Guidebook ASME Section VIII.pdf

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

by

James R. Farr Wadsworth, Ohio

Maan H. jawad Nooter Corporation St. Louis, Missouri

ASME Press

New York

2001

Copyright © 200 I The American Society of Mechanical Engineers Three Park Ave., New York, NY 10016

Library of Congress Catetogtng-tn-Publtcattcn Data

Parr, James R. Guidebook for the design of ASME Section R. Farr, Maan H. Jawad.-2nd ed. p. em.

vm pressure vessels/by

James

Includes bibliographical references and index, ISBN 0-7918-0172-1 I. Pressure vessels-Design and construction. 2. Structural engineering. 1. Jawad, Maan H. II. Title. TA660. T34 F36 2001 681 '.76041-dc21 2001046096 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 plio! written permission of the publisher.

Statement [Tom By-Laws: The Society shall not be responsible for statements or opinions advanced in papers . . . or printed in its publications (B7.1.3) 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 rms 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 TIlAT 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. For authorization to photocopy material for internal or personal use under 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.eopyrighLeom.

To our children, Katherine, David, Susan, Nancy, and Thomas Jennifer and Mark

Cover Photo Courtesy of Nooter Corp.

PREFACE TO SECOND EDITION The ASME Boiler and Pressure Vessel Code, Section VIII, is a live and progressive document. It strives to containthe 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 minimumrequired thickness. This was the firstreduction in this design margin in 50 years andwas based upon the many improvements in material properties, design methods, and inspection procedures during that time. Chapters and parts of chapters bave 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 arc extensive and some are minor. Some of the examples in this book have changed completely and some remainunchanged. This book continuesto 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

v

ACKNOWLEDGMENTS We are indebted to many people and organizations for their help in preparing this book, Special thanks are given to the Noorer Corporation, fellow Committee Members; and to former coworkers for their generous support during the preparation of the manuscript. We also give thanks to Messrs. Greg L. Hollinger and George B. Komora for helping with tbe manuscript, and to our editor Ray Ramonas at ASME for having great patience and providing valuable suggestions.

vii

CONTENTS Preface Acknowledgments List of Figures List of Tables

~

:

;:;;.;;.;;;.:;..;;;;:;;.;;;;;;;;;;"";;;;;"';;;;;;;;;;':";"';;"'.";",;.;;,;

. ;;;;.,;,." •.".,."",.;,.;;.,;,;,;.•; . ,,,..,, . ., .

Chapter 1 Background Information 1.1 Introduction 1.2 Allowable Stresses ., 1.3 Joint Efficiency Factors 1.4 Brittle Fracture Considerations 1.5 Fatigue Requirements 1.6 Pressure Testing of Vessels and Components 1.6.1 ASME Code Requirements 1.6.2 What Does a Hydrostatic or Pneumatic Pressure Test Do? 1.6.3 Pressure Test Requirements for VIlJ-l 1.6.4 Pressure Test Requirements for VIII-2 Chapter 2 Cylindrical Shells 2.1 Introduction 2.2 Tensile Forces, VIII-I 2.2.1 Thin Cylindrical Shells 2.2.2 Thick Cylindrical Shells 2.3 Axial Compression 2.4 External Pressure 2.4.1 External Pressure for Cylinders with Dolt :2: 10 2.4.2 External Pressure for Cylinders with Dolt < 10 2.4.3 Empirical Equations 2.4.4 Stiffening Rings 2.4.5 Attachment of Stiffening Rings 2.5 Cylindrical Shell Equations, Vlll-2 2.6 Miscellaneous Shells 2.6.1 Mitered Cylinders 2.6.2 Elliptical Shells

'"

. .. . . . . . . .. .. . . . . . .

,

Chapter 3 Spherical Shells, Heads, and Transition Sections 3.1 Introduction 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 , 3.2.2 External Pressure in Spherical Shells and Pressure on Convex Side of Hemispherical Heads . 3.3 Spherical Shells and Hemispherical Heads, VIII-2 ix

. . .. .. . .. .. .. .. .. ..

. .

v vii xiii xvii

1 1

2 3 9 19

22 22

22 23 24

27 27 27 27 33

36 42 43

46 47 48 50 53 54 54 55

.. . .

57 57 57

.

57

.

61

64

x Contents 3.4

3.5

3.6

3.7

3.8

Ellipsoidal Heads, VIII-I 3.4.1 Pressure on the Concave Side 3.4.2 Pressure on the Convex Side Torispherical Heads, VlIl-1 3.5.1 Pressure on the Concave Side 3.5.2 Pressure on the Convex Side Ellipsoidal and Torispherical Heads, VnI~2 Conical Sections, VIII~ 1 3.7.1 Internal Pressure 3.7.2 External Pressure Conical Sections, VIII-2

.. . .. . . . . . . . .

"

Chapter 4 Flat Plates, Covers, and Flanges 4.1 Introduction 4.2 Integral Flat Plates and Covers 4.2.1 Circular Flat Plates and Covers 4.2.2 Noncircular Flat Plates and Covers 4.3 Bolted Flat Plates, Covers, and Flanges 4.3.1 Gasket Requirements, Bolt Sizing, and Bolt Loadings 4.4 Flat Plates and Covers With Bolting 4.4.1 Blind Flanges & Circular Flat Plates and Covers 4.4.2 Noncircular Flat Plates and Covers 4.5 Openings in Flat Plates and Covers 4.5.1 Opening Diameter Does Not Exceed Half the Plate Diameter 4.5.2 Opening Diameter Exceeds Half the Plate Diameter 4.6 Bolted Flange Connections With Ring Type Gaskets 4.6.1 Standard Flanges 4.6.2 Special Flanges 4.7 Spherically Dished Covers 4.7.1 Definitions and Terminology 4.7.2 Types of Dished Covers

. .. . .. .. .. .. .. . . . .. .. .. . . .. .. ..

Chapter 5 Openings 5.1 Introduction 5.2 Code Bases for Acceptability of Opening 5.3 Terms and Definitions 5.4 Reinforced Openings-General Requirements 5.4.1 Replacement Area 5.4.2 Reinforcement Limits 5.5 Reinforced Opening Rules, VIII~l 5.5.1 Openings With Inherent Compensation 5.5.2 Shape and Size of Openings 5.5.3 Area of Reinforcement Required 5.5.4 Limits of Reinforcement 5.5.5 Area of Reinforcement Available 5.5.6 Openings Exceeding Size Limits of Section 5.5.2.2 5.6 Reinforced Opening Rules, VIII-2 5.6.1 Definitions 5.6.2 Openings Not Requiring Reinforcement Calculations 5.6.3 Shape and Size of Openings 5.6.4 Area of Reinforcement Required 5.6.5 Limits of Reinforcement 5.6.6 Available Reinforcement

. . .. . .. . .. . .. . . .. . .. . . .. .. .. . ..

, "

65 65

67 68 68 71 72

74 74 85 95 101 101 101 101 104 105 105 106 106 107 107 107 108 108 109 118 124 125 125

133 133 133 134 134 134

134 136 136 137 137

140 140 lSI 153 153 153 ISS ISS 155

157

Contents

5.7

5,6.7 Strength of Reinforcement Metal .." ,.,."." 5.6.8 Alternative Rules for Nozzle Design Ligament Efficiency Rules. VIII-l

",,'

, ,

,,' ,.,., ",.""

".."."""""

"..".",

Chapter 6 Special Components, Vlff-1

6.1 6.2

6.3

6.4

6.5

Introduction " ,.,., , , ,' ", Braced and Stayed Construction 6.2.1 Braced and Stayed Surfaces ,., 6.2.2 Stays and Staybo1ts Jacketed Vessels 6.3.1 Types of Jacketed Vessels , , 63.2 Design of Closure Member-for.Jacket to Vessel "; , 6.3.3 Design of Openings in Jacketed Vessels " "." Half-Pipe Jackets 6.4.1 Maximum Allowable Internal Pressure in Half-Pipe Jacket 6.4.2 Minimum Thickness of Half-Pipe Jacket Vessels of Noncircular Cross Section 6.5.1 Types of Vessels " 6.5.2 Basis for Allowable Stresses 6.5.3 Openings in Vessels of Noncircular Cross Section 6.5.4 Vessels of Rectangular Cross Section

,

,

. . . .. . , , ,, ""'" . . ,.., .. , .. .. , .. ..

" "

Chapter 7 Design of Heat Exchangers

7.1 7.2

7.3

7.4

Introduction Tubesheet Design in Ll-Tube Exchangers 7.2.1 Nomenclature " 7.2.2 Design Equations for Simply Supported Tubesheets 7.2.3 Design Equations for Integral Construction 7.2.4 Design Equations for Integral Construction With Tubesheet Extended as a Flange Fixed Tubesheets , 7.3.1 Nomenclature , 7.3.2 Design Equations Expansion Joints .

Chapter 8 Analysis of Components in

8.1 8.2 8.3 8,4 8.5

VIII~2

Introduction Stress Categories Stress Concentration Combinations of Stresses Fatigue Evaluation

References

,

Appendices Appendix A·-Guide to VIII-1 Requirements ,... Appendix B-Material Designation Appendix C-Joint Efficiency Factors Appendix D-F1ange Calculation Sbeets Appendix Be-Conversion Factors Index

, ,

,

,

,

, ..

xi

157 157 164

169 169 169 169 172 173 174 175

179 181 181 182

186 187 187 187

197

.

201

. .. .. .. . .. . . . ..

201

. . . . . ..

233 233 233 239 240 245

201 201

207 209 212 213 213

217 230

249

.

,..................................................................

..

25] 253 255 277 283 285

LIST OF FIGURES Figure

Number i.i 1.2

Welded Joint Categories (ASME VIII-I) Category C Weld

. .

Some Governing Thickness Details Used for Toughness (ASME VIII-I)

.

ELl 1.3 E1.2 1.4 1.5 1.6

1.7 2.l 2.2 2.3 2.4 2.5 E2.8 2.6 2.7 E2.l3 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.l3 3.8

3.9 3.10 3.11 3.12 4.1 4.2

Impact-Test Exemption Curves (ASME VIII-I) . Charpy Impact-Test Requirements for Full Size Specimens for Carbon and Low Alloy Steels With Tensile Strength of Less Than 95 ksi (ASME VllI-I) .. Reduction of MDMT Without Impact Testing (ASME VIII-I) . Fatigue Curves for Carbon, Low Alloy, Series 4XX, High Alloy Steels, and High Tensile Steels for Temperatures Not Exceeding 700'F (ASME VIII-2) . Comparison of Equations for Hoop Stress in Cylindrical Shells

.

5 7 8 11 14

15 16 17 20 28

29 30

Chart for Carbon and Low Alloy Steels With Yield Stress of 30 ksi and Over, and Types 405 & 410 Stainless Steels . C Factor as a Function of R!T (Jawad, 1994) .

39

Geometric Chart for Cylindrical Vessels Under External Pressure (Jawad and Farr, 1989) Some Lines of Support of Cylindrical Shells Under External Pressure (ASME VIII-I)

.. .

40 43 45

Some Details for Attaching Stiffener Rings (AS"ME VIII-I) Mitered Bend , Elliptical Cylinder

.. . .

38

49 , "

,

51

55 56 59 62 66

69 70 73

75 78 79 83

92 95 Inherent Reinforcement for Large End of Cone-to-Cylinder Junction (ASME VIII-2) Values of Q for Large End of Cone-to-Cylinder Junction (ASME VllJ-2) Inherent Reinforcement for Small End of Cone-to-Cylinder Junction (ASME VIII~2) Values of Q for Small End of Cone-to-Cylinder Junction (ASME Vill-2) Some Acceptable Types of Unstayed Flat Heads and Covers .. Multiple Openings in the Rim of a Flat Head or Cover With a Large Central Opening

xiii

.. . . .

96 97 98 99

.

103 109

xiv

List of Figures

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

Ring Flange Sample Calculation Sheet Welding Neck Flange Sample Calculation Sheet Reverse Welding Neck Flange Sample Calculation Sheet Spherically Dished Covers With Bolting Flanges (ASME VlII-l) Example Problem of Spherically Dished Cover, Div. 1 Reinforcement Limits Parallel to Shell Surface Chart for Determining Value of F for Angle Determination of Special Limits for Setting t, for Use in Reinforcement Calculations Example Problem of Nozzle Reinforcement in Ellipsoidal Head, Div. 1 Example Problem of Nozzle Reinforcement of 12 in. X 16 in. Manway Opening, Div. 1 Example Problem of Nozzle Reinforcement of Hillside Nozzle, Div. 1 Example Problem of Nozzle Reinforcement of Hillside Nozzle, Div. 1 Example Problem ofNozzle Reinforcement With Corrosion 'Allowance.Triv. ·1··

5.5

Nozzle Nomenclature and Dimensions (Depicts General Configurations Only) Limits of Reinforcing Zone for Alternative Nozzle Design Example Problem of Nozzle Reinforcement in Ellipsoidal Head, Div. 2 Example Problem of Nozzle Reinforcement of 12 in. x 16 in. Manway Opening, Div. 2 Example Problem of Nozzle Reinforcement of Series of Openings, Div. 1 Typical Forms of Welded Staybolts Typical Welded Stay for Jacketed Vessel Some Acceptable Types of Jacketed Vessels Some Acceptable Types of Closure Details Some Acceptable Types of Penetration Details Spiral Jackets, Half-Pipe and Other Shapes Factor K for NPS 2 Pipe Jacket Factor K for NPS 3 Pipe Jacket Factor K for NPS 4 Pipe Jacket Vessels of Rectangular Cross Section Vessels of Rectangular Cross Section With Stay Plates Vessels of Obround Cross Section With and Without Stay Plates and Vessels of Circular Section With a Stay Plate , Plate With Constant-Diameter Openings of Same or Different Diameters Plate With Multidiameter Openings Example Problem of Noncircular Vessel, Div. 1 Various Heat-Exchanger Configurations (TEMA, 1999) Some Typical Tubesheet Details for If-Tubes (ASME, 2001) . Tubesheet Geometry . Effective Poisson's Ratio and Modulus of Elasticity (ASME, 2001) .. Chart for Determining A (ASME. 2001) Fixity Factor. F (ASME, 2001) Some Typical Details for Fixed Tubesheet Heat Exchangers (ASME, 1995) Z" Z,. and Z.. versus X, (ASME, 2001) Values of Q3 Between 0.0 and 0.8 Values of Q3 Between -0.8 and 0.0 Bellows-Type Expansion Joints Flanged and FIued Expansion Joints

5.6

E5.5 E5.6 E5.7 6.1 6.2 6.3 6.4

6.5 66 6.7 6.8 6.9 6.10 6.11

6.12 6.13 6.14

E6.8 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 E8.1 8.1

E8.4 8.2 8.3 A.1

. . . . . .

e

. .. .. .. .. .. ..

.. .. .. .. .. . . . . . . .. . . . .

llO ll5 119 125 128

135 138 139 141 143 146 147 149 152 152 154

158 159 162

165 171 171 174 176

180 182

183 184 185 188 190

Cross . . . .. .

191 192 193 198 202

203 205

206 .

208 209 214 221

.

. .

222

.. . , . .

223 231 231

238 Linearizing Stress Distribution Model of a Finite Element Layout in a Flat Head-to-Shell Junction Fatigue Curves for Carbon, Low Alloy, 4XX High Alloy, and High Strength Steels for Temperatures Not Exceeding 700°F (ASME VllI-2) Cyclic Curves

240 243

. .. .

..

246 247 252

List of Figures

Cl C2 C.3

CA C5 C6 C7

c.S C.9 C.lO C.ll C.12 C.B C.14 C.15 C.16 C.l7 C.18 C.19 C.20.E D.1 D.2 D.3

DA D.5 D.6

Fig. Fig. Fig. Fig. Fig. Fig.

D.l-Ring Flange With Ring-Type Gasket D.2-Slip-On or Lap-Joint Flange With Ring-Type Gasket D.3-Welding Neck Flange With Ring-Type Gasket DA-Reverse Welding Neck Flange With Ring-Type Gasket D.5-Slip-On Flange With Full-Face Gasket D.6-Welding Neck Flange With Full-Face Gasket _

. .. . . .

.

xv

255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 277

278 279 280 281 282

LIST OF TABLES Table Number

i.i

1.2 1.3 104

1.5 ELl

1.6 1.7 2.1 3.1 3.2 3.3 304 E3.14 6.1

6.2 6.3 8.1 8.2 8.3 8.4

8.5 8.6 E8A B.1 B.2

B.3

BA B.5

Criteria for Establishing Allowable Stress Values for VIII-l (ASME II-D) Criteria for Establishing Design Stress Intensity Values for VITI-2 (ASME II-D) Stress Values for SA-5IS and SA-516 Materials . . " . Allowable Stress Values for Welded Connections , "., Maximum Allowable Efficiencies for Arc- and Gas-Welded Joints , ,," , . Stress Categories , , ,..,.., . Assignment of Materials to Curves (ASME VIII-I) Minimum Design Metal Temperatures in High Alloy Steels Without Impact Testing Tabular Values for Fig. 2.4 . . Factor K o for an Ellipsoidal Head With Pressure on the Convex Side .... Values of.6. for Junctions at the Large Cylinder Due to Internal Pressure Values of n. for Junctions at the Small Cylinder Due to Internal Pressure . Values of 11 for Junctions at the Large Cylinder Due to External Pressure Allowable Stress and Pressure Data , . Example of Pressure Used for Design of Components .. Closure Detail Requirements for Various Types of Jacket Closures Penetration Detail Requirements .. ,'.., Primary Stress Category , . . Structural Discontinuity, ,.......... .. , . Thennal Stress " ..... Stress Categories and Their Limits (ASME VII1~2) . Classification of Stresses (ASME VIII-2) Some Stress Concentration Factors Used in Fatigue Summary of Finite Element Output , . Carbon Steel Plate " . Chrome-Moly Steel Plate Specifications, SA-387 .... Chrome-Moly Steel Forging Specifications, SA-182 Chrome-Moly Steel Forging Specifications, SA-336 .." .." ........ Quench & Tempered Carbon and Alloy Steel Forgings, SA-50S

xvii

,

.

"

.

3 3 4 5 6 9 10 18 37

67 76 77

85 96 173 179 181

234 234

235 236 237

239 244

253 253 253 254 254

CHAPTER

1 BACKGROUND.. INFO.RMATION 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 land 2, tbe following designation will be used from here on to facilitate such references. ASME Section VIII, Division 1 Code will be designated by VIII-I. Similarly, VIlI-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. Eqnations and paragraphs referenced in each of these divisions will be called out as they appear in their respective Code Divisions.

Many design rules in VIIl-l and VIII-2 are identical. These include flange design and external pressure requirements. In such cases, the rules of VIII-I 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 VlIl-1 that pertain to various components of pressure vessels.

Section VIll requires the fabricator of the equipment to be responsible for its design. Paragraphs UG22 in VIII-l and AD-110 in VIIl-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 hecause they are not applicahle 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-l give guidance for some design requirements. VIII-2, paragraph AD-I 10 and the User's Design Specifications mentioned in AG-30l provide the loading conditions to be used by the manufacturer.

Many design rules in VIII-I and VIII-2 are included in the Appendices of these codes. These rnles are for specific products or configurations. Rnles 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 he transferred to the Mandatory section of the Code after a period of nse and verification of their safety and practicality. However, guidance-type appendices will remain in the Non-Mandatory section of the Code.

The rules in VIlI-I do not cover all applications and configurations. When rules are not available, Paragraphs U-2(d), U-2(g), and UG-IOl must he used. Paragraph U-2(g) permits the engineer to design components in the absence of rules in VIll-i. Paragraph UG-lOI is for allowing proof testing to estahlish maximum allowable working pressure for components. In VIII-2 there are no rules similar to those in

UG-lOI, since VIII-2 permits design by analysis as part of its requirement'. This is detailed in Paragraphs AD-JOO(b), AD-J40, AD-ISO, aud AD-l60 of VIlI-2. 1

2 Chapter 1

1.2 ALLOWABLE STRESSES The criteria for establishing allnwahle stress in VIII-I are detailed in Appendix P of VllI-I and Appendix I ofIJ-D and are summarized in Table 1.1. The allowahle 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 helow the creep and rupture values. The controlling allowahle stress for most bolts is 1/5 the tensile strength. The minimum effective tensile stress at elevated temperatures is ohtained 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 hy the factor 1.1. The effective tensilestress is then equal-to the lessor ·of this-quantity-or-the-minimum tensile stress· at . room temperature given in ASTM. This procedure is illustrated in example 4.1 of Jawad and Farr (reference 14, found at hack of hook). The 1.1 factor discussed above is a constant estahlished hy the ASME Code Committee. It is hased 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 hetween room temperature and 500"F or higher for carhon steels; and the adjustment of minimum strength data to average data. Above approximately 5000F or higher the allowahle stress for carbon steels is controlled hy creep-rupture rather than tensile-yield criteria. Some materials may not exhibit such an increase in tensile stress, hnt the criterion for 1.1 is still applicahle to practically all materials in VIlJ-1. Table 1.I also gives additional criteria for creep and rupture at elevated temperatures. The criteria are hased on creep at a specified strain and rupture at 100,000 hours. The 100,000 hours criterion for rupture corresponds to ahout eleven years of continual use. However, VIII-I does not limit the operating life of the equipment to any specific numher of hours. The allowahle stress criteria in VJJl-2 are given in JI-D of the ASME Code. The allowahle stress at the design temperature for most materials is the smaller of 113 the tensile strength or 2/3 the yield stress. The design temperature for all materials in VJJJ-2 is kept helow the creep and rupture values. Tahle 1.2 summarizes the allowable stress criteria in VIII-2. A sample of the allowahle stress Tables listed in Section ll-D of the ASME Code is shown in Tahle 1.3. It lists the chemical composition of the material, its product form, specification numher, grade, Unified Numhering System (UNS), size, and temper. This information, with very few exceptions, is identical to that given in ASTM for the material. The Tahle 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 Tahle 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 hetween listed stress values, hut is not permitted to extrapolate heyond the puhlished values. Stress values for components in shear and bearing are given in various parts of VllJ-l, VIll-2, as well as II-D. Paragraph UW-15 of VIII-I and AD-132 of VIJI-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 puhlished in Section ll-D. VIII-] 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 JI-D for the material. This difference in the definition of temperature in VIJI-I and VIIl-2 can he substantial in thick cross sections suhjected to elevated temperatures.

Background Information

3

TABLE 1.1 CRITERIA FOR ESTABLISHING ALLOWABLE STRESS VALUES FOR VIII-1 (ASME II-D) Below Room Temperature

ProduetlMaterial

Tenshe Slrell9th

Wrought or cast ferrO\l5 and nonferrous

~

Welded p"pe or tube, ferrous and nonferrous

~Sr a.s

Room Temperatureand Above

Yield SlrenSth

'1, s-

Tensile Strength

~

VI >< 0.85 s,

~S 3.5

% s,

!2 SrRr '-'

a

"

Yield Strength

r

'I, S"R y

F."SR",

'f,

(F.... "

Or O.qSyR y [Note nll

~SrRr

';; ><

O,85Sy

'5

i em, Rate

Stress Rupture

>< O.85S yRy cr 0.9 x O.85S yRy

O.85lSR",

I

().85 R",,"

1.05,

{O.8 " O.851S R",",

0.85S e

[Note u»

NOTE: (i) . Twcisets of allowable stress values-may be provided.tn .Table·lA .for austenntc.rnatenals and in .Ta,bl e.1B.f{}rspec:ific nonferrous alloys~

The lower values are not specifically identified by a footnote, These lower values do not exceed two-thirds of the minimum yield strength at temperature. The higher alternative allowable stresses are ldentifled by a footnote. These higher stresses may exceed two-thirds but do not exceed 90% of the minimum yield strength at temperature. The higher values should be used only where slightlyhigher deformation is not in itself objectionable. These higher stresses are not recommended for the design of flanges or for other strain sensitive applications.

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 SRaVQ = 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 Be = 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

TABLE 1.2 CRITERIA FOR ESTABLISHING DESIGN STRESS INTENSITY VALUES FOR VIII-2 (ASME II-D) Product/Material Wrought or cast, ferrous and nonferrous

Tensile Strength 1/3

s,

Yield Strength

% SyRyor

2/~ Sy

.!.:.! s-e,

O.9S yRy[Note

3

Welded pipe or tube, ferrous and nonferrous

0.85 3

s-

{I,I x 0.85j S R 3

T T

0.85 1.5

s,

u»

0.85 SyRyor 1.5 (0.9 x 0.85)S yRy

[Note (Ill NOTE: (1)

Two sets of allowable stress values may be provided In Table IA for austenitic materials and in Table 18 for specific nonferrous alloys. The lower values are not specifically identified by a footnote. These lower values do not exceed two-thirds of the minimum yield strength at temperature. The higher alternative allowable stresses are identified by a footnote. These higher stresses may exceed two-thirds but do not exceed 90% of the minimumyield strength at temperature. The higher values should be used only where slightly higher deformation is not in itself objectionable. These higher stresses are not recommended for the design of flanges or for other strain sensitive applications.

1.3 JOINT E}'FICIENCY 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 radiograpby, with few exceptions. VIII-I, 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-l and by the user. The degree of examination influences the required 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-I vessels have variable

4 Chapter 1 TABLE 1.3 STRESS VALUES FOR SA-515 AND SA-516 MATERIALS Line

Nominal

Product

Spec

Alloy

Classl

DesigJ

Cond.!

Size/

Temper

Thick, in.

Group P·No

No.

No.

Composition

Form

28

CS

Plate

SA·515

70

K03101

2

29

CS

Plate

SA·516

70

K02700

2

No.

Type/Grade

UNS No.

External

Applic. & max. Temp. Limits

Min.

Min.

Tensile

Yield

(NP

Line

Strength

Stress

(SPT ~ Supports only)

No.

ksi

ksl

III

VIII-1

No.

Notes

28

70

38

1000

700

1000

CS·2

G10,S1,T2

29

70

38

850

700

1000

CS·2

G10,S1,T2

~

Pressure

Not Permitted)

Chart

Maximum Allowable Stress, ksi, for Metal Temperature, "F, Not Exceeding Line No.

-20 to 100

150

200

300

400

500

600

650

700

750

800

850

900

28

20.0

20.0

20.0

20.0

20.0

20.0

19.4

18.8

18.1

29

20.0

20.0

20.0

20.0

20.0

20.0

19.4

18.8

18.1

14.8

12.0

9.3

6.7

14.8

12.0

9.3

6.7

Note: G10, 81, T2 are described in ll-D and pertain to metallurgical information

factors of safety, depending on the degree of radiographic examination of the main vessel joints. As an example, fully radiographed longitndinal 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 tbe 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 dne to a nonradiographed joint results in a 43% increase in the required thiekness over that of a fully radiographed joint. ASME VIII-I 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 hentispherical 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-l 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-I. 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.


5

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

Type of Stress

Stress Value

tension shear tension shear shear shear bearing

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

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

*$ .~.. allowable stress tor vltt-t construction

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

Type of Stress tension shear tension shear shear bearing

Stress Value O.58 m*

o.ss,

0.75Sm O.758 m

o.es, S,

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

*5 m = stress intensity values for VIIl-2 construction

FIG_ 1.1 WELDED JOINT CATEGORIES (ASME VIII-1)

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. Thns, a Category C weld, which identifies the attachment of a flange to a shell, can be either fillet, comer, or butt welded, as illustrated in Fig. 1.2. The Joint Efficiency Factors apply only to tbe butt-welded joint in sketch (c). The factors do not apply to sketches (a) and (h) since tbey 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 tbe component. As an example, tbe 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.

6 Chapter 1

TABLE 1.5 MAXIMUM ALLOWABLE JOINT EFFICIENCIES'" FOR ARC- AND GAS-WELDED JOINTS Degree of Radiographic Examination b - -ca Joint Joint Type -Description Limitations Category Full' Spot' None No. (1)

Butt joints as attained by double-welding or by other means which will obtain the same quality of deposited weld metal on the inside and

None

A,B,C&D

1.0

0.85

0.70

(a) None except as shown in (b) below

A,B,C&D

0.90

0.80

0.65

(b) Circumferential butt joints with one plate offset, see UW-13(c) and Fig. UW13.1(k).

A, B &C

0.90

0.80

0.65

A, B& C

NA

NA

0.60

A

NA

NA

0.55

NA

NA

0.55

B

NA

NA

0.50

C

NA

NA

0.50

(a) For the attachment of heads convex to pressure to shells not over 5/8 in. required thickness. Only with use of fillet weld on inside of sneus, or

A&B

NA

NA

0.45

(b) For attachment of heads having pressure on either side. To shells not over 24 in. inside diameter and not over 1/4 in. required thickness with fillet weld on outside of head flange only.

A&B

NA

NA

0.45

outside weldaurtaces tc

agree with the requirements of UW-35. Welds uSing metal backing strips which remain in place are excluded.

(2)

Single-welded butt joint with backing strip other than those included under (1)

(3)

Single-welded butt joint without use of backing strip

Circumferential butt joints only, not over 5/8 in. thick and not over 24 in. outside diameter

(4)

Double full fillet lap joint

Longitudinal joints not over 3/8 in. thick Circumferential joints not over 5/8 in. thick

(5)

Single full fillet lap joints with plug welds conforming to UW-17

(a) Circumferential joints- for attachment of heads not over 24 in. outside diameter to shells not over 1/2 in. thick (b) Circumferential joints for the attachment to shells of jackets not over 5/8 in. in nominal thickness where the distance from the center of the plug weld to the edge of the plate is not less than 1-1/2 times the diameter of the hole for the plug.

(6)

Single full fillet lap joints without plug welds

Notes: (1) The single factor shown for each combination of joint category and degree of radiographic examination replaces both the stress reduction factor and the joint efficiency factor considerations previously used in this Division. (2) See UW-12(a) and UW-51. (3) See UW-12(b) and UW-52. (4) Joints attaching hemispherical heads to shells are excluded. (5) E = 1.0 for butt joints in compression. (6) For Type NO.4 Category C joint, limitation not applicable for bolted flange connections.

Background Information 7

.AI

, I

,I

I

,

I

I

I

I

I

""

I

I

,

(b)

,I I

. I

I

...

I

FIG. 1.2 CATEGORY C WELD

8 Chapter 1 Example 1.1 Problem Determine the category and Joint Efficiency Factor of the joints in the heat exchanger shown in Fig. E1.l. The channel side is spot radiographed. The longitudinal seam, b, is a single-welded butt joint with a backup

COVER

I 0

CHAN~EL

PASS PARTITION

d

At

b

u

\

f

9

C

h

r

SHELL

J

n .....

=6

..

TUBE

~

~

1

"il u

JAC KET

~ ~

F m

I

I .......-.:::f"HEAD

.....

0

SKIRT

FIG. E1.1


9

bar. The shell side is not radiographed. The longitndinal and circumferential seams m and I are singlewelded butt joints witb backup bars. The jacket longitudinal seam, n, is a single-welded butt joint witbout backup bar. Solution The joint categories of tbe various joiuts can be tabulated as given in Table ELI: TABLE E1.1 STRESS CATEGORIES Location

Joint

Category

Joint Efficiency

(a)

Ohannel-to-ftanqe connectlon

C

Does not apply

(b)

Longitudinal channel seam

A

0.80

(e)

Channel-to-tubesheet weld

C

Does notapply

(d)

Nozzle-to-channel weld

D

Does not apply

(e)

Flanqe-to-nczzle neck

C

Does not apply

(f)

Pass partitlon-to-tubesheet weld

None

Does not apply. See also

UW~15(c)

and

UW-18(d) of VIII-1 (9)

Tube-to-tubesheet weld

None

(h)

Shell-to-tubesheet weld

C

Does not apply

(i)

Jacket bar-to-inner-shell weld

None

Does not apply

OJ

Jacket to bar-to-outer-shell weld

None

Does not apply

(k)

Nozzle-to-jacket weld

D

Does not apply

(I)

Longitudinal shell seam

A

0.65

(m)

Head-to-shell seam

B

0.65

(n)

Longitudinal jacket seam

A

0.60

(0)

Sklrt-tc-head seam

None

Does not apply. See also UW-20 of vul-t

Does not apply

1.4 BRITTLE FRACTURE CONSIDERATIONS Both VIII-I and V111-2 require tbe designer to consider brittle fracture rules as part of tbe material and design selection. The rules for carbon steels are extensive and are discussed. first. VIII-l has two options

regarding toughness requirements for carbon steels. The first is given in Paragraph UG-20(f) and allows tbe designer to exempt tbe material of construction from impact testing when all of the following criteria are met: 1. The material is limited to P-No. I, Gr. No. I or 2.

2. 3. 4. 5. 6.

Maximum thickuess of 1/2 in. for materials listed in Curve A in Table 1.6. Maximum thickness of I in. for materials listed in Curves B, C, and D of Table 1.6. The completed vessel shall be hydrostatically tested per UG-99(b), (e), or 27-3. Design temperature is between - 20°F and 650°F. Thermal, mechanical shock, and cyclical loadings do not control the design.

The above requirements are intended for relatively thin carbon steel vessels operating in a service that is neither severe in thermal and pressure cycling nor in extreme cold temperatures. Vessels of low alloy

10 Chapter 1

TABLE 1.6 ASSIGNMENT OF MATERIALS TO CURVES (ASME Vm·1) GENERAL NOTES ON ASSIGNMENT OF MATERIALS TO CURVES: (a) Curve A applies to: (1)

(2:)

all carbon and all low alloy steel plates, structural shapes, and bars not listed In Curves S, C, and 0 below; SA-lIb Grades WeB and wee If normalized and tempered or water-quenched and tempered, SA-217 Grade Web If normalized and tempered or water-quenched ami tempered,

(b) Curve 8 applies to:

(ll SA-2lb Gradf! WCA if normalized and tempered or water-quenched and tempered SA-21b Grades WeB and wee for thicknesses not elCceedlng 2: tn., if produced to fine grain practlce and water-quenched and tempered SA-217 Grade WC9 if normalized and tempered SA·285 Grades A and B SA-414 Grade A SA-SIS Grade 60 SA-Sib Grades 65 and 70 If not normalized SA.b iau not normalized SA-662 Grade B If not normalized; (2) except lor cast steels, all materials of Curve A if produced to fine grain practlce and normalized which are not listed in Curves C and D below; (3) all pipe, fittings, forgings and tubing not listed for Curves C and D beloW; (4) parts permitted under us-n shall be included in Curve B even when fabricated from plate that otherwise would be assigned to a dlllerent curve. (c) Curve C Ul SA-IB2 Grades 21 and 22 if normalized and tempered SA-302 Grades C and D SA-33u F21 and F22 If normallzed and tempered SA-387 Grades 21 and 22 If normalized and tempered SA-Slu Grades 55 and 60 II not ncrmauzeo SA-533 Grades Band C SA-bul Grade A; (2) all material 01 Curve B If produced to fine grain practice and ncrmaltaed and not listed for Curve D below. (dl Curve 0 SA-203 SA-SOB Grade 1 SA-51u if normalized SA-524 Classes 1 and 2 SA-537 Classes 1, 2, and 3 SA·bI2 if ncrmallzed SA...f:,b2 If normalized SA-BB Grade A SA-738 Grade A with Cb and V deliberately added in accordance with the provisions of the material specification, not cctoer than _20°F (_29°C) SA-BB Grade B not colder than _20°F (_29°C) (e) For bolting and nuts, the 10llowlng Impact test exemption temperature shall apply:

Bolting Impact Test Exemption Temperature, OF

Spec. No.

Grade

SA-193 SA-193

B5 B7 (2Y~ in. dia. and under) (Over 2Y2 in. to 7 In., Incl'>

SA-193 SA-193 SA-307 SA-320 SA-325 SA-354 SA-354 SA-449 SA-540

B7M B16

-'0

·55 -40 ·55

-'0

-'0

B L7, L7A, 17M, L43

impact tested

"a

·'0

Be

o

BO

+20

·'0

623/24

+10 Nuts

Spec. No.

Grade

Impact Test Exemption Temperature, of

SA-194

2, 2H, 2HM, 3, 4, 7, 7M, and 16

-55

SA-540

823fB24

-55

(f) When no class or grade is shown, en classes or grades are included. (g) The following shall apply to all material assignment notes. (l) Cooling rates faster than those obtained by cooling ill air, followed by tempering, as permltted by the material specification, are considered to be equivalent to normalizing or normalizing and tempering heat treatments. (2} Fine grain practice is defined as the procedure necessary to obtain a fine austenitic grain size as described in SA·20.

NOTES: Tabular values for this Figure are provided in Table UCS-66. (2) Castings not listed in General Notes (a) and {b} above shall be impact tested.

(l)


11

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 nsually given by the user, while the calculated MDMT is obtained from VIlI-1. The calculated MDMT is kept equal to or colder than the specified MDMT.

2

Section x-x tg1 "" tA tff2 "" tA (seamless) or ta (welded)

--<1.(8)

Butt Welded Components

II

I

tc

!r1I +1 I--tc

2

1

tB

110 1-1'" .

<1. I I

tg 1 "" the thinner of tA or tc

tg3 = the thinner

tff2 ""the thinner

of tBor tc

NOTE: Using tgl' t g 2. and tg 3' determine the warmest MDMT and use that assembly.

oftAortB

8S

the permissible MDMT for the welded

(bl Welded Connection with Reinforcement Plato Added

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

12 Chapter 1

tg 1 =

t:

Cl. (For

'A

0

rt-.

tg1 "" 4" (For ~ welded welded

or nonwelded)

or nonwelded) tg2 =

tc

The governing thickness of

0

is the greater

of f g 1 or tQ2

(c) Bolted ~Iat Head or Tubtnlheet and Flange

(d) Integral Flet Head or Tubesheet

Cl.

tg 1

'A f:\ . = T {For 0./ welded or nonweldedl

trn

'" thinner of

fA

or fa

The governing thickness of

0

is the greater

of t0 1 or tg2 (8)

Flat Head or Tubesheet With 8 Corner Joint

FIG. 1.3 (CONT'D)

BackgroundInformation

13

Pressure part

tg1 = thinner of fA or fa (ft Welded Attachments

8S

Defined in UCs.&6(a)

FIG. 1.3 (CONT'D)

3, The temperature obtained from Fig, 104 may be reduced in accordance with Fig, 1.6 if fhe component operates at a reduced stress, This is detailed in Paragraph UCS-66(b) of VIlI- 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

fhat brittle fracture does not occur regardless of temperature level. 4, The rules in VIlI-l also allow a 30'F reduction in temperatnrc below that obtained from Fig, 104 when the component is post-weld heat tteated but is not otherwise required to be post-weld heat treated by VIlI-l rules, The toughness rules for ferritic steels with tensile properties enhanced by heat treatment are given in Paragraph UHT-6 of VIlI-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 VIlI-I. The permissible Minimum Design Metal Temperature for base material is summarized in Table 1.7. Similar data are given

in VIII-I for fhe 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 VIlI-2 are different fhan those in VIII-I. However, the concepts of exemption curves and Charpy impact levels are similar in VIlI-2 and VIII-I. 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.

14 Chapter 1

Example 1.2 Problem Determine the Minimnm Design Metal Temperature, MDMT, for the reactor shown in Fig. E1.2. Let the shell, head, pad, and ring material be SA-516 Or. 70 material. Flange and cover material is SA-105. Pipe 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,5 I6 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 Or. 70 material. Using Fig. 1.4 and a governing 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 = - J7 op . Head

For a I in-thick head, SA-516 specifications permit a non-normalized material. Thus, from Table 1.6. Curve B is used. Using Pig. 1.4 aud a governing thickness of 1.0 in., we get a minimum temperature of 300P. The ratio of reqnired thickness to actual thickness is 0.86/1.0 = 0.86. Using Pig. 1.6 for this ratio, we obtain 14°F. Hence, MDMT = 30 - 14 ~ 16°P.

t=I.0'

3' X

'J,:;'

RING 4' RAD, 2~'

300" WN FLANGE

X 2' PAD SA 105 -~"ci-'-

4' SCH. 40 PIPE

FIG. E1.2

--

~ ~ I t=2~'

Background Informarion

140

120

100

I I

II I

I I

80

A~

I

40

20

o - 20

- 40

- 55 - 60

-80

I

1/ / / ~

--"""'" J--

Y /'

II

J

/

! II /

i~ /

V

V

---of

/'

/

8

.....-.>

/' I

I 60

15

....

..........

.»

~

~

s,

.....V

~ """

....

--

~

V 1/ i

--, 1I

0.394

-- f-- -

- - ... .,.-- - - - - - - -

--

--

TP'" "Sj,n. requlT 2

3

4

--

6

Nominal Thickness, in. (Limited to " in. for Welded Construction)

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

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 ntinimum temperature of 15°F. Since stresses cannot be established from VlIl-l rules, the MDMT = IsoF.

Pad The material will be normalized since it is 2.00 in. thick. Curve D of Table 1.6 is used. Prom Pig. 1.4 and a governing thickness of 2.0 in., we obtain a minimum temperature of - 5°F. Since stresses cannot be established from VIIl-l rules, the MDMT = - SOp.

16 Chapter 1

! I

0.3~4 in.

50

l I

I 40

I

Minimum specified ylelclm~ngth

I

I

"0

•E •i);

I

'0

•~

-S '0

30

•

~

E ~

I

./

£ 20

./

!

./

I

15

I

10

!

i.>

55 ksi

' / ...V

I I I

~

"o:>

65 ksi

./

I

~

I

/

./

.

50 ksi

~~

~ ~.... ~

45 ksl

,38 ksi

I

I I I

I I o

1.0

2.0

;;., 3.0

Maximum Nominal Thickness of Material or Weld, in.

GENERAL NOTES: (a) Interpolation between yield strengths shown is permitted. (bl The minimum impact energy for one specimen shall not be less than 2}3 of the average energy required for

three specimens. le) Materials produced and impact tested in accordance with SA-320, SA·333, SA-334, SA-350, SA-3S2, SA·42Q and SA-765 do not have to satisfy these energy values. They are acceptable for use at minimum design metal temperature not colder than the test temperature when the energy values required by the applicable specification are satisfied. Idl For materials haVing a specified minimum tensile strength of 95 ksi or more, see UG"84(cH4)(bl.

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)


1.00

.,ro 0

a: ~

>

.~

E

0.80

!

5

"\

-, I'\.

'"

~

5

~

ro

"0

0.60

c m

E 0

z ~

<, r-,

~

til

~ 0040 I 0.35

.s •IJJ (;

"i

0.20

a:

0.00

<,

r--

v/c //// t:-;//c 'l//: f"////: '///'///'/// '///'/// V// V/C V/C t:-;//'/// rr>>

~~ ~~ ~ ~~ l% ~~~~~~~~ ~ ~ ~~ l% ~

.;.

17

'///////V///'///////X///:~

~/ See UCS-66Ibl(3) when ratios are 0.35 and smaller ~ ////'////V/// ' / / / ' / / / ,///////X///,

%§ ~~~~ ~ ~~~~~ ~~~ o 20

40

60 80 of [See UCS-66Ib)J

100

120

140

Nomenclature (Note references to General Notes of Fig. UCS~66"2.) t r = required thickness of the component under consideration in the corroded condition for all applicable loadings [General Note (2)J, based on the applicable joint efficiency £ (General Note (3)], in. t n = nominal thickness of the component under consideration before corrosion allowance is deducted, in. c = corrosion allowance, in. E* = as defined in General Note (3). Alternative Ratio = $* E* divided by the product of the maximum allowable stress value from Table UCS~23 times E, where 5* is the applied general primary membrane tensile stress and E and E* are as defined in General Note (3).

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

Nozzle Neck From Table 1.6, Curve B is to be used for a nozzle neck of 0.258-in. thickness. From Fig. lA, minimum temperature is - 20'F. The ratio of required thickness to actual fhickness is 0.08/0.258 X 0.875 = 0.36. Using Fig. 1.6 and this ratio, we get 130'F. Hence, MDMT = - 20 - -130 = - 150°F.

18 Chapter 1

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

Austenitic chromiummanganese-nickel stainless steel (200 series)

Is material • Austenitic ferritic duplex steel with t « 3/8 in. • Ferrltic chromium stainless steel with t < 1/8 in. • Martensitic chromium stainless steel with t « 1/4 in.

Flange Since the flange is ANSI BI6.5, it is good to -20 oP.

Cover Prom Pig. 1.3(c), the controlling cover thickness is 2.5/4 = 0.625 in. Curve B applies for this material, and the MDMT = 5°P. Therefore, the MDMT for this reactor is governed by the head with a value of 16°P. A colder value can be obtained by impact testing the various components. Thus, assuming a specified MDMT of -15°P is required, then the head, stiffener, pad, and cover need impact testing.

Background Information 19

1.5 FATIGUE REQUIREMENTS Presently, VIII-l does not list any rules for fatigne evaluation of components. When fatigue evaluation of a component is required in accordance with UG-22 or U-2(g) of VIII-I, the general practice is to use the VIII-2 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 II1-H are followed for general guidance. Other fatigue criteria, such as those given in other international codes and ASME B31.3, may also he considered as long as the requirements of U-2(g) of VIII-l 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.l 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 requited if all of Condition A or all of Condition B is satisfied, as noted below.

Condition A

Fatigue analysis is not required for materials witb 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 tbe following chart in order to transform them to equivalent cycle number. Metal Temperature Differential, OF

50 51 101 151 251 351

Factor

o

or less to 100 to 150 to 250 to 350 to 450

1 2 4 8 12

Higher than 450

20

d. Number of temperature cycles in components that have two different materials where a difference in the value (uJ - (2)b.T exceeds 0.00034. Where, U is the coefficient of thermal expansion and AT is the difference in temperature. Condition B

Fatigue analysis is not requited when the following items (a) through (f) are met:

20

Chapter 1 a. The number of full range pressure cycles, including startup and shutdowu, is less than the uumber of cycles determined from the appropriate fatigue chart, Fig. 1.7, with an S, 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)(S,1Sm), where P is the design pressure, S, 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(l/3)(SI Sm)' S is defined as S = S, taken at 10' cycles when the pressure cycles are 0; 10'. S = S, taken at actual number of cycles when the pressure cycles are > 106 c. The temperature difference between adjacent points during startup and shutdowu does not exceed S,I (2E
coefficient of expansion or modulus of elasticity aud that do not exceed the quantity S,I [2(E,
ror

.11

E",30 x 10'p$", rL... (1) NOTES!.

_

12l Interpolate for UTS 80-115 kll. (3) Table 6--110.1 contains tabu1eled values and 8 formula for an accurate interpolation of

the" curves.

-

::

a

~

/

Fo' UTS .. 80 k$i

~--

...... 1'-.--

For UTS 1'5·130 k,i

-- '-:"" ---

r I

I

""I

I

,

I "

"

,

I

1111111

Numb~, (If

,

I

I

--- -- --

I II "

I

I

I'";'ii"

eveh'l

. 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

The third criterion for nozzles with nonintegral reinforcement is given in Paragraph AD-160.3 of VllI-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"P are reached in two hours. The maximum /IT between any two points during start-up is 250"P. At normal operation, the /IT is negligible. At shutdown, the maximum /IT is 100oP. 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 X 10- 6 in.lin.l"P and that for stainless steel be 9.5 X 10- 6 in.lin./"F. Solution

Prom 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. Prom the chart, the 250 0P difference in temperature during start-up corresponds to 4 cycles. The 1000P difference in temperature during shutdown corresponds to I cycle. Thus total equivalent cycles due to temperature in one year is (4 + I) 6 = 30 cycles. d. At nozzle attachments, the quantity (9.5 X 10- 6 - 6.5 X 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), (e), and (tl) = 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 oP. 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 sbell 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 0P are reached in eight hours. The maximum flT between any two points during start-up is 60 oP. At normal operation, the /IT is negligible. At shutdown, the maximum /IT is 50 oP. Determine if the shell and heads are adequate for 100,000 cycles without the need for fatigue analysis. Prom II-D, the coefficient of expansion for carbon steel is 7.25 X 10- 6 in.lin./"P and the modulus of elasticity is 27.3 X 106 psi. Use Pig. 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) 1.7 for this value gives a fatigue life of 4200 cycles. b. This condition does not apply.

=

54,800 psi. Using Pig.

22

Chapter 1 c. From Fig. 1.7, with 100,000 cycles, the value of S, = 20,000 psi. The value of S,/(2Ea) = 20,000/(2 X 37.3 X 106 X 7.25 X 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 AT 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 lA, determine the required thickness of the shell and heads for 1,000,000 cycles without the need to perform fatigue analysis.

Solution From Fig. 1.7, with a cycle life of 1,000,000, the value of S, = 12,000 psi. From Condition B, subparagraph (a), the maximnm stress value for the shell is (12,000/3) = 4,000 psi. The needed shell thickness = 0.5 X 18,000/4000 = 2.25 in. The required head thickness = 0.25 X 18,000/4000 = 1.13 in. The maximum AT at start-up or shutdown cannot exceed S,/(2Ea) = 12,000/(2 X 27.3 X 106 X 7.25 X 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-l rules, except those tested in accordance with

the requirements of UG-IOI, are required to pass either a hydrostatic test (UG-99) or a pneumatic test (UG100) of the completed vessel before the vessel is U-stamped. Pressure vessels that are designed and constructed to VIIl-2 rules also are required to pass either a hydrostatic test (Article T-3) or a pneumatic test (Article T-4) before the U2-stamp is applied. Each component section of the ASME Boiler and Pressure Vessel Code has a pressure test requirement thatcalls 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 meaut to detect major leaks, while others feel that there sbould 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 errorhas 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

23

Pressure Test Requirements for YIn·1

1.6.3.1 Hydrostatic Test Requirements. A hydrostatic pressure test is the preferred test method. A pneumatic test or a combination of pueumatic/hydrostatic test is conducted only when a hydrostatic test caunot 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 detemtined 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 testpressure 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. Also, a liquid pressure relief valve set at 1 1/3 times the test pressure is recommended in the pressure test system. After the test pressure is reached, the pressure is reduced to the test pressure divided by 1.3, and 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 exantined 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 fornting 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 dne 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 ofUW-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 tbe vessel sball be l.l 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 Class-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 Reqnirements 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 UCl, 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

24 Chapter I pressure equal to or greater than 30 psi. Fnr Part UCD, the test pressure shall be 2.0 MAWP. With these changes, the remaining rules of UO-99 are followed. 1.6.3.5 Test Requirements fnr Vessels Built to the Rnles of Part ULT. Alternative rules for the design and construction of vessels to operate at cold temperatures as low as - 320 0 P 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 WO°F. When the vessel is desigued and constructed to Part ULT rules, special hydrostatic testing requirements are . n~(;essfl1"Y . due .to .t~~.fa(;tt1l~t. . the . lllateri:ll.. is.str0ll~t3r . at.. desi~n . . t~lTIP~ratllre .thaIl.at . . aIIllJieIltt~st 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: 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 bydrostatic 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 ant less than 1.1 MAWP, a pneumatic test shall be conducted using the rules of UO-lOO, but omitting the adjustment for the allowable tensile stress ratio. A vessel to be installed vertically may he tested in the horizontal position, provided the test pressure is applied for 15 minutes at not less than 1.4 MAWP, including a pressure equivaleut 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 MA WP 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 UO-1OI of VlIl-l 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 establi.shed by a proof test according to the reqnirements of UO-1OI(d) of VIII-l 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 V1II·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 valne 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 MA WP). Similar to the alternative in VIII-I, 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 membraue stress intensity Pm of 90% of the tabulated yield strength S; at test temperature;


25

b. A calculated primary membraue plus primary bending stress intensity F; + P, not to exceed tbe limits given below: [Pm [Pm

+

+ Pb] '" 1.35 S" when Pm'" 0.67 S"

Pb] '" 2.35 S, - 1.5 p.. when 0.67 S, < P; '" 0.90 S,

(1.1) (1.2)

1.6.4.2 Pneumatic Test Requirements. A pneumatic test is permitted only when one of tbe 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 iu the vessel shall be 1.15 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 desigu 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 P; of 80% of the tabulated yield strength S, at test temperature; b. A calculated primary membrane plus primary beuding stress intensity Pm + Ph not to exceed the following limits:

+ F.] '" 1.20 s; when r; '" 0.67 S,

(1.3)

+ Pb] '" 2.20 S, - 1.5 p.. when 0.67 S, < P; '" 0.80 S,

(1.4)

[Pm [Pm

CHAPTER

2 CYLINDRICAL SHELLS 2.1 INTRODUCTION The rules for cylindrical shells in VIII-l and VIII-2 take into consideration internal pressure, external pressure, and axial loads. The rules assume a circular cross section with uniformthickness in the circumferential and longitudinal directions. Design reqnirements 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-2Cg) of VIII-l and, AG-lOOCb) and (d) of VIII-2. The design and loading conditions given in VIII-I are discussed first in this chapter, followed by the rules in VIII-2.

2.2 TENSILE FORCES, vm.i The governing equations and criteria for the design of cylindrical shells under tensile forces are given in several paragraphs of YIIl-I. The tensile forces arise from various loads such as those listed in Paragraph UG-22 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 eqnation for the required thickness in the circumferential direction, Fig. 2.1Cal, due to internal pressure is given as t = PR/ CSE - 0.6P),

when

t

< O.5R

or

P < 0.385SE

(2.1)

where E= P= R= S=

Joint Efficiency Factor internal pressure internal radius 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

=

SEI/ (R

27

+ O.6t)

(2.2)

28 Chapter 2

,, --1, --,,,

, --' .. "

,

(bl

FIG. 2.1

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 and Jobnston, 1992), given by t

~

(2.3)

PRISE

The difference is in the additional term of O.6P in tbe 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 a.1R. This is demonstrated in Fig. 2.2 for circnmferential 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 Lame'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 = PRI(2SE

+

OAP),

with

t

< O.5R

or

P

< 1.25SE

(2A)

Cylindrical Shells 29

20.0 -I-r 10.0H-

~ SIP

l\

i~~ ........, 1"-r-, -e-,

to

-~

--..........

.........

--

I--~

--

---;;

Eq.(2.12) Eq.(2.1 )

1"--... ~

1--k Eq.(2.3)

.10

to

1.5

2.5

2.0

3.0

3.5

4.0

(R+t) I R FIG. 2.2 COMPARISON OF EQUATIONS FOR HOOP STRESS IN CYLINDRICAL SHELLS

or in terms of pressure, P

~

2SEtl(R - OAt)

(2.5)

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

=

PRI2SE

(2.6)

Equations (2.1) and (2.4) are in tenus 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 i~ expressed in

30

Chapter 2

terms of the outside radius Ro. This equation, which is obtained from Eq. (2.1) by substituting (Ro - t) for R, is given iu VIIl-I, Appendix 1, Article 1-1, as I

=

PRo/(SE

+

with

O.4P),

I

< O.5Ro

or

P < 0.385SE

P = SEI/(R a - O.4t)

(2.7) (2.8)

VIIl-1 does not given 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 PRol (2SE

+

l.4P)

(2.9)

P = 2SEI/(R o - 1.4/)

(2.10)

I =

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 platematerial as well as weld seams. The Joint Efficiency Factor for the overall thickness of a layered vessel is calculated from the ratio E

a

= (:l:

E,I,)II

-,

-,

a

-+-+--1-+-+--+--+++++

FIG. 2.3

(2.11)

Cylindrical Shells 31 where E = overall Joint Efficiency Factor for the layered cylinder E, = Joint Efficiency Factor in a given layer t = overall thickness of a layered cylinder t, = thickness of one layer The rules in VIII-l assnme that the longitudinal welds in varions 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 fl. 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-I. The qnantity 0.3855E = 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 ~

t ~ [PR/(SE - 0.6P)]

+ 0.125

48.125 in.

+ corrosioo

[100 X (48.125)/(20,000 X 0.85 - 0.6 X 100)] + 0.125 ~

0.41 in.

The calculated thickness is less than O.5R. Thns, Eq. (2.1) is applicable. A check of Eq. (2.4) for the required thickuess in the longitndinal 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. bas a shell thickness of 2.0 in. Determine the maximum 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 X 0.85 X 2.0/ (60 + ~

556 psi

0.6 X 2.0)

32 Chapter 2 For the longitudinal direction, the maximnm pressure is obtained from Eq. (2.5) as p

~

2 X 20,000 X 0.85 X 2.0/(60 - 0.4 X 2.0)

~1]49psi

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 VIII-I. It has an outside diameter of 8 It 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 X 48/(17,500 X 1.0 + 0.4 X 450) + 0.125 1.222 + 0.125 1.35 in.

From Eq. (2.10), the maximum allowahle axial pressure is p

~

2 X ]7,500 X 1.0 X 1.222/(48 - 1.4 X 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 = 'IT(R; - R2 ) ~

1T(48' - 46.778')

= 363.9 in.'

Hence, total allowable force in cylinder during operation is F

~

474.0 X 363.9

= 172,500Ib

Cylindrical Shells 33 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, t, = 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. Solntion

This problem must be solved by trial and error. Let E = 1.0. Then from Eq. (2.1), I

=

1400

~

5.87 in.

x

72/(18000

x

l.0 - 0.6

x

1400)

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 ~

x

l.00

+ 1 x 0.65)/24

0.985

Using this Joint Efficiency Factor, recalculate the required thickness: t

x

~

1400

=

5.97 in.

72/(18000

x

0.985 - 0.6

x

1400)

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

2.2.2 Thick Cylindrical Shells The VIlI-l 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-I(d) of VIlI-1. AB the ratio of II R increases beyond 0.5, the tbickness 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 Lame's equation, is given by SE = P(R~

+

where Ro and R are outside and inside radii, respectively. By substituting the relationship R o = R this expression, Eq. (2.12) becomes t = R(Z,n - 1)

= (SE

+

P)/(SE -

+

t

into

(2.13)

where

z

(2.12)

R') I (R;, - R')

P)

34

Chapter 2

Equation (2.13) is used in Appendix 1-2 of VIII-l to determine the required thickness in thick cylinders for the conditions t > O.5R or P > O.385SE. This equation can also be written in terms of pressure as P

~

SE[(Z - 1)1(Z

+

(2.14)

I)J

where

+

t)IRJ'

t>

O.5R

Z ~ [(R

For .1ongitudinaJ .Stre:ss, t = R(ZII2 -

1),

with

or

P> 1.25SE

(2.15)

where ~

Z

(PIS£)

+1

Equation (2.15) can be written in terms of pressure, P, as P

~

SE(Z -

(2.16)

1)

where

+

Z ~ [(R

t)IRJ'

The thick cylinder expressions given by Eqs. (2.12) throngh (2.16) can be expressed in terms of ontside radii as follows. For circumferential stress,

with

t

>

O.5R

or

P > 0.3855E

(2.17)

where Z

~

(SE

+

P)I(SE - P)


~

SE[(Z -

1)1(Z

+

(2.18)

1)]

where Z ~ (Rol R)'

=

[Rol(R o -

t)J'

For longitudinal stress with t > O.5R or P > 1.25SE, (2.19)

Cylindrical Shells 35 where Z

= (PIS£) +

1

or in terms of pressure, P, P = SE(Z -

(2.20)

1)

where Z = (RoIR)'

= [Ro/(R a -

t)]'

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 resnlting 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 psi, and E = 1.0. Assume a corrosion allowance of 0.25 in.

=

10.000 psi, R

=

18 in., 8

=

20,000

Solution The quantity 0.3858£ = 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 x 1.0

+ 10,000) I (20,000 X 1.0 - 10,000)

= 3.0

t = R(Z'12 - 1) = (l8.25)(3.0~' =

Total t

13.36

- 1.0)

13.36 in.

+

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?

36 Chapter 2 Solution The quantity 0.385S£ = 7700 psi is greater than the design pressure of7650 psi. Thus, Eq. (2.1) is applicable. t = PRI(SE - 0.6P)

= 7650

x

18/(20,000

x 1.0 - 0.6 x 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 =

x

1.0

+ 7650) I (20,000 x 1.0 - 7650)

2.239

t = R(Z;/2 - 1)

18(2.239 0.5

-

1.0)

8.93 in.

This comparison demonstrates the accuracy of the "simple-to-use" Eq, (2.1) over a wide range of Rlt 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 II in. Solution The thickness of 6.50 in. is greater than O.SR. 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 (12 2

+

5.52)/(12' - 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 allowable tensile stress, using a Joint Efficiency Factor of 1.0, or the allowable compressive stress, whichever is less. TIle allowable tensile stress controls thick cylinders, while the allowable compressive stress controls thin

Cylindrical Shells 37 cylinders. The procedure for calculating the allowable axial compressive stress in a cylinder is given in Paragraph UG-23 of VIII-l and is based on a theoretical equation with a large LID ratio (Jawad, 1994). It consists of calculating the quantity A

~

O.12S/(R o / t)

(2.21)

where A = strain

Ro

= outside radius of the cylinder

t = thickness

andthenusingastress",straindiagramfurnishedbytheASMEtodeterminethepermissibleaxial.compressive stress, B. The ASME plots 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 hotizontal axis, and a stress, B, along the vertical axis. The majority of the materials listed in the stress tables of II-D or VlII-l construction have a corresponding External 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

~

(2.22)

AE/2

TABLE 2.1 TABULAR VALUES FOR FIG. 2.4 OF 300

500

A 0.100 0.765 0.800 0.900 0.100 0.200 0.300 0.400 0.500 0.25 0.100

-04 -03

0.100 0.663 0.900 0.100 0.250 0.300 0.800 0.100 0.150 0.200 0.272 0.100

-04 -03

-·02

-01 +00

-02

-01

+00

B, psi

OF

0.145 +03 0.111 +05 0.114 0.119 0.123 0.150 0.163 0.170 0.172 0.178 0.178

700

0.135 +03 0.895 +04 0.965 0.101 +05 0.121 0.124 0.143 0.147 0.155 0.163 0.170 0.170

800

900

A 0.100 0.559 0.100 0.300 0.100 0.250 0.100

-04 -03 -02

0.100 0.499 0.100 0.150 0.200 0.300 0.300 0.100

-04 -03 -02

0.100 0.247 0.100 0.150 0.200 0.300 0.800 0.300 0.100

-04 -03 -02

-01 +00

-01 +00

-01 +00

B, psi 0.124 +03 0.665 +04 0.808 0.101 +05 0.122 0.139 0.139 0.114 +03 0.569 +04 0.717 0.805 0.849 0.897 0.124 +05 0.124 0.104 +03 0.444 +04 0.605 0.689 0.742 0.795 0.927 0.112 +05 0.112

38 Chapter 2

GENERAL NOTE: See Table 2.1 for tabular values.

--

.- .-

25,000

U3~t

20.000

500 F

18,000

I I

./

/

,,

700 F 800 F I

16.000 14,000

I

900 F

12.000

!-

10,000

9,000

~

8,000

0

7.000

~

~ ~

6.000

E

I-+-+++-++E

m D

29.0

x We <,

ar.o x 10~

I-+-+++-++E",

CL+,H-t+1f-++++-+-H-+t+t--/--i-H-I,-++t+1-l I

3,500

f--,:::oj~~++I+.\+l-+-+++-+-t-++I++-+-+-H-H+++1-l

3.000

r---.

24-5 X 10

1 20,8 x 1(1

I 1111

5,000

-++t+H--I~H+-++++++t---+-+-++++-I+t-H 4.000

1

E 22.8 X 101 1-- 1--+-1--+-1-+Em e

ml£

I

2345678923456789

.00001

0.0001

3456789

0.001

0.01

3

456789

2,500

0,1

FACTOR A

FIG. 2.4 CHART FOR CARBON AND LOW ALLOY STEELS WITH YIELD STRESS OF 30 ksl AND OVER, AND TYPES 405 & 410 STAINLESS STEELS

where E = modulus of elasticity of the material at design temperatme

The modulus of elasticity, E, in Eq. (2,22) is obtained from the actual stress-strain diagrams furnished by the ASME, such as those shown in Fig, 2,4, It is of interest to note tbat 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 correspoudiug to a straiu, A, of 0,1 in.rm. is 17,8 ksi, This is half the actual yield stress of 35,6 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 = EI16(R olt)

This can also be written as B

= O,0625EI(Rolt)

(2.23)

Cylindrical Shells 39 The theoretical equation for the critical axial buckling stress used by the ASME is given by
(2.24)

O.6E/(R o / t )

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
(2.25)

O.6CE/(R o / t)

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 Ral t ratios to an unconservative value of 1.0 for large Roll ratios. This fact should be considered when designing cylinders with large diameter to thickness ratios. VIIl-l 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. Deterntine 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 hase and 800eF at the top.

1.0

0.8 0.6

C 0.4 0.2

lh

I\~I

,~,. 'i, ~~~. . .1. R .... ~

- C= 1-0.90l1-e I I I I

400

16

t

r

• • -. 1-- .s: .-- r--• f - oj ...

d~

.~

~<

u

. c

II 800 1200 1600 2000 2400 2800 3200 3600 4000

R/t FIG. 2.5 C FACTOR AS A FUNCTION OF RfT (Jawad, 1994)

40

Chapter 2

I

i

I

32 psf

>-----

p,f

-1------

i

r--

Lin

20

N

psf

FIG, E2.8

Solution Assume t = 3/8 in. Axial force

=

Axial compressive stress

60

+ 251 = 311 kips

=

force/area of material in skirt

= 311,000/-rr =

X 96 X 0.375

2750 psi

The wind moment at the bottom of skirt using a vessel projected area of 8 It is M

=

32 X 8 X 36 X (36/2 + 34 + 26) + 24 X 8 X 34 X (34/2 + 26)

+ 20 X 8 X 26 X (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

~

Mell

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 / ('ITR~t) = 1,053,632 X 12/('IT X 48' X 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°P' 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 x 8 X 36 X (36/2 + 34 + 26 - 16) + 24 X 8 X 34 x (34/2 + 26 - 16)

+ 20 X 8 X 10 X (10/2)

= 571,392 + ~

176,256 + 8000

755,648 ft-lb

42 Chapter 2

Bending stress

~

755,648 X 12/('iT X 482 X 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 thickness is adequate at the top of the skirt,

7,000 psi. Thus, the selected

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, outof-roundness considerations, and handling factors.

Example 2,9 Problem What is the allowable compressive stress in an internal cylinder with Do = 24 ft, temperature = 900'F? Use Fig, 2,4 for the External Pressure Chart,

t

= 3/16 in" and design

Solution From Eq, (2.21), A = 0,125/(12010,1875) =

0.0002

From Fig, 2.4, this A value falls to the left of the cnrve for 900'F, Therefore, Eq. (2,22) must be used, The value of E is obtained from Fig, 2,4 as 20,8 X 106 psi for 900'F, Hence, allowable compressive stress B is B

= 0,0002

X 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 bnckling of cylindrical shells under external pressure are extremely cumbersome to use directly in

Cylindrical Shells 43 design (Jawad, 1994). However, these equations can be simplified for desigu 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 VTII-l and Vill-2 are identical for external pressure. Accordingly, references in this section are made to paragraphs in Vill-l only.

2.4.1 External Pressure for Cylinders with Dolt

2:

10

The ASME uses plots to express the lowest critical strain, A, in terms of the ratios LIDo and Dolt of the cylinder, as shown in Fig. 2.6. The designer calculates the known quantities LIDo and D,1t and then uses the figure to determine buckling strain, A. To correlate buckling strain to allowable external pressure, the designer uses the 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 Vill-I Paragraph UG-28 for determining the allowable external pressure for cylinders with Dolt ratios equal to or greater than ten consists of the following steps: 1. Assnme a value of t for the cyliuder. 2. Calculate the quantities LIDo and Dolt. 3. Use Fig. 2.6 with the ca!cnlated values of LIDo and Dolt and establish an A value.

e

LIDo

-•

, e e

e

,

e

s

...

'..

A

FIG. 2.6 GEOMETRIC CHART FOR CYLINDRICAL VESSELS UNDER EXTERNAL PRESSURE (Jawad and Farr, 1989)

44 Chapter 2 4. 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)/(D olt)

(2.26)

6. When A falls to the left of the curves, the value of P is determined from p

~

2AEI3(Dolt)

(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 VIII· 1 to define the effeetive length of some commonly encountered cylinders. The effective length of cylinders with spiral stiffeners or variable thicknesses between supports is not addressed by VIII·\. 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.

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"F? Use Fig. 2.4 for the External Pressure Chart

Solution Try t = 0.50 in. Then LIDo = 4.0 and Dol t = 120.00. From Fig. 2.6, we obtain a value of A Then from Fig. 2.4, B = 3000 psi. The allowable pressure is obtained from Eq. (2.26) as

=

0.00022.

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 LIDo = 4.0 and Dolt = 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 X 0.00017 X 27.0 X ]0'/(3 X 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 I 8 in.


r:

T

T 1

hl3

h/3

L

L

L

h/3

-l h

(a - 1)

18- 21

Ibl

[Notes OJ and (21]

[Note (311

---i

T r rt 1 1 1 L

L

te -

1)

h

T

Ic - 2) {Notes Il} and (21J

T L L

L

Idl

tel [Note (311

r: L

1 IfI [Note 13H

NOTES:

(1) When the cone-to cylinder or the knuckle-to-cylinder junction is not 8 line of support, the nominel thiokness of the cone, knuckle, or toriconical section shall not be less than the minimum required thickness of the adjacent cylindrical shell. (2} Calculations shall be made using the diameter and corresponding thickness of each section with dimension L as shown. (3) When the cone-to-cylinder or the kmrckle-to-cvlinder junction is a line of support, the moment of inertia shall be provided in accordance with 1-8.

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

46 Chapter 2 The allowable compressive hoop stress is then

19.1

x

3010.375

1530 psi

2.4.2 External Pressure for Cylinders with Dolt < 10 When Dol I is less than 10, the allowable external pressure is taken as tbe smaller of tbe valnes determined from the following two equations: [2.167/(D olt) - 0.0833] B 2S Po' ~ Dolt [1 - lI(Dolt))

(2.28)

(2.29)

wbere B is obtained as discussed above. For values of (Dol I) of less than or equal to 4, the A value is calculated from A ~ 1.1/(Do l t)'

(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 Cbart 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 Dol I = 10 to about 1.67 for Dol I = 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 I 3 s;

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°F. The allowable stress from tension is 17,500 psi. Solution Calculations give LIDo = 3.60 and Doll = 4.0. And since Dolt = 4.0, Eq. (2.30) must be used. Hence, A ~ 1.1/(4.0)'

= 0.0688 From Fig. 2.4, B

17,800 psi.

Cylindrical Shells 47 From Eq. (2.28). P"

~

[(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 nsed. From Eq. (2.29), P"

~

(2 X 32,040/4.0)(1 - 1/4.0)

= 12,020 psi

Therefore, the allowable jacket pressure in accordance with VUl-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 VUl-I yet.

2.4.3 Empirical Equations It is of interest to note that Fig. 2.6 can only be used for (Dolt) of up to 1000. Larger values are not

permitted presently by the ASME. One approximate equation (Jawad, 1994) that is frequently used by designers for large (Dol t) ratios was developed by the U.S. Navy and is given by P

~

0.866E/(LlDo)(Dolt)25

(2.31)

where E P

~ ~

modulus of elasticity, psi 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 VU1 does not give any method for calculating the allowable compressive stress due to combined effect of vacuum and axial loads. One such method is given by Bergman (Bergman, 1955). It uses an equivalent external pressure to account for the axial compression effect on external pressure. Another method that is used to combine axial and external pressure is that of Gilbert (Gilbert and Polani, 1979). This method uses an interactive equation sintilar to the one used for calculating the buckling of beam columns. Example 2.12 Problem Solve Example 2.10 using Eq. (2.31). Solution From Example 2.10, t = 0.375 in., LIDo = 4.0, Dolt P

~

160.00. andE = 27,000 ksi. Theu from Eq. (2.31),

~

0.866 X 27,000,000/(4.0)(160)25

~

18.1 psi

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

48

Chapter 2

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 sbortening 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 suchrings is determined from

I,

=

[D~L,(r

+

(2.32)

A,IL,)A] 114

or

+ A,/4)A]/l0.9

(2.33)

where Is = required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in." I' = required momeut of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about s their combined ueutral axis, in.' The effective length of the shell is taken as l.lO(D ot,)!l 2. L, = 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.? t = minimum thickness of the shell, in. t, = nominal thickness of the shell, in.

To design the stiffening ring 1. Assume first an area, A" of the stiffening ring and calculate the available moment of inertia. I, or Is. 2. Calculate B from the equation B = 0.75 [PDo/(l

+ A,IL,)]

(2.34)

3. Use the appropriate External Pressure Chart and determine an A value. 4. If B falls below the left end of tbe 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 pressnre is 12 psi, and the design temperature is lOOoP.

I

I

Cylindrical Shells 49 Solution

Try t = 1/4 in. Then Dolt = 240.0 and LIDo 0.00016, and from Eq. (2.27),

= 12(10 +

2.5/3)/60

2.17. From Fig. 2.6, A

P = 2 X 0.00016 X 29,000,000/3(240)

= 12.9 psi

Thus, a shell thickness of 1 14 in. is adequate. For the stiffeniug ring, try a 3 X 1 14 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 bd31 12. Thus, I = 0.25 X 3.0'/12

= 0.56 in."

: I I

l-r-r--r-

I

I I

5' 0.0. I

~

I I I I

-

I I I I

(

-..

.. '"

I I

I I

.671'

r---

I

~

"

• :

; FIG. E2.13

tJ

50

Chapter2

From Eq. (2.34), B = 0.75[12 X 60.0/(0.25 =

+ 0.25 X 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 X 2110/29,000,000

= 0.000146

From Eq. (2.32),

t,

= [60.0' x 140.0(0.25 =

+ 0.25 x 3.0/140.0) 0.000146] /14

1.34 in.'

Since this number is larger than the actual moment of inertia of the ring (0.56 in.'), 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),

t; = =

[60.0' x 140.0(0.25 + 0.25 x 3.01l40.0) 0.000146]/10.9 1.72in.4

The effective centroid of the shell-ring section, Fig. E2.13(h), is h = [4.26 X 0.25 X 0.125 =

+ 0.25 X 3.0 (1.5 + 0.25)]/(4.26 X 0.25 + 0.25 X 3.0)

0.796 in.

The actual moment of inertia is

t

= 4.26 X 0.253/12 ~

+ 4.26 X 0.25 X 0.671' + 0.25 X 3.03/12 + 0.25 X 3.0 X 0.954'

0.006 + 0.480 + 0.563 + 0.683

= 1.73 in.'

Thus, using the composite section results in a 1/4 in. X 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. DO-30 of VIII-I, which is reproduced in Fig. 2.8. The welds must he able to support a radial pressure load from the shell of PL,. 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 PL»o. This shear load is arbitrary and is based on the assumption that if the rings buckle, bending moments

Cylindrical Shells

('..- .f'1 p--..,..

I 2in.min.

In-Line

St8f98red Intermittent

Intermittent Weld

ContinuousFillet Weld One Side, Intermittent Other Side

Wald

S '" St external stiffeners

S -c 12 t internal stiffeners

t

(al

(el

(bl

w

fdl

fal

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

51

52 Chapter2 occur and generate shear forces. VITI-l 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-!.

Example 2.14 Problem Calculate the required size of the double fillet welds attaching the stiffening riug shown in Fig. E2.13(b) of Example 2.13 to the shelL Let the allowable tensile stress of SA 285-C at loo'F be 15,700 psi. Solution The radial load, FI, on the rings is equal to PL,. PL, = 12.0 x 140

=

1680 lb I in. of circumference

Allowable tensile stress in the fillet weld from Table 1.4 is 0.55S weld is

8635 psi. The total load carried by

Total load = number of welds attaching ring X size of weld X allowable stress =2XWX8635

Hence, the required weld size, W, is W = 1680/ (2 x 8635) =

0.10 in.

Use 2 1/4 in. continuous fillet welds, in accordance with the minimum requirements of UG-30(t). Shearing force, V, on the weld is V

=

om PL,Do

= 0.01 x

12

x

140

x

60

= 1008lb

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 -r = VQ/lt

wbere Q is at the location of the weld, as shown in Fig. E2.13(b), and is given by Q

= 4.26

x 0.25(0.796 - 0.125)

= 0.71

in.'


Hence, T

=

1008

=

827 psi < 8635 psi

x

0.71 I 1.73(2

x

0.25)

2.5 CYLINDRICAL SHELL EQUATIONS, VIII·2 The equation for the design ofthin cylindrical shells is given in Section AD-20l ofVIII-2 and is expressed as t

where P= R= 5= t=

= PRI(S -

O.5P),

for

P

<

OAS

(2.36)

internal pressure inside radius allowable stress at design temperature required thickness

This equation is derived from the basic equation for thin cylindrical shells, t = PRIS, 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 O.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

=

(O.5PR

+

(2.37)

F) I (S - 0.5P)

For values of P > OAS, the ratio of t IR increases beyond the scope of Eq, (2.36) and the design eqnation in VIIl-2 is based on the plastic theory of thick cylindrical shells (Prager and Hodge, 1965). This equation is given by t = R(eP'S - I)

for

P

>

OAS

(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

=

1.54 in.

x

30/(20,000 - 0.5

x

1000)

54 Chapter 2

(b) Since P

> OAS, Eq. (2.38) applies. Then, t = 30(e85OO120000 =

-

I)

15.89 in.

2.6 MISCELLANEOUS SHELLS 2.6.1

Mitered Cylinders

Mitered cylinders, Fig. L9,are used in nozzle connections, transition sections, and reducers. Neither VITI-l nor VIII-2 give design rules. The piping code (ASME B31.3, 1993) gives design equations for various miters. The basic equation for the allowable pressure in a shell with a single miter is P = (SEt/R){l/[1

+

0.643(Rlt)"2 tan

P = (SEtIR){I/[1

+

1.25(Rlt)'12 tan

en

for

e<

22.5'

(2.39)

ell

for

e>

22.5'

(2.40)

where E = Joint Efficiency Factor P = allowable pressure, psi R = radius of the shell in accordance with Fig. 2.9, in. S = allowable stress t = shell thickness, in. a = angle of change in the direction of the miter joint, Fig. 2.9 e = «t ; 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., a 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. Solntion From Eq. (2.1) for a straight cylinder, t

= 500 x =

Try t

=

1 in.

e = 40/2

24/(15,000

x

0.85 - 0.6

x

500)

0.96 in.

= 20'. Thus, Eq. (2.39) governs.

P = (15,000 x 0.85 x 1.0/24) (I I [I

= 248 psi

+ 0.643(24/1.0)112 tan

20])

Cylindrical Sliells 55

FIG. 2.9 MITERED BEND

Tliis is inadequate. Try

t ~ 1.8125

in.

P = (15.000 X 0.85 X 1.8125/24){lI[1 + 0.643(24/1.8125)'/2 tan 20])

= Use t

520 psi> 500 psi

113/ 16 in.

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). Tlie hasic equation for hoop stress is expressed as (2.41)

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. = angle as defined in Fig. 2.10

56 Chapter 2

,.....--------.....,- - -

::::;;;;;;--r--~,

- - - - -...11--.....

, FIG. 2.10 ELLIPTICAL CYLINDER

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 For

q,

~

100 psi, S = 20,000 psi, and E = 0.85

= 0", Eq. (2.41) gives t

=

100 X 50.125' X 46.125'/20,000 X 0.85(50.125' sin' 0 + 46.125' cos' 0)3;2 + 0.125

~

0.320 + 0.125

= 0.45 in.

For

q,

= 90", Eq. (2.41) gives

t = 100 X 50.125' X 46.125'/20,000 X 0.85(50.125' sin' 90 = 0.250 =

+ 46.125' cos' 90)3;2 + 0.125

+ 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-l 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 cortical and toriconical shapes, The design rules for most of these shapes differ significantly in VIII-l and VIII-2, This difference is due to the design approach used in developing the equations for VIII-l 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 Pressnre in Spherical Shells and Pressure on Concave Side of Hemispherical Heads The required thickness of a thiu spherical shell due to internal pressure is listed in Paragraph UG-27 and is given by t

~

PRI(2SE - Oo2P),

when

t

< 0.356R

or

P < O.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 I (R

57

+ 0.2t)

(3.2)

58

Chapter 3

Notice the similarity between Eq. (3.1) and the classical equation for the membrane stress in a spherical shell (Beer and Johnston, 1992), given by (3.3)

t = PRI2SE

The difference is in the additional term of 0.2? 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 determiniug the thickness in such spheres is given in (API 620, 1990). In some instances, the outside radius of a shell is known rather than the inside radius. In this case the governing equatiou is obtained from Eq. (3.1) by substituting (Ro - t) for R. The resulting equation is given in VlII-l, Appendix 1, Article 1-1, as t = PRol (2SE

+

0.8P),

1< 0.356R o

with

or

P

<

0.665SE

(3.4)

or P

=

2SEI/(Ro - 0.81)

(3.5)

As the ratio of tlR increases beyond 0.356, the thickness given by Eq. (3.1) becomes noneonservative. This is similar to the case for cylindrical shells discussed in 2.2. The ASME VIIl-I eqnation for thick spherical shells is given by I = R(f't; -

1)

(3.6)

where Y = 2(SE

+

P)/(2SE - P)

Equation (3.6) is used in Appendix 1-3 of VIII-I to determine the required thickness in thick spherical shells for the conditions where I > 0.356R or P > 0.665SE. This equation can also be written in terms of pressure as P = 2SE[(Y -

1)/(Y

+

(3.7)

2)]

where Y = [(R

+

I)IR)'

The thick shell expressions given by Eqs. (3.6) and (3.7) can be stated in terms of outside radii as with

t

>

O.356R

where Y = 2(SE

+

P)/(2SE - P)

or

P> 0.665SE

(3.8)

Spherical Shells, Heads, and Transition Sections

59


= 2SE[(Y -

Il/(Y

+ 2)J

(3.9)

where Y = (RoIRl' = [Rol(R a - I)J3

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 abc and def are subjected to concave pressure and Eqs. (3.1) through (3.9) may be used. Paragraph UGe32(f) of Vlfl- I 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 e70 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 tbe hemispherical beads if the allowable stress is 20,000 psi? Solution The quantity 0.6658E = 11,300 psi is greater than the design pressnre of 100 psi. Thus, Eq. (3.1) applies. The inside radius in the corroded condition is equal to

b

1l------IIc A

SHELL A

d II-----llf

INTERMEDIATE HEAD

e SHELL B

B

SKIRT

FIG, 3.1

60 Chapter 3 R = 48

+ 0.125

= 48.125 in.

The total head thickness is t

= PR / (25£ =

100 X (48.125)/(2 X 20,000 X 0.85 - 0.2 X 100) + 0.125

= 0.142 =

- 0.2P) + corrosion

+ 0.125

0.27 in.

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 x 20,000 X 0.85 X 1.0/(60 + 0.2 X 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-I. 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 hemispberical head if the allowable stress is 19,700 psi. Solution From Eq. (3.4), the required head thickness is t = 450 X 48/(2 X 19,700 x 1.0 = 0.543

+ 0.125

= 0.67 in.

+ 0.8

X 450)

+

0.125


61

Example 3.4 Problem Calculate the required hemispherical head thickness of an accumulator with P S = 15,000 psi, and E = 1,0, Assume a corrosiou allowance of 0.25 in.

=

10,000 psi, R

=

18 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 =

+

2(15,000

P)/(2SE - P)

x

1.0

+ 10,000)/(2 x 15,000 x 1.0 - 10,000)

= 2.5

t

= R(y"3

- 1)

=

(18.25)(2,5'!3 - 1.0)

=

6.52 in.

Total head thickness = 6.52 + 0.25 = 6.77 in. 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-I, This taper, however, is impractical to make in this case since the thickness of the head is abouttwo-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-I and consists of calculating the quantity A = 0.125/(R o / t)

(3.10)

where A = strain R o = 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 Po

=

B/(Ro/t)

(3.11)

62 Chapter 3

t - 6.77"

..--t =13.61"

FIG. E3.4

If the calculated value of A falls to the left of the stress-strain line in a given External Pressure Chart, then P, must be calculated from tbe equation P, ~ O.0625E/(R o / 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) ofVllI-1. Tbis is illustrated in Fig. 3.1. For an applied internal pressure in compartment B, the hentispherical head def is subjected to convex pressure and Eqs. (3.10) and (3.11) may be used.


63

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), Po = 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 liue in Fig. 2,4, we have to use Eq. (3.12). Po = 0.0625 ~

x

20,800,000/(150/0.25)'

3.6 psi

Since this value is less than 15 psi, a larger thickness is needed. Try t A

~

0.125/(150/0.50)

= 0.00042

= 0.50 in.

=

64

Chapter 3

From Fig. 2.4, B = 4500 psi. And from Eq. (3.11), Po = 4500/(15010.50)

=

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 giveu in Paragraph AD-202 of VIII-2 as I = 0.5PRI(S -

P < 0.48

when

0.25P),

(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 In [(R +

P > 0.48

when

1)1R] = 0.5P/S,

(3.14)

This can be written also as t

= R(eO.5PIS

-

1),

when

P

> O.4S

(3.15)

When meridional forces, F (for instance, wind and earthquake loads) are present on the head, then Eq. (3.13) is modified as follows: I = (0.5PR

+ F)/(S

- 0.25P),

when

P < O.4S

(3.16)

where F = Meridional force, lh/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.l 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 a1lowahle external pressure on spherical shells in VIll-2 are given in Paragraph AD-320. The rules and factors of safety are identical to those given in VIII-!. Similarly, the rules in VIII-2 for calculating the allowable pressure on tbe convex side of hemispherical heads are given in Paragraph AD-350.1. They are identical to those given in VIII-I.


65

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 PIS = 0.5. Since this ratio is larger than 0.4, Eq. (3.15) must be used. t

=

20 (e°.5

=

5.68 in.

x 10,000/20,000 -

1)

3.4 ELLIPSOIDAL HEADS, VIU·l 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 spberical radius of 0.9D and a knuckle radius of O.17D, as sbowu 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 Vlll-L Tbe thickness is obtained from the following eqnation: t

= PD / (2SE -

0.2P)

(3.17)

or in terms of required pressure, P

~

2SEt/(D

+ 0.2')

(3.18)

where D = inside base diameter E ~ Joint Efficiency Factor

P = pressure on the concave side of the head S = aJlowable stress for the material t = tbickness of the head Ellipsoidal heads with a radius-to-depth ratio other than 2:I may also be designed to the requirements of VIII-I. The governing equations are given in Appendix 1-4 of VIII-I as t = PDK/ (2SE - 0.2P)

(3.19)

where K = (116)[2

+

(D/2h)']

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 I of Vlll-I. Equation (3.19) can be expressed in terms of the required pressure as P

= 25Et/ (KD +

0.2t)

(3.20)

66

Chapter 3

= 2:1

D/2h

o

o (b) FIG. 3.2

These equations can also be written in terms of the outside diameter, Do> Thus,

=

+

2P(K - 0.1)]

(3.21)

P = 2SEtl[KDo - 21(K - 0.1)]

(3.22)

t

PDoKl[2SE

or in terms of required pressure

It is of interest to note that VIII-I does not give any PIS limitations for the above eqnations. Nor does it have any rules for ellipsoidal heads when the ratio of PIS is large.


67

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 UG-33 of VIII -I. The required thickness is the greater of the two thicknesses determined from the steps below. 1. Multiply the design pressnre on the convex side by the factor 1.67. Then use this new

pressnre 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 procednre given for spherical shells in Section 3.2.2. The procednre consists of calculating the quantity A

= 0.125/(KoDolt)

(3.23)

where A = strain Ko

Do t

= function of the ratio D o/2ho and is obtained from Table 3.1 =

=

outside base diameter of the ellipsoidal head thickness

Then, using a stress-strain diagram similar to Fig. 2A, determine the B value. The allowable pressure is calculated from P,

~

BI(KoDolt)

(3.24)

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

= 0.0625E/(KoDolt)'

(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 lOO°F, concave pressure of 200 psi, convex pressnre of 15 psi, allowable stress is 17,500 psi, and joint efficiency of 0.85. The head is made of low-carbon steel.

TABLE 3.1 FACTOR Ko FOR AN ELLIPSOIDAL HEAD WITH PRESSURE ON THE CONVEX SIDE Dol2h o Ko Dol2ho

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

68 Chapter 3

Solution For Concave Pressure

From Eq. (3.19), with K = (1/6)[2

+

(2.2)2] = 1.14,

t = (200 X 216.0 X 1.14)/(2 X 17,500 X 0.85 - 0.2 X 200)

=

1.66 in.

For Convex Pressure

1. First calculate the pressure and thickness. P

=

1.67 X 15

=

25.1 psi

t = (25.1 X 216.0 X 1.14)/(2 X 17,500 X 1.0 - 0.2 X 25.1)

= 0.18 in.

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

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

From Fig. 2.4, B = 12,000 psi. P,

=

12,000/(0.99 X 219.32/1.66)

=

91.7 psi

Thus, minimum t = 1.66 in.

3.5 TORISPHERICAL HEADS, VIII-l 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-l rules, in accordance with Paragraph UG-32(e). The most commonly used F&D heads can be approximated by a spberical 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

=

0.885PLI (SE - O.IP)

(3.26)


I·

o

69

·1 L=O

FIG. 3.3

or in terms of required pressure, P = SEt I(0.885L + O.lt)

(3.27)

where E = L = P= S= t

Joint Efficiency Factor inside spherical radius pressure on the concave side of the head

allowable stress for the material = thickness of the head

Torispherical heads with various spherical and knuckle radii may also be designed to the requirements of VIII-I. The governing equations are given in Appendix 1-4 as t

=

PLMI(2SE - 0.2P)

(3.28)

70 Chapter 3 where M ~ (114)[3

+

(Llr)'I']

and Llr varies between 1.0 and 16.67. The 1.0 ratio corresponds to a hemispherical shell. The M equation is given in Article l-4(d) of Appendix 1 of VIII-I. 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, La, as t

~

PLoM/[2SE

+ P(M

- 0.2)]

(3.30)

2SEt/[MLo - t(M - O.2)J

(3.31)

or in tenus of required pressure, P

~

The theoretical membrane stress distribution in the circumferential, N e, and meridional, N' directions in shallow heads due to internal pressure are shown iu Fig. 3.4. Both the circumferential and meridional stresses at the crowu of the head are tensile with a maguitude of S ~ Pa' / 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 - (alb)']. This compressive stress, which is not considered by Eq. (3.28), could canse buckling of the shallow head as the ratio of Dlt 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 IS, = (0.33

+ 5.5rID)(t1L) + 28(1 - 2.2rID)(tlL)' -

Q

FIG. 3.4

0.0006


71

where

D

= base diameter of head, in,

L n P r

= spherical cap radius, in. = factor of safety

S, t

= design pressure, psi = knuckle radius, in.

= yield stress of the material, psi = thickuess, in.

This equation normally results in a thickuess 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-l states that the maximum allowable stress used to calculate the required thickuess 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 thickuess 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 Kof) oExample 3.9 Problem Calculate the required thickness of an F&D head with an inside base diameter of 18 ft, design temperature of lOooF, interual (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 from Eq. (3.28) t

~

X

216

13.0 in., and M = (1/4)[3

(200 X 216.0 X 1.77)/(2 X 17,500 X 0.85

+ (216/13)112]

0.2 X 200)

= 2.58 in.

For Convex Pressure 1. Find the pressure and the thickuess. P

=

1.67 X 15

=

25.1 psi

t = (25.1 X 216.0 X 1.77)/(2 X 17,500 X 1.0 - 0.2 X 25.1) ~

0.27 in.

1.77, we get

72 Chapter 3 2, Let t = 2.58 in, Then Outside radius = 216 + (2

X

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. P, = 14,000/(221.16/2.58)

163 psi> IS psi Thus, t = 2.58 in.

3.6 ELLIPSOIDAL AND TORISPHERICAL HEADS,

vm-z

The required thickness for ellipsoidal as well as torispherical heads is obtained from Paragraph AD-204 and Article 4-4 of Vlll-Z. The procedure utilizes a chart, Fig. 3.5, which takes into consideration the possibility of buckling of thin shallow heads, as discnssed in the previous section, The design consists of calculating the quantities PIS and riD first and then using Fig. 3.5 to obtain the quantity t/L, and thus t. The thickness for 2:1 ellipsoidal heads is obtained by using the riD = 0.17 curve, while the thickness for a standard F&D head is obtained by using the riD = 0.06 curve. Figure 3.5 is plotted from the following equation: (3.32) where A =A, + A 2 + A3 A[ = -1.26176643 - 4.5524592 (riD) + 28,933179 (riD)' A 2 = [0.66298796 - 2.2470836 (rID) + 15.682985 (rID)2] [In(P / SJ] A] = [0,26878909 X 10- 4 - 0.42262179 (rID) + 1.8878333 (rID)2][ln(PlS)]' 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 Vlll-2 and then VlII-1 rules. S = 20.000 psi for VIIl-2, and 15,000 psi for VIIl-1.

73

Spherical Shells, Heads, and Transition Sections 0.10 0.09 0.08

/-

0.07 0.065 0.06 0.055 0.05

1/1

7

~ !-

7TI V//

/ WI, 0

0.045

~

0.04

Z

0.035

z

~/. 0.025

£!j

0.02 ,10

0.015

\

1\

0.20 0.01

PIS

\

0.009 0.008

\

»;:

\ 1\

~~~

/, IX//, V/, 10: l/jW "z)) Ih W

«

\ \-:; 0~ «;; 0: ~

\

r~

~ @ ~W

@ ~ ~ :fj :;f~ ~ W

0.005

0.002

0.06\

1\

\

0.006

0.003

\

0.15

~/,

~

0.10

\

0.007

0.004

k%

0.17 (2:1 Ellipsoidal head)

h

L'

IA ~ ~

~, ,

t

I

0/2

--:J

oat heads use L '" 0.90 to calculate tiL

,

I

N

ci

M

0

0

1

0

s0

0

'"00 ci

80

ci

~

0 0

ci

co 0 0

ci

0 '" 0

in

0

ci ci

tiL FIG. 3.5

- -

--

-

- !-

0.001 0 0

---

-

NOTE: For 2: 1 Elhpsoi-

I'.

--

0

ci

N

'" 0 N

M

0

ci

ci

0

0

in

M

0

ci

sci ..'"ci '"ci 0

0

74

Chapter 3

Solution

Using VIIl-2 rules, PIS

= 0.002

tiL = 0.003 Using VIIl-l rules and Eq. (3.26),

and

and

riD = 0.06

From Fig. 3.4,

t = 0.885 =

x

t

= 0.003

x 168

= 0.50 in.

40 X 168/(15,000 X 1.0 - 0.1 X 40)

0.40 in.

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

3.7 CONICAL SECTIONS, VIll-l 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. Vlll-l provides rules for the design of the junctions. These rules differ for internal and external pressure.

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

where

(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 CY./(D + 1.21 cos CY.)

(3.34)

Equations (3.33) and (3.34) can also be expressed in terms of outside diameter as t = PDal [2SE cos CY.

+ P(2 - 1.2 cos CY.)]

(3.35)


75

,

Portion of a cone

rt

t t L

D,

(al

(hi

DL,

-.-

DU

DL

Le

t

DL D,

D" D"

Iel

(dl

(.1

FIG. 3.6

p

~

2SEt cos at/[D a - t(2 - 1.2 cos at)]

(3.36)

Equations (3.33) to (3.36), which are applicable at any angle at, are limited by VIII-I to ex :5 30". When the angle at exceeds 30", then VIII-I requires a knuckle at the large end, as showu 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 coneto-shell 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 I of VIII-I. The required area is obtained from A" ~ (k Q, R,fS, E,) (l - /)./ at) tan at

where, A r L = required area at the large end of the cone, in. E, = Joint Efficiency Factor of the longitudinal joint in the cylinder

(3.37)

76 Chapter 3

E; = E, = E, = k=

QL = R, = S, = S, = S, =

y= LI. =

modulus of elasticity of the cone, psi modulus of elasticity of the reinforcing ring, psi modulus of elasticity of the cylinder, psi 1 when additional area of reinforcement is not required y I S,E,. but not less than 1.0 when a stiffening ring is required axiaJload at the large end, lb/ in., including pressure end-load large radius of the cone, in. allowable stress in the cone, psi allowable stress in the reinforcing ring, psi allowable stress in the cylinder, psi S,E, for the reinforcing ring on the shell S,E, 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 (3.38)

where AeL = available area at the junction, in.' t = minimum required thickness of the shell, in. t; = nominal cone thickness, in. t, = minimum required thickness of the cone, in. t., = 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 I of VIll-1. The required area at the small end of the cone is obtained from A"

=

(3.39)

(k Q, R,IS, E,)(l - LI.! c) tan a.

where A" = required area at the small end of the cone, in," Q, = axial load (including pressure end load) at small end, Ib I in. R, = small radius of the cone, in. LI. = angle obtained from Table 3.3

TABLE 3.2 VALUES OF LI. FOR JUNCTIONS AT THE LARGE CYLINDER DUE TO INTERNAL PRESSURE LI, deg.

0.001 11

0.002 15

0.003 18

0.004 21

PIS,E, LI, deg,

0.006 25

0.007 27

0.008 28.5

0.009' 30

P/SsE1

NOTE: (1) I:!.. = 30° for greater values of PISsE1 .

0.005 23


77

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 A"

=

0.78(R,I,)'" [(t, - I)

+

(3.40)

(I, - 1,)/eos e]

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 a exceeds 30°, VIII-I 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 = PLMI(2SE - O.2P)

where M = (1/4)[3

+

(Llr)1/2]

L = D;!2 cos a D,

=

inside diameter at the knuckle-to-cone junction cos a) inside knuckle radius, in.

= D - 2r (I -

r =

Eqnation (3,41) can be expressed in terms of the required pressure P

= 2SEII(LM + 0.2t)

(3.42)

Equations (3,41) and (3,42) can also be written in terms of the outside diameter, Do, as

= PL,MI[2SE +

P(M -- 0.2)J

(3.43)

P = 2SEII[ML, - !(M - 0.2)]

(3.44)

t

or in terms of required pressure,

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 D, ratber than on the shell diameter. ASME VIII-I does not give rules for the design of knnckles (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

TABLE 3.3 VALUES OF b. FOR JUNCTIONS AT THE SMALL CYLINDER DUE TO INTERNAL PRESSURE 0.002 4 PfSsEj ~,deg.

0.08 24

NOTE:

(1) A = 30° for greatervalues of PI SsE1

0.005 6 0.10 27

0.010 9 0.125'

30

0.02 12.5

0.04 17.5

78 Chapter 3

O2 tf

01 I

ex

t

•

s

.I I

FIG. 3.7

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 If

~

(180/""r)[P(K,

where E = Joint Efficiency Factor K, = 0.125 (2r + DJl' tan" - ""lTr'/360 K, = 0.28D,(D,t,)'12 K, = 0.78Ks(Kst,)'12 K4 = 0.78tJK,t,)1I2 K, = 0.55t,(D,t,)1I2 K, = [D, + 2r(1 - cos a)]/2 cos a P = internal pressure, psi

+ K, + K,)/1.5SE - K, - K,]

(3.45)


79

S = allowable stress, psi t, = thickness of the cone, in. If = thickness of the flue, in. I, = thickness of the shell, in. a = 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.l 1. Let axial compressive load at cone vicinity from mounted equipment 50 kips.

=

Allowable stress, psi Joint Efficiency Factor Modulus of elasticity, ksi Pressure, psi

Small Cylinder

Cone

Large Cylinder

Reinforcing Ring

15,000 0.85 27,000 100

16,000 l.0 29.000 100

17,500 0.85 25,000 100

13,000

• Rs

= 5'-0'

I

•

. FIG. E3.11

30,000

80

Chapter 3

Solution

Small Shell The required thickness from Eq. (2.1) is t

=

100 X 60/(15,000 X 0.85 - 0.6 X 100)

= 0.47 in.

Use t = 1/2 in. Cone

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

Use t

=

5/8 in.

Large Shell Again, using Eq. (2.1), we get t = 100 X 7 X 12/(17,500 X 0.85 - 0.6 X 100)

= 0.57 in. Use t = 5/8 in.

Large Cone-to-Shell lunction Assume that a reinforcing ring, if needed, is to he added to the shell. Then from Eq. (3.37), we calculate the stiffness ratio, k, as k

=

17,500 X 25,000,000/(13,000 X 30,000,000)

= 1.12

The axial loads are given by QL = PRLI2 - axial equipment load

=

100 X 84/2 - [50,000/(27l'84)]

=

4105 Ib/in.


81

Next, we need to calculate the need for reinforcement in accordance with Table 3.2. PlS,El = 100/17,500 X 0.85

= 0.0067 From Table 3.2, Ii. = 26.4". Reinforcement is needed since a The amount of reinforcement is calculated from Eq. (3.37): A'L

=

28".

=

(1.12 X 4105 X 84/17,500 X 0.85)(1 - 26.4/28) tan 28

=

0.79 in.'

The available area in the shell and cone due to excess thickness is calculated from Eq. (3.38):

=

(0.625 - 0.57)(84

= 0.040 = 0.59

X

0.625)112

+ (0.625 - 0.60)(84

X

0.625/c08 28)112

+ 0.193

in.'

0.79 - 059

The additional area needed at the large junction rolled the hard way.

0.20 in.' Use a 2 in. X 1/4 in. bar

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

=

1.04

X

27,000,000/(13,000

X

30,000,000)

The axial loads are equal to Q" = PR)2 - axial equipment load

=

100 X 60/2 - [50,Ooo/(27l'60)]

=

2867 Ib/in.

The need for reinforcement is obtained from Table 3.3. PI $,£, = 100/15,000 X 0.85 =

0.0078

82

Chapter 3

From Table 3.3, Ll = 7,68°. Since this is less than 28°, reinforcement is required in accordance with Eq. (3.39), A" = (1.04 X =

2867 X 60115,000 X 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 mnst calculate the required thickness of the cone at the small junction. This information is needed because the cone thickness used so faris 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 X 2 X 5 X 12/[2 cos 28(16,000 X 1.0 - 0.6 X l00)J

= 0.43 in. From Eq. (3.40), the available area is A" = 0.78(60 X 0.50)<12 [(0.50 - 0.47)

=

+

(0.625 - 0.43)/C08 28J

1.07 in.2

The additional area needed at the small junction = 5.41 - 1.07 = 4.34 in.' Therefore, use a 4.5 in. X 1 in. bar rolled the hard way. Example 3.12 Problem Determine the required thickness of the cone, knuckle, fine, 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 X 60/(16,000 X 0.85 - 0.6 X 100) = 0.44

Use t

in.

= 1/2 in.

Large Shell From Eq. (2.1), t

=

100 X 7 X 12/(16,000 X 0.85 - 0.6 X 100)

= 0.62 in.

Use

t

= 5/8 in.


Rs == 5'-0"

•

'Ii'Thick 4" Radius

I

• I I I

I • I •

I

10" Radius

•

FIG. E3.12

Knuckle at Large End The thickness of the knuckle is obtaiued from Eq. (3.41). D; ~ 7

x 2 x 12 - 2 x 10(1 - cos 25)

= 166.13 in.

L ~ 166.13/(2 X cos 25) = 91.65 in.

M

~

(1/4)[3 + (91.65110)11']

=

1.51

83

84

Chapter 3 t

=

100 X 91.65 X 1.51/(2 X 16,000 X 0,85 - 0,2 X 100)

=

0.51 in,

Use t = 9/16 in. Cone

From Eq. (3.33), with D = D, = 166.13 in., t = 100 X 166.13/[2 cos 25(16,000 X 0.85 - 0.6 X 100)] =

0.67 in.

Use t = 11/16 in. Flue

From Eq. (3.45), the required thickness of the flue is K, = 0.125(2 X 4

K,

K6

K3

K"

s;

+ 120)'

tan 25 - 25-rr4'/360

=

951.51

=

0.28 X 120(120 X 0.50)'"

~

260.26

=

[120

=

66.62

=

0.78 X 66.62(66.62 X 0.6875)>12

~

351.67

=

0.78 X 0.6875(66.62 X 0.6875)>12

~

3.62

+ 2 X 4(1 - cos 25)]/2 cos 25

= 0.55 X 0.50(120 X 0.50)'" ~

2.13

Spherical Shells, Heads, and Transition Sections t

~

85

(l80/251T4)[100(95J.51 + 260.26 + 35J.67)/1.5 X 16.000 X 0.85 - 3.62 - 2.13J 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

Cone

Thickness Diameter Length

t of cylinder Do of cylinder

t, = (t of cone)(cos a) D L = outside large diameter of the cone L< = (L!2)(1 + D,/D r), where L is obtained from Fig. 3.6.

L of cylinder

(3.46)

0.47) (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 VlII-1. The required area is obtained from (3.49)

where all terms are the sarue as those in Eq, (3.37) and !l 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

fl.L

= 0.55(DLO'" (t, + tJ cos a)

(3.50)

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 An from the equation (3.5t)

TABLE 3.4 VALUES OF!l FOR JUNCTIONS AT THE LARGE CYLINDER DUE TO EXTERNAL PRESSURE P/S,E, A, deg.

a 0

0.002 5

0.005 7

0.010 10

0.02 15 0.15 40

P/S,E, 1:., deg.

0.04 21

0.08 29

0.10 33

0.125 37

P/ S,E, 1:., deg.

0.20 47

025 52

0.30 57

0.35 1 60

NOTE: (1) 1l = 60° for greater values of PISsE1

86 Chapter 3

where

As = area of the stiffening ring LL = effective length of the shell L, = effective length of the cone [L 2 + (RL - R,)2]1/2 L = axial length of the cone 2. Calculate the quantities M = (Rr - R!)/(3R, tan e) + L,/2 - (R, tan ,,)12

(3.52)

and F L = PM + (axial forces other than pressure)(tan a)

(3.53)

3. Calculate B from the equation B = O.75(F,D,/An)

(3.54)

4. Enter the appropriate EPC and determine an A value. 5. If B falls helow the left end of the temperature line, calculate A from the equation (3.55)

where Ex is the smaller of Eel En or £8" 6. Calculate the moment of inertia from one of the following equations: Is = ADrAn/14

(3.56)

or

I;

= ADi,AnI1O.9

(3.57)

where I, = required moment of inertia, Fig. 2.8(a), of the cross section of the ring ahout its neutral axis, in."

I;

= required moment of inertia, Fig. 2.8(h), of the cross section of the ring and effective

shell about their combined neutral axis, in." 7. The required momeut 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 I of Vlll-I. The required area is obtained from A" = (kQR,IS.,E,) tan "

(3.58)


87

The area calculated from Eq. (3.58) must be furnished at the juuction. Some of this area may be available at the junction as excess area. TIlls excess area can be calculated from the equation A" = 0.55(D,t,)Jl2[(t, - t)

+

(I, - t,) 1cos 0:]

(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 A" from the equation (3.60)

where A, = area of the stiffening ring L s = effective length of the shell L, = effective length of the cone = [L' + (RL - R,)'pt' L = axial length of the cone

2. Calculate the quantities N = (Rr - Rj)I(6R, tan 0:)

+

+

(R, tan 0:)12

(3.61)

+ (axial forces other than pressure)(tan 0,')

(3.62)

L,12

and F, = PN

3. Calculate B from tbe equation B = 0.75(F,D,IA,,)

(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 (3.64)

where Ex is the smaller of En En or E; 6. Calculate the moment of inertia from one of the following equations; (3.65)

or (3.66)

where I, = required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in."

88 Chapter 3

I;

= required moment of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about their combined neutral axis, in." 7. The required moment of inertia must be greater than the furuished one.

When the cone is fianged and flued, then the required thickness of the cone is determined as before, except that Eq, (3.48) is replaced by L, = r, sin ts:

+

(4/2)[(D,

L, = r,(D"IDL) sin a:

L, = [r,

+

+

+

(L,/2)[(D,

r,(D,,1D u)] sin rx

+

+

+

(3.68)

for sketch (d) in Fig. 3.6

DL)IDJ

(L,/2)[(D,

(3.67)

for sketch (c) in Fig. 3.6

D,)1D1"J

DJlDuJ

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 IS 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 temperatnre is lOO°F.

Allowable stress, psi Joint Efficiency Factor Modulus of elasticity, ksi External P, psi Effective L

SmaU Cylinder

Cone

Large Cylinder

Reinforcing Ring

15.000 0.85 27,000 15 10 ft

16,000 1.0 29,000 15

17,500 0.85 25,000 15 20 It

13,000

Solntion

Small Shell From Example 3.11, use I

=

1/2 in. Do

=

2(60

+

0.5)

=

121 in.

Then LIDo = 0.99 and Doll = 242. From Fig. 2.6, A 0.00038. From Fig. 2.4, B = 5500 psi. From Eq. (2.26), P = (4/3)(5500)1(242) = 30.3 psi> 15 psi

By trial and error, it can be shown that for I

=

0.40 in., P

=

15 psi.

30,000


Large Shell From Example 3.11, try

I

89

= 518 in. Do

=

2(84

+ 0.625) = 169.3 in.

Then LIDo = 1.42 and D,It = 271. From Fig. 2.6, A = 0.00021. From Fig. 2.4, B = 3050 psi. From Eq. (2.26), P = (413)(3050)1(271) = 15.0 psi

Cone From Example 3.11, use t = 518 in. Do = 2(84

From From From From

+ 0.625)

=

169.3 in.

Eq. (3.46), I, = 0.625 cos 28 = 0.552 in. Eq. (3.47), o; = 169.3 in. Fig. E3.11, L = (7 - 5)ltan 28 = 3.76 ft. Eq. (3.48), L, = (3.76

x 1212)(1 + 1211169.3) = 38.68 in.

Then L,ID L = 0.23 and DLI t, = 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)1(307)

By trial and error, it can be shown that for

I =

=

56.5 psi> 15 psi

0.33 in., P = 15 psi.

Large Cone-to-Shell function 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 X 25,000,000/(13,000 x 30,000,000)

=

1.12

The axial loads are given by Qr. = PR 1./2

~

axial equipment load

-15 X 84.625/2 - [50,000/(2'IT84.625)] -728.7 Ib/in.

90 Chapter 3

Next, we must calculate the need for reinforcement in accordance with Table 3.4. PlS,E, = 15/(17,500 X 0.85) =

0.001

From Table 3.4, Li = 2.5°. Hence, reinforcement is needed. The amount of reinforcement is calculated from Eq. (3.49); A,L = (1.12 X 728.7 X 84.625/17,500 X 0.85) {l -

(0.25)[15 X 84.625

+

728.7)/728.7](2.5/28)} tan 28

= 2.32 in.'

The available area in the shell and cone due to excess thickness is calculated from Eq. (3.50). A,L = 0.55(84.625 X 0.625)H2 (0.625

+

0.6251 cos 28)

= 5.33 in.'

Therefore, the junction is inherently reinforced and no additional rings are required. Next, determine the required moment of inertia at the largejunction needed for external pressure. From Eq. (3.51), and assuming a ring area of 1.0 in', An = (240)(0.625)/2

+ (3.67

X 12)(0.625)/2

+ 1.0

= 89.8 in.

From Eq. (3.52), M

= (84.625' =

- 60.52)/(3 X 84.625 tan 28)

+ (240/2) - (84.625 tan 28)/2

123.44

From Eq. (3.53), FL

=

15 X 123.44

=

1901.6

+

(50,000/27r84.625)(tan ex)

From Eq. (3.54), B

= 0.75(1901.6 =

X 84.625/89.8)

1344

From Eq. (3.55), A = 2 X 1344/25,000,000

= 1.08 X 10-'


91

From Eq. (3.57),

I;

~

1.08 x 10- 4 X 84.625' x 89.8/10.9

= 6.37 in.'

A trial run indicates that the availahle moment of inertia is inadequate without a stiffening ring. Assume that a 2 in. X 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 Xl

= [(5.64

x

+

0.625'/2)

(5.64

X

+ [2 X 0.5(-1.0)J/[(5.64 x = (1.10 = 0.59

+ 4.67 -

1.00)/(3.53

0.625

x

0.625)

+ (5.64 X

+ 3.53 +

2.65/2) 0.625)

+ (2 X

0.5)]

1.0)

in.

X,

= 2.65/2

X,

= 1

- 0.59

+ 0.59

~

0.74

in.

= 1.59 in.

The moments of inertia of the cone about its two major axes are h ~ 5.64' X 0.625/12

= 9.34 in.'

I; ~ 0.625' X 5.64/12 = 0.12 in.' I"i"j = 0.0

The moment of inertia of the cone around an axis y through its centroid is Iy

=

1-; sirf a + ly cos' a + 2l;y sin a cos a

= 9.34

sin' 28 + 0.12 cos' 28 + 0.0

= 2.15 in.'

The total moment of inertia of the composite section = (l of shell about its neutral axis) + (area of shell) (distance to composite section neutral axis)' + (l of cone about the y axis throngh its centroid) + (area of cone) (distance to composite section neutral axis)' + (l of the stiffener about its neutral axis) + (area of stiffener) (distance to composite section neutral axis).' I ~ 0.625.3 X 5.64112

+ (5.64 X

+ 2.15 + (5.64 X

0.625)(0.59 - 0.625/2)'

0.625)(0.74)'

+ 2.0' X 0.5112 + (2.0 X 0.5)(1.59)'

92 Chapter 3

y I

I

· · ·

O.55 ..m;:tS 5.64'

-J1-0.625·

5.64' 2' X

\.2' RING

I

5.64'

I I

• I

,•

O.55 JOL tc 5.64'

2.65'

,

(a)

y

(bl 1:1

'.55~ 4.76'

2.23"

,I

. I I

O.55 JOsts I 4.26'

.II , I

.

Os

4.,26" •

"I

II

r

XI (d)

(e)

FIG. E3.13

= 0.12 =

+ 0.27 + 2.15 + 1.93 + 0.33 + 2.53

7.33 in.'

Hence, use 2 in. X 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 X 27,000,0001(13,000 X 30,000,000)

=

1.04


93

The axial loads are equal to Q,

=

PR,/2 - axial equipment load -15 X 60.5/2

[50,000/(21160.5)]

- 585.3 Ib/in.

Reinforcement is required in accordance with Eq. (3.58). A,.,

585.3 X 60.5/(15,000 X 0.85)] tan 28

= [1.04 X

= 1.54 in.?

From Eq. (3.59), the available area is A,.,

= 0.55(2

X

60.5

X

0.5)"2[(0.5 - 004)

+

(0.625 - 0.33)/ c0828]

= 1.86 in.'

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), An = (120)(0.5)/2 + (3.76 X 12)(0.625)/2 + 0.0 =

44.1 in.

From Eq. (3.61), N

=

(84.6252 - 60.52)/(6 X 60.5 tan 28) + 120/2 - (60.5 tan 28)/2

= 62.06 in.

From Eq. (3.62), F, = 15 X 62.06

+ (50,000/21160.5)(tan 28)

= 1000.8 Ibs/in.

From Eq. (3.63), B

= 0.75(1000.8 =

1030 psi

X 60.5/44.1)

94 Chapter 3 From Eq. (3.64), A

=

2

=

7.63 X 10-'

1030/27,000,000

X

From Eq. (3.66),

I;

=

7.63 X 10- 5 X 60.5' X 44.1/10.9

=···1;13··in.4

The available moment of inertia is obtained from Fig. E3.l3(c) and (d). The neutral axis is at

x, = [(4.26 X 0.5'/2) + (4.76 X 0.625 X 2.23/2)]/[(4.26 X 0.5) + (4.76 X 0.625)]

X2

=

0.75 in.

=

0.365 in.

The moments of inertia of the cone about its two major axes are I, = 4.763 X 0.625112 = 5.62 in:

I;

= 0.625'

X

4.76/12

=

0.10 in.'

Ii; = 0.0

The moment of inertia of the cone around an axis y through its centroid is Iy = Ix sin' a + 1y cos' a = 5.62 sin' 28 =

I

= 0.53

+ 2I-;.y sin a

+ 0.10 cos' 28 + 0.0

1.32 in.'

X 4.26112 + (4.26 X 0.5)(0.75 - 0.5/2)'

+ 1.32 + (4.76 X 0.625)(0.365)'

= 0.04 + =

cos a

0.53 + 1.32 + 0.4

2.29 in.'

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


95

3.8 CONICAL SECTIONS, VIll-2 The required thickness of a conical shell in VIIl-2 suhjected to iuternal pressure is obtained from Eqs. (2.36) throngh (2.38). These equations, which are for cylindrical shells, are used with the radius R perpendicnlar to the surface of the cone, as shown in Fig. 3.8. VIIl-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 a when the PIS value is known. If the actnal angle a 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 actnal angle a is larger than that obtained 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.

I

•

I

• I

•

I

• I

•

R

I

•

I

FIG. 3.8

96 Chapter 3 0.02

I

,

I

, ..I I

NOTE: CurvelJOV8med bV maximumstress intensitY at surface (primarily due to axial bending 11"*1. limited to 3 S

=

0.012 0.010 0.008

./

0.006 Adoquote

0.005

PIS

~

/

.-

.... Not

I

,-

,

V

/V /\nc_Th~kn"

0.004

Permitted

~

-

i

-:

0.003

0.002

0.001

10

15

20

25

30

35

Maximum Angle, a' FIG. 3.9 INHERENT REINFORCEMENT FOR LARGE END OF CONE-TO-CYLINDER JUNCTION (ASME VIII-2)

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 mles and factors of safety in VlII-2 are identical to those in VIII-I.

Example 3.14 Problem Determine the required thickness of the cone, the two cylinders, and the area at the cone-to-cylinder junctions shown iu Fig. E3.1J. 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

Large Cyli nder

Reinforcing Material

15,000 100

16,000 100

17,500 100

13,000 100


97

2. 2

~

"'\

'"

-, "

2.0

1.•

\. NOTE: Curva gowrned by maximum ,tr." intenlity.t surface (primarilv due to .'liel bending ltre,,) limited to 3$

1\".30 .....

~

'\.

-. "\. -, ""-. "-, I"'" r-, -. <,.......

\.

'," at

__ T

~deg.

Q 1.6

'<;

1.2

i'-..

'" <;

~

1

'

317

2.~!

l- r

t,

I,

l~ f-----

<,

<,

..

,

<,

"

r-,

---*-A , ,, ./

'"<, -,

",-

---1 "

2.~\ - - \ ') cos

r-, """ <, ~" ~. r-, J".., r-.... " ...... ........

0:

>.-/,

./

1.0 0.001

0.0015

I, 1,

Q

....

10 .....

0.003

0.002

,

I

<,

<,

0.004

0.005

0.006

0.008

0.01

0.015

PIS 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 X 60.0/(15,000 - 0.5 X 100)

= 0040 in.

Use t = 7116 in. Cone

From Eq. (2.36), with R = 7 X 12/cos 28 = 95.14 in., the cone thickness is calcn1ated as t

Use t = 5/8 in.

"

=

100 X 95.14/(16,000 - 0.5 x 100)

=

0.60 in.

0.02

98 Chapter 3 0.10 0.08 0.06 0.04

./

Adequate\ 0.02

V

./

PIS

1/

0.010

\

In"eat'e

Thi'kne~

0.008 0.006

r

0.004

:/

I

0.002

NOTE; Curve governed bv membrane stress intensitY (due to average circumferential tension stress and average radial compression stress) limited by 1.1 Sm at 0.5 "radius X t either side of junction; where radius 'sIRS+~tl2l cn evt lnder side and ( R t 121f cos a 0., •cone side

s+..

I

I

I

0.001

o

2

6

4

Maximum Angle,

10

8

UO

FIG. 3.11 INHERENT REINFORCEMENT FOR SMALL END OF CONE·TO·CYLINDER JUNCTION (ASME VIII-2)

Large Shell Again, nsing Eq. (2.36), we get t = 100 ~

Use t

=

x

7

x

12/(17,500 - 0.5

x

100)

0.48 in.

1/2 in.

Large End of Cone-to-Shell function The PIS ratio is PIS = 100/17,500 = 0.0057

From Fig. 3.9, maximum u = 19°. Hence, reinforcement is required. From Fig. 3.10, Q required thickness at the large end of the cone-to-shell junction is t, = Q

x

t ratio of shell stress to reinforcement stress (must be equal to or greater than 1.0)

1.30 X 0.48 X 17,500113,000 = 0.84 in.

Use t

7/8 in., distributed as detailed in Fig. 3.10.

1.30. The


3.4~

NOTE: Curves governed by membrane stress intensitY ldue to average circumferential teneton stress end average radial compression stresst limited by 1.1 Sm at 0.5 J radius X tr either side of junction, where radius is (R s + tr / 2) on cylinder side and IRs + tr / 21 cos

I

f\

I

a on ccne stde.

99

-+-+-+-1

I I II

t

3.0 ~-+~-t---+-+-l.......jH-++t----.l-_~_.L--'--'--'-+ti I ',-0' a· .... 30i\: \ /' t2.6 [ \

25~

'\

$'

"\r-. ","'-" ~

- , 20d..: \ " ,

2.2

"'

Q _

15~

~

1.4

~

5d...

I"'"

)/'

.....

r-.....'

0.002

I

tt-

-,-rG\

'"

"."

I"--

~\:4Y.::'~ i

...

i'-............. t-... .... ~

14.r.;;-R 1

. v..,', :

~

L e-\ IS: '"-"\ ',-

_L_''pf4-~ \ t

,

_

-"I,I---~~--l

t--

0.003

0.008 0.01

0.005

0.02

0.03

0.05

0.07

0.1

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

Small End of Cone-to-Shell Junction Here the PIS ratio is PIS

~

100115,000

~

0.0067

From Fig. 3.11, maximum (Y = 1.5'. Hence, reinforcement is required. From Fig. 3.12, Q required thickness at the small end of the cone-to-shell junction is t,

x

=

2.15

=

0.99 in.

0.40

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

x

15,000/13,000

2.15. The

CHAPTER

4 FLAT PLATES, COVERS,

ANI> FLANGE.s 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 ohround, 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 reqnire 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-I 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-I and in AD-530 of VIIl-2. Also, design requirements are given in Nonmandatory Appendix AA of VIIl-I for the design of tubesheets with multiple openings. For the design of flat plates and covers which are attached hy bolting that causes an edge moment due to the gasket and bolt loading actiou, 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 VIIl-I and VIlI-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 VITI-I.

4.2

INTEGRAL FLAT PLATES AND COVERS

Since the design rules of circular flat plates and covers is the same in VITI-I and VIII-2, only the references in VITI-I will be nsed. Design rules for noncircular flat plates and covers are given only in VIII-I.

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)'12 101

(4.1)

102 Chapter 4

where d = effective diameter of the fiat plate, in. C = coefficient between 0.10 and 0.33, depending on comer details P = design pressnre, psi S = allowable stress at design temperature, psi E = butt-welded joint efficiency of the joint within the fiat plate t = minimum required thickness of the fiat plate, in. For further description of the terms given above, refer to Fig. 4.1 (a) through (i), or Fig. UG-34 of VIII-I. Depending on the shape and welding details of the comer, a value of C is selected. A value for E, the buttweld joint efficiency within the fiat 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 ofE depends on the degree of NDE' performed. It is not a weld efficiency of the head-to-shell comer joint! Example 4.1 Problem

Using the rules of UG-34 of VIII-I, determine the minimum required thickness of an integral fiat 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 comer details conform to Fig. 4.1 sketch (b-2) assuming tbat m = I. Solution

From Fig. 4.1, sketch (b-2), C

= 0.33

d

= 24,

d,

=

24

X m

=

0.33(1)

+ (2)(c.a.)

= 0.33

= 24.375 in.

From Eq. (4.1), t, = (24.375)[(0.33)(2500)/(17,500)(1)]'''

= 5.29 in.

t, = to + 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, t; must be calculated using Eq. (2.1). t, = PRJ(SE - 0.6P)

=

(2500)(12.1875)/(17,500 x 1 - 0.6 x 2500)

lNDE is the nondestructive examination of the butt joint.

=

1.904 in.

Flat Plates, Covers, and Flanges

tttt

;... Center of weld

,,..y~

Lttt\l ~ t

ts

time

r .. 3t

'!'

Taper

min.

I

d

I

..L-[J'.'"

t,mm -2t$

Tangent

I

.

C" 0.17

C-O.l10r C" 0.10

"1ft: 1:'T .....y!-

Tengen, ..., :,hne

',"3'

d ' m i n ." 0.25 t$ for t 1 1/2·

mm.

l!

-

~

r .. 3t t

.

In,

$ /.

d

but need not be greater than 3/4 in.

C"O.33m C min ... 0.20

0"

d

~

t

,

.

~

t

d

0.71,

J

r,

leI

min. nor less than 1.25 t$

f,

.l

T

..~ Continuation of shell optional

d

,hen' Projection ,,"yoM weld

l'"

d

t t o o

o

C .. 0.13

'.

'r

'M oeee no, ........"

a7t .';.<

LA

rmn.

C .. 0.30

~~I

t w <02

,e"., mm,

_.

C"O.20orO.13

·'" "ITi fiff fi m 1~11

leI

Cemer of lap

I

0 375 in min·' forts" 1·1/2 in,

T

t(

d

I too-

~ J

103

t

~s optional.

Bevel optional

- 45 deg. max.

Sketches leI. ttl, and (91 circular covers. C'" O.33m, C min. '" 0.20

Id}

Ie)

IfI

See Fig. UW·13.2 sketches Ie} to {gl. inclusive. for details of welded joint t s not lets than 1.25t,

---j

.L

"rITl d

•

See Fig. UW·13.2 sketches la) to (91. inclusive, for detail of outside welded joint

~ dJ h

~

O'.l.rnt· f"'

f:J ,

f

d

t

I

0 o

C = 0.33

0

C"0.33m Cmin." 0.20

Ih}

_.rmt

1.1

.Reteiniog ,in.

Im d

_

C"O.3D

c = 0.3

C- 0.3

:

(Use Eq, (21 or (511

I Use Eq. (21 Or (511

Ii)

Ik)

T h...... ,in.

I '

C" 0.30

1m}

C" 0.30 lo}

Inl

Seal weld

30 deg. mm.

rrs. -+---1- t+I-

.S.

fZl~~~.~1:

C" 0.75

C" 0.25

Ipl

NOTE. When pipe threads are used, set Table 00-43

1.1

C - 0.33

1<1

deg.max'~3/4tmin.

mu\.t1"rort$ wtllchever is greater

:. .

1

, -oT d

.

0.8ts min,

r.s

C" 0.33

lal

FiG. 4.1 SOME ACCEPTABLE TYPES OF UNSTAYED FLAT HEADS AND COVERS

G

104

Chapter 4

For t > 1 \12 in., rmi. ~ 0.25/,

= 0.25(1.904)

~

0.476 in.

4.2.2 Noncircular Flat Plates and Covers When the fiat plate or cover is sqnare, rectangular, elliptical, obround, or any shape other than circular, the

minimum required thickness is calculated in the samemanner as for a circular plate,except for the addition 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

~

(4.2)

3,4 - (2Ad/D) '" 2.5

Using this value of Z, the minimum required thickness of an integrally attached noncircular fiat plate is calculated from the following equation: (4.3)

/ = d(ZCPlSF/"

where all terms are the sarue as in Eq. 4.1.

Example 4.3 Problem Using the rules in UG-34 of VIII-I, determine the minimum required thickness of the fiat end plate of a rectangular box header which is 8 in. X 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 (I) 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(8116)J = 2.2

(3) Using Eq. (4.3), t

~

8[(2.2)(0.33)(350)/(15,000)(1)]'''

~

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.


105

Solution (I) Rearrange Eq. (4.3), above, to solve for Z as follows: Z

= SEt'ICPd' =

(15,000)(1)(0.5)' 1(0.33)(100)(8)'

= 1.78

(2) Rearrange Eq. (4.2), above, to solve for D as follows: D = 2.4dl(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.

4.3 BOLTED FLAT PLATES, COVERS, AND FLANGES Bolted connections are used on pressure vessels because theypermit easy disassembly of components. The bolted connection may consist of a flat plate or so-cal1ed blind flange, a loose-type flange, or an integraltype 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). These methods were 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-I 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-I.

4.3.1 Gasket Requirements, Bolt Sizing, and Bolt Loadings Appendix 2 of VIII-l 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 f1ate plate or blind flange, loose-type flange, and integral-type flange. Loadings arc 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-I. The vaiues of m and y in the table are nonmandatory, and different values of m and y, wbich 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 fol1owing procedure:

1. The basic gasket seating width, b.; is selected from Table 2-5.2 of VIII-I. 2. When b., :5 1/4 in., b = bo, and when b., > 1/4 in., b = O.5(bo) 1I2 From b, the values of G and he can be determined.

106

Chapter 4

4.3.1.2 Bolt Sizing and Bolt Loadings. The required bolt load for operating condition, Wm1, is determined as follows: Wm1

= H + H, = O.785G'P +

(2b x 3.14GmP)

(4.4)

The reqnired bolt load for gasket seating condition, Wm" is determined as follows: W", = 3.14bGy

(4.5)

Once the required bolt loads are determined, the required bolt area for each loading coudition can be calculated as follows: (4.6)

and (4.7)

The total required cross-sectional area of the bolts, Am' is the greater of A m1 or A m2 • The bolts are selected so that A b , the actual bolt area, is equal to or greater than Am. The bolt load used for the design of flanges, W, is theu determined from the following. For operating condition, (4.8)

For gasket seating condition, W = O.5(A m

+ Ab)S,

(4.9)

wbere S, 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: (4.10)

4.3.1.3 Check for Gasket Crushout. Although not considered in VIII-I, it is prndent to design against crushout of the gasket by determining the minimum gasket width using the following: (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-I and Table AG-150.1 of VIII-Z. Staudard flanges are acceptable without further calculations for diameters aud pressure!


107

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 tbe circular flat plate is calculated by using the following equation:

+

, = d[(CP/SE)

(1.9WhGISEd')]112

(4.12)

where the definitions of terms are as given in 4.2.1 and the determination of Wand h G 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 minimnm required thickness of the noncircular flat head is calculated by using the following equation: t = d[(ZCP/SE)

+

(6Wh G/ SELd 2)Jll2

(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-I. Single, small openings which do not exceed the size limits UG-36(c)(3)(a) & (b) and are not 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: A,

~

O.5dt,

(4.14)

where

AT d

= required area of reinforcement, in.'

= 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 averagediameter, Eq. 4.14 for the minimum required thickness of a plate with a single opening may be used. If the spacing between pnirs of openings is less than twice the average diameter, but not less than 1V4 the average diameter, the amount of reinforcement is calculated by

108 Chapter4 adding the required reinforcement of the pair to no less than 50% of the required reinforcement hetween the pair. If the spacing is less than I V4 the average diameter, U-2(g) applies. In all cases, the width of the ligament hetween 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-I, as an alternative to the rules for reinforcement of a single opening, the following procedure may be used to determine the minimum required 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-I), (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)1n

(4.15) (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 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-I. 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 required 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 v'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 v'2 rule.

4.6 BOLTED FLANGE CONNECTIONS WITH RING TYPE GASKETS As described earlier in section 4.3, Appendix 2 of VIIl-I 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

Flat Plates, Covers. and Flanges 109

u" l~

p • spaclhll. ceMel{O-¢IIhter. be\Wfth Ope
IIIf.,

FIG. 4.2 MULTIPLE OPENINGS IN THE RIM OF A FLAT HEAD OR COVER WITH A LARGE CENTRAL OPENING

of the gaskets and determination of the bolt sizes, bolt loading, and loading moment arms are obtained in the sarne 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 sarne 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 ringand-shell or ring-and-pipe 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.

110 Chapter 4 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 fillin 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 Vlll-l, determine the minimum required thickness of a ring flange, shown in Fig, E4.5, with the following desigu 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. l.D. X 1.0 in. wide; No corrosion allowance.

1

2.000 1'5L

'SA -105' 13.15"1.0." ,"wid,c,. SA-32S' Gr-i Gc l>,i'+(ZX'')-(2J<.~~ l5',d

flong_ Mat.rlol &oiling Material

No Cer re ~j"o .... 4

COi'tosion AUowanee

&011;"'0

-

1"1,100 :ZOO, >'00

A.

O,d,," Temp., Sit

2(),100rf<

N

.:.. (;%.,..'/4 _

W

20,106 rs-'

3",500

Atm. Temp., S.

w.., _

CONDITION

5

LOAD

-b'll"Gy

HI'

+H=

HO=W.I-H_

200,5'00

he _ .5fC- Gl

Hr=H-Hll=

100, '00

HG=W

=

liT - .5( hD

':;102,5'00

.)

+ fIG}

1,,_ = .51C- Ol

o ,"f-c j2A. cZ2.S'~ 26,5" ho B'

6

S:~81 I H-~ 1-",. _~Hr 20@ \ ~ ;8.'t! I "

I

12J5"

G'. 1>'0'1)

21.5

'/:>,'

MOMENT

~

=

4.815 ?>,12'l

Me

= Haho

'i,,0l-

Mr

=

Mo

=

~.12q

SHAPE CONSTANTS

H,nr

M,

K = AlB

+

1,1'.1.\5,000

Hohll ~

~

141,100 'l?>1.2.00 2, L{ 'l'i, 000

2,OQ'O,000

• 2.018

r= 2,812-

If bolt ,pacinG ucuch 20 I, multiply ~ 8011 IPOein; M. end ~ in I equotionl by: 2a I

"H

~:

no;

:;o:r w..,/S. or w..l/S~_ 8,3 or = 2S,2 ZO~ \ q> = .5("' .. + A.15. _ ><'2,,"00

M. S.OI'''';

15',OQ3 \0,000 3,0

A.. _

LEVER ARM

X

hD _ .51C

7:11' / 4

G y m

'5~;'G, 0 00

2S'5, '100

H.

Operating

l

- 2b...G mf'_

11,&DOr" w.., J

1.0

LOAD AND BOLT CALCULATIONS

H,

A'm. Temp..

N

•

Ro.i sed. f ..
S,. 20 oaD " .i.

OeNonTemp.. S,., Flange

3

fACE

GASKET

Sf'ira.\ Wo... Kd. "lIe"'..I, .j:; ~.... f; \ \ed.,st.;,,,le...~(

65'0 OF

De$i;n Temp,rat"".

il I-

2

DESIGN CONDmONS

0.';0" Prellur•• ,.

7

+

OPERATING

.

~ ~='fi,l{

.=

t=

OF

SEATING

GREATER

.,

,=~

M.r ~~ gj

SlaB

•

•

•

COftIputed_ _ _ Dote

Chechd

FIG, E4.5 RING FLANGE SAMPLE CALCULATION SHEET

Number

)


111

Solution (1) The allowable tensile stress of the bolts from II-D at gasket seating and operating conditions (design temperature) is S, = S, = 20.2 ksi. (2) The allowable tensile stress of the flange from II-D at gasket seating is Sf' = 20.0 ksi and operating conditions is Sf' = 17.8 ksi. (3) The diameter of the gasket's line-of-action, G, is determined as follows: bo

= NI2 =

0.5 in.

G = 13.75

(4) With N = I, b of bolts are

=

0.3535, m

=

+

and

= 0.5(bO) 1I2 = 0.3535

(2 X I) - (2 X 0.3535)

=

3.0, and y

H

b

=

in.

15.043 in.

10,000, the bolt loadings and the number and diameter

= (7f/4)G'p = (7f14)(15.043)'(2000)

Hp

=

355,5001b

=

2b'ITGmp

=

2(0.3535)7f(15.043)(3.0)(2000)

= 200,500

lb

= 355,500

+ 200,500

=

556,000 Ib

Wm2 = 7fbGy

=

7f(0.3535)(15.043)(10,000)

=

167,1001b

or

Wm2/S, = 556,000/20.2 = 27.5 in.'

Using 20 bolts of I V2-in. diameter, Ab = actual bolt area = 20(1.41) = 28.2 in.2

= 0.5(27.5

+ 28.2)(20,200)

112 Chapter 4

Wo

=

562,6001b

=

Wm1

= 556,000

lb

(5) Using Eg. (4.10), the gasket crushout width is

=

(28.2)(20,200)/2(10,000)(7T)(l5.043)

= 0.6 in. < 1.0 in. actual

(6) The total flange moment for gasket seating condition is:

Flange Load

=

HG

Wo

=

562,600 lb

Lever Arm hG = 0.5(C - G)

=

3.729 in.

Flange Moment

= (562,600)(3.729) = 2,098,000 in-Ib

(7) The total flange moment for operating condition is: Flange Loads HD

(7T/4)B'p

=

= (7T/4)(l2.75)'(2000)

= HG

255,4001b

= H; =

200,500 lb

Flat Plates, Covers, and Flanges H; = H - H p = 355,500 - 255,400 = 100,100 Ib

Lever Arms hD

=

0,5(C - B)

= 0.5(22.5 hG = 0.5(C -

- 12.75)

h, = O.5(h D

=

4.875 in.

G)

0.5(22.5 - 15.043)

=

=

+

= 3.729 in.

hG)

0.5(4.875 + 3.729)

=

4.302 in.

Flange Moments

= (255,400)(4.875) = 1,245,000 in.-lb

= (200,500)(3.729) = 747,700 in.-lb

= (100,100)(4.308) = 431,200 in.-lb

=

2,424,000 in.-lb

(8) Shape factor from Appendix 2 of VIII- I for K

=

AlB

=

26.5/12.75

=

2.078

From Fig. 2-7,1 of VIII-I, Y = 2,812, (9) The minimum required thickness of the flange is the larger tmi" of: For gasket seating condition:

'mi"

=

[(MGsY)/(Sj)l)]in

=

[(2,098,000)(2.812)/(20,000)(12,75)] in

=

4,81 in.

113

114 Chapter 4

For operating condition:

[(2,424,000)(2.812)/(17,800)(12.75)]1n ~

5.48 in.

Example 4.6 Problem Using the rules of Appendix 2 of VIII-l, determine the minimum required thickness of'aweldingneck flange, of the type shown in Fig. E4.6 with the following design data: Design pressure = 2000 psi;

Design temperature

=

650oP;

Flange material is SA-lOS; Bolting material is SA-325 Gr. I; Gasket is spiral-wound, fiber-filled, stainless steel, 13.75 in. J.D. X 1.0 in. wide; No corrosion allowance. Note: This flaoge has facing details, gasket size, and bolting that are the same as those given in Example 4.S, except that this is a welding neck flange instead of a ring flange.

Solution (I) The allowable tensile stress of the bolts from IT-D at gasket seating and operating conditions (design temperature) is S, = S, = 20.2 ksi. (2) The allowable tensile stress of the flange from IT-D at gasket seating is Sf' = 20.0 ksi and operating conditions is Sf' = 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. M GS

~

2,098,000 in.-lb

(5) The total flange moment for the operating condition is:

Flange Loads H D ~ (-rr/4)B'p

= (-rr/4)(10.75)'(2000)

He

~

181,5001b

~

H, =

200500lb

H, = H - H o = 355,500 -

181,500 = 174,000Ib


115

FACE

1.0 b G

'."i", M.,.,i., CorrOI;on AUO"'<>tlc~

SA-32'S Gr.1 6- e, 1:'.'5 "2-.101<\!>.O~'" N Q Col"yo'$ ioV\. 4 LOAD AND 10LT CALCULATIONS

CONDITION

5

LOAD

H.

Operating

I4

'l!"Il I P

He -W.. l-ti

-

Hr =H - tlo =

LEVER ARM

X

.. I~l, 70 :!-OO

0

IILj,o<:>O

3.0

MOMENT

ho _ 11+ .5g\

4. II!>

hG -

'!>.l'2.

hr

.5lC -Gl

= ..5ll! +

I!'

+ h(;1

=

Mo

'"0, \ 1'1 ,100

Hohp

00

4. 02

2,343,600

M. Hc=W

he;

= .51C -

Gl

3.12'1 K AND HUB fACTORS

STRESS CALCULATION-Operating

«

S,. S,.

Tong. fIg., ST

'i '{QO \4 4~O

_m.Y/r' -2S. -

:;·.~'·.51SH + S~)or.51SI!

+ Sr)

-

STRESS CALCULATION

1.5 51. 51.

S,. S,.

long. !i
fm(l/Ag,1l

Radial fIg., Sa -

{3ma/Afl

2·~"5

A/,

f

V

$;"

z

\."q

r~j.,

y

2."1..'1

sr/s, h.

...}89.

7

Tang. fIg" ST -11\0 V!I'-ZSH

F/h~

"!>."l>1S 3-'11'\

d

Vhog.1

u

0.114

207.7l:>

11.0

+1

{3 = 4/3 te "t

'>4'i'1I--"-1+--B' 10.1 5 "

1.0 e _

STRESS FORMULA FACTORS

a _Ie

9,· 3.,,15"

0.04

2.',I

u

$ealing

t. '\0(" 0.91

hl"~

1.315

+I \·25,,",

_a/T

rno _ Mo/S _ operollng_ mo ::::: MQ/E

= ,eating =

%IB 000

+

If bolt .padng flxceedt 20 f, multiply mo end flIQ in above equaliont by:

..

Bolt ~pgc;ing 2" f

+

~

GwW Taylor·Bonney DiViSion.

"4.0"

LI.--.!-t--'--~,..-'

" Compvted'

001.,,

Ch.,ckeO

Number

FIG. E4.6 WELDING NECK FLANGE SAMPLE CALCULATION SHEET

_

116 Chapter 4

Lever Anns hD

=R +

0.58,

= 2.5 +

= 4.188 in.

0.5(3.375)

ho = 0.5(C - 0) = 0.5(22.5 -

+ 8, +

hr = O.5(R

15.043) = 3.729 in.

hG)

+ 3.375 + 3.7285)

= 0.5(2.5

~

4.802 in.

Flange Moments

Mo

=

(181,500)(4.188)

=

Ho

~

(200,500)(3.729) = 747,700 in-Ib

~

(174,000)(4.802) = 835,500 in-lb

X

~

760,100 in-lb

ho

Mo = MD + Mo

+ MT

= 2,343,000 in-Ib

(6) Shape factors from Appendix 2 of VIII-I for K are K

~

AlB

~

26.5/10.75

~

2.465

From Fig. 2-7.1 of VIII-I, T

=

1.35 Z

8,/80

~

~

1.39

y

~

3.375/1.0

2.29

=

U

3.375

= [(1O.75)(1.0)J'" = 3.279

hlhD = 6.25/3.279 = 1.906

~

2.51


117

From Appendix 2 of VIII-I,

F

= 0.57

e

= 0.04 f =

V

1.0

= F/h o

= 0.57/3.279 = 0.174

d = (UlV)hog~ = (2.51/0.04)(3.279)(1)2 = 205.76

= 2,098,000 in.-Ib and Sf' = 20.0 ksi and M, = 2,343,000 in.-Ib and Sf' = 17.8 ksi. Since at gasket seating condition, the moment is smaller and tbe allowable stress is larger, only the operating

(7) M GS

condition is calculated.

Assume a flange thickness of t = 4.0 in. L = [(Ie =

+

1.256

1)/(7)

+

+ (I)'/a]

0.311

=

1.567

Longitudinal hub stress:

= (1)(2,343,200)/ (1.567)(3.375)'(10.75)

=

12,210 psi

Radial flange stress: SR = [(4/3)le

+

1]M,ILt 2B

=

(1.928)(2,343,200)/ (1.567)(4)'(10. 75)

=

16,760 psi

Tangential flange stress: Sr

=

[(YM,/ ,'B) - ZSR]

= {[(2.29)(2,343,200)/(4)'(l0.75)] - (1.39)(l6,760)} =

7900 psi

I

I

118

Chapter 4 Combined stresses: O.5(SH

0.5(SH

+

+

SR) =

S,) =

0.5(12,210 + 16,760) = 14,490 psi

0.5(12,210 + 7900) = 10,060 psi

(8) Allowable stresses: SH ,; 1.5Sf:

12,210 psi < 26,700 psi

S, ,; Sf:

16,760 psi < 17,800 psi

S,'; Sf:

7900 psi < 17,800 psi

0.5(SH

+

SR) '" Sf:

14,490 psi < 17,800 psi

O.5(SH

+

S,) ,; 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.

4.6.2 Special Flanges Rules for special flanges with different geometry and/or loading are given in Appendix 2 of VITI-I. Included are: 2-9 for split loose flanges, 2-10 for noncircular shaped flanges with a circular hore, 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 VITI-I. They are similar to standard flanges, except some of the loads on the flange ring cross section may he applied at different locations and in a reverse direction, possibly causing a reverse moment. Vlll-I has chosen to use the term aD 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 DA (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-I. For both analyses, a special limitation of the geometry is given. When K :5 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 :5 2. Example 4.7 Problem Using the rules of Appendix 2 of VIIl-l, determine the minimum required thickness of a reverse welding neck flange, shown in Fig. E4.7, with the following design data:


1

G50" f
""'''' T......."'"

..",.,"'••••,., Conoliof\AllowaN:.

flong.

GASKET

fACE

SriV'Q.\ w."w.\. \.oeo1e;1:...\,

2 1 0 0 0 YIA

"_g. Mat.rial

•

2

DESIGN CONDmONs

o.MllI n Pr••..,r •• ,.

tiM.f;lI...!., st.'.W!SHt"" I, \;'15" I .P ?t \ " w'.o\e. G, \?1S"+Z-.1d1, 1<;.O~3U

SI\·~25 ur.1 No Cor'(0i'U:I1'''L 4

o.lign Temp., 5,. \1.e-OD,~

Rll.iSE.d..f~

b G


W..1

20000 "'.

119

\00

brGy

'~1

2b.c.' -

1.00, SOD

:;.~. W.2/S.

.A.

~.'3 or Z1.5 20@ 1Y-z" 4>

or W.l/S. _

.. - 28.:L

i11---+-:------:=;;:,-'~~;=.-r--="=-----;;'~~~-I-"'--'="-:------;;~;:-T*-'----1 OelignTemp.,S. '1o,'ZOO"IA H _GlTPI4 ~S5'. . '500 w -.51A.+~ls.- '502>bOO A.... T...... S..

:I

Bolting

H.

-

r.::.::c::-:::::c::=~"-;;:?'5+C-:--=-::..:.:;.c.:.-=--_;:;;'zT~_SE_+"-'::..::=-'..;""''-=-=~''-''='----AIm. Temp••S.

CONDITION

'2t> 200PfA

W.I - H,

LOAD

+ H_

X

5'"5'fvJooO

=

LEVER ARM

MOMENT

5 H,_W••-H= H,_H

H.=

1.00,500 -llbb,f.jOO

'._.S(C-G) '5.12.'1 h,-.'(C-'t G)_ \.1'1\

1'!1100

""_H,h, M,_H,,",

-SU"OOO

-

Seating STRESS

CALCULATION-Dperat~·ng

6

K AND HUB FACTORS

1.5 5,.

S,.

9 1.5

00

h/h. F

=

v

s,

-

I

-

S,.

•

Tong.F1Q.. ST- moYaN -ZSIl.IO,67te+lj/.B=

Z

\.C.1

-

,)

.....

S,. S,.

RadiaIFIIlI·.S.

Pmcnt·

Tal'lQ. Fig., Sot

may.N

Tang. fIg .. St {AT 8'l

=7

?',1 eo ffJ.. L\,.'-\O

Z5R(O.67Ie+l)/,B

[V

f'f;'"

~~:(lt~·te)J =- \1,0~Of#.

- \·1,41 1.3Mo 91/go

1.0

•• - v;;:g;; ".9~O STRESS FORMULA 7

,

a. .=-te + 1 .8=.4/3t_+ 1

r ;;:; «ITt! 6 _,'/d

l_Y+5 m.

h:

0

L[>

hr+Hr

1.~I'2.S"--.gof4--': 2.2..~"S'

20- I ~'~8""

::

A"

2(,.5"

\.0

O.I.\\q 0.851

All....l>l•

S,. S,. S,.

0.<;<;0 O.I~I

Fill•

]

MO'l

_M./8'

d

u~ h I V 090

f6.~~3

FACTORS

- 4.0 = I. !>2'! = I. ~"'1 = 1.11$

-

1.1:11

2.'10"-

-

~ ~ Comput.d

Check.d

_

Oal.

Number

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

_

120 Chapter 4

Design pressnre = 2000 psi; Design temperature = 650 o P; Flange material is SA-I05; Bolting material is SA-325 Gr. I; Gasket is spiral-wound, fiber-filled, stainless steel, 13.75 in. LD. X 1.0 in. wide; No corrosion allowance. Note: This flange bas 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 (I) The allowable tensile stress of the bolts from lI.D at the gasket seating and operating conditions (design temperature) is S, = S» = 20.2 ksi. (2) The allowable tensile stress of the flange from lI-D at the gasket seating is Sf' = 20.0 ksi and operating conditions is Sft> = 17.8 ksi. (3) The diameter of the gasket line-of-action, bolt loadings, bolt number and diameter, and the crushont width are the same as in Steps 3-5 of Example 4.5. (4) The total flange moment for the gasket seating coudition is the same as in Step 6 of Example 4.5. MGS

=

2,098,000 in.-Ib

(5) The total flange moment for operating condition is:

Flange Loads HD

=

('IT/4)B'p

=

('IT/4)(22.875)'(2,000)

= H,

=

~

~

556,000 - 355,500

821,900 Ib

200,500 Ib

H - Hn

= 355,500 - 821,900

~

-466,400Ib

Lever Arms ho = ~

=

0.5(C + g, 0.5(22.5

+

2go - B)

1.8125 - 2 X 1.8125 - 22.875)

-1.094 in.

he = 0.5(C - G) ~ 0.5(22.5 -

= 3.729 in.

15.()43)


+

hr = 0.5[C - 0.5(B

G)]

= 0.5[22.5 - 0.5(22.875 =

+ 15.043)]

1.771 in.

Flange Moments

=

(821,900)( -1.094) = - 899,200 in-Ib

=

=

(200,500)(3.729) = 747,700 in.-lb

(-466,400)(1.771)

=

-826,000 in-lb

- 977,500 in.-lb

Use the absolute value in the calculations. (6) Shape factors from Appendix 2 of VIII-l for K are K

=

AlB' = 26.5/13.25 = 2.0

From Fig. 2-7.1 of VIII-I; assuming fL = 0.3: T

=

1.51

Z

=

a, = (11K'){l

Y

1.67

+

[3(K

= 2.96

+

+

=

0.857

=

1.241

fL)(Z -

=

3.26

1)(1 - fL)]/1TY)

= 0.419 T, = [(Z

U

fL)] a,T

121

122

Chapter4

~

1.366

= [(26,5)(1,8125)]'" ~ 6,930

From Appendix 2 of VIII-I: F

~

0,909

f

V = 0550

= 1,0

e = Fl h, = 0,909/6,930

d

~

~

0,131

(U,IV)h,g;

= 56543

= 2,098,000 in-Ib and Sf' = 20,0 ksi and M, = 977,500 in-Ib and Sf" = ]7,8 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,

(7) M GS

L ~ «(Ie

= 1,778

+

+

1)/T,]

+

(l"ld))

Ll31 = 2,909

Longitudinal hub stress:

= [(1)(2,098,000)] I [2,909)(1,8125)'(13,25)J ~

16,570 psi

Flat Plates, Covers, and Flanges 123

Radial flange stress: SR ~ ([(4/3)te

+

I]MGsl/Lt'B'

~

[(1.699)(2,098,000)] I [(2.909)(4)'(13.25)]

~

5780 psi

Tangential flange stress: 5,

~ [(YRMGs/t'B') -

2S,(0.67te

+

I)]/~

~

{[(1.24I)(2,098,000)1 (4)'(13.25)] - [(1.67)(5,780)(1.351)] 11(1.699)

~

4610 psi

Combined stresses: 0.5(SH

+

S,)

O.5(SH

+

S,) ~ 0.5(16,570

~

0.5(16,570

+

5,780)

+ 4610)

~

1l,180 psi

= 10,590 psi

Tangential flange stress at B':

S; ~

(M Gslt 'B')[Y - [2K '(0.67te

+

1)/(K' -

I)L])

[(2,098,000)1 (4)'(13.25)]{2.96 - [2(2)'(1.351)1 (3)(2.909)]) 17,040 psi

(8) Allowable stresses: SH'" 1.5Sf

:

16,570 psi

<

26,700 psi

S, '" Sf:

5780 psi

< 17,800 psi

S, '" Sf:

46JO psi

<

17,800 psi

0.5(SH

+

S,) '" SJ'

ll,180 psi

< J7,800 psi

0.5(SH

+

S,) '" Sf:

10,590 psi

< 17,800 psi

S; '" 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 perntits 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).

124 Chapter4 One of the basic differences with a full-face gasket is that a reverse moment is generated from tbat part of the gasket loading outside the bolt circle. Most often, a full-face gasket is used where the m aud 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 aud from operation to be fairly low, and consequently, only a nominal flauge 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-I 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 flauges. Gasket loadings usually are small, as most gaskets are of the self-sealing type. Inorder 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-I.

Classification of Assemblies Class I: Class 2: Class 3:

A pair of flanges which are identical except for the gasket groove A pair of nonidentical flauges in which the inside diameter of the reducing flange exceeds half the bolt circle diameter A flange combined with a flat head or a reducing flauge with au inside diameter that is small and does not exceed half the bolt circle diameter

Categories of Flanges Category I: Category 2: Category 3:

An integral flange or an optional flange calculated as an integral flange A loose-type flange with a hub that is considered to add strength A loose-type flauge with or without a hub-or au optional type calculated as a loose where no credit is taken for the hub

type~

The analysis of au 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, He, aud its moment arm, he, 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 WmI

=H +

He + He

(4.17)

4.7 SPHERICALLY DISHED COVERS Rules are given in Appendix 1-6 of VIII-I 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-l are approximate and may be conservative because they do not take into account the discontinuity condition which exists at the intersectiou of the ring and head aud that would distribute forces aud moments between the two parts relative to theirresistauce (stiffness). The Code procedures and equations assume the entire loadings at the intersection are taken by the bolting ring alone. The rules of VIII-I permit the use of a discontinuity analysis, should the designer so choose.


125

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. P = internal design pressure or MAWP, psi S = maximum allowable tensile stress value, psi T = flange thickness, in. M, = total moment applied to the ring, in.-Ib 13, = angle formed by the tangent to the center line of the dished cover thickness at its point of intersection with the flange ring 13,=carc sin [B/(2L + tl] A = outside diameter of the flange, iu. 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.

Not less Thin 2r and in No

CaI8 Less Th.n

1/2 in.

~,;:::;:=r=;

lL

L

Knuckle \ Radiu.

[U+","".."Jf'\

Gasket Loooo Fie... Typo

-r

Ring'

Integr.' F'• • Type

Ge.ket Shown

te)

r-".

I'

----1/2 A

f-----l/2A

!•PreferablY

T> t

1/2 C

1 2tmin.

I 1/2 c--+-l L

\

1/28--+-.i

(e)

------+t

1'~---1/41A + 8/--01 Point of H 0 Full Penetration Weld • Action t

T

~~i HD L

Hr

T

1

\ Shown as Welded. Smooth Weld Both

Centroid

~-\._-1

Ring G_ket Shown

O.7tmin.

Ibl

Skle.

~1/28

Use Any Suitable f----l/2 C----\--'; Type of Gnket

tdl

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

126 Chapter4 4.7.2.1

Ring and Head of Uniform Thickness.

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

(al Head thickness is

= (5PL)/(6S)

t

(4.18)

(b) Flange ring thickness using a ring gasket is T = ([M,/SB]/(A

+ B)/(A - B)]}'"

(4.19)

(e) Fiange ring thickness using a full face gasket is

+

T = 0.6([PlS]/[B(A

4.7.2.2

B)(C - B)/(A - B)]}'/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) Fiange ring thickness using a ring gasket and round bolt holes is T

=

+

Q

([1.875M,(C

+

B)]/[SB(7C - 5B)]}'/2

(4.22)

+

B)/(7C - 5B)]

(4.23)

where Q

=

(PL/4S)[(C

(e) Flange ring thickness using a ring gasket and bolt holes slotted through the edge is ([1.875M,(C

+

B)]/[SB(3C - B)]}'/2

(4.24)

Q = (PL/4S)[(C

+

B)/(3C - B)]

(4.25)

T = Q

+

where

(d) Flange ring thickness using a full-face gasket and round bolt holes: T = Q

+

(Q'

+

[3BQ(C - B)/L]}'/2

(4.26)

+

(4.27)

where Q = (PL/4S)[(C

B)/(7C - 5B)J

(e) Flange ring thickness using a full-face gasket and bolt holes slotted through the edge T = Q

+ (Q' +

[3BQ(C - B)/L]}'"

(4.28)

where Q = (PL/4S)[(C

+

B)/(3C - B)J

(4.29)


4.7.2.3 Ring and Head. (a) Head thickness:

127

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

(4.30) t = (5PL)/(6S) (b) Flange ring thickness: The flange ring thickness is determined by combining the circumferential ring stress and the tangential bending stress, as follows: (i) Circumferential ring stress

(2) Tangential ring stress

=

S,

=

(4.31)

PDI2T

= S, = YM,/T 2B

(4.32)

(3) Total ring stress = S2 = S, + S, = (2S/ T)(F) + (S/ T')(J)

(4.33)

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

=F +

(F T

+

J)'/2

(4.34)

-

B') V2I/[8S(A - B)l

(4.35)

+

B)]/[SB(A - B)l

(4.36)

where F = [PB(4L 2

and J = [M,(A

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 M; is determined by combining the moments from

bolt loading and gasket loading with the moment caused by the intemal pressure loading at the head-taring intersection. When M; 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-lOS, and the head material is SA-SI6 Or. 70. The dish radius is 0.9 LD. Solution (i) The allowable tensile stress for SA-SI6 Gr. 70 at 6S0'F is S, = 18.8 ksi and for SA-lOS at 100'F is Sf' = 20.0 ksi and at 6S0'F is Sf' = 17.8 ksi.

128 Chapter4

A = 26.5" 0.5(A + B) = 18.625"

Full-penetration weld

T 1

L

T

Centroid

B = 10.75" C = 22.5" FIG. E4.8 EXAMPLE PROBLEM OF SPHERICALLY DISHED COVER, DIV. 1

(2) The dish radius, L, of the spherical head segment is L = 0.9S = 0.9(10.75) = 9.675 in.

(3) The minimum required thickness of the head segment, using Eq. (4.17), is Ih

=

5PLl6S,

= (5)(2,000)(9.675)/(6)(18,800)

= 0.858 in.

Assume that the thickness of the head is T, = 1.0 in. (4) The head-to-ring angle, i3;, using the equation given in VUl-l, 1-6(h) is 13,

= arc sin[SI(2L + =

I)J

arc sin{IO.75/[2(9.675) + I])

(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-Ib

(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, H" due to the internal pressure load on the spherical head.

Flat Plates, Covers, and Flanges HD

~

181,500 in.-Ib

HG

~

200,500 in.-lb

129

H, = 174,000 in.-Ib

=

(181,500)(1.607)

=

291,700 in.-lb

Lever Arms hD = 0.5(C - B) = (22.5 - 10.75) = 5.875 in.

ho = 3.729 in. h, = 4.802 in.

h, is obtained by trial, using an assumed flange thickness and the perpendicular head thickness at the head-to-ring intersection. The thickness of the head parallel to the flange ring face is:

= (1)(1.1778) = 1.178 in.

h, = 0.5(T - tp )

Assume that the thickness of the flange ring is T = 5.375 in. h,

~

0.5(5.375 - 1.178)

~

2.099 in.

Flange Moments M D = H D X hD = (181,500)(5.875) = 1,066,000 in.-lb

~

(200,500)(3.729) = 747,700 in.-lb

4

~

(174,000)(4.802) = 835,500 in-Ib

=

(291,700)(2.099)

=

2,037,000 in-Ib

= 612,300 in-Ib

(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: F = [PB(4L' - B')U'J/[8SiA - B)J =

{(2,000)(10.75)[4(9.675)' - (l0.75)'J'n}/[8(20,000)(26.5 - 10.75)]

= 0.137

J

+

~

[MGs/(S")(B)][(A

=

[2,098,000/(20,000)(10.75)][(26.5

B)/(A - B)J

+

10.75)/(26.5 - 10.75)J

= 23.079

T = F

+

(F'

= (.137) =

+

J)'"

+ [(.137)' + 23.079J'"

4.943 In.

For operating condition: F ~ [PB(4L' - B')U'J/[8Sf"(A - B)J =

{(2,000)(1O.75)[4(9.675)' - (10.75)'J"'j/[8(17,800)(26.5 - 1O.75)J

= 0.154


.I

=

[M"/(S~)(B)][(A

=

[2,037,000/(17,800)(10.75)][(26.5 + 10.75)/(26.5 - 1O.75)J

+

131

B)/(A - B)J

= 25.177 T = F

+

(F'

= (.154) +

+

.1)1"

[(.154)' + 25.177Jw

= 5.174 in.

Since this is less than the 5.375 in. assumed, the thickness is acceptahle. 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. Altbough UG-22 of VIII-l and AD-110 of VIII-2 require that both types be considered when evaluating loadings, in VlII-l rules are given only for calculating the primary membrane stresses.

5.2

CODE BASES FOR ACCEPTABILITY OF OPENING

VIII-l 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 tbe 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 openiugs, 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-I 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-S of VIII-2 contains reinforced opening rnles 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-S, an opening may be considered subject to the requirements of Appendices 4 and S of VIII-2. 133

134 Chapter 5

5.3 TERMS AND DEFINITIONS Many terms and definitions used for openings, reinforcements, and ligaments are the same for VIII-I 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 db d-, d; = diameters or chords of various or adjacent openings D = inside diameter of the cylindrical shell D" = outside diameter of the cylindrical shell t" = nominal wall thickness of the nozzle t = nontinal wall thickness of the shell t, = height of the reinforcement base (see Fig. UG-40 of VITI-l)

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 fiat 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 fiat heads needs to be only half the area required for cylindrical shells and formed heads. Themethod presented for determining any needed reinforcement examines theregionaround theopening for available areas to carry the primary stress around the opening. Since stress is related to the load and crosssectional area, areas: aresubstituted when makingcalculations. Placementandlocation of the replacementarea 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 externalpressure requirements and may be called t" instead of t.. (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 deterntines 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,

Openings 135 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 is equal to O.5d + t + t, 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.

I, - t,

,

I: I,, t

.4~~

I

d

; ;

-

d O,5d

+ t + t"

d

t

d

O,5d

+ t + to

FIG. 5.1 REINFORCEMENT liMITS PARALLEL TO SHELL SURFACE

136 Chapter 5

For Section VIll-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/ fl, where fl = 1.285/ (rt)112 for f1 = 0.3. When this vertical limit was set by the ASME Code committee years ago, the assumption was made that rlt of 10 was appropriate. This gave L

~

1/ 13

~

(rt)'" / 1.285

~

(0.lr 2)''' / 1.285

= 0.246r

~

2.46t

(5.1)

where rand 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-!, it remains the smaller

of 2.5T or 2.5T, where T is the nominal shell or head thickness and T, is the nominal uozzle wall thickness. For VIII-2, instead of assuming a fixed value of r/t, the actual values of rand 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,

vm-i

In calculating the nozzle reinforcement requirements for VIII-I, it should be recognized that many of tbe requirements weredevelopedyearsago, basedon the information availableatthattime.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 iu 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 :s 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 (d, + d,). (4) When two openings in a cluster of three or more are used, their centers shall be no closer than: (5.2) (a) For cylinders and cones: (l + 1.5 cos
= angle between the line connecting the centerlines of the two openings being considered and the

longitudinal axis.

Openings 137

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-l are limited to the following sizes: 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-l 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

= dt.F

(5.4)

where t, = 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-I6.1, where F = 0.5(cos 28 + 1) is permitted 8 = angle of the plane being examined from the longitudinal plane The value of the F-factor vs. 8 is plotted in Fig. 5.2. The F-factor corrects the minimum required thickness for all planes between 0° (the longitudinal plane) and 900 (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 = ds,

(5.5)

where d = chord length at the midsurface of the thickness required, excluding the excess thickness available for reinforcement t, = minimum required thickness of a seamless shell on the plane being examined If the longitudinal plane is being examined, the value of t, is determined by Eq. (2.1). If the circumferential plane is being examined, t, is determined by 0.51, or by Eq. (2.4).

138 Chapter 5

1.00

....

J

0.95

• _

~Ongitudinal

,/

8

shell axis

~

0.90

0.85

0.80

..."-

.

0

'" iO

::J

0.75

> 0.70

,

~,

0.65

T

0.60

" 0.55 I""ho..

0.50

o

10

20

30

40

50

60

70

80

Angle 8. deg.• of plane with longitudinal axis

FIG. 5.2 CHART FOR DETERMINING VALUE OF FFOR ANGLE 6

90

Openings 139 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 (5.6)

A = dt.

where d = diameter or chord dimension of the opening t, = minimum reqnired thickness of the spherical shell or formed head (l) When the opening and its reinforcement are entirely within the spherical part of a torispherical head [see Fig. 5.3(a)], r.is the minimum required thickness for a torispherical head using M = L (1) When the opening is in a cone or conical shell, t, 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 [see Fig. 5.3(b)], t, is the minimum required thickness of a seamless spherical shell of radius KID, where D is the shell diameter and KI is given in Table UG-37 of VllI-l.

spherical part· special limit for t r

(al Limits for Torispherical Head

0.8D = special limit for t r

h

~--------D

(b) Limits for Ellipsoidal Head FIG. 5.3 DETERMINATION OF SPECIAL LIMITS FOR SETTING CALCULATIONS

t, FOR USE IN REINFORCEMENT

140 Chapter 5

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 + t,.

5.5.4.2 Perpendicular to Shell Surface. vertical limits are the smaller of:

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

(1) 2.5t

or (2) 2.5t,

+

t..

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, A" is the larger of: A, = (2d - d)(Et - Ft,)

(5.7)

or A, = [2(O.5d + t

+

t,) - dJ(Et - Ft,)

(5.8)

(2) Area available in nozzle wall, A 2, is the smaller of: A 2 = (5t)(t" - tm )

(5.9)

or (5.10)

Example 5.1 Problem Using the rules of VIII-I, determine the reinforcement requirements for an 8 in. l.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 fnll penetration weld. The inside diameter of the head skirt is 41.75 in. The head material is SA-5l6 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.

Openings 141

d = 41.76" ASCU-LIMIT OF REINFORCEMENT

FIG. E5.1 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT IN ELLIPSOIDAL HEAD, DIV, 1 Solution (1) The allowable tensile stress for both SA-5l6 Gr. 70 and SA-I06 Gr. C at 400°F is 20.0 ksi. Therefore,

I =

1.0.

(2) Using UG-32(d), the minimnm required thickness of a 2:1 ellipsoidal head without an opening is: t, = (PD)/(2SE - 0.2P)

=

(700 x 41.75)/(2 x 20000 x 1.0 - 0.2 x 700)

=

0.733 in.

Nominal thickness used is 0.75 in. (3) According to Rule (3) of t, 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, t, is the thickness required for a seamless sphere of radius K]D, where D is the shelll.D. and K is 0.9 from Table UG-37 of VlIl-1. For this head, the opening and its reinforcement shall be withiu a circle with a diameter ofO.8D = (0.8)(41.75) = 33.4 in. (4) The radius is R = K,D = 0.9(41.75) = 37.575 in.

This radius is used in UG-32(f) to determine the t, for reinforcement calculations as: t,

=

(PR)/(2SE - 0.2P)

=

[(700)(37.575)]/[2(20,000)(1.0) - 0.2(700)] = 0.625 in.

(5) Using UG-27(c)(I), the minimum required nozzle thickness is:

=

[(700)(4)]/[(20,000)(1.0) - 0.6(700)]

=

0.143 in.

Nominal thickness used is 1.125 in.

142

Chapter 5

(6) Limit parallel to head surface:

x

or (O.5d + t + t,), whichever is larger.

= d ~

+

8 in. or (4

+

.75

1.125 = 5.875 in.)

Use X = 8 in. (7) Limit perpendicular to head surface: Y

=

2.St

~

2.5(.75)

or

2.5tn, whichever is smaller. ~

1.875 in. or 2.5(1.125) = 2.81 in.

Use Y = 1.875 in. (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 VIll-1 is

+

A, = dt,F

2t,t,F(1 - f,)

= (8)(.625)(1)

+a

= 5.00 in.'

(10) Reinforcement area available in the head is A, = d(Et - Ft,) - 2t,(Et - Ft,)(1 - f,,)

1.0, the second term becomes zero. Therefore, A,

~

d(t - t,)

= (8)(.75 -

.625) = 1.00 in.'

(II) Reinforcement area available in nozzle is A,

~

2Y(t, - tm)

=

(2)(1.875)(1.125 - .143)

=

3.68 in.'

(12) Reinforcement area available in fillet welds is:

=

2(.5)(.75)' = 0.56 in.'

Openings 143 (13) Total reinforcement area available in head, nozzle, and welds is:

=

1.00 + 3.68 + .56

=

5.24 in.'

Area available of 5.24 in.' is larger than the area required of 5.00 in.' (14) Determination of weld strength and load paths: According to UW-15(b). weld strengtb and load path calculations for pressure loading are not required for nozzles which are like the one shown in Fig. UW-16.I(c) of VIII-I. Since this nozzle is similar to that detail, no calculations are required.

Example 5.2 Problem Using the rules of VIII-I. determine the reinforcement requirements for a 12 in. X 16 in. opening for a manway as shown in Fig. E5.2. The man way 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 tbe manway forging. The LD. 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 I quality factors are E = 1.0.

14t" 00 0

AI

U

\' M

--

/

II

_373-

I

V/AJ

l ~ .~ W/#/-Y///

~~

•

I

I

__ c- - -

D

-

0

-

=.B I

I

•

"'. ""! '"

-

!

..,1",

m

X=12"

2X

=

Y

=

1.87 5"

-rl Y =

t = 0.75-

1_8 75-

l,

X=12" 24"

ABCD-LIMIT OF REINFORCEMENT FIG. E5.2 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF 12 in. x 16 in. MANWAY OPENING, DIV. 1

144

Chapter 5

Solution

(I) The allowable tensile stress for both SA-516 Gr. 70 and SA-lOS at4000p is 20.0 ksi. Therefore, (2) Using UG-27(c)(I), the minimum required thickness of the shell is:

f,

= 1.0.

t, = (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 rio equations for determining the minimum required thickness of an elliptical shell in VIII-I, 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:

=

[(700)(8)2(6)2] 1{(20,000)(1.0)[(8)2(1)2 + (6)'(0)']3I2}

= 0.373 in. NoJl1inal thickness used is 1.375 in. (4) Examination of the longitudinal plane. (a) Limit parallel to shell surface whichever is larger.

x

= d =

or

(O.5d

+

I

+

I,)

12 in. or (6 + .75 + 1.375

=

8.125 in.)

Use X = 12 in. (b) LiJl1it perpendicular to shell surface Y

=

2.5t or 2.5tm whichever is smaller.

= 2.5(.75) = 1.875 in. or 2.5(1.375) = 3.437 in. Use Y = 1.875 in. (e) Reinforcement area required by UG-37(c) of VIIJ-I is A, = dt.F

+

2t,I,F(1 - 1,)

=

(12)(.749)(1.0) + 0

=

8.988 sq. in. whcn j;

1.0.

(d) Reinforcement area available in the shell is: A, = d(E,1 - FI,) - 21,(E,1 - Ft,)(1 - 1,,)

Openings

When!,t

145

1.0, the second term becomes zero; therefore, At = (12)(.75 - .749) = 0.012 in.'

(e) Reinforcement area available in nozzle is:

Outward: =

2(1.875)(1.375 - .373) = 3.758 in.'

Inward:

= 2(1.875)(1.375) = 5.156

in.'

(f) Reinforcement area available in fillet welds is: A, = 2(.5)/,/ =

2(.5)(.75)' = 0.562 in.'

(g) Total reinforcement area available from shell, nozzle, and welds is:

=

0.012

+

3.758

+

5.156

+

0.562 = 9.488 in.'

Area available of 9.488 in.' is larger than required area of 8.988 in." (5) Examination of the circumferential plane. The opening has a 16 in. dimension on this [liane, but F = 0.5 and j, = 1.0. (a) Reinforcement area required by UG-37(c) of VIII-I is:

= (16)(.749)(0.5) = 5.992 in.'

(b) Total reinforcement area available from shell and nozzle is: AT =

> 5,992 in.'

Area available of >9.488 in.' is larger than area required of 5.992 in.' 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 arid load path calculations for pressure loading are not required for nozzles like the one shown in Fig. UW-16.1(c) of VUI-l. Since this nozzle is similar to that one, no calculations are required.

146 Chapter 5 Example 5.3 Problem Using the rules of VIII-I, determine the reinforcement requirements for a 5,625 in, LD, 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 LD, of the shell is 41.875 in, The shell material is SA-516 Or, 70, and the nozzle material is SA-106 Or. B. The design pressure is 700 psi, and the design temperamre is 400°F. There is no corrosion allowance, and all joint efficiency1quality factors are E = 1.0, Solution (I) At 400°F, the allowable tensile stress for SA-516 Or. 70 is 20,0 ksi and for SA-106 Or. ksi, Therefore.j, = 17,1/20,0 = 0,855 (2) Using UG-27(c)(1), the minimum required thickness of the shell is: I, ~ (PR)/(SE ~

it is 17,1

0,6P)

[(700)(20,9375)J/[(20,000)(1.0) - 0,6(700)J ~ 0,749 in,

Nominal thickness used is 0,75 in, (3) Using UO-27(c)(1), the minimum required thickness of the nozzle is: t.,

~

(PR,)/(SE - 0,6P)

~

[(700)(2,8125)J/[(17,100)(1.0) - 0,6(7oo)J

~

0,118 in,

Nominal thickness used is 1.5 in,

8.622" 00

:I

I

A

•

~e II.!

B

d- ,.625" I

). f;-. .;.

~

~////0"//';:; •

x·

'"M

r0OJ

---lIT.

~.625"

tit"

I

~---

x·

'r. W"

0.118"

y

=

1.875"

c

0.75"

k~ .... "'''..

W/#.i.% r0c.

t

,.625"

2X- 11.2$"

ci

'"

II

II

~

DA

BCD• 1I t.I IT 0 F R £INFOR(EME NT

c

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

Openings 147

t.= 0.749" t" = 1.5"

t.n = O.11S"

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

(4) Examination of the longitudinal plane: (a) Althongh allowable stresses of the shell and nozzle are different, since the nozzle weld is a full penetration weld abutting the shell wall.j, = 1.0 for both A, and A, audj; = 0.855 for A2 . (b) Liutit parallel to shell surface:

x= =

d

or (O.5d

5.625 in. or

+

t

+

(2.81

t n) , whichever is larger.

+ .75 + 1.5 = 5.062 in.)

Use X = 5.625 in. (c) Liutit perpendicular to shell surface: Y = 2.5t =

or 2.5t n, whichever is smaller.

2.5(.75)

=

1.875 in. or 2.5(1.5)

=

3.75 in.

Use Y = 1.875 in. (d) Reiuforcement area required according to VIII-I, UG-37(c) is:

=

(5.625)(.749)(1.0)

= 4.213 in.'

(e) Reinforcement area available in the shell is: A, = (2X - d)(t - t,) =

(11.25 - 5.625)(.75 - .749)

=

0.005 in.'

148

Chapter 5 (f) Reinforcement area available in the nozzle is:

~

2(1.875)(1.5 - .118)(.855)

~

4.431 in.2

(g) Reinforcement area available in fillet welds is:

= 2(.5)(.75)' = 0.562 in.2

(h) Total reinforcement area available from shell, nozzle, and welds is:

~

.005 + 4.431 + .562

~

4.998 in.'

Area available of 4.998 in.' is larger than area required of 4.213 in.' (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) hased on the midpoint of the minimum required thickness of the shell, t; = 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:

= (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

Area available of 4.998 in.' is larger than the required area of 4.575 in.' Since AT based on the longitudinal plane is larger than A; even without any consideration of an increase in the limit parallel to the shell surface due to the iucreased 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.I(a) of VIII-I. Since this nozzle is similar to that one in detail, no calculations are required. Example 5.4 Problem Using the rules of VIII-I, determine the reinforcement requirements for a 6.0 in. I.D. nozzle wltich is located in a cylindrical shell, as shown in Fig: E5.4. The nozzle abuts the vessel wall and is attached by a fun penetration weld. The J.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

Openings 149

d

T. = 1.37S"

=6.0"

A 0= 6.2S"

~"= 0.189"

I

T = 1.12S"

FIG. ES.4 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT WITH CORROSION ALLOWANCE,

DIV.1

allowance is 0.125 in.• and all joint efficiency I 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.j, = 17.11 17.1 = 1.0. (2) Using UG-27(c)(1), the minimum required thickness, In of the shell is: R = r + ~

1,_ = (PR)/(SE -

15.0

c.a.

+

0.125 = 15.125 in.

0.6P)

= [(1000)(15.125)]/[(17,100)(1.0) - .6(1000)] = 0.917 in.

T, = 0.917

+

0.125 = 1.042 in.

Nominal thickness used is 1.125 in. (3) Using UG-27(c)(1), the minimum thickness. t.; of the nozzle is:

R,

=

r. + c.a.

=

3.0

+

0.125

=

3.125 in.

150 Chapter 5

~

[(I000)(3.125)J/[(17,100)(1.0) - 0.6(1.000)]

~

0.189 in.

T;

= 0.189

+

0.125 = 0.314 in.

Nominal thickness used is 1.375 in. (4) Limit parallel to the shell surface:

x= ~

d

or

(O.5d

6.25 in. or

+

t

+

[3.125

tn) , whichever is larger.

+ (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 2.5tm 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:

~

(6.25)(0.917)(1.0)

~

5.731 in.'

(7) Reinforcement area availahle in the shell is: A,

= (2X ~

- d)(t - ',)

(6.25)[(1.125 - 0.125) - 0.917J

~

0.519 in.'

(8) Reinforcement area available in the nozzle is: A,

~

2Y(," - tm)

~

(2)(2.5)[(1.375 - 0.125) - 0.189]

(9) Total reinforcement area available from shell and nozzle is:

~ 0.519

+ 5.305

~ 5.824 in.'

This is larger than area required of 5.731 in.'

~

5.305 in.'

Openings 151 (10) Determination of weld strength and load paths. According to UW-15(b), weld strength and load path calculations for pressure loading are not reqnired for nozzles of the type shown in Fig. UW-16.I(a) of VIII-I. 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 reqnirements shall be met in addition to those listed in sections 5.5.3, 5.5.4, and 55.5,

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

5.5,6.2

~

(5. t 1)

0.67 dt.F

Limits of Reinforcement Parallel to Shell.

The horizontal limits are the larger of:

(I) 0.75d or (2) 05d + I + I,.

5.5.6,3 Limits of Reinforcemeut Perpendicular to Shell. as those given in section 5.5.4.2.

The vertical limits are exactly the same

5.5.6.4 Stresses When Nozzle Radius/Shell Radius 0; 0.7. In VIII-I, a method is provided for examining membrane and bending stresses in radial nozzle openings when the nozzle radius I shell radius 0; 0.7. Other alternatives given in Code Case 2236 may also be used. This method is as follows: Membrane stress, Sm, and bending stress, S; 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. (I) 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: (5.12)

Case A: Case B:

s; ~

P(R(R, + t, + (R.t)"'] + R,[I + (R,mt,)U2]}IA,

(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{[(R,)'/6]

+ RR,e)

a = e + t/2

s,

~

Mall

(5.14) (5.15)

(5.16)

152 Chapter 5

Rn

e

ea,aB Nonie With Integral

Case A Nozzle With Reinforcing Pad

Type Reinforcament

GENERAL NOTE: When any part of a flange is located within the greater oftha JR';;';r;;+ teor 1etn + te limit as indicated in Fig. 1·7-1 or Fig. 1-7-2 Case A, or the greater of ./Rnm tn or 16t n for Fig, '-7-' or Fig. 1-7-2 Case B. the flange may be included as part of the section which resists bending moment.

FIG. 5.4.1

R n

/~

" L T- --- ,- ---1;,-1 ...........

f4-

t

Neunat exts

L

of shaded area

:I:

e

L

r

J

te + 16t n

e

;;;d--+

~~'6r--1--r-rr r _+1--" --* t I

Shell center line

~ Nozzle center line

Case A Nozzle With Reinforcing Pad

case B Nozzle With Integral Type Reinforcement

GENERAL NOTE: When erw part of a flange is located within the grealer of the )Rnm ttl + te or 16tn + fa limit as indicated in Fig. 1-7-1 or Fig. 1-7-2 Case A, or the greater of~or 16t n for Fig. 1-7-1 or Fig. 1-7-2 Case B, the flange may be included as part of the section which resists bending moment.

FIG. 5.4.2

Openings 153 where in addition to the definitions in section 5.3, A, = shaded area in Fig. 5.4.1 or Fig. 5.4.2, Case A or Case B, in.' I = moment of inertia of the shaded area abont the neutral axis, in.' 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 S» = bending stress calculated by Eq. (5.16), psi.

(3) Calculated stresses are compared with allowable stress, S, as follows:

S; + Sb

S

1.5S

5.6 REINFORCED OPENING RULES, VIII-2 Although there is a close similarity between the reinforcement rules in VllI-1 and VIIl-2, there are some differences. Reinforcement limits and spacing 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 vm-2 may be used.

5.6.1 Definitions Symbols and terms used in opening calculations in VllI-2 are d = diameter or chord dimension of the opening in the given plane, in. t,

=

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. t.; = minimum required thickness of the nozzle, in. t;

nominal thickness of the nozzle, in. = mean radius of the nozzle, in, ri = transition radius between the nozzle and the vessel, in. fp = nominal thickness of the connecting pipe, in. =

r = r + O.5t" In

K = 0.73r, when a transition radius r, 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 VTII-2.

5.6.2 Openings Not Reqniring Reinforcement Calcnlations No reinforcement calculations are required when the following criteria are met for circular openings: (1) Single openings with d=, = 0.2(R m t)112 and two or more openings within a circle with a

diameter

S;

2.5 (Rmt) l l2. The sum of the diameters is

S;

0.25 (Rm t ) l l2.

154 Chapter 5

.'90deg.T -

Alternate nouHt to pipo transrtion

1.1

'3

'-t---looi /m -11--0111

,-+-......'" r

t

r

L'

rm.r+o.5t'J '2

-+.. . . . . ++-""'1

t

Shell thickness

Ibl lei Alternate with fillet, transition

r

/m tn

Alternative with pad plata and fillet

'. 'dl

FIG. 5.5 NOZZLE NOMENCLATURE AND DIMENSIONS (DEPICTS GENERAL CONFIGURATIONS ONLY)

Openings 155 (2) Center-to-center spacing <': 1.5(d1 + d,). (3) Center-to-edge distance of another local stressed area, where PL is greater than US" is equal to 2.5 (R mf ) l J2

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) The ratio of the large to small dimension of the opening is limited to 1.5. (2) 'Iheratio dID :5 0.50. (3) The arc length between centerlines of openings is limited to no less than (a) 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 circumfereutial 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 Vill-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:

5.6.5 Limits of Reinforcement 5.6.5.1

Parallel to the Shell Surface

For 100% of the Required Reinforcement The borizontal limits are the greater of: (1) d or (2) O.5d + t

+

fo'

For 2/3 of the Required Reinforcement Area The horizontal limits are the greater of:

+ O.5(Rm f)lI' O.5d + t + fo'

(I) O.5d (2)

or

156 Chapter 5 5.6.5.2

Perpendicular to the Shell Surface

For the Nozzles Shown iu Fig. 5.5(a) and (b): When h < 2.5t, + K, the perpendicular limits are the greater of (1) 0.5(rmt,)112

+

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 "" 2.5t, + K, the perpendicular limits are the greater of

or (2) 2.5t, In either case, the limit shall not exceed 2.5t. For the Nozzle Shown in Fig. 5.5(c):

e "" 30 deg., the perpendicular limits are the greater of

When 45 deg, >

or (2) L' + 2.5tp

When

In either case, the limit shall not exceed 2.5t. e < 30 deg., the perpendicular limits are the greater of

or

(2) 1.73x + 2.5t, 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(rm t,)[/2

+

t,

+

K

+ K

or

(2) 2.5/,

+

t,

In either case, the limit shall not exceed 2.5t.

Openings 157

In all cases, t, :'5 1.5t and :'5 1.73W, where W = width of the reinforcing element, in. t, = 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 he limited to a material which meets the following criteria: (a) (b) (c) (d) (c) (f)

Area in excess of that required to carry the primary membrane stress, Area of nozzle wall in excess of that required to carry the primary membrane stress, Weld metal within the shell and nozzle wall which required PQR, Full-penetration weld joining weld pad to nozzle neck, Other weld areas meeting requiremeuts of AD-570, Metal meeting the following requirement: [(a, - a,.)

(5.17)

L>Tj '" 0.0008

where, e, = mean coefficient of expansion of the reinforcing metal, in.! in. OF cc, = mean coefficient of expansion of the vessel metal, in. / in. OF 6..T = operating temperature range from 70 0 P to operating temperature, OF

5.6.7 Strengtb of Reinforcement Metal (a) S"/ S, 2': 0.8,

(b) For S"/S, > 1.0, use S"/S, = 1.0 maximum where S; = design stress intensity value of the nozzle material, ksi S, = 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 he perpeudicular to shell. (b)The reinforcement will have all integral construction using corner fillet radii.

5.6.8.2

Required Reinforcement Area, A r •

The required minimum reinforcement area related to

d/(Rt,)l/2 is:

<0.20 2::0.20 &
In Cylinders

In Spheres & Heads

None except rz required {4.05[d/(Rt r ) 112] 112 - 1.81sdi, O.75dt,

None except r2 required {5AO[d/(Rt,)I12]I12 - 2A1 }dt, dt, cos
158 Chapter 5 5.6.8.3 Limits of Reinforcing Zone. Metai included in meeting the reinforcing area requirements shail be located in the zone boundary shown in Fig. 5.6.

Example 5.5 Problem Using tbe rules of Article D-5 of VIII-2, determine the reinforcement requirements of an 8 in. J.D. nozzle which is centrally located in a 2:1 ellipsoidai head as shown in Fig. E5.5. The nozzle is iuserted through the head and attached by a full penetration weld. The inside diameter of the head skirt is 41.75 in. The head materiai is SA-516 Gr. 70. and the nozzle materiai is SA-106 Gr. C. The design pressure is 700 psi, and the design temperature is 500'F. There is no corrosion allowance.

Plane of NouN and

V_A... """"'" in Cylindricol Shells

fa) Reinforcing Zone Limit (1)

Lc • 0.945 ffr IR)2J3 R for nozzles in cylindrical shells.

(2)

Ln· 1.26 ItrlRl2/3 [R frlR + O,5)} for nozzles in heads.

(31 The center of Lc or L n is.t the junctureof the outside surfaces of the shell and nozzles of thickneues t f and rrlt' Zone bound....

R

+ Tfllftll'VOf'lO

Ib) Rein_ Atu

I t I Hatched are. represent ..... ilable reinforcementarea A., 121 Metal are. within the zone boundary. in IXcessof the .... formedby the intersection of the basic shells, shallbe considered as contributing to the fllQuired area At· The basic shells are defined as having inside radius R. thickness tr,inside radius r, thickness tN r

P''''I

Cyltndrtcal Sheela

I

{4} In constructions where the lone boundary passes thrau;h • uniform thickneu wall segment. the zoneboundary may be considered .. Lc or Ln through the thickness.

All PI. . . . " ' "

(31 The available reinforcement area A, shall be It lean equal to A,/2 on each side of the noule center line and in every plane containingthe nozzle axis.

R

\ FIG. 5.6 LIMITS OF REINFORCING ZONE FOR ALTERNATIVE NOZZLE DESIGN

Openings

2l4.t

159

SA-106 C

= 2.5"

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

Solution (I) The allowable stress intensity for SA-5l6 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.l, the minimum required thickness of a 2:1 ellipsoidal head is: P /S

= 700/20,500

~

0.034

which gives II L

L

~

I

~

0.021

0.9D = 0.9(41.75) = 37.575 in. =

0.021(37.575)

~

0.789 in.

Nominal thickness used is 1.0 in. (3) Using AD-20l(a), the minimum required nozzle thickness is: 1m ~ PRI(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 = ~

Use X = 8 in.

d or (O.5d + t

8 in. or (4

+

til)' whichever is larger

+ 1 + 1.25

=

6.125 in.)

160 Chapter 5 (h) For 2/3 of required reinforcement: K'

+ O.5(Rmt) 1/2 or (O.5d + t +

= r

= [4

+

0.5(38.075 X 1)1/2

=

tn) , whichever is larger

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: h < 2.5t, + K

(i)

or (ii)

h '" 2.5t,

+

K

(h) h = 2.50 in. 2.5t, = (2.5)(1.125)

2.81 in.

=

K = 0.25 in. 2.5t,

+K=

+

0.25

=

3.06 in.

+

2.5tp

+

K, whichever is larger,

2.81

Limit (5)(a)(i) applies where h < 2.5t, + K (c) Determine the perpendicnlar limit as follows: Y

= 0.5(r",£")112

+

k

or 1.73x

but not more than 2.5t nor (L + 2.5tp) = 0.5[(4.5625)(1.125)]'" = 0

=

+ 2.5(1.125) + 0.25

2.5(1)

= (4) =

+ 0.25

=

= 1.383 in.

= 3.063 in.

2.5 in.

+ 2.5(1.125)

= 6.813 in.

2.5 in.

(6) Reinforcement area required according to AD-520 of VTII-2 is:

=

dt,F

=

(8)(0.789)(LO)

=

6.312 in.'

2/3A, = 2/3(6.312)

=

4.208 in.'

A,

(7) Using the 100% limit, reinforcement area available in the head is

Openings A,

~

(r -

~

(1.0 - 0.789)(2 X 8 - 8)

161

te)(2X - d) ~

1.688 in.'

(8) Reinforcement area available in the nozzle is A,

~

2Y(t, -

~

2(2.5)(1.125 - 0.132)

1m )

~

4.965 in.'

(9) With the 100% limit, total reinforcement area available in the head and nozzle is:

~

1.688 + 4.965

~

6,653 in.'

Area available of 6.653 in." is larger than area required of 6.312 in.' (10) With the 2/3 limit, reinforcement area available in the head is: A,

~ (I -

~

t,)(2X' - rI)

(1.0 - 0.789)(2

X

7.085 - 8)

~

1.302in.'

(11) With the 2/3 limit, total reinforcement area available in the head and nozzle is AT ~ A,

+

AT = 1.302

+ 4.965

~

6.267 in.'

Area available of 6.267 in.' is larger than the required area of 4.208 in.' Example 5.6 Problem Using the rules of VIII-2, determine the reinforcement requirements for a 12 in. X 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 tbe 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-lOS. The design pressure is 700 psi, and the design temperature is SOoop. There is no corrosion allowance. Solution (I) The allowable stress intensity for SA-516 Or. 70 is 20.5 ksi, and for SA-lOS it is 19.4 ksi. An adjustment of t: ~ 19.4/20.5 = 0.946 is required for A,. (2) Using AD-201(a), the minimum required thickness of the shell is: t, = (PR) / (SE - 0.5P) ~

[(700)(20.9375)]/[(20500)(1.0) - 0.5(700)]

~

0.728 in. Nominal thickness used is \.0 in.

1.25" 0.385"

AI

:!

! I

/

1~/MI0/,

y =

2lz' t " 1"

~~////,Y///

'\

I

D

Y =

2lz'

-

X=12"

X=12" I

2X 24" ABCD=L~mit of Re~nforcement FIG. E5.6 EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF 12 in. x 16 in. MANWAY OPENING, DIV. 2

(3) The manway forging is elliptical. Since there are no eqnations for determining the minimum required thickness of an elliptical shell in VIII-2, oue ueeds 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:

=

[(700)(8)'(6)']/ {(l9,400)(1.0)[(8)'(l)2

+

(6)'(0)2]312}

= 0.385 in. Nominal thickness used is 1.0 in. (4) Examination of the longitudinal plane:

(a) Determine the limit parallel to the sbell surface. (i) For 100% of reinforcement area required:

x

=

d or

= 12 in.

(O.5d + t + t,J, whichever is larger

or (6 + I + 1 = 8.0 in.)

Use X = 12 in. (ii) For 2/3 of reinforcement area required: X' = r

= Use 8.315 in.

+

O.5(R",t)l!2

or

(O.5d

[6 + 0.5(21.4375 XI)''']

+t+ =

tn) , whichever is larger

8.315 in. or 8.0 in.

Openings 163 (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

< 2.51. + K or

(b)

h '" 2.51.

+

K

(ii) h = 2.50 in. 2.51. = (2.5)(1.0)

2.50 in.

~

K = 0.25 in.

2.5t.

+

K

~

Limit (5)(a)(i)(a) applies where h < 2.51. (iii) Determine the perpendicular limits

2.75 in.

+

K

Y = O.5(rmt,y n + K or 1.73x + 2.5tp + K, whichever is larger, but not more than 2.5t nor (L + 2.5t p ) =

0.5(6.625 X 1.0)'12

= 0

+ 0.25

+ 2.5 X 1.0 + 0.25

=

2.5(1) = 2.50 in.

~

3

=

2.50 in.

+ 2.5(1.0)

=

=

~

1.537 in.

2.75 in.

5.50 in.

(c) Reinforcement area required is

For 100% reinforcement: A,

~

dtP

+

2',1,(1 - j;)

= (12)(0.728)(1)

+ 2(1.0)(0.728)(1 - 0.946)

~

8.814 in.'

For 2/3 A: 2/3A, = 5.876 in. (d) Using the 100% limit, reinforcement area available in the shell is: A,

~

(I - 1,)(2X - d)

= (1 - 0.728)(2 X 12 - 12)

(e) Reinforcement area available in tbe nozzle wall is:

~

3.264 in.'

164

Chapter 5 Outward:

A'l ~ 2(2.5)(1.0 - 0.385)(0.946) ~ 3.075 in.'

A" ~ 2(2.5)(1.0)(0.946) ~ 4.730 in.'

Inward:

(f) Total reinforcement area available from the shell and nozzle is:

= 3.264

+ 3.075 + 4.730

= 11.069 in.'

Area available of 11.069 in.' is larger than area required of 8.814 in.' (g) Using the 2/3 limit, reinforcement area available in the shell is A1

~ (I -

0.728)(2 x 8.315 - 12) ~ 1.259 in.'

AT ~ 1.259

+ 3.075 + 4.730

=

9.064 in.'

Area available of 9.064 in.' is larger than area required of 5.876 in.' (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 A, ~

drF

=

(16)(0.728)(0.5)

~

5.824 in.'

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

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

5.7 LIGAMENT EFFICIENCY RULES, VIII·! Section VIII-l 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 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 cousidered from any uozzle 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 e, p' d, and t. Equations were developed to make the determination of the ligament efficiency an easy task. These equatious and a plot of the curves are given in Figs. UG-53.5 and UG-53.6 of VIII-I. Calculations should be made using both the reinforced opening method and the ligament efficiency method: the thinner required plate thickness should be selected.

Openings 165 Example 5.7 Problem

Using the reinforcement requirements give in UG-37 through UG-42 of VIII-I, determine the minimum required thickness of a 36 in. LD. 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-21O Gr. C. There is no corrosion allowance. The openings are not located in or near any butt weldedjoint. Solution

(1) The allowable tensile stress for both SA-516 Gr. 70 and SA-21O Gr. C at 500'F is 20.0 ksi. Therefore, f, = 1.0. (2) Using UG-27(c)(l) and assuming a searoless shell with E = 1.0, the minimum required thickness of the shell for reinforcement calculations is I, = (PR / (SE -

0.6P)

= [(600)(18)]/[(20,000)(1.0) - 0.6(600)J = 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 = 21, = 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 + I), whichever is larger. X = 2.50 in. or (1.25 + 1.25 = 2.50 in.). Use X = 2.50 in.

2'\4"

2'\4"

All holes 2Y2" dia. LONG. AXIS

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

166 Chapter 5 (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: A, = dt,F = (2.5)(0.550)(1.0) = 1.375 in.'

(c) Reinforcement area available in the shell is: A,

=

(spacing - d)(t - t,)

~

(4.5 - 2.5)(1.25

0.550)

~

1.400 in.'

(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 centerto-center spacing is p' = [(3)' + (2.25)2Jlf2 = 3.75 in. And, = tan:" (3/2.25); = 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-I, F = 0.68 for = 53.13'. (b) Reinforcement area required is

e

e

e

A,

~

dt,F

= (2.5)(0.550)(0.68)

~

0.935 sq. in.

(e) Reinforcement area available in the shell is A,

=

(spacing - d)(t - Ft,)

~

(3.75 - 2.5)(1.25 - 0.68

x

0.550)

~

1.095 in.'

(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-I, UG-37(c)(l), the minimum required thickness of the nozzle with E = 1.0 is: tm

~

(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 167 Example 5.8 Problem Using the rules of VIII-I, 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 ~ (1' - d)/1' ~

(4.5 - 2.5)/(4.5) = 0.44

(3) Determine the equivalent longitudinal ligament efficiency from the diagonal efficiency using Fig. UG-53.6 of VIII-l as follows: p' = p

~

3.75 in.; e

~

53.13"; d = 2.5 in.; 1'ld = 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 t, from UG-27(c)(l) as follows: t, = (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-I, 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. ForVIII-2,procedures aregiven for design by analysis of componentswith 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 VITI-I as follows: (I) Braced and Stayed Construction (UG-47 through UG-50, and UW-19)

(2) Jacketed Vessels (Appendix 9) (3) Half-pipe Jackets (Appendix EE) (4) Vessels of Noncircular Cross Section (Appendix 13)

6.2 BRACED AND STAYED CONSTRUCTION Rules for hraced and stayed construction are in UG-47 through UG-50 and in UW-19 of VIII-I. 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 memhers. Depending on the numher of ties, the thickness of braced and stayed surfaces may he less than wheu the surfaces are not stayed, hecause the loading is now resisted by both bending moments and bending strength and hy tensile strength of the stays.

6.2.1 Braced and Stayed Surfaces For hraced and stayed surfaces tied together with threaded-end or welded-in stayholts of uniform diameter and symmetrical spacing, the following formulas apply for determining the minimum required thickness or internal design pressure: t = p(P/SCl'"

P

=

t'SC/p'

where t = minimum required thickness of the stayed plate, in. P = internal design pressure (or MAWP), psi 169

(6.1) (6.2)

170 Chapter 6

S = maximum allowable stress value in tension at design temperature, psi p = maximum pitch hetween 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 stayholt measured on the outside of the stayholt 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 hefore 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 OAp 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 cao be attached, other than outer cylindrical or spherical plates, is 5116 in. (b) When two plates are stayed together and oniy one requires staying, the C value is set by the plate requiring staying. (c) Maximum pitch for threaded-end staybolts is 8-lI2 in. (d) Wben 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 I 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 I5dn where d, 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 treattoent; see US-40(f) of VITI-I. 6.2.1.3 Welded Stays for Jacketed Vessels. the vessel meets the following criteria:

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

(a) Vessel design pressure is ,,; 300 psi. (b) Required thickness of the plate does not exceed lI2 in. (c) Minimum fillet weld size is not less than the plate tbickncss. (d) 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 tbe bole in the plate is I lI4 in. 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 VIll-1 are met:

Special Components, VIII-1

171

Roundanchor block

0.7 r min.

1'1

Ihl

101

Idl

tmin.

momp,,,, penetration

lei

Diameter used 10 satisfy UG-50 requirements

ttl

Diameter used to

satisfy UG-50

r ; nomina! thickness of the thinner stayed plate

requirements

Ihl

191

FIG. 6.1 TYPICAL FORMS OF WELDED STAYBOLTS

... ,..-- ...

II:~lII

/D', --'"

I

, ., ,

I

'\

I t I I

\.'.....__. _....<,/

I \

!

~ = 1% in.

rnax.

I Min. width

2dmin.

Stay ber

e

d I

FIG. 6.2 TYPICAL WELDED STAY FOR JACKETED VESSEL

(a) A welded attachment is made by fillet weld aronnd the edge of the opening. or when the plate thickness with the opening is OS; 3/16 in. and the hole diameter is OS; I in., the hole may be 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 DO-lOl 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.

172 Chapter 6 (c) For a plain plate welded with one of the methods listed below, the minimnm required thickness or the maximum allowable workiug pressure shall be determined by Eq. (6.1) or (6.2) using a value of C = 3.0. The welding procedures are: (I) 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 tbe area supported by the stay times the internal design pressure (MAWP).

6.2.2.2

Minimum Reqnired Area of a Stay

(a) The minimum required area of a stay at its ieast 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 buttweld 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 I7 of VIII-I, 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 I /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.

Special Components, VIII-I

173

Using Eq, (6.2), P = [(0.25)'(12,900)(3)/(5)'J = 96.8 psi

Example 6,2 Problem A flat plate is stayed by welded-in stays. The design pressure is liS 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 (I) The allowable tensile stress from IT-D for SA-516 Gr. 60 at 100°F is 17.1 ksi.

(2) If the plate thickness is > 7116 in., C = 2.2. If the plate thickness is s: 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)/(l7,100)(2.2)]112

= 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 lise.

6.3 JACKETED VESSELS Jacketed vessels, as considered in Appendix 9 of VIII-I, 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 ontside 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 waJl or the jacket wall which is greater than the individual loading, the combined loading is considered within the scope of VIII-I. 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.

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

174 Chapter 6

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 I-Jacket of any length confined entirely to the cylindrical shell Type 2-Jacket covering a portion of the cylindrical shell and oue head 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

--+-.. ,-

-,,

I

I :

-T ,

T

!

I

L

,,

I

I

'-

L

I

I

.....L

./

I

-'-

I I I

I.

,

L

I

I I

Type 4

I I

I

,

I I,

I, I ./

I

-

TVpe3

I

I

.....L

Type 3 - Jacket covering a portion of head.

Type 4 - Jacket with addition of stay or equalizer

rings to the cylindrical

I

'"

cylindrical shell.

shell and one head.

I

,

Type 1 - Jacket of sny length confined entirely to

Type 2 - Jacket covering 8 portion of cylindrical

I

I

./

"

I

, I L , ,I I

I

~

,-

/1. --,,=-

I I

I

I L

-L.

,r-

I

-t--

I,

Type 2

,

,-

I,

I

TVpe'

-, ,-

I

I

I

I,

_i..

--t--

... r--~, ,

/.

"

-, /"

./ TypeS

shell portion to reduce

effective length. Type 5 - Jacket covering cylindrical shell and any portion of

eitherhead.

FIG. 6.3 SOME ACCEPTABLE TYPES OF JACKETED VESSELS


175

6.3.2 Design of Closure Member for Jacket to Vessel 6.3.2.1

Nomenclature.

The symbols and terms used to design jacket closures are

t, = nominal thickness of the inner vessel wall, in.

t., = minimum required thickness of the jacket wall, in. = minimum required thickness of the closure member, in. nominal thickness of the closure member, in. Ij = nominal thickness of the jacket wall. in. t; = nominal thickness of the nozzle wall, in, r = corner radius of the torus closure, in. Rs·=··outside·.radiusofthe.innervessel,·.in. R, = inside radius of the jacket, in. R; = 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 hetween 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. f rc

te =

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: ;:;::: t rj ; r 2: 3fc; t-. :=; 5/8 in.; Y 2: O.7fc I,,;=: Ifj; r;=: 31,; 1,,:5 5/8 in.; Y;=: 0.831, I" ;=: Ifj I" ;=: 0.707j(PlS)lf2; permits fillet weld with throat;=: 0.71, Iff;=: 0.707j(P /S)ll2; requires groove weld with throat of If Iff [from VIII-I, UG-32(g), Eq. (4)] ;=: I fj; :5 30° Iff;=: larger of 21f} or 0.707j(P/S)1I2; I f} :5 5/8 in.; Y;=: smaller of 0.751f or 0.751,; Z;=: I} I,,;=: larger of 21fj or 0.707j(P/S)lf2; Y;=: smaller of 1.51, or LSI, I,,;=: 1.414[(PR,j)/S]lf2;j = {[(2SID/(PR)] - 0.5(1, + Ij )) ; Y;=: smaller of 1.51, or 1.51, Meet the details and dimensions shown in the sketch I f } :5 5/8 in. I f } :5 5/8 in.; welds shall meet the details shown in Fig. 6.4, sketches (i-I) or (i-2), which is the same as Fig. 9-5 in VIII-l [13] For conical and toriconical jackets shown in Fig. 6.4, sketches (f-I) through (f-3) and (g-l) through (g-6).

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

t rc

e

(6.3) (6.4) (6.5) (6.6) (6.7) (6.8)

(6.9) (6.10) (6.11) (6.12) (6.13)

176 Chapter 6

TVpo 1 Jackets

dlmimslon

temln.

,,
:;-Tb

MIn. throBt

TVpes 2 and 4 Jackets

• t

c

Mln.2fc but need net exceed 1/210. fc

-1,6f C

,..

a.83f min. c r min. '" 3f c

(Elongated to

maintain min. threat dimension)

lal 1.25

f

c

min. 1.25fC min.

rmln." 3fc

o

max. '" BOdeg.

see Nete (1) to

See Note (1) to sketch e-u

sketch (b·l)

t;

r

(b.31

(b·l)

NOTE (sketch b-fl: (11 Cloture Bod ,heH one piece cceerruencn or full penetration butt weld. Backing strip may be used.

b

() max. = 30 deg.

t,

~

-Jv--l.f--'v-~=-'i

rcmin.

t

R,:..-J (f·l)

(f-21

FIG. 6.4 SOME ACCEPTABLE TYPES OF CLOSURE DETAILS

Special Components, VIII-l

O.7Y min.

O.7Y min.

(d·l)

(d·2)

y

Backing strip may be used

(e-2)

FIG, 6,4 (CONT'D)

177

178 Chapter 6 Weld Detail Pet Fig. UW-13.2 (d)"

Weld Detail Per Fig, UW·13.2 (el.

,, , ,

min.

,,

.

tJ nun.

1-7-_,),146 de•.

t-~+-fi

~ , Backing Strip

_..J

2tl min.

Weld Detail Per

Fig. UW-13.2 (e)

May be Used

'I

Y""s+b

1.·21 11·31 _ 0.7 tlmln.

1

J-- 't min.

,...,P-tr ~_ __ '--'---1

Not Less

'e

. :::r. • . ~

-:1- -~'

Than

II

"Plug Weld Per UW·17

-.'_t t

/ min.

't

0.83

min, Throat Dimension" fJ

_'t

tl min. ----,

./

,

;>

1.5 tl (Elongated to

30 deg, max.

Maintain min. Throat Dimension)

_ tj

R· --

Rj ---

J

1.-61

1.-6'

Sea Welding Details (1·1) and 0-2)

Torispherlcal ellipsoidal and hemispherical heads (0. D. of jacket head not greater than 0, D. of vessel head, or I. O. of jacket head nominally equal to O. D. of vessel heedl

See Details

Ihl

If·'1 to 1f·31 and

A

A

'3

'3

5/8 in.

'3

\.

\

'r: 5/8 in. max,

'r: 5/8 in. max,

max.

A>B

(/')

B

'i: [i-1 tal)

I"'~_-/ ,

/-~

A

"see Note

V ·See Note

B

_"

li-1 (b)J A=8

0·2)

A
NOTE: *Full Penetration Welds

Conical and Toriconical

Ikl

FIG. 6.4 (CONT'D)

III

/

\I


179

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(6-1) Fig. 6.4(6-2) 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-S) Fig. 6.4(g-6) Fig.6.4(h) Fig. 6.4(k) Fig. 6.4(1)

Type 1

Type 2

[1] [3]

[2]

[3J [4J [6J

[7J [7] [7] [7J [8J [8J [8J [10] [10J [10J [11J [11J

[11]

Type 3

Type 4

Type 5

[2]

[3J [3J

[3]

[3J [3J

[5J

[SJ

[S]

[3J [3J [SJ

[9J [9J [9J

[9J [9J [9J

[9J [9J [9J

[9J [9J [9J

[10] [10J [10J [11J [11J [11J

[10J [10J [10J [11J [11] [11J [12]

[10J [10J [10] [11J [11] [11J

[10] [10J [10J

[3J

[11J [11] [11J

[13J [13J

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

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 innervessel 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 VlIl-1 shall be considered.

6.3.3.1 Openings in Inner Vessel. The design of openings in the inner vessel shall be according to therules given in Chapter 5 for combinationsofloadings due to internal pressure, external pressure (vacuum), or both. No consideration shall be made for cross-sectional area from the jacket and closure.

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] Jacket welded to nozzle wall, with details as shown in Fig. 6.5(a) [2] t~ determined for the shell under external pressure [3] i; ~ tu

(6.14)

180 Chapter 6

t Vessel Wall

t,

I

Attach per Fig. UW-13.2 leI. If) or Ig}

,

t

I, I

t,

I I

, I' I--b \, t1'

Jacket WaH

I

Attach per Fig. UW-13.2 tel, ttl or (g)

I

I,

• .L iF.'IiFt=

:j1~~'

I

'j

B~'king Strip

Backing Strip May be Used

Mav be Used

1i'"'2fjmhi;

a»

tj min.

(.1

t I

Iel

Ibl

Fig.

\

\ tj

,,

Full Penetration

'e2

a e 2tj

Butt Weld Backing Strip May be Used

Backing Strip May Be Used

b '" tj C =- 1.25tc 1 1.·1)

[d}

Co

,I

. Attach per Fig, UW-13.2 (e), If} or Ig)

~~b

a'" 2t c2 b

»

t e2

C'" 1.25 tel

(8-21

______ R p

----~ (fl

FIG. 6.5 SOME ACCEPTABLE TYPES OF PENETRATION DETAILS


181

TABLE 6.3 PENETRATION DETAIL REQUIREMENTS Special Requirements

Penetration Detail

[1] [2J

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(1)

[3]

[2] [4J [4] [5]

NOTE:

Indicates special penetration detail requirements.

[4] t"l determined for shell under external pressure t,,2 determined from the following formulas: When no tubular section exists between jacket and torus, t-a

= Pr/(SE

-

O.6P)

(6.15)

When tubular section exists between jacket and torus, t-a

= PRp/(SE - O.6P)

(6.16)

where E = weld efficiency from Table UW-12 of VIII-l 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.

6.4 HALF·PIPE JACKETS The rnles in Appendix EE of VIII-l 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 other combinations of pressure loading that need to be considered separately from these rules.

Some combinations are (I) (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 as follows: P'

= F/K

(6.17)

182

Chapter6

]

R

R

r

R

R L

FIG. 6.6 SPIRAL JACKETS, HALF-PIPE AND OTHER SHAPES

where pi

=

maximum allowable internal pressure in the jacket, psi

F = J.5S - S' (but F shall not exceed J.5S), psi (6.18) S = maximum allowable tensile stress at design temperature of the vessel, psi S' = actual longitudinal stress in the shell or membrane stress in the head due to internal pressure and other axial forces, psi. When axial forces are negligible, S' shall be taken as PR2t. When the combination of axial force and pressure stress (PR2t) 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

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 = (P,r)/(O.85S, - O.6P,)

where T = minimnm thickness of the half-pipe jacket, in. r = inside radius of the jacket, in. (see Fig. 6.6)

(6.19)

Special Components, VUH

183

'000 9 8

7 6

5 4

3

2 Shellthickness

100 9 8

3/16 in.

.-

7 6

5 4

lILn.

V

........ 31B in.

2

........

1/J in.

-"

'0

9

7

3/4 in.

6

1

5

I.

1

4

In.

3

2

z In.

1 30

40

50

60

70

80

90

100

110

120

130

140

150

160

170

D

FIG. 6.7 FACTOR K FOR NPS 2 PIPE JACKET

s, = allowable stress of tbe jacket material at design temperatnre,

psi

P, = 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.

184

Chapter 6

1000 9 8 7 6 6

4 3 Shell thickness

I

2

100 9

8 7 6

5

V

4

.,.

3 2

3f1~ in.

-- ---

vlin. 318 in.

1fd in.

V 3/4 in.

I

10 9 B 7 6 6

1 in.

4 3

2 in.

2

1 30

40

50

60

70

80

90

100

110

120

130

D

FIG. 6.8 FACTOR K FOR NPS 3 PIPE JACKET

140

150

160

170

Special Components, VIII-1 185 1000 9

8 7 8 6

4 Shell th1ickness

3

1

3116 in, 2

l-- ~

1/}in.

i""" 100 9

8 7 8

3/8 in.

4

....

3

.......

2

....... 10

}

I

.--

6

1/2 in.

-

3/4 in.

J

1 m.

8 7 8 6

4

2 in. 3

2

1

30

40

50

60

70

80

90

100

110

120

130

140

150

160

170

D

FIG,6.9 FACTOR K FOR NPS 4 PIPE JACKET

Example 6.3 Problem Using the rules of Appendix EE of VIII- I, determine the minimnm required thickness of a cylindrical shell with an interual design pressure of 350 psi and an externally attached NPS 4 schedule lOS half-pipe jacket at an internal design pressure of 500 psi in noncyclic service. The inside diameter of the shell is 36 in. and

186 Chapter 6

E = 1.0. The design temperature is 100°F. The shell material is SA-5l6 0r.70. The jacket material is SA53 Or.SIA. There is no radiography of any jacket joints and no corrosion allowance.

Solution (1) From II-D, aliowable 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 UO-27(c)(1) of VIII-l is I,

= (PR)/(SE - 0.6P)= (350)(18)]/[20,000 X 1.0 - 0.6 X 350]= 0.318 in. Nominal thickness used is

I

= 3/8 in.

(3) From Fig. 6.9 with D = 36 in. and t = 3/8 in.,K = 40. S'

= PRI2t=

(350 X 18)/(2 X 0.375)

=

8400 psi

P' = F/K= (1.5 X 20,000 - 8400)/40 = 540 psi> 500 psi

(4) Assuming NPS 4 Schedule lOS, Ij

= 0.120 X 0.875

rj = [(4.5/2) -

=

0.105 in.

0.105] = 2.145 in.

(5) Minimum thickness of the half-pipe jacket using Eq. (6.19) is t,j = [(500)(2.145)J/[0.85 X 13,700 - 0.6 X 500] = 0.095 in.

< 0.105 in. actual

(6) Minimum fillet weld size is 0.120 X 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 lOS; 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 Vlll-l 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,

Special Components, VIII-!

187

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-I, 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 witb 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-I.

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 Il-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: (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 tl,e allowable tensile stress at design temperarure 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 wben greater deformation is acceptable where 0.90 times tbe 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 operting 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 es; are applied only to the plates in which the openings are located. When both em and eb are less than E, the membrane sttess and bending stress are calculated on the gross area of the section using E = 1.0. Those membrane and bending stresses are tben divided by em and ei, respectively, to get the stresses based on the net area. When both em and e; 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.

188 Chapter 6

r

"

! a,

a

T

hi'

a

p== -..

h/2

r

-

h/2

1 d·

d·

+

M,

M

i

hi. H

I+-

•

:.-

T

"

"

111

'21 {See Notes IH and (4)]

[see Notes (11 and (4)1

Pitch distance to next

L,

L,

-J

D

dl

reinforcing member

'I

-+

C

a

P dj

+

hi'

h,

A

L,

+

-I-

'31

........

'.

hi'

'41

FIG. 6.10 VESSELS OF RECTANGULAR CROSS SECTION


'01

pi,

R INot.11l1

I

t:

A

LL,,'

t

t

.ore It I

n...~

_

'2· '1 "

(6b)

FIG. 6.10 (CONT'D)

\ \

'.

.1

btl the _ I n eight placoa..

(6al

,,

189

190 Chapter 6

"

t

0

N

•

P

h N

0

P:;t

h

M

St!JV

'3

P

I

h Stay

'3

-..

, -~-

f-.

t

h

'.

M

'2

I-

h

P~

M

Stay

".,

" "

t

te'

"

•, 'I

P:;;

+',

StAv

t

N

P~

P

"

Stay N

h/2

0

«n -l-

"

'2

'2

" Stl'l¥

"

--

.

+ '2

...-

h/2

0

..

i+-!!..,

-

'2

SIRy

Stay

t

~

,

P=1

"

'I

t91

(101

FIG. 6.11 VESSELS OF RECTANGULAR CROSS SECTION WITH STAY PLATES

!-

Special Components, VllI-!

191

-- r' R---B . '2 Plteh dl,tlW'left to M_t

d,

-t-1---4-

,..Inforc"",·membtr

l~.

I

I

L,

P"

I- ---

'21

"I

"

'2

StB¥

A

Plate member

L,

131

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

192 Chapter 6 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: D, = O.5(d, + d,)

(6.20)

The ligament efficiencies then are (6.21)

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.

14------p

1 4 - - - - - - P -----.....;

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

Special Components, VIII-l

193

!+-------p--------..{

FIG. 6.14 PLATE WITH MULTIDIAMETER OPENINGS

For each multi-diameter opening, an equivalent diameter, DEl, DEl., etc., is determined as follows: (6.22)

This calculation is repeated for an adjacent tube, and tbe results are then combined to obtain an equivalent diameter for the combination of the two multidiameter openings as follows: (6.23)

If d, and d-a are equal, DE = d.. The ligament efficiency for the membrane stress is: em

=

(6.24)

(p - DE)/p

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=

:l: AX/:l:A

(6.25)

I = :l: AX'

(6.26)

where :l: AX = b oTo(O.5To + n + T, + .. + T,)

+

b,T,(O.5T,

+

b,T,(O.5T,)

+ .. +

7~)

(6.27)

194

Chapter 6 (6.28)

x

~ SAX/SA

(6.29)

Also,

(6.30)

1

~

(boTb)1l2 + (b,Tl)1l2 + (b2Tll1l2 + .. + (b,T;)1l2

+ boTo(O.5To + T, + T2 + .. + T, - X)' + b,T,(0.5T, + T, + .. + T. - Xl + b2T,(0.5T, + .. + T. - Xl2

+

(6.31)

b.T.(X - 0.5T.)'

c

=

largerofXort - X

(6.32)

The width of a ligament is:

(6.33) Other properties are:

~

D.5f

(6.34)

(b,JJ)/1I2

(6.35)

e

1

~

And solving: ell

~

(b,f)

(6.36)

P - [(61)/(t2e)]

(6.37)

With

DE

~


195

The ligament efficiency for bending stress is:

e, = (p - DE)/p

(6.38)

Example 6.4 Problem Using the rules of Appendix 13 of VIII-I, determine the membrane and bending ligament efficiencies in a pressure vessel of square cross section, in which H = h = 6 in. and t 1 = t-i = 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), calcnlate the ligament efficiencies as foliows: 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. centerto-center 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 foliows: 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 foliows: em = ev

=

(3 - 1.375)/(3)

=

0.542

Example 6.6 Problem Using the rules in Appendix 13 of VIII-I, 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 ali of the multidiameter openings are

do = 1.625 in. d, = 1.5 in. d, = 1.375 in.

To = 0.125 in. T, = 1.125 in. T,

=

0.25 in.

1%

Chapter 6

Solution Using Eq. (6.22), determine the equivalent diameter, d.; of the opening as follows: d,

=

[(1.625)(0.125) + (1.5)(1.125) + (1.375)(0.250)J/(1.5)

=

1.490 in.

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

Exampie 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. To = 0.125 in. T, = 1.125 in. T, = 0.25 in.

do = 1.625 in. d, = 1.5 in. d, = 1.375 in.

b"

=

3.5 - 1.625 = 1.875 in. 1.5 = 2.0 in. 3.5 - 1.375 = 2.125 in.

b, = 3.5 -

b,

=

Then ~ AX =

1.875 X 0.125(0.0625 + 1.l25 + 0.25)

+ 2.0 X 1.125(0.5625 + 0.25) + 2.t25 X

0.25(0.125)

= 2.2314

:s A

=

(1.875 X 0.125) + (2.0 X 1.125) + (2.125 X 0.25)

=

3.0156

X = (2.2314)/(3.0156) = 0.7400

I

=

(1/12)[(1.875)(0.125)' + (2.0)(1.125)' + (2.125)(0.25)3J

+ 1.875 X 0.125(0.0625 + 1.125 + 0.25 - 0.74)'

+ 2.0 X 1.125(0.5625 + 0.25 - 0.74)' + 2.125 X 0.25(0.74 - 0.125)' = 0.5672

Special Components, VIII-1 197 c

~

DE ~

larger of 0.74 or (1.50 - 0.74)

3.5 - [(6)(0.5672)]/(1.5)2 (0.76)

~

~

0.76 in.

~

1.526in.

(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 sectionis one with a rectangular cross section. Basic equations are given here for that cross section, which is shown in Fig. 6.10, sketch (1). Eqnations for other cross sections are given in Appendix 13-7 of VIlI-l. (l) Membrane stress: (a) Short-side plate:

(6.39) (h) Long-side plate: (6.40)

S; = PH/2tz

(2) Bending stress (a) Short-side plate: (S')N ~ (PeIl2I,)! -

1.5H 2

+

(Sb)Q ~ (Ph 2eIl2I,)[(l

h2[(1

+

+

a.'K)/(1

a.'K)/(1

+

+

K)])

(6.41) (6.42)

K)]

(b) Long-side plate: (S')M ~ (Ph2e/12I,)( -1.5

+

(Sb)Q ~ (Ph'cIl21 2)[(1

[(1

+

+

a.2K)/(1

a.'K)/(1

+

+

K)]

K)Jj

(6.43) (6.44)

(3) Total stress (a) Short-side plates: (ST)N

= Eq.

(ST)Q

~

(6.39) + Eq. (6.41)

(6.45)

Eq. (6.39) + Eq. (6.42)

(6.46)

Eq. (6.40) + Eq. (6.43)

(6.47)

+ Eq. (6.44)

(6.48)

(h) Long-side plates: (ST)M =

(ST)Q = Eq. (6.40)

198 Chapter 6 Example 6.8 Problem A noncircular cross section vessel has outside dimensions of 7.25 in. X 7.25 in. and is 0.625 in. thick, as shown in Fig. E6.8. Material is SA-516 Or. 70. Design temperature is 650°F, and design pressure is 150 psi. There is no corrosion allowance. All bull 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 Vill-I? If not, what thickness is required? Solution Following the rules in section 6.5.4, t = 1'1 = 12 = 0.625 in. h = H = 7.25 - 2(0.625) = 6.00 in. J = bd 3112 = (l )(0.625)3112 = 0.0203 in.'

(I) The membrane and bending ligament efficiencies according to Eq. (6.21) are as follows:

= (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

7.25"

~I

3.5"

f•

r--

J

L

0.625 ••

FIG. E6.B EXAMPLE PROBLEM OF NONCIRCULAR VESSEL, DIV. 1

Special Components, VIII-l (b) Bending stress at midpoint of the side, using Eq, (6.41) or Eq. (6.43) aud (Sb)N =

(X

=

I and K

=

199

I, is

[(150)(6)'(0.3125)112(0.0203)][ -1.5 + (I + 1)/(1 + I)] = 3460 psi

(c) Bending stress at the comer, using Eq. (6.42) or Eq. (6.44) and (Sb)Q =

(X

=

1 and K

=

1, is

[(150)(6)'(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 comer is: Sm + (Sb)Q = 720 + 6930 = 7650 psi (f) From lI-D, allowable stress for SA-516 0r.70 at 650'F is S = 18.8 ksi. Allowable design stresses are:

SE = (18,800)(0,277) USE

=

=

(1.5)(18,800)(0.277)

5200 psi =

7810 psi

(g) Calculated stresses vs. allowable stresses: S; sSE: 720 psi < 5200 psi S; + S, ,; USE: 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-I. Details of tube-totubesheet welds are given in UW-20 and Appendix A of VIII-I. The rules for tubesheet design are given in Appendix AA of VIII-I 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 expansion joints are given in Appendix CC of VIII-1 and those for bellows-type expansion joints are given in Appendix 26 of VIII-I. Other rules that govern the construction ofheat exchangers are given in the Tubular Exchanger Manufacturers Association standards (TEMA, 1999). These rules govern design, tolerances, baffle construction, and other details of heat exchangers not listed in VIII-I. Design and construction of expansion joints are given in the Expansion Joint Manufacturers Association Standard (EJMA, 1998) and Appendix 26 of VIII-I. The rules govern bellows-type expansion joints and include design, fabrication tolerances, and fatigue. Design of the 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-I and so is the stiffening effect of the attached tubes. 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. The theoretical design equations in this case are extremely complicated. VIII-l rules include numerous assumptions and simplifications in order to provide practical design equations. Design equations for tubesheets of U-tube and fixed heat exchangers are given in the remainder of this chapter.

7.2

TUBESHEET DESIGN IN U-TUBE EXCHANGERS

VIII-I gives the design equations for fonr types of tubesheets. The four types are simply supported, fixed, extended as a flange and welded to the channel side, and extended as a flange and welded to the shell side. These four types 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. The design of the tubesheet is lengthy but straightforeword.

7.2.1 Nomenclature The following nomenclature is used in the design equations of tubeshccts with If-tubes. The nomenclature is based on Fig. 7.2 for tubesheet configuration and Fig. 7.3 for tubesheet geometry. The tube pattern layout 201

202 Chapter 7

FRONT END

pJL

I E

It.

-- - .....c---

L,

ID

I ONE PASS SHELL

L.

CHANNEL

F

AND REMOVABLE COVER

~

m-}·------·u---... w

G

" " t'..I_CJ

~I hm--l---n ~

.~ t::

.."

~r

H

SHEET AND REMOVABLE COVER

~

-e

:..~~

S?

~

._~--

J

10U

TUBE SHEET

~~~ liKE "8" STATIONARY HEAD

N

'~IT FIXED TUBE SHEET

LIKE "N" STATIONARY HEAD

TUBE~

K

I'

'.1_1-'

SPECIAL HIGH PRESSURE CLOSURE

I

T

~5IT

OUTSIDE PACKED FLOATING HEAD

W

4\'\\

.''-..::>.....:>...

_-_-~_t.:'~.d==::

FLOATING HEAD

WITH BACKING DEVICE

T

'~\'\,,,

:,,~f~{'.L.~-

PULL THROUGH FLOATING HEAD

,, ,,

1

1.

KETIlE TYPE REBOllER

U

U

~

U-TUBE BUNDLE

P

'iJ-: r~

p

5

T

~L~ " ,I

M

DIVIDED FLOW

SHEET AND REMOVABLE (OVER

" 1:--

nxso

LIKE "Ali STATIONARY HEAD

spur HOW

I

N

CHANNel INTEGRAL WITH

--i- I] T

:::'J

dl.lC

T

1. DOUBLE

CHANNel INTEGRAL WITH TUBE-

D

~~IT

SPliT FLOW

BONNET (INTEGRAL COVER)

...

L

flXEO TUBE$HEET

,J

",,..""' ""'" ••.o ..", ON" __

~

TWO PASS SHEll WITH LONGITUDINAL BAfflE

"","----~-

B

C

REAR END HEAD TYPES

SHEll TYPES

51ATIONARY HEAD TYPES

X

~I

~ J]

CROSS fLOW

W

~ EXTERNALLY SEALED FLOATING TUBESHEET

FIG. 7.1 VARIOUS HEAT-EXCHANGER CONFIGURATIONS (TEMA, 1999)

Design of Heat Exchangers Channel \ .

.--

....-

r

,,r--"l,

1

Shell ",.

I

,(

II

P, :::;

:=P

s

hfJ ...............

" -

i"'-= h

G

A A ,(

t

~

L:J '--

1......1

(a) Typical Simply Supported If-Tube Tubesheet Arrangement

Channel

's

Shell

Ds

(b) Typical Integral U-Tube Type Tubesheet Arrangement

FIG. 7.2 SOME TYPICAL TUBESHEET DETAILS FOR U-TUBES (ASME, 2001)

203

204

Chapter 7 Chann.I ' \

r:-

\~-

--

rr:

r--t

r-

I ~

A

PI

C

G

A

----

~P,

==::

-..

n

Dc

l -

D,

-

I.

tc

t

t-

-L..:..-

........I

(c) Typlcallnlegral Channel-Tubesheel If-Tube Tubesheet Arrangement

Channel

t,

(d) Typlcallnlegral Shell-Tubesheel U-Tube Tubesheet Arrangement

FIG. 7.2 (CONT'D)

Shell

Design of Heat Exchangers

205

00(j)

ooobo

JP 0,0 . : ocFCDoo 00000 000 FIG. 7.3 TUBESHEET GEOMETRY

is normally on a triangular or square pattern. Subsequent equations will refer to the following symbols and definitions: c, = tubesheet corrosion allowance on the tube side, in. C = bolt circle diameter of flange, in. D I = diameter of the cylinder which is integral with the tuhesheet (either D, or D,), in. Do = equivalent diameter of outer tube limit circle, in. = 2r o + de D, = inside shell diameter, in. D, = inside channel diameter, in. d, :::= nominal outside diameter of tube, in. d* = effective tube hole diameter, in.

d* =

MAX

{[d' - 2t, (~)(%)

p} [d, - 2t,1}

E = modulus of elasticity for tubesheet material at design temperature, psi E, = modulus of elasticity for channel material at design temperature, psi E, = modulus of elasticity of cylinder which is integral with the tubesheet (either E, or E,), psi E.• = modulus of elasticity for shell material at design temperature, psi E, = modulus of elasticity for tube material at design temperature, psi E* = effective modulus of elasticity of tubesheet in perforated region, psi, obtained from Fig. 7.4 1 - v*

Fe =

E*/E In K

G = diameter at location of gasket load reaction, in.

h = tubesheet thickness, in. h g = tube-side pass partition groove depth, in. h, = required tubesheet thickness, in.

K= MAX [ (DI ;, 2/'). (~) ]

206 Chapter 7

0.7

0.'

0.5

i;'"

0.4

0.3

0.2

Np

so.ic 0.25

0.1

0.1

0.50 ~.OO

:!OO.10

o

0 0.1

0

0.2

0.3

0.4

0.5

o

0.'

0.1

0.2

0.3

0.4

0.6

0.5

p'

p'

v* (Equilateral Triangular Pattern)

E*JE {Equllat&fal Triangular Pattern}

0.4

0.8

, III

W_"'P

~.Oo

, !1.00

~:TI

o.3

II

III 0.25

0.5

'"'"

~

0.4

0.2

I

0.15

0.3

I

",p

~

SO'.lO 0.2

0.25

0.50

'~2.0C 0.1

0 0

0.1

0.2

0.3

0.4

p'

0.5

~

0.1

, I

II

o 0.'

o

g).10

0.1

0.2

03

[@ 0.4

0.5

p'

E* IE {Square Pattern}

FIG. 7.4 EFFECTIVE POISSON'S RATIO AND MODULUS OF ELASTICITY (ASME, 2001)

0.8


207

itt = expanded length of tube in tubesheet (0 :5 itt :5 h), in. MAX [(a), (b), (c), ...] = greatest of a, b, c, ... P = MAX [(P'd)' (P,d)J PG = pressure acting on the side of the tubesheet which is gasketed (either P'd or P'd)' psi PI = pressure acting on the side of the tubesheet which is integral (either P'd or P,,), psi P s = shell-side internal design pressure, psi. For shell-side vacuum use a negative value for P;

PI = tube-side internal design pressure, psi. For tube-side vacuum use a negative value for PI' Psd = most severe shell-side coincident design pressure

= MAX rep,), 0]. When either tube-side vacuum exists or differential pressure design is specified by the user, use P'd = MAX [(P, - P,), OJ.

Pld

most severe tube-side coincident design pressure = MAX [(P,), OJ. When eithershell-sidevacuum exists or differential pressure design is specified by the user, use PM = MAX [(P, - P,), 0]. =

P = tube pitch, in. p* = effective tube pitch, in. p

(

I _ 8r, UL) 1l2

TfD;

r; = radius to outermost tube hole center, in. S = allowable stress for tubeshect at tubesheet design temperature, psi S, = allowable stress for tube material at tubesheet design temperature, psi. For a welded tube, use the allowable stress for an equivalent seamless tube. S, = allowable stress for shell material at design temperature, psi S, = allowable stress for channel material at design temperature, psi Sf = allowable stress for cylinder which is integral with the tubesheet (either S, or S,), psi II = thickness of the cylinder which is integral with the tubesheet (either t, or I,), in. t, = nominal tube wall thickness, in, t, = shell thickness, in. tc =

channel thickness, in.

U; = largest center-to-center distance between adjacent tube rows, but not to exceed 4p, in. W = flange design bolt load, Ib ex = tube pattern factor = 0.39 for equilateral triangular pattern = 0.32 for sqnare pattern )" = factor from Fig. 7.5 fL = basic ligament efficiency = p - d, p fL* = effective ligament efficiency

p* - d* p*

v* = effective Poisson's ratio in perforated region of tubesheet, obtained from Fig. 7.4 p = tube expansion depth ratio = itt/h, (0 :5 P :5 I)

7.2.2 Design Equations for Simply Supported Tubesheets The general geometry of the heat exchanger is usually establisbed by the process engineer. The inside diameter of the shell and nnmber of required tubes to handle the heat-transfer requirements are obtained

208 Chapter 7

10 0

.,/'

500

./

-""" v/ V V v/V-""" ~ /'

v/ ~~V V ~ V

/

10

/

7

/

./

/

/

//

/

/

/

.,-/

//

'/

I I V// '//

I

ViJ V~ /'

//; ~

1.0

~

1 1/

I

'IJ

'/ 1/

iIJ

If! 0.1

""2.335.!L(~)1~ hb Dz

o

0.1

0.2

0.3

0.4

0.6

2.825

r:

hb

0.8

0.9

(lLf(EL 3.631(!Lf-

0.7

hb

2t[

FIG. 7.5 CHART FOR DETERMINING A (ASME, 2001)

1.0

200 100

60 20 10


209

by the process engineer. Design of the tubes is based on cylindrical equations as outlined in Chapter 2. Tubesheet design is governed by one of two equations. The first is based on bending of the tubesheet, and the second is based on shear of the tnbesheet around its periphery. The bending eqnation is given by

h, = 0.556

(

g)

) In

",(IfilL*)

C.5:*S

G

(7.1)

While the shear eqnation is given by (7.2)

7.2.3 Design Equations for Integral Construction Equations (7.1) and (7.2), with slight modification, are also applicable to integral tubesheets. The thickness hi calculated from Eq. (7.1) is reduced by the stiffness of the shell. This is accomplished by using a reduction factor, FJ, obtained from Fig. 7.6. Also, the gasket diameter, G, in the various expressions is replaced by the inside diameter as follows.

h

bs =

0.556 (

~:)

(

0""" )

D"

)

1.::*S F in

(7.3)

I

(7.4)

where,

h b., = tubesheet thickness for bending due to shell-side pressure. = tubesheet thickness for bending due to tube-side pressure.

hill

1.0

<, .......

0.9

<,

0.8

<,

0.7

0.6

0,01

0.02

0.03

0.04

0.05

tm/D m

FIG. 7.6 FIXITY FACTOR, F(ASME, 2001)

0.06

0.07

210

Chapter 7

and F 1 is obtained from Fig. 7.6 by using

o;

~

(D,

+

D,)/2

+ (t, + tJ/2

Example 7.1 Problem A U-tube heat excbanger with a fixed tubesheet has details as shown in Fig. 7.2(b). Check the thicknesses of the shells, tubes, and tubesheetif the design data are as shown in the table below. Design Pressure Temperature

Shell material Joint efficiency for shells Tube material Tubesheet material S of shells and tubesheet material S of tube material

Additional Design Data: D, = 59.0 in. t, = 0.035 in. E = 28,000 ksi e = 0.39

D, = 52.0 in. r, = 23.0 in. E, = 27,000 ksi UL = 3.5

Shell Side

Tube Side

50 psi 300'F SA 516-70 0.70 SA 688-304 SA 266-Cl 2

700 psi 3QO°F SA 516-70 1.0

psi 16,100 psi

20,000 psi

20,000

I.,

=

0.4375 in.

t, = 1.125 in. d, = 1.0 in.

I' = 1.25 in.

S = 20,000 psi assume a trial h

S, = 16,100 psi 7.625 inch

Solution Channel Shell From Eq. (2.2), P = 20,000 x 1.0 x 1.125/(26 ~

+ 0.6 x 1.125)

843 psi> 700 psi

Shell-Side Shell From Eq. (2.2), P

~

20,000 x 0.7 x 0.4375/(29.5 + 0.6 x 0.4375)

= 206 psi > 50 psi

Tubes Check the I in. 0.0. X 0.035 in. thick tubes for internal and external pressure. For internal pressure: from Eq. (2.8),

Design of Heat Exchangers 211 p

x

~

16,100

~

1160 psi

1.0

x

0.035/(0.5 - 0.4 X 0.035)

This pressure is greater than the applied internal pressure of 700 psi. For external pressure: The 50 psi external pressure does not control in this case,

Tube Sheet Tube Side: The following factors have to be determined in order to calculate the required thickness in accordance with Eqs. (7.3) and (7.4) fJ, ~

p

~

-

1.0)/1.25 = 0.2

317.625 = 0.393

Pv ~ 1.25/(1 - (8 X 23 X 3.5)/(1T(47)2)05 = 1.312 d*

= d,

2t, = 1.0 - 2(0.35)

-

=

0.930

or d* = d, .- 2t, (E,IE)(S,IS)p ~

1.0 - 2(0.035)(27,000/28,000)(16,100/20,000)

= 0.979

use d* = 0.979 fJ,*

=

(p* - d*)/p*

o;

=

(59

=

(1.312 - 0.979)/1.312

+ 52)/2 + (0.4375 + 1.125)/2

t; = (0.4375'"

=

=

0.254

56.281

+ 1.125''')"5 = 1.166

tmlDm = 0.021

From Fig. 7.6, F[ = 0.99 The maximum pressure on the tubeshcet is from the tube side. Hence, the required thickness due to bending is given by Eq. (7.4). hb, = 0.556(52/47)039(}90''') X 52 X (700/(1.5 X 0.254 X 20,000»°5 (0.99)

=

0.556 X 0.947 X 52 X 0.303 X 0.99

=

8.21 inch.

The required thickness due to shear is given by Eq. (7.2)

212

Chapter 7 h,

= 700 =

X 47 I (3.2 X 0.22 X 20,000)

2.57 inch.

Tube Side: By inspection, the shell-side pressure condition does not control due to low design pressure. Hence, tbe required thickness of the tubesheet is 8.21 inch.

7.2.4 Design Eqnations for Integral Constrnction With Tubesheet Extended as a Flange The design equations for tubcsheets extended as flanges must include the effect of the bolt load in the flange. They also include the effect of tube-side and shell-side pressures. The Vlfl-Lprocedure consists of calculating equivalent moments given by

MG =

MI

2:~o (1 - ~)

we = -

2'ITDo

(1

G)C

- -

+ PG ~l

(go - 1) (~; + 1)

D; (D2 16 Do

-

PI -

-

-

) (D7 - + 1) D;

1

(7.5)

(7.6)

From these two moments, we can determine the effective moments

MG M,

-

1

D; 32 FaPa (7.7)

+ r,

and

(7.8)

as well as

M, = M,

+

~1 (3 +

v*)PG

(7.9)

M(3 + v*)P,

(7.10)

and M, = M2

-

D2

If we define M, to be the larger of M3 and M 4 , then the required thickness of the tubesbeet, h, is the larger of

(7.11)

or

Design of Heat Exchangers h

PD,

213

(7.12)

a = 3.2fLS

Equation (7.11) does not take into consideration the stiffness of the attached shell. When the thickness of the attached shell is large, then its stiffness could be utilized to reduce the thickness of the tubesheet. The design procedure is based on a trial and error basis and begins by using the thickness from Eq. (7.11) to calculate a factor F given by 1 - v* E*/E

F~---

(E-A+lnK E 1

)

(7.13)

Applying F obtained from this equation in lieu of F" to calculate M, through M4 in Eqs. (7.7) through (7.10). Then calculate the bending stress in the tubesheet from the equation (7.14)

The designer iterates through Eqs. (7.13) and (7.14) with various tubesheet thicknesses until the value in Eq. (7.14) is as close as possible to the quantity 1.5S. Using the new value of tb calculate the tubesheet to shell junction stress as follows

ITb

(7.15)

DO)

6 ( --.J..M--"'P t} '+' 2 32 [

(7.16)

(7.17)

Tbe stresses from Eqs. (7.16) and (7.17) are limited to the quantity 1.5S. If they are larger, then the tubesheet and/or the shell thickness must be increased and the iteration started from the beginning.

7.3 FIXED TUBESHEETS The terms "fixed tubesheets" apply to those beat exchangers in which two tubesheets are used with the tubes acting as stays and the 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 tubesbeets. They are simply supported, fixed, and flanged and are shown in Fig. 7.7. The design of such tubcsheets 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; = radius of the perforated region, in. a; = radial channel dimension, in. as = radial shell dimension, in. A = outside diameter of tubesheet, in.

214

Chapter 7

t

8e

t

....-Sheli

...----.

Channel """"""

r as

P,

-

h _ g

e=p,

h ;..... --... 1-

-

t +

-

J

A 2

Os

t

--! - ro

+-

e

.

hi

t

tt

s

(a) Tvplcal Two-Side Integral Type Tubesheet Construction

A as 2

Channel'

Shell

Ib) Typical Shell-Side Integral and Channel-Side Gasketed Type Tubesheet Construction With Tubesheet Extended as a Flange

FIG. 7.7 SOME TYPICAL DETAILS FOR FIXED TUBESHEET HEAT EXCHANGERS (ASME, 1995)


0,

(e) Typical Shell-Side Integral and Channel·Side Gasketed Type 'rubesbeet Construction With Tubesheet Not Extended as a Flange

r-

-

-

-

0,

as

P,

.;;. 2

[Lps

-

hg :::::

!

Channel

- -

Os

,-- I--

-

~L

h

r

I, Jt 1..... ........ - =I

t +'s

'-Shell

ld) Typical Two-Side Gasketed Type Tubesheet Construction

FIG, 7.7 (CONT'D.)

215

216

Chapter 7

c, = tubesheet corrosion allowance on tubeside C = bolt-circle diameter, in. d, = tube outside diameter, in. d* = effective tube hole diameter, in. D; = inside channel diameter, in. D, = inside shell diameter, in. E = modnlus of elasticity for the tubesheet materialat T, psi E, = modulus of elasticity for the channel material at T" psi E, = modulus of elasticity for the shell material at T" psi E, = modulus of elasticity for tube material at T" psi E* = effective modulus of elasticity of tnbesheet in perforated region, psi h. = tubesheet thickness, in. k, = tube-side pass partition groove depth, in. k, = height of expansion joint outside of shell envelope, in. .I = expansion joint factor (.I = 1.0 if no joint). K, = axial rigidity of expansion joint, total force/elongation, lb/in. e,., = expanded length of tube in tubesheet, in. (0 :5 f", :5 k) L, = tube length between onter tubesheet faces, in. L = L, - 2k, tube length between inner tubesheet faces N, = number of tubes p = nominal tube pitch, in. P, = shell-side design pressure, psi (use a negative value for vacuum) P, = tube-side desigu pressure, psi (use a negative value for vacuum) r, = radius to outermost tube hole center, in. S = allowable stress for tubesheet material at T, psi S, = allowable stress for channel material at T" psi S, = allowable stress for shell material at T" psi S, = allowable stress for tube material at T" psi t, = channel thickness, in. t, = shell thickness, in. t, = nominal tube wall thickness, in. T ~ tubesheet design temperature, OF T' = tubesheet metal temperature at the rim, OF T, = channel design temperature, OF T; = channel metal temperature at the tubesheet, OF T, = shell design temperature, OF T"m = mean shell metal temperature along shell length, OF T; = shell metal temperatnre at the tubesheet, OF T, = tubes design temperature, OF Tt,m = mean tube metal temperature along tube length, OF UL = largest center-to-center distance between adjacent tube rows, in., but not exceed 4p (U, = 0.0 if no pass partition) W = channel flange design bolt load, lb Cls.m = mean coefficient of thermal expansion of shell material at Ts,m, in. / in. / OF (X/,m = mean coefficient of thermal expansion of tube material at Tt,ml in.lin.l°F c ' = mean coefficient of thermal expansion of tubesheet material at T', in./in./°P ex; = mean coefficient of thermal expansion of channel material at T;, in.lin./op CY.,~ = mean coefficient of thermal expansion of shell material at T:, in. / in. / OF 'Y = axial differential thermal expansion between tubes and shell 11 = flexural efficiency

Design of Heat Exchangers 21.7 '" = basic ligament efficiency

'"* =

effective ligament efficiency

v v*

Poisson's ratio of tubesheet material effective Poisson's ratio in perforated region of tubesheet Poisson's ratio of channel material Poisson's ratio of shell material Poisson's ratio of tube material

= =

Vc = Vs = VI =

p = e",lh, tube expansion depth ratio (0

7.3.2

:5

P :5 I)

Design Equations

The design of fixed tnbesheets is complicated by the fact that the thickness is needed in order to calculate the stress. Accordingly, the thickness is assumed, and the stress is calculated. If the stress is high, a new thickness is assumed, and the process is repeated. The procedure consists of the following steps:

In this step, geometric parameters are obtained. It is assumed that the material and the diameter and thickness of the shell, bonnet, and tubes are known. Also, the tube layout and tube-to-tubesheet junction are assumed to be known. Then the following parameters are obtained:

L.. *_' _-::-=d_* !J.* = P

p*

a, a,

p, = -

a, a,

Pc = -

218 Chapter 7

where

1(, = shell axial stiffness

t,. (D.. + tJ E.r L

'IT

K,

=

tube axial stiffness 'IT t r

(dt

t r) E1

-

L

V

2 ~ = 3 (I - v,.)

yD

s

+

s

2

t't

'

If the shell is gasketed, set

13,

= O. If the channel is gasketed, set

1 J=--

1 + Ks KJ

J = 1.0 if there is no expansion joint

'1

x a

E*

= -

1 - v2 1-

X

E

r-

= [24 (1 _ V*2) N E, t, (d ll) I E* L h"

v = ...2.-3 [13. Ii E.• a, (1 Eh 1 - v; Q _ p, 1-

where

1

1 -

+

(1)Zm

a;]'"

13,

=

o.


219

I-V'[ (A) ]

~-'lj-In

2a,

+V

and Z is obtained hom Fig. 7.8.

In this step the equivalent pressures are determined. Assume a tubesheet thickness, h, and calculate the following parameters: Q

_ (Zd + Q, Z,)

x:

Q

_ (~ + Q, Zm) 2

X:

zt -

2

Z2 -

0 for configuration a in Fig. 7.7

"Yb =

= Pc - C/(2a o ) for configuration b

=

Pc for configuration c

=

Pc - p, for configuration d

~* ~ IS

"YO'

=

*_ "Yc

-

0' I-' s

p' (1 + ~, h) s 6 (l __ v;)

t2 ,

(p; - 1) (p, - I) * 4 - "Is Q2

I-'c

t'

3

c pc

= (p; -

=

l~

+

IHp, 4

"Yc

U

(I

+

~, h)

6 (1 _ v;)

+ I) _ pi - p, + ,,* 2

Ie

(p, - I) z"JX: 1 + q,Zm

the average temperature of the unperforated rim T, is

T

,

=

T'+T'+T' s c 3

T*=T;+Tr ,

T*c

2

~

T~+Tr -2-

220 Chapter 7 For conservative values of Pt and Pt , use T, = T',

Pt = E, t,

a,

= u (p'! 'Y'f -

P"I*

(x

P' = s

' PI

ra; (n -

+

=

P, =

(

1

Xr

P'!

Ti,

"In

v + 2p; v _ p; K',l

r

JK,,(

+ JKAQZl +

+

V,

2

2ao

Po - 1)

I _ (I - J) ~ (D, + h;l) P

JK"t

S

N(d, - t,) (d, - 2t,)

+

and T;

70) - a' (T, - 70)J

N(d, - t,) d, 2a;

s

r7 = T',

lK,f

D,

a;

S

I ) P JK I &,1

Q J X (P; - P;

+

P,

Z2

+

P"

+

e; + Prim)

Step 3 In this step, the effective pressnre and stress are obtained. First calculate Q3' Then with known values of X, and Q3, use Figs. 7.9 and 7.10 to determine F m• Then calculate the stress

2(Ft "Ie

+ P! "I; + e, 'Ys + P'f "'In + ~ 'lTa~

P,(I

Tubesheet Stress The bending Stress is given by = IT

(1.5 *F

m

!L

)

(~)2 P h _ h;

e

+

q,Zm)

Design of Heat Exchangers 0.80

gLng

0.70

~

0.60

a >i

I

~

\

0.50

*

",e 0'.40

~

J.

z;

Curves Zd' Zm'ar. valid for 0.4. They are sufficiently accurate to be used for other values of V--,

!'-J..

\

'\

Zd 0.30

\.

~

0.20

~c

I

~

I I

~

Zm

~

0.10

A

,,~ --l

, o

o

~ -~

4

2

8

FIG. 7.8

X, (ASME, 2001)

where h; = MAX [(h g

-

c,), (0)] for pressure load only

h; = 0 for pressure +

thermal load combined, or thermal load only

For pressure loads, [o ] :5 L5S. For pressure and/or thermal loads, [o ] :5 3S. The sbearing stress is given by

"

rv:

6

z; z; and z; VERSUS

-

10

12

221

222 Chapter 7 0.7

0.6

\ _\ \ \\ ~ \\ ~ \' t\..

0.5

0.4

"\

0.3

...

\,

O:J ~ o.7

!\"-

O:J - o.

I

O:J = o.6

"' l\'-

......

0.2

O:J = o.8

!

O:J = o. 4

!

\.\"'- .....

O:J = o.3

~ <;

0.1

<;

o

1.0

2.0

3.0

O:J

-

4.0

J.2 I

O:J - O. I

O:J = o. o 5.0

6.0

7.0

8.0

9.0

10.0

11.0

12.0

13.0

14.0

15.0

16.0

X.

FIG. 7.9 VALUES OF Q 3 BETWEEN 0.0 AND 0.8

where

f'

p -

d,

~--

'r :s;

p

O.8S

Tube Stress The axial stress is given by

where

Fq = (Zd

+

Q3 Z,) X:12

ICJ"t,o I s: fit s,

If ITt,o is negative, the tubes must be checked against buckling. The maximum permissible buckling stress limit S"bl for tubes is given by the followiug equatiou.


o. 4

223

0,= ..0.8

I

0,= ..0.7

I

0,= ..0.6

0.3

I

0, = -0 .5

I

......

0,=..0 .4

\ \

0,=..0 .3

\\

r-......

0.1

I

i\

<,

\

I

0,=..0 .2

~ ........

I

0,=..0 .1 I

0, = o. o

o 1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

11.0

12.0

X,

FIG. 7.10 VALUES OF Q, BETWEEN -0.8 and 0.0

When Cc

e

i

::::;-,

r

St,bk

When Cc

=

e

i

>-, r

NOTE: S,.b' shall not he greater than S,.

where

8 y, 1

=

yield stress for tube material at Tn psi.

F1 x

,

1T

2

E

--',

(f~",)

13.0

14.0

15.0

16.0

224

Chapter 7

r=

Vd; +

(d, - 2 t,)2

4

equivalent unsupported buckling length of the tube, in. The largest value considering unsupported tube spans shall be used.

e=

unsupported tube span, in.

k = 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. F, = factor of safety given by F, = MAX[(3.25 - 0.5 Fq ) , (1.25)]

NOTE: F, need not be taken greater than 2.0. Shell Stress

VD, t, adjacent to the tubesheets. Using

The shell shall have a thickness t, for a minimum length of 1.8 the calculated values of X, and Q3, calculate QZ2*' (2, + Q3 2 m ) X; 2

Then calculate memhrane, bending, and total stresses in the shell due to the joint interaction, using the following eqnations:

12 (1 + 13, h). (§.) (13, t,) (~) PQ * .

,

1 (.:. PS I-v;

~ ---.

The value of

0',

E

2··

-

hJ X~·

. 11·

V s fT"m -t,

as

+.

p*) ,

R2 1-'.,

,

Z1

a.21

must be kept below 3.0 S,

Channel Stress

The channels shall have a thickness t, for a minimum length of 1.8 vr;:t; adjacent to the tubesheets. Calculate membrane, bending, and total stresses in the channel due to the joint interaction, using the following equations:


225

The value of (T, must be kept below 3.0 S,. In complying with the above three steps, the designer must consider all loading combinations, such as shell- and tube-side pressures, metal temperatures, and various mechanical loads. TEMA (TEMA, 1999) gives a table of various load combinations, which may be evaluated by the designer.

Example 7.2 Problem A fixed-tubesheet heat exchanger, with the tubesheet extended as a flange, has details as shown in Fig. 7.7(c). The shell has an expansion joint, as shown in Fig. 7.7(b). Check the thicknesses of the shells, tubes, and tubesheet of the fixed-tubesheet heat exchanger. The design data are as shown in the table below. Design

Tube Side

Shell Side

Pressure Temperature Shell material Joint efficiency for shells Tube material Tubesheet material S of shells and tubesbeet material

370 psi 3000 P

110 psi 3000P

SA 516-70 1.0 SA 249-316L SA 516-70

SA 240-316L 0.85

20,000 psi for shell

16,700 psi for channel

20,000 psi for tubesheet 14,200 psi

S of tube material

Additional Design Data D, = 46.0 in. tl = 0.083 in. E = 28,000 ksi UL = 0

a,

= 9.0

f =

X 10- 6 in.lin.lF 36 in.

46.0 in. 21.5 in. E, = 27,000 ksi

D,

r,

=

=

€I.X = 3.0 in. L = 264 in. Sy.1 = 19,000 psi

t, = 1.0 in. 1.875 in. E, = 27,000 ksi 6 (XI = 9.0 X 10in./in.lF N, = 1835

p =

Solution Channel Shell

From Eq. (2.2), P = 16,700 X 0.85 X 0.4375/(23 + 0.6 X 0.4375) =

267 psi> 110 psi

t, = 0.4375 in. d, = 1.5 in. E., = 28,000 ksi a, = 6.26 X 10- 6 in.lin.lF J = 1.0

226

Chapter 7

Shell-Side Shell From Eq. (2.2), p

~

20,000 X 1.0 X 1.01(23.0 + 0.6 X 1.0) = 847 psi> 370 psi

Tubes Check the 1.5 in. a.D. X 0.083 in. thick tnbes for internal and external pressure. For internal pressure: from Eq. (2.8), P

=

16,700 X 1.0 X 0.0831(0.75 - 0.4 X 0.083)

=

1934 psi

This pressure is greater than the applied internal pressure of 110 psi. For external pressure: External pressure calculations (made using appropriate External Pressure Chart from the VIll-1 code) resulted in P = 525 psi, which is greater than the applied external pressure of 370 psi. Tubesheet A trial tubesheet thickness of 4.5 inch was initially investigated. It met all of the stress criteria except an overstress in the tubesheet-to-shell junction. After a number of trials, a thickness of h = 6.75 inches will be checked.

Step 1: The following parameters are calculated

=

p

d*

~

~

lih

3.0016.75

= 0.444

[1.5 - 2(0.083)(27128)(14.22120)0.444]

1.450 in.

or, d* = 1.5 - 2(0.083)

~

1.334 in.

use d* = 1.450 in.

Qo

~

21.5 + 1.512 = 22.25 in.

p*

=

1.8751(1 - (2 X 21.5 X 0)1('IT(22.25)2)))v2

fL* ~ (1.875 -

1.450)11.875

+ 1.012

=

=

1.875

0.227

a,

~

46.012

p,

=

23.5122.25

a,

~

46.012 + 0.437512 = 23.219

p,

=

23.219122.25

x,

~

1 - 1835((1.5 - 2(0.083»12(22.25)2 = -0.649

~

=

23.50

1.0562

=

1.0436


x,

~

1 - 1835(1.512(22.25»)2

K, = 'JT(1.0)(46

+

~

-1.085

1)(28 X 10')/264 = 15.6604 X lO'

=

K, ~ 'JT(0.083)(1.5 - 0.083)(27 X 10')/264

K,., = 15.6604 ~, ~

B,

lO'/(1835)(37,788)

X

(3(1 - 0.3'))"'/«(46 - 0.3'»)'14/«(46

=

227

~

37,788

0.2258

+ 1.0)/2)(1.0»'" = 0.2652

+ 0.4375)12)(0.4375))'"

~

0.4033

J = 1.0

hlp

=

From Fig. 7.4, v* = 0.42

6.75/1.875

=

3.6

=

£*1£

and

0.19.

E* = 0.19 X 28.0 X 106 = 5.23 X 10' N

x,

~

~

0.19(1 - 0.3')/(1 - 0.42 2 )

= 0.21

[24(1 - 0.42 2)(1835)(27 X 0.083 X (1.5 - 0.083) X 22.25')/(5.23 X 264 X 6.75')]'" (

= 3.404

v=

(2/(28 X 6.75'))[«0.2652 X 1.0' X 28 X 23.5)/(1 - 0.3'»)(1

+ ~

«0.4033 X 0.4375' X 27 X 23.219)/(1 - 0.3 2)(1

(2/(28 X 6.75'))[191.7600(4.3923)

+

+

+

0.2652 X 6.75

+

0.4033 X 6.75

+

0.4033' X 6.75'/2)]

23.2667(7.4277)]

= 0.2358

=

+

«1 - 0.3 2)/0.21)[ln(48.0/(2 X 22.25)

From Fig. 7.8, Zd

=

0.04, Z,

Q,

~

=

0.08, Z;

=

0.2358]

~

1.350

0.43.

(1.0562 - 1 - 1.35(0.08))/(1

+ 1.35(0.43»)

~

-0.033

Step 2: The effective pressure is calculated as follows

'I

~

Q"

~

(0.04

+

(-0.033)(0.08))(3.404)'/2

~

2.50

Q,2

= (0.08

+

(-0.033)(0.43)(3.404)'/2

~

4.42

[9 X 10-'(300 - 70) - 6.26 X 10-'(300 - 70)](264)

'Yb = 0

0.2652' X 6.75'/2)

=

0.166

228

Chapter 7

'It = 0.2652'(1.0)'(1.0562)3(1 + 0.2652(6.75»/6(1 - 0.3') 'I,

=

(1.0562' - 1)(1.0562 - 1)/4 - 0.0423

'If

~

0.4033'(0.4375)'(1.0436)3(1

'I,

=

(1.0436' - 1)(1.0436

U

=

+

(0.08

=

0.0423

-0.0407

+ 0.4033(6.75»/6(1 - 0.3') = 0.0241

+ 1)/4 - (1.0436 3

-

(1.0562 - 1)(0.43))(3.404'/(1 T, = (300

~

+ 300 + 300)/3

n

= (300

+ 300)/2

= 300

n

= (300

+ 300)/2

= 300

1.0562)/2

+

+ 0.0241 = 0.0294

1.350(0.43»

~

8.85

= 300

Pf = «27)(0.4375)/23.219)/[9(300 - 70) - 6.26(300 - 70)] = 320.61 Pf = «28)(1.0)123.5)/[6.26(300 - 70) - 6.26(300 - 70)] = 0.0 Pi

=

8.85[0(0.0423) - 320.61(0.0241)]

P;

=

[-1.085

=

-68.38

+ (1835(1.5 - 0.083) 1.5/2(22.25)')(0.3) + 2(1.0562)'(0.3)/0.2258

- (1.0562' - 1)/(1.0)(0.2258) - [(1 - 1)/(1.0(0.2258»][(23.5)(0)/22.25'][(46

+

0)/46])

P;

=

(-1.085

+ 1.1818 + 2.9643 - 0.5118 - 0)370

P;

=

[-0.649

+ ((1835)(1.5 - 0.083)(1.5 - 2 X 0.083)/2(22.25)')(0.3) + 1/(1)(0.2258)] 110

P;

~

531.38 psi

p,

=

«1835)(37,788)17r(22.25)')0.166

p"

=

-

Prim

=

-C

[(0)/(27r(22.25)')](8.85)(0) 8.85[370( -0.0407)

P, = «1.0)(0.2258)/[1.0

+

=

=

=

943.24 psi

7400.96

0

+ 110(0.0294)]

= 104.65

+ (1.0)(0.2258)(2.51 + (1.0562 - 1.0)(4.42)]} (943.24 - 531.38

7400.96 - 68.38

+

0

+

104.65)

= (0.2258/1.6228)(7849.09) = 1092 psi


Step 3: Determine the actual stress in various components as follows Q, = -0.033

+ {2[(Jl0)(0.0294) + (320.61)(0.0241) + (370)( -0.0407) + (0)(0.0423)]

+ (0)(0)l7r(22.25)2)/(1098)(1 + (1.350)(0.43» Q3

=

-0.033 + [2(3.234 + 7.7267 - 15.059 + 0) + 0]/1735.39

Q3

=

-0.033 - 0.0047

From Fig. 7.10 with X,

=

-0.0377

= 3.404 and Q, =

~0,0377,

we get F",

= 0.065

Tubesheet Stress o,

=

(1.5 X 0.065/0.227)(2 X 22.25/6.75)2(1092)

=

20,385 psi < 3S

I" = (1.875 - 1.5)/1.875 = 0.2 7 =

=

0.8S = 16,000 psi

(11(2 X 0.2»(22.25/6.75)(1092) = 9000 psi < 16,000 psi

Tube Stress F,

=

(0.04)

+

(-0.0377)(0.08»(3.404'/2)

=

2.4828

S", = {[370(-1.085) - 110(-0.649)] - 1092(2.4828)}/«-0.649) - (-1.085»

=

(-401.450 + 71.390 - 2711.22)/0.436

=

-6975 psi

Since this stress is in compression, buckling allowable stress must be determined. r

=

[1.52 + (1.5 - 2(0.083»2]'"/4

kIlr

=

(1.0)(36)/0.5018 = 72

C,

=

[(2)(7T)2(27,000,000)/19,000]'" = 167.5

F,

=

3.25 - 0.5(2.4828)

=

= 0.5018

2.024

or F, = 1.25

use F,.

=

2.0 by definition

S,,", = (19,00012.0)[1 - (36/0.5018)/(2)(167.7)] = 7465 psi> 6975 psi

229

Chapter 7

230

Shell Stress

Q,,*

~

[0.08

+ (- 0.0377)(0.43)](3.404)'/2

= 4.2823

«.; = (22.25' X 1092)/(2)(23.5)(1.0) + (110)(23.5)/(2)(1.0)

+

«370 - 110)/2)(22.25/1.0562 X 1.0)(1.0562' - 1.0)

11,566 'Y,b

~

+ 1293 + 316

= 13,175 psi

12(1 + (0.2652)(6.75)/2)(28/28)(0.2652 X 1.010.21)(22.253/6.75 3 X 3.404')(1092)(4.2823)

+ (1 -

0.3')~'[370

- 0.3(13175)(1.0/23.5)

12(1.8952)(1)(1.2629)(0.2668)(1092)(4.2823)

",.m

=

35,830

=

(T,.,

=

+ OJ(0.2652)'(23.5)'

+ (1.0989)(201.8) (0.0703) (552.25)

+ 8609 = 44,440 psi 13,175

+

=

44,440

57,615 psi < 60,000 psi

Channel Stress

",.m

= (110)(23.219)/(2(0.4375) = 2918 psi

(T,.b

= -12(1

+

+ (1 -

(0.4033)(6.75)/2)(27/28)(0.4033

0.3')~'[(l

- 0.3/2)110

x

0.4375/0.21)(22.25 3/6.75 3 X 3.404')(1092)(4.2823)

+ 320.61](0.4033)'(23.219)'

-12(2.3611)(0.9643)(0.8402)(0.2668)(1092)(4.2823) - 28,640

",.m +

+ 39,916

+ 11,276

'Y,.b

=

2918

38

=

3 X 16,700

=

=

~

+ (1.0989)(414.11)(0.1627)(539.122)

11,276 psi

14,194 psi

50,100 psi

7.4 EXPANSION JOINTS Two types of expansion joints are covered in VIIJ~ 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. 7.11. They inclnde unreinforced as well as reinforced construction. The required thickness of an unreinforced bellows due to pressure is obtained from the equation t = P(d

+

w)18(1.l4

+

4w/q)

(7.18)

where t, d, w, and q are defined in Fig. 7.11. The stress in the point due to expansion is obtained from any structural analysis method. VITI-I also gives rules for the design of reinforcing rings as well as fatigue analysis. The second type of expansion joints are flanged and flued expansion joints, shown in Fig. 7.12. They are discussed in Non-Mandatory Appendix CC of VITI~ 1. The method of analysis for pressure and expansion is not explicitly given in VIII-I, but is left to the discretion of the designer. Some simplified equations are given in VIII-I for fatigue analysis. However, more experimental data are needed to substantiate these equations.

Design of Heat Exchangers 231

f

ddlam.

(al Unfeinforced Bellows

I

A

/~_' A, I \ f

Reinforcing rings

L,..A

I

V

A-A Equaliling ring

I End equaliling ring

A, GENERAL NOTE: Nominal r" 3tm

(b)

~f=r

t

Reinforced Bellows

FIG. 7.11 BELLOWS·TYPE EXPANSION JOINTS

(a) Flanged Only

(b) Flanged and Flued

FIG. 7.12 FLANGED AND FLUED EXPANSION JOINTS

dd"m.

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 componentsdueto loads otherthanpressure. In performing the stressevaluation,the designermust determine the maximum stress at a given point or location. When compnter 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, TIlls is necessary in order to compare each of these components to a corresponding allowable stress given in Vill-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 VIlI-2.

8.2 STRESS CATEGORIES Stress in any component and location is classified by VIIl-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 tbe 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 VIlI-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 Vlll-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 VlII-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 failnre. An example of secondary stress is the bending stress that develops at the attachment of a body fiange to the shell. This attacbment 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

I

Primary Stress

I

I I

Membrane

I

I

I

I

I General

I Bending

I

I

I

Local

.I

I

I

Examples of a local primary

A genera! primary membrane stress is one that is so distributed in the structure that no redistribution of load occurs as a result of yielding.

membrane stress are the membrane

stress in a shell produced by external load and the moment at a permanent support or at a nozzle connection.

An example is the stress in a circular cylindrical shell due to internal pressure.

An example is the bending stress in the central portion of a flat head due to pressure.

-

TABLE 8.2 STRUCTURAL DISCONTINUITY

I

Structural Discontinuity

I

I

I

Gross Structural Discontinuity

I

This is a source of stress or strain intensification that affects a relatively large protion of a structure and has a significant effect on the overall stress or strain pattern. Examples of gross structural discontinuities are head-to-shell and tlanqe-to-sheil junctions, nozzles, and junctions between shells of different diameters or thicknesses.

I

Local Structural Discontinuity

I

This is a source of stress or strain intensification that affects a relatively small volume of material and does not have a significant effect on the overall stress or strain pattern or on the structure as a whole. Examples of local structural discontinuities are small-nozzle-to-shell junction, local lugs, and platform supports.

I

Analysis of Components in VIII-2 235 TABLE 8.3 THERMAL STRESS Thermal Stress

I

I This is a self-balancinq stress produced by a non-unltorrn distribution of temperature or by different thermal coefficients of expansion. Thermal stress is developed in a solid body whenever a volume of material is prevented from assuming the size and shape that it normally should under a change in temperature.

I

I

General

I General thermal stress is classified as a secondary stress. Examples of general thermal stress are 1, Stress produced by an axial termperature distribution in a cylindrical shell. 2. Stress produced by the temperature difference between a nozzle and the shell to which it is attached. 3. The equivalent linear stress produced by the radial temperature distribution in a cylindrical shell.

I I

Local

I

I Local thermal stress is associated with almost complete suppression of the differential expansion and thus produces no significant distortion. Examples of local thermal stresses are 1. Stress in a small hot spot in a vessel wall. 2. The difference between the actual stress and the equivalent linear stress. 3. Thermal stress in a cladding material.

The third category of stress defined in VIIl-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 Tahle 8.2; and local thermal stress. as defined in Table 8.3 VIIl-2 establishes limits for the tbree 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 Parr, 1989). VIIl-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 fiat 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 sectious a-a and b-b, and what are the allowable stress intensities at these locations? Solution

Section a-a From Table 8.5, fiat 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

236

Chapter 8

TABLE 8.4 STRESS CATEGORIES AND THEIR LIMITS (ASME VIII-2) Stress Category Description

Primary General Membrane Average primary

(For ex-

stress across

amples,

solid section, Excludesdiscon-

see Table 4-120.11

tinuities and

concentrations. Produced only by mechanical loads.

Local Membrane

Average stress across any solid section, Considersdiscontinuities but not concentrations.

Produced only by mechanical loads~

Bending Component of primary stress proportional to distance from centroid of solid

Seeon ary Membrane plus Bending Self-equilibrating

III Increment added to primary or secondto satisfy conary stress by a continuityof structure. centration (notch).

stress necessary

Occurs at structural discontinuities. Can be caused by mechan-

section. Excludes discontinuities and

ical load or by differerifiallher'" mal expansion.

concentrations; Produced only by mechanical

Symbol

(2) Certain thermal stresses which may cause fatigue but

not distortion of ve!i!iel·snape~

Excludes local

loads.

stress concentrations. P,

[NOI. (31]

Peak

Combination of stress components

Q

F

I

, ,I

I I I

and allowI

able limits of stress intensities.

I

I

I

t..,..- __ J.

I

•

I

I I I

I

I

I

•

I

L-----...,.I+----

J

I

I Use design loads - -

-

_

Use operating loads

GENERAL NOTES, (a) The stresses in Category Q are those parts of the total stress which are produced by thermal gradients, structural discontinuities, etc., and do not include primary stresses which may also exist at the same point. It should be noted, however, that a detailed stress analysis typically gives the combination of primary and secondary stresses directly and, when appropriate, this calculated value represents the total of Pm (or PL) + Pb + Q and not Q alone. Similarly, if the stress in Category F is produced by a stress concentration, the quantity F is the additional stress produced by the notch, over and above the nominal stress. For example, if a plate has a nominal stress intensity, S, and has a notch with a stress concentration factor, K, then Pm = 8, Pb = 0, Q = 0, F = Pm (K - 1) and the peak stress intensity equals Pm + Pm (K ~ 1) = KP m• (b) the kfactors are given in Table AD·150.1. NOTES, (1) This limit applies to the range of stress intensity (see 5·110.3). The quantity 38m is defined as three times the average of the tabulated 8 m values for the highest and lowest temperatures during the operation cycle. In the determination of the maximum primary-plus-secondary stress intensity range, it may be necessary to consider the superposition of cycles of various origins that produce a total range greater than the range of any of the individual cycles. The value of 38m may vary with the specific cycle, or combination of cycles, being considered since the temperature extremes may be different in each case. Therefore, care must be exercised to assure that the applicable value of 38m for each cycle, and combination of cycles, is not exceeded except as permitted by 4-136,7. (2) Sa is obtained from the fatigue curves, Figs. 5-110.1, 5-110.2, and 5-110.3. The allowable stress intensity for the full range of fluctuation is 2Sa (see 5-110.3). (3) The symbols Pm, PLo PI» Q, and F do not represent single quantities, but rather sets of six quantities representing the six stress components 0'1' Ui, 0'" 'Tff, "if> and 1'11-

Analysis of Components in Vlll-2

237

TABI.E 8.5 CI.ASSIFICATION OF STRESSES (ASME VIII·2) Vessel Compoftfllt

Cylindrical or spherical shell

Any shell or head

Dished head or conical head

Flat head

Type of Stress

Internal pressure

Gt!Oeral membrane Gradiern··thtough plate thickness

Axial thermal gradient

Membrane Bending

Q Q

Junction with head or flange

Internal pressure

Membrane Bending

Q

Any section across entire vessel

External load or moment, or intemal pressure

General membrane averaged across full section. Stress compcnent perpendicular to cross section

External load or moment

Bending across full section. Stress compcnent perpendicular to cross section

S,hj!lI.plilter.emote from discontinuities

P,

r;

e;

Near rczrle or other opening

External load moment, or in~ temal pressure

Local membrane Bending Peak (fillet Or comer>

Q

Any location

Temp. diff. between shell and head

Membrane

Q

Bending

Q

Crown

Internal pressure

Membrane Bending

P,

Membrane Bending

Q

Membrane Bending

P,

Membrane Bending

Q [Note (2)]

Knuckle or junction to shelf

Internal pressure

Center region

Internal pressure

Junction to shell Perforated head or shell

Classification

Origin of Stress

Location

Internal pressure

Typical ligament in a uniform pattern

Pressure

Isolated or atypical ligament

Pressure

P,

F

e; P t [Note u)]

r: P,

Membrane (avo thru cross section) Bending (avo thru width of !ig., but gradient thru plate) Peak

p.

Membrane Bending Peak

Q

P,

F F F

238 Chapter 8

TABLE 8.5 (CONT'D) Vessel Component Nozzle

location

Origin of Stress

Type of Stress

Cross section perpendicular to

Internal pressure

General membrane

nozzle axis

Nozzle waH

Cladding

Any

or external load

(av. across full section).

p.

or moment

Stress component perpendicular to sectlon

See 4-138

External load or moment

Bending across nozzle section

See 4-138

Internal pressure

General membrane Local membrane Bending

P", See 4-138 P, Q

Peak

F

Differential expansion

Membrane

Q Q

Bending Peak

A",

Any

F F F

Equivalent linear stress [Nole {4Jl

Q

temperature distribution

Nonlinear portion

F

Radial

[Note (.)}] Any

p.

Membrane Bending

Differential

expansion

Any

Classification

of stress distribution

Any

Stress concentration

F

(notch effect)

NOTES: (1) Consideration must also be given to the possibility of wrinkling and excessive deformation in vessels with large diameterto-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 p/}; 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.

67!l' THK,

e

b_

I' THK.

48' 1.0,

FIG. E8.1

_b

Analysis of Components in

VIII~2

239

primary membrane stress intensity, Pm, is equal to S; (20 ksi). The allowable primary bending stress intensity,

Ph, is equal to 1.5Sm (30 ksi). Section b-b From Table 8.5, flat heads develop local primary membrane stress, PL , and secondary stress, Q, at the junction with the sbell 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 primary membrane plus secondary stresses (h + 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 sbown in Table 8.4.

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 juuction with the shell due to internal pressure. Similarly, PL due to design pressure is equal to 1.5Sm (30 ksi). The quantity PI. + 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 S, in determining the fatigne life of this section.

TABLE 8.6 SOME STRESS CONCENTRATION FACTORS USED IN FATIGUE

Location

Stress Co ncentratlon

Fillet welds NOZZle in spherical segment NOZZle in cylindrical segment Tangential nozzle in cylinder Backing strips

4.0 2.2 3.3

Bolts

4.0 5.0

Crackllke defect

5.5 2.0 membrane 2,5 bending

VIII-2 Paragraph 5-112 4-612 4-612 4-614 AD-412.1 5-122 5-111

240 Chapter 8

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 peale 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 prograrmned 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, IT" ITI, ITh, and three shearing stresses 'Trj, Trh. 'Tlh' From these six stresses, the designer can obtain three principal stresses (Tj, (Tz, CT) by using the classical equation

(T

= (aj

+

(Tj)/2 ± [(O'i - uY/4

+

'T~PI2

min

'Thlcknoss: I, (trOll

Oulslde Membrane.Plus·Bendlng plates :l: 8m1l'

• • •

SCL

lAembran. (PIma'

FIG. 8.1 LINEARIZING STRESS DISTRIBUTION

(8.1)

Analysis of Components in Vill-2 241

The maximum stress intensity defined in VIII-2 is the absolute value of the larger of the following values

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 lb 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

Section b-b Bending moment in the axial direction = 5988.0 in.-lb Bending moment in the tangential direction = 1796.4 in.-Ib Shearing force in the radial direction = 2602.3 lb Membrane force in the axial direction = 9600 lb Membrane force in the boop direction = 0 (assunting 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.) Deterntine the stress values at sections a-a and b-b in accordance with the VIIl-2 procedures and compare them with the allowable stresses. Solution

Section a-a The membrane stress is P« = forcelt = 2602.3/6.375

~

410 psi

~

13,150 psi

The bending stress is P,

~

6MIi'

~

6 x 89052/6.375'

242

Chapter 8

From Table 8.4, ~

Allowable Pm Allowable P;

+ P,

~

20,000 psi> 410 psi

30,000 psi> 13,560 psi (13150

+ 410)

Section b-b The axial membrane stress at design pressure is ~

foreelt = 960011.0 = 9600 psi

The axial membrane stress at operating pressure is P«

~

(operating Pldesign P)(foreelt)

=

8400 psi

~

(700/800)(9600/1.0)

The axial bending stress at operating pressure is Q = (operating Pldesign P)(6Mlf) ~

~ (7001800)(6 X 5988.0/1.0')

31,440 psi

From Table 8.4, Allowable P; Allowable Pm

+

Q

~

~

20,000 psi> 9600 psi

60,000 psi> 39,840 psi (8400

+ 31,440)

The hoop membrane stress at design pressure is Pm

~

foreelt

= 0/1.0

~

0 psi

The hoop membrane stress at operating pressure is P;

~

(operating Pldesign P)(forcelt)

~

0 psi

~

(700/800)(0/1.0)

The hoop bending stress at operating pressure is

Q

~

(operating Pldesign P)(6Mlf)

= 9430

psi

~ (700/800)(6 X 1796.4/1.0')

Analysis of Components in VIII-2 243

From Table 8.4, Allowable P; = 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. E8A. Three different loading conditions were calculated. They were pressure, mechanical.vandthermal 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.

AXISYMMETRIC FINITE ELEMENT MODEL

7 r- i f CL

IrL

~V

% .S tress Classification Line

rscu r

h

L..

FIG. E8.4 MODEL OF A FINITE ELEMENT LAYOUT IN A FLAT HEAD-TO-SHELL JUNCTION

244

Chapter 8

TABLE E8.4 SUMMARY OF FINITE ELEMENT OUTPUT

seL Membrane Stress, psi Loads

0'"

Pr Me Th

-900 400 200

6200 1000 100

11400 500 100

100 -600 -800

COMBINATION 1

Pr + Me

-500

7200

11900

-500

COMBINATION 2

Pr+Me+Th

-300

7300

12000

-1300

Pressure Mechanical

Thermal

(not applicable; not used)

act, Membrane Plus Bending Stress, psi
Pressure Mechanical Thermal

0',

0'"

Pr Me Th

-2000 400 200

2100 1500 9200

11000 -700 3700

100 -600 -800

COMBINATION 3

Pr + Me

-1600

3600

10300

-500

COMBINATION 4

Pr+Me+Th

-1400

12800

14000

-1300

SeL Peak Stress, psi 0',

Pressure

Pr Me Th

Mechanical Thermal

Pr + Me

COMBINATION 5 COMBINATION 6

Pr

+

Me

+

Th

0',

0'"

0 850 -200

-100 2400 2000

0 -1250 1900

100 -1350 700

850

2300

-1250

-1250

650

4300

650

-550

NOTES: COMBINATIONS computed for each component of stress. Calculations of Principal Stresses and Stress Intensities made AFTER combinations. r = in-plane component in radial direction I = In-plane component in longitudinal direction h = out-of-plane component in hoop direction

Solution Primary Membrane Stress Table 8.4 indicates that primary membrane stress is produced by mechanical loads ouiy. Thus, in Table E8.4 under the Membrane Stress part, ouly 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, "1 and "2, in this plane are calculated from Eq. (8.1), while the third principal stress is "h. The three principal stresses become

Pressure, psi Mechanical, psi Pressure plus mechanical, psi

0',

0',

6200 1370 7230

-900 30 -530

11,400 500 ll,900

Analysis of Components in VIII-2

245

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 Secondary Stress Table. SA indicates that secondary stress is produced bymechanical and thermal loads, The. FE results indicate that there is a shearing stress in the r.l plane. Thus the two principal stresses, 0"1 and 0"2, in this plane are calculated from Eq. (8'\), while the third principal stress is CJh' The three principal stresses become 0",

Pressure, psi Mechanical, psi Pressure plus mechanical, psi Pressure plus mechanical plus thermal, psi

2100 1760 3650 12,920

~540

140 -1650 ~1520

0",

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

8.5 FATIGUE EVALUATION When a fatigue evaluation is required in accordance with AD-160 of VIIl-2 or by the user or a qualified engineer, it shall be performed in accordance with the requirements of Appendix 5 of VIIl-2, The number of cycles are evaluated from fatigue charts such as the one shown iu 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. 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. E8A, Use Fig. 8.2 for a fatigue chart,

246 Chapter 8

,

00

ll-

I'

J

!- Nom (1)E~30>:lO"p8i.

l2f lll19rPOlate for UTS 8(1-1' S bi-

(3) Tabla 5-110.1 eontBi", tfbul-'ed vallHll and II form",l. lor lin IIl;CUrat.. lnlllrpolsllon of ttl_ eurve •.

i-

,

'......

~L

S., psi

Fo' UTS" SO k'i

~

~-

.......

s

""" ~

Fo' UTS 115·130 ~.i

.....

I-~ .....

.....

iI

"

"

I

I

I I

I

I

I

I I I "

•

I

I

-..... -..... --"" I

I

I

I";"

1I1I 00'

Numberof Cycles

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)

Solution Table 804 indicates that the peak stress must be combined with the membrane aud bending stresses to determine fatigue life. Peak Plus Secondary Stress From Table E8A, 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: 0', Pressure, psi Mechanical, psi Pressure plus mechanical, psi Pressure plus mechanical plus thermal, psi

-2000 1250 -750 -950

11,000 -1950 9050 10950

2000 3900 5900 7900

200 -1950 -1750 -1050

The three principal stresses become

Pressure, psi Mechanical, psi Pressure plus mechanical, psi Pressure plus mechanical plus thermal, psi

0',

+0',

0'.

2010 4930

--2010 220 -280 -1070

11,000 -1950 9050 10,950

6400 8020

Analysis of Components in VIII-2

P.T -'::'

I C . = ~

_ Trme

(0)

P.T -,T,me

~_ _--..J.-

(b)

P.T f-

~-----_r--:T,me

(e)

FIG. 8.3 CYCLIC CURVES

247

248 Chapter 8

And the maximum stress intensity values are giveu 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 S"

=

13010/2

= 6500 psi

From Fig. 8.2, with S" equal to 6500 psi, the maximum uumber of cycles is

> 1,000,000.

REFERENCES API, 1996, American Petroleum Institute, Recommended Rules for Design and Construction of Large, Welded, Low-Pressure Storage Tanks, API 620, Washington, D.C., API. ASCE, 1998, American Society of Civil Eugineers, Minimum Design Loads for Buildings and Other Structures, ASCE 7-98, New York, ASCE. ASME, 2001a, American Society of Mechanical Engineers, Boiler and Pressure Vessel Code, Section VIII, Division I, Pressure Vessels, New York, ASME. ASME, 2001b, American Society of Mechanical Eugineers, Boiler and Pressure Vessel Code, Section VIII, Division 2, Alternative Rules for Pressure Vessels, New York, ASME. ASME, 1999, American Society of Mechanical Engineers, B31.3, Chemical Plant and Petroleum Refinery 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 Ylll-Division 2 of the ASME Boiler and Pressure Vessel Code-Guide to Alternative Rules for Pressure Vessels, New York, ASME. Beer, F. P., and Johnson, Jr., E. R., 1992, Mechanics of Materials, New York, McGraw Hill. Bergman, E. 0., 1955, "The Design of Vertical Vessels Subjected to Applied Forces," Transactions of the ASME, New York, ASME. Flugge, W., 1967, Stresses in the Shells, New York, Springer-Verlag, Gilbert, N., and Polani, J. R., May 1979, Stability Design Criterion for Vessels Subjected to Concurrent External Pressure and Longitudinal Compressive Loads, New York, ASME. ICBO, 1997, International Conference of Building Officials, Uniform Building Code, Whittier, CA, ICBO. Jawad, M. H., 1994, Theory and Design of Plate and Shell Structures, New York, Chapman and Hall. Jawad, M. H., and Farr, J. R., 1989, Structural Analysis and Design of Process Equipment, New York, John Wiley & Sons. G+ W Taylor-Bonney, Bulletin 502: Modem Flange Design, 7th Edition, Southfield, MI, G+ W. Peterson, R. E., 1974, Stress Concentration Factors, New York, John Wiley & Sons. 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 Mannfacturers Association, Inc., Standards of Tubular Exchanger Manufacturers Association, 8tb ed., Tarrytown, NY, TEMA.

249

250 References

Waters, E. 0., Wesstrom, D. B., Rossheim, 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

252 Appendix A

1----------------------------------------------------------------------------------------------1

i

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:

.1 ''

QUICK REFERENCE GUIDE

, ,, :

: ''' '

ASME PRESSURE VESSEL CODE (SECTION VIII, DIVISION 1)

, :

fUll fACE CA$KtT,APPX. '-62-'

SP+lERICALlY OISH£O COVERS,

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H(MI$f>HERICAI.. HEAD, PRESSURES, t.Je' H, l!NEOIJAl THICKNESSES, UW-B no. U'I'/-9, U\II'13, I'IC. UW-13.1 SHtLVTHlCKNtSS: \I(F16'; PRESSlJRES IIH.UG·V ExT.\1G-28 APPX.I·l,I-Z AF'fIX.L·3 APPX.l-l TO l.-~ STI,fENlNG RINGS. UO-211. \10-30, APPX. L-~

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STUDDED CONNECTIONS, UG-43. UW-15. flC. UW-16.1.1JW-15

~~l~~~:~~~~:~& Ufp~~: ~?i.J~:~?t

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COlolPRESsrON lImG, AWX, 1·~. 1'6 /2 APEX ANC\.E UG·J2 CONICAl. HHOS. PRtSSlIIltS, EXT. l){;·J3. AWl(, 1.-6 INT. UG-n. UG-J6, flO. UG·36.A?PX. '-4, 1-5 SIoIAlL WEl.OED 'ITnNGS. UG-II, UG-4J. \11'1'·15. UW·15.FLC. UW·15.1, FlO. UW·16.2 TtJREAO£D OPENtNCS.UG-431~J HEAO ATTACHMENT, UW-13, FIG. UW-IJ.1 nLLET WEWS. UW-18, UW-35. TAaLE !JW'12 KNUCKLE RAOllIS.UG·n. UCS·79

FLUID OPENINGS. l!G·J6. flG,UG-J6 YOKE. UG·11-----~_ STUDS, NUTS.WASHERS.UC-12. UO-1'"\, UCS-lD. UCS-ll.UNI'-12. UNF-1J

:, _T lREATMENI UC'55, UW·IO. UW-40. U'·J1. UCS-56. TA!lLE UCS'S6, UCS-19l
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THE PURPOSE OF THIS GUIDE IS TO ILLUSTRATE SOME OF THE lYPES OF PRESSURE VESSEL CONSTRUCTION WHICH ARE PROVIDED UNDER SECTION VIII,DIVISION " OF THE ASME CODE ISSUED IN JULY 2001,AND TO FURNISH DIRECT REFERENCE TO THE APPUCABLE RULE IN THE CODE. IN THE EVENT OF A DISCREPANCY, THE RULES IN THE CURRENT EDITION OF THE CODE SHALL GOVERN.

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ApPENDIXB MATERIAL DESIGNATION TABLE B.l CARBON STEEL PLATE Nominal Composition

Current

Formerly

C-SI C-SI C-SI

SA-515 Gr. 60 SA-515 Gr. 65 SA-515 Gr. 70

A-201 Gr. B A-212 Gr. A A-212 Gr. B

C-SI C-Mn-SI C-Mn-SI C-Mn-SI

SA-516 SA-516 SA-516 SA-516

A~201

1/2 Cr-1/5 Mo-V 3/4 NI-1/2 Cr-1/2 Mo-V

SA-517 Gr. B SA-517 Gr. F

Gr. 55 Gr. 60 Gr. 65 Gr. 70

Gr. A-201 Gr. A~212 Gr. A-212 Gr.

A fine B fine A fine 8 fine

Yield Stress, ksi

Tensile Stress, ksi

32 35 38

60 65 70

30 32 35 38

55 60 65 70

100 100

115 115

grain grain grain grain

T1A T-1

NOTE: Old "Fire Box" quality steel more closely corresponds to modern-day pressure vessel steels, while "Flange" quality contains more impurities and may not be as homogeneous,

TABLE B.2 CHROME-MOLY STEEL PLATE SPECiFICATIONS, SA-387 Nominal Composition 1/2 Cr-1/2 Mo 1 Cr-1/2 Mo 1 1/4 Cr-1/2 Mo-SI 2 1/4 Cr-1 Mo 3 Cr-1 Mo 5 Cr-1/2 Mo

Current

Formerly

2 12 11 22 21 5

A B C D E (SA·357)

Yield Stress, ksi Ci. 1 ~ Ci. 1 ~ Ci. 1 ~ Ci. 1 ~ Ci. 1 ~ CI. 1 ~

33, 33, 35, 30, 30, 30,

Ci. 2 ~ Ci. 2 ~ Ci. 2 ~ ci 2 ~ Ci. 2 ~ CI. 2 ~

45 40 45 45 45 45

Tensile Stress, ksi Ci. Ci. Ci. Ci. Ci. CI.

1 ~ 1 ~ 1 ~ 1 ~ 1 ~ 1 ~

55, Ci. 2 ~ 55, Ci. 2 ~ 60, Ci. 2 ~ 60, Ci. 2 ~ 50, Ci. 2 ~ 60, CI. 2 ~

70 65 75 75 75 75

NOTE: Each grade of SA-387 is available in two classes of tensile strength levels, which depend on heat treatment. Class 1 is the lower-strength material and generally is the material that has been annealed. Class 2 is the higher-strength material that has been normalized and tempered or quenched and tempered.

TABLE B.3 CHROME-MOLY STEEL FORGING SPECIFICATIONS, SA-182 Nominal Composition 1/2 Cr·1/2 Mo 1 Cr·1/2 Mo 1 Cr-1/2 Mo 11/4Cr·1/2Mo 11/4Cr·1/2Mo 2114 Cr-1 Mo 2 1/4 Cr·1 Mo 3 Cr·1 Mo 5 Cr·1/2 Mo 5 Cr-1/2 Mo

Current

Formerly

Yield Stress, ksi

Tensile Stress, ksi

Gr. F2 Gr. F12, CI. 1 Gr. F12, CI. 2 Gr.F11,CI.1 Gr. F11, CI. 2 Gr. F22, CI. 3 Gr. F22, CI. 1 Gr. F21 Gr. F5 Gr. F5a

Gr. F2 Gr. F12b Gr. F12 Gr.F11b Gr. F11 Gr. F22 Gr. F22a Gr. F21 Gr. F5 Gr. F5a

40 30 40 30 40 45 30 45 40 65

70 60 70 60 70 75 60 75 70 90

NOTE: Several Chrome-Moly forgings have the same nominal composition but vary in strength and/or chemical requirements.

253

254

Appendix B

TABLE B.4 CHROME-MOLY STEEL FORGING SPECIFICATIONS, SA·336 Nominal Composition 1 Cr-1/2 Mo 11/4Cr-1/2Mo 11/4Cr-1/2Mo 11/4Cr-1/2Mo 2 1/4 Cr-1 Mo 21/4 Cr-1 Mo 3 Cr-1 Mo 3 Cr-1 Mo 5 Cr-1/2 Mo 5 Cr'1/2 Mo

Current

Formerly

Yield Stress, ksi

Tensile Stress, ksi

Gr. F12 Gr.F11,CI.1 Gr.F11,CI.2 Gr.F11,CI.3 Gr. F22, CI. 1 Gr. F22, CI. 3 Gr. F21, CI. 1 Gr. F21, CI. 3 Gr. F5 Gr. F5a

CI. F12 CI.F11b CI.F11 CI.F11. CI. F22a CI. F22 CI. F21a CI. F21 CI. F5 CI. F5a

40 30 40 45 30 45 30 45 36 50

70 60 70 75 60 75 60 75 60 80

NOTE:Several Chrcrne-Molyfcrqinqs havethesamenominal composition but vary instrength and/or chemical requirements.

TABLE B.5 QUENCH & TEMPERED CARBON AND ALLOY STEEL FORGINGS, SA-50B Nominal Composition C-Si C-Mn-Si 3/4 Ni-1/2 Mo-1/3 Cr-V 3/4 Ni-1/2 Mo-Cr-V 31/2 Ni-1/2 Mo-1 3/4 Cr-V 31/2 Ni-1/2 Mo-1 3/4 Cr-V 31/2 Ni-1/2 Mo-1 3/4 Cr-V

Current

Formerly

Yield Stress, kst

Tensile Stress, ksi

Gr. 1 Gr.1A Gr. 2, CI. Gr. 3, CI. Gr. 4, CI. 1 Gr. 4, CI. 2 Gr. 4, CI. 3

CI. 1 CI. 1. CI. 2 CI. 3 CI. 4 CI. 4. CI. 4b

36 36 50 50 85 100 70

70 70 80 80 105 115 90

ApPENDIXC JOINT EFFICIENCY FACTORS JOINT EFFICIENCY FACTORS SECTION VIII DIVISION 1

C

B

A

B

D

SEAMLESS 1+--1 ELLIPSOIDAL OA

TORISPHERICAL

A TYPE 1 BUTT WELD FULL RT B TYPE 1 BUTT WELD FULL RT SEE UW·11 (a)(5)(a)/UW·12(a)/TABLE UW·12(a) E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E=1.00 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR) E=1.00 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD FULL RT B TYPE 1 BUTT WELD SPOT RT SEE UW·11 (a)(5)(a)&(b)/UW·12(a),(b)&(d)/TABLE UW·12(a)&(b) E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E=1.00 SEAMLESS HeAD CALCULATIONS (QUALITY FACTOR) E=O.85 LONGITUDINAL STRESS CALCULATIONS FIG. C.1

255

256 Pcppendix C

JOINT EFFICIENCY FACTORS SECTION VIII, DIVISION 1

c

A

A

D

SEAMLESS

+Ii-! HEMISPHERICAL

A TYPE 1 BUTT WELD FULL RT SEE UW·11 (a)(5)(a)/UW·12(a)1TABLE UW·12(a) E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E=1.00 HEAD CALCULATIONS (JOINT EFFICIENCY) E=1.00 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD IN SHELL FULL RT A TYPE 2 BUTT WELD HEAD TO SHELL FULL RT SEE UW·11 (a)(5)(a)/UW·12(a)1TABLE UW·12(a) E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.90 HEAD CALCULATIONS (JOINT EFFICIENCY) E=O.90 LONGITUDINAL STRESS CALCULATIONS

FIG. C.2

Joint Efficiency Factors

257


C

A

A D

A

SEAMLESS

+f-l HEMISPHERICAL

A TYPE 1 BUn WELD IN SHELL FULL RT A TYPE 2 Bun WELD HEAD TO SHELL SPOT RT SEE UW-11 (a}(5)(a)&(b)/UW·12(a),(b)&(d)/TABLE UW-12(a)&(b) E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.80 HEAD CALCULATIONS (JOINT EFFICIENCY) E=O.80 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 Bun WELD IN SHELL NO RT A TYPE 2 Bun WELD HEAD TO SHELL NO RT SEE UW-12(c) lit (d)/TABLE UW-12(c) E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.65 HEAD CALCULATIONS (JOINT EFFICIENCY) E=O.65 LONGITUDINAL STRESS CALCULATIONS

FIG. C.3

258 Appendix C


C

A

A D

A

SEAMLESS

~0...-4 HEMISPHERICAL

A TYPE 1 BUTT WELD IN SHELL SPOT RT A TYPE 2 BUTT WELD HEAD TO SHELL SPOT RT SEE UW-11 (a)(5)(a)&(b)/UW-12(b)&(d)/TABLE UW-12(b) E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.80 HEAD CALCULATIONS (JOINT EFFICIENCY) E=O.80 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD IN SHELL SPOT RT A TYPE 2 BUTT WELD HEAD TO SHELL NO RT SEE UW-11 (a)(5)(a)&(b)/UW-12(~),(c)&(d)/T ABLE UW-12(b)&(c) E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.65 HEAD CALCULATIONS (JOINT EFFICIENCY) E=O.65 LONGITUDINAL STRESS CALCULATIONS

FIG. C.4


259


C

B

B

D

SEAMLESS +lH ELLIPSOIDAL OR TORISPHERICAL

A TYPE 1 BUrr WEI..Il SPOT FIT B TYPE 1 BUrr WEI..Il SPOT FIT SEE UW·11 (a)(5)(a)&(b)/UW.12(b)&(d)/TABLE UW-12(b) E:O.85 SHELL CALCULATIONS (JOINT EFFICIENCy) E:1.00 HEAD. CALCULATIONS (QUALITY FACTOR) E:O.85 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUrr WELD SPOT FIT B TYPE 1 BUrr WELD NO FIT SEE UW·11 (a)(5)(a)&(b)/UW-12(b),(c)&(d)/TABLE. UW-12(b)lIt(c) E:O.85 SHELL CALCULATIONS (JOINT EFFICIENCy) E:O.85 HEAD CALCULATIONS (QUALITY FACTOR) E:O.70 LONGITUDINAL.. STRESS CALCULATIONS

FIG. C.5

260 Appendix C


C

B

A D

B

SEAMLESS

A TYPE 1 BUTT WELD SPOT RT B TYPE 2 BUTT WELD SPOT RT SEE UW-11 (a)(5)(a)&(b)/UW-12(b)&(d)/TABLE UW-12(b) E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY) E=I.OO HEAD CALCULATIONS (QUALITY FACTOR) E=O.80 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD SPOT RT B TYPE 2 BUTT WELD NOT RT SEE UW-11 (a)(5)(a)&(b)/UW-12(b)(c)&(d)/TABLE UW-12(b)&(c) E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.85 HEAD CALCULATIONS (QUALITY FACTOR) E=O.65 LONGITUDINAL STRESS CALCULATIONS

FIG. e.6


261


C

B

A

D

B

SEAMLESS ......f--l ELLIPSOIDAL OR TORISPHERICAL

A TYPE 2 BUTT WELD NO RT B TYPE 3 BUTT WELD NO RT SEE UW·11(a)(5)(a)&(b)/UW·12(c)&(d)ITABLE UW·12(c) E=O.65 SHelL CALCULATIONS (JOINT EFFICIENCY) E=O.85 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR) E=O.60 THE LONGITUDINAL STRESS CALCULATIONS A TYPE 2 BUTT WELD SPOT RT B TYPE 3 BUTT WELD FULL OR SPOT RT SEE UW·11(a)(5)(a)&(b)/UW·12(a),(b)&(d)ITABLE UW-12(a)&(b) E=O.80 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.85 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR) E=O.60 LONGITUDINAL STRESS CALCULATIONS

FIG. C.7

262 Appendix C


C

B

A

D

B

SEAMLESS .~I-j ELLIPSOIDAL OR TORISPHERICAL

A TYPE 1 BUTT WELD NO RT B TYPE 1 BUTT WELD NO RT SEE UW-12(c)&(d)/TABLE UW-12(c) E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.85 HEAD CALCULATIONS (QUALITY FACTOR) E=O.70 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD NO RT B TYPE 1 BUTT WELD SPOT RT SEE UW-11(a)(5)(a)&(b)/UW-12(b),{c)&(d)/TABLE UW-12(b)&(c) E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCY) E=1.00 HEAD CALCULATIONS (QUALITY FACTOR) E=O.85 LONGITUDINAL STRESS CALCULATIONS

FIG. C.B


263


C

A

B

D

B

SEAMLESS ......--i ELLIPSOIDAL OR TORISPHERICAL

A TYPE 1 BUTT WELD NO AT B TYPE 6 SINGLE FILLET WELD SEE UW-11 (a)(5)(a)&{b)/UW·12{d)/TABLE UW-12{C) E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCy) E=O.85 HEAD CALCULATIONS (QUALITY FACTOR) E=O.45 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD SPOT AT B TYPE 6 SINGLE FILLET WELD SEE UW-11 (a){5)(a)&{b)/UW·12{b)&{d)/Table UW-12{b)&(c) E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCy) E=O.85 HEAD CALCULATIONS (QUALITY FACTOR) E=O.45 LONGITUDINAL STRESS CALCULATIONS B TYPE 4 DOUBLE FILLET WELD SAME AS B TYPE 6 EXCEPT E=O.55 LONGITUDINAL STRESS CALCULATIONS

. FIG. e.9

264 Appendix C


c

B

B

D

SEAMLESS

+1-1 ELLIPSOIDAL

SEAMLESS SHELL

OR TORISPHERICAL

SEAMLESS SHELL B TYPE 1 BUTT WELD FULL RT SEE UW-11(a)(5)(a)/UW-12(a)&(d)/TABLE UW-12(a) E=1.00 SHELL CALCULATIONS (QUALITY FACTOR) E=1.00 HEAD CALCULATIONS (QUALITY FACTOR) E=1.00 LONGITUDINAL STRESS CALCULATIONS SEAMLESS SHELL B TYPE 1 BUTT WELD SPOT RT SEE UW·11 (a)(5)(a)&(b)/UW-12(b)&(d)/TABLE UW-12(c) E=1.00 SHELL CALCULATIONS (QUALITY FACTOR) E=1.00 HEAD CALCULATIONS (QUALITY FACTOR) E=O.85 LONGITUDINAL STRESS CALCULATIONS

FIG. C.10


265


C

B

B

D

SEAMLESS 4-1---4 ELLIPSOIDAL

SEAMLESS SHELL

OR TORISPHERICAL

SEAMLESS SHEL.L. B TYPE 6 SINGL.E FIL.L.ET WELD SEE UW·11 (a)(5)(a)!UW·12(d)/TABL.E UW-12(c) E=O.85 SHELL CAL.CULATIONS (QUAL.ITY FACTOR) E=O.85 HEAD CAL.CUL.ATIONS (QUAL.ITY FACTOR) E=O.45 L.ONGITUDINAL. STRESS CAL.CUL.ATIONS B TYPE 4 DOUBL.E FIL.L.ET WEL.D SAME AS B TYPE 6 EXCEPT E=O.55 LONGITUDINAL STRESS CAL.CULATIONS NOTE: INTENT IN INTERPRETATIONS·VOLUME 20·JUL.Y,1987 REVISIONS APPEAR IN THE 12/31/87 ADDENDA

FIG. C.11

266 Appendix C

JOINT EFFICIENCY FACTORS SECTION VIII, DIVISION 1 C

A

B

B

D

~~

SEAMLESS ELLIPSOIDAL OR TORISPHERICAL

A ERW PIPE B TYPE 1 BUTT WEL.O SPOT RT SEE UW·11 (a)(5)(a)&(b)/UW.12(b)&(d)/TABLE UW.12(b) E=1.DO SHELL. CALCULATIONS (QUALITY FACTOR), NOTE 1 E=1.00 HEAD CALCULATIONS (QUALITY FACTOR) E=0.85 LONGITUDINAL STRESS CALCULATIONS. NOTE 2 A ERW PIPE B TYPE 1 BUTT WEL.O NO RT SEE UW·11 (a)(5)/UW.12(c)&(d)/TABLE UW.12(c) E=0.85 SHELL. CALCULATIONS (QUALITY FACTOR), NOTE 1 E=O.85 HEAD CALCULATIONS (QUALITY FACTOR) E=0.70 LONGITUDINAL STRESS CALCULATIONS, NOTE 2 NOTE 1: THE QUALITY FACTOR SHOWN IN THE SHELL CALCULATIONS IS IN ADDITION TO THE CATEGORY A JOINT FACTOR (E=0.85) INCl.UDED IN THE ALLOWABLE STRESS VALUE OBTAINED FROM THE APPLICABLE STRESS TABLE. NOTE 2: DIVIDE THE Al.l.OWABLE STRESS VAl.UE OBTAINED FROM THE APPLICABl.E STRESS TABLE BY 0.85 BEFORE APPLYING A JOINT EFFICIENCY IN THE LONGITUDINAl. STRESS CALCUl.ATION (eg, SEE NOTE 26, TABLE UCS.23)

FIG. C.12


267

JOiNT EFFICiENCY FACTORS SECTION VIII, DIVISION 1

C

B

A D

SEAMLESS EL.lIPSOIDAL. OR TORISPHERICAL.

D TYPE 1 BUTT WELD SEE UW·11(a)(5) FULL RT IF PART IS DESIGNED E:1.00 RT NOT REQUIRED FOR tr or trn CALCULATIONS E:1.00 E:EFFICIENCY IF OPENING THROUGH CATEGORY A WELD D FULL OR PARTIAL PENETRATION CORNER WELD RT NOT REQUIRED E:1.00 RT NOT REQUIRED FOR tr or trn CALCULATIONS E:1.00 E:EFFICIENCY OF THE BUTT WELD PENETRATED

FIG. C.13

268

Appendix C


C

B

A

o

B

SEAMLESS

+f-l ELLIPSOIDAL OR TORISPHERICAL

B&C TYPES I OR 2 BUTT WELD IN NOZZLES/COMMUNICATING CHAMBERS RT NOT REQUIRED TO CALCULATE trn RT MAY BE REQUIRED FOR: SERVICE RESTRICTION MATERIAL THICKNESS USERIDESIGNATED AGENT

FIG. C.14


269


C

•

B po

~C

1 D

.

A

C-'

•

FLAT HEAD

C IS PARTIAL OR FULL PENETRATION CORNER JOINT FOR ALL CASES E FOR C IS NOT ESTABLISHED BY CODE RULES A A A A A

TYPE 1 BUTT WELD FULL RT - E=1.00 TYPE 1 BUTT WELD SPOT RT - E=O.85 TYPE 1 BUTT WELD NO RT - E=O.70 ERW BUTT WELD-E=1.00 (JOINT EFFICIENCY IN STRESS VALUE) SEAMLESS· E=1.00

NOTE: FLAT HEAD FORMULA HAS BUILT-IN STRESS MULTIPILER REGARDLESS OF TYPE OF JOINT OR EXAMINATION. A FACTOR FOR E WILL APPLY IN THE FLAT HEAD FORMULA IF A CATEGORY A JOINT DOES EXIST IN THE HEAD

FIG. C.15

270 Appendix C


• C

~C

t

D

B '-

A

•

C-'

TUBE SHEET OR FLANGE

A TYPE 1 BUTT WELD FULL RT C TYPE 1 BUTT WELD FULL RT E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E=1.00 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD FULL RT C TYPE 1 BUTT WELD SPOT RT E.1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E.. O.85 LONGITUDINAL STRESS CALCULATIONS

FIG. C.16


JOINT EFFICIENCY FACTORS SECTION VIII. DIVISION 1

c

• ~c

1D

B

.

A

C

TUBE SHEET OR FLANGE

A TYPE 1 BUTT WELD FULL RT C TYPE 1 BUTT WELD NO RT E=O.85 SHELL CALCULATIONS (QUALITY FACTOR) E=O.70 LONGITUDINAL STRESS CALCULATIONS A TYPE 1 BUTT WELD SPOT RT C TYPE 1 BUTT WELD NO RT E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY) E=O.70 LONGITUDINAL STRESS CALCULATIONS

FIG. C.17

271

272 J\ppendix C


c

• ~C

1D

B

• A

C

-

TUBE SHEET OR

FLANGE

A TYPE 1 BUTT WELD FULL RT C WELD FULL OR PARTIAL PENETRATION CORNER JOINT E:1.00 SHELL CALCULATIONS (JOINT EFFICIENCY) E FOR C IS NOT ESTABLISHED BY CODE RULES A TYPE 1 BUTT WELD SPOT RT C WELD FULL OR PARTIAL PENETRATION CORNER JOINT E:O.85 SHELL CALCULATIONS (JOINT EFFICIENCY) E FOR C IS NOT ESTABLISHED BY CODE RULES A TYPE 1 BUTT WELD NO RT C WELD FULL OR PARTIAL PENETRATION CORNER JOINT E:O.70 SHELL CALCULATIONS (JOINT EFFICIENCY) E FOR C IS NOT ESTABLISHED BY CODE RULES

FIG. C.18


273

JOINT EFFICIENCY FACTORS SECTiON Viii, DIViSION 1

c

. El

~C

T 0

. A

TUBE SHEET OR

C.....

F!..ANGE

A ERW PIPE C TYPE 1 BUTT WELD SPOT RT E=1.00 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1 E=0.85 LONGITUDINAL STRESS CALCULATIONS, NOTE 2 A ERW PIPE C TYPE 1 BUTT WELD NO RT E=0.85 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1 E=0.70 LONGITUDINAL STRESS CALCULATIONS, NOTE 2 A ERW PIPE C TYPE 2 BUTT WELD NO RT E=0.B5 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1 E=0.65 LONGITUDINAL STRESS CALCULATIONS, NOTE 2 A ERW PIPE C FULL OR PARTIAL PENETRATION CORNER JOINT E=1.00 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1 E FOR C IS NOT ESTABLISHED BY CODE RULES, NOTE 2 NOTE 1: THE QUALITY FACTOR SHOWN IN THE SHELL CALCULATIONS IS IN ADDITION TO THE CATEGORY A JOINT FACTOR (E=O.B5) INCLUDED IN THE ALLOWABLE STRESS VALUE OBTAINED FROM THE APPLICABLE STRESS TABLE. NOTE 2: DIVIDE THE ALLOWABLE STRESS VALUE OBTAINED FROM THE APPLICABLE STRESS TABLE BY 0.B5 BEFORE APPLYING A JOINT EFFICIENCY IN THE LONGITUDINAL STRESS CALCULATIONS (eg. SEE NOTE 26, TABLE UCS-23).

FIG. C.19

274 Appendix C

EXAMPLE CALCULATION FOR JOINT EFFICIENCY FACTORS SECTION VIII, DIVISION 1

c

B

A D

B

DESIGN SPECIFICATIONS 175PSI AT 600F 60F AT 175PSI SHELL MATERIAL SA-414A SHELL DIAMETER 40" S=12,800 FROM SECTION II PART D-TABLE 1 CATEGORY A TYPE 1 NO RT E=0.70 FROM TABLE UW-12 COLUMN (c) ELLIPSOIDAL HEAD MATERIAL SA-285A HEAD SKIRT DIAMETER 40" S=12,300 FROM SECTION II PART D-TABLE 1 CATEGORY B TYPE 1 SPOT RT JOINT EFFICIENCY E=0.85 FROM FIG. UW-12 (d) QUALITY FACTOR E=1.0 FROM UW-12(d) NOZZLE MATERIAL SA-53A ERW NOZZLE DIAMETER 14" S=11,700 FROM SECTION II PART D-TABLE 1 CATEGORY B TYPE 1 NO RT JOINT EFFICIENCY E=0.85 FACTORED IN STRESS VALUE QUALITY FACTOR E=0.85 FROM UW-12 (d) & (e) NOZZLE FLANGE SA-105 300 LB SLIP ON B16.5

FIG. C.20.E

~_~

SEAMLESS ELLIPSOIDAL


CIRCUMFERENTIAL STRESS IN SHELL UG-27(c)(1) 1= PRISE-0.6P 1= 175 X20/12,800X 0.70-0.6 X 175 t =3,500/8855 1= 0.395" USE 0.500" GO TO FIG. UCS-66 SA-414A IS IN CURVE B 60F IS ABOVE CURVE - NO IMPACT TEST REQUIRED NOT NECESARY TO CALCULATE LONGITUDINAL STRESS IN SHELL CATEGORY B SPOT RT ·E=0.85 IF CALCULATED ELLIPSOIDAL HEAD UG-32(d) 1= PD/2SE-0.2P t= 175 x40/2 X 12,300X 1.0-0.2x 175 t 7,000/24,565 t=0.285 USE 0.375 GO TO FIG. UCS-66 SA-285A IS IN CURVE B 60F IS ABOVE CURVE· NO IMPACT TEST REQUIRED

=

CIRCUMFERENTIAL STRESS IN NOZZLE UG-27(c) t= PRISE-0.6P t= 175 x7/11 ,700 X 0.85-0.6 X 175 1= 1 ,225/9480 1=0.129° USE 0.210° GO TO FIG. UCS-66 SA-53A IS IN CURVE A 60F IS ABOVE CURVE· NO IMPACT TEST REQUIRED

FIG. C.20.E (CONTD.)

275

ApPENDIXD FLANGE CALCULATION SHEETS FLANGE CALCULATION SHEETS Blank fill-in calculation sheets are given for the following types of flanges: Sheet D.l - 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.S - Slip-on flange with full-face gasket Sheet D.6 - Welding neck flange with full-face gasket

1

2

DESIGN CONomONS

3

fA.CE

GA.SKET

N_ b G !.oIling MOhHial

4

COl',olion "110","01\0

LOAO AND BOLT CALCULATIONS

Duign Temp.. Sf.

W .. z

AIm. Temp., Si.

Hp

= 2brGmP =

Delign Temp., Sb

H

.....G'lrP/4 -

AIm. Temp., S.

W .. l

=

CONDITION

LOA.O

5 He -W.,

H-

Hr-H-Hp-

b.Cy

Hp

+

H

=

x

lEVER ARM ho -

SIC -

he -

.5(C

M.OMENT Mp

B)

Gl

hI - .S! ho

=

Me -

+ j'cl

Hpho -

Helle -

Mr =Hrf,r

=

M. Hc-W

he - .SIC -

6

Gl

M.

-

SHAPE CONSUNTS

J(

11 boll,poclI''i1 u.ceedl 2<;1

M. ond MiJ

7

=

A(B

+ to mulhply _ I &011 ,pacing ~" 20 +,

In I "quollOf'lI by

OPERATING

~-

1=

GR~TER

,== ~ ~:~

SEATING

t

f

= 'J ~

• . ._ _.....

Cc>mputed

0.·'

1

C!welted

Numb.,

)

FIG. 0.1 RING FLANGE WITH RING-TYPE GASKET

277

278 Appendix D

0.. 1

Desir

2

DESIGN CONDITIONS

"0.__,

FAa

GASKET

3

l'

N

•

r....p.toIVf.

0

flaftge MOI.rio!

,

lolling MOlerlol

•

w.o,ion

Flange

II ~.

4

Al1owon<.

Bolling

W.,

AI.... Temp., 51.

H.

Oelign T.mp~ 51>

H

Aim. remp., S.

W.,

CONDITION

5 OJnraJ.ing

8

S,. S" S,. s"

LOAD

. S" S,.

,.

S,.

+H LEVER

X

Ho

W.,

H

ho

.5(C

Ho -

h,

.sre + "1 + j'GI

Hc=W

R+91

=

he

~

.SIC -

Gl

-

M,

M.

~

"f'

-

gl/g.

Rodigl fig., 51

tJmG/A/1

h.

a

t-

I

F,

"

U h.g.1

V,


1 ~9O'

A " W

~g,"

-+

~

It + 1 = .4/3 te +

, ==

-r

x

tljd

,+,

(flo_KG/

••

I

«rt

KGI

II II

OPERATING SEATING.

+ /,

If boll fpO(i"O .XU.df 20 multiply 11I0 o"d 1II(l in ebeve equolionf by:

,~

1--8'

t"

-C'

I

b:4= '801t$

~

I

Ho

~R"

-

e=h,

-/8g.

7

== "'0Y/t":- zs• .sts, + SrI

::.,~" .SIS.'! + 5.] or

-l>

h/I>.

F,

U

Sealing

mol}"",)

IH

-

V,

tong. Hub, SH

1.

~

K and HUB FACTORS

V

STRESS CALCULATION

i

~

Z

::••;;.srs, + 5.} or.5 (SH + 5r)

Tong. fig., Sf

Hlh,

M,

,

lS,

I1php

Me - Hchc

r

Rodiol Fig., S~ - tim. /'io.P m. YIII

,1,\.0

6

m./)..gll

Tong. FI"., 5r _

MOMENT

~

01

STRESS CAlCULATION-operating long. Hub, Sit

'RM

-

hI)

9~' 1.5

w- .5(A... + At.}S.

GlrP/4 __ N,

::-:t W.J/S. or W .. ljSb

A.

....

_ 2bll"Gmf'_

~$I"I.4

All_.. b1. $I....

1.5

CALCULATIONS

brGy

No

Ht_H

SeQllnll

ecu

tOAD AND

Oesign remp_, Sf.

He

H, G' Comput.d

Oat.

C....ck.d

Numb..

FIG. D.2 SLlp·ON OR LAp·JOINT FLANGE WITH RING·TYPE GASKET

&olt ,pod"" 2Q

+'

Flange Calculation Sheets

1

2

DESIGN CONI)fTIONS

3

FAa

GASKET

D.Jion "e.wr., ,

N

D.tiOn r.llIJ:lerolure

b G

,

Flonoe Motorial

•

loltinO Material Cotlo,ior\ Allowonco

4

..

Dctsign romp., 5#.,

W.. ~ _ bll'Gy

AIm. Temp., S,.

H,

= 2bll'GmP =

H

_ Gl ll'PI 4

W.,

=H,.

h I' ,J

Flong,

Do,;on lohmg

Tf!mp~

51.

Atm. Tomp., S,

CONDITION

5

lOAD AND BOLT CALCULATIONS

lOAD

s..:.ol;n'il

H, _

1.5 S,.

hG -

.5(C -Gl

+ 0'

Long. Hub, Sa _ Im,/).,;,'

S,.

:;'~, ".S(SH

frno/}.Ql"

So.

Rodiol FIg., S.

PIno/At'

S,.

Tong. flg., S'l' _frio Y-/t"-ZS.

S"

:;":t .SIStI + S.l or ,S!SIl + 5,}

t

h'

t"

I

=

-

M.

-

Mo

=

-

f

-

, _ FII>,

F' 1 ~'-R'

.../ ~

~ :-ho

,-91<

=

yu h"g.l

_I.

=

4/3 I, + I

,

-r - alT = Il/d

,

-

+1

-'Y+~

= -

=

-

_ Mol B _ Operaling_

"

==

Mo/B

== 'ealing =

+

If bolt spacing eJl:eeeds 20 f, mllltiply m. and mo in above equalionl by:

I

I

;Bolfs

Hr

,I

Boll IpaeinQ 20

+I

.

HO

t~ +

-


, "~

me.

B'

d

.fBi.

7

-

=

-

V

gl/g.

...

-

hlh, F

-

,

noting

Mr_Hrhr

K AND HUB FACTORS

U

-n~' ..E'

-

z

STRESS CALCULATION

-

MG - HanG -

T

+ Sd or.s ISH + Sri -

!.orlg. Hub, Sa

+hcl

MOMENT

Mo - Hphp

K _ AlB

-

Radial Fig., Sa _

9 AI,_obl, $I, ...

= =

6

STRESS CALCULATION-Op'l'Otillg

S,.

1.5 S,.

11:+ .50'

hG = .51C - G)

13m../).,t" Tong. Fig., ST _m,Y/t" -IS._

S,.

_

=

lEVER ARM bQ _

h, -.S(R

=

Hc=W

8 .."o.....bI. 50,...

A~lS,

+H=

-

He;; -W.. l-H Hr _H

-

W _ .S{A.. +

-

X

fip.~ffB~PI.l~

Operating

.

A.. _ :;'.:" W",~/S. Of W .. l/S~_

-

G'

ComplIted

Dote

Checked

Numbe.

I-f<;

FIG. 0.3 WELDING NECK FLANGE WITH RING·TYPE GASKET

279

280 Appendix D

1

2

DESIGN CONDmONS

GASKET

fACE

DeliVR P'.,wre, ,.

H

0

O.sign Temp.raW••

G

.,

RCl""_ MOlorial

lolling Mot.rig!

4

COHo,ion Allowonc;.

5,.

il !j

FIORile

W., - bTCy

A. A. _

01uign Temp., S.

H

== G'TPjJ =

W _ .5{)...

Aim. Te..,p., S.

W., _ lip

CONDITION

LOAD Hp,..,.d J ,. /

O/Hroting

:;0;;'

_ 2bll"GmP_

A'm. remp., Sf.

tolling

5


H.

Design Temp.,

+

LEv ER ARM

hp,=·,5(c:.+9'

Ha_W.. 1 H_

0,

Hr_H_Hp_

hT

:::::

.5(C

GJ

.5(C -

B

8 Al'o-ob!
t

G)

MOMENT

~

BI_

211~

=

MIJ _ Hpllo -

-

M,

STRESS CAlCUlATION-operating tong. Hub, 5/1 _fmol>'QI 1

5,.

Rodicll fjg., S.

5" 5,.

Tong. Fig., S-r_ m.YRIt' -ZSR(O.67Ie+I}/f3_

.lJ1....

1,5 St.

=

Tang_ fIg, 5T (AT 8') = ~;

"to.

2K' (1 (K'

[ V

STRESS CALCULAnON

+ ';. tel

Radial Fig., SR

f3ma/At'

S,. S,.

Tong. Fig., 5T

maYl
I) A

~

a.

S"

Tong. Flg.• $or(AT8)=_ y

U,

(K'

"

H

1) A

"f

W

~"o{-h(;j

I

t=

~"'~

7

~

1<---",

U~ h

1_ '-,

~

f3_~/3Ie±l

-

'Y _ alTl<

-

0= ,'/d

~

•

).=1+6 m. = M.!S'

= Mo/S'

--

~

~

~

e= n

HT

,.---B =

I0c +."

V 09"

-

a=te± 1

mo

d


, G=

I

~

'" l-

h'

;Bolfs

--

,,'"

i

go

+ v. Ie) ]

J

g,/90

h, 2K' (1

'J

F/n.

a,V a, U

V-

' " r:

,

.

(Z +11) T (Z _II) aft

"

lSI< (O.67te+ 1) 1{3

-

V

1~+3(K±ll(l K' r,ry

Sealing

-

hfh.

,

U

~~oft •.5{S" + SR)er .5(51/ + 5"';

->

-

-

,

-

Long. Hvb, SF! _ fma/A91"

5,.

t

K AND HUB FACTORS

Z

]

=

""

K _ A(B'

$or) -

-

jM·1

~

T

fjm./).,1

~~e:t .5 (5H + S1I) Ot.5 (SH +

$0,...

G)

Ho'< -

Mr - Hrhr

6

S,.

5"

9

Ihe = .5(C -

=

HG=W

W .. ,JS~_

+ AJ,)S. =

~~f ABSOLUTE VALUEJMollN All Sl/8S~~U{NT CALCULATIONS Sealinll

0<

H_

X

.( ".,.

W.l/S.

A=

Comp",tad

Oola

C"'~kad

l'l",mbat

FIG. D.4 REVERSE WELDING NECK FLANGE WITH RING·TYPE GASKET

Flange Calculation Sheets

I

2

DESIGN CONDITIONS

3

GASKET

0.1li9R ", ••w,e. ,.

G

c

O••igR rempe,alu,.

,

tc

.

flange Mat.riol

4

I &oItingMaterial

.

flonQe

:i:,

""

kohing

o. ..;n Temp., Sf.

H, _

+ H'n

b?TGy

A. g,eol., of W.,jS. a' W.,/S. A._

~b'1l'"GmP_

+ A.}S._

Aim. Temp., Sf•

HI, _ .o/.to H,_

W_.5(A..

Oe.ign Temp.. 5"

H _ G""IrP/,,_

H'o:t _ .0/.'0 b...Gy _

Aim. Temp.• S.

WN1_ H+ Hp

CONDITION

LOAD

5

Ho _ ..., ,21'/4_

Operating

H,

H

+ H'p_ LEVER

X

H, _

II, _ SIR

ARM

=

=

"ll-Il+"l

+ g. + j.cl

MOMENT

Mo - Hpho -

-

M,

"rl"

M. LEVER ARMS

hI>=

REVERSE MOMENT

8

+

q-

no .- ----

h" _

h'lI

1I1.. /).,;g,"

Radial FIg., S. _

Pm./).,,"

_

6 r

Ton;. Fig., ST-m"Y/t'-ZS.-

Z

.S(5H + 5.) or.5 ISH

+ 5,) -

Y

Sf.

.""

511""

t'('1I"C

"Ih.

-

-

f,

-

-

Y,

h. _

Vag.

i

~Eo

~+-hO

ho

~

~

i .. -

•~HG

+

,-

=

,

-aIT

=

"'" tl/d

-

X _"(+6 _ Moll

-

4/3 Ie

m~

=

=

+ t, mllitiply ~ Salt ,pocing ~

."

Co --.--~~

A

"'-h,

H.tH G

"'"

9,'

!

-+ h"G ...-h G ~$

+I

Il bolt $padng ~llc.ed$ 20 m. ill oOOn .qllOtiOM by:

W

'0

L

Bo

f4-R'

t-

f--

1

_Ie

~

,

== g,Y, ""g.l

=

STRESS fORMULA fACTORS

, -

•

d

-

7

I+-k

e==h:

=

Illig.

nd,)

= f,

U

RADIAL STRESS AT 80LT CIRCLE

::::

K AND HUI fACTORS

AI'

X

=

:;.,,~

-

M1J = HlIho/ l

+ h'lI

STRESS CALCUlATION-Dperating Long. Hub, III

-

, (A-C)(2"+C)= hlr= 6(C+A)

hlr

!oI. ...

S,. S,. S,. S,.

8Xle + 6(8 C)

HI>=W-H=

"'''O'o'obl.

1.5

(e

Ill!"

LOAD AND IOLT CALCULATIONS

W.,

Co<,OI;on AHowonc;e

,..

T

G'

Compvted

Date

Ch.ecked

N",,,,b.'

FIG. 0.5 SLIP-ON FLANGE WITH FULL·FACE GASKET

281

282 Appendix D

1

2

DESIGN CONDITIONS

3

GASKET

G b

De.;o" Pt•• wt., ,. D.Ii"" Te....p .rolu'.

4

MOI.riol

:1;

Aim. Temp.,

+

H'(lT _

H, _ 2b'lTGmP

Deli;" Temp., S,.. flange

LOAD AND BOU CALCULA nONS

W .... _ b'ITGy

Couo,ion Allowonu

2hq 8)/..

•

flon". MoleriQI ~linQ

C (e

S,.

W_.5(A.. +A..)S._

l!1-,----t:----:-:~.:.::_t_--+:_::==~=-----h____=~_;'__;=:_--------1 H'or _ ·"I.. brCy

:G

Celign Temp., 5"

"

Aim. Temp., S"

CONDITION

X

LOAD

LEVER ARM

MOMENT

Htl,.,....r!JPIl, ..,ho.,..,.R+QI 5 Operating I-'=-=..::=-.c:=--=----------f''-':..::,'-'''---:--:-=-------F=--::=-=----------I Hr-H Ho_ hr-.S(R+",+hGIMr - tit"r -

LEVER

ARMS REVERse MOMENT

h _(C-BX 28±qQ6{8+C)

h'G:::::(A-C)(2A+Cl6(C+A)

Ho=W-H:::::

6

STRESS CAlCULAT10N-operating 1.5 Sf.

m.. /All'·

s,.

Radial fig., Sll

pm./Xt"

s.,

Tong. Fig., ST-m"Y/r'-ZSJ;-

s.,

:; ...:.; .5(5/1

Sf.

+ 5.1 or .5 ISH + SrI

-

RADIAl STRESS AT sou CIRClE SUl>_ 6Mo = "('lTC /'Id,)

f(

AND HUB FACTORS

-

h/h"

-

F,

z

-

v,

v u

-

J;1I!l1o

-

J( _

long. Hub, Stl -

A/a

F,

• =;;:-

~

ho = .,Ilgo =

7

STRESS FORMUlA FACTORS

, -

a

_1.+1

13 _ l'

.4/3 Ie

=

+

I _

= a.IT

If bolt Ipacing ."teeedl 20 m. in abo.... equationl by:

+ " multiply

1k>!t ,padn" 20 +,

t,

c= Compul.d

_

Cn.ck.d

FIG. 0.6 WELDING NECK FLANGE WITH FULL·FACE GASKET

0019'

Numb.'

_

ApPENDIXE CONVERSION FACTORS CONVERSION OF U.S. CUSTOMARY UNITS TO SI UNITS! Multiple U.S. Customary Units

inches feet square iuches (in.') square feet (ft") U.S. gallons cubic feet (ft.'} pounds, mass pouuds, force psi, pressure bars Btu ksi-x/in., fracture toughness

OF

By Factor 0.0254 0.3048 0.0006452 0.092903 0.003785 0.02832 0.4536 4.448 6,894.8 100,000. 1,005.056 1.099 X 10' t, = (t, - 32)/1.8

'Por other conversions, see ASTM E 380.

283

To Get SI Units

meters meters m-

m' m' m' kilograms Newtons Pascals Pascals Joules Pa-x/rn

°C

INDEX Categories, welded joint, 5, 6 Charpy impact-test requirements, 16

A A value, 43-44 Allowable stress, 2-3, 5, 6 ASME Boiler and Pressure Vessel Code, Section. VIII, v,seealso Section VIII Assignment of materials to CUI\!es, 10 Axial buckling stress, critical, 39 Axial compression, cylindrical shells under, 36-42

Chrome-moly steel forging specifications SA-182,253 SA-336,254 Chrbrile~m()lysteelplatespecificatibns,

253 Circular flat plate, 101 Circumferential membrane stress, 28 Circumferential stress, 34, 70 Closure design details, 175-178 Cold temperatures, 24 Compensation, inherent, openings with, 136 Component analysis in VIII-2, 233-248 Compression axial, cylindrical shells under, 36-42 cone-to-shell junction at large end of cone in, 75-76

B B value, 43-44 Beams on Elastic Foundation, 134 Bellows-type expansion joints, 230, 231 Bending moment, 41 Bending stress, 41-42 ligament efficiency for multi-diameter openings for, 193-197 Blind flanges, 106-107 Boiler and Pressure Vessel Code, Section VIII, ASME, v, see also Section VIII Bolt loads, 106 Bolt sizing, 106 Bolted flanges, 105-106 connections with ring type gaskets, 108-124 Bolted flat head, 12 Bolted flat plates and covers, 105-106 Bolting, flat plates and covers with, 106-107 Bolting rings, 124 Braced and stayed construction, 169-173 Braced and stayed surfaces, 169-172 Brittle fracture, 9-13 Buckling, of cylindrical shells, 42-43 Buckling equation, 38 Buckling stress, critical axial, 39 Butt joints, 6 Butt welded, 5, 7 Butt welded components, II

cone-to-shell junction at small end of cone in, 86-87 Compressive stress, 70 Cone-to-cylinder junction large end of inherent reinforcement for, 96 values of Q for, 96 small end of inherent reinforcement for, 98 values of Q for, 99 Cone-to-shell junction large, 80-81, 89-92

at large end of cone in compression, 75-76 in tension, 85-86 small, 81-82, 92-94 at small end of cone in compression, 86-87 in tension, 76-77 Conical sections external pressure on, 85-94 internal pressure on, 74-85 VIII-l,74-94 VIII-2, 95-99 Conical shells, 74 Conversion factors, 283 Comer joint, 12 Corner welded, 5, 7 Corrosion allowance, 31, 32

c C value, 102 Carbon steel plate specifications, 253 Cast Ductile Iron, rules of Part VCD for, 23-24 Cast Iron, rules of Part VCI for, 23 285

286 INDEX Covers flat, see Flat plates and covers spherically dished, 124-131 Creep, 2 Creep rate, 3 Critical axial buckling stress, 39 Critical strain, lowest, 43

Crown radius of ellipsoidal heads, 67 Curves, assignment of materials to, 10 Cylinders effective length of, 44 elliptical, 56 Cylindrical shells, 27_56 under axial compression, 36-42 buckling of, 42-43 equations, VIII-2, 53-54 external pressure on, 42-53 hoop stress in, 29 lines of support of, under external pressure, 45 mitered, 54-55 openings in, 137 under tensile forces, 27-36 thick, 33-36, 53 thin, 27-33, 53

D Design rules, 1 Design temperatures, 2, 4 Dimpled and embossed assemblies, welded stays for, 170-172 Double full fillet lap joint, 6

E E (Joint Efficiency Factors), 3-6, 255-275

Earthquake forces, 27 Edge moment, 101 Effective length of cylinders, 44 EJMA (Expansion Joint Manufacturers Association) Standard, 201 Elastic foundation theory, 134, 213 Elasticity, modulus of, 38, 206 Ellipsoidal heads crown radius of, 67 pressure on concave side of, 65-66 pressure on convex side of, 67-68 VIII-l,65-68 VIII-2,72-74 Elliptical cylinders, 56 Elliptical shells, 55-56 Empirical equations, 47 Enameled vessels, 23 EPC (External Pressure Charts), 37-38 Excess area, 77

Expansion Joint Manufacturers Association (EJMA) Standard, 201 Expansion joints, 230, 231 External pressure on conical sections, 85-94 on cylindrical shells, 42-53 on ellipsoidal heads, 67 on torispherical heads, 71 lines of support of cylindrical shells under, 45 reinforced opening design for, 134 in spherical shells, 61-64 External Pressure Charts (EPC), 37~38

F F-factor, 137, 138 Factor 1.1, 2 Fatigue, stress concentration factors used in, 239 Fatigue curves, 20, 246 Fatigue evaluation, 245-248 Fatigue requirements, 19-22 F&D heads (Flanged and Dished heads), 68 Figures, list of, xiii-xv Fillet welded,S, 7 Fixed tubesheets, 213-230 design equations for, 217-225 details for, 214, 215 Fixity factor, 209 Flange calculation sheets, 277-282 Flange connections, bolted, with ring type gaskets, 108-124 Flange rings, 124 Flanged and Dished heads (F&D heads), 68 Flanged and flued expansion joints, 230, 231 Flanges, 12, 77 blind, 106-107 bolted, see Bolted flanges flat-face, 124 full-face gasket, 123 integral, 109, 110, 125, 126 lap-joint, see Lap-joint flanges loose, 109, 125 optional, 109, 125 reverse, 118-123 ring, 109 slip-on, see Slip-on flanges standard, 109-118 welding neck, see Welding neck flanges Flat-face flanges, 124

287

Flat head bolted, 12 integral, 12 Flat plates and covers, 101-108 bolted, 105-106 with bolting, 106-107 circular, with bolting, 106-107 circular integral, 101-104 integral, 101-105 multiple openings in rims of, 109 noncircular, with bolting, 107 noncircular integral, 104-105 openings in, 107-108 unstayed, 103 Flues, 77 Forces

earthquake, 27 meridional, 64 tensile, see Tensile forces Fracture, brittle, 9-13 Full-face gasket slip-on flange with, 281 welding neck flange with, 282 Full-face gasket flanges, 123

G Gasket crushout, check for, 106 Gasket design requirements, 105 Gaskets full face, see Full-face gasket ring type, see Ring-type gasket Geometric parameters, 217 Glass-lined vessels, 23 Group numbers, 2

H Half-pipe jackets, 181-186 maximum allowable internal pressure in, 181-182 minimum thickness of, 182-184 Head configurations, 57 Heads conical, 74 ellipsoidal, see Ellipsoidal heads hemispherical, see Hemispherical heads spherically dished, 124 tori spherical, see Torispherical heads toriconical, 77 Heat exchangers configurations of, 202 design of, 201-231 design rules for components of, 201 example, 8

tubesheet, see Tubesheets If-tube, see Uetube exchangers Hemispherical heads, 57 pressure on concave side of, 57-61 pressure on convex side of, 61-64 thickness, 60-61, 63-64 VIII-I,57-64 VIlI-2, 64-65 High alloy steels, without impact testing, minimum design metal temperatures in, 18 Hillside position, 146

Hoop stress, 4 basic equation for, 55 in cylindrical shells, 29 Hydrostatic, term, 22 Hydrostatic test, 22 for VIII-I, 23 for VIII-2, 24-25

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, 87-88 Inherent compensation, openings with, 136 Inside radius, 58 Integral flanges, 109, 110, 125 Integral flat head, 12 Internal pressure, 27 on conical sections, 74-85 maximum allowable, in half-pipe jackets, 181-182 reinforced opening design for, 134 in cylindrical shells, 27 in spherical shells, 57-61 in eliiptical shells, 55 in torispherical shells, 68

J Jacket closure bars, 4 Jacket penetrations, 179-181 Jacketed vessels, 173-181 design of closure member for, 175-179 openings in, 179 types of, 174 welded stays for, 170-171 Jackets half-pipe, see Half-pipe jackets

288 INDEX

spiral, 182 Joint categories, welded, 5, 6 Joint Efficiency Factors (E), 3, 4, 6, 255-275

Mitered cylindrical shells, 54-55 Modulus of elasticity, 38, 206 Moment of inertia, calculating, 87-88

N K K factor, 181-186 Knuckle, thickness required for, 77 Knuckle radius, 65

L l\ factor, 207 ~208 Lame's equation, 33 Lap-joint flanges, 109 with ring-type gasket, 278 Layered vessels, 30-31 Ligament efficiency for constant-diameter openings, 192 for multi-diameter openings for bending stress, 193-197 for membrane stresses, 192-193 Ligament efficiency method, 133 Ligament efficiency rules, VIII-I, 164-167 Loads, 1 bolt, 106 radial, 52 shear, 50 vacuum, 47 wind,27 Longitudinal stress, 34 Loose flanges, 109, 125

M Mandatory rules, 1 Material designation, 253-254 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, 28 ligament efficiency for multi-diameter openings for, 192-193 in spherical shells, 58 Meridional forces, 64 Meridional stress, 70 Minimum Design Metal Temperature (MDMT),11 in high alloy steels without impact testing, 18 reduction of, without impact testing, 17

Non-Mandatory rules, 1 Noncircular cross section, vessels of, 186-199 Noncircular flat plates, 104, 107 Nozzle connections, 54 Nozzle design, alternative rules for, 157-164 Nozzle nomenclature, 154 Nozzle reinforcement, 140-151 of series of openings, 165-165

o Obround cross section, vessels of, 183, 191 Openings, 133-168 code bases for acceptability of, 133 constant-diameter, ligament efficiency for, 192 in cylindrical shells, 137 exceeding size limits, 151-153 in flat plates and covers, 107-108 with inherent compensation, 136 in jacketed vessels, 179 multi-diameter, ligament efficiency for, see Ligament efficiency for multidiameter openings multiple, in rims of flat heads or covers, 109 nozzle reinforcement of series of, 165-166 reinforced, see Reinforced openings in spherical shells, 139 terms and definitions for, 134 in vessels of noncircular cross section, 187,192-197 Optional flanges, 109 Outside radius, 58

p P numbers, 2 Part Dcn for Cast Ductile Iron, rules of, 23-24 Part DCI for Cast Iron, rules of, 23 Part DLT rules, 24 Peak stress, 133, 235 Penetrations, jacket, 179-181 Plate diameter opening diameter does not exceed half, 107-108

289

opening diameter exceeds half, 108 Plates, flat, see Flat plates and covers Plates on elastic foundation, 213 Pneumatic, term, 22 Pneumatic test, 22 for VIII-\' 23 for VIII-2, 24 Poisson's ratio. effective, 206 Pressure on concave side of ellipsoidal heads, 65-66 of hemispherical heads, 57-61 of torispherical heads, 68-71 on convex side of ellipsoidal heads, 67-68 of hemispherical heads, 61-64 of torispherical heads, 71-72 external, see External pressure internal, see Internal pressure maximum, for thickness, 27 Pressure-area procedure, 77 Pressure boundary, 133 Pressure test requirements for VIII-I, 23-24 for VIII-2, 24-25 Pressure testing, 22-25 Pressure vessels, 22 Primary stress, 133, 233, 234 Process cyclic curves, 247

Q Q factor, 95 Quality factor, 255, 259-266, 271, 273 Quality Factors, 3 Quench and tempered carbon and alloy steel forging specifications, 254

R Radial loads, 52 Radius-to-depth ratio, 65 Reactor, example, 14 Rectangular cross section, vessels of, 188-190, 197-199 Reducers, 54 Reduction factor, 209 References, 249-250 Reinforced openings area of reinforcement available for, 140-151 area of reinforcement required for, 137-139 general requirements, 134-136 limits of reinforcement for, 140 rules

VIII-l,136-153 VIII-2, 153-164 shape and size of, 137 Reinforcement, nozzle, see Nozzle reinforcement Reinforcement limits, 134-136 Reinforcement plate, welded connection with,l1 Reverse flanges, 118-123 Reverse welding neck flange with ringtype gasket, 280 Ring flanges, 109 with ring-type gasket, 277 Ring girders, 4 Ring-type gaskets bolted flange connections with, 108-124 reverse welding neck flange with, 280 ring flange with, 277 slip-on or lap-joint flange with, 278 welding neck flanges with, 279 Rings, stiffening, see Stiffening rings Rupture, 2, 3

s Secondary stress, 233 Section VIII background information, 1-25 Divisions 1 and 2, 1 Section VIII-1, pressure test requirements for, 23-24 Section VIIl-2, pressure test requirements for, 24-25 Sections, conical, see Conical sections Shear loads, 50 Shearing stress, 52 Shells conical, see Conical sections cylindrical, see Cylindrical shells elliptical, 55-56 spherical, see Spherical shells 51 units, conversion of U.S. customary units to, 283 Single full fillet lap joint, 6 Single-welded butt joints, 6 Slip-on flanges, 109 with full-face gasket, 281 with ring-type gasket, 278 Special components, VIII-I, 169-199 Spherical radius, 65 Spher-ical shells external pressure in, 61-64 internal pressure in, 57~61 large, 58 membrane stress in, 58 openings in, 139

290 L'lDEX

thick,58 VIII-I, 57-64 VIII-2, 64-65 Spherically dished covers, 124-131 Spiral jackets, 182 Standard flanges, 109-118 Stays and staybolts. 172-173 welded, see Welded stays Stiffening rings, 48-50 attachment of, 50-53 designing, 48-50 Strain, 37 critical, lowest, 43 Stress allowable, 2-5 bending, 41-42 circumferential, 34, 70 classification of, 237-238 combinations of, 240-245 compressive, 70 critical axial buckling, 39 hoop, see Hoop stress longitudinal, 34 membrane, see Membrane stress meridional, 70 peak, 133, 235 primary, 133,233,234 secondary, 233 shearing, 52 tensile, 2 thermal, 233, 235 Stress categories, 233-239 and limits, 236 Stress concentration, 239 Stress multipliers, 3 Stress rupture, 3 Stress-strain diagrams, 37-38 Structural discontinuity, 234

T Tables, list of, xvii TEMA (Tubular Exchanger Manufacturers Association) standards, 201 Temperatures cold,24 design, 2, 4 metal, minimum design, see Minimum Design Metal Temperature Tensile forces, cylindrical shells under, 27-36 Tensile strength, 3 Tensile stress, 2

Tension cone-to-shell junction at large end of cone in, 85-86 cone-to-shell junction at small end of cone in, 76-77 Thermal stress, 233, 235 Thickness details governing, used for toughness, 11-13 hemispherical heads, 60-61, 63-64 maximum pressure for, 27 minimum, of half-pipe jackets, 182-186 required for knuckle, 77 Threaded-end stay construction, special limitations for, 170 Threaded-end stays, special limitations for, 172 Torispherical heads pressure on concave side of, 68-71 pressure on convex side of, 71-72 shallow, 68 VIII-I, 68-72 VIIl-2,72-74 Toughness, governing thickness details used for, 11-13 Toughness rules, 13 Transition sections, 54, 57 conical, see Conical sections Tube patterns, 206 Tubesheet attachments, 4 Tubesheet design rules for, 201 in U-tube exchangers, 201-213 Tubesheets, 12 fixed, see Fixed tubesheets integral, design equations for, 209-213 simply supported, design equations for, 207 types of, 201 Tubular Exchanger Manufacturers Association (TEMA) standards, 201

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, 283

v Vacuum loads, 47 Venting, 23

291 Vessels jacketed, see Jacketed vessels layered, 30-31 of noncircular cross section, 186-199 of obround cross section, 187-191 of rectangular cross section, 188-190, 197-199 VIII-l requirements, guide to, 252

w Weld efficiencies, 3 Welded attachments, 13 Welded connection with reinforcement plate, 11 Welded-in stay construction, special limitations for, 170, 171

Welded joint categories, 5, 6 Welded stays for dimpled and embossed assemblies, 170-172 for jacketed vessels, 170, 171 Welding neck flanges, 114, 115 witb full-face gasket, 281 with ring-type gasket, 279 Wind loads, 27 Wind moment, 40

y Yield strength, 3 interpolation between, 16

z Z factor, 104

Guidebook ASME Section VIII.pdf

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