Volume Mass (weight) Force (load) Bending moment Pressure Heat Units Temperature Fracture toughness 1

To Get SI Units meters (m) meters (m) meters2 (m2) meters2 (m2) meters3 (m3) meters3 (m3) kilograms (kg) newtons (N) newton-meters (N-m) pascals (Pa) pascals (Pa) joules (J) degrees Celsius (°C) Pa sq.rt.meters (Pa✓m)

For other conversions, see ASTM E 380 and Appendix 33 of VIII-1

285



INDEX Butt joints, 6–7 Butt welded, 5–8 Butt welded components, 12

A A value, 42 Allowable stress, 2, 3, 4, 5 ASME Boiler and Pressure Vessel Code, Section VIII, v, see also Section VIII Assignment of materials to curves, 10 Axial buckling stress, critical, 38 Axial compression, cylindrical shells under, 35–41

C C value, 98 Carbon steel plate specifications, 255 Cast Ductile Iron, rules of Part UCD for, 23–24 Cast Iron, rules of Part UCI for, 24 Categories, welded joint, 6–7 Charpy impact-test requirements, 16 Chrome-moly steel forging specifications SA-182, 255 SA-336, 256 Chrome-moly steel plate specifications, 255 Circular flat plate, 97 Circumferential membrane stress, 28 Circumferential stress, 33, 67 Closure design details, 165–168 Cold temperatures, 24 Compensation, inherent, openings with, 128 Component analysis in VIII-2, 233–248 Compression axial, cylindrical shells under, 35–41 cone-to-shell junction at large end of cone in, 72–73 cone-to-shell junction at small end of cone in, 81–83 Compressive stress, 67 Cone-to-cylinder junction large end of inherent reinforcement for, 90 values of Q for, 91 small end of inherent reinforcement for, 92 values of Q for, 93

B B value, 41–42 Beams on Elastic Foundation, 126 Bellows-type expansion joints, 230, 231 Bending moment, 39 Bending stress, 39–40 ligament efficiency for multi-diameter openings for, 182–185 Blind flanges, 102 Boiler and Pressure Vessel Code, Section VIII, ASME, v, see also Section VIII Bolt loads, 101 Bolt sizing, 101 Bolted flanges, 101–105 connections with ring type gaskets, 104–118 Bolted flat head, 12 Bolted flat plates and covers, 102–103 Bolting, flat plates and covers with, 102 Bolting rings, 118 Braced and stayed construction, 159–163 Braced and stayed surfaces, 159–162 Brittle fracture, 9–14 Buckling, of cylindrical shells, 41–42 Buckling equation, 38 Buckling stress, critical axial, 39

287


288

Index

Cone-to-shell junction large, 76–77, 84–87 at large end of cone in compression, 72–73 in tension, 80–81 small, 77–78, 87–89 at small end of cone in compression, 81–82 in tension, 73–74 Conical sections external pressure on, 80–89 internal pressure on, 71–80 VIII-1, 70–89 VIII-2, 89–93 Conical shells, 70 Conversion factors, 284 Corner joint, 13 Corner welded, 5, 8 Corrosion allowance, 30, 31 Covers flat, see Flat plates and covers spherically dished, 124–131 Creep,2 Creep rate, 3 Critical axial buckling stress, 38 Critical strain, lowest, 41 Crown radius of ellipsoidal heads, 64 Curves, assignment of materials to, 10 Cylinders effective length of, 42 elliptical, 53 Cylindrical shells, 27–54 under axial compression, 35–41 buckling of, 41 equations, VIII-2, 51 external pressure on, 41 hoop stress in, 29 lines of support of, under external pressure, 43 mitered, 52–53 openings in, 128–129 under tensile forces, 27–35 thick, 33–35, 51 thin, 27–32, 51

D Design rules, 1 Design temperatures, 2, 4 Dimpled and embossed assemblies, welded stays for, 160–162 Double full fillet lap joint, 6

E E (Joint Efficiency Factors), 4–7, 257–277 Earthquake forces, 27 Edge moment, 101 Effective length of cylinders, 42 EJMA (Expansion Joint Manufacturers Association) Standard, 189 Elastic foundation theory, 126, 208 Elasticity, modulus of, 37, 191 Ellipsoidal heads crown radius of, 64 pressure on concave side of, 62–63 pressure on convex side of, 63–64 VIII-1, 62–65 VIII-2, 68–70 Elliptical cylinders, 53 Elliptical shells, 53–54 Empirical equations, 45 Enameled vessels, 24 EPC (External Pressure Charts), 36–37 Excess area, 72 Expansion Joint Manufacturers Association (EJMA) Standard, 189 Expansion joints, 229–231 External pressure on conical sections, 80–89 on cylindrical shells, 41–51 on ellipsoidal heads, 63 on torispherical heads, 67 lines of support of cylindrical shells under, 43 reinforced opening design for, 126 in spherical shells, 59–61 External Pressure Charts (EPC), 36–37

F F-factor, 129–130 Factor 1.1, 2 Fatigue, stress concentration factors used in, 239 Fatigue curves, 21, 245 Fatigue evaluation, 244–247 Fatigue requirements, 19–22 F&D heads (Flanged and Dished heads), 65 Figures, list of, xvii-xix Fillet welded, 5, 9 Fixed tubesheets, 208–223 design equations for, 212–217 details for, 210–211 Flange calculation sheets, 279–284 Flange connections, bolted, with ring type gaskets, 104–120


Index 289

Flange rings, 120 Flanged and Dished heads (F&D heads), 65 Flanged and flued expansion joints, 229–231 Flanges blind, 102 bolted, see Bolted flanges flat-face, 118 full-face gasket, 117, 118 integral, 105, 118 lap-joint, see Lap-joint flanges loose, 105 optional, 105 reverse, 113–117 ring, 105 slip-on, see Slip-on flanges standard, 105–112 welding neck, see Welding neck flanges Flat-face flanges, 118 Flat head bolted, 13 integral, 13 Flat plates and covers, 97–100 bolted, 102 with bolting, 103 circular, with bolting, 103 circular integral, 97–100 integral, 97–101 multiple openings in rims of, 109 noncircular, with bolting, 102 noncircular integral, 99–100 openings in, 102–104 unstayed, 97 Flues, 74 Forces earthquake, 27 meridional, 61 tensile, see Tensile forces Fracture, brittle, 9–14 Full-face gasket slip-on flange with, 283 welding neck flange with, 284 Full-face gasket flanges, 117

G Gasket crushout, check for, 102 Gasket design requirements, 101 Gaskets full face, see Full-face gasket ring type, see Ring-type gasket Geometric parameters, 191 Glass-lined vessels, 24 Group numbers, 3

H Half-pipe jackets, 168–175 maximum allowable internal pressure in, 169–171 minimum thickness of, 171–172 Head configurations, 55 Heads conical, 70 ellipsoidal, see Ellipsoidal heads hemispherical, see Hemispherical heads spherically dished, 118 torispherical, see Torispherical heads toriconical, 73 Heat exchangers configurations of, 190 design of, 189–232 design rules for components of, 189 example, 8 tubesheet, see Tubesheets U-tube, see U-tube exchangers Hemispherical heads, 55 pressure on concave side of, 55–59 pressure on convex side of, 59–62 thickness, 58–59, 60–62 VIII-1, 55–61 VIII-2, 61–62 High alloy steels, without impact testing, minimum design metal temperatures in, 18 Hillside nozzle position, 129 Hoop stress, 4 basic equation for, 53 in cylindrical shells, 29 Hydrostatic, term, 23 Hydrostatic test, 23 for VIII-1, 23–25 for VIII-2, 25–26

I Impact energy, minimum, 16 Impact-test exemption curves, 15 Impact-test requirements, Charpy, 16 Impact testing high alloy steels without, minimum design metal temperatures in, 18 reduction of MDMT without, 17 Inertia, moment of, calculating, 82 Inherent compensation, openings with, 128 Integral flanges, 97–100 Integral flat head, 13 Internal pressure, 27 on conical sections, 71–80


290

Index

Internal pressure (continued) maximum allowable, in half-pipe jackets, 169–171 reinforced opening design for, 126 in cylindrical shells, 27 in spherical shells, 55–59 in elliptical shells, 53 in torispherical shells, 65

J Jacket closure bars, 164–165 Jacket penetrations, 166–167 Jacketed vessels, 163–167 design of closure member for, 164–165 openings in, 165 types of, 163–164 welded stays for, 160–162 Jackets half-pipe, see Half-pipe jackets spiral, 171 Joint categories, welded, 6–7 Joint Efficiency Factors (E), 3, 4, 6, 7, 257–277

K K factor for pipe jacket, 170–175 Knuckle, thickness required for, 73 Knuckle radius, 62

L Lamé’s equation, 33 Lap-joint flanges, 105 with ring-type gasket, 280 Layered vessels, 30 Ligament efficiency, non-circ. vessel for constant-diameter openings, 181 for multi-diameter openings for bending stress, 182–185 for membrane stresses, 181 Ligament efficiency method, 125 Ligament efficiency rules, VIII-1, 154–157 Loads, 1 bolt, 101 radial, 49 shear, 49 vacuum, 45 wind, 39 Longitudinal stress, 33 Loose flanges, 101, 105, 118

M Mandatory rules, 1 Material designation, 255–256 Materials to curves, assignment of, 10 MAWP (maximum allowable working pressure), 23, 24 Maximum allowable working pressure (MAWP), 23, 24 MDMT, see Minimum Design Metal Temperature Membrane stress, 24 circumferential, 27 ligament efficiency for multi-diameter openings for, 181–185 in spherical shells, 55 Meridional forces, 61 Meridional stress, 67 Miscellaneous transition sections, 93 Minimum Design Metal Temperature (MDMT), 12 in high alloy steels without impact testing, 18 reduction of, without impact testing, 17 Mitered cylindrical shells, 52–53 Modulus of elasticity, 37, 191 Moment of inertia, calculating, 81–82

N Non-Mandatory rules, 1 Noncircular cross section, vessels of, 175–187 Noncircular flat plates, 99, 102 Nozzle connections, 52 Nozzle design, alternative rules for, 148–154 Nozzle nomenclature, 126, 144 Nozzle reinforcement, 136–141

O Obround cross section, vessels of, 179 Openings, 125–157 code bases for acceptability of, 125 constant-diameter, ligament efficiency for, 181 in cylindrical shells, 129 exceeding size limits, 142–143 in flat plates and covers, 102–104 with inherent compensation, 128 in jacketed vessels, 165 multi-diameter, ligament efficiency for, see Ligament efficiency for multidiameter openings multiple, in rims of flat heads or covers, 104 nozzle reinforcement of series of, 155–156


Index 291

reinforced, see Reinforced openings in spherical shells, 129 terms and definitions for, 126 in vessels of noncircular cross section, 180 Optional flanges, 105 Outside radius, 56

P P numbers, 3 Part UCD for Cast Ductile Iron, rules of, 24 Part UCI for Cast Iron, rules of, 24 Part ULT rules, 24 Peak stress, 125, 234 Penetrations, jacket, 166–167 Plate diameter opening diameter does not exceed half, 103 opening diameter exceeds half, 104 Plates, flat, see Flat plates and covers Plates on elastic foundation, 213 Pneumatic, term, 23 Pneumatic test, 23 for VIII-1, 24 for VIII-2, 25 Poisson’s ratio, effective, 208 Pressure on concave side of ellipsoidal heads, 62–63 of hemispherical heads, 55–59 of torispherical heads, 65–67 on convex side of ellipsoidal heads, 63–65 of hemispherical heads, 59–61 of torispherical heads, 67–68 external, see External pressure internal, see Internal pressure maximum, for thickness, 27 Pressure-area procedure, 74 Pressure boundary, 125 Pressure test requirements for VIII-1, 23–24 for VIII-2, 25–26 Pressure testing, 22–26 Pressure vessels, 22 Primary stress, 125, 233, 234 Process cyclic curves, 246

Q Q factor, 95 Quality factor, 257–277 Quality Factors, 5

Quench and tempered carbon and alloy steel forging specifications, 256

R Radial loads, 49 Radius-to-depth ratio, 62 Reactor, example, 14 Rectangular cross section, vessels of, 180–181, 185–187 Reducers, 52 References, 249–250 Reinforced openings area of reinforcement available for, 132 area of reinforcement required for, 129 general requirements, 126–128 limits of reinforcement for, 126 rules VIII-1, 128–144 VIII-2, 144–154 shape and size of, 128 Reinforcement, nozzle, see Nozzle reinforcement Reinforcement limits, 131–132 Reinforcement plate, welded connection with, 12 Reverse flanges, 113–117 Reverse welding neck flange with ring-type gasket, 282 Ring flanges, 105 with ring-type gasket, 279 Ring girders, 8 Ring-type gaskets bolted flange connections with, 101 reverse welding neck flange with, 282 ring flange with, 279 slip-on or lap-joint flange with, 280 welding neck flanges with, 281 Rings, stiffening, see Stiffening rings Rupture, 1–4

S Secondary stress, 233 Section VIII background information, 1–26 Divisions 1 and 2, 1 Section VIII-1, pressure test requirements for, 23–24 Section VIII-2, pressure test requirements for, 25–26 Sections, conical, see Conical sections Shear loads, 49 Shearing stress, 51


292

Index

Shells conical, see Conical sections cylindrical, see Cylindrical shells elliptical, 53–54 spherical, see Spherical shells SI units, conversion of U.S. customary units to, 285 Single full fillet lap joint, 8 Single-welded butt joints, 8 Slip-on flanges, 105 with full-face gasket, 283 with ring-type gasket, 280 Special components, VIII-1, 159–187 Spherical radius, 62 Spherical shells external pressure in, 59–61 internal pressure in, 55–59 large, 55 membrane stress in, 55 openings in, 129 thick, 56 VIII-1, 55–61 VIII-2, 61–62 Spherically dished covers, 118–124 Spiral jackets, 171 Standard flanges, 105–112 Stays and staybolts, 162–163 welded, see Welded stays Stiffening rings, 46–49 attachment of, 49–51 designing, 46–49 Strain, 36 critical, lowest, 41 Stress allowable, 2–5 bending, 39–40 circumferential, 33, 67 classification of, 236–238 combinations of, 239–244 compressive, 67 critical axial buckling, 38 hoop, see Hoop stress longitudinal, 33 membrane, see Membrane stress meridional, 67 peak, 125, 234 primary, 125, 233, 234 secondary, 233 shearing, 51 tensile, 2 thermal, 233, 235

Stress categories, 233–239 and limits, 236 Stress concentration, 239 Stress multipliers, 3 Stress rupture, 3 Stress-strain diagrams, 36–37 Structural discontinuity, 235

T Tables, list of, xxi TEMA (Tubular Exchanger Manufacturers Association) standards, 190 Temperatures cold, 24 design, 2, 4 metal, minimum design, see Minimum Design Metal Temperature Tensile forces, cylindrical shells under, 27–35 Tensile strength, 3 Tensile stress, 2 Tension cone-to-shell junction at large end of cone in, 80–81 cone-to-shell junction at small end of cone in, 73–74 Thermal stress, 233, 235 Thickness details governing, used for toughness, 12–14 hemispherical heads, 58–59, 60–62 maximum pressure for, 27 minimum, of half-pipe jackets, 171–175 required for knuckle, 73 Threaded-end stay construction, special limitations for, 160 Threaded-end stays, special limitations for, 162 Torispherical heads pressure on concave side of, 65–67 pressure on convex side of, 67–68 shallow, 65 VIII-1, 65–68 VIII-2, 68–70 Toughness, governing thickness details used for, 12–14 Toughness rules, 14 Transition sections, 52–54 conical, see Conical sections Tube patterns, 197–198 Tubesheet attachments, 8 Tubesheet design rules for, 189 in U-tube exchangers, 189–208


Index 293

Tubesheets, 9, 189 fixed, see Fixed tubesheets types of, 190 Tube-to Tubesheets Acceptable types of strength welds, 224 Locking mechanism, 228 Methods of attaching, 223 Push-out tests, 227 Tubular Exchanger Manufacturers Association (TEMA) standards, 190

U U-tube exchangers, tubesheet design in, 201–213 Unified Numbering System (UNS), 2 UNS (Unified Numbering System), 2 Unstayed flat plates and covers, 103 U.S. customary units to SI units, conversion of, 285

of obround cross section, 179 of rectangular cross section, 175–187, 197–199 VIII-1 requirements, guide to, 251–253

W Weld efficiencies, 4 Welded attachments, 14 Welded connection with reinforcement plate, 12 Welded-in stay construction, special limitations for, 160 Welded joint categories, 6–7 Welded stays for dimpled and embossed assemblies, 160–161 for jacketed vessels, 160–161 Welding neck flanges, 114, 115 with full-face gasket, 284 with ring-type gasket, 281 Wind loads, 39 Wind moment, 39

V Vacuum loads, 45 Venting, 24 Vessels jacketed, see Jacketed vessels layered, 30 of noncircular cross section, 175–187

Y Yield strength, 3 interpolation between, 16

Z Z factor, 100


Guidebook for the Design of ASME Section VIII Pressure Vessels 3th Edition - James R. Farr-Maan H. Jawad (L) 'I'

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