Process Equipment

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STRUCTURALANALYSIS AND DESIGN

OF PROCESS EQUIPMENT

Mqon H. JowodNooter CorporationSt. Louis, M issouri

Jomes R. FqrrBabcock & Wilco.r Company

Barberton, Ohio

A Wiley-lnterscience Publicqtion

JOHN WILEY & SONS

Brisbone Toronto Singopore

Copyright O 1984 by hhn Wilev & Sons, Inc

All righis reserve{]. Publishcd simultaneously in Canada

Reproduction or transiation ()f any part oi this work

hcyond that permitted by Secton 107 or 108 of ihe

It)?6 linited States Copyrighl Act wrthout lhe permrssron

,,1 rlr .i't)\rfi!hl owner is unl.rwlul Requests iot

| ,"' ,1,,, !,, lrrrhcr infomati,)n sbould be addrcssed lo

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To Our Wives,

Dixie and Barbara

PREFACE

We wrote this book to serve three purposes. The first purpose is to providestructural and mechanical engineers associated with the petrochemical industrya reference book for the analysis and design of process equipment. The secondis to give graduate engineering students a concise introduction to the theory ofplates and shells and its industrial applications, The third purpose is to aidprocess engineers in understanding the background of some of the design equa-tions in the ASME Boiler and hessure Vessel Code. Section VIII.

The topics presented are separated into four parts. Part 1 is intended tofamiliarize the designer with some of the common "tools of the hade." ChapterI details the history ofpressure vessels and various applicable codes from aroundthe world. Chapter 2 discusses design specifications furnished in purchasingprocess equipment as well as in various applicable codes. Chapter 3 establishesthe strength criteria used in different codes and the theoretical backgroundneeded in developing design equations in subsequent chapters. Chapter 4 in-cludes different materials of construction and toughness considerations.

Part 2 is divided into three chapters outlining the basic theory of plates andshells. Chapter 5 develops the membrane and bending theories of cylindricalshells. Chapter 6 discusses various approximate theories for analyzing heads andtransition sections, and Chapter 7 derives the equations for circular and rectan-gular plates subjected to various loading and support conditions. These threechapters form the basis from which most of the design equations are derived inthe other chapters.

Part 3, which consists of flve chapters, details the design and analysis ofcomponents. Chapters 8 and 9 derive the design equations established by theASME Code, VI[-l and -2, for cylindrical shells as well as heads and transitionsections. Chapter 10 discusses gaskets, bolts, and flange design. Chapter llpresents openings and their reinforcement; Chapter l2 develops design equationstor support systems.

Part 4 outlines the design and analysisof some specialized process equipment.Chapter 13 describes the design of flat bottom tanks; Chapter 14 derives the

ftitAct

cquations for analyzing hest transfer equipment. Chapter l5 describes the theory

of thick cylindrical shells in high-pressure applications. Chapter l6 discusses the

stress analysis of tall vessels. Chapter 17 outlines the procedure of the ASMECode, VI[-l, for designing rectangular presswe vessels.

To simplify the use of this book as a reference, each chapter is written so that

it stands on its own as much as possible. Thus, each chapter with design or other

mathematical equations is written using terminology frequently used in industryfor that particular type of equipment or component discussed in the pertinent

chapter. Accordingly, a summary of nomenclature appears at the end of most ofthe chapters in which mathematical expressions are given.

In using this book as a textbook for plates and shells, Chapters 3, 5,6 md7form the basis for establishing the basic theory. Instructors can select otherchapters to supplement the theory according to the background and needs of thegraduate engineer.

In deriving the background of some of the equations given in the ASMEBoiler and Pressure Vessel Code, attention was focused on Section VIII, Di-visions 1 and 2. Although these same equations do occur in other sections of the

ASME Code, such as the Power and Heating Boilers, no consideration is given

in this book regarding other sections unless specifically stated'

MAAN JAWAD

JAMES FARRSaint Louit, MissouriBarberton, OhioSeptember 1983

ACKNOWLEDGMENTS

We are indebted to many people and organizations for their help in preparing this

book. A special thanks is given to the Nooter Corporation for generous support

rluring the preparation of the manuscript. Also a special thanks is given to the

American Society of Mechanical Engineers for supplying many of the illustra-tions used in this book and also to the American Petroleum Institute and the

Tubular Exchangers Manufacturers Association.We also give thanks to Messrs. W. D. Doty, G. Hays, G. G. Karcher, T. W.

[,odes, H. S. Olinger, and R. F. O'Neill for reviewing the manuscript, and to

Mr. W. H. Schawacker for supplying many of the photographs.

We would also like to extend our appreciation to Mrs' Y. Batteast for typingportions of the manuscript.

M. J.

PART I

Chopter Il.l1.2

CONTENTS

BACKGROUND AND BASIC

CONSIDERATIONS

Hisiory ond Orgonizotion of Codes

Use of Process Vessels and Equipment

History of Pressure Vessel Codes in the United

States

Organization of the ASME Boiler and Pressure

Vessel Code

Organization of the ANSI B31 Code for Pressure

Piping

Some Other Pressure Vessel Codes and Standards

in tie United States

Worldwide Pressure Vessel Codes

ReferencesBibliograPhY

3

4

l314

14

l5l5t616

1.3

1.4

1.5

1.6

8

9'r0

ll

Chopter 2 Selection of Vessel, Specificotions'

Reports, ond Allowoble Slresses

Selection of Vessel

Which Pressure Vessel Code Is Used

Design Specifications and Purchase Orders

Special Design Requlrements

Design RePons and Calculatjons

Materials' SPecifi cations

2.1

2.22.32.42.52.6

CONTINT!

2.72.82.92.10

2.11

2.12

Chopter 3

Dcsign Data tbr Ncw MaterialsFactors of Safety

Allowable Tensile Stresses in the ASME Code

Allowable Extemal Pressure Stress and AxialCompressive Stress in the ASME Boiler and Pres-sure Vessel Code

Allowable Stresses in the ASME Code for PressurePiping B31

Allowable Stress in Other Codes of the WorldReferences

Strength Theories, Design Criierio, ondDesign Equotions

Strength Theories

Design CriteriaDesign Equations

Stress-Strain Relationships

Strain-Defl ection Equations

Force-Stress Expressions

References

Bibliography

Moteriqls of Construction

Material Selection

4,l.l Corrosion

4.1.2 Strength

4. 1 .3 Material Cost

Nonferrous Alloys4.2.1 Aluminum Alloys

't7

17

t7

l9

22

22

26

3.1

3.23.33.43.53.6

29

303l33

33

35

39

42

43

45

4646495253

53

3J5656606l63

68

Chopter 4

4.1

4.2

4.2.2 Copper and Copper Alloys4.2.3 Nickel and High-Nickel Alloys4.2,4 Titanfum and Zirconium Alloys

4.3 Ferrous Alloys4.4 Heat Treating of Steels

4.5 Brittle Fracture

4.5. I ASME Presssure Vessel Criteria

4.64.7

4.5.2 'l'heory ol' Brittle Fracture

4.5.3 Hydrostatic Testing

4.5.4 Factors Influencing Brittle Fracture

Hydrogen Embrittlement

Nonmetallic Vessels

References

Bibliography

ANAIYSIS OF COMPONENTS

Slress in Cylindricol Shells

Ends5.3.3 Pressure on Ends OnlyThermal Stress

5.4.1 Uniform Change in Temperature

5.4.2 Gradient in Axial Direchon5.4.3 Gradient in Radial DirectionNomenclature

References

Bibliography

CONTENTS xlll

7074

75

76

7778

79

8l

83

116lr8119

124127

r30137

r38139

PART 2

Chopfer 5

5.1

5.2

5.3

5.4

Stress Due to Intemal Pressure 84Discontinuity Analysis 925.2.1 Long Cylinders 965.2.2 Short Cylinders lO7Buckling of Cylindrical Shells I 145.3.1 Uniform Pressure Applied to Sides Only 1145.3.2 Uniform Pressure Applied to Sides and

Chopter 6 Anolysis of Formed Heods ond TronsitionSections

6. I Hemispherical Heads

6.1 . I Various Loading Conditions6.1.2 Discontinuity Analysis

6.1.3 Thermal Stress

6.1.4 Buckling Strength

141

142146r52158

159

xiv CONTENTS

6.26.36.4

Chopter 7

7.1

7.27.37.4

PART 3

Ellipsoidal Heads

Torispherical Heads

Conical Heads

6.4.1 Unbalanced Forces at Cone{o-CylinderJunction

6.4.2 Discontinuity Analysis

6.4.3 Cones Under Extemal Pressure

Nomenclature

References

Bibliography

Stress in Flot Plotes

Introduction

Circular Plates

Rectangular Plates

Circular Plates on Elastic Foundation

Nomenclature

References

Bibliography

DESIGN OF COMPONENTS

163

167

r68

169

172

175178'r80

t8t

183

184

184

193

197

200201

201

203

20520620821822623r

235238240240241

Chopter 8 Design of Cylindricol Shells

8.1 ASME Design Equations

8.2 Evaluation of Discontinuity Stresses

8.3 ASME hocedure for Extemal Pressure Design

8.4 Design of Stiffening Rings

8.5 Allowable Gaps in Stiffening Rings

8.6 Out-of-Roundness of Cylindrical Shells underExternal Pressure

8.7 Design for Axial Compression

Nomenclature

References

Bibliography

Chopier 9 Design of Formed Heods ond TronsifionSeclions

Introduction

ASME Equations for Hemispherical Head

Design

ASME Design Equations for Ellipsoidal and

Flanged and Dished Heads

9.3.1 Ellipsoidal and Torispherical Heads

Analysis due to Intemal Pressure

9.4.2 Conical Shells under External Pressure

9.4.3 ASME Simplification of DiscontinuityAnalysis due to External Pressure

Nomenclature

References

Bibliography

CONTENTS xv

243244

247

249

25626r

261

265266267

9.1

9.2

9.3

under External Pressure 255

9.4 ASME Equations for Conical Head Design 256

9.4.1 ASME Simplification of Discontinuity

Chopter l0l0.lro.2

Bfind Flonges, Cover Ploles, ond Flonges 269

Introduction 270

Circular Flat Plates and Heads with UniformLoading

ASME Code Formula for Circular Flat Heads

and Coversr0.3

10,4 Comparison of Theory and ASME Code Formulafor Circular Flat Heads and Covers withoutBolting

10,5 Bolted Flanged Connections

10.6 Contact Facings

1O.7 Gaskets

10.7.1 Rubber O-Rings

10.7.2 Metallic O- and C-Rings

10.7.3 Compressed Asbestos Gaskets

10.7.4 Flat Metal Gaskets

10.7.5 Spiral-Wound Gaskets

274

276

278278279281

281

281

282283285

CONTENTS

1O.7.6 Jacketed Gaskets

10.7.7 Metal Ring Gaskets

10.7.8 High-Pressure Gaskets

10.7.9 Lens Ring Gaskets'10.7. I0 Delta Gaskets

10.7.1I Double-Cone Gaskets

I0.7. l2 Gasket Design

10.8 Bolting Design

10.9 Blind Flanges

10. 10 Bolted Flanged Connections with Ring-TypeGaskets

l0.l I Reverse Flanges

10. l2 Full-Face Gasket Flange

10. l3 Flange Calculation Sheets

10, l4 FlatFace Flange with Metal-to-Metal ContactOutside of the Bolt Circle

10.15 Spherically Dished Covers

Nomenclature

References

Bibliography

285285285286287288290292

294

298307310317

317324330332332

335

336338

343346

349

359

368

379

383

Chopter I I Openings, Nozzles, ond Externol[oodings

General

Stresses and Loadings at Openings

Theory of Reinforced Openings

Reinforcement LimitsI I .4. I Reinforcement Rules for ASME.

Section II I .4.2 Reinforcement Rules for ASME,

Section VIII, Division Il l.4.3 Reinforcement Rules for ASME,

Section VIII, Division 2

I I .4.4 Reinforcement Rules for ANSUASME831. I

I L4.5 Reinforcement Rules for ANSI/ASME83 t.3

ll.lI 1.2

I 1.3'| 1.4

I I.5I 1.6

1t.7

CONTENTS xvii

Ligament Efficiency of Openings in Shells 387Fatieue Evaluation of Nozzles under Internal

Chopter l212.1

12.2

Pressure

Extemal Loadings

11 .7.1 Local Stresses in the Shell or Head

I 1.7.2 Stresses in the Nozzle

Nomenclature

References

Bibliography

Vessel Supports

Introduction

Skirt and Base Ring Design

12.2.1 Anchor Chair Design

12.3 Design of Support Legs

12.4 Lug-SupportedVessels

12.5 Ring Girders

12.6 Saddle Supports

Nomenclature

References

Bibliography

PART 4 THEORY AND DESIGN OF SPECIALEQUIPMENT

Chopter l3 Flot Bottom Tonks

13.1 Introduction

13.2 API 650 Tanks

13.2.1 Roof Design

13.2.2 Shell Design

13.2.3 Annular Plates

13.3 API 620 Tanks

13.3. I Allowable Stress Criteria

I 3.3.2 Compression Rings

13.4 ANSI 896.1 Aluminum Tanks

13.4. I Design Rules

392394394407

415416417

421

422423434438442443449456456457

459

461

462462462470476482487

490496496

-

xviii coNTENrs

13.5 AWWA Standard D100

References

BibliograPhY

Chopter 14 Heql Tronsfer Equipmeni

l4.l TYPes of Heat Exchangers

14.2 TEMA Design of Tubesheets in U-Tube

Exchangers

14.3 Theoretical Analysis of Tubesheets in U-Tube

Exchangers

14.4 Background of the ASME Design Equations for

Tubesheets in U-Tube Exchangers

14.5 Theoretical Analysis of Fixed Tubesheets

14.6 TEMA Fixed Tubesheet Design

l4'6'l Local Equivalent Pressure

l4'6'2 General Equivalent Pressure

14'6'3 Relationship Between Local and

Equivalent Pressure

14.7 ExPansion Joints

Nomenclature

References

BibliograPhY

Chopfer 15 Vessels for High Pressure

15.l Basic Equations

15.2 Pres$essing of Solid Wall Vessels

15.3 Layered Vessels

15.4 Prestressing of Layered Vessels

Nomenclature

Biblio$aphY

Chopter 16 Toll Vessels

l6.l DesignConsiderations

16.2 Earthquake Loading

16.3 Wind Loading

16.3'1 Bxternal Forces from Wind Loading

498499

499

501

502

505

508

514519523523527

533537

537

538539

541

541

543547

558562563

565

566567

573573

CONTENTS

I 6.3.2 Dynamic Analysis from Wind Effects16.4 Vessel Under Intemal Pressure Only16.5 Vessel Under Internal Pressure and Extemal

Loading

16,6 Vessel Under External Pressure Only16.7 Vessel Under External Pressure and External

Loading

References

Bibliography

Chopter 17 Vessels of Noncirculor Cross Section

17,1 Types of Vessels

17.2 Rules in Codes

17.3 Openings in Vessels with Noncircular CrossSection 601

17.4 Ligament Efficiency for Constant DiameterOpenings 601

17.5 Ligament Efficiency for Multidiameter OpeningsSubject to Membrane Stress 603

17,6 Ligament Efficiency for Multidiameter OpeningsSubject to Bending Stress 606Design Methods and Allowable Stresses 610Basic Equations 612Equations in the ASME Code, VIII-I 619Design of Noncircular Vessels in Other Codes 626I 7. 10. I Method in Swedish Pressure Vessel

Code 627I 7. 10.2 Design by Lloyd's Register of Shipping

Rules 630References 633Bibliography 633

577581

595

596601

585588

591

593593

17.717.817.9t7.to

APPENDICES 635

Appendix A Guide to Various Codes 636Appendix B Sample of Heat Exchanger Speciflcation Sheet U6Appendix C Sample of an API Specification Sheet 648

-

II CONIENTS

Appendix D Sample of a Pressure Vessel Design Data Sheet

Appendix E Sample of Various Materials for Process Equipment

Appendix F Required Data for Material Approval in the ASME

Section VIII Code

Appendix G Procedure for Providing Data for Code Charts for

Extemal Pressure Design

Appendix H Corrosion Charts

Appendix I Various ASME Design Equations

Appendix J Joint Efficiency Factors

Appendix K Simplified Curves for Extemal Loading on Cylindrical

Shells

Appendix L Conversion Tables

INDEX

652668

675

678683

686

689698

PART

BACKGROUNDAND BASIC

CONSIDERATIONS

CHAPTER

HISTORY ANDORGANIZATION OF CODES

-OtD TIMERS [(lop) Courtesy Bobcock & Witcox Compony, (bol|or,) ( iuroly ,",r,,, , ,"r,,,,r,,,1

2

-Y

HISTORY AND ORGANIZATION OF CODES

I.I USE OF PROCESS VESSELS AND EQUIPMENT

'I'hroughout the world, the use of process equipment has expanded considerably.ln the petroleum industry, process vessels are used at all stages of processing oil.At the beginning of the cycle, they are used to store crude oil Many differenttypes of these vessels process the crude oil into oil and gasoline for the con-

surner. The vessels store petroleum at tank farms after processing and, finally,scrvc to hold the gasoline in service stations fol the consumer's use. The use ofDroccss vessels in the chemical business is equally extensive. Process vessels are

uscd everywhere.Prcssure vessels are made in all sizes and shapes. The smaller ones may be

no larger than a fraction of an inch in diameter, whereas the larger vessels may

be 150 ft or more in diameter. Some are buried in the ground or deep in the

occan; most are positioned on the ground or supported on platforms; and some

lctually are found in storage tanks and hydraulic units in aircraftThe internal pressure to which process equipment is designed is as varied as

thc size and shape. Intemal pressure may be as low as I in water gage pressure

to as high as 300,000 psi or more. The usual range of pressure for monoblock

construction is about 15 to about 5000 psi, although there are many vessels

designed for pressures below and above that range. The ASME Boiler and

Itcssure Code, Section VIII, Division t*, specifies a range of intemal pressure

liom 15 psi at the bottom to no upper limit; however, at an intemal pressure

abovc 3000 psi, the ASME Code, VIII-I, requires that special design consid-

crations may be necessary.r However, any pressure vessel that meets all the

rrquircrncnts of the ASME Codc. regardless of the intemal or external design

prcssuro. rnay slill bc acccptcd by thc authorized inspector and stamped by the

nrlrnrllclurcr with thc ASMI'l ('rxlc syrttbol. Some other pressure equlpment,srrch as Al'l'' sl(nagc t Dks. rrriry bc dcsigned and contain no more intemalpf('ssur( llriur lhitl gcncrirlc(l l)y lllc sllllic hcird of fluid contained in the tank.

I,2 HISIORY OF PRISSURE VESSET CODES IN THT UNITED STATES

llrt(,rt1lr llr( lrlr' lS(X):, ;rrrrl lrrtlv ltX)O\. (\l)losiotls in boilers and pressure

vcsscls rlcrc lr({tr{nt /\ lrrctrllx lrorlt t trplosiott tlrr thc Mississippi River:,1{rlrlx);rt .\rtlt,ttt,t.t '\1rrrl .'/ lStr5. rcsttllctl itt thc boat's sinking within 20

nrnrt(.\,rr,l tlrrr|..rtlr,,l |')l)ilr,(,llr(r\JtoittlllrotrtcaliertheCivilWar.Thistyper,l r rrtrr,,tr,rlrlrl r.trltttttr'rl un,rl)irl( (l rrrlo tlrc clrr'ly 1900s. In 1905, a destructive

, rlrl,,.r,,rr (,1 .r lr, lrlr( l!,rl(-r rrr ir sllrr'' lltellrly in Brockton, Massachusetts (Fig.

I l r. l rlllrl ''Il rr ,'r'l( . rrrlrrr, rl l l / otlrcls. and did Xi400,000 in property damage

'1,' rlL, r,\r \'.Alt r,rl, \'lll l,rrrrl VIII .'. rsrrrie(l lo (lcscribc thc ASME Boilcr and I'rcsstrrc

Vi....tl(,trit ,, l',," \'ftl ffl\, r'r l. /,,,'r.vt( V, rfry'.r, and l)ivisitttl2, Alk'r'ttttiK |tttll li'r/!,11r,, l, \ \, /,

Fisure l.l Firerub€ boiler explosion in sho€ focrory in Brockron, Md!3ochuseits in 1905. (Courlesy HorrfordSt@m Boiler Inrpection ond Insurdn.e Co., Horrford, Cr.)

6 HISIORY AND ORGANT/N rION Of CODTS

Irr l(X)(r, l'r.llre'cx;rkrsi.rr irr . rlrr)c llrel.ry i'l,yrrrr. Massirclrrtsc.s, r.cs.ltcdirr dcalh, injrlry, a|ld cxtcnsivc propcrty darragc. Aticr this accidcnr, the Massa_clrusctt$ governor directed the fbrmation of a Board of Boiler Rules. The first setof rules for the design and construction of boilers was approved in Massachusettson August 30, l9O7 . This code was three pages long-!-

In 1911, Colonel E. D. Meier, the president of-the American Society ofMechanical Engineers, established a committee to write a set of rules tbr thedesign and construction of boilers and pressure vessels. On February 13, 1915,the first ASME Boiler Code was issuid. It was entitled ,,Boiler ConstructionCode, 1914 Edition." This was the beginning of the various sechons of theASME Boiler and Pressure Vessel Code, which ultimately became Section 1,Power Boilers.3

^ The first ASME Code for pressure vessels was issued as ,,Rules fbr the

construction ofUnfired Pressure Vessels,', Section VIII, 1925 edition. The rulesapplied to vessels over 6 in. in diameter, voiume ove. 1.5 ft3, and pressure over30 psi. In December 1931, a Joint API_ASME Committee wis ibrmed todevelop an unfired pressure vessel code for the petroleum indusiry. .l.he firstedition was issued in 1934. For the next 17 years,iwo separate unfiied pre;surevessel codes existed. In 1951, the last API_ASME Code ;as issued as a separaredocument.a In 1952, the two codes were consolidated into one code_the ASMEUnfired Pressure Vessel Code, Section VIII. This continued until the 196gedition. At that time, the original code became Section VIII, Oivislon I , pres_sure Vessels, and another new part was issued, which was Seciion VI II, Division2, Alternatiye Rules for pressure Vessels.

The ANSUASME Boiler and pressure Vessel Code is issued by the AmericanSociety of Mechanical Engineers with approval by the American'National Stan_dards lnshtute (ANSI) as an ANSI/ASME document. One or morc sections ofthe ANSI/ASME Boiler and pressure Vessel Code have been established as thelegal requirements in 47 of the 50 states in the United Str,", ,,",f in all theprwinces of Canada. Also, in many other countries of the worlti, the ASMEBoiler and Pressure Vessel Code is used to construct boilcrs arrc pressurevessels.

In the United States most piping systems are built to the ANSI/ASME Codefor P.ressure Piping B3l . There are a number of different piping couc sectronsfor different types of systems. The piping section that i" ,ir".i tiu. boiiers incombination with Section I of the ASME Boiler and pressure Vcsscl (ixle is the!o09 fo1!-o1er Piping, 831.1.5 The piping secrion thar is olicn uscrt withSection VIII, Division I , is the code for

-Cheniical piant and lretnricLrrrr t{clinery

Piping, 831.3.6

I,3 ORGANIZATION OF THE ASME BOILER AND PRESSUREVESSET CODE

The ASME Boiler ancl pressure Vessel Code is clivided into many sectrons,divisions, parts, and subparts. Some ofthese sections relat",u

" ro"lrti. tina of

T].4 ORGANIZATION OF THT ANSI 83 ] CODI] IOR PRISST'RE PIPINO 7

cqUipl c|l{ irrrtl ir;lrlielrliorr; olllcrs fctalc lo sl)ccilic Illillcliltls all(l tlrclll{xls l()f

applicatiOn rn(l cot)trol ol cclt'tiprnctrt; lnd tlthcrs rclate ttt care !lnd inspoctioll ()l

installed cquipnrctrt. 'l'hc tirllowing sections specifically relate to boiler and

pressure vessel design and constructlon:

Section I. Power Boilers (one volume)

Section IIIDivision 1. Nuclear Power Plant Components (7 volumes)

Division 2. Concrete Reactor Vessels dnd Containment (one volume)

Code Case Class I Components in Elevated Temperature Service (tn

N-47 Nuclear Code Case book)

Section IV, Heating Boilers (one volume)

Section VIIIDivision 1. Pressure Vessels (one volume)Division 2. Alternative Rules for Pressure Vessels (one volume)

Section X. Fiberglass-Reinforced Plastic Pressure Vessels (one vol-ume)

A new edition of the ASME Boiler and Pressure Vessel Code is issued on July

I every three years and new addenda are issued every six months on January I

and July l. A new edition incorporates all the changes made by the addenda to

the previous edition; it does not incorporate, however, anything new beyond that

coniained in the previous addenda except for some editorial corections or a

change in the numbering system. The new edition of the code becomes manda-

tory when it appears. The addenda are permissive at the date of issuance and

become mandatory six months after that date.

Code CasesT are also issued periodically after each code meeting They

contain permissive rules for materials and special constructions that have not

been sufficiently developed to place them in the code itself. Finally, there are the

Code Interpretations8 which are issued every six months These are in the form

of questions and replies that further explain items in the code that have been

misunderstood.

I.4 ORGANIZATION OF THE ANSI 83I CODE TOR

PRESSURE PIPING

In the United States the most frequently used design rules for pressure piping are

the ANSI 83l Code for Pressure Piping. This code is divided into many sections

for different kinds of piping applications Some sections are related to specific

sections of the ASME Boiler and Pressure Vessel code as follows:

HISTORY AND ORGANIZATION OF CODES

R!1.1. Power Piping (which is related to Section I)

F.31.2. Fuet Gas Piping (which may be related to Section VIII)

831.3. Chemical Plant an(l Petoleum Refnery Piping (which may be

related to Section VIII)R31.4. Liquitl Petroleum Transporting Prping (which may be related to

Section VIII)831.5. Refrigeration Piping (which may be related to Section VIII)831.7, Nuclear Power Piping (which has been discontinued and incorpo-

rated into Section III)B31,8. Gas Transmission and Distribution Piping Systems (which may be

related to Section VIII)

The ANSI B31 Piping Code Committee prepares and issues new editions and

addenda with addenda dates that correspond with the ASME Boiler and Pressure

Vessel Code and addenda. However, the issue dates and mandatory dates do not

always correspond with each other.

I.5 SOME OTHER PRESSURE VESSEL CODES AND STANDARDS IN

THE UNITED STATES

In addition to the ANSVASME Boiler and Pressure Vessel Code and the ANSI

B31 Code for Pressure Piping, many other codes and standards are commonly

used for the design of process vessels in the United States. Some of them are:

ANSUAPI Standard 620. "Recommended Rules for Design and Construction

of Large, Welded, Low-Pressure Storage Tanks," American Petroleum Insti-

tute (API), Washington, D.C.

ANSVAPI Standard 650. "Welded Steel Tanks for Oil Storage," American

Petroleum Institute, Washington, D.C.

ANSI-AWWA Standard D100. "Water Steel Tanks for Water Storage"'

American Water Works Association (AWWA), Denver, Colorado'

ANSVAWWA Standard D101. "Inspecting and Repairing Steel Water Tanks,

Standpipes, Reservoirs, and Elevated Tanks, for Water Storage," American

Water Works Association, Denver, Colorado.

ANSI 896.1. "specification for Welded Aluminum-Alloy Field Erected Stor-

agc Tanks," American National Standards Institute, New York'

lll, (A4. Standartl for Conk ner Assemblies.lor I'P-Gas, 4th ed, Under-

wlitcrs Laboratories. Nolthbrook, Illinois.

I.6 WORLDWIDE PRESSURE VESSEI CODES q

Stanlarh of Tubular Exchanger Manufacturers Association, 6th ed., Tu_bular Exchanger Manufacturer's Association, New york.Standnrds of the Expqnsion Joint Manufacturers Associ(ltion, 4th ed. , Exoan_sion Joint Manufacturer's Association, New york.

I.6 WORI-DWIDE PRESSURE VESSEL CODES

In addition to the ASME Boiler and Pressure Vessel Code, which is usedworldwide, many other pressure vessel codes have been legally adopted invarious countries. Difficulty often occurs when vessels are designed in onecolntry, built in another country, and installed in still a different country. Withthis worldwide construction this is often the case.

The following list is a partial summary of some of the various codes used indifferent countries:

Australia.

Australian Code for Boilers and Pressure Vessels, SAA Boiler Code (SeriesAS 1200): AS 1210, Unf.red Pressure Vessels and Class 1 H, pressareVessels of Advanced Design and Constuction, Standards Association ofAustralia.

Belgium.

Code for Good Practice for the Construction of Pressure Vessels, BelgianStandard Institute (IBN), Brussels, Belgium,

France.

Constructton Code Calculation Rules for Unfred pressure Vessels, SyndicatNational de la Chaudronnerie et de la Tuyauterie Industrie e (SNCT), paris,France.

Germany.

A.D. Merkblatt Code, Carl Heymanns Verlag KG, Koln/Berlin, FederalRepublic of Germany.

haly.

Itqlian Pressure Vessel Code, National Association for Combustion Control(ANCC), Milan, Iraly.

t0HISTORY AND ORGANTZATION OF CODES BIBTIOGRAPHY ll,ltpun.

,lqnnt'st' l)tt,.t,rurt Vt,l;scl Code. Ministry of Labor, published by JapanlJoilcr Associution. Tokyo, Japan.Juyuu'ts-t' Standarrl, Construction of pressure Vessels,JIS B g24j, publishedby the Jupan Srandards Association. Tokyo, Japan.

"- - e' 'J' Y'

Jap,ayle High pressure Gas Control Law, Ministry of International Trade1i1,1,".t":,ry.Jibtished by rhe rnstitution for sffi

"r riigi;;Jrr*" c",tngtneering, Tokyo, Japan.

Netherlands.

f,:|;:#i:"*-e vessets. Dienst voor het stoomwezen, The Hague, the

Sweden.

Swedish Pressure Vessel Code,-Tryckkarls kommissioner, the Swedish pres-sure Vessel Commission, Stockholm, Sweden.

United Kingdom.

British Code 85.5500, British Standards Institution, London, England.

More complete details. discussions of factors of safety. and applications ofthe codes mentioned are given in Section 2.7. e summ,lry of iti. p_ug.upt,which.ar.e appticable for ihe various,.reqrl;il ;'r#J # ,ti .o0., ur.oaround the world is given in Appendix A.

REFERENCES

f. ASME Boiler and pressure Uo*] "aT.:

,:".,,:n. Unr, Division l, pressure Vessets,ANSVASME BpV-Vm_1, Americar2. Apr srandard 620, "R""",*";;;'R';:"i."iiirT"firi,"ffilffi;TiJi:i,*Li.;,

.;: j::::"[iJ:** tanks," ANsr,/Apr srd. 620,;.;;;";-;;;;;; i,i.tr"iot", wu,r,ing_

l. ASME Boiler and pressure Vesset Code, Section_|, power Boileru, ANSTASME BPV_I,_ nlll*,*n Sociery of Mechanicat Engrneers. New york, 1983.

" lilHy3,::ff '.{iX::#l#f#*y:::k-!y ^Pa'r:teu!

Liq.uids and Gases, 5th ed.,1951. gmeers and American petroleum Institute, New york,

S, ASME Code for hessure pipinq BJl.ol Mechanicar Engineers, Niw-york, 73l;;"'0"'ANSL/ASME

B31 l' American societv

6. ASME Code for hessure pioins B3l, g!t:m:cal. ptant and petroleunt Refinery piping,

- ^NSvASME 83t.3. American Siciety of Mechanical ;;C;;, ;u;;"lo.k,r,,*0.7' A_SME Boije-r and hessure vesser code,

-cod? cases, Boirers antr pre,rrrre y€$dh, AmericanSocicty of lvlechanicaj En8incers, Ncw york, 19g3.

8. ASME Boiler and pressure Vessel Code, _fu terpretations, (isstred every six months), AmedcanSociety of Mechanial Engineers, New york.

BIBTIOGMPHY

' Steel Tanks for Liquid Stoege', in Steel plate Engifieerin| Data, Vol. l, 1976 ed., American Ironand Sreel lnslirute, Washingron, D.C.

is

CHAPTE R 2SELECTION OF VESSEL,

SPECI FICATIONS, REPORTS,AND ALLOWABLE STRESSES

l3

l4 SttECTlON OF VESSIL, SPECIFICAIl()N". rtlr",lrr'., nND ALLOWABLE STRESSES

2.1 SELECTION OF VTSSI I

Although nrlrrly lttr l t. ,,'rrlrl,rt, 1,, llr( \( lL'clion of pressure vessels, the twobasic r.r;rrirr.rrfrrt,, tlr,rt ,rll,, t tlr, ,( [.r lion are safety and economics. Manyit(.Drs i||r. r rr,,rrI r,,l rr,tr,r', rrrrrtcrials' availability, corrosion resistance,lrltllrrl,, rrr, rrl,tlr r11, . .rr,l rrrrrgnitudes of loadings, location of installationrr, lr,lprl, ( rnl I,r.r,l'rt' ,"r,t r.rrr'(lrquake loading, location of fabrication_(shoD"r 1., l,lr t", rrr,,r ,,t \i.,,s(.1 installation, and availability of labor supply at the

\l rrt, rrr, r, ,r'.rrr1' rrsc of special pressure vessel in the petrochemical and otherrrr,lrr rl, , rtr. ;rvrilability of the proper materials is fast becomrng a maJor1,r,,t,1,,,' I lr(. nrost usual material for vessels is carbon steel. Many other special_r,,, l r r,rr{ rlls iLre also being used for corrosion resistance or the abilily ro conmln.r tlrrrr I wrthout degradation of the material's properties. Substitution of materialsr'. I x (.vl lent and cladding and coatings are used extensively. The design engineerrrrrrst lrc in communication with the process engineer in order that all materialsrrsctl will contribute to the overall integrity of the vessel. For those vessels thatrctluire field assentbly in contrast to those that can be built in the shop, proper(luality assurancc must be established for acceptable welding regardless;f iheadverse condilions under which the vessel is made_ provisions must be estab_lished for ftrrliography, stress relieving, and other operations required in thefield.

For thost. vcssels that will operate in climates where low temperatures areencounlcr((l r)f contain fluids operating irt low temperatures, special care mustbe takc rr Ir crrsure impact resistance of the materials at low timperatures. Toohlirirr tlrs l,r()l)crty, the vessel may require a special high-alloy steel, nonferrousrrrirlcrirrl, rrr some special heat treatment.

2.? WHICH PRESSURE VESSEL CODE IS USED?

'l lrc lrrst consideration must be whether or not there is a pressute vessel law atllrc lo( irt ion of the installation. If there is, the applicable iodes are stated in thel:rw. ll thc jurisdiction has adopted the ASME Code, Section VIII, the decisionrrrly bc narowed down to selecting whether Division I or Division 2 is used.

. I'here are many opinions regarding the use of Division I versus Division 2,but the "bottom line" is economics. In the article ,.ASME pressure_Vessel Code:Which Division to Choose?",r the authors have listed a number of factors forconsideration. Division I uses approximate formulas, charts, and graphs insimple calculations. Division 2, on the other hand, uses a complex methocl offbrmulas, charts, and design-by-analysis which must be describcd in ir stressreport. Sometimes so many additional requirements are addcd lo tltc rriuirnumspecifications of a Division I vessel that it might bc rnorc ccorrorrrir.rrl to supplylu I)ivision 2 vcssel and lake advantage of thc highcr itlL)rvrl)l(. strrsscs.

2.4 SPECIAL DESIGN REQUIREMENTS

2.3 DESIGN SPECIFICATIONS AND PURCHASE ORDERS

Currently, the only pressure vessel code, exclusive of the ASME Code, III-l-NB, Nuclear Vessels, which specifically requires formal design specifications as

part of the code requirements is the ASME Code, VIII-2, Alternative Rules forPressure Vessels. This code requires a User's Design Specification to be pre-pared and certified by a registered professional engineer experienced in pressurevessel design. This certification by the professional engineer is given on theASME Manufacturer's Data Report, Form A- 1. The manufacturer is responsiblefor retaining the User's Design Specification for five years.

For other codes and standards, design specifications and design requirementsare not well defined. For the ASME Code, VIII-1, there is no specific statementthat any design specifications are required. The only indication of some sort ofdesign specifications is the list of minimum loadings in UG-22 that is consideredfor all construction . Sectron l, Power Eoilers, is less definitive on what loadingsare necessary to consider and what shall be included in a design specification orpurchase order. PG-22 of Section I states that loadings that cause stresses to gohigher than 107o above those stresses caused by internal design pressure shall beconsidered. The Manufacturer's Data Report, Form U-1 for the ASME Code,V I-1, requires many items to be listed, which means that most of the basicdesign information must be given in a design specification or purchase order.Although some codes help the purchaser regarding what data are needed forinclusion in the design specifications, this is usually done by mutual agreementbetween the purchaser and the manufacturer.

"For those process vessels that do not have a "suggested" list of items in designrequirements and specifications as part of code requirements, it is necessary toestablish them in the purchase order or contract agreement. The contract infor-mation is supplied by the purchaser or user with the manufacturer's help as towhat is needed and what shall be considered. Some design standards help theuser and manufacturer by offering fill-in forms that specifically list the require-ments for designing a process vessel. Design specification forms for a heatexchanger built to the standards of the Tubular Manufacturers Associationz aregiven in Appendix B and lor an API Srandard 650 Storage Tanki are given inAppendix C. It is always necessary to maintain a document containing designspeciflcations so that a permanent record is kept for reference. Often on a largeprocess vessel, some loadings from attached or supported equipment are notknown until after the job has started.

2.4 SPECIAL DESIGN REOUIREMENTS

In addition to the standard information required on all units, such as designpressure, design temperature, geometry, and size, many other items of infbrma-tion are necessary and must be recorded. The (xrrrosion and erosion amounts arc

l5

16 sfl,tcTtoN Ot Vtssll, st,tctt tcaTtoNs, RfpoRTs, AND AU-OWABLE STRTSSES

l, lx' *,u,.,, rrrrrl rr srrrtirlrlt. r'irlcri.l uld method of protection are to be noted. Thelyl)c (,l lllrirl tlrrrl will lrc t,0|llainctl, such as lethal, must be noted because ofthercqltitc(l slx\.ili(.rk.sigrr tlctaiis. Supported position, vertical or honzontat, ands[pl)oll lor.rrtiorrs rlusl bc listed as well as any iocal loads from supportedcrltip,rc,t rrrrtl piping. Site locatiorr is given so that wind, *o*, una

"u.tnquut"lcquircntcots ctrn lre determined. Impact loads and cyclic requirements are alsoinclurlcd.lirr thc ASME Code, VIII-2, a statement as to whether or not a tatigue

:::'.r,:::'.-"111r'llo according.to AD_160 is given. rf u rutilu" analysis isrc(lurrc(t. lhe specitlc cycles and loadings will be given. In addiiion, the designspccilications state whether or not certain loadings ire sustained or transrent. Theallowable stresses vary with the type of loadinls.

2.5 DESIGN REPORTS AND CATCULATIONS

T:,1YE ,C"1.. .VII.2. requires a formal design report with rhe assumptionsrn.the User's Design Specification incorporated in the stress analysis calcu_lations. These calculaiions are prepared and certified by a registered professionalengrneer experienced in pressure vessel design. As with the Usir,s DesignSpecification, the Manufacturer's Design Report is mandatory and thecertification reported on the Manufactu.".i Datu Repo.t. This is kept on file bythe manufacturer for five years.

- For vessels not requidng design reports, the manufacturer has available forthe- Authorized Inspector's review those necessary calculations for satisfyingU-2(g) or other design formulas. The pressure vessel design sheets shouldcontain basic design and materials data and at least the basic calculations ofpressure parts as given in the design formulas and procedures in the applicable

:_od^. onT.nd1d_fg. a simple vessel, an example of calculation sheets rs given

ll ilp"yiT D. This example depicts only those calculations that are requiredfor the Authorized Inspector and for construction. Other vessels may requrernuch more extensive calculations depending upon the complexity and con_(raclutl greements.

2.6 MATTRIALS' SPECIFICATIONS

All crxles itnd standards have materials, specifications and requirements de_sclibirrg whirl rrralcrials are permissible. Those material, tirut *"i"r_rtt"O *ittir sp(.( rli( ((xlc arc cither listed or limited to the ones that have aliowable stressvrrlrrts liivcrr. l)upcnding upon the code or standard, permitted rnatenas tor apirrtit rrliu plxt.ss vcsscl are limited. For instan"., o;i.;";;, Jin an se orljll (lcsif nirrior crr bc uscd in ASME Boiler and piersir" V"rr"i-Cot "rnr,_"_:]:lil...Y:::,:t l:,lf ::l',t

SI) specifications are the same u';;;, B specifi-flltlotl rr lltc ASIM Stirrrtlirltls a On specific instances, certain materiais thatItttvc lrt'rr rr.rlrril( r'r'r(r to sonrc other spccification, such as the DIN standard..

2.9 ATLOWABLE TENSITE STRESSES IN THE ASMI CODE 17

may be recertified to an SA or SB specification for an ASME certified vessel.Depending upon the contract specifications, permissible materials for construc-tion are given in lists such as that shown in Appendix E.

2.7 DESIGN DATA FOR NEW MATERIALS

When design data, such as allowable stresses, are requested for a new material,that is, one not presently in the code, extensive information must be supplied tothe Code Committee for evaluation. The ASME Code Committee lists thisinformation to develop allowable stresses, strength data, and other requiredproperties for accepting a new material into the code. Each section of the codecontains an appendix listing these requirements such as the one for the ASMECode, VIII-I, in Appendix F. The code also provides data to establish extemalpressure charts for new materials; this is given to those who want to establishnew external pressure charts. The required information is given in Appendix G.It is the person's responsibility requesting the addirion to supply all the dataneeded to establish those properties required in the code.

2.8 FACTORS OF SAFETY

In order to provide a margin of safety between exact formulas, which are basedon complex theories and various modes of failure , and the actual design formulasused for setting the minimum required thicknesses and the stress levels, a factorof safety (FS) is applied to various materials' properties that are used to set theallowable stress values. The factors of safety are directly related to the theoriesand modes of failure, the specific design criteria of each code, and the extent tox.hich various levels of actual stresses are determined and evaluated.

2.9 ALLOWABTE TENSILE STRESSES IN THE ASME CODE

As previously discussed, the basis for setting the allowable stress values or thedesign stress intensity values is directly related to many different factors de-pending upon the section of the code used. The criteria for setting allowabletensile stresses for each section of the ASME Boiler and Pressure Vessel Codeare as follows:

For Section I, Power Boilers, the ASME Code, YIll-l , Pressure Vessels, andSection III, Division 1, Subsections NC, ND, and NE, except for bolting whosestrength has been enhanced by heat treatment, the factors used to set the allow-able tensile stresses are summarized below.

At temperatures in the tensile strength and yield strength range, the least of:

1. j of the specified minimum tensile strength.2. j of the tensile strength at remperarure.3. ! of the specified minimum yield strength.

I8 SEI.TCTION OI VESSEL, SPECITICATIONS, REPORTS, AND AttOWABtE STRESSES

4. r{ ol thc yicld strength at temperature (except as noted below where 90Zois uscd).

At temperatures ip the creep and rupture strength range, the least of:

l, l00qa of the average stress to produce a creep rate of 0.0l per l000 hours(l7o in 105 hour).

2. 67Ea of the average stress to produce rupture at the end of 100,000 hours.3. 80Vo of the rninimum stress to produce rupture at the end of 100,000

hours.

,_ In the temperature range in which tensile strength or yield shength sets the

allowable stresses, higher allowable stresses are permitted for austenitic stainlesssteels and nickel-alloy materi-als where gleater deformation is not objectionable.

!9h:l*,the criterion of I yield strength at lemperature may be increased to9oVo,yield strength at temperature. However, the factor ! spicified minimumyield strength is still maintained.

For the ASME Code, VIII-I, bolting material whose slrength has been en_hanced by heat treatment or strain hardening have the addition; criteria of (l) jof the specified minimum tensile strength and (2) t of the specified minimumyield strength.

For the ASME Code, VIII-2, and Section III, Division 1, Subsection NB andNC-3200 of Subsection NC, the factor used to set the design stress intensityvalues for all materials except bolting is the least of:

1. i of the specified minimum tensile strength.2. ] of the tensile strength at remperarure.3. .2 of the specified minimum yield strength.4. J of the yielded strength at temperature except as noted in the tbllowing

paragraph.

Higher design stress intensity values are permitted for austenitic stainlesssteels and nickel-alloy materi€ls where greater deformation is not objectionable.In this_ case, the criterion of J yield strength at temperature may be increased toas high as 90Vo yield strength at temperature or any value beiween ! and gOVoyield strength at temperatue depending upon the acceptable amount of defor-mation. However, the factor of j specified minimum yield strength is stillmaintained.

There are two criteria for setting bolting design stress intensity values in theASME Code, VIII-2. For design by Appendix 3, the criteria are the same as forthe ASME Code, VI -1, because these values are used for the tlcsign of boltsfor flangjs. Ior design by Appendix 4 of the ASMII (ixlc. VIII_2, and bySectirrn III, Division -l

, Slbsdition NB ancl NC-32(X) ot' Sutiscc.riirn IrtC. thecrilcria lirr setting bolting design stress intcnsity vitlucs urc thc lesscr of the

2.IO ALLOWABLE EXTERNAI PRESSURE STRESS AND AXIAI. STRESS I9

following: (1) | of the specified minimum yield strength and (2) j of the yieldstrength at temperature.

For Section IV, Heating Boilers, the criterion for setting the allowablestresses is much more simple: (1) I /5 of the specified minimum tensile strength.

2.IO ALTOWABLE EXTERNAL PRESSURE STRESS AND AXIALCOMPRESSIVE STRESS IN THE ASME BOILER AND PRESSUREVESSEL CODE

Within the ASME Boiler Code, simplified methods are given to determine themaximum allowable external pressure and the maximum allowable axial com-pressive stress on a cylindrical shell without having to resort to complex ana-lytical solutions. Various geometric values are contained in the geometry chart,whereas materials' properties are used to develop the materials charts.

Allowable stresses in the materials charts are based on the followine criteriaFor cylindrical shells under external pressure, the least of:

l. 33Vo of the critical buckling stress with a factor of 807o for tolerance.2, 33Va of the specified minimum yield strength and yield strength at tem-

perature.

3. 67Vo of the average stress to produce a creep rate of 0.01%/1000 hours(17ol 100,000 hours).

4. IOOVo of the allowable stress in tension.

- For spheres and spherical portions of heads under extemal pressure, the least

OI:

l. 25Eo of the critical buckling stress with a factor of 607o for tolerance.2. 25Va of the specified minimum yield strength and yield strength at tem-

perature.

3. 507o of the average stress to produce a creep rate of 0.017o/1000 hours(17ol100,000 hours).

4. IOOVo of the allowable stress in tension.

For cylindrical shells under axial compression, the least ol

l. 259o of the critical buckling stress with a factor of 5OVo for tolerance.2. 50Vo of the specified minimum yield strength and yield strength at tem-

perature.

3. 1007o of the average stress to produce a creep rate of 0.017o/1000 hrs( l7ol 100,000 hours).

4. ljQVo of the allowable stress in tension.

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22 STI.TCIION Ot VTSSTI-, SPTCIFICATIONS, REPORTS, AND ALLOWABIE STRESSES

2.I I ALLOWABLE STRESSTS IN THE ASME CODE FOR PRESSUREPIPING 83I

'I'hc ullowrrblc sircsscs given in various sections of the ASME 831 Code forl\'csnulc I'ipirrg urc sinrilar to the corresponding sections of the ASME Boilernrtl l\'cssurc Vcssel Code; however, in some sections, the basis is different. Inthc (lxlc lirf Power Piping B31.l, the allowable tensile stresses are set by thesrrrrrc crilcria as used for ASME Code, Section I. In the Code for Chemical plantrn(l llctrolcum Refinery Piping B31.3, the allowable tensile stresses for otherth n bolting are set on a similar basis as used for ASME Section VIII, Divisionl, sxcept a factor of i is substituted for j on the tensile strength. The factor ofi on yield strength is used in both codes. This makes 831.3 in the tensile andyield strength range is similar to Division 2 and in the creep and rupture strengthrange similar to Division 1.

2.12 ALLOWABLE STRESS IN OTHER CODES OF THE WORLD

Throughout the world, various factors of safety are applied to materials' data toestablish allowable shesses for the design of boilers, pressure vessels, andpiping. For the temperature range to that temperature where creep or rupture setsthe allowable stresses, the universal factor for setting allowable stresses is basedon yield strength. In some countries, a factor is applied to sets of data that havebeen established from many tests; in others, the data are determined by the lowyield point or the high yield point. In still other countries, the actual data for thecomponent being designed have its yield strength determined by tests . The actualdata of the part are then factored into the design formulas. Not all countrieschoose to use the ultimate tensile strength as a criterion for setting allowablestresses. When they do, the factor of safety between various countries rs some-times very different. In order to show these differences, a discussion followsregarding the allowable stress basis of several different countries.

The ierms, symbols, and definitions used are as follows:

UTS : ultimate tensile strength (either specified minimum or data at designtemperature)

y5 = yield strength (either specified minimum or data at design tem-perature)

R = stress to cause rupture in 100,000 hoursC : stress to cause total creep or creep rate in 100,000 hours

na : not applicablen : none or not used

2.12 ATLOWABTE STREss IN OTHER CODES OF THE WORI.D

Australia

rs

23

The rules used for the design of boilers ald pressure vessels set by the StandardsAssociation of Australia are called the SAA Standards Series AS 1200. Thefactors of safety used to set the allowable stresses for the various sections are:

RUTS

AS 1210_1977Pressure Vessels

Class lH-1979AS 1228-1980Boilers

Belgium

nn

The Belgian rules issued by The Belgian Standards Institute (IBN) permir amixture of code rules from various other countries. The allowable stressesdepend usually upon the codes used. However, the basic allowable smesses areset as follows:

42.4

2.7

1.6*1.5

1.5

1.6na

1.5

UTS ys

' Boilers

Liquid gas

Air receivers

Pressure vessels

Czechoslavakia

3.22.7

1.6

1.6

1.5

n

n

nn 1.8 n nVarious factors at designer's choice

Czechoslovakian rules are extensively detailed for all types of vessels withdifferent allowable stresses used for intemal pressure as compared with extemalpressure. For the design of boilers and pressure vessels, the allowable stressesare established by the least of:

*l.5 at temDerature.

24 SELECTION OF VESSEt, SPECITICATIONS. REPORTS, AND ATIOWABTE STRESSES

UTS

Intemal pressure----cylindersHeated wall nUnheated wall nCastings n

All wallsCastings

1.61.62.0

n 2.0n 2.5

Extemal pressure-cylinders without stiffening rings

1.6l.)2.0

2.02.5

1.01.01.25

1.351.7

F rance

The French rules for pressure vessel design establish the allowable shessesdepending upon the inspection and the compliance with the SNCT (SyndicatNational de la Chaudronnerie et de la Tuyauterie Industrielle) rules. The allow-able stresses are established by the least of:

SNCT w/insp.

SNCT w/insp.and analysis

Not SNCTw/insp.

Not SNCTw/o insp.

2.7

Gennany

The three types of pressure vessels recognized by the regulatory agencies inGermany are fixed pressure vessels, mobile pressure vessels, and boilers. Theallowable stresses are established according to a specific DIN (German IndustrialStandard) certificate or by testing. The allowable stresses are set by variousfactors of safety on the yield strength with a constant factor of safety on ruptureas follows:

UTS

v,5

1.0

1.0

1.0

1.0

1.6

1.6

1.6

1.6

1.6

1.6

1.5

1.9

2.4

ys

Steel (not cast)DIN 3

DIN2orlTested

nn

n

nn

n

I .:)

1.8Ll

1.51.51.5

2.I2 ALTOWABLE STRESS IN OTHER CODES OF THE WORID

ys RUTS L

Cast SteelDIN 3 nDIN2or 1 nTested n

Aluminum and seamless copperDIN 3 nDIN2or I nTested n

2.02.5l.)

l 5

1.51.5

nnn

nnn

Italy

The Italian rules for boilers and pressure vessels establish the allowable stressesby the least of the following:

UTS

J.f4.02.5

1.51.51.5

Boilers 2.7Pressure vessels n

UTS

1.6 1.01.5 1.0

1.6

1.5*1.6

Japan

The Japanese government does not have one group to develop their designcodes. Rather have several different rules for designing boilerJ and pressurevessels. Although they are published by different groups and generally writtenby the different groups, all codes are under the control of the Japanese govern-ment. The basis for setting allowable stresses is the least of:

rs

Boilers

Pressure vessels

Altemativepressure vessels 3

*1.5 used when YS/IS > 0.85.

'whcrcT: YS/TS < O.7.

1.6 1.67 av.l 25 min.

2/(1.6 - jI 1.5 av.I .25 min.

1.0

1.5

1.0

26 SETECTION OF VTSSEI, SPTCIFICATIONS, REPORTS, AND ATTOWABI.E STRESSES

Sweden

The Swedish rules for the design of boilers and pressure vessels set the allowablestresses using only the yield strength and the rupture strength as follows:

UTS ys C

Unilctl Kingdom

The British rules for the design of boilers and pressure vessels are collectivelycalled British Standards. The basis for settine the allowable stresses is the leastof:

UTS ys

1.51.5All

R

Boilers-BS 1113

Pressure vesselsBS 5500Carbon steelStainless steel

2.7

2.352.5*

1.5 I _.'

I _J

l.J1.51.5

l.

RTTTR.ENCES

Srrrrlcn, A. M., and J. R. Mase, "ASME Pressure-Vessel Code: Which Division to Choose?",('hrt\k\tl lit|ineering, January ll, 1982.

lnrthorlt oJ luhular Exchanger Manufacturers.Asroc., 6th ed., Tubular Exchanger Manu-lrrllrrrr As$oci0lbn, White Plains, N.Y., 1978.

rl,JJ [l lcnr|t(rrlrtrr.

REFERENCES

ANSUAPI Standard 650, Welded Steel Tanks for Oil Storage, 7th ed., American PetroleumInstitute, Washington, D.C., 1980.

1982 AnnuaL Book of ASTM Standards, Afieican Society for Testing and Materials, Philadel-phia, Pa., 1982.

DIN Standa (Deutsche Normen Dll,lr, Herausgegeben vom Deutschen Normeruusschu(D,VA), Berlin, Gemany.

27

- - 2f l+u)'Fxy

i,=#n(* .,&*)

r1 = -q.C-/afu*razn1" i 2 (l -p'J \ayz ,,'I

L. t" a_w

^J i 211*u; a*aY

Th6ori6s, €riter;o, ond bosic equorions.

29

CHAPTER 3STRENGTH THEORIES,

DESIGN CRITERIA, ANDDESIGN EQUATIONS

30 sTRINGTH THEORIES, DESIGN CRITERIA, AND DISIGN TQUATIONS

3.I STRENGTH THEORIES

ln the design of process vessels and pressure equipment, two basic modes offailure may be assumed: elastic failure based on the theory of elasticity andplastic failure based on the theory of plasticity. Except for thick-walled vessels,

elastic failure is usually assumed for the design of pressure vessels. It is consid-ered to occur when the elastic limit of the material is reached. Beyond this limit,excessive deformation or rupture is expected. These limits are usually measuredin terms of tensile strength, yield strength, and, to some degree, rupturestrength.

Of the many theories developed to predict elastic failure, the three mostcommonly used are the maximum principal stress theory, the maximum shear

stress theory, and the distortion energy theory. The maximum (principal) stress

theory considers failure to occur when any one of the three principal stresses has

reached a stress equal to the elastic limit as determined from a uniaxial tensionor compression test. The maximum shear stress theory (also called the Trescacriterion) considers failure to occur when the maximum shear stress equals theshear stress at the elastic limit as determined from a pure shear test. Themaximum shear stress is defined as one-halfthe algebraic difference between thelargest and smallest of the three principal stresses. The distortion energy theory(also called the maximum strain energy theory, the octahedral shear theory, and

the von Mises criterion) considers failure to have occurred when the distortionenergy accumulated in the pad under stress reaches the elastic limit as deter-mined by the distortion energy in a uniaxial tension or compression test.

Engineers have known for some time that the maximum shear stress theoryand the distortion energy theory predict yielding and fatigue failure in ductilematerials better than does the maximum stress theory.r However. the maximumstress theory is easier to apply, and with an adequate safety factor it givessatisfactory designs. But where a more exact analysis is desired, the maximumshear stress theory is used.

Two basic theories of strength are used in the ASME Boiler and hessureVessel Code. Section I,2 Section IV,3 the ASME Code, VI[-1, and Section III,Division 1, Subsections NC,4 ND,5 and NE6 use the maximum stress theory.Section III, Division l, Subsection NB7 and the optional part of NC, and theASME Code, VtrI-2, use the maximum shear stress theory.

In the two sections of the ASME/ANSI Code for Pressure Piping 83l that are

used primarily with the ASME Boiler and Pressure Vessel Code, both ANSI83 l. l6 and 83 1. 3e use the maximum stress theory. 83 1.3 is unique in that it usesthe maximum stress theory but permits allowable stresses to be established onthe same basis as the ASME Code, VIII-2, which requires use of the maximumshear stress theory. The other sections of 831 also use the maximum stresstheory. They require that in addition to the stresses caused by intemal and

cxternal pressures, tiose stresses caused by thermal expansion of the piping are

to he considered.

3.2 DESIGN CRITERIA 3I

3.2 DESIGN CRITERIA

The design criteria for both Sections I and IV basically call for determining theminimum wall thickness that will keep the basic circumferential stress below an

allowable stress level. Additional rules and charts are included for determiningthe minimum thickness of various components. However, in general, a detailedstress analysis is required only for special designs. Sections I and IV recognizethat local and secondary stresses may exist in some areas of pressure vessels;

design details, however, have been established to keep these stresses at a safe

level with a minimum of stress analysis investigation.The design criteria of the ASME Code, VI -1, and Section III, Division l,

Subsections NC except NC-3200, ND, and NE, are similar to those for SectionsI and IV except that the ASME Code, VI[-I, and Section III, Division l,Subsections NC, ND, and NE require cylindrical shell thickness calculationsbased on both the circumferential and the longitudinal directions. The minimumrequired thickness may be set by stresses in either direction. In addition, theASME Code, VIII-1, permits the combination of primary membrane stress andprimary bending stress to go as high as 1.5 S at temperatures where tensile and

yield strength control and 1.25 S at temperatures where creep and rupturecontrol, where S is the allowable tensile stress values.

The design criteria for the ASME Code, VIII-2, provide formulas and rulesfor thd?nore common configurations of shells and formed heads for temperatureswhen the allowable stress criteria do not exceed the yield strength and tensilestrength range. Requirements include detailed evaluations of actual stresses incomplex geometries and with unusual loadings, especially if a cyclic loadingcondition exists. These calculated stresses are assigned various categories and

subcategories that have different allowable stress values as multipliers of thebasic allowable stress intensity value. The various categories and subcategoriesare:

Primary stresses, including general primary membrane stress, local primarymembrane stress, and primary bending stress

Secondary stresses

Peak stresses

Primary stress is caused by loadings that are necessary to satisfy the laws ofequilibrium between applied forces and moments. Primary stesses are not

self-limiting .

Secondary stress is developed by self-constraint of the structure. Its basic

chamcteristic is that it is selfJimiting. That is, rotation and deformation ordeflection take place until the forces and moments are balanced even though

some pennanent geometric changes may have taken place.

Lastly, peak stress is the highest stress condition in a structure and is usually

32 STRTNGTH THEORIIS, DESIGN CRITERIA, AND DESIGN EQUATIONS

due to a stress concentration caused by an abrupt change in geometry. This stressis important in considering a fatigue failure because of cyclic load application.

In general, thermal stresses are considered only in the secondary and peakcategories. Thermal stresses that cause a distortion of the structure are catego-rized as secondary stresses; thermal stresses caused by suppression of thermalexpansion, but may not cause distortion, are categorized as peak stresses.

Potential failure modes and the various stress limits categories are related.Limits on primary stresses are set to prevent deformation and ductile burst. Theprimary plus secondary limits are set to prevent plastic deformation leading toincremental collapse and to validate using an elastic analysis to make a fatigueanalysis. Finally, peak stress limits are set to prevent fatigue failure due to cyclicloadings.

The basic stress iniensity limits for various categories relating to an analysisaccording to the ASME Code, VIII-2, and Section III, Division 1, SubsectionNB, and optional Part NC-3200 of Subsection NC are:

Stress Intensity CategoryAllowable

Value

FactorBased on

YieldStrength*

FactorBased onTensile

Strength*

General primary membrane(P,) ks,

Local primary membrane(P") UKS^

himary membrane plusprimary bending (PM + Pd liks.

Primary plus secondary(PM+PB+Q)

3s, -!c

s) +s"

s, +s,

25, S-1(

In the ASME Code, VIII-2, and Section III, Division 1, optional Part NC-3200 of Subsection NC, a factor of ft is applied to various loading combinationssomewhat related to whether or not the loading is sustained or transient. Thelaotors are k = 1.0 for sustained loads including dead loads and pressure;k - 1.2 for sustained load plus wind or earthquake loads; t = 1.25 for hydro-$tiltic tcsts; and k - 1.15 for pneumatic tests.

'I'hc dcsign criteria for Section III, Division l, Subsection NB, are verysinrillr lo thoso for the ASME Code, VIII-2, except there is less use of designlirrrrrrrlrrs, culvcs, tnd tables, and greater use of design by analysis in Section IIL'l'h(. cfllcgorics ol slrcsses and stress intensity limits are the same in both sec-liorrs.

+AiiurriflI lhrt I | .O. ,\,,, (lcsiSn strcss intensity valuc fbr Section III, Division l, SubsectionNll, n[(l thc |

't{ i',nrl pIr I {rl S hsr(.li()n NC, and thc ASMts Codc, VIII-2 (psi), S" = yicld strength

(plri). url ,\, ultirrxrtc k nsil(. slfrJrgth (psi)

3.4 STRESS-STMIN REIATIONSHIPS

3.3 DESIGN EQUATIONS

Once the allowable stresses are set, the basic design equations must be devefoped. The design of process equipment is based on the assumption that thematerial generally behaves elastically at the design pressure and design tem-perature. Accordingly, most of the equations are derived from the theory ofelasticity and shength of materials basis.

3.4 STRESS-STRAIN RETATIONSHIPS

The stress-strain relationship at any point within a homogeneous, isotropic, andlinearly elastic body that is subjected to a system of forces is obtained from thetheory of elasticity. Referring to Fig. 3.1, the stress-strain relationship is givenby

1.e,: ELo,-

p(oy + ozl)

t.er =

ELor- ploz 'r o^)J

Ie, : ;lo,- tt(o, I o)l (3.1)- I1- -

rs 2(1 + 1t),DGE

2(l + 1t)^lv = ---V- rn

2(1 + 1t).t/,- =

-

i--L

Or, in a different form.

(r+ tt)(1 -zp.)

(1,+p.)(1 -2tt)

[e,(l - pc) +

[e,(1 - p) +

p,(e, + e,)]

p.(e" + e")l

(l + p)(1 -Ii^,

: " ltl2(l + 1.t)

T,

- [€,(1 -LIL)

pr,)+p(e.+er)] (3.2)

34 STRENGTH THIORIES, DTSIGN CRITERIA, AND DESIGN EQUATIONS

' f1---> <-f3

t2

I@I t"lt4

(b)

II

,v

Figure 3.1 Slrcls rerulianr ot o point wirhin o homogeneous, isotropi<, ond lin€orly eldsri. body.

F.,r, = zG; tL)

Ev-Tn : ;:;---,----.._L\r t trL)

where eb ey e,: axial strain in the.r, y, and z-directions, respectively

oo o, oz = axial stress in the .r, y, and z-directions, respeptively

"y,!, yr,, y- = shearing strain in the -r, y, and z-directions, respectively

re, T

rz, rp : shearing stress in the.r, y, and z -directions, respectively

E = modulus of elasticity of material (psi)

G = shear modulus of material (psi)

1,c : poisson's ratio

In most pressure vessel applications, the values of o,, rr,, and r,, are relativelysmall compared with o, and or. Hence, they are normally ignored and the

3.5 STRAIN-DEFTECTION EQUATIONS

equations reduce to

35

Or, in a different form,

1.e,: E\ox - psr)

t.q= i\ar- Po')

€,: Elo' + stl

2(1 + tt)f"y = --E- r"t

o, = ---! ,1u" * *rsL-p-

o,:;\1e, + p'est- lL-

c,:0E

',=76+ iil*

(3.3)

(3.4)

3.5 STRAIN-DEFLECTION EQUATIONS

Figure 3.2 is cross section of a pressure vessel wall. It undergoes an extensionin the niddle surface of €o due to stretching plus extension due to bending. Theoriginal length lr at a distance z from the middle surface is given by

t,=adr-1\

The final length l2 after extension is

tz: dstt + .*r(r - 4)\ r./

whereas strain is given by

lr- lt,": l,

36 STRENGTH THEORIES, DESIGN CRITERIA, AND DESIGN TQUATIONS

Fisure 3.2 Cross s€ction of o sh€ll woll subie€ted to str€rchine ond bendins lodds'

Substituting the values of lr and lz into the above and deleting all small terms

results in

( t l\€.: €or _ ,\,:_ i) : e0,_ z. x\

where 1, is change in curvature. Similarly,

/ r l\€n : €ou -,\4- i) = es - z' xt

Substitution of the above two equations into Eq. 3.4 gives

Fo,: ,-:--Lr" + peo! - z(y" + trt'yt)

t-lt-tiq- ,--l€vt + l"r*- z(Xr+ PX')t- lL-

Nolr llrirl llr(' cx|)tcssirttt f, is related to the deflection by the expression

dzw / dx2x'=tt+kt"4'hffn

(3.5)

3.5 STRAIN.DEFTECTION EQUATIONS 37

However, because the quantity dw fdx is smal! compared with unity, the expres-

sion above becomes

d2w . d2wX': 77 a;to Xt = 7F

Hence, Eq. 3.5 may be written as

(3.6)

(3.7)

(3.8)

The shearing strain-displacement relationship can be obtained from Fig. 3 3.

The quantity 7," is shown in Fig. 3'34 and can be expressed as

"l'Y:"loq+a+P

where 7qry is the shearing stress due to in-place forces and d and B are due to

twisting moments. Also, from the figure,

. (d/ d\'ldv dud-srna -__6-: dy

(dD/?x)dx 0aIJ-srnP- d, =A

*=T+1^+ peo,-,(#. - *fu)lE I ldzw, drr\'lot: T7 *.leb

+ Pew - '\dy, - It dr') l

du 0af,t: Ioq, dy- a,

and

From Fig. 3.30, which represents the middle surface, the rotation is given by

-@w I Ai. The minus sign indicates counterclockwise rotation. As a result ofthis rotation, any point at a distance z from the middle surface will have a

deflection of

dw

dx

38 STRENGTH THEORIES, DESIGN CRITERIA, AND DESIGN EQUATIONS

z tD,

Figura 3.3 Sh€or dolormdliom of o unit cro$ .ection.

dteD = -Z--dy

Similarly,

-\,/./,?--zt,--'-,..'7 |t _----,- tu..-..

-

|_-,l-_

Hence Eq. 3.8 becomes

3.6 FORCE-STRESS EXPRESSIONS

And Eq. 3.4 becomes

^, :^'^ - )2-:----:-$y rvt -- dx b

39

(3.9)

3.6 FORCLSTRESSEXPRESSIONS

The force-stress relationship for the cross section shown in Fig' 3 4a can be

exDressed as

",=c("* -*#)

In the majority of cases, the quantity z/r is small with respect to unity and can

thus be dlsregarded. Also, substituting Eqs. 3.7 and 3.9 into 8q 3 10 gives

*,= f',,",(r - z\a,

N,= [n( - )a"n"=[,"(t-i)a"

u,=1,,,\t-z)a'

u.= [".'lt - z)a"

u,= J",'lr - i)a,

,, = -[ ,,'lt - 1)a"

u*= [,,",(t-4;,

(3. 10)

40 STRENGTH THEORITS, DESIGN CRITERIA, AND DESIGN EQUATIONS

,/

N.

N,

Et= r r(€0r+ /,€q,r-lJ'

Ft: .--------t (€0) + p€0r.)

l- lL'

1'u,Et2(l + 1t)

N,}

3.6 FORCE.STRESSEXPRESSIONS 4l

(3.1 r),.:&(#.#)u,=ffi\(*tu.

,*tu)

,. Eilt - tL) drwl2ll - lt2t ax dy

Example 3.1 Stresses are to be determined at the inside comer of an opening

in a cylindrical shell by applying strain gages at the location. The cylindricalshell is carbon steel with E : 29.9 x 106 psi and p : 0.3. The strain readings

from the three gages are €,: +360 x 10-6; €): +180 x l0 o and e' =-230 x 10-6. What are the stresses in the three principal directions at the

opening?

Solutian. Using the equations given under Eq. 3.2, the stresses are determined

AS

,oqo,: ;#1Q60X0.7) + 0.3(180 - 230)l : 13'630 psi( r.Jrw.+.,

,qq",: -*l(180)(0.7) + 0.3(360 - 230t1 = 9499 O.;' t r.JJ(u.+,

?qoo = "'' tr-?10rr0.7) + 0.3(360 + 180)l : 60 psi I(1.3x0.4)" --""'

Exanple 3.2. What are the stresses in the two principal directions of the

cylindrical shell with the o, = gt

Solution. Using the simplified equations given under Eq. 3.4, the stresses are

determined as

', =ffioso + 0.3 x 180) = 13,6oo psi

o, =ffi{rto + 0.3 x 360) : 9460 psi r

42 SIRTNOTH THEORI€S, DTSIGN CRITERIA, AND DESIGN EQUATIONS

Problems

3.1 Strain gages are attached to the surface of a tube subiected to internalpressure. The gages lie along the circumferential and l,ongitudinal axes.The tube is carbon steel with t = 29.9 x 106psi, 1.r,

: 0.3, and the stressat the surface in the circumferential direction is 17,500 psi. What are thestrain gage readings in the two directions?

Answer: e,: *498 x 10-6

€i: +117 x 10-6

3.2 In the tube of Problem 3.1, what is the strain in the z -direction? Usins thatanswer and the other answers in Problem 3.1, what are the calcrilatedstresses in the three directions?

Anst'er: a" = 17,500 psi

o" = 8,750 psi

o"=o

REFERENCES

7.

Criteria of the ASME Boiler and Pressure Vessel Code for Design by At],',ltsis in Sections IIIand VIII, Division 2, American Society of Mechanical Engineers, New york, 1969.ASME Boiler and Pressure Vessel Code, Section I, Power Boilers, ANSVASME BPV-I,American Society of Mechanical Engineers, New york, 1983.

ASME Boiler and Pressure Vessel Code, Sectionly, Heating BoiIeB, ANSVASME BPV-IV,American Society of Mechanical Enginers, New York, 1983.

ASME Boiler and Pressure Vessel Code, Section III, Division l, Subsection NC, Class 2Components, ANSUASME BPV-m-I-NC, American Society of Mechanical Engineers, NewYork, 1983.

ASME Boiler and Pressure Vessel Code, Section III, Division l, Subsection ND, Crllrr.iComponents, ANSL?ASME BPV-III-1-ND, American Society ofMechanical Engineers, NewYork. 1983.

ASMII Boiler and Pressure Vessel Code, Section III, Division l, Subsection NE, Class MC(i)tnpt,nt',ttr, ANSTASME BPV- I-l-NE, American Society ofMechanical Engineers, NewY,nk, 1983.

ASMIj lllrilcr and Pressure Vessel Code, Section III, Division 1, Subsecrion NB, CIaJJ 1( t'ntt\'tl' t.t, ANSI/ASME BPV- I-l-NB, American Society ofMechanical Engineers, NewY(n (. l()8lASMf i ( \nk. li,r I'rcssurc PipingB3l, Power Prping, ANSLIASME 83l.1, Amedcan Societyol Mrrhfiri(nl li[8inccrs, New York, 1980.

ASMf (irtfc lix Prcssure Piping B3l, Chemical plant and petroleum Rertnery pipin|,ANSI/AliMll ll I I L Amcr'can Soc'ety of Mechanical Engineers, New york, 19g0.

BIELIOGRAPHY 43

BIBTIOGRAPHY

Brownell, L. E., and E. H. Yoi{Irlg, Process Equipment Design, John Wiley, New york, 1959.

FattWI, J, H., Engineering Design, John Wiley, New York, l9&.Ha0ey, J. F., Theory and Design of Modern Pressure Vessels,2nd ed., Van Nostrand-Reinhold,

Princeton, N.J.

Seely, F. B. and J. O. Smith, Advanced Mecfuinics of Makriak,2nded., John Wiley, New york,1952.

CHAPTE R 4MATERIALS OF

CONSTRUCTION

Top: Metologroph of titanium wetd. Boltom: Tonrolum_ctdd veset (Courtesy of rhe Nooter Corp., Sr. l"ouis,

44

46

4.I MATERIAL SETECTION

MATERIALS OF CONSTRUCTION

The vast majoriry of vessels are constructed of ferrous and nonferrous alloys'

Ferrous alloys are defined as those having more than 50Vo iron They are used

in the eSME Code, VI[-l and 2, and include carbon and low-alloy steels'

stainless steels, cast iron, wrought iron, and quenched and tempered steels'

Nonfenous alloys include aluminum, copper, nickel, titanium, and zirconium

The ASTM designates all ferrous alloys by the letter A and all nonferrous alloys

by B. ASME uses the prefixes SA and SB, respectively ln most cases the ASME

unO eSfV specifications are identical. However, vessels built to the ASME

Code usually refer to the ASME specifications'Nonmetilic pressure vessels may also be constructed to the ASME Code'

Recently, ASME Section X was published to include fiberglass-reinforced plas-

tic (FRij vessels. Details of construction are given in Section 46' Concrete

vessels are also being considered by the ASME However, no specific rules are

available at this time.Selecting materials that are adequate for a given process is complicated and

depends on many factors such as corrosion, strength, and cost'

4.1.1 Corrosion

Corrosion,whichisdefinedasthedeteriorationofmetalsbychemicalaction'is probably the single most important consideration in selecting materials A

stigtrt ctrange in the chemical composition of a given -environment

can sig-

nifi'cantly cilange the corrosive behavior of a given metal This is illustrated in

epp""Oi* H, i,trich lists various environments and their effect on different

ferrous and nonferrous alloYs.

In a new chemical process, it is prudent to determine the factors that affect

the corrosion and then run tests on various materials in order to select the most

suitable one. Figure 4.1 shows an example of a heat exchanger.used in "city

water" service. The corroded tubesheet is made of carbon steel and the un-

corroded tubes are made of copper' Another example shown in Fig 4'2 is a

titanium lubesheet after exhibiling crevice corroslon'

fn nighty corrosive environments, every phase of the pressure. vessel fabri-

.ution pio""r, rnurt be evaluated for corrosion' Items such as buming' forming'

weldini, stress relieving, and polishing must be.considered Figure 4 3 illus-

trates a"Hastelloy C chu-te with corrosion in which marking Points were left on

rnu,"tlut Ou.ing h"ut treatment. Conosion at points of high stress along the break

lines can also 6e seen. In Figure 4.4, a Carpenter 20 tube shows knifeline attack

at a plug weld (shown by arrow) in a bayonet tube used in hydrofluoric acid

service.The cleanliness and finish of the inside surface of a pressure vessel before its

opcrltion ilrc vsry imPortant in preventing subsequent .c.onoslon in service'

Many uscrs tcquirc spccial clcaning proccdutcs ol'the insitlc surlace Prior to lts

insl;rllaliott.

Fisure 4.1 Cotroded corbon sleel lubesheet. (Courl$y of the Nooter Corp., St. touis, Mo.)

Frrruro 4.2 Corroded rirqnaum tubesheet. (Courlesy ol the Nooler CorP, Sr' touis, Mo )

Fisure 4.3 Hosrelloy C chure. (Courtesy o{ the Nooter Corp., Sr. rouk, Mo.)

48

Fisure 4.4 Crock in o Corpenter 20 lube weld. (Court6y of Nooier Corp., Sr. toui., Mo )

4.I MATERIAT SITECTION

4.1.2 Strenglh

The strength level of a material has a significant influence on its selection for a

given application. This is especially true at elevated temperatures where theyield and ultimate strength are relatively low and the creep and rupture behaviormay control the allowable stress values. In the ASME Code, VIII-l, the criteriafor allowable stress at elevated temperatures take into account both the creep and

rupture behavior as discussed in Section 2.4. In applying the ASME criteria forallowable stress as given there, the following procedures are used.

Specified Minimum Yield Stress

In obtaining the minimum yield shess of a given material, test data are plotted

at vadous temperatures as shown in Fig . 4 . 5 . A smooth trend curve is then drawn

though the averages of the data for individual test temperatures. The specified

minimum yield stress curve is obtained by applying to the yield trend curve the

ratio of the specified minimum value, as given in the material specification, to

the trend value of 80"F.

temperature,"F

Figure 4.5 Tensile ond yield 3irengrh.

49

o 200 400 600 800 1000 1200 1400 1600

!0 MATERIATS OI CONSTRUCTION

Sptcllled Mlnlnum 'l'ensile Siress

'fhe tensilc trend curve is determined by the same method as the yield trend curveincluding_ th-e,.ratio factor. The specified minimum tensile striss is arbitrarilytaken as I l07o of the tensile trend curve, as illustrated in Example 4. 1

Creep Rate

In order to establish the creep rate of lVo /lffi,O}} hours, data are plotted asshown in Fig. 4.6. Interpolation and extrapolation may be needed to establishthe creep rate for various temperature levels.

.0001

Figur€ ,{.6 Cre€p strengrh.

4.I MATTRIAI. SETECTION

Rupture Slrength

Test data are normally plotted as shown in Fig. 4.7. In some cases the data needto be extended to 100,000 hours and must be done with extreme care to extraD-olate accurately.

Example 4.1. A user is requesting code approval for a new material that hasa minimum specified tensile stress of 120 ksi and a minimum specified yieldstress of 60 ksi at room temperature. Tensile and yield values for various heatsand temperatures are shown in Fig. 4.5. Creep and rupture data are given in Figs.4.6 and 4.7, respectively. What are the allowable stress values at 300 and 1200'Fbased on criteria siven in Section 2.4?

1,000

Llle llrs.

Figur€ 4.7 Rupture strengrh.

5l

52

Solution

MATERIATS OF CONSTRUCTION

Allowable Stress at 300'F

l. From Fig. 4.5, average tensile stress = 130 ksi.Tensile stress reduced to minimum : l3O x 120/140: 111 ksi.Specified minimum tensile stress : 111 x 1.10 : 122 kst.Maximum stress to be used cannot exceed 120 ksi.Allowable stress based on tensile stress : 120/4 = 30 ksi.

2. From Fig. 4.5, average yield sffess : 60 ksi.Yield stress reduced to minimum : 60 x 60/75: 48 ksi.Allowable stress based on yield stress : 48 x 6 = 32 ksi.

3. From Figs. 4.6 and 4.7 it is apparent that creep and rupture are not a

consideration at 300'F.

4. Therefore, maximum allowable stress at 300"F = 30 ksi

Allowable Stess at 1200"F

1. From Fig. 4.5, average tensile saess : 112 ksi.Tensile stress reduced to minimum : lI2 x 120/1,4O: 96 ksi.Specified minimum tensile stress = 96 x 1.1 : 106 ksi, which is less

than maximum allowed of 120 ksi.Allowable stress based on tensile stress = 106/4 : 26.5 ksr.

2. From Fig. 4.5, average yield shess = 54 ksi.Yield stress reduced to minimum = 52 x 6O/7t: 42 ksi.Allowable stress based on yield stress = 42 x 6 = 28 ksi.

3. From Fig. 4.6, creep stress for 0.O1Vo rn 1000 hours = 15 ksi.Allowable stress based on creep = 15 ksi.

4. From Fig. 4.7. stress to cause rupture at 105 hours = 22 ksi.Allowable stress based on rupture = 0.67 x 22: 14.7 ksi.

5. Therefore, maximum allowable stress at 1200"F = 14.7 ksi. I

4. I .3 Moteriol Cost

llt'r'rrrrsc costs of materials vary significantly, the designer must evaluate mate-riirl (osl vcrsus other facton such as corrosion, expected life of equipment,Ivrrilirl'ility ol material, replacement cost, and code restrictions on fabricationrrrrl rr'1ririrs. n summary of the cost of some frequently used materials is givenrn l irlrlt ,l. | . With the large difference in cost, the designer should consider thelrx l(xs ltfior' 11r sclccting a given material.

4.2 NONFERROUS ATLOYS

TqblE 4.1 Approximote Cost ofMqteriqls Used in Pressure VesserConslruction

Cost inType Dollars/lb

0.30

0.7 5

0.90-2.501.50

t.'7 5

4.00

5.00

6.00

15.00

15.00

20.00

250.00

Carbon steel

Low-alloy steel

Stainless steel

AluminumCopper, bronze

IncoloyMonel

Inconel

Hastelloys

TitaniumZirconiumTantalum

'As of January 1983.

4.2 NONFERROUS ATLOYS

The 1983 ASME Section VIII Code, VI[-1, lists five nonfenous alloys for codeconstruction: aluminum, copper, nickel, titanium, and zirconium. These alloysare normally used in corrosive environment or elevated temperatures whereferrous alloys are unsuitable. Nonferrous alloys are nonmagnetic except forcommercially pure nickel which is slightly magnetic.

4.2.1 Aluminum Alloys

Aluminurn alloys have a unique combination of properties that make them usablein process equipment applications. They are nonmagnetic, light in weight, havegood formability, and have an excellent weight-strength ratio. Aluminum sur-faces exposed to the atmosphere form an invisible oxide skin that protects thernetal from further oxidation. This characteristic gives aluminum a high resis-lance lo corroslon.

Aluminum alloys have a systematic numbering system as shown in Table 4.2.'l he specification number also designates the various product forms. For exam-plc, SB 209 applies to plate products, and SB 210 applies to drawn seamless tubelxrtlucts. The first digit of the alloy designation number indicates its majorrllloying element as shown in Table 4.2.

All aluminum alloys are categorized by ASME specification number, alloy

54 MATERIATS OF CONSTRUCTION

Tqble 4.2 Aluminum Alloy Designotion

EXAMPLE+

designation, and temper designation as shown in Tables 4.2 and 4.3. Some ofthe terms in the tables are defined as follows:

Annealing.

Consists of heating the material to a given temperature and then slowlycooling it down. The purpose is to soften the material in order to remove coldworking stress.

Norm.alizing.

Consists of heating rnaterial to a temperature slightly higher than the an-

nealing temperatue and then cooling at a rate that is faster than annealing.

S o lution H eat Tr eatin g.

Heat heating at a temperature high enough for the alloys to be randomlydispersed.

Stahilizing.

A low temperature heating to stabilize the property of an alkry

,.ffi"ffi#*EffidF-Jfit<,6lLlfY.

"'ffi,#P.*,fffr4ffi..ffisffi*t' ffi.,,ffi*.ffi.ffi";^318H:" ,oPt-tcar td€

'' H&.,ffi'#^E lH,ffia'.nn5.

fFEATAa-E ILLOY.

5. tAtN ADV$J'AA IS fl,.l tT ls t€^I-lnElr a-E.

7, TLTRA HIgl S'FE €7H FFCENTTE!'' M)' USEDIN ffi.

C ST'TGFLITES(n^|?{ sElra-EssNEfr-TttaF4t5, A4ASa $aPElicEr€E €EF a |€,/17E(U{A'GF ltE

ffiffi6-nEDIE I TUTOFffi'T€S

DEGREE G C@,ITF|OL OFTE]FER OR TO IENTIFYA STECIAL s,ET G MECH-

NONIERROUS ATTOYS

Toble 4.3 Temper Clossificotion For Aluminum Alloys

Stain Hardening.

Modification of metal structure by cold working resulting in an increase instrength with a loss in ductility.

Thermal Treating.

Temperature treatment of an alloy to produce a stable temper.

4.2.2 Copper ond Copper Alloys

Most copper alloys are used because of their good corrosion resistance and

machinability. They are also homogeneous as compared with steel or aluminum

and thus not susceptible to heat treatment' Their strength, generally speaking,

may be altered only by cold working. The alioy designation system serves to

identify the type of material as shown in Table 4.4. Alloys 101-199 are normallya high grade copper with very few alloys added. Alloys 201-299 normally refer

b brass products that are mainly copper and zinc. Alloys 501-665 are bronze

products composed of copper and elements other than zinc. Other properties ofcopper alkrys are also sbown in the Table.

Jf

2- At'tt.€ALm3- nLUT TO.J HT TI]EN

"- #?,'ffif fi? ,HEN NATLTRALLYAGED TO sTAfuE COI'IDIT ION

5- ARTIFICIALLY AGED o/v-Y6- SLUT lc',l HT TIIEN

ART]FICIALLY AGED

a.ffiiig,rri;p:::ffi!*r9- SOLUT IO'I HT, ARTIFICIALLY

AoEO. T'/FN COLD IItrT<EDO- ARTIFICTALLY AGED TTEN

COLD V@KEO

56 MATERIALS OF CONSTRUCTION

Tqble 4.4 Copper AlloYs

Alloy Designation of CoPPers

l0l-19920tJ99301-399401-499501-599

601-64564s-665666-699701J30

Coppers

Copper-zinc alloys (brass)

Copper-zinc-lead alloys (leaded brass)

Copper-zinc-tin alloy (tin brass)

Copper-tin alloy (phosphor bronze)

Copper-aluminum alloys (aluminum bronze)

Copper-silicon (silicon brcnze)

Miscellaneous copper alloys

Copper-nickel alloys

Cold-Worked TemPer Designations

Approximate 9o ReductionbY Cold Working

Qua(er Hard

Half hard

Three-quarte$ hard

Hard

Extra hard

Spring

Extra spring

r 0.9

20.7

29.4

31 .l50.0

60.5

68.7

Most copper alloys are distinguishable by their color except for Cu-Ni alloys

that tend to lose their color as the amount of Ni is increased

4.2.3 Nickel ond High-Nickel Alloys

Nickel and high-nickel alloys have excellent corrosion and oxidation resistance

that makes thJm ideal for high temperature applications with corrosive environ-

ments. Products are normally called by their commercial names rather than their

ASME designation number as shown in Table 4 5'

4.2,4 Titonium ond Zirconium Alloys

Titanium and zirconium alloys are used in process equipment subjected to severe

environment. In the ASME Code, VIII-1, unalloyed titanium is listed for grades

I , 2, and 3, and alloyed titanium is listed for grade 7. Two zirconium grades also

given in the Code are unalloyed alloy 702 and alloyed alloy 705'

The modulus of elasticity for both titanium and zirconium is about half that

ol stccl. Also the coefficient of thermal expansion of both is about half that ofstccl. I'hc dcnsity o1'zirconium is slightly less than stcel' whereas the density of

lilrr|lirrrr is irlrottt 0.5ti lirttcs thltt ol stccl.

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i

MATERIATS OF CONSTRUCTION

4.3 FERROUS ALTOYS

Iron alloys with carbon content of less than 2qo arc known as steels and thosewith more than 27o are known as cast iron. Steels are further divided into those

with carbon content of more than 0.87o, called hypereutectoid steels, and those

with carbon content of less than 0.87o, known as hypoeutectoid steels. Moststeels used in pressure vessel applications have a carbon content of less than

0.47o. Steels with carbon content of over O.4Eo are very brittle and hard to weld.Cast iron used in pressure vessels dates back to the nineteenth century.

However, because cast iron is very brittle and because it cannot be rolled,drawn, or welded, its use in pressure vessels presently is limited to complicatedcomponents and configurations. The ASME Code, VIII-I, also imposes lim-itations on the pressure and temperature ranges and the repair methods.

Steel alloys can be produced with a wide variety of alloying elements. Someof the common elements and their effect on steel products are shown in Table4.6. The ASME Code, VIII-1, divides steel alloys into the following categories:

Carbon Steels.

These are widely used in pressure vessels. They have mainly silicon and

manganese as the main alloying elements and are limited in temperatureapplications to about 1000'F.

Low-Allny Steels.

These are essentially cbromium (up to l07o), Molybdenum, and nickel-alloy

Tqble 4.6 Effecl of Alloying Elements in Sfeel

AdvantagesElement Advantases Element

AluminumChromium

Restricts grain growth

Incrcases resistalca tocorrosion and oxidationIncreases hardenabilityAdds strength at hightemperature

Counteracts sulfurbrittlenessI ncrcases hardenability

l{ iscs grain-coarseningl('l|ll)crilturc( illrrlcrircls tcndencyt(,wirr(l lcnrpcr brittlenessIinllnrl(cs corrosionl( sislirtlec

Strengtheqs annealedsteelsToughens steels

Improves oxidationreslstanceIncreases hardenabilityStrengthens steel

Prevents formation ofaustenite in highchromium steels

Prevents localizeddepletion of chomiumin stainless steel duringlong heating

Increases hardenabilityRcsists tcmpcring

MirrUlocsc

Molylrk rrrrrrr

Titanium

Vanadium

4.4 HEAI TREATING OF STEETS 6I

steels. lhese elements enhance the steel for high temperature applications and

in hydrrigen service.

High-Alloy Steels.

These are commonly refened to as stainless steels. They have mainly chro-mium (over lOVo), nickel, and molybdendm alloys. The basically three types

of stainless steel used in process equipment are as follows:

M artensitic Stainless Steels.

This group includes type 410, which has a low chromium content of slightlyabove lTEo. They behave like steel, are magnetic, heat heatable, and difficultto fabricate.

Feffitic Stainless Steels.

This group includes types 405 and 430. They are magnetic but not heat

treatable.

Aust enitic Staink s s Ste e ls.

This group includes all 200 and 300 series and are chromium-nickel andchromium-nickel-manganese steels. They arc nonmagnetic and not heat

heatable.

4.4 HEAT TREATING OF STEELS

The lattice structure of steel varies from one form to another as the temperature

changes. This is illustrated in Fig. 4.8. Between room temperature and 1333"F,

the steel consists of what is known as "ferrite and pearlite." Ferrite is a solidsolution of a small amount of carbon dissolved in iron. Pearlite, which is shown

in Fig. 4.9, is a mixture of ferrite and iron carbide. The carbide is very hard and

britde.In Fig. 4.8 between lines A1 Qower critical ternperature) and A3 (upper critical

temperature) the carbide dissolves more readily into the lattice that is now called"ferrite and austenite." Austenite is a solid solution of carbon and iron that isdenser than ferrite.

Above line 43 the lattice is uniform in property, with the austenite the main

structure. The actual temperature for this austenite range is a function of the

carbon content of the steel as shown in Fig. 4.8.With this brief description, we can now discuss various heat treatments of

carbon steel.

Normalizing.

This consists of heating the steel to about 100"F above the upper critical lineAi and tben cooling in still air. The purpose is to homogenize the steel

structurc and oroduce a harder steel than the annealed condition

MATERIATS Of CONSTRUCTION62

'ri

=

PERCENT CARAON OF WEIGHT

Fisure 4.8 lron-iron corbide equilibrium didsrom'

ANNEALED

PEARLITE

NORI'ALIZEDPEARLITE

Figure 4.9 P€arlile 3tructure.

llrrr r'orrrisls ol lreating the steel to about 50'F above the upper critical line

,4, lrrrl tlrlrr ltttttittt cortling slowly. The purpose is to refine the grain and

rtrlttr'r srtllttr'ss

634.5 BRITTTE FMCTURE

Posweld Heat Treating-

It consists of heating to a temperature below the lower critical temperature-rJil- f- rft" purp;se of reducing the fabrication and welding stress and

softening the weld heat-affected zones'

Quenching.

The rate of cooling of steel after heat treating is very important in e-stablishing

ifr" ft.an"t. of st"eel Some steels such as SA-517 obtain most of their high

,t "ttgtft

Uy qt"*fting. The rate of cooling.depends on manY.factors such as

qu"nittlng -iAiu., tlmperature of quenching medium' and size and mass of

the part.

Tempering.

Ouenched steels are very brittle. In order to increase toughness' they are heat

;;;; il"* ;.;d then cooled to produce the desired propertv of high

strength and good toughness.

4.5 BRITTTE FRACTURE

Pressure vessel componenllconstructed of ferrous alloys occasionalllfail:ht-

i*tir;at"d,=A;".-"p:.-Iormalopqr3tingtdlFPeratureatalnressurewellilliiffi=0.;-;ftJG d*-i i"ir'* $"Jutt! o'"u'' ut to* !;'i-{'n"'1r iooffilrir;d uy incorporati4g.ufg1e fracture -coriiiderations

at the d€sgn

ffi ffirf .;Oiiticaiion requlreO vaiies from the simplest state-of-

il::* ;;d;dTi;,rt""t"iii.o.pri*tei mathematical analvses Both extremes

;; ;i;ii;,h;;i;isure uessel'designer, and their application ^depends

on the

amountofinformationavailableandtherequiredreliabiljtyolaglvencom-ponent.

Charpy V-Notch Test (Cv)

The Cv test is the simplest and most popular method of qualitatively determining

ffi'rrl."--i-Grl^cF*-TASTM- A-370_-and consistf of impact teslrn_g 9-n:sl:dJp.j"r_Tglllg: lljStaken from a spbcifib location of a product lorm. l he speclmen ls srucK wlln a

falling weighi(Fig.4.l}b) and the energy required to fracture it at various

tempJratures is recorded. Figure 4.11 shows two typical plots of the temperature

versus absorbed energy. The magnitude of measured energy, shape of energy

curve, and appearance of the cross section of tested specimens are all significant

in evaluating material toughness.

(a) srnolno specruel

ffil fn',n| | '-*nn

(c) pencert or sxeln rnlctuneAFTEB TESTII'IG.

Fig'rre ,1.10 Chorpy V-notch sp€c'men.

=sI

z

I

(b) resr ann*rormeu

cuRvi(A) CURVE(s)

64

Fisure 4.ll Cv En€rgy tronlilion €urye..

4.5 BRITTTE FRACTURE

The magnitude of energy level at a given temperature varies with differentsteels, as shown in ASTM 4-593. An energy level of 15 ft-lb is consideredadequate for 4-283 steel at room temperature. However, such a level is exceed-ingly low for .4-387 steels. Recognizing this fact is imperative in specifyingenergy requirements for various steels at different temperatures.

The slope of the energy curve in Fig. 4.11 gives the rate of change of steeltoughness with increasing temperature. At the bottom shelf of the curve, thesteel is very brittle as indicated by the cleavage appearance of the tested specimen. Failure is normally abrupt. At the upper shelf, material fails in shear and

- the cross section has a dull area. Failure bicurs after excessive yielding. [ow]--.r-o"ttgm;f"@Slt W,q sharp increase in toughness as the temperaiure increases,

as sho_wnin_Curve,4 qf Fig. 4 1 l. @-. as-sharu!_!J_ rfrrvx g-rn-\E:-.4.1t. This slight increase in toughness makes the

Cv test impractical to use in high strength steels.of dull and areas in the cross section of tested

nil ductility temperature.The nil ductility transition (NDT) temperature shown in Fig. 4.11 is of

significant importance when considering low strength steels. This temperature isbelow which the fracture appearance of steel changes from part shear to com-plete cleavage. Thus, this temperature is below which vessels with low strengthsteel must not operate without a detailed fracture evaluation.

The Cv tests give a good qualitative indication of fracture hends. TlpylQlot,however. eive anv coffelation between enersv and stress levels. Such informa-tion G-neeh wriirJ i-strisianaiviii is ftooila. roitrti.-iJ*n. other methodswere devised such as the drop weight test {DWT) established by the U.S. NavalResearch. Laboratory .

Drup We@ht Test (DWT)

The DWT procedure is given in ASTM E-208 and consists of welding a brittlebead on a test specimen. The bead is then notched and the specimen impactl$ted at various temperatures. The NDT temperature is obtained when thespecimen does not break upon impact.

In testing the specimens, deflection can be limited such that the stress atfailure does not exceed the yield value. Thus, a direct correlation is establishedbetween the NDT temperature and yield stress. Such information is used inconstructing the fracture analysis diagram (FAD).

Fracture Analysis Diagratn (FAD)

The FAD is one of the earliest applications of brittle fracture rules to fail-safedesigns. The results obtained from the curve are very conservative but require

The(Fig. 4.10c).

66 MATERIATS OF CONSTRUCTION

the minimum in engineering analysis. A simplified version of the diagram forlow strength steels is shown in Fig. 4.12 and indicates the types of tests requiredto construct the diagram.

Point A is obtained from the DWT and it establishes the location of the NDTtemperature with respect to yield stress. The crack arrest temperature (CAT)curve, developed by the Naval Research Laboratory is obtained by runningexplosive tests on sample plates at various temperatures and observing the crackpattem. From such tests the fracture tear elastic (FTE) point is determined at thetemperature at which the crack pattem changes from bulge and fracture to bulgeand partial fracture, as shown in Fig. 4.12. The fracture tear plastic (FTP) pointis obtained when the crack pattem changes from bulge and partial fracture tobulge and shear tears. The FTE point also locates the yield stress with respectto temperature, whereas the FTP point locates the ultimate stress.

Below point A in Fig. 4.12, fracture does not propagate regardless of thetemperatue as long as the stress is below 5-8 ksi. Between points A and I otherstress lines are drawn to correlate various stress levels. These lines are obtainedfrom the Roberson test, which consists of impact testing a specimen that is

TEMP+

" FLAT"FRACTIJRE

lft\ ItL) I"BuLGE"

&SHEARTEABS

tetL_________J

,BULGE'&

PARTIALFRACTURE

lKlMBULGE

&FRACTURE

0{DT+60"F) ( DT+120'F)

Figurc 4.12 Frocluro onolFis diogrom.

(NDT+r20'F)

-/

4.5 BRITTIT FRACTURE 67

stressed to a certain level and heated from one side to create a temperaturegradient as shown in Fig. 4.13.

Figure 4.14 shows the complete fracture analysis diagram. The range of flowsizes at various stress levels has been obtained from experiments as well as

experience. The experiments consisted of using large spheres of good impactmaterial and replacing portions of them with a notched brittle material. Thespheres were then pressurized to a given stress level at the NDT temperature ofthe brittle material. The size of the notch was varied with different stress levelsto obtain the range in the figure.

mpAcT

ERiPFigure 4.13 Diosrom of specimen urd in Roberrson crock-drrBr bsr.

YIELDsrnEss

lslzT

ll

= riloT (NI'I+3o'F) (NDT+60'F)TEMe-.-

Fisuro 4.14 Generdliz€d frocture onolysis d,osrom.

MATERIATS OF CONSTRUCTION

ln using Fig. 4-14, the following limitations must be considered:

1. It applies only to low carbon steels.

2. It is valid for thicknesses of less than 2 in. Larger thicknesses requirespecial evaluation and it has been proposed that the FTE temperature forthicknesses over 6 in. should be taken as NDT + 120"F rather thanNDT + 60'F. The FfP temperature should be NDT * 210"F instead ofNDT + 120'F. This indicates that for thick sections, Fig. 4.14 is on the

unconservative side and the safe operating temperature should be greater

than those indicated by the figure.

Example 4.2. A low-carbon steel material with NDT temperature of 15"F is

used in a pressure vessel. What is the rninimum safe operating temperature forsuch material?

Solution. Because no stress level is given, the minimum stress is assumed atyield. Entering Fig. 4.14 at yield stress, the CAT curve is intersected at the FTEpoint. Moving vertically, a temperature of NDT + 60"F is obtained. Thus, theminimum safe opemting temperature is 75'F.

If stress concentrations are assumed in the vessel and the stress level is beyondyield at some areas, then a conservative design is at the FTP point. In this case,

the safe operating temperature is NDT + 120'F, or 135'F. I

Example 4.3. A low-carbon steel vessel with an NDT temperature of -20oFis to have a start-up temperature of 0'F and a stress level of one-half yield. Isthe start-up iemperature safe?

Solution. From the CAT curve in Fig. 4. 14 the minimum safe temperature is

at NDT + 30" or 1ffF for a stress of one-half yield. Thus, start-up temperatureis on the unsafe side because it is less than 10'F. If start-up temperature iscritical, the shess will have to be decreased or a better impact material se-

lected. I4.5.1 ASME Pressure Vessel Criterio

The ASME Code, VIII, uses a different approach for preventing brittle fracturein pressure vessels with carbon steel construction. Division 1 prohibits the use

of some carbon steels below -20'F and requires impact testing of all others thatare subjected to temperatures below -20"F, with some minor exceptions. Di-vision 2 uses a more refined approach that takes into account the effect ofmaterial type, thickness, and temperature. Figure 4.15 is a simplified version ofthe code approach. It exempts some tough materials of a given thickness fromimpact testing when the service temperature is above a specific value given inthe filure.

4,5 BRITTLE FRACTURE

or11+22+3THICKNESS

GRoUP A: Nonnallzed SA-442 over 1,0 ln.orinrl ized 5A-516 And sA-662

GRoUP 8: SA-442 over 1.0 In. I'lhen l{ot Nonnallzed5A-516 Up To 1,5 In. Thick

GRoUP C: SA-442 Up To 1.0 In, Thlck

GRoUP D: All Carbon And Loll Alloy Steels Not Listed Above.

Figure,l.l5 lmpdci test exemption curvos for some corbon sreels. (Co'rrte.y of the Americon Society ofM€€honicol Ensineers.)

The above approach, although different from the FAD concept, is a practicaltool for preventing brittle fracture without requiring elaborate analysis; it isbased on test data.

The FAD and ASME criteria are applicable to low-carbon steels where theeffect of temperature is prominent. Toughness of higher strength steels or mate-rials such as stainless steels, aluminum, or titanium is independent of tem-perature. Thus, a different approach based on the mathematical theory of frac-ture toughness is used in establishing adequate toughness.

Example 4.4. A 3.O-in.-thick pressure vessel is made of SA-533 Gr. B mate-rial with an NDT temperature of 0'F. The design temperature is 50"F and thedesign membrane sffess is three-fourths of yield. What are the code fracturerequirements of this vessel if it is constructed in accordance with (a) Division Iof Section VIII, (b) Division 2 of Section VIIL

69

60

50

,4O":l 30

Cl 20

El 10

El o4l-l -1oulFt -20-30_40

-50-60

MATTRIALS Of CONSTRUCTION

Solulian. (a) Because the temperature is over -20"F, Division I does notrequire any analysis.

(b) From curveD ofFig. 4.15, the minimum temperature that exempts impactrequirements for a 3.0-in.thick vessel is 120'F. Thus a Cy test is manda-tory. I

4.5.2 Theory of Brittle Frocture

Basically the brittle fracture theory assumes that stress at the vicinity of a crack(Fig. 4. 16) due to a load applied perpendicular to the direction of crack is givenby the following expressions:

".: #(*":)('.: #(*":)('

. 0 30\-stn-stn-l'... . -... ,, I- -/,a30\-i srn; srn , ," -/

K, 1.0 0 3a\T4= ./^ | Sln; cos; cos r,vzTIT\ z z z/

where a,, o], 7ry = stress components at a point (ksi)

r, 0 = polar coordinates from tip of crack

Kt : fracine toughness factor (ksiV-in)

The fracture toughness factor K1 is a function of applied load as well as the

Figure 4.16 Elostic srres, distribulion neor rhe tip of

tft

4.5 BRITTTE TRACTURE

configuration of the body and crack. Thus K1 can be expressed by

&=oF (4.l)

where F : crack shape factor.

Unstable crack propagation occurs when the value of K1 reaches a criticalvalue K1c, which is a function of the properties of the material. Temperaturevariation could have a drastic effect on the value of K1c. as is the case with lowstrength carbon steels.

Some published K/c values are given in Table 4.7. Experimental deter-mination of the Krc factors is described in ASMT E-399 and is rather costly toestablish.

Values for the crack shape factor F are normally obtained from the theory ofelasticity. Because of the complexity of such analysis, only a few cases aresuited for practical use. Some of them are shown in Table 4.8.

Materials in general lose their toughness as the yield strength increases. Onemeasure of toughness is the ratio K16 f or. Ratios larger than I .5 indicate toughmaterials, whereas lower ratios indicate more brittle materials. A study ofK16/o,and Eq. 4.1 reveals that the defect factor F has to be very small when o" is highand K1q is low. In other words, very small defects in high strength materials canlead to catastrophic failures.

Fracture theory is one the most accurate methods presently available forevaluating maximum tolerable defect size. The main drawback is the difficulttask of obtaining I</c factors for different materials. Economics might dictate asimplified approach like FAD or the ASME criteria with a small permissibledefect size rather than a fracture theory approach that might allow a largertolerable defect.

Relatiaiship Between K1g and Cy

Determination of K1c values is tedious and expensive especially for low strength

Toble 4.7 Some Approximofe K7a Volues

JK/c (ksi\-inJ

Material - 300"F -200'F - 100"F

7l

A302-Gr. B

,A.5l7-Gr. F

A203 Gr. A norm.

A203 Gr. A Q and TA.533 Gr. B

HV-80

25

34

38

35

55

34

44

50

83

40 78

48

7'7

46

Tqble 4.8 Shope Foclors for Common Configurolions

case l: Flovl in a sheet of

Case 2:

Clse 3:

inf

l"

I n terna l ci rcul ar flow in

toItllll

Internal clrcular flow in a

F=zVT

inlte vrldth.

a sheet of finite wi dth

r l1'2F=Fr.L# tan @wl

thick pl ate.

F=V ra

( Ref. 2, p. 49 )

( Ref. 2, P

3, P. 39 )

o s0 )

( Ref.where "a" is radlus'of crack

73

Tqble 4.8 Shope Foctors for Comrnon Configurolions (Coniinued)

and oy ls yleld

Case 4: Internal elliptic flow jn a thlck plate

3t . If .a2E-'g_ E_

nor axi s and "

notcn.

axl sl,ihere

La se 5

u2a' is

: Singlo

the ni

e edge

II

6 E- ( Rer'

is and "2c" is the maior

F=CVir.a

c=1i1'iF+o227+ roa(i)'-zasz(if

+4272/ \4t+l\r /

( XET 328 )

case 6: Elliptlcal surface flow.

F=

where "2c" is cra

materlal stress,

ck length' "a" is crack depth, o lsstress. ( Ref.

actual

31 5 )

- o.ztzo2f or2

74 MATERIALS OF CONSTRUCTION

steels. Various methods relating the,K/c factor to the relatively inexpensive Cytest have been suggested. One empirical method proposed by Rolfer and Barsomconsisted of preparing two equations for correlation purposes. One equationrelates the Cy and K1s values at the transition temperature region, whereas theother equation is applicable at the upper shelf region. Thus, for the hansitionregion,6

Krc:155Cvtl2

whereas for the upper shelf range,

(4.2)

(4.3)

where Cv is in ft-lb, o" in ksi, and K1g in ksivG. Equations 4.2 and 4.3 are formedium strength steels such as 4.517 Gr. F and 4302 Gr. B.

4.5.3 Hydroslotic Testing

Hydrostatic testing of a pressure vessel is the best available method for deter-mining maximum tolerable defect size. Thus, if a thick pressure vessel ishydrotested at a pressure that is 50% greater than the design pressure, the criticalK1 is given by Eq. 4.1 as

K1g = oF

Assuming an intemal defect represented by case 3 of Table 4.8, the maximumK16 immediately after hydrotesting is

/ I;\Krc: t.5 r.(r\/;J

Maximum defect size r at the design pressure is given by

: "

(,

or

- - t t<^

Hence, a crack that is discovered after hydrotesting can grow 2.25 times itsoriginal size before causing failure. This fact illustrates the importance ofhydro-testing and is based on a hydrostatic temperature that is the same as the lowestoperating temperature of the vessel.

f&Y:4ofQ-o.r)\ov./ \sr /

"',('fu) -l1f/

4.5 BRITTTE FRACTURE

4.5.4 Foctors lnfluencing Britlle Froclure

Many factors such as torch cutting, arc strikes, and cold forming affect the britdefracture behavior of metals and should be considered in fabricating pressurevessels. Torch cutting or beveling of the plate edges may lead to hard and brittleareas. In cases where this condition is underdesirable the plate should be heatedto minimize this effect. Grinding the edges eliminates the hard surfaces.

Arc strikes can create failure by brittle fracture especially if the strike is madeover a repaired area. It is desirable to grind and repair all arc strikes beforehydrotesting, especially at low temperatures.

Cold forming of thick plates may lead to fracture in areas with stress raisersor plate scratches. All stress raisers should be ground off to minimize theireffect. Hot-forming substantially improves the situation because it increases theNDT temperature and thus prevents brittle fracture.

Example 4.5, A titanium pipe (ASTM 8265 Gr. 5) with a 2.375-in. outerdiameter and a 0.154-in. wall thickness has an actual stress of 30 ksi. a vieldstress of 120 ksi, and K1q : 40 ksiVin at a given temperature. The pipecontains a flow of depth 0.05 in. and length 0.25 in., which is similar to case6 of Table 4.8. What is the maximum internal pressure the pipe can hold?

Solutian. From a conventional strength of material analysis, the pressurerequired to yield the pipe is given by Fig. 5.4 as

75

Using fracture toughness approach, maximum stress is

KrcF

From case 6 in Table 4.8,

1.12\4;Xo.o5)

^ a\Rl - Rl) 120(1.88'z - 1.034':)' Rl - Ri r.88r - r.034,

= 0.359

Hence, o : 40/0.359 - 111.4 ksi

^ tll.4(t.88)2- t.0341maxr: ll8' r l034': : 14Jl(sl

'l'hcrclirre liacture toughness criteria control the design. I

NONMETATTIC VESSEIS

Hydrogen ,orliol pretlu?c, P.r. i

Fisure 4.17 (Courtesy of the Americon Pelroleum lBrirute.)

4.7 iIONMETALLIC VESSETS

Rules for fiberglass-reinforced plastic (FRP) pressure vessels are covered in

Section X of the ASME pressure vessel code. Construction of FPR vessels is

divided into four classes: the contact molding, bag molding, centrifugal casting,

77

76 MATERIAIS OF CONSTRUCTION

Examnle 4'6. An A302-B material with a yield stress of 50 ksi is to be used

irl ".!*"* "".r"1. The Cv value is 15 ftlb and an examination of the percent

#"i^il;; .rott ...,i* of tested specimens indicate a temperature in the

transition range. Ultrasomc examination of the plarc uncovered an elliptical

il;;;;d. fit" surface that is 0.375 in long and 0 25 in deep How safe is

the vessel if the operating stress is I yield?

Solution. From Eq 4'2' K1g = 15(15)0 5 = t9 ksiVin-

From Table 4.8, case 4,

\/;ajzrF = ;ts+l;ts,,oI25/orus" - "-"

Hence, from Eq' 4'1, o-: 19/0'46: 41 ksi Actual stress : 'l

x So = r:i.fl*fit"rt it l_es's,rtan the critical brittle stress Therefore' operation of the vessel

is safe unless the defect grows in size l

4.6 HYDROGENEMBRITTLEMENT

Essentially, the two different methods by which hydrogen can embrittle steels

are:

l. Hydrogen decarburization' In this case hydrogen penetrates the steel

and combines with the carbides in the structure (Fig' 4 8) to form meth-

ane gas. This gas accumulates in the space of the original carbide and

;;il;t ;p p;";t" that leads to cracking' This process normally acceler-

ates with an increase rn temperature ind in operating -pressure'

One

method of minimizing hydrogen attack is by using Cr-Mo steels' Here

the carbides are in solution wiih ttre cr or Mo and do not readily combine

*ittt th" hyd.og"n. The type of steel to be used in a giv€n combination

of te-perutu."hd p,"ttot" it normally determined by the Nelson chart

in Fig. 4.17.

2. Hydrogen attock. Researchers have observed that hydrogen attacks-'

"Jatuin"r"gions of a pressure vessel at temperatures below 200'F when

they have-high hardniss zones in the range o1200 Brinelland higher' The

exact mechanism is not known exactly, but it is believed that the hydro-

een is attracted to hard regions with higher stressed zones. Accordingly,

;;;;;;;q-t" soft h;-affected zones with a Brinell hardness below

200 to avoid hydrogen attacks at low temperatures'

1C;-Yr Mo Stccl

MATERIATS OF CONSTRUCTION

and fi lament-winding Processes.In the contact-mol,cling process reinforcements and resins are placed in a cast

mold and cured at room temperature. Vessels constructed by this process are

limited to a design pressure of 150 psi.

In the bag-molding process a pressurized bag is used to compress Prerolled

fiberglass "!hnd".t

and heads preforms against an outer heated mold The

uessels conjtructed by this process are also limited to 150 psi pressure'

In the centrifugal casting process, the cylindrical sections are formed from

chopped fiberglasi strands and a resin system in a mandrel, which is spun to

oroduce a suitible laminate and heated to cure the resin system Pressure vessels

constructed by this method are also limited to 150 psi design pressure'

In the filamenrwinding process, filaments of glass and resin are wound in a

systematic manner to form various components. The ASME code limits the

pressure range to 1500 psi for filament-wound vessels with cut filaments and to

i000 psi for filament-wound vessels with uncut filaments'

FRi' vessels normally operate at low temperatres. The ASME Code, Section

X, limits the temperature iange between a minimum of -65oF and a maximum

of 150'F. Also, because the modulus of elasticity is about 1 x 103 ksi, special

care must be exercised in designing various components Because of this and

because different fabrication processes produce different strength vessels, the

ASME Code states that in order for a given vessel to be adequate, a prototype

must be cycled 100,000 times between zero and design pressure and then burst

at a Dressure not less than six times the design pressure'

REFERENCES

Pellini, W. S., "Principles of Fracture Safe Design-Part l" ln Pressure Vessels and Piping:

Design and A'',,lysis, Vol. l, American Society of Mechanical Engineers' New York' 1972

Tetelman. A. S., and A. J. McEvily, Jr' Fracture of Structural Materidlt, John Wiley '

New

York, 1967.

Rolfe. S. T., and J. M Ba$om F/(r cture anal Fatig e Control in Structures' Prentice-Hall'

Englewood Cliffs, N.J., 1977.

Riccardella, P. C., and T. R. Mager, "Farigue Crack Growth in hessunzed Water Reactor

Vessels" in Pressare ye ssels and Piping: Disigtl and Analysis, Vol l, American Society of

Mechanical Engineers, New York, 1972.

Wessel, E. T., andT. R Magel, "Fracture Mechanics Technology as Applied toThick-Walled

ttu"t"- pa".*," Vessels" in Pressure vessels and PipitlS: DesiSn and Analysis' vol l'American Society of Mechanical Engineers' New York, 1972

Roberts, R., and C Newton, "lnterpretive Repoft on Small-Scale-Test Conelations with f''DaA," Welling Rcs\rch Council' Bulletin 265, New York' | 981'

BIBTIOGRAPHY

BIBLIOGRAPHY

Aluminum Standards aM Data, Alrmin'um Association, Washington, D.C., 19'19.

Alner, S. H.,Introduction to Phlsical Metallurgy, McGraw-Hill, New York, 1964.

Nichols, R. W -, Pressule Vessel Engineering TechnologJ, Applied Science Publisherc, England,t91l .

Thielsch, H., Defects and Failures in Pressure Vessels akd PipinS, R. E. Krieger, New York,1965.

PART2ANALYSIS OF

COMPONENTS

8l

CHAPTER 5STRESS IN CYLINDRICAL

SHELLS

Atr'0|t|h,t|.otcy|lndr|co|lh!|hinochomicolP|on|'(cour|g,yofE.|.dUPontdaNomoursondco.)

83

11

where p = g€ssure

Z = length of cylinder

o, = hoop stress

r = radius

I = thickness

The stain ee is defined as

and from Fig. 5.2,

PL.2r:ZoeLt

final length - original lengthoriginal length

2t(rIw)-2rr€0=

---;-z1fr

3TR!IS IN CYTINDRICAI gHErrS

(s.1)

(s.2)

5.I STRESS DUE TO INTERNAL PRESSURE

Thc classic equation for determining stress in a thin cylindrical shell subjectedto pressur€-is obtained from Fig. 5.1. Summation of forces perpendicular toplane ABCD gives

Prot= T

w

Flgor.5.l Fr€c-body diogrom ol d rylindricol $ell lubi.ctod to intarnol prerlure.

3,I STRI3E DUE TO INTTRNAT PRESSURE

Figur€ 5,2

Also,

where

the.radj4 deflection of a cylindncal shell subjected to internal pressure isobtained by subsdrudng rhe quantity e6: osf E in6 Eq. 5.i. ilnce for thincvllnoers

85

dwar (s.3)

(s.4)

E = modulus of elasticity

Equations 5.1 and 5.4 give accurate results when r/r > 10. As r// decreases,however, a more accurate expression is needed because the stress distributionthrough the thickness is not uniform. Recourse is then made to the,.thick shell,,theory' first developed by Lame. The derived equations are based on the forcesand stresses shown in Fig. 5.3. The theory assumes that all shearing stresses arezero due to symmetry and thal a plane section that is normal to thjongitudinalaxls

-betore pressure is applied remains plane after pressurization. In other

words, e1 is constant at any cross section.

_. A relationship between oi and_o1 can be obtained by taking a free_body

diagram of ring dr as shown in Fig. 5.30. Summing ior"", ii the verticaldirection and neglecting higher-order tdrms, we then h-ave

ae- o,

A second relationship is obtained from Eqs. 3.2, which is written as

= do,dr

\. o,II(l + rr)(1 - 2*rl+(t - 1t) + p,(e, + e)l

(5.5)

86

(l + p)(1

E

[er(l-p)+pr(e,+e)] (s.6)

(1 +pXl- .r.^ [€r(l - ;r) + g.(e, + e)l

Substituting Eqs. 5.2 and 5.3 into the first two expressionssubstituting the result into Eq. 5.5 results in

dzw ldw w ---'-t -r - --- - -- = var rar r-

A solution of this equation is

of Eq. 5.6

(s.7)

whcre A and B are constants of integration and are determined by first substi-tuting Eq. 5.7 into the first one of Eq. 5.6 and then applying the boundaryconditkrns

o,: -Pi at r = ri

rnd

o!

la'Fisuro 5.3 Cro5s i€.rion of

tb)o tfiick cylindricol :hell.

and

.B

o, = _p,

ljxorrulon 5.7 then becomes With or known, Eq. 5.8 for the deflection of a cylinder can be expressed as

5.I STRESS DUE TO INTERNAT PRISSURE 87

w = -t .€1 . *n--t=Alr'(t - t, - ztt\(Er? - p,r|) + r?r1{o + D(n - p")l (5.8)

Once w is obtained, the values of o6 and d; are determined from Eqs. 5.2, and5.3, and 5.6 and expressed for thick cylinders as

_ _ pr? - p,r? + (pt - p")(r7r3/r2)aa =

----- 4:i- (5.9)

4rZ - nrl + (n - n)blrl/rz),-" - ,l

where o; = radial stress

od : hoop stress

p, = intemal pressure

p, : extemal pressure

4 : inside radius

r, = outside radius

r = radius at any point

The distribution given by Eq. 5.9 of the shesses through the thickness of acylinder due to iniernal and extemal pressues is shown in Figs. 5.4 and 5.5,respectively.

A comparison between Eqs. 5.1 and 5.9 is shown in Fig. 5.6. The figureillustrates the adequacy of Eq. 5. I for r,/4 ratios of less than or equal to 1.1 (or,conversely, rt/t > lO).

The longitudinal stress in a thick cylinder is obtained by substituting Eqs. 5.2,5.3, and 5.8 into tle last expression of Eqs. 5.6 to give

oI= EeI+2p(\? - !"31r;- ri

This equation indicates that o1 is constant throughout a cross section because €r

is constant and r does not appear in the second term. Thus, from Fig. 5.7 theexpression ot can be obtained ftom statics as

(5.10)4r? - P"r'z.

,2_.?

6=tr,..t!id- )- f;-Ji \ J' I

"'=-++r(#)

Ar Inne. surr.ce +=t+)=-t

.Ai Ourer surraco +="#

ft=oFigurc 5.4 Slrca! dishibdion in o rfiick cytirder du. to inrornol prcrlur€.

,,=#e#)

At fnnersurtace "r=ffift=o

aroorersurfaca "r-#ifl9=_r.oP"-

Figuro 5,5 Str€ls dithibt ion in o thick cylin&r due l,o enornol pr.3sur€.

88

Fi9ur6 5.6 Comporison ot tornulor for hoop sire33 in o cylindricol 5hcll.

90 STRESS tN CYr"tNDRtCAt SHtrrs

l(Ptri - P,,r!,)lt - zr.t) + (h - t',,)rlrl,(l + rt)

-

Lr\r; - ri)(5. 1 1)

The deflection pattem for external and intemal pressures is shown in Fig. 5.8.

Example 5.1. The inside radius of a hydraulic cylinder is 12.0 in. What is therequired thickness if P : 7500 psi and a6 = 20,000 psi?

Solulinn. From membrane Eq. 5.1

Pr 7500 x 12t=-: oo 20,000

= 4.50 in.

lnternal Pressur€

*=!"il+##tdl

Max.!t At Inner Surlace

w.=E:tfl2:lll'E(ro"-r;")

*=-& . t 'tr'(r - zr) * r,"(r *,., )Ie.r (r;- r,')

Max.q At Inner Surrace w,=-B:dJiJ2:t)EGTil

External Pressure

Figure 5.8 Rodiol deflftrion du€ lo inlernol ond exrernol prelsore.

ot _rl+ r?

P rl- r!

5.I STRESS DUE TO INTTRNAI PRESSURT

From Lame's equation (see Fig. 5.4)

: 5.80 in.

Hence, the error of using Eq. 5.1 in this case is 22Eo. I

Example 5.2. A cylinder has an inside radius of 72.0 in. and an internal

pressure of 50 psi . What is the required thickness if the allowable stress is 1 5 ,000psi, 1.r

: 0.3, and E = 30 x 10o Psi?

Solution. From membrane Eq. 5.1,

50x72 : 0.24 rn.

9l

15,000

From Eq. 5.4,

50 x 722 : 0.0360 in.30x106x0.24

Using the thick shell theory, we obtain the required thickness from Fig. 5.4 as

'): O.24 in.

and from Fig. 5.8,

(5q?2f | 2-0.3 \' : 30 x rgu \tf* - rV1

= 0.0305 in.

Examples 5.1 and 5.2 indicate that Eq. 5.1

sreater than 10. I

loe + p, = /l "\ / --------= -

is adequate when the ratio rlft is

3TRI3S IN CYIINDRICAI SHETIS

Problcms

5,1 A cylinder with an inside diameter of 24 in. is subjected to an intemalpressure of 10,000 psi. Using an allowable stress of 25,000 psi, determinethe required thickness.

Answer: t:6.33 in.

5.2 A cylinder has an inside diameter of 12 in. and an outside diameter of 18

in. Determine the maximum intemal pressure that can be applied if themaxirnum allowable stress is 20.000 psi.

Answer: p1 = 7690 psi

5.3 A cylinder is subjected to an external pressure of 5000 psi and an internalpressure of 2000 psi. If 11 = 15 in. and rz = 19 in., what is the maximumcircumferential, longitudinal, and radial stresses? At what location do theyoccur?

Answer: o6: -17,900 psi at inner surface

o, = -5000 psi at outside surface

at = -9960 psi uniform tkough thickness

5.2 DISCONTINUITY ANALYSIS

All the previous equations were based on the assumption that the cylinder is freeto deform under pressure. In practical applications, the cylinder is attached toend closures that reshain its deformation. Other items such as stiffenirg ringsand internal bulkheads affect the cylinder deformation and introduce localstesses. These local shesses can be evaluated by a dicontinuity-type analysisusing the general bending theory of thin cylindrical shells. The theory assumesthat the loads are symmetric around the circumference and that the thickness ofthe shell is small compared with its radius. It is also assumed that the in-planeshearing forces and moments are zero. The problem then reduces to that ofsolving the forces shown in Fig. 5.9. The relationship between these forces canbe obtained from statics. Hence, from Fig. 5.9,

>4:0

ff*,ar=o

5.2 OISCONTINUITYANAIYSIS

l{o dx

(b)

Figore 5.9

which indicates that N, must be a constant.

Let

N,:0

Also,

>4=0

93

dx

+ 4!!.|i ax

dox dx-?t

d?'+at=rdxr

Similarly,

2M,=o

(s.12)

94

or

3TRE5S IN CYIINDRICAT SHETIS

(s.l3)

(5.14)

*-a.=oax

Deleting Q, from F4s. 5.12 and 5.13 gives

N, d2M,

- + ---:--l : rr ta-

This equation has two unknowns, N6 and M,' Both unknowns can be expressed

in t".-, of,h" O"flection w. The relationship between M' and w is given by Eq'

3.1I as

D, Et'= rro=E (s'15)

Because the rate of change of deflection in the a-direction is zero due to

symmetry, the above two equations reduce to

".= -r(**)(s. 16)

Mo=

and

Me = FM, (s.17)

'l'hc cxprcssion for No is derived from the axial and hoop strains' In refening to

liil, 5.9, thc uxitl strain is given by

",: -r(r,tt. - *t")

duax

-'(#..?)

",: _*(#)

5.2 DISCONTINUITYANAIYSIS

and the hoop strain as

wee= -i

Substituting expressions 5.18a and 5.18b into Eq. 3.11 gives

95

Substituting Eqs. 5.16 and 5.19 into Eq. 5.14 yields

Also,

y,=.Et ;1r,r p,e6)=ot' - lt-

EtNs:7 ,\es+ Pe)r-p-

u4*r(*\:-,r' \dx+ /

du /w\=

: lLl- |dx \r/

(5.18b)

(5.19)

(s.20)

Defining

".ilno: Lt -Jlt

- lL-)4r2D r2t2

the differential equation becomes

(5.21)

)(5.l8a)

where p is a function of .r.

ffi*oon*=-+

96 tTRlSt tN CYUNDRTCAT SHlrtg

5.2.1 Long Cyllndorr

One of the most practical applications of Bq. 5.21 is for long cylinders subjectedio end shears and moments as shown in Fig. 5.10. The force and deformationdistribution at any point r along the cylinder due to O0 and M0 can be obtainedfrom Eq. 5.21 with P = 0. Hence,

aw;i+4p.w:o

A solution of this equation can be expressed as

' '"- = ::lg l -:1. **!.gz;it -

p*t + - y-tc-'-e$ €L-L. 9r-t* -ff)- _J-.gD

By observation we can conclude that the deffection due to @6 and Me approacheszero as .r aplnoaches infinity. Thus the constants C1 and C2 must be set to zeroand the solution becomes

The constants C: and Ct can be evaluated from the boundary conditions

(s.22)

M..=o=

Mo= -'e;1.="

;i = e-tu(q cos B, a Ca sin

Figur.5.l0

5,2 DISCONTINUIIY ANAIYSIS

and

Hence,

a.l._,= n"= -,(#)1."

.'.'*-"".--",^ _ -lct=;;7i(Qo + BmsliLP tl

;

'Ms\ a=,L*--- ?E:p" - '"" l

Equation 5.22 then becomes

w = ]6rl?tvtobin Bx - cos F.r) - O cos Fx].

By defining

{* = e-e(cos Br + sin Br)

!* = e-&(cos Br - sin pr)

Cs = s-$t "o"

PDs = e-tu

"in Pthe deflection and its derivatives can be expressed as

, :;fu<oro"* + escp)

#= *1roU""ctu+ S.AP)

t:=$aor"os+zeoD,,)

# = jour,De'- enBB,)

Values of Ap,, BB,, CB,, ar]d Dg, are given in Table 5.1.

Using the terminology of Eqs . 5 .23 and 5 .24 , the expressions for M,are represented by

i7

(s.23)

(s.24)

arfi Q,

98 sTRtss tN CYUNDRtCAt SHEttS

Tqblr 5.f Vqfurr of functiqng Ap* Bp- Cp. Dp,

Cp" DB,Bu,B,

00.05

0.10

0.15

0.20

0.30

0.40

0.50

0.55

0.60

0.80

l_00

1.20

1.40

1.60

1.80

2.00

2.50

3.00

3.5

4.05.06.0't.o

1.0000

0.99760.9907

0.9'19'7

0.9651

0.926'7

0.8784

0.8231

0.79340.76280.6354

0.5083

0.3899

0.2849

0.1959

0.12340.0667

-0.0166-0.0423-0.0389-0.0258-0.0045

0.0017

0.0013

1.0000

o.9025

0.8100

0.7224

0.6398

0.4888

0.3564

o.24150.1903

0.1431

-0.0093-0.1108-0. 1716

-0.201l-o.2077-0.1985-0.r794-0.1149-0.0563-o.0177

0.0019

0.0084

0.0031

0.0001

1.0000

0.9500

0.9003

0.8510

0.80240.707'7

0.61740.5323

o.49190.4530

0.3131

0.1988

0.1091

0.0419

-0.0059-0.0376-0.0563-0.0658-0.0493-0.0283-0.0120

0.0019

0.00240.0007

0.0000

0.0475

0.0903

0.1286

o.1627

0.2189

0.2610

0.2908

0.3016

0.3099

o.3223

0.3096

0.2807

0.24300.2018

0.1610

0.1231

0.0491

0.0070

-0.0106-0.0139-0.0065-0.0007

0.0006

M"=(5.25)

Q,=QBMnDy-QoB*)

The relationship between M,, M" Q, N, w, and d for various boundary condi-tions is shown in Table 5.2.

Exarnple 5.3. A long cylindrical shell is subjected to end moment M0. Plot thevalue of M, from F, : 0 to p, : 4.0. Also, determine the distance .r at whichthc moment is abort 7Vo of the original applied moment M0.

Solution. From Eq. 5.25, M,: MoAs.The values of Ak are obtained from Table 5. 1 and a plot of M, is shown in

_r(+\\ax- / =

fi<zou,ea + 2eoDB,)

5,2 DISCONTINUITYANATYSIS 99

Toble 5.2 Vqrious Disconlinuity Funclions

e

t4^

Ne

qo

- ilo,F;tlo---F

-2 oB2.t

0

Qq

2 B3A

2Bz,D

0

26. r. Qo

Qo

0

zB2.o.to

E.t.Ao

46 3. 0, A0

0

0

28. D. eo

0

282 .D.e o

0

Ne

Qx

-llo 8o",F.o ""

t4o C^E;r'- b"

zlioB2.r.Br,

-26. tito , DBx

0o C-

28'D

-0oiPo ^p,

q9 . D-B FX

28. r. Qo. CBx

Q6. Bgx

Ao(2Cax-8Bx)

z8do ( ABx - cBr)

zB2.o,t4lrr -cr,

!1. ao ( 2ce " -8."

4B3D,aJ B6x- DBx)

? ttu'-tu''

eo {ABr( - 2CBx)

?$ D'ed DB, -aB")

:.t.00(c^ -8" )F-3- p'

tB2. D. oo ( 2DBr+B B)(

'clockwise moments and lotation arcpositive at point 0. Me = tir|..

positive at point 0. Outward forces and deflections arc

Fig. 5.11. From Table 5.1are about 2.00.

or

the values of Bx at which M, is equal to 7Vo of Mp

Bx = 2.00

and

.r: 1.56\G

The significance of the quantity 1.56Vr . l is apparent from Fig. 5. I 1. It

2

e

(l)

shows

r00 3TR!35 lN CYllNORlCAL Sl{llts

that 0 momcnt epplicd at thc cnd dissipatewery rapidly and reduces as much as

94% ofthe original momcnt atr = l.56Vrt. This indicates that any other forceapplied ot that distance x can be analyzed without regard to the applied momentMo' IExample 5.4. A long cylinder is subjected to end shear ps. Plot the value ofM, as a function of C0 from Fx : 0 to Fx = 4.0 and derive the location of themaximum value of M,.

Solutinn, From Eq. 5.25,

,'=*o^Referring to Table 5. 1 for values of D p,, a plot of M,/ (Qo/B) can be constructedas shown in Fis. 5.11.

q'ry

Figura 5.1I

and

(l) I

5.2 DISCONTINUITYANATYSIS 't0l

To find the maximum moment M,, the above equation can be differentiatedwith respect to r and equated to zero. Hence,

*, = 3.e-& sinA

ff = o = ftr-Ou* sin Bx * Be-tu cos Bx)

and maximum moment is given by

, _ o.34Qo

PrEtlw\==:- "lexr lr-lz r-F-\ r/

pr=,E' ,(Y* ur,\r - rr'\r /

and

Solving for e, and w gives

I

Example 5.5. Determine the expression for the deflection of a long cylinderwith end closures due to intemal pressure p.

Solutinn. For intemal pressrre p, the axial force it 7'1, = pr/Z and the hoopforce is Ne = pr. Also from Eqs. 3.11 and 5.18,

Pr (l - 2tt)''Et2

'#('- t)

t02 STNESS IN CYTINDRICAT SHELTS

trlxomple 5.6. A still'cning ring is pllced anrund u cylinder at a distance

rcmoved from the ends us shown in Fig. 5. 12. The cylinder has a radius of 50.0in., a thickness of 0.25 in., and is subjected to an intemal pressure of 100 psi.

Assuming E : 30 x 106 psi and & = 0.3, find

(a) The discontinuity stress in the shell with the ring assumed to haveinfinite rigidity .

(b) The discontinuity stress in the shell and ring if the ring has a thicknessof 0.375 in. and a depth of 4.0 in.

Solution.

(a) A free-body diagram of the shell-to-ring junction is shown in Fig. 5.13.Because the ring is assumed to have infinite rigidity, the deflection due topressure must be brought back to zero by a force pq. Also, because the slope atthe shell-to-ring junction is zero (due to symmetry), a moment M0 must beapplied at the junction to reduce the slope created by force Qs to zero . From Fig .

5. 13,

deflection due to P - deflection due to Qe * deflection due to Mo = g

The deflection due to P is obtained from Eq. I in Example 5.5, whereas thedeflections due to M0 and Qs are obtained from Eq. 5.24. Hence

tl(r-E\ -4 +!!=:oE.r \' 21 2B3D zB'D

4" x 3/8" ri ng

tigure 5.12 Me = 96.4 in.-lb/in.

5.2 DISCONTINUITYANATYSIS-+rrtr" I

oo {odo

r03

Figuro 5.13 Sign conv6ntion oi poini 0, clockwiss 0 ond i4" ore + , outword w dnd Q. oro -

From Eq. 5.15,

D = 0.00143 E

and from Eq. 5.20,

B=03636

Hence the deflection compatibility equation becomes

Mo- 2'75OQo= -321'39

The second compatibility equation gives

(1)

rotation due to O0 - rotation due to Me = 0

Qo-ZBUo:g e)

Solving Eqs. I and 2 gives

Mo : 321'4 in.Jb/in'

The maximum longitudinal stress is given by

.=N.ry= 40,900 psi

The maximum hoop moment is given by Eq. 5.17 as

-, Etw"r

But because w = 0, N6 is equal to zero and the maximum hooP stress is

*:T*,",

O) The shell deformations are expressed as follows:Due to R

, - p,-'- -1-',

", - ='#l'- i)

{

L'=o *-*-- I

?Y'oo'

l0.l

Tho hoop force Nc is glvcn by Eq. 5.19 as

Due to Mo,

tTRttt tN cruNDRrcat sHttls

Mn

':t9o-Mo

@

The ring deformations are expressed as follows:Due to p,

pr(r + d/2\n=- dE

0=0

Solving Eqs. 3 and 4 yields

!.2 DISCONTINUITY AI,IATYSIE

Due to 00,

Due to Mo,

The deflection comPatibilitY is

2Qor(r + 2)

'= - bdE

0=0

w=0d=0

t0t

(3)

and

(4)

Mo = 152'0 in-lb/in.

Oo : 110.6 lb/in.

wp - wQo* ,*]"' - ,o + *roof^"

,i( - t) - #. #=,*#!".'3e#2

Mo - 4.0@o: -296'8

Sirnilarly,

f0, + ilao- 9al"t. = lflp + ilMo - 0oJn*

2BMs-20=g

pr , 2QorO,: -- r ---:---

aDa

= 1250 + 7370

= 8620 psi

Maximum longitudinal stress in cylinder is

Pr . 6MnO'= - - ---A-zt I-

_ 100x50 x6x t5?2 x 0.25 (0.25).

: 24,600 psi

t06

lloop stress in ring is

Hoop force at discontinuity is

where

and hoop stress at discontinuitv is

STRESS tN CYUNDRTCAT. SHfttS

w=lt,p-wgo*wyo

=P"(,-E\- Q, * uoE \' 21 2B3D 2B2D

_ M7,500E

.. E (0.25)(447 ,sOO / E)

'", = -- 50

= 2238 lb/in.

No . 6Meoe=----1----r-T T-

2238 , 6(0.3 x rsz)0.25 0.252

= 13,300 psi

5.2 DISCONTINUITYANATYsIS

whereas away tiom discontinuity

Proe:::20'000Psi

Problams

5.4 A long cylindrical shell is welded at one end to a rigid bulkhead such thatthe deflection and rotation due to applied pressure are zero. If r = 36 in.,I = 0.5 in., p : 240 psi. lt: 0.3, and E = 29 x 10" psi, what is themaximum lonsitudinal stress?

. Pr 3PrAnswet: a= n'T -:E-zt tv3\t _ p2)

: 40,015 psi

The shell in Problem 5.4 is welded to a thin bulkhead such that only the

deflection is zero due to applied pressure. What is the maximumdiscontinuity stress?

. Pr 0.966PrAnsweri o= z *;'Fyt - u31

= 18,740 psi

5.2.2 Short Cylinders

It was shown in Eq. 1 of Example 5.3 that the applied edge forces in a long

cylinder dissipate to a small value within a distance of l.56Vrr. This basic

behavior enables the designer to discard the interaction between applied loads

when they are far apart. As the cylinder gets shorter, the assumption of long

cylinders does not apply and constants C1 and C2 in Eq. 5.8 must be considered.

Consequently, Eqs. 5.23 and 5.24 have to be modified to include the effect ofall four constants. Equation 5.8 may be rewritten in a different form as

w = Ar sin Bx sinh Bx * 42 sin Pt cosh Bx

*A3 cos Bx sinh pt + A4 cos Br cosh B-r 6.26)

and a solution obtained for various boundary conditions.The most ftequent application of this solution is in the case of edge forces and

deformations as shown in Table 5.3. Many practical problems can be solved

with the aid of Table 5.3 by itself or in conjunction with Table 5.2

t07

f,.f,

o!

=I0

o

o

c

3

(t

-9.oqF

=

t ----TI ,rlt_,Y

"--l

108

5,Tlris -:-

dld,-------, --

Jr ulu v:{vdle | ++F:E $16 T

-i=._, : d<'Fi €1. s

'=-

's":ss)sdl6 dl6 dl6 dl6 -:-| + + | u'16ys:gvrdl6 sl6 dl6 dl6 ;s| + | | dl6:ss)9;S+

dlu dlu d'lu dlu s,, L--;-,!t,:

-l-F.ils ' P '"'",o,l$ l-l$ dl.q R d

r-li- r"-rS r--=-'1 | .-=-t>-r-rS+s I s 6le --i -dlu if | + -Hlu-+9lri>-s-s "i rJlr.r 51s I

dl.: s | | ."T..t l-. tTlri :g r---'J,v ' v,

=;+ dt6, E ssls slS-E- h t

:s,-----..-- dl6UIU :9r $16

5ddld +

t>.:s,1,+<'< (a,

S .--- .ir ,:--:- --r d|U } - ":"f- f-"i-clo *dl6<ilr-, * r f, -i-.ir .. s tilrir s ",- dlu +>' s '-i- dl6 +riltiril6 * + f,i -i- ..i.. >- dld

;' )J L-J (JlU €>,' _ ,u ru,

s >" --.: dlu -dlddlu F f SEc:i R $l* i€lcq

€

".A.E.E66aak k,al ,5

l+rid.d.&&d.d{de.eAP)'eAAes.d.&6.{&ds.iiifriEEEE5r!666altltllllllllllll5ss>":lss:s

ca ac,

aaca cc- + |

.= .E aa cq

,,99ail I = a€€€€.'€'da6daaltlillllllll.: .l ,i .1 ,1 rt

1.2 otscot{TtNutTY ANALYS|S t09

Example 5.7. Derive Np for the case of applied bending moment M6 at edger = 0 for a short cylinder of length l.

Solution: The four boundary conditions are:At.r:0

_,(*\ =,"\ax-,/

atx: I

_,(*\:o\ax'/

From Eq. 5.26, the second derivative is given by

jj : Z|tte, cos Bx cosh Bx * 42 cos p"r sinh Bx - A3 sin pr cosh pr

-,(*\ = o\dx' /

-,(*\: o\ax- /

- Aa sin Bx sinh pr)

whereas the third derivative is expressed as

,1J u,# = Z!3lAr(cos px sinh pr - sin pt cosh Bx)

+ A2(cos Bx cosh Bx -sin pr sinh pr)

- A(sin pr sinh pr + cos pt cosh pr)

. - .A+(sin pr cosh Bx * cos F-r sinh pv)l

Substituting Eq. I into the first boundary condition gives

. _M^q, :

2D B,

Substituting Eq. 2 into the second boundary condition gives

Az: Atand from the third and fourth boundary conditions the relationships

(l)

(2)

il0

and

3TRT33 IN CYIINDRICAI 9HEITS

Mo lsin Pl cos Pl + sinhB/coshB/= Dtr \--GF7t:;tnT--

N, : Et*r

Ft= 1(Ar sin ft sinh ft * Az sin Fx cosh Fr

A3

are obtained.From Eq. 5.19,

Using the values of A1, A2, A3, Aa obtained above and the terminology of Table

5.3, Eq. 3 reduces to

c, t'r f C"N, = "' ;i*; | -sin pr sinh p1 + ::1sin Bx cosh Bx + cos Pv sinh Fx)r U)15'| Lt

C,- frcos Fx cosh Pr I

or

n, = ZrMoF'( -v, * !v, - !v,\ r\ ct Lt ,/

Example 5.8. Determine the maximum shess at point A of the thin cylinderin Fig. 5.lzla. I*t p' = 9.3.

Solution. A free-body diagram of junction a is shown in Fig. 5.140. The

deflection at point A in the thick cylinder due to P is obtained from Table 5.3

by letting Br equal to Bl. Hence,

-r lgtu _ Qu. - Qu\*o: ifrEV-,"' - c,u c,"u)

For B' = 0.7421, Dt = 0.01145 E, znd Bx = Bl' theobtained:

+ A3 cos Br sinh Bx + A+ cos Fr cosh Fr) (3)

following terms are

-M" / sin2 61 + sinh'? B/e^ = zaB, \.l"[tB, -]t"t}

I.2 DISCONTINUITY ANAI.YSIS

or from Tables 5.2 and 5.3 with r : 0,

"r I

l_* ,,t-

w: toolrn

The deflection compatibility equation at point A is

ltl

(

(b)

Figure 5.1,1

Cr = O.2028 Cz = 1.1164

Ct = 1.5444 Cr : 0.5481

C5 = 0.4568 Ce = 0.6596

U=o.272r yl=0.4006

Vq = 1.4984 V: 0 8707

%=0.5986 V=0.949s

Thus the expression for wo due to P is given by

=6,

tt2 STRESS IN CYI.INDRICAT SHEII.S

and with B2 : 1.0495 nd D2 = 0.00145 E, the equation becomes

144'O2P - 288'S2Qo + 436'59M0 = 2983OQo - 313 'O7Mo

or

5 '2O6Mo - 4'M7Qo: -P

The rotation at point A alue to P is obtained ftom Table 5.3 as

^ p //- r' a. \A = _l:!U +

=u4 +;Vl"P 2q1Dt\ct ' L1 Lt /

or

t, =*n';t'

The rotation compatibility equation at point A is

t')=o = t'

Hence,

429.33P , Mo lrct\ - Qo (cs *co\-{-

-l

-

| -E - zp,o,\c, )' zB?o'\cr' cr,/

or

(1)

-Mo , Qo= -9244

v6,

a1a aa P

T + 896.45M0 + 436.5980 = -657.r3Mo + 3l3.wQo

which reduces to

l2'58M0 + Qo = -3J '48 P

Solvlng Eqs. I and 2 yields

Mo = -0.27 P

0o = -0.08 P

5.2 DISCONTINUITY ANALYSIS

Hence, maximum axial stess is

I t3

" =ry =6Q-'?77:\ = 2s.e P psi0.252

Thb circumferential bending moment is given by

Mo = drI" = 0.08 P

The circumferential force N6 is given by Eq. 5.19 as

-- Etwwe= _T

= 4 {wo^ - wroyr 'to

Et2l-o.og P -o.27 P\R \ zBiDz 2giDz /

= lezt.tu + 84.53) = 2.53 Pu' -'

No.6MoOO : --:' -r ----;--t

_ 2.53 P ., 6(0.08_P)

0.25 0.25'

= 17.80 P

Probbms

5.6 Find the discontinuity stess in the figure shown due to an inlemal pressureof 375 psi. I*t E : 29 x lff psi and p = 9.3.

Probbm 5.6

(2)

I l4 STRESS IN CYTINDRICAT SHEtts

5.3 BUCKTING OT CYLINDRICAL SHELLS

Most cylindrical shells are subjected to various compressive forces such as dead

weight, wind loads, earthquakis, and vacuum. The behavior of cylindrical shells

undlr ih"se compressive iorces is different from those under intemal pressure'

In most instancei, the difference is due to the buckling phenomena that render

cylindrical shells weaker in compression than in tension'

Sturm3 used the system of fbrces in Fig. 5.15 to establish the buckling

characteristics of cylindrical shells subjected to extemal comp-ressive forces'

From the figure he derived a relationship between strains and deflections Using

this relatioiship and Eq. 3. 1 l, he obtained a system of equations .that relates

forces and moments to deflections. These equations together with the equi-

librium equations determined from Fig. 5.15 result in the four basic differential

equations for the buckling of cylindrical shells'

5.3. I Uniform Pressure Applied to Sides Only

For this case, the hoop force is

Ne=-Pr+f(x'0)

where/(x, g) is a function of .x and 0, which expresses the variation ofN0 from

tt" an".uge uutue. Wtten the deflection w of the shell is very small,/(x, d) is also

u.ry ,-u- . Similarly, the axial and shear stresses are expressed by

N,:0+g(ir'0)Ne':O+h(x'0)N'p=0+i(x'0)

Substituting these expressions into the four basic differential equations for the

buckling oi cylindrical shells, and using boundary conditions for simply sup-

oorted J.rds. tire solution for the elastic buckling of a simply supported cylindri-

cal shell due to uniform pressure applied to sides only is given by

c.: rB(!\ 6.27a)

where P., = buckling Pressure

E : modulus of elasticitY

/ = thickness

D" = outside diameter

K=KtI "'(?)

5,3 BUCKIING Of CYI.INDRICAL SHELTS 5

rvr6" + aSoe

r"re +affoo

re+affae

*$oervp" +$ae

..2n. - :=:;---------;1r"'z[N2,\2 - prti - lt - l]3r \t - lt-)

- a * l.+ trlNz[l + (i - t)(z- tt)]-AA

.,2L2 : --n=d-r1

where N : number of lobes as defined in Fig. 5.16

n'r' , ,

N2 L2

N2L2 ,

E-r-: poisson's ratio

l}]

r.r"=+ So'a @o"

lrx + Sox

Fisuro 5.15

I l6 STRISS IN CYLINDRICAT SHETLS

/nlfi"T$d\J/,1,

Cdgor Sitnph Suppod€dSymm.rd.ol Ahour C

Eds6 rirdSymn€rri.ol Aboul

c

Figur. 5.16

F=N2-r*l-t-----Ld- Aa rlt - 1tzlt)ta

{N'?[l + (^ -r)(2- p)]- 1]

(l P,r\ I'{(t - $}tctt - pl + (1 + ri\ + a + 1 + p}t\ Ltl )

1 = moment of inertia : 412

A plot of the ( value in Eq. 5.27a based on the first two terms of expression Fis shown in Fig. 5.17.

5.3.2 Uniform Pressure Applied to Sides ond Ends

The values of No, N,e, and N6 are the same as those for pressure applied to sides

only. The value of N, is given bY

*^=+rge.ol

Substituting those expressions into the four basic differential equations for the

5.3 EUCKUNG OF CYUNDR|CA| SHttrS

Fisure 5.r7 corop,e c..n"""" * -*o.il'Ji'i; r,.",,,.".",,r.,.nry, edses 3impry supporred;

buckling of cylindrical shells, and using boundary conditions for simply sup-ported ends, the solution for the elastic buckling of a simply supported cylindri-cal shell due to uniform pressure applied to sides and ends is given by

Pn = ln, for side pressure only) E *1;r"u.tr r\t | \u, tLL t

"= *(;l (s.27b)

where ": ",.

&(Dj)

^:: ^tF ll7rl2L\F

A4 : A?;---------;;--' -F + (n'r'/2Lt)

A plot of the K value in Eq. 5.27b using the first two terms of expression Fis shown in Fig. 5.18.

tt7

0r4 0.6 I 2 4610 20 40 60 100 m0

r l8 STRESS IN CYLINDRICAT SHEtts

ttl4610 r00 200

vatues of IFisure 5.lS Collopse coefficienls of round cylinders wirh pr$3ur€s on sides ond ends, edges 3implv

supporred; p, : 0.3.1

5.3.3 Pressure on Ends OnlY

For this case,

Ne=0+f(x'0)N" = -t' + g(x' 0)

No'=O+ h(x' 0)

N'o = 0 +i(x' 0)

and the four differential equations are solved for the value of P", For small values

of tfr, the critical compressive longitudinal stress or can be expressed as

ttr20 ,10 6006 I

r l-r0*

o,:n': o.6oE(t)'t\r/

t- : q.or

(5.28)

f,xample 5.9. A cylindrical shell with r : 30 in is simply supported at the

cntl.. il L : l0 ft and t : A315 in., find the critical buckling pressure for a

runilirlrr applictl pressure to sides and ends. Let E : 29 x 10" psi'

Solttlirtlt

2: root

5.4 THERMAT STRESS

From Fig. 5.18, K - 16 and from Eq. 5.28,

I l9

IP". : ( r6)(2e,ooo,r*,(q#)

P". = ll3 psi

5.4 THERMAL STRESS

If a cylinder is subjected to a uniform change in temperature and is allowed todeform freely, no thermal stress is produced. Any restraint that prevents freedeformation produces thermal stress. The amount of restraint affects the stress

level. Figure 5.19 illustrates bar restrained in one and then in two directions. InFig. 5.19a the bar is fixed in the.r-direction only and is subjected to an increase

in temperature. To determine thermal stress, the restraint is first removed and the

bar allowed to deform due to change in temperature. A force F is then appliedto produce the same but opposite deformation. The thermal stress in the bar can

be calculated from the compatibility equation

&:5o

where 6r : deformation due to temperature

6r : deformation due to force F

Substituting the values of $ and E7 in the compatibility equation gives

o = -q LTE for a uniaxial case (5.29a)

where o = stress (positive values indicate tension)

d : coefficient of thermal expansion

AI : change of temperature (an increase is taken as positive)

E = modulus of elasticity

If the same bar is restrained in two directions as shown in Fig. 5.19, the(lcli)nnations due to 4 and 4, are calculated in the same manner as the uniaxialtlolirrmation. The two compatibility equations then become

t20 sTRESs IN CYTINDRICAT SHEttS

il----------'1--1\ ti\____J.i'l*-'-l L-,,

(")

A comparison between Eqs. 5 -29a nd 5.29b indicates that a higher stress

level is obtained when the number of restraints are increased Hence, for a bar

with r.r. - 0.3. a stress increase of 437o results when the number of restraints

incrcasus lirrrn one to two. Another interesting feature of Eqs' 5 '29a and 5 29b

is tlrrl thc thormal stress is independent of thickness and length'

lirlualion 5.29b oan also be obtained from the theory of elasticity Hence, ifc rv AT is substituted into the first two expressions ofEq (3'3), the fo owing

tcsttlls:

+&rH.lN\F\Nl--|l.,l

]--t i{ li-'fI L,.6t o.llE

T----l

L

d = coeff. ol .rp.hsjon (b)dr- ch.nse 1n tenp. (positlve Hhen tenp. lncre.s.s)

T = Polssot's rrtlo Figur.5.r9

&-6r'+/'6"r=0&+p8r,-6"r=0

where g, = proisson's ratio

6i", = deformation due to force 46p, : deformation due to force F,

Solving the above two simultaneous equations gives

l---lrlll-I rl+llt

o;- -

TI

_JL-l*1,*'

a LTE1- tL

for a biaxial case (5.29b)

'r t N

5.4 THERMAT STRESS l2l

from which the expression

Ut : o' : --tl LTE' t- p

is obtained.In a triaxial case, the thermal stress can be determined easily from the theory

of elasticity. Substituting e: a LT in the first three expressions of Eq. 3.1results in

(5.29c)

Example 5.10. An intemal stainless steel rod is welded to the inside of acarbon steel vessel as shown in Fig. 5.20. If the coefficient of thermal expansionis9.5 x 10 6 in./in.'F for the rod and 6.7 x 10-6 in.iin."F for the vessel, whatis the stress in weldA due to a temperature increase of400'F? Use E = 28 x 106

psr.

Solution. Weld A is essentially subjected to a uniaxial stress. Hence,

max 0: E La AT

= (28 x 109(9.s - 6.7)(10 9(400): 31,400 psi I

-laLT = "

(o"- Lro)

_,|a LT = -; (tr, - p,rr,)

-a ATEo,: ct = ", = | _ Zl" for a triaxial case

Fisure 5.20

122 STRESS IN CYLINDRICAT SHELLS

l.)xumple5.ll.Anintcrnu|stuinlcsstrayiswcltlctllrrthcinsitlctr|.acarbtlnstccl vcssel as shown in l'ig 5 2la. ll'the coeflicicnt ol'thermal exprnsion is

g.5 x l0 6 in./in."F for the tray and 6.7 x l0 r' in./in.'F fbr the vessel, what

is the stress in the weld due to temperatunj increase of 400"F? Use E : 28 x 106

psi and pr : 0.3.

Solution. A conservative answer can be obtained by assuming the tray attach-

ment to be rigid. Because the weld is subjected to both hoop and axial stresses'

it can be treited as a biaxial condition Hence, from Eq' 5 29b

-(28 x 106)(9.5 - 6.7X10-1(400)o=@: _44.800 psi

If a more accurate result is desired, then a discontinuity-type analysis can be

performed. In referring nFig.5.2lb, and due to symmetry, the equations ofcompatibility and equilibrium can be written as

(a) ft=6r(b) 0r=0(c) 0z=0(d) )F=0

From the above four equations, the four unknowns 4, Fz, Ft' and Mo can be

obtained.Equation (a) can be written as

ollMo

fn 'l/-fr\_-/ .*F3,h--Fz

I i,o

@l(b)

Fisure 5.21

: deflection of shell due to temperature -l 4 - Mo

F,r F, M(a,,)(AD(r) +'#tt - p): (a,,)(An(r) - rrn- tifi {t)

Equations (b), (c), and (d) can be written as

Ft Mo

-I Mo-n2PrD - BD:

u

Ii+F +4:0From Eq. 2, \:2FMoFrom Eq. 3, n: 2BMoFrom Eq. 4, 4: -4FMo

and Eq. 1 becomes

^l#. #] = 1a", - a,,)(Af(r) (5)

5,4 THERMAL STRESS

dellection of tray due to temperature + Fl

Assuming the thickness of the cylinder is I = 0.1875 in. and using other givenvalues, the value of { from Eq. 5 is

E - -618 lb/in. or o: 618 psi compression in weld

The value of618 psi is significantly lower than the conservative value of44,800psi obtained from Eq. 5.29b because of the ffexibility of the cylinder. If thethickness of the cylinder is r : 3.0 in., then Eq. (5) gives

n = Zl,20O lb/in. or o:21,2O0 psi compression in weld

This value indicates that as the cylinder gets thicker, the stress approaches thatof Eq. 5.29b. In fact, if the cylinder is taken as infinitely rigid, then Eq. (1)becomes

(a,,XAO(r) - p.) : (a-XA?.)(r)

and the equation yields 4 : -44,800 lbiin. or o = 44,800 psi compression inthe weld, which is the same as that obtained from Eq. 5.29b. I

(2)

(3)

(4)

* H<t

t24

5.4.1 Uniform Chonge in Temperolure

STRESS IN CYIINDRICAI. SHELTS

A unitbrm change of temPerature in a component usually results in a thermal

stress both at and adjacent to the component. The magnitude of the stress is a

function of many factors such as geometry, degree of restriction and temperature

variation. The stress can normally be determined from a free-body diagram ofthe various components. The following examples illustrates this point

Example 5.12. A pipe at 10'F is partly filled with liquid at 40"F and gas at

250"F as shown in Fig. 5.22a. What is the maximum thermal stress if a =6.5 x 10 6in./in.'F, E = 30 x 106psi, and p = 0.3?

Solution. A solution can be obtained by taking a free-body diagram at the

gas-liquid boundary as shown in Fig. 5.22b. Compatibility at the interface

requires that the deflection in (a) equals the deflection in (D). Hence' from Eq.\)L

H" M^ HN Mo(axA7i)(r) - ffi* ffi= t")tAl'ttn * rp'o* ,Bo

from which

Hn = @)(Lr1- AAX/)@)(D)

: (6.s x 10-6)(240 - 30)(6X1.4843f(5366)

= 144 lb./in.

,=$

lri| fieas lt

I "--{';"'' --14t" "o' ,--. { ' li| ( 2 )Liquid lli-li

(b)

Figwe 5.22

Hence,

Ne = 827 lblin.

5.4 THTRMAI. STRESS

Also

N.: 2565 lb/in. at interface

t25

M0 can be obtained tiom the second compatibility equation whereby the skrpcin F\g. 5.22a at the interface is equal to the slope in (b). Or

Mo __!, _Mo, Hs

BD 296: pD- 2B'D

and Mo : 0.The circumferential force in the pipe due to I1o is obtained from Eqs. 5. 19 and

5.24:

t't':4' H: c"r 2BtD "w

Maximum value of Cp. is obtained from Table 5.1 as 1.0. Hence,

M": O

and

t\6<Max o: o.rrr- = 20.500 psi

The maximum bending moment due to I{0 was derived in Example 5.4 as

"":Y "Stress due to bending moment at Bx = r/4 is

6M 6 (o'34H0\ = l2,ooo psio':v:vx p /

Deflection due to flq zt Bx = n/4 is obtained from Eqs. 5.23 and 5.24 as

o.322Ho

2B"D

B'=x

t26

end circumfcrcntial $trcss is

sTxEss lN cYllNDRlcAt SHELl"S

827 ^oo= #+ 0.3 x 12.000 : 10,200Psi

Thus maxrmum stress occurs al rnterface with magnitude of 20,500 psi. I

Example 5.13. Determine the bending stress in a cylinder fixed at one edge

(Fig. 5.23a) due to a uniform rise in temperature of 200"F.

c = 6.5 x 10-6 in./in..F

E:30xl06psip=o'3

Solution. Radial deflection of cylinder if ends are free is

w : (a)(Afl(r): 0.039 in.

From Fig. 5.23a the rotation at the end is zero because the cylinder is fixed.

Hence

M" O^

BD zB,D

Qo = 2PMo

/1\-lMo

Hs

Fisur€ 5.23

(b) {! * aB'*:EJ$tlx- fl,)(5.30b)

5.4 THERAAAT STRESS 127

ln the second compatibility equation the deflection due to temperature plusmoment plus shear is equal to zero. Or

o.$s + #_#:oand

Mo: (0.039)Qp2D)

: (0.039)(2X0.46941 (42.e30)

= 738 in.-lb/in.

6Mo = :+ = 70,800 psi I

5.4.2 Grsdient in Axiol Direction

The stress in a cylinder due to a thermal gradient f, in the axial direption can beobtained by first subdividing the cylinder into infinitesimal rings of length dr.Hence, the radial thermal expansion due to 4 in each ring is given by (c)(4)(r).This expansion can be eliminated by applying an external force { such that

deflection due to P, = deflection due to I

Hence

and

E= ",r,

p, = Etan

r

-Prae: -l : -taT.Because the cylinder does not have any applied loads on it, the extemal forceP, used to reduce the deflection to zero must be eliminated by applying an equaland opposite force in the cylinder. Hence, Eq. 5.21 becomes

(5.30a)

Itt STRTSS rN CYIINDRICAL SHEttS

Thc total strcss in thc cylindcr is dctcrmincd from Eqs. 5.30a and 5.30b.

Example 5.14. A vessel that operates at 800oF is supported by an insulatedskirt. The thermal distribution in the skirt is shown in Fig.5.24a.Ifthe top andbottom of the skirt are assumed fixed with respect to rotation, what is themaximum stress due to temperature gradient? d = 7 x 10-o in./in.'F, p =0.3,8= 30 x 106psi.

Solutian, The equation for linear temperature gradient is

T,:4+4 iT*

The temperature change can be expressed as

{a)

Tx rv=(EJ, )x=-sx

A'lr t

t\t\

J_/

(dFhut. 5.21

Tb= 2odF

t-aodF

r n-T=rx

and the circumferential stress due to ring action obtained from Eq. 5.30a is

5.4 THIR 'tAt STRISS

E4uation 5.30b gives

and w reduces to

From Eq. 5.19

(l)

e").

",=+:*(T) (2)

.="u$(Lf),*",which upon substituting into the differential equation gives

t,=# and cr=9

w=rq.

129

,,= -u"(T),

onfu * o.o* =";;(*),

A particular solution takes the form

Adding Eqs. I and 2 results in

oe= 0

which means that for a linear distribution the thermal stress along the skirt iszeto.

The slope due to axial gradient is given by

- dw lT" - T,\u = --:- = ral -----=; Idx\r/

rt0 3TRr93 lN CYI'INDRlCAt gHEtts

Bccausc thc cnds arc fixcd against rotation, a momcnt mu$t be applied at the ends

to reduce 0 to zero. From Eq. 5 24

lr, - r^ -Mo'"\ , )= B,

u,= -oolff)<n- n

Since B = 0.2142, D = 2.74'1 x 106 '

Mo = 742 in'-lblin'

and

o = ,1450 psi

5.4.3 Grodient in Rodiol Direciion

Thermal stress in a cylinder due to a temperature gradierrt in-the radial direction

can be obtaned from ttre theory of elasticity' Hence, Eq' 3 can be written as

,,=!f",- tL(oo+ c)f + ar

*: jl", - P(at + o)f + uT

u=|b,- p(a,+ os)l+ ar

where aT is the suain due to temPerature change' If the temperature is sym-

^"iii" *i,tt respect to 0, all shearing stresses are zero and the radial and hoop

$truins can be exPressed as

I

dww',=6 "=i

(5.31)

6 32)

(5.33)

or

I de9\e,: ee + r\a, )

(5.34)

I.4 THIRMAT 3TRI33

Substituting Eqs. 5.31 and 5.32 into 5.34 gives

/ do.\ p do,o,: {rs* r\*)- t_*to,trdr-oo)

. (r-- J",' . ^,(#)(s.35)

In Section 5. I it was shown that the equilibrium of an element in a cylinder (Fig'

5.3) can be expressed by the equation

",- ''='(*\

Solving Eqs. 5.35 and 5.36 for a gives

dlt d , .f -E ,, , ..,-(41\,kl;ftt,'",t|: T.;;"tr * rt,\i) (s'37)

Solving the above differential equation and applying the boundary conditions

|lr

(5.36)

o. l = 0

i="4l=0

gives

Ea-l + P(r'z - r? ['" rra, - [' rrar\o. = -:-------'-- _--' | - tt' r' \16 - ri J., J,, /

Frorn Eq. 5.36

E<x r + u' (rl! :j ['. r,a, - [' rra, - rr\ (s.38)oa= t - ttz I \rj- r? J,, ""' J,," /

and from Eq. 5.31 for a cylinder umestrained in the:-direction,

Ea / 2 r,-, -\ot = _-_l_-;__ ,l trar_ tl- L- p\ri-rih, /

From these equations, some cases can be derived.

t32 STRTSS IN CYTINDRICAL SHTLLS

Cav L Llnear'l'hcnnul l)islrihulion

[,irr thin vcssels, a stcady-state condition produces linear thermal dishibutionthrough the thickness that can be expressed as

r - r'L-J

where d : inside wall temperature relative to outside wall temperature.

Substituting f into Eqs. 5.38 gives

EaT, f (r2 - r\zrt + r.)rt-:

-l

' r'(l - 1t\ | 6(ri'r r")

EeT f (r2 + r?)(2rt + r")nr-r-\t - pl L otrr r ro,

LAIi I zrt+ ro r"-rl' (l - pll3(r1 + r,) r"- r,l

6(ri - r")

2\r3 - rl) - 3r"(r'z - r!)l^

| \J.J')o\r, - ro) I

2(r3-rl)-3r"(r2-r!)

Figure 5.25 is a typical plot of o,, ae, and cz. The plot indicates that o, isrelatively small compared with o6 and o,. For all practical purposes, o6 and o.are equal.

The maximum values of op and ol occur at the inner and outer surfaces. From8c.5.39.

"/ffi

And for thin wall cylinders. Eq. 5.43 reduces to

5.4 THERMAI STRESS

-EaT,f 2r,, + r, lov= o,- l, . .l| - pLJtr. -f ritlEaT f r"+ 2rt1ue=u,=. l-l| - p LJtro -f riJ l

For thin wall vessels, Eq. 5.40 reduces to

- EaTtCfe = C, = ,r,l _!\r tL)

EaTce: o,: ^-Ltt - lL)

t33

for inside surface

(s.40)

for outside surface

for inside surface

for outside surface

(5.41)

(s.42)

for inside surface

(s.43)

for outside surface

Case 2. Logarithmic Thermal Distribution

In thick vessels, a steady-state thermal condition gives rise to a logarithmictemperature distribution that can be expressed as

t: "/lu l"')

\ln r" - ln r,/

and upon substitution of this expression in Eqs. 5.38 results in

Again disregarding or as being small compared with o6 and o", Eqs. 5.42have a maximum value of

'(;)l#cr.l'-'^? - *('. #)'(;)]

u?T:,, r1, - 2Lnk-,t'i,,n/tr)l- p,ntro/nL r ri-r; \/,/l

ErxT f, 2rl ,lr"\f

-t

r - --------- lnt -ll- pJ tntro/ rJ L 16 - ri \r, / l

2(l

z(r

2(r

z(rEaT I. 2,1 . /a\l

- t"t t"lJ,)L' ;:'l' 'n\-/l

t34 STRESS tN CYUNDRTCAL SHfl"t"S

I')tt'li lirr insidc surlacetft: (r: - ,zt<l _ tl

EqToe: o, = Z\l _ lD lor outslde sunace

which are the same as those for the linear case.

Case 3. Complex Thermal Distribution

ln many instances such as transient and upset conditions the temperature distri-bution through the wall of a vessel cannot be represented by a mathematical

expression. In this case a graphical solution can be obtained from the thermal

stress. From Eq. 5.38

- - eq lt-rr,lf.'t'z[,.rrar-!f ,ror-r]oe: l- *l-rz, - r, 1, r- J,, IFor a cylinder where the thickness is small compared to the radius, the firstexpression in the brackets can be expressed as

I r trilrt2 p --,- : 2r Ji: Trdr-----';-'------- | IrAr :

r; - ri J,, ntr; - ril= mean value of the temperature (5.44)

distribution through the wall

'l'he second expression can be expressed as

I trc-, 2n fiTrdr-l rror: ---': ,r' J,, 27Tr -

: one half the mean value of the temperaturedistribution from the axis of the vessel to r

However, because the temperature distribution from the axis tocxpression 5.45 for all practical purposes can be neglected. Hence,

cxoressed as

/5 45\

| 15 Z€rO,

oo can be

(5.46)Ea .^oo: --\1n - I)I- IL

4n : mean value of temperature distribution through the wall

f : bmDerature at desired location

wncrc

Fisure 5.26

5.4 THERMAI. STRTSS

From Eq. 5.38 it can be seen that cr, can also be expressed by Eq. 5.46.

Example 5.15, A thin cylindrical vessel is heated by ajacket from the outsidesuch that the temperature distribution is as shown in Fig. 5.26.1f E : 27 x 106psi, a : 9.5 x 10-b in./in.'F, and p, : 0.28, determine (a) maximum thermalstress using Eq. 5.40 and (b) maximum thermal stress using Eq. 5.41.

Solution. (a) 4: 400 - 700 = -300'F. Hence at inside surface

-(27 x t01(9.5 x tO-ox-300)/2(lJ) - t0\- (t - 0.28) \3(13 - tO)i

= 55,800 psi

and at outside surface

t35

u = -- ,1 _ 92s1 \ :tr: _ rot

= -51,000 psi

(b) For inside surface

(-27 x 109(9.5 x l0 6)(-300)

'': 2(r - or2s)

: 53,400 psi

and for outside surface a = -53,400 psi. I

Example 5.16. A pressure vessel operating at 300'F is subjected to a shortexcursion temperature of 600'F. At a given time, the temperature distribution in

(27 x 106)(9.5 x 10 6)(-300) lt3 + 2 x 10

(l-028, \3(13-lo)

It6 3TR!!t tN CYUNORICAI sHEtLs

thc wall is shown ln Fic..3.27 . Find thc msximum thcrmal stress at that instance.

Lct rr = 0.3, E = 3dx 106 psi, and a = 6.0 x 10-6 in./in."F

Solutlon. This problem can be visualized as a biaxial case where the inner

surfacs heats quickly while the rest of the wall remains at 300'F' Using Eq.

5.29b results in

(6 x l0-6x600 - 300x30 x 106)o:ffi= -77,100 psi

which is extremely high and is based on very limiting assumptions. A more

realistic approach is that based on Eq. 5.46. The mean temperature is obtained

from Fie. 5.27 and tabulated as follows:

Locations as

Ratios of ThicknessTemperature

T Area

00.1

0.20.30.40.50.60.7

0.80.91.0

And I' = 356"F.From Eq. 5.46, at inner surface,

600

460

400

370

340

320

310

305

300

300

300

53.043.0

38.5

35.5

33.0

31.530.8JU.J

30.030.0

>355.6

_ - (30 x 106)(6.0 x 10-6)/?56 - 600)

= -62.700 psi

ond Bt outcr surface

., _ (30 x 196x6:0_x 10-6)(356 _ 3oo)" l-0.3= 14,2100 psi

NOMIIiICTATURI

It is of interest to note that the high sness occurs at the surface only. Thus atone-tenth of the thickness inside the surface, the stress is

_ _ (30 x 10ux9I_!o-1res6 _ 460)t= r-o!3 \JJu 'uu''

= -26,700 psi

The high stress at the inner surface indicates that local yielding will occur. I

NOMENCTATURE

Fl3l'l = -----=:-" l2(1 - tt2)

D, = outside diameter of cylinder

E = modulus of elasticity

K = constant

L : length of cylinder

M, = axial bending moment

t37

tisur' 5.27

rtt

Mp ' hoop bcnding momcnt

p = pressure

P", = buckling pressure

P/ = internal pressure

Po = extemal pressure

O, = shearing force

r : radius of cylinder

4 : inside radius

ro = outside radius

, = thickness of cylinder

T = temperature

AT = iemperature change

p = deflection

c = coefficient of thermal expansion

B

6

€t

€t

e0

0

lL

o

Ot

or

Oe

= deflection due !o temperature change

= longitudinal strain

: radial strain

= hoop strain

= rotauon

= proisson's ratio

: stress

= longitudinal stess

= radial shess

: hoop sftess

ETRI'I IN CYTINDRICAT SHTTT!

REFERENCES

l. Murphy, G., A 'anced Mechanics of Materials,

ll4-t2t.McGEw-Hill, New York, 1946, pp.

IIIIIOORAPHY

2. American Society of Mechanical Engineers Boiler ond Prcssure vessel Code, Sdct/.rn yrrl-Rules for Constructiotr of Pressure Vessels, Division 1, ANSI/ASME BVP-VI -l,NewYork,1980 Edition.

3. Strum, R. G., "A Study of lhe Collapsing Pressue of Thin-Walled Cylinde$," University ofIllinois Bulletin, Vol. XXXIX, No. 12, 1941.

BIBI-IOGRAPHY

The Amedcan Society of Mechanical Engineers, Ptessure Vessel and PipinS DesiSn-ColkctedPapers 197-1959, New York, 1960.

Baker, E. H., L. Kovalevsky, F. L. Rich, Stn ctarulAnalysis ofShells, Mccraw-Hill, New York,1912.

Flugge, W., Stesses in Shells, Springer-Verlag, New York, 1960.

Gibron, J. E., Linear Elasrtc Theory oJ Thin SherrJ, Pergamon hess, New York, 1965.

Hefey,I. F -, Theory and Design of Moder PresJrt? y?JtetJ, Van Nostrand-Reinhold, Princ€ton,N.J., 1974.

Timosheoko, S., S. woitrowskey-Krieger, frreory of Plates 4r1d Shelrs, Mccraw-Hitl, New York,1959.

t39

Con|co|hrodconsistingof|woeccgn|liccones.(co',rtesyoftheNool€rcorPorgtion,sr.LoUi!,Mlo.)

r40t4l

CHAPTER 6ANALYSIS OF FORMED

HEADS AND TRANSITIONsEcTtoNs

142 ANAI.YSIS OT TORMTD HEADS ANO TRANSITION SICTIONS

6,I HEMISPHERICALHEADS

The required thickness of hemispherical heads is determined from a free-bodydiagram as shown in Fig. 6.1. Hence

Pm2 = Zmo

', Pri-2t

where a= membrane stress

P : pressure

r = radius

t = thickness

This equation which assumes uniform stress distribution through the thick-ness, is adequate for relatively thin heads. As the thickness increases withrespect to the radius, this assumption becomes invalid. Hence, a more accurate

formulation is needed, which is obtained from the "thick head" equations.

From symmetry, it can be demonstrated that at any point in a hemihead

subjected to uniform pressure,

From Eqs. 3.1,

(6.1)

(t6.2)I.

e, = E\ot - zlto,)

FisiJr€ 6.1

6.I HEMISPHERICATHIADS

1...eo: ELt

t - lL)1o4, - pa,)

t43

(6.3)

(6.6)

The strain displacement relationship is the same as that derived from cylindri-cal shells:

and

dw

Hence, expressions 6.2 and 6.3 become

(t - QfrQo6) - vfiWS - o, * 21t'o6 -- o (6.4)

Figure 6.2 shows an inflnitesimal segment of a spherical head Summation offorces in the radial direction gives

(6.s)

w'f

,",= -l(fi)<,'"t

i12

5r-

Solving Eqs. 6.4 and 6.5, we obtain

ld\l' , 'r

rl; ll+ +(rro,) | : 0\dr / Lr- qr I

Its solution is expressed as

The boundary conditions are given by

o,= -n al r: ri

and

a,: -n at r:ro

Solving the boundary conditions forA and B and substituting into Eqs. 6.5 and

6.6, we obtain

l.l{ Al'lAtYlll Ol lOR illD H!AD3 AND TnANSl?lON SICTIONS

r?P' / rl\ r3P / 't\06: oo= -:ll + -l - -;l=lt +:+lr; - ri \ zr'/ r; - /i \ 2r./

Equation 6.7a can be simplified for the following cases:

Case 1. Internal Pressure Only

max o.= _e atr=ri

max oo = or: | :?.'?r* ')r,]n at r = riLzlr; - ri)l

Case 2, Exlernal Pressure Only

maxot= -P. at r:r3 rlkrnrxa6= ot= - ZE:;, atr=ri

(6.7a\

(6.7b)

[6.-dor \\ ' ' d. /

Figure 6.2

",:f+('-*)-*('-*

(6.7c)

6.I HIMISPHERICATI{TAD3

Jo/f I

fisurs 6,3

A comparison between Eqs. 6.1 and 6.7 is shown in Fig. 6.3.

Example 6.1. A hemispherical head with r = 15.0 in is subjected to an

internal pressure of 4000 psi. If the allowable stress is 23,000 psi, find the

required thickness from Eqs. 6.1 and 6.7.

Soltttion. From Eq. 6.1,

,:+- (4000x1s)

2(23,000)

= 1.30 in.

From Eq. 6.7,

rln l. ,11o.: _-i_-1 | I _r : " Iri,- riL zril

ANAIYSIS OT TORMED HTADS AND TRANSITION STCTIONS

= 16.31

t : 1.31 in.

6.1.1 Vorious Looding Conditions

Occasionally, hemispherical heads are subjected to a variety of loadings such as

wind forces, snow and dead loads, and agitator and equipment reactions. The

membrane stresses induced by such loads usually are obtained from "thin shell"

membrane theory that assumes that the loads are carried by membrane action

rather than bending moments.When referring to Fig. 6.4a, the middle surface of a shell is taken as a surface

of revolution. This is generated by the rotation of a plane curve about an axis

in its plane. This generating curve is called a meridian. An arbitrary point on the

middle surface of the shell is specified by the particular meridian on which it isfound and by giving the value of a second coordinate that varies along the

meridian and is constant on a circle around the shell's axis. Because these circles

are parallel to one another, they are called the "paralled circles."The definition of r, n, 12, and @ are shown in Fig. 6 4a. The radius 11 is

mcasuretl from point 0, which is the center of curvature of the meridian; 12 is

mcasurctl fiom the z-axis and is normal to the meridian. The parallel circle isdolincd by r.

lirrrrr lrig. 6.4a,

r = 12 sin $ds = r, d.6

lrigruc (r.4/r is a free-body diagram of a section of a surface of revolution.

Srrrrrrrrirrg lirrces parallel to the tangent at the meridian and simplifying by

th.lctirg lt:rrns of higher order, we obtain

-!, gN,pt - r,(q# -r,Npcos d + 16',) = gt4l, \ do /

I

2(15t(4000) + 2(23.000X l5 )3

2(23,000) - 4000

6.I I{EMISPHERICAIHEADS 147

f :re SrN0

cls:f1 d0

dr - ds .Cos P

r'r6e+ ffi dQ

N@+tl do

Noo + *F de

++ do

Figure 6.4

Summation of forces in the direction of parallel circles gives

,i a:*Ut=n\12rt

For the majority of pressure vessel applications, the loads are symmetric with

respect to the axis of revolution. Hence, all derivatives with respect to 0 in

expression 6.8 and 6.9 can be deleted. Shearing stresses due to torsion are small

comDared with other stresses. Thus expression 6.9 can be deleted completely'

Tire last equation of equilibrium is obtained by summing the forces in Fig'

6.4b perpendicular to the middle surface:

j6<,N*l -,,(#* rlNep cos 6 * r,',) :0 (6.9)

(6.8)(6.10)

I4g ANAI.YSIS OT TORMTD HTADS AND TRANSITION SICTIONS

Substituting oxprcssi()n 6. l0 into 6.tl givcs

I fr 1No: . -,,, , I I r, r2(P, cos {- Pasin Q)s\nQdQ + Cl (6.11)12 sln- @ LJ I

The right-hand side of Eq. 6.11 is equal to the sum of all the N6 forces arounda circle of angle d. Therefore we can solve Nd at any given location { bysumming all forces in the {-direction. Once N6 is obtained, Nd can be deter-mined from expression 6. 10.

For a spherical shell, 11 = 12 = r. Hence, expressions 6.10 and 6.11 can besimplified as follows:

N5 I N6= P,r

" t. I (6 12)No : ^,-, .l ltP, cos @ - P6 sin @) sln $ d$ + C Isln - @ LJ I

The displacement for various loading conditions is derived from Fig . 6 . 5 . Thetotal change in length AB is

u:4ra6 - w d6taQ

The strain is therefore

1/do w\ea=-l-;--l\ \aQ rr./(6.13)

)(-

i-d0

Fisurs 6.5

'fhc change in r,, is given by

Ar' = Dcos A -:=srn @

and the strain is expressed as

,r=!1ucos@-wsin@)

6,I HEMISPHERICAT HEADS I49

(6.14)

Substituting ro = 12 sin { and Eqs. 3.3 into expressions 6.13 and 6.14 gives

(6.15). h.__lt:ucotO- EtlNe f"Na)

Equations 6.15 can be solved for the deflections once N, and Nc are estab-

lished from Eqs. 6. 12. Table 6.1 shows the solution of Eqs. 6. 12 and 6. 15 forvarious loading conditions.

Example 6.2. Determine the forces in a spherical shell due to snow load.

Solutian. From Fig. 6.6,

P, -- -P" cos2 Q

P:P,cosdsind

From Eq. 6.12

rlN^ = +-l f(-p, "os' S - P" sin2@ cos @) sin A dQ + Cl' sln- q)LJ I

-p"rf r I=,.'i' ;l l(cos' d + sin'z d) sin 6 cos 0 dO + C Ism- qlJ I-a-P", lt. ",1 ^l:rtt}Lts'n-@l+c.l

.- -P.r 4rC/vr :

-

- -

2 sin 0

As 4 approaches zero, the second expression of Nd approaches infinity unless

C is set to zero.

;.t'

!l

.trrr

r^r

<f ! ,9,/+ --,{

- ..l+

= +k!

| | -lc

le

-i,+

.f;|*

+

\j/

l5lr50

I

!l !la oo-l dl

eo

o.

+

!lrl

r-;-1-l'1

-l+

'.t ..lN*il

!te.tr

f---l

+

NI

+

@e

!'6

o

.tq.

c

o

oo

o!

oI{t

IIoo

-o

152 ANAI.YSI! OI TO|.MTD H!ADs AND TRANSITION SICTIONS

Thus, for C = 0

From Eq. 6.12,

-P.;

rle: -r.r cos' q + \

Yr= -P.r(cos2 Q - L)

= -4cosz|

6, L2 DiscontinuitY AnolYsis

'l'hc rnctnbrane analysis discussed in the previous section fails to Sive adequate

rcrult$ whon the loais are localized or when the hemispherical section is attached

to lnothcr shell that acts differently under certain loads' In these cases the

hcnding moments must be considered in the analysis' In Figure 6'7 it is seen that

lix u givcn krading condition, the membrane and the bending moments can be

con$id;rcd us shoin. Proceeding as before where both the free-body forces and

Figor€ 6.6

6.1 HtMllpHlRlcAt l'llAol ltl

the comDatibility equations are taken into consideration, a complicated numb€r

"i.i*oii-*ut Oifierential equations result' The solution of these equations is

imoracticat. however, withouisome simplifications. By assuming symmetric P.

forces only, the differential equations for a spherical shell reduce to

#r.#*rQ- o(cotz O+ tD= -#and

(6.16)

(6.17)

#**Affi-o<""eQ- rt)=Eto

EilBRAI{E FORCES

8 EIIDI NG FORCES

Fig'.rre 6.7

lla Af{Al,Ygl ot fotMrD HIADI AND TRANIITION SlCrlONS

whcrc 0 is thc anglc of rotation and is given by

By a rigorous analysis Gibsont has shown that in Eqs. 6' 16 and 6.17 only the

higher-order terms are significant in most usual pressure vessel applications.

Accordingly, the equations reduce to

^ a ldwr raQ

d2 o _ -Qrzdo' D

(e: ",,dE'

Eliminating 0 from Eqs. 6.19 and 6.20 gives

(6.18)

(6.1e)

(6.20)

(6.2r)

where

ffi++*o=o

/ ,\2,\a=3(l-r1l;/

Nr- -Qcot0

N" = --9-do

M^:2(+\- r \dQ/

Mo = pMo (6.23)

(6.22)

The solution of Eq. 6.21 can be expressed as

g: g^o(c1cos A0 + c, sin,\@ + e-^o(ct cos,\d + c4 sin ld)

Once the value of O is determined for a given loading and boundary condi-tbns, the other quantities can be obtained from

6.t HEITATSPHERICAI HIADI ll!

0 = slope

| /d'zo\EVG)

w : radial deflection

r ..,I F;lt\ e - luYA)

The solution of Eq. 6.23 for various common loading conditions is given inTable 6.2.

Example 6.3, Calculate the head discontinuity forces of the head-to-shelljunction shown in Fig. 6.8a. Let p : 3gg psi and /, = 0.3.

Solutian. From Fig. 6.8b and Table 6.1 the deflection in the head due topressure rs

_ (300x50)'?rl _o?lE(0.50) '-

_ 1,050,000

E

P126o :

=(l - ,r) sin

Tqble 6.2 Approximote Force qnd Deflection funcfions for Sphericol Segments

I,//-r-\Hoftflff)--xo'<l"L/

a

lO

lr

6

-f e-rYsln4ocos ( rr+"/+)no

2re-r1sln{o(cosr.r)tio

Vi*e-rYst nqrotocos { ry+nllltto

L-rYsi n0osi n (ry ) Ho

to f""-rt"t no" [zr"t no"o"rvtT(-rF*""o"""t rv*r+l]J

1; t-z'F^2u-tt" rnro"i n (r.r+r/a)l

?I "-rtrtn(rr)ru

z16.f. "-^t"o,

( rr*nrl)no

-?1 e-11cotosi n ( ry)ro

y6-"- r'r" i n1r1*nll lto

,ao lzre- ry I-,6- rs tiocos{ ry+r/a)-+ucos0sinrYl,l

ro /-1r,3 "-rt.o"rr\rT \-- /

It6 AltlAtYlll Of iotlillo HIAD! At'lD ltAilllTloN slcTloN3

Figirr€ 6.8

8nd

- O.3)(lqoJ

12.038

6,:Wl^ouo=-2f4

From Table 6.2,

8Ho+Mo=-995.17 (1)

6.r HlMrsPHtRtcat HtADs lt7

6,"=s#6

,rr: -3#ro

Similarly, the deflection in the shell due to pressure is obtained from Eq. I ofExamnle 5.5 as

tt\_;,,='#('

and from Table 6.2.

B=

= 0.1818

E/ | n\3D=-ij:j_=0.0916Era\t - u.J-)

:ffi,,-0.,r,_ 637,500

E

u,":,--#4r"

eo" = aEEm

6rr=yM,

0,"=gfMo

total deflection of head = total deflection of shell

r.r+4pq * Tr" * Y*" = q# _ ryH" * Yr,

OT

I58 ANAI.YSIS OI FORMED HTADS AND TRANSITION SICTIONS

Sirnilurly,

rotation of head : rotation of shell

-579.65.. 2t9.tt .. 165.15.. 60.05..._-t uo - E *o: - -E-nrt --iwo

rlo = -0 818 Mo

and from Eq. 1,

Mo : 179'4 lb-in'/in

and

Ilo = - 146.8 lb/in.

From Table 6.2, N0 at discontinuity is

)A2 prN.: 2tH" +':::-M^ + -2

: z(r2.o3s)(-146." * ?g?qg@-' !9PNe= -3534 + 1040 + 7500

No = 5006 lb/in'

and

y^=A=7500 lb/in.

Mo : 179.4 lb-in'lin.

and

Me : 53 '8 lb/in' I6.1,3 Thermol Sfress

'l'hc dcrivation of thermal stress in a spherical segment due to radial distributionol ternpcrature can be derived similarly to cylindrical shells (see Section 5.4.3).Thc meridional and circumferential stresses in a sphere are the same due to

6.I HEMISPHIRICAIHEADS

symrnetry, expressed as

ZEa lr'-r| r,.^,, f -,,\O,:-l-ltr-Ar-ltr-Arltt - p)r'\r; - ri ,, ,, /

159

(6.24a)

Eu l2r3+rl r"-,, f,^,, ,-\06: ao: ^l .. I Tr.dr + | Tr.dr - r.Tl \6.24b)tt- p)r- \r;-ri J,, J\ /

Example 6.4. Determine the circumferential thermal stress on the inside sur-face of a hemispherical head subjected to an inside temperature of 600'F andvarying linearly to a temperature of 400"F at the outside surface. Let 11 :30in.,12 : 40 in.,E : 30 x 106psi, o : 7.0 x 10-6in./in.'F, and g, : 6.3.

Solution. The temperature distribution across the thickness can be expressedAS

r=6oo_ Lg_rs

or

T=1200-2Or

The first integral in Eq. 24b gives

f; r,,a, = J* trzoor, - zor3tdr

= 6.O50.000

The second integral is zero because the limits of integration at the inner surfaceare both ri. Hence Eq. 6.24b gives

(30 x 106)(7 x l0 6)

(1 - 0.3X30t

[2(3ot + (30t.. ^-^ ^^^. ^^ .ltffi ro'o5o'ooo) - (3ofll2oo - 2o(rnj

= -32,800 psi I6.1.4 Buckling Strength

The buckling equations developed by Von Karman and Tsien2 are the basis ofthe design equations developed by ASME. Von Karman's equations, which aresubstantiated by tests, give a more accurate prediction of buckling strength of

160 ANAI.YSIS OT FORMTD HTADS ANO TRANSITION SICTIONS

splrcricul sccli(nls tlurl thosc dcvclopcd carlicr by l.luggc, Timoshenko, andothers. Von Karrnan antl 'l'sicn took the out-ot-roundness imperfections intoconsideration. 'l'hey also used the energy equations as a basis for derivation.Refering to Fig. 6.9, it can be shown that the strain energy due to the extensionof the sphere is given by

u,: Er'(!\o f fYJ^- r)'sin d/d (6.2s)' \r/ Jo \cos a /

Similarly, the strain energy due to bending is expressed as

,, _ -_,/,\ T (p -., ,fcos0 d0 ,\') , /sin 0 .\,1 .,U7: Lr"l-l; I srn Ql--- _ -ll+l --tlld6

6.26)\r/ tz Jo lcos @d@ / \srn A / |

The potential energy of the external pressure p is given by

-rBU1: Prit I P sin'?dcan 0 - tan O) cos Q dg (6.27)JO

The total energy of the system is the sum of Eqs. 6.25, 6.26, and 6.27 . Hence

U=Ut+U2+Ur (6.28)

Equation 6.28 can be simplified by assuming B to be small. By neglecting termsof higher order and expanding the sine and cosine functions in a power series,liq. 6.28 becomes

U E(t/rt rp ,^,-:-# | (0. - 6'\6d6Tr" 4 Jo

.ry rUffi-'l . G- 'flaoo+PrQz@-Q)dQ (6.29)

'l'hc solution of Eq. 6.29 is obtained by the Raleigh-Ritz method by finding ancxprcssion of the deflection that satisfies the boundary condition

0:0 at 4:0s:B at Q=B

6.I HTMISPHERICAI.HEADS t6l

(6.30)

(6.32)

Such an expression can be written as

o:alr-c,1"-4)lL \ p'/lwhere 0 is the slope and is related to the deflection by

a: , fu @ - o) dO (6.31)JO

From Eqs. 6.30 and 6.31 the value of C, can be determined to be

^46wt -;F

Substituting Eq. 6.30 into 6.29, the energy expression becomes

It / 14\ / Ci Ci\ Et, ^,^, pBo_-: l+186lCl _ ir-al - "" B2ct _,_L cin' \bur/ \ 7-i) - l8tP-ci-pc'This expression can be minimized by taking its derivative with respect to Cr andequating the derivative to zero. This gives

or 4 /a\f-. ..ls\t/r /.^6, _^\trlr),]a =

ros \;/L'zr - 63\i)F - lot7 * 'o)'p-l (6'33)

where

Pro:'

and D is obtained from Eq. 6.32.

Figure 6.9

I62 ANAI,YSIS Of fORMTD HEADS AND TRANSITION SICTIONS

A plot ol l!. 6..|.1 is shown in lrig. 6. l(). 'l'hc nrinirnum value ol ljq. 6.33crn bc lirund by taking thc derivittivc with respect to B and equating the result(o zero:

or 4/D\l + (3/280)t6/r),Et 5\r/l + t24/3st(6/tP

(6.34)

which is shown as a dashed line in Fig. 6.10. This figure illushates the effectof 6/t on the buckling shength of spherical sections. The minimum value ofbuckling strength is obtained from the figure as

!=o.z+ at 9:9.35tt t

The value of 0-24 can be reduced if the strain energy due to membrane stressbefore buckling is considered. Therefore, if Eq. 6.25 is modified to include thisstrain energy and if the revised expression is substituted into Eq. 6.29, the

1.2

1.1

t.o

o9

o.8

o.7

o.6

o.5

o.4

o.3

o.2

o.1

.o

64.rc%

Fisure 6.10

('REt

\l--\

\t, '#;*t, I 20,l

.15

,r)

5\,7

I (\

Eltve op,i

6.2 ETTIPSOIDAL HTADS

dil'lerentiation results in an expression whose minimum value is

fi = o.rtt

Experimental values have shown that the minimum value obtained is of the orderof

r63

Dxample 6.5. What is the required thickness of a hemispherical head subjec-ted to an external pressure of 15 psi? Let r : 96 in., E : 27 x l0o psi, andfactor of safety (FS) = 10.

Solution. From Eq. 6.35, with o", : (FS)o and o: Pr/Zt

ff: o.tzs

(,t)'"''(;) : ' '*

,ro: 2

Ne: +Prz=j;-

(6.35)

(6.36)

= 0.45 in.

6.2 EttIPSOIDAt HTADS

The governing equations for the design of ellipsoidal and torispherical heads areobtained from expressions 6. l0 and 6. I l. For internal pressure, P. : P, P6 : O,

and the two equations give4

I

(15X96)110)

0.25(27 x 106)

14 18

We can write Eqs. 6.36 in terms of the major and minor radii a and b. Using

l6tl ANAl,Ytlt Ol lOt,UlD HtADt AND TRANSII|ON SICTIONS

fisure 6.ll Ellipsoi&l heod.

the notations of Fig. 6.11, we then obtain

a2b2(az sin2 Q I b2 cosz E1zlz

n2''

1a2 5i1z Q r b2 cos2 qlrlz

Expressions 6.36 then become

^, -Po' I^6- 2 @rJfI? AT brrxf, O,7t

,, _ Po' b' - (a2 - b2) sin2 E2b2 (a2 sin2 6 i b2 cos2 q1r/z

The radial deflection rll and meridional deflection ll due to intemal Dressureare given by

(6.37)

GIu'-,l;t

Pa'nD^2

n W'(Fcosv)

ulc'- l] and

6.2 rruPsotDAt HEADS

where

ffi''r)"G= \/l + [(a/ br2 - 1] cos2 7

At any given point on the ellipse given by.re and y6, the angle { can be obtainedIrom

"i"Q=ffiA plot of Eq. 6.37 in Fig. 6. 12 shows that for ellipsoidal heads with a/, rarios

over 1.4, the hoop stress at d = 90'is in compression. The curves indicate thatthis compressive force increases as the head gets shallower. Design of headsbased on these high compressive membrane forces iends to give ulha-conservative answers. This is because discontinuity forces tend to lower themaximum compressive sFess which results in more realistic desims. The ASMECode uses such an approach in the design of elliptic and torisfherical heads.

Example 6.6. A2; I ellipsoidal head is subjected to an intemal pressure of100 psi. If a = 48 in. and r = 0.5 in., detemine the hoop and meridional stressat6=9A'.

Sol.ntion. With d : 90', Eq. 6.37 becomes

nr=t

= ryEq = 24oo lb/in.

%=r# = 48oo psi

,j

,1'"(i.{|,(tl -, - *-y, - *,(t)

I

166 ANATYSIS OF TORMED HEADS AND TRANSITION STCTIONS

elozlq

Similarly,

h/.Fis'rr€ 6.12

No=+(2br-ar)zo-/lno.|r4Rl

= -ffitz x 24, _ 482)

: -4800 lb/in.

- 4Rfno' = -(rf = -9600 psi I

and

6.3 TORISPHTRICAI. HEADS 167

6.3 TORISPHERICALHEADS

In formulating the discontinuity equations for torispherical heads at the cylinderjunction, two assumptions must be made. First, the ratio a// must be over 30.Second, all deflections dissipate rapidly away from the junction. With these twoassumptions (known as Geckeler's approximations), the discontinuity analysisof a torispherical head near a cylinder junction is similar to that for a cylindricalshell. Hence, the governing equations are (see Section 5.2.1)

-! + 4Baw : O (6.38)

where

B=

Equation 6.38 is similar to Eq. 5.21 for cylindrical shells except that in Eq.6.38 the quantity p is a function of 12 that is variable along the meridian. Thisrequires numerical integration of all moment, force, deflection, and slope ex-pressions at angles less than @ : 9g'.

If a discontinuity force is applied at the edge as shown in Fig. 6. 13, Eq. 6.3gyields the following values.3

.:#rru,n,-FoB^Md.

u, : 7, : z1,f;c ^oo

- FoBp,Mo)

n = Gr"^n, + zBoDe,Mi (6 3e)

*r=ffi<-oupo+ Ap"Mo)

Me = t&10

Fisuro 6.13

I6E ANATYIIT OI IORMTD HTADS AND TRANSITION STCTIONS

whcrc AA = ., rr'(cos ps + sin Bs)

Be, = e &(cos Bs - sin Bs)

Cs = s-8" t.,t PDp" = a-Fs rin B"

.Rt - u\rh= \l--a-p = poisson's ratio

6.4 CONICAL HEADS

The sfress distribution in a conical head can be obtained from Eqs.6.11. From Fig. 6.14 with d constant,

r=ssind/t=o

12= stard

Redefining N4 as N, and p4 as p", Eq. 6.10 becomes

and Eq. 6.l1 becomes

Na : p,s tan d

rv" = + I ,0,- o,tan d)r dr

6. l0 and

(6.40)

Fisure 6.1,{ 2 cos c!

6.4 CON|CAT HEADS t69

The forces and deflections obtained from Eq. 6.40 due to some typical loadingconditions are shown in Table 6.3.

Exampfe 6.7. A conical shell with d : 45" and base diamerer of g ft issubjected to an intemal pressure P Find the expressions for N6 and N".

Solutinn. For intemal pressure, N" : 0 and lr'e = p. Therefore, from Eq.6.40,

Ne : Ps tan 6r

p-

cos a

P .2N"=;*d,;+c

s=0 N"=0

(1)

Also

at

Hence

and

c= 0

Ps tan a'

PrlV":;-: ,zcosa

6,4,1 Unbolonced Forces ot Cone-to-Cylinder Junction

Thejunction of cones-to-cylinders must always be considered as part of the conedesign because of the large stresses that occur there. By referring to Fig. 6.15,the force 1{, at point 0 was found in Example 6.7 to be

I

PR

6.4 CONICAT HEADS t7l

(b)

Figure 6.15

Sumrnation of vertical forces at point 0 gives

2zrRV = PnR2

or

V =PR2

Since V is the resultant of components N" and A, it follows that 11 is an inwardforce with magnitude

.. PR t^n d,= 2

I

I

od

d -o\nP

d o I at4(J NI

. 6t .t. Etro , Fr olsrA AI AI

^ r--:---r:+! ilN

!

'IR

itlloE - Ol

v; -l t 3l6

oldIE N I

- - ;lo ;l .l olpo ol olN olq 'l't *" a,

""1-

a f' drN

-3"i

| -o r:JI ld d -

I (/lla o - ot-oN | .1.^ u ltldd old alN o olN(JI I E OIFc tF Qtia . lo) ilo N le Ull

of | 6 | . , tP

^" ? | ".liz@@Fo

+

t

d.

o

o

oI

o

olt

oooop

-o

(irlt

oF

170

172 ANALYSI3 O' TORM!O HIAOS AND TRANSITION SCCTIONS

This tbrce H must bc resisted by ring action at the junction The required area

of the ring is given by

PR2 tairr a' (6.41)

where A : required ring area

P : intemal pressure

R = radius at base of cone

o' = one-half the apex angle

o : allowable compressive stress of ring

Example 6.8. What is the required area of the compression ring at the cone-

to-cylinder junction in Example 6.7? Let P : 20 psi and the allowable stress inthe ring 10,000 psi.

Solution. From Eq. 6.41,

20x48'?x1.002 x 10,000

= 2.3O in2 .

6.4.2 Discontinuity AnolYsis

The derivation of the discontinuity expressions for conical shells is similar to that

for cylinders. The resulting moment and force equations for conical shells are

expressed in the more complicated Bessel function terms. However, approxi-

mate solutions for various edge loading conditions can be expressed in simple

form as shown in Table 6.4. In this table,

.HR(f

2o

I

^tIJ='-,' srn 4)

-

o==

ts

+

Nl@l x xNll-l

ttN->l *'l * e-NI

=I

9cl .-l € =.._:- 61 ,-

ol 4. ol @IE

' lo -l!o,lNN l@tl.{

=

i-:-r

I .\r

l-->+

qI

F^ x Fl+

o

o

.olt @l to .l 0,

I NIS'tN !l IN--> N | -_>

=cg6:

=11.o *.

173

t74 ANAIYSIS Of fORMEO HEADS AND TRANSITION STCTIONS

I

Figure 6.16

ExamDle 6.9. Calcuiate the maximum longitudinal and circumferential

*"r... i" the cylinder shown in Fig 6' 16 due to intemal pressure P

Solution. From Fig' 6.16

f+F:H (1)

where from Section 6.4.1,

Pr tan a'

'l'he deflection compatibility between the cylinder and cone is given by

dclloction of cylinder at junction due to M and /

: deflection of cone at junction due to M and F (2)

Sirrrilutly,

r()trti()n ol cylinder at junction due to M and/

= rotation of cone at junction due to M and F (3)

llsirrl f rrlrlcs 5.2and6.4 and solving Eqs 1,2, and3 result in the following

6.4 CONICAT HEAD5 t/5

exprcssrons:

where

f:HuF:H(1 -U)

/r\M: HIiIVZ\p/

H: Pr lan d'2

cos2 o'(3 * cos2 c')i + cos'zc'(6 + cos2 a')

_.uvz=o*"o"2"';

The maximum longitudinal stress due to M and pressure is expressed as

* : ?(t t - 4 sss v,,,4 "" "')

whereasthemaximumcircumferentialstressduetoMandpressureisgivenby

6.4.3 Cones under Exlernol Pressure

The solution of the buckling of a conical section subjected to external hydrostatic

oressure is normally obtained by energy methods The resultant equation is very

ffi;;#ili; the iterativi prociis needed for lhe solution s Experimental

,"."-"ft "o-paring

the buckling equations of conical and cylindrrcal shells has

.iro*oiftut tfti, U*lling of a conicai shell is similar to the buckling of cylindrical

r-n"fi.-*ift "

length eq-ual to the slant length of the cone and a radius equal to

,h" un"aog" radiris of curvature of the cone Research has also shown that the

qr""tiiy O - Dr/Dr) has a significant influence on the buckling of a cone'

176 ANAI.YSIS OF fORMtD HTADS AND TRANSITION SICTIONS

Accortlingly, thc buckling oquitti(n ol a cono citrl bc cxpresscd as

*,:"('-';)

where p is the modified buckling equation of a cylindrical

f(l - DtlD) is a function of Drf Dz.

A simplified equation for the buckling of a cylindrical shell" is

p 2.42E (t /2r)25=6_EfnlL/r,-o.4sen84

shell and

(6.42)

For most applications, the second quantity in the bracketed denominator is small

compared with the first one and can thus be neglected' Based on this, the

buckling equation of a cone (Fig. 6.17). may be wrinen as

4, _ 2.6(cos a')25(r)15./, _ 4\E tl1,Dt-rDtl21'5rU or/

!V"'

Fisure 6.17,. : T (' * noq)

= 24 3't5 in.

6,4 CONTCAT HEADS 177

'Ihc magnitude ofthe tunctionl(l - Dtf D) canbe deternrined thetlrctically

Based on this plus the "scatter" band of experimental data, a value of l 0 was

used for the function at Dr/D2 of 1.0 (cylinder). The function changes linearly

to a value of 0.8 for DtfDz of zero (full cone). Thus the buckling equation

becomes

p

ti

The second bracketed expression can be approximated by the quantity

2

r+ND,

Using a factor of safety (FS), the allowable external pressure on a cone is given

by

P, - 2.6(t"/D2)25

E (FS)(Le/Dz)

P. = allowable external pressure

E : modulus of elasticity

t"=tcosd7"=11/2)(t + Dt/D2)

Dl : diameter at small end of cone

D2 : diameter at large end of cone

(6.43)

Example 6.10, Design the cone shown in Fig. 6.18 for an extemal pressure

of 15 osi. Let FS = 4.0 and E = 30 x 166 psi.

Solutinn. a' = 30.96" and cos a' : 0.858.

I70 ANATYSIS OT FORMED HEAOS ANO TRANSITION SICTIONS

From Eq. 6.43,

,": o,(wo-1a,ly'

=,,(u?,#i:,'*i"l': 0.20 in.

I : """""""': : U.zJ ln.cos a'

and

I

NOMENCTATURE

a : major radius of ellipse

b : minor radius of ellipse

lt:Et3/12(l-ttz)/)r : diameter at small end of cone

D r =60"

Fisur€ 6.1S

NOMENCI.ATURE

/)z = diameter at large end of cone

C : modulus of elasticity

L = axial length of cylinder

L":(h/2)(1 +Dt/D2)

/ = axial length of cone

l' : slanted length of cone

M = bending moment in hoop direction

M = bending moment in meridional direction

N, = axial force in cone

N = force in hoop direction

N = force in meridional direction

P : pressure

P" = allowable extemal pressure

4 = intemal pressure

P, : external pressure

P, = radial pressure

P" = axial pressure in cone

P = meridional pressure

Q : shearing force in head

r : radius

rr = radius of curvature as deflned in Fig. 6.4

12 = radius of curvature measured from axis of symmetry

4 = inside radius

r, : outside radius

s = distance along the slanted length of cone, measured from apex

? = temperature

t : thickness

t" = tcos a'v = axial deformation

w : radial deformation

179

I8O ANATYSIS OT IORMED HEADS ANO TRANSITION STCTIONS

* coellicient ol thcfl)lal oxpansion

= one-half the apex angle of a cone

a

a

e=w for cylinders

for cones

7 =7r/2-Q6 = deflection measured perpendicular to axis of symmetry

A : rotation

^ =\yto=-6i/Fp = poisson's ratio

o = stress

o- = critical buckling stress

o, : radial stress

.r, = longitudinal stess in cone

od = hoop stress

od : meridional sfess

S = angle as defined in Fig. 6.4

R,EFERENCES

Gibson, L E., Linear Elastic Theory of Thin SherrJ, Pergamon Press, New York, 1965.

von Kaman, T. and Hsue-Shen Tsien, "The Buckling of Spherical Shells by ExtemalP.essue" in Ptessure Vessel and Piping Detign: Collected Papers 197-1959, AfieicanSociety of Mechanical Engineers, New York, 1960.

Coates. W. M., '"The Stale of Shess in Full Heads of Pressure Vessels" in PressureVessel anlPiping Design: Collected Papers 1927-1959, American Society of Mechanical Engineers,New York. 1960

Baker, E. H. et al., Srell Arnlysis Manual, NASA CR-912, National Aelonautics and SpaceAdminisb"ation. Washingto!, D.C., 1968.

Jawad, M. H., "Design of Conical Shells Under Extemal Loads," Jounal of Pressure VesselTechnology, Vol. lO2, 1979.

R^etz, R. Y., An Experimental Investigarton of the Strength of Small-Scale Conical ReducerSections Between Cylindrical Shells under Extenal Hydrostatic Pressure, V. S. Department ofthe Navy, David Taylor Model Basin, Report 1187, February 1959.

t.

BIBTIOGRAPHY I8I

BIELIOGRAPHY

Bilfington, D. P., Thin Shell Conctete Structures, Mcclaw-Hill, New York, 1965.

Flugge, W,, Stresses in Shells, Springer-Verlag. New York, 1967.

Timoshenko, S., and S. Woinowsky-Kieger, Theory of Plates and Shells, McGraw-Hill, NewYo*, 1959.

I

CHAPTE R7STRESS IN FLAT PLATES

V'nnn^ flot plote3. (Court$y of lhe Nooier CorPorotion, Sr' Louis, Mo )

t83

l8.l 3TR!SS lN fLAT ptATrS

7.1 INTRODUCTION

lli:4gl * very common in grgcesl .quiq-"nt.Qiqrgql

such areas as nozzle covers, bulk heads, ani tuUesilOets, Ttrereasru\,rr .ucas .ts rozztc covers, DulK neads, and tubesheets, -whereas rcefanCularglles.are used as segnrylq! trays, baffles, and in rectangulir'p."rGffi;i;.This chapter presenB-6rieTdescription of the theoretical iackground of circularand rggjqggkUllates.

The theory of symmetric bending of laterally loaded plates is generally basedon the following assumptions:

l. Thickness of plates is significantly smaller than the least lateral dimen-sion of the plate.

2. Loads are applied perpendicular to the middle surface of the plate.3. No forces are imposed in the middle surface.4. Lines perpendicular to the middle surface before deformatron remain

perpendicular to the deformed middle surface.5. These lines are inexlensible.6. These lines remain straight lines.

These assumptions form the basis for developing the bending theory of platesand apply to plales where buckiing is not a consideration.

7.2 CIRCULAR PLATES

The relationship between the radius of curvature and the deflection of a circularplate is obtained from Eq. 3.6 as

1 d,2w

;= 77or in terms of the terminology of Fig. 7.1,

! =d2w : -dQrn dx,2 dr (7 .t)

'fhc second radius of curvature is also obtained from Fig. 7.1. Line,4g is them(lius of curvature r, of all points at a distance r forming a cone:

sind-at=I

Using the sign convention that clockwise angles and moments are positive and

7.2 CIRCUI.AR PI.ATES t85

downward deflections are positive, the relationship becomes

6 1dwfr f rdr

The moment-curvature relationship is based on Eq. 3.11 and is given by

Substituting Eqs. 7.1 and 7.2 into this expression gives

M,: -D(+ * L+\\4r- r dr./

= D(!+ * *9\\/ dr r/

(7.2)

(7 .3a)

(7.3b)

Sinilarly,

M,: -D(!+ * p+\\r ar dr/

= -o(9 *

Fisure 7.1

1 1\- -r p-l

dd\p-, Iar/

(7.4)

t86 STRESS tN r[AT ptAnS

For a unilorrnly bodcd platc, thc tbrces acting on an clement are shown inFig. 1.2u. Taking moments uboul a-n gives

(M,r d0) - (". - ff*)<, + dr) d0 + z(u, a,.t) - " *(i)

/ )n \ /,{.\-lO + 7 drl\r + dr) dol+l = o

\ 4r / \z/(7 .s)

Ttle qvant\ty Mt dr d@/2 is the component of M, perpendicular to axis d to a as

shown in Fig. 7.2b.Disregarding higher-order terms, Eq. 7.5 can be reduced to

M,+#r-M,'rQr:0

(b)

tiswe 7.2

(1 .6)

(Mr1!!r

(o+5F dr)

7.2 CIRCUTAR PTATES

Substituting Eqs. 7.3 and 7.4 into 7.6 gives

d1w I d2w 1dw O_--=:drr r dr2 12 dr D

or

dlrdld,/,)\1 O-:-|_-|r-||:=drlr dr\ dr / I D

Similarly, substituting Eqs. 7.3 and 7.4 into Eq. 7.6 gives

d20,ldo o -oa*-;ar-7: D

187

Equations 7 .7 and 7 -t are the basis differential equations for the bending ofcircular plates due to symmetric loading. Equation 7.7 can also be written interm of the local load as

(7.7)

(7.8)

(7.e)

Once w is determined from Eqs. 7.7, then the moments are obtained from Eqs.7 .3 arrd 7.4. The shearing force is determined from Eq. 7.6, and is exDressedas

o: D(+.!+ - \+\\ dr- r ar- r'ar/ (7.10)

Example 7.1. Derive the moment expression for a uniformly loaded, simplysupported circular plate of radius a. For g, : 0.3, plot the moment diagram anddetermine the maximum deflection. rotation. and stress values.

Solution. From Fig. 7 .3, the sheat Q at any radius r is givenby 2mQ : nnz Por

l,lii,rr1= -g

I d [,d l! d (,a\.ll = cr drl drlr dr\ dr/l) D

Pr

188 STRlSl lN rtAT PLATIS

Fisure 7.3

Therefore, from Eq. 7.7,

dltdldw\l Prd,l;E( d, ) ): b

Integrating both sides gives

dl dw\ Prl ^d,\' d,)

:5 - '"A second integration gives the expression for the slope that is given by

^ dw Pr3 Crr Cz-drl6D2r

and the third integration gives the deflection w as

Pr4 C,rzw:_L+++C2lnr+,C,64D4-

(1)

At the center of the plate, r = 0 and the slope is zero due to symmetry. Hence,liom liq. l, f,2: Q.

At r : a, moment M, = 0 and Eg. 7.3 gives

Cr: Paz /3 + u\-8D\l+,,/

At r a, thc deflection is zero and Eq. 2 gives

,,=#(T#-')

7,2 CIRCUI.AR PI.ATES

Hence, the deflection as expressed by Eq. 2 becomes

Pra r2 13 + p\ Pa2 Paa /6 + Zu \r4':_ t_!_t _llsD 4\l+prl8D 64D\t+p 'lor

o '15+u '\w: -)-(az - ,tll'; :, * .a2 - r2lUD'- \l+p Iand

^ dw Pr(, 3+p,.\- dr l6D\' l+p')

The maximum deflection occurs in the middle where r = 0. Hence

Paa /5 + p,\ Pa4 15 + u\ 12(l - 42tnj*, =_t | = _t ____________:_ | ,

64D\t + pl 64 \r + p/ Et3

andwithp=0.30

"'-"" - o'696!aq,

Et3

The maximum rotation occurs at the edge where r = d. Hence,

^ -Pat -3^, (l - rr)d'- = 8r,(r + p) : tP"' -i

and with /.r = 0.30

s^^= -t.os EtY

The moment expression is obtained by substituting Eq. 3 into Eqs. 7.3 and7 .4. Hence

,,=*rt + 1t)(a2 - 12)

u, = ftbt{z + p.) - r2(1 + 3p.)l

t89

(3)

(4)

(s)

(6)

r90 tliltt rN trAT PtATtt

A plot of Eqs. 5 and 6 for g. = 0.3 is shown in Fig. 7.4. The plot indicat€s thatthe maximum moment occurs in the center and is given by

M*=3#

6M l.24Pa2T- I-

Example 7,2. Derive the moment expression for a uniformly loaded circularplate of radius a that is fixed at the edge. For trr, = 0.3 plot the moment diagamand determine the maximum deflection and stress values.

Solution. From Example I,

Pr4 Ctz ^ ,.:64D- -T-- Lz lll / -f L:

2n

{{'o

00

I'h

Figoro 7.4 Momeni diltriburion for simply suppord plolo.

I

(l)

lt.

and

7.2 CIRCUTAR PTATIS t9l

(2)

At the center of the plate r = 0 and the slope is zero due to symmetry. Hence,from Eq. (2), C, = 0.

At r = a, the slope is zero and from Eq. 2,

-Pa2Cr=

Also at r = a, the deflection is zero and from Eq. l,

Ct:6

Hence, Eq. 1 becomes

pra pa2r2 paa

64D 32D 64D

= L1^z - 'z1zuD'"The maximum value of deflection occurs at / = 0.

- Pr3 C,r C,H=-+-+i-l6D2r

u,: f,bT + p) - r2(3 + tL)l

u, = llt tt + tLt - 121t + 3p))

8D

(3)

Paa Pa4 l2(l - tr-z)w=-'64D&

and for P = 0.3,

r,v.*=0.171r9)\Lt" /

The moment expression is obtained by substituting Eq. 3 into Eqs. 7.3 and7.4. Hence,

(4)

and

(s)

tt2 $iltt tlit llaT ptATtt

l.flgure 7.5 ,nom€nt didrihtion tor f,or ft(€d plote.

A plot of Eqs. 4 and 5 for p = 0.3 is shown in Fig. 7.5. The plot indicates tlntthe maximum moment occurs at the edge and is given by

-PazM,*"=

6M -0.7 SPazo*,=v_---v-

Problems

7.1 Determine the maximum deflection, slope, and bending moment for asimply supported plate subjected to edge moment Me.

16tlffi

and

T

.,.

7.3

7.3 RTCTANOUTAR PrAlt3 t9g

^ azMnAnsweri mo( lr = # at centerzD\t + l.L)

^ aM"max 0 = -_ft

at edgeu\r 'f p)

max M, = Mo throughout plate

m^x M, = Mo throughout plate

A ctcular plate is fixed at the edge and is at an ambient temperature of70T. What is the maximum stess if the top surface is heated to alemperature of l70T and the bottom surface is cooled to a temperature of-30T?kta = 9 x 10-6 in./in. T,r:0.5in.,a:60in.,8:30 xlffpsi,p=9.3.

. aLTEanswer. o =n _ 1.,

= 38,600 psi

Determine the maximum moment in the circular plate shown in Fig. 7.6if a = 4 ir., b = 2 in., p : 0.3, and P : 100 psi.

Answerz M' : 384.6 in.-lb/in.

Fieur. 7.6

7.3 RECTANGULAR PTATES

In developing the differential equation for circular plaies, the shearing shess wasignore.d because the load was symmetric with respect to 0. In rectangular platesunder uniform loads, the shearing stress interacts with the normal shesses in the.r- and y-directions and thus cannot be ignored. This results in a more compli-cated differential e4uation than that for circular plates. In addition, the solutionof th€ differential equation of rectangular plates is more elaborate and involvesthe use of Fourier series. Because of this, only the case of a simply supported

Ifi lnl$ il ftAT ftATtt

rrctangulsr plate loadcd throughout its surface is discussed here. Nonsymmetricloadings and boundary conditions other than simply supponed result in quitecomplicated solutions that are beyond the scope of this book. The examplesgiven in this section are intended to give the reader a concept of the generalbehavior of rectangular plates and the difference between them and circularplates.

If an infinitesimal section is removed from a rectangular plate, the forcesacting on it will occur as shown in Fig. 7.7. Summation of forces in the z-axissrves

q(x, y)dx dy - Q, dy + Qt dx +

This equation can be reduced to

(a. ft,,)*=0

(a.. ff *)', -

q(x,yt+#.#=, (7 .1,r)

Summing mornents around the "r-axis and deleting all quantities of higher ordergives

m, *$$o'

Figwe 7.7

7.3 RKTANOUTAR PIAT!3 19!

W-+-Q'y=o (7.12)

(7 .t3\

(7.t4)

(7.15)

(7.16)

(7.18a)

(7.18b)

(7.18c)

dQ" =

d,g _ atu,dy dy- dxdy

Similarly, summing moments around the y-axis and deleting all quantities ofhigher order gives

#.ry-e"=odo, d2M, azM,,ar-: d* - atfr

Substituting Eqs. 7 .13 andT .15 into7.1l and using M"t = -MofromEq.3.1lgives

, d,M, 2d2M* a2M,q\x, y,-r -;z - -;i + ir, = u

The differential equation relating deflections and applied loads is obtained bysubstituting Eq. 3.11 into 7.16 and obtaining

daw 2daw 1aw ak. v\

-

-l-

-

-t

-

: :-:---:--:-:-ax4 dx2dy2 dy4 D

(7 .17)

which is the differential equation of the bending of a rectangular plate subjectedto lateral loads.

For any grven loading and boundary conditions, the deflection p can beobtained frorn Eq. 7.17. Tlie bending.moments can then be determined ftorn Eq.3.1I as

".: _,(#. _?)

^: -'(#.,#)

u* = o !-'Y^' dxdy

Itc filttt tt{ il.tT tta?tl

and tho shcarfng forccs Q, and p, arc determined from Eqs .7 ,12,7 .14, and3.lla8

n.- -,*(#.?)

"= -,&(#.?)

Example 7.3. Detennine the maximum moment in a simply suppofied rectan-gular plate of length a and width D if the applied load is expressed as

q = qssinTF*T)

I*t a = 6O in., b : 25 h., eo = 3psi, and p = 0.3.

Sohttian, Assume y to be of the form

w=C txl rrY\rr;\rr 7/This expression satisfies the boundary conditions of w = O and M : 0 at all fouredges. Substituting this expression into Eq. 7.17 gives

r-= 4o" Dr4(1/ az + l/b\2

and the deflection expression becomes

w : DFn# +Ilbry't # ('* ?)Substituting this expression into Eqs. 7.18, gives

,,= aOffilnG,. *) sinr sin|

u = Qo /.]|l ..,- -t \ ^'- Tx -,- n!'""

: }(FaI7FfV-j - F) sln - srn ,M,:dffi"*Y"o"|

Thc maximum value of M, and M, occur when .r = a/2 and y = b/2. A

(7. l9a)

(7.19b)

7.4 CIRCUIAR PIATIS ON IIASTIC IOUNDATION I97

comparison of the denominator in parentheses in the expressions for M, and M,indicates that M) will always give a larger value of M for the given values of aand b. Accordingly, the maximum value ofM is given by

M*=atrffitm@.*)M^."==,?'o,==fq!*a\

1r2(1/ffi2 + | /25\2\602' 252f

= 145.1 in.lb/in.

The maximum value of M,o occurs when.r = 0 and y = 0 or r. Hence, themaximum valae of M, is given by

M,r3.0(0.7)

12(1/602 + r/25r,6q25)= 40.2 in.lb/in. I

7.4 CIRCUIAR PLATES ON ETASTIC FOUNDATION

Many tubesheets of heat exchangers are designed as plates on elastic foundationas discussed in Chapter 14. The solution of the differential equation of a plateon elastic foundation involves Bessel functions. The four Bessel functions usedin this section are

- t -l)ix4iZ@) = 6sr1r1 = >. *, -?"7'[(2])ll''- (x/2\o (x/2\8 (x/2\t2=t-E--+o. 61, -"'

S. (-1\jx4j'Ll\xt = Det\x) = 47W* ,X

_ (x/2)2 , (x/2)6 (r/2)'o- --l?---L- -f?-- "'23ft) = _ ? *"r(,) =

t+ _ 1lr, * r^i . a<.t]

z1i = ar.,k) =z+ .1lr, * ,"5.t,<,t]

r9r ltiltt ll{ fLaT ptATll

whcrc

" = /lY - 99/rY * 4(?1r\" - . . . .

" - \z) 31'z \2) 51'? \2),, _ 4{z\ (x\ _ d(4) /.rY _r d(6) /r\u + . . . .h:-';-\1) --+r\1) -?tr\r/ - "

I I l*.......*lQ\n)= t*t*1-z "

"t = 0.577216

The limits of the Z functions as 'r --> 0 and as x :+ co are given in Table 7' 1'

The table also shows the limits for the first derivatives of Zt tltroryh Zq.

The relations between the various derivatives of the Z functions are as fol-lows:

Toble 7.1 Limits of Zfunctions

FunctionLimit as

.r ---+ 0Limit as

Z{x)

z2@)

h6)

h(x)

&(x)dx

il.(x)dx

&'(x)

1.0

_x2--7-

0.5

2t^+

-x'-:7lo

-xzx, yx-Itr=

2

nx

dx

E+(x)dx

f cos K

-f sin K

r, sin rl,

-z cos ry'

1fu(cosr-sinr)

;!(cosr+sinr)

ft@osi,- sinrli

1fu(cosg+sin0.1xI =:eXD""'=' !2tx - !2

xnv86

,t, = j- +i'v26

7.4 C|RCUIAR PtAftS ON ltAgTrC TOUNDATTON t99

d2z,k) | EtG)Zzlx) - - ---jax- x a.x

dzz2\x) 1 fl,2@)-----:--;-=-zt\x)-- ,ax- x ax

d'zA(x) lfi,l?)----=-- = L4\x) - - -'--:-dx' x dx

d2z4@) | dZlx)-----:--t-=-23\r)-- ,ax- x ax

The force exerted by an elastic foundation on a ciruclar plate due to deflectionof the foundation is expressed as

P=Kow

where p = foundation load

w = deflection of foundation

Ko = stiffness of foundation

: modulus of elasticity of foundatior/depth of foundation.

The differential equation of a circular plate on an elastic foundation can beobtained by modifying Eq. 7 .7 as

i,lli,(,#)l=ryThe solution of this equation is expressed as

(7.19)

w = Cfl(ar) 'f C2Z2@r) * C3Z3(ar) i CaZa@r)

where ": ^{EJD21 - Za : Bessel functions

Ct - Qa : constants of integration

Example 7.4. Detemfne the maximum deflection in a circular plate on anelastic foundation subjected to a concentrated load F in the center of the plate.

Solution. From Table 7.1 it is seen that as r approaches infinity, Z and Zz alsoapproach infinity. Therefore, Ct and C2 must be set to zero. Thus

w:CtZt(ar)iCaZa(ar\

100 tTmll rt{ tuT ptaTtt

and

0 = + = Ct a Zi@r) + Ca a Zi@r)-dr

As r atrrproaches zero, 0 must be zero due to symmetry. But from Table 7. l, Ziapproaches infinity as r approaches zero. Hence, Ca must be set to zero. Thus

w = CzZz(ar)

dw* = CtaZS@r)

d2w ^.f I Ijj=c*'lzo{or)--zi@nld3w ^"f I I7j - Aa'zlozi@r\ -:zKar)]

Substituting these derivatives into Eq. 7.10 and equating this to F' gives

^F4azD

' =

'fi6t'<*>

and

and

F8rr2D

NOMENCI-ATURE

Et3;:;:----------;:tz\t - p-)'

= modulus of elasticity

= stiffness of foundation= modulus of elasticity of foundation/depth of foundation

= radial moment in circular plates

E

Kq

M,

UEUOORAPHY 201

Mt = tangential moment in circular plates

M, = moment in .r-direction of rectangular plates

My = moment in y-direction of rectangular plates

M,r = shearing moment

P = applied pressure

O = Shearing force

applied load

r = radius

T = temperature

t = thickness

w = deflection

21 to Za = Bessel functions

a = */EJDpoisson's ratio

REFERENCE

l. Tiomoohenko, S., and S. Woircwsky-l<liege\ Theory of phtes and Shelts, McGmw-Hill,New Yort, 1959.

BIBTIOGRAPHY

Hetenyr,M., Beams on Etastic Foandation, University of Michigan Pr€ss. Atrn Arbor, Michigan,1964.

McFarland, D, E., Smith, B, L,, aDd Bernhan, W. D., Analyris of plates, Spaftan Books, NewYork, 1972.

Szilard, R., Theory and Awlysis of Prorer, hedice-Hall, Englewood Cliffs, {.J., 1974.

PART3DESIGN OF

COMPONENTS

203

CHAPTER 8DESIGN OF

CYLINDRICAL SHELLS

2052tJ4

a,' ,'l,rrrr-n rower used by o ferlilizer monufocturer' (Courtesv of lhe Nooter CorPorolion' St tot,,is' Mo )

X

206 DlgloN or cyuNDRrcAr. sHEr.rs

Cylindricul vcsscls ure very liequently used in the petrochemical industry. Theya.re easy to fabricate and install and economical to maintain. The requiredthickness is generally controlled by intemal pressure, although in some instancesapplied loads and extemal pressure have control. Other factors such as thermalstress and discontinuity forces may also influence the required thickness.

8.I ASME DESIGN EQUATIONS

A simplified equation was developed by the ASME Code, Vltr-l, for deter-mining the required thickness of a cylinder subjected to intemal pressure. It isa simplification of Eq. 5.3 and gives accurate results over a wide range of r2/r1.This equation is expressed as

(8.1)PR

where t: required thickness

P = inlemal pressure

R = inside radius

S = allowable stress

E : joint efficiency factor

sE - 0.6P

A comparison of Eqs. 8.1 and 5.9 is shown in Fig. 5.6. It indicates the widerange of applicability of Eq. 8.1. The ASME Code, Vltr-l, has, however,limited the use of Eq. 8.1 to t less than or equal toR/2 and pressure less or equalto 0.0385 S. Various forms of Eq. 8.1 are shown in Appendix I together withan altemate equation that expresses the thickness in terms of Re rather than R.

The factor E in Eq. 8.1 is an efficiency factor and its magnitude depends onthe extent of radio$aphy performed at the various seams of the cylinder. Appen-dix J illushates the effect of radiography of various seams on the values of E as

established by the ASME Code, VI[-l.In Section VIII, Division 2, of the ASME Code, the equation for required

thickness is based on the stress at an average radius. Hence,

^ P(R + t/2)t

PR

s - 0.5P(8.2)

8.I ASMI DESIGN EOUATIONS 207

As the pressure increases above 0.4 S, Division 2 uses plastic analysis (seeSection 15.l) to obtain

r =ftn/R+r\\Ri (8.3)

Example 8.1. A pressure vessel with an inside diameter of 50.0 in. is subjec-ted to an intemal pressure of 100 psi. Using an allowable stress of 17,500 psi,find the required thickness according to Section VIII, Division 1. Assume thatall circumferential and longitudinal seams are double-welded butt joints and arespot radiographed.

Solution. From Appendix J a value of, = 0.85 is obtained. From Eq. 8.1,

PRt=-=- SE - 0.6p

100 x 25

17,500x.85-0.6x100r = 0.17 in. I

Exarnple8.2. A seamless cylindrical shell with an outside diameter of 30.0 in.is butt-welded to seamless ellipsoidal heads. The circumferential seams are notx-rayed. Find the required shell thickness ifthe allowable stress is 15,000 psi andthe intemal design pressure is 250 psi. Use Section XIII, Division 1 rules.

Solutinn. From Appendix J, with a value ofE = 1.0 (seamless shell), allow-able circumferential stress must be reduced to 8070 since the circumferentialseams are not x-rayed.

From Appendix I, the required thickness equation in terms of outside radiusis given by

PR

SE + O.4P

250 x 15

(15,000 x 0.80)(1.0) + (0.4 x 250)

= 0.31 in.

Probhms

8.1 An ASME pressure vessel with an inside diameler of4 ft has a seamlessshell. The head{o-shell seams are partially radiographed. Find the re-

t

206 DtStON Of CYt"tNORrCAt" SHE|TS

quircd thickness il'thc alkrwablc stress is 20,()00 psi and the design prcs-sure i$ 2900 psi.

Answer: t: 3.81 in.

8.2 What is the maximum allowable pressure that can be applied to a cylindershell with an outside diameter of 6 ft, thickness 1.25 in., and an allowablestress of 17,500 psi? Let E : 0.85.

Answer: p = 524psi

8.2 EVATUATION OF DISCONTINUIW STRESSES

In Chapter 5 we showed how stresses are evaluated at different locations due tothermal and mechanical conditions. The magnitude of these stresses must bekept below a given allowable shess. This allowable stress is established in theASME Code, VIII-2. The designer has to establish first whether the stress is ata local or a gross shuctual discontinuity, as defined in Fig. 8.1. Next the stress

STRUCTURAL

DISCONTINUITY

GROSS STRUCTURAL

DISCONTINUITY

LOCAL STRUCTURAL

DISCONTINIJITY

A source of stress or stfainintensification which affectsa relatively lafge portion ofa s truc ture and has a

slgnlficant effec t on the

overall s tres s or strainpattern. Exahples of gross

structural d I sconti n ui ti es are

Icad-to-shel'l and flange-to-!hcll junctlohs, nozzles, and

.lunctions between shells of(llfferent d i ameters or

A source of stress or stralnintenslf{cation which a ffec ts a

relati v€ly small volune ofmat€rial and does not have a

significant effect on the overall

stress 0r strain pattern or on the

s truc ture as a l1lhole.

Flsuro 8.1 Slrucrurol Dirconlinuities (Coortesy ot tho Ameri.on Socisiy ol r.te.honicol Engineers,)

8.2 TVATUATION OF DISCONTINUITY STRISSIS

is categorized as a primary, secondary, or a peak stress as shown in Figs. 8.2and 8.3. In Fig. 8.4 is a description of the two categories of thermal stress. Oncethe stress categories are established, the stresses at a vessel's different locationscan be classified as in Table 8. l

Table 8.2 shows the allowable stress for various stress caiegories. Applicationof Table 8.2 to various stress categories is given in the following example.

Example 8.3. Calculate the stress at points A, B, and C of the vessel in Fig.8.5. LetR : 60 in., L= 2.0625 in., 11, = 1.9313 in., P : 500 psi, .S. =15,000 Psi, P = 0.3, Eo = 30 x 106 psi.

STRESS

CATEGORIES

PR II4ARY SECON DARY

A stress developed by the constraint

of a structure. Secondary stress isself-limiting, Local yieldinq and

minor distortions can satisfy the

condltlons l'hich cause the stress tooccur and failure fron one application

of the stress is not to be expected,

Examples of secondary stress are geheral

thermal stress and bendi ng stress at a

gross structural discontinuity.

Peak stress does not cause any noticeable distortion

and is objectlonable only as a possibl€ source of a

fatlgue crack or a brlttle fractufe. Examples of peak

stress ar€: I ) thernrl stress in austenitic steel

cladding of carbon steel vessels, 2) thermal stress

li the vrall of a vessel caused by a rigid change in

teFperature of the contalned fluid. 3) the stress at

a I ocal structural discontinuity.

Fisure 8.2 Str$s Colesories (Courr$y of rhe Americon So.iery of rvle€honicdl Ensineer3.)

2lo DtslGN of cYLlNDRlCAt SHtLLS

tRlt4ARY STRtSS

stress is not classlfied as a primary stress'

A stress developed by the lmposed loading

|lhich is necessary to satisfy the la$s of

equilibrium, The basic characteristic of

a primary stress is that it is not self_

limiting. Primary stresses |,lhich considerably

exceed the yield strength l,lill result in failure

or at least in gross dlstortlon. A therrial

BENDINGI'4E14BRANE

An exanple ls the bending

staess ln the central

portion of a flat head due

membrane sttess is the menbrane

stress in a Shell Produced bY

external loadr dnd monent at a

perFanent suppoft or at a

noz2le connection.

nn e*'ampte of a local PririarY

stress is one which is so

distributed ln the structure

that no fe-distributlon of'load occurs as a result of

yielding, An exaniPle is the

stress in a circular cYlinder

due to internal Pressure.

general prima rY membaane

Figure 8.3 Primory Slre$ Cotosoria3 (Courtelv of lt'e Ahericon Socisiv of M3€honicol Eneine€rs )

Solution

Point A. From Fig. 8'6 and Eqs. 5.9 and 5' 10,

oo = 14,300 Psi

o'=0Psior = 7150 Psi

The stress differences are

14,300 | psi

-7150 | psi

8.2 EVAI.UATION OF DISCONTINUITY STRESSES 2I I

or o = | -71501 psi

Hence, maximum stress = 14,300 psi. From Tables 8.1 and 8.2, the maximum

stress for a general primary membrane sffess is

S, : 15,000 psi > 14,300 Psi O.K

Point B. Frcm Fig. 8.7 and Eqs. 5.3 and 5.4'

ae : 14,800 psi

o., : -500 psi

or = 7150 psi

TIlE RI4AL STRESS

normally sho!ld under a chang€ in tefiperature.

A self-balancing stress produced by a non-uniform

distribution of tenperature or by differing thermal

coefficients of expansion. Thermal stress is developed

in a solid body whenever a volume of material is

prevented from assu'ning the si2e and shaPe that it

Local thermal stress uhich

is associated wi th almost

conpl ete suppression of the

differential expansion and

thus produces no significant

distortion. ExanPl es of

I ocal thefmal stresses a|"e:

I ) stress in a snall hot sPo

in a vessel v/al l.

2) the difference betl/leen th

actual stress and the

equlvalent I i near stress.

3) the thermal stress in a

cladding material.

General thermal stress is classified

as secondary stress. ExamPles of

general thermal stress are:

I ) stress produced bY an axial

tempera tufe distribution in

a cylindrical shell.

2) stress produced bY the

tempe rature dIf ference

betqeen a nozzle and the

shell to t,Jhich it is attached.

3) the equivalent linear stress

produced by the radial tenperature

distribution in a cylindrical shell

Figure 8.,{ Ihermdl Stress Cot€gori6 (Courr$y of lhe Aftericon Socierv of Mechdnicol Ensineers )

Tobb 0.1 Clorrlflcotlon of Slrun (Rrforrncr l)

0

o

a

a

o

{av. &dslull sclio.t

o

a

212

Tobb 8.2 Slreu Cotrgorlrr qnd Llmltr of Strcrr Intrnrlfy (Rrf l)

SirartEe.dlns

!o Pnhe.y or eR0n6.

-

U* &5iEn load3

---- Us op.aungloads lPr +P. +Q +F H s. ). , \-/

tt-I--,--.---t----*-1--1 | itP. F{ s.) I I i i :T\-/ | | | i I

I | | r-.+ - - -;.-r- /'\ I- --l- iI P. Hr.5 s,) llPr+PD+O+ rs. ) '-l- \__/ l---l- \__/ :L----T-------r i !

.- -r- /-\ r :

lP(-PD H1,5 s-) L__---*f ----.t

-

\--'l I

Fisure 8.5

213

2ta Dltloltl Of CYtlNoRlCAt 3H!rr3

.-....._"

Fisur.8.6

=._"a

n, oe

Figur€ 8.2

The stress differences are given by

r0_, = 115,3001 psi

4_r : l_2650lpsi6,_, = l-7650lpsi

Hence, the maximum shess = 15,300 psi, which is about 27o above the allow-able sffess of 15,000 psi.

- Point C . The discontinuity forces at point c are shown in Fig . g . g . From this

ngure,

8.2 EVATUATION OF DISCONTINUIW STRESSES

r'" = +e) - 500 x !'0-x-2 0625 = 7734 in lb/in'

Mt=M,+N,e=M,+7734

r, : 60 + '#:6l.olt3 in.

rh = 60 + +! = 60.5157 in.

B=

o, = ,^'.!o'1 ,, = o.so35Etz\t - p-)

215

Also,

3(1 - p') = 9.8465,,1h/

Fisure 8.8

216 Dt]Ot{ of cful{DflcAt 3H!U.3

Thc lirst cornputibility cquution is given by

defl€ction of shell = deflection of head

or

wo * wlsn ! wy" * wql^,"n"n: wp I w9 I wy1l6, **where for the shell

Pr? m2,987It'- =

-" Et, Es

N,e _ 366,454t*" = zB'o,= k47.3822M,."" : ___Eo _

''': 4- - 4r3'4s69Q. zp"D, Eo

and for the head

Prl . |,242.851we=EA\t - lt)= E"

*^ = 2'QrA - 1155.56650

' Eoh Eo

2Mhi2 188.o22M, t.454.162.ur: -EJo = E, r -E-Substituting these values into Eq. 1 gives

M' + r1.1563Q = -10,150.55 (2)

The second compatibility equation is obtained from

rotation in shell = rotation in head

or

0N*+ 0M" - 0ol.*.u : -Iun- Lal^rc"a

where for the shell

(l)

8.2 EVATUATION OF DISCONTINUIW 5TRE55IS

^ N,e 83,991aw"= D,= h^ r0.86M,ar, =

En

217

- 47.3822QEo

and for the head

^ 4A3Mh 6l.t86lM, . 473,213

E6t1r6 Es Es

^ zA,Q 188.022QO^= -' Eoh E,)

Substituting these values into Eq. 3 gives

M"+ 1'.9521Q= -7733.99

Solving Eqs. 2 and 4 gives

Q = -262'55 lb/in'

M' : -7221.47 in.-lblin.

Mr: 512.53 in.lb/in.

totaly = l$y

The actual forces are shown in Fig. 8.9.

o2p'D

0s:

(4)

(3)

Fisurs 8.9

2ll DlltoN of cYuNDRtcAt SHttts

noop sress ar point C =!3y:35,000psirs

axial stress at point C : ff : ZSOO psi

axial bending stress at point C : Y = 720 pri

circumferential bending stress at point C = 220 psi

These stresses are divided into two categories in accordance with Table g.2:

l. Local membrane stress (Pr)

o6 = 35,000 psi

o1 : 7500 psi

a, = -500 psi

maximum stress difference : 35,500 psi

From Tables 8.1 and 8.2 the maximum allowable local membrane stressis equal to

1.5S, = 22,599 < 35,500 psi overstressed

2. I-acal membrane plus secondary stress (pr + O)

a6 = 35,000 + 220 = 35,220 psi

ot = 7500 + 720 = 8220 psi

o;: -500+0 : -500 psi

maximum stress difference = 35,720 psi

From Tables 8.1 and 8.2 the maximum allowable local membrane olussecondary stress is equal to

3S. = 45,000 psi > 35,720 OK I8.3 ASME PROCEDURE FOR EXTERNAL PRESSURE DESIGN

A comparison of Figs . 5 . 1 7 and 5 . I 8 indicates that the buckling due to pressureapplied to sides and ends is more critical than the pressure applied to the endsonly. Accordingly, the ASME code, VIII, uses Eq. 5.17 as the basis for design.This equation is modified to take into consideration nonelastic bucklins and

8.3 ASM€ PROCEDURE FOR EXTERNAL PRTSSURE DTSIGN

cxpresses the basic relationship as

219

,,: + = i*'(L)'Defining A = e""

(8.4)

Equation 8.4 is plotted as shown in Fig. 8.10. Hence, for any given value ofL,D", and, t, a value ofA can be determined from Fig. 8.10.

The allowable compressive stress in the elastic region can be determined fromthe equation

AEo-FS

where FS is factor of safety and E6 is modulus of elasticity. Or in terms ofDressure

,2AEg

(D,/'XFS)

In the plastic region, ASME uses quasi-stresFstrain curves similar to those inFig. 8.11 to determine plastic buckling. These curves are plotted on log-loggraphs with a factor of safety of two for stress. Because the stress-strain curvesdiffer for different temperatures, a number of curves for different temperaturesare plotted in Fig. 8.11. Hence, allowable stress is given by

(factor I from chart)(factor of safety of chart)o=T (8.s)

or

2"8o=FS

If allowable pressure is needed, then

oe x/ tY^: a= ,1;"1

^ zto-D. 4tB

FS (D,)

8lolgv:l

$ 33933 BssFF e ?3s 3

g;a.>

E.nJR

- 5:.c fqfiAtt55c;.:.'-a9

E.E

- s 5Fq 9 fl-

; f :('+Tet.:J I'= bti;+;bEEE9

4 -Fo'd g9aE 8.i; !a

tl

-8 -6 -6

:l V5- 11-'t

\ \ \\

\

39PP9

:* R F R R

221220

II

II

i

g

-6

Ej

E

E

E

€

'6

I

E

daoE

5g

!""I .."

j6

olsloN oF cYt"rNDRrcAL sHEtrs

P : 4

----!-@"/t)FS

ASME uses a factor of safety of 3.0 for buckling of cylindrical shells subjectedto lateral and end extemal pressures. Hence, for elastic region (D./t > l0),

2AEo

3(D,/t)(8.6)

and for elastic or plastic region (D./t ;- lO),

P = .t4B,t3(D./t)

where A = factor determined from Fig. 8.10 and is equal to e",

B : factor determined from Fig. 8.11

D, : outside diameter of cylinder

Eq = modulus of elasticity

P = allowable extemal pressure

r : thickness of cylinder

For D,/t values less than 10, ASME uses a variable factor of safety that rangesfrom 3.0 for values of D"/t = 10 to a factor of safety of 2.0 for values ofD"/t = 4-O. This reduction occurs because for very thick cylinders, bucklingceases to be a consideration and the allowable values in tension and comoressionare about the same . Hence, for D"f t < 10 the allowable value of P is tfie lowerof the quantities Pr and Pz given below.

For D./t < lO,

^: (T# - 00833)'

^ 2ol. I \"= DJr\'- oJ,)

(8.7)

where o = two times the allowable stress in tension or 0.9 times the yieldstress of the material, whichever is less.

Note also that for values of D"/t < 4, the following equation can be used:

8.3 ASME PROCEDURI FOR EXTERNAI" PRESSURE DTSIGN 223

(E.lr )l.l

The ASME procedure for the design of cylindrical shells under extemalpressure is complicated because of the various parameters that must be consid-ered. A summary ofthe procedure is shown in Fig. 8. 12 as an aid to the designer.

Example 8.4. The length of a cylindrical shell is 15 ft, outside diameter l0 ft,and is constructed of carbon steel with minimum yield strength of 36,000 psi.The shell is subjected to an extemal pressure of 10 psi. Find (a) the requiredthickness using ASME factor of safety and (b) the required thickness using afactor of safety of 2.0.

Solution. (a) Assume

t : 6 rn.

Then

4:zzo L=1.2st '-- D.

From Fig. 8.10, factor A = 0.00018. From Fig. 8.11, modulus of elasticity arroom temperature is 29,000,000 psi. Hence, from Eq. 8.6

p _ (2x0.009_l-8..1!Z? x 106) : r0.9 psi o.K.(3)(32u)

A check is needed to ascertain that buckling is in the elastic rather than the plasticregion. FromFig.8.ll with A = 0.0018, a value of B:2600 psi is obiainedin the elastic region of the curve. Hence, the above solution ;f l0.g DSi isadequate.

ur.r:]in.6

(b) For a factor of safety 2.0, assume

4=38at

\D./t)'

t = 0.3125 in.

I

Then

Cd loulot. Oo/t

l. Oo./t, l..r Lhar 4 ?

o = idtrr" Aanft.t L 4 oolculolar L lo

l. L./Do gF.ot.r. thdr 50 ?

l. L./Do 1... ttrar O.O5 "

Coloulot.! A ad u..

E rt,r- not.rlol. ofi.r.i, vtLh A

Ext rrd eD..oi |ot.l.nE r€tir-. I lfltE lzchlol ls crd..cd B

I. A lo r.lght off oa .c61. ?

t. A to l.it, otfroO .col. ?

Ia Oo./t ar .lO ?

O.l.rmln 2Sl |fl.|-.St l. ol lor. l. ll..tr... fFon 9.$..c1| on C

I. 2Sl, < O.96U ?

I. Pol < Poa ?

Figure 8.12 The ASME method for daerminins moximum ollowobte oxternot pr€ssure on cvtinders.

224

From Fig. 8.11,

8.3 ASME PROCEDURE FOR EXTERNAI PRESSURE DESION 223

I.rom F'ig. 8. 10, factor A : 0.00014 and from Eq. 8.6

p : zAEo ASME factor of safety' 3(D"/t) specified factor of safety

_ 2(0.000t_41_ll?9 x 106 x 3.0

= 10.6 psi O.K.

Returning to Fig. 8.1 1 with A = 0.00014, it is seen that the value from the chartindicates an elastic behavior.

ur",:jin. rlo

Example 8.5. A cylindrical shell with length 18 ft and an outside diameler 6ft is constructed of carbon steel with a yield stress of 38,000 psi. Determine thethickness needed to resist an external pressure of 300 psi.

Solution. I€t r = 1.25 in.

L: z.o 4: st.oD.t

From Fig. 8.10, factor A : 0.00095. From Eq. 8.6 with Eo :29 x 106 psi,

2x0.00095x29x106p3 x 57.6 = 320 psi

Now check for plastic region. From Fig. 8.11, factor B : 12,000 psi in theplastic region. Hence, the first of Eq. 8.6 cannot be used. From the second ofEq.8.6,

P-4x12'oon,: 3x57n = 278 psi inadequare

Try r: 1.375 in. Then

!=sr.o and A:o.oo11

226 Dl3tON Of CYUNDR|CAL SHEttS

lJ = 12,400 psi

dx It donP: '=^:" = 316 psi O.K.' 3 x 52.4

Use r = 1.375 in. IProblems

8.3 The thickness of a l3-ft diameter reactor is 5.50 in. and its effective lenethis l8 ft. Ifthe design temperature is 900'F, what is the maximum allowa6leexternal pressure?

Answer: P = 300 psi

A vessel has a 15-ft outside diameter and effective length of 6 ft. If it issubjected to 15 psi extemal pressure, what is the required thickness (to thenearest l/ 16 in.) if the design iemperature is 300'F?

Answer: t = 7/16in.

A jacketed pressure vessel with an intemal diameter of 12 ft is subject€dto an interni pressure of 400 psi and a jacket pressure of 200 psi . The shellthickness is controlled by the intemal pressure using an allowable tensilestress of 15,000 psi at 800T with an E factor of 1.0. Determine therequired stiffener spacing from Figs. 8.10 and 8.11.

Answer: L = 16.2in.

A distillation tower is subjec0ed to a vacuum of 15 psi. lf D" = 9 ft,t = 0.75 in., and stiffener spacing = 8 ft, what is the maximum permis-sible lemperature?

Answer: T : 900"F

8.4 DESIGN OF STIFFENING RINGS

In deriving Eq. 5.17 for the maximum strength of a cylindrical shell undercxternal pressure, it was assumed that the ends of the shell were simply sup-F)rted. For this to be true, stiffening rings, flanges, and so on (Fig. 8.13) arenccded as lines of supports. These supports are assurned to carry all the load thatthc shell carries due !o external pressure. By refering to Figs. 8. 13 and 8.14,Iotal force in stiffener is

8.6

E

9

E

'6

Ep

EE

i5 a,

,r&.oge;AF

I

i-FI

.JI

+I

-JI

+-l'

JI

+

;+-I

227

DESTON Ot CYltNDRtCAt SHfl.tS

s ti ffener

Figure 8.I.t

PD.L = 2F

PD"L

The stress in the shaded area of Fie. 8.14 is

Io- '- - PD.- L(t + A,/L) 2(t + A,/L)

In this equation it is assumed that the area A" of stiffening ring is ,,smeared', overthe total length z.

Using the terminology of Fig. 8.11 and expression 8.5,

28'FS

The expression for strain is

o

8,4 DESIGN TOR STIFFENING RINGS

12I

,:4P

': FS PP"2 2(t + A,/L)

With a factor of safety (FS) of 3.0 expression 8.5 becomes

-3PD"a:or,aatra

I 2116

U;

E l1ti Itf=-:- A Di,G + A,/L)L

The stiffening ring must also be checked against buckling. The classical expres-

sion for the buckling of a ring due to external pressure is

(8.8)

(8.e)

Es D')"A + A'/L)L

With e", = A, the expression for l becomes

. DzL\t + A,/L)AI: a

This equation can be used in conjunction with Eq. 8.8 and Fig. 8.10. In doingso, a hial A, is normally selected and I is calculated from Eq. 8.8. Using the

value ofB, which already incorporates a factor of safety of 3.0, a value ofA is

obtained from Fig. 8.11. With this A, the required moment of inertia is calcu-laled from expression 8.9

Because the stability of the stiffening ring is essential in calculating the shellstability, a higher factor of safety is used by ASME in the stiffening ringcalculations as compared with shell calculations. With a factor of safety of 3.5,exoression 8.9 becomes

. DZL(I + A"/L)At:-- A-(8.10a)

230 DIS|ON Ot CYUNDRTCAT SHEItS

The svuilable / obtuined liom Eq. tt. lOa must be lower than the availabletmoment of inertia of the stiffening ring. This inertia is calculated without

considering the contribution of the adjacent cylinder. If the composite 1of thering and the effective cylinder are considered, then a penalty of 28Zo is appliedto Eq. 8.10 and a new expession given by

,, _ DzLa + A"/L)' - lo.9

(8. r0b)

must be used. The actual f is calculated from the available rins and shell areasas shown in Fig. 8.15.

Example 8,6. A long cylindrical shell is constructed of carbon steel with ayield stress of 38,000 psi and a radius of 36 in. If the stiffeners are spaced at a4-ft interval, calculate the required shell thickness and the size of stiffening ringsfor an external pressure of l0 psi at 100"F.

Solution. l*t

r : 0.1857 in. = 0.67D-- = tt+t

From Fig. 8.10,

A = 0.00028

From Fig. 8.11,

A : 21000 psi

a

Fisure 8.15

.ing..By comparing expressions 8. 1la and 8. I lb, it can be concluded that in order

for the column aD to be as strong as the plate, its length must be about one-half

8,5 ATIOWABI.E GAPS IN STITFENING RINGS 23I

Hence,

4 /4ooo\r=;t ,* l: l0.4psiJ\--.,

Use a shell thickness of3/16 in. and try a 3 x 3 x 3/16 angle stiffening ring.

A, = 1.09 in.2 and I = 0.9622 in.a

r /t0x72)R = : ----------l-"

: 4 (u187 5 + l og/4a: z)/u PSr

From Fig. 8.11, A : 0.00018.

, _722 x 48(0.18?5 + 1.09/48 x0.00018)A

: 0.67 in. a OK

8,5 AITOWABLE GAPS IN STIFIENING RINGS

Gaps in stiffening rings are normally provided to allow for drainage of vesselcontents or permit piping and other internals to extend through the ring. Exam-ples of various gap arrangements are shown in Fig. 8.16. The maximum allow-able gap can be calculated by assuming the distance between points a and b ofFig. 8. 17 as a simply supported column of length l. The maximum buckling loadthat can be applied to this column is given by

^ lf'I1olI'

The strength of the column must be equal to or greater than the cylindrical shell.For large diameter shell the curvature is small and the buckling streneth of theshell approaches that of a simply supponed flat plate. The riinimuit criticalbuckling strength of the simply supported plate abcd (Fig. 8.17) loaded in thecircumferential direction is given by

_ 4i2Enl 4T2EJI'c, =;r=- _ = ----=-

t-tr - lt-t t'

t

(8.1 la)

(8.1 lb)

Grp lnor to.tc..d I rtm.rth. hrckn.r. ot rh. $.lt ptlrd

Thii.&iion rha h64 momsnt of i..ni!r€qutr€d tor.i^q unl.r r€qun€m€ns ot

L.ngth of.ny g.p in u6eppo.r.d .lt nor o.r..sd l.ngrh ol .c ih*. in Fig. UG.29_2

Tht larion lhatt h!v. mfr@r oti...ri. r.qui4d to. nng.

Figur' 8 1 6 voriou orrong€rnenrs oI rrittuning ring. for cyrindri.or vessers 3ubiecd ro exrernor preasure(Ref. 2).

S]

_l- Typ. ot onttrucrio. _+

1xa- ̂';

232

8,5 ATTOWABTE GAPS IN STIFFENING RINGS 233

Figure 8.17

that of the plate. Therefore, the maximum gap length must be one-half that ofa buckling lobe length of a cylindrical shell. In referring to Fig. 5.16 themaximum gap length can be expressed as

-_l _ ttD.g_;_-;; (8. l2)

ASME has developed curves that are based on Eq. 8.12. These curves areshown in Fig. 8.18. A comparison, however, between Eq. 8.12 and Fig. 8.18indicales some differences. These differences are due to the fact that Fig. 5.18,which is used with Eq. 8.12, is plotted using the first two terms of expressionfl in Eq. 5.27a whereas Fig. 8.18 uses all terms. Therefore, the results of Fig.8.18 are more accurate.

Example 8.7. What is the maximum gap allowed in a stiffening ring of a shellwith D. = 7.O ft, L : 15.0 ft, and r : 1.0 in.?

Solution

'-: t+t

L = a.zsr

-d!

.s

E3$-E

;as!5=

bet-5;IJ=_-@io5

;

E

8I 889 S R

"":#ii3,6+Id=

"+r3

*"""p;'L;

d;:; ";:

g gfig g F

rTog'reu1lq1 - neurrg epcrng

234

and

for cylinders

lrronr F'ig. 5.lll, N

G = 16.8 in.

8.6 OUT.OF-ROUNDNESS OF CYLINDRICAL SHELLS UNDEREXTERNAL PRESSURC

In the fabrication of cylindrical shells, slight out-of-roundness invariably results.This is due to forming, welding, or postweld heat-treating operations. Normally,intemal pressures tend to minimize out-of-roundness, whereas extemal pres-sures tend to increase it. Because of that, and to prevent failure, extra pre-cautions must be taken in fabricating shells that are subjected to external pres-sures.

In Fig. 5. 16 it is assumed that the shell is approximated by a series of columnsconnected end to end. The length of each column is one-half a lobe length, or

, = nDo-2N

The slendemess of each column is expressed by the ratio //r where r is the radiusof gyration. Since r is eqtal to t2/\/12 in a shell wall. the equation becomes

t _ 5.44r N(t/D.)

The eccentricity of each column in expressed by e. If it is assumed that theeccentricity ratio e/r affects the strength of a column in the same way as theshell, it can be concluded that

for columns

8.6 OUI-OF.ROUNDNESS OF CYTINDRICAT. 5HIItS

= 4 and fiom Eq. tl.12,

G ::+ : o.2oD^(4X4)

T

elrr

e 5.44------ca-t/y12 N(t/D.)

DESION OF CYI.INDRICAI. SHELLS

!: c't N(t/D)

(8. r 3)

(8.14)

where C1 : 1.57.

Experiments have shown that for constant tfD, ratios, the value of eftincreases with an increase in Lf D.ratios. This, however, applies only in the caseof intermediate length shells. For this range, Eq. 8.13 was found to give ade-quate results. As the length gets longer, tests have shown than an increase inL/D"has no influence on ef t. On the other hand, tests have shown that as thelength decreases, the value of e/l increases slightly. Thus, Eq. 8.13 needsmodification to take into account the two extreme cases. Thus. disreeardins theincrease in e/t as L/D. decreases. an empirical equation of the fo;

9: Ct ..'- . nt N(Jf,j- "'"is found satisfactory. In using the values of C1 and C2 as obtained from tests, theequation becomes

C : o, o,'=8 + o.ol5Nt N(t / D")

Figure 8.19 is a plot of this equation.

Example 8.8. Calculate the maximum out-of-roundness allowed in a cylindersubjected to extemal pressure with D, : 5.0 ft, Z - 14.0 ft, and t : 0-75 in-Compare the result with that obtained from Fig. 8.19.

5.60 D- = totR

Solution

237

@EoB

q QE

. E=q 66o €o

ct €.0P6

dF

E

'i3{

,9d

II uI I

V I/

tl / I

( / /

//,1

"1/t 4

ls

ilI tt-*:1If

/ I I/ I

/ / I I I I

/

I

3.ssps F s 8 t I tt3R8rTog rcurralql; lraueqq epnrng

2il D|lrol{ o; cYuNDilcAr tH[t3

From Fig, 5.18, N = 3. From Eq. 8.14,

f nnrr -l

"=Lffi)+o'or5(3)J(o7s)e : 0.39 in.

From Fig. 8.18,

e : O.53t

e = 0.40 in.

8.7 DESIGN FOR AXIAT COMPRESSION

For axial compression, Eq. 5.29 rnay be written as

T

.".=*Lo

or

A:lt :99Eot RJt

The allowable strain can be expressed as

FS(R,/r)

A large factor of safety is normally used in this expression because a slightout-of-roundness can significantly reduce the critical sfain. ASME uses anapproximate factor of safety of 10.0. Hence, the expression for allowable com-pressive stress in the ASME is for elastic range,

NOMINCIATURI 239

(8.1s)

For plastic range,

ot= B

where B = is obtained from Fig. 8.ll using A = 0.125/(R,/t).

Example 8.9. A cylindrical tower is constructed of stainless steel 410 mate-rial. Its radius is 6 ft and thickness is 0.5 in. Determine the maximum allowablecompressive stress at room temperature.

Solution

From Fig. 8. 11,

Eo = 29 x 106 psi

For elastic region,

0.062510o,: - n/t

o,=fifo.oazs x 29 x 106)

: 12,600 psi

A =W: o.ooo87

&=3=roo, u.)

For plastic region,

and

A = Il,fiX) psi <- use

NOMENCTATURE

A = shain obtained ftom extemal pressure charts

A" = area of stiffening ring

I

210 DtStON Oi CyUNORtCAt SHH"|"S

I = stress magnitude in a cylinder due to external pressure

D, = outside diameter of cylinder

6 = joint efficiency

E6 : modulus of elasticity

F : peak stress as denned in Table 8.2

/ = moment of inertia of stiffening ring

1' = combined moment of inertia of stiffening ring and adjacent shell

L = effective length of shell

P = intemal or extemal pressure

P6 = primary bending stress as defined in Table 8.2

P1 = pnmary local membrane stress as defined in Table 8.2

P- : primary general membrane stress as defined in Table 8.2

Q : secondary membrane plus bending stress as defined in Table 8.2

R = inside radius of cylinder

R, = outside radius of cylinder

S = allowable tensile shess in the ASME Code, VI[-1

& = alternating stress in the ASME Code, VIII-2

S, : allowable tensile stress in the ASME Code, Vm-2

t : thickness of cylindrical shell

REFERCNCES

ASME Boiler and Pressue Vessel Code, Section Vm, Division 2, Alrs rnative Rules-pressureVessels, ANSVASME BPV-Vn-2, American Society of Mechanical Engineers, New york,1980.

ASME Boiler and Pressure Vessel Code, Section VItr, Division l, prcssure Vesrels,ANSUASME BPV-Vm-I, American Society of Meahanical EDgineers, New york, 1980.

EIBTIOGRAPHY 241

BIBTIOGRAPHY

Windcnburg, D, F., "Vessels under Extemal Pressure," in Pressure Vessel and Piping Desiqtl:

Collected Papers 1927-1959, American Society of Mechanical Engineers, New York, 1960.

tk)lt, M., "A Procedule for Determining the Allowable Out-of-Roundness for Vessels under

External Pressure," in Pressure Vessel and Piping Desiqfi: Collected PaPers 1927-1959,

American Society of Mechanical Engineering, New York, 1960.

Inside surfoce of o vessel hecd. (Courtosy of the Nooler Corp.. St. to'rir, Mo.)

243

CHAPTER

DESIGN OF FORMED HEADSAND TRANSITION SECTIONS

9.I INTRODUCTION

A large variety of end closures and transition sections are available to the desisnengineer. Using one configuration versus another depends on manv factors suihas method of forrning, material cost, and space ."ihi"tionr. Some frequentlyused heads are:

Flanged Heads.

These heads (Fig. 9.1c) are normally found in vessels operating at lowpressures such as gasoline tanks, and boilers. They are also used-in high-pressure applications where the diameter is small. Various details for theirdesign and construction are given by the ASME Code, VI[-l.Hemispheical Heads.

Generally, the required thickness of hemispherical heads due to a grventemperature and pressure is one-half that of cylindrical shells with equivalentdiameter and material. Hemiheads (Fig. 9.10) are very economical whenconstrucied of expensive alloys such as nickel and titanium----either solid orclad. In carbon steel, hemiheads are not as economical as flaneed and dishedheads because of the high cost of fabricalion. Hemiheads are iormally fabri-cated from segmental "gore" sections or by spinning or pressing. Segmentalgore hemiheads are economical in thin, large diameter equipment or thick,small diameter reactors. Because hemispherical heads are thinner than cylin_drical shells to which they are attached, the transition area between the headand shell must be contoured so as to minimize the effect of discontinuitystress. Figure 9.2 illustrates the hansition requirements in the ASME Code,vm.

Elliptical and Torispherical (Flanged and Dished) Heads.

These heads are very popular in pressure vessels (Fig. 9.lc and d). Theirthickness is usually the same as the cylinder to which they are attached. Thisreduces considerably the weld build-up shown in Fig. 9.2. Thus, because therequired thickness in areas away from the knuckle region is less than thefurnished thickness, the excess can be advantageously used in reinforcingnozzles in these areas. Many mills can fumish such heads in various di_

2U DISION OT IORMED HTADS AND TRANSITION SECTIONS

amelers and thicknesses that are competitive in price.In a true elliptical head the radii of curvature vary between a jacent points

along a meridian. To simplify the calculations and fabrication, the ASMECode established the following various approximations. A 2: 1 ellipticalhead can be assumed to consist of a spherically dished head with a radius of9OVo and a knuckle radius of 17go of the shell diameter to which thev areattached, as shown in Fig. 9.3. The smallest knuckle radius allowed for aflanged and dished head is 67o of the shell diameter and a spherical radius of1007o of the shell diameter.

9.I INTRODUCTION 245

--r-.].'

_1 ,

(TORISPHERIcAL)

( g) IlISCEL!ANE0Us

Figure 9.1 commonly usad lormed closurc heods.

Conical and Toriconical Heads.

These heads shown in Fig. 9.1e and / are used in hoppers and towers as

bottom end closures or as transition sections between cylinders with differentdiameters. The cone-to-cylinder junction must be considered as part of thecone design due to the high unbalanced forces at the junction. Because of

_T

(a) FLANGED(b) HEI.lISPHERICAL

(c) ELLIPTICAL (d) FLAIIGED & DISHED

(e) C0NICAL (f) T0Rl c0lt IcAL

II

lilc

l- '".ca<1l2las-.hl

rb

F

taper./, may inctudethe width ol the weld-

{cl (dl

Fisure 9.2 H€od-to{hell iunction. (Courrety of the Americon Societ}' of ldechonicol Ensi'ieers)

{a) ASiIE 2:1 Head

I

<112k,-.hl

T246

(b) ASI.IE Flanged and 0ished Head

fisure 9.3

where

(9.2a)

9,2 ASME EQUATIONS fOR HEMISPHERICAI HIAD DISIGN 247

thcse high fbrces, the ASME Code, Vlll-1, limits the apex anglc to a max-

imum of 30" when the cone is subjected to intemal pressure. Above 30'adiscontinuity analysis is done or a toriconical head used to avoid the un-

balanced forces at the junction.

Miscellaneous Heads.

Many chemical processes require unusual vessel configurations. The heads ofsuch vessels can have an infinite number of contours. One such contour is

shown in Fig. 9. lg. The design of these heads is very complicated and there

are no simple methods of analysis. Experience, proof testing, and sophisti-

cated analyses are generally used to determine required thicknesses.

9.2 ASME EQUATIONS FOR HEMISPHERICAT HEAD DESIGN

The ASME Code, VI[-l, has combined Eqs. 6.1 and 6.7 for internal pressure

into one simplified equation:

PR (e.1)2SE - O.zP

where t: required thickness

P = intemal pressure

.lR = inside radius

S : allowable stress

B = joint efficiency

This equation with E : 1.0 is plotted in Fig. 6.3 and it approximates the more

complicated Eqs. 6.7 over a large range of r'fri- Othet forms of Eq. 9. 1 are

shown in Appendix I.For external pressure, Eq. 6.35 is taken as the basis for the ASME Code

equations. Defining e". : A, r : R,, and modulus of elasticity as Es, Eq. 6.35

can be written as

0.t25Es R"/t

. 0.125" R"/t

A = critical shain

R, = outside radius

t : thickness

214 DISION Of FORMED HTADS AND TRANSITION STCTIONS

The critical strcss in a sphcrical section is given by

P",R.a*: Zt

or

P., = 2% =4+R"/t R./t

and the allowable pressure is expressed as

zAEo(e.4)

FS(R,/')

Using factor of safety (FS) : 4.0 and substituting Eq. 9.2a into 9.4 gives

p _ o.0625Eo- (R./ t)'

where P = allowable extemal pressure

Eq = modulus of elasticity

The ASME procedure for determining the allowable external pressure for aspherical section is to detemine fust the A value from Eq. 9 .2a. The allowablepressure can then be obtained by referring to a stress-strain chart similar to theone shown in Fig. 8.11. IfA falls in the elastic region, then P is calculated from84.9.2b. If ,4 falls in the plastic region, however, a value ofB is determinedfust from the chart. The allowable pressure is then calculated from exoression9.3 as:

(e.3)

(e.2b)

(e.5)

(9.2c)

P" = 20-

FS(&/r)

Substituting B : o",/2 inl.o Eq. 9.5 gives

p= B- R./t

where B = factor determined from Fig. 8.11.

Equations 9.2 form the ASME basis for determining allowable external pres-sure for sDherical sections.

9,3 ASME DESION EQUATIONS 249

lixamplc 9.1. Using the ASME criteria, determine the allowable cxternrlprcssure on a spherical shell with R, = 60 in. and t : 0.5 in. Use the 300'F linein Fig.8.ll.

Solution, From Eq. 9.2a,

n l rsA:ffi=0.00r

From Fig. 8. I 1, with A : 0.001, a plastic behavior exists and Eq. 9.2c must be

used. Hence, from Fig. 8.11

B : 11,000 psi

and

":*@ = nt o" I

9.3 ASME DESIGN EQUATIONS FOR ELTIPSOIDAI AND FLANGEDAND DISHED HEADS

The general solution of Eq. 6.39 is very cumbersome because 12 is a variablefunction. However, the stresses obtained from Eq. 6.39 are important becausethey can be added to the membrane stresses of Eq. 6.37 which results insignificantly reducing the total stress at the vicinity of the junction. Therefore,this equation can be advantageously utilized by the designer in reducing therequired head thickness. The ASME used this fact in developing design param-eters for ellipsoidal and torispherical heads.

A study was mader with ellipsoidal heads to determine the effect of the ratioa/b on the stress level at the head-to-shell junction for a constant ratio 32 of head

thickness t to shell radius r. The study indicated that the point of maximum stressin the head changes with a change of afb.Forheads shallower than a/b of 2.5,the maximum stress is in the hoop direction at the outside surface of the knuckleregion and is in compression as shown in Fig. 9.4. For ratios of a fb between 2.5and 1.2, the maximum stress occurs at the junction and is a hoop tensile stress.

The stiess magnitude for various ratios ofa/b is shown in Fig. 9.4. A simplifiedequation used by the ASME Code, VIII-I, approximates the theoretical stressratios of Fig. 9.4 for values of af bbetween2.6, which is the maximum allowedby the code, and 1.0 for a spherical head:

- : il'. (-tl] (e.6)

;t$

DISION Oi TORMID HEADS AND TRANSIIION SICTIONS

at6

Fisur€ 9.4 l,{oximum ske5s in ElliFoidol Heo&. (Ref. t, p. l3,t).

A plot of this equation is also shown in Fig. 9.4. The ASME Code uses the Kvalues given by Eq. 9.6 to determine the required stress ratios needed in obtain_ing thickness of ellipsoidal heads. This is accomplished by multiplying thecalculated thickness of a cylindrical shell with diameter D by'K valuis. Hince,for ellipsoidal heads,

(e.7)zSE - 0.2P

where t : thickness of ellipsoidal head

P = internal pressure

D = diameter of shell to which head is attached

l< = sfiess intensity factor obtained from Eq. 9.6

S : allowable stress

6 = joint efficiency

For torispherical heads, tests conducted by H<ihnr and others have shown thatthe stress at the knuckle area due to internal pressure reaches the yield value longbefore the spherical region does. Hrjhn plotted an empirical equation that cor_relates well with available test data, as shown in Fig. 9.5. To ivaluate Hcihn'sempirical curve, Fig. 9.5 shows anotler curve that indicates the stress rn an

___-_ Lh.or.t, loo I vo lLr..tmot. votu.. etvon bU Eq.{9.O1

lle hooD etr€e6 of

I le nerldlonol EtroBs otde eual'oc€ of knucklo o-eo

I

1 rffz,tf;r] i

e

d

-6

g

lrcdoqo-

d9;i"P:.{6iE5

,^=

6odoq-:doact

ci

3*3533ss€J1s ur\oJc 01 sseJ,ls ur^rr r xour Jo o|1ou

o x-o*r

-E, tt.o.- o

---.'ia *,- tJrxo0-oxl(

-ti 6to ", o.l

IEL_4tl-!i-;;-F;-3

q-

:sEiE,6

l0^6il-\.

-:Ztlli-l

C.

(-lI't{

-_!

-jlI

l'o -l.!tot9.aIO-

.lLi:L -)o

-l

li't^.,lJ]a)tolts

roo (.l-

'l_.o-o

t.

ttl

//

rt

251

equivBlent ellipsoidal head whose thickness is equal to the shell thickness. Thiscurve indicates that Hrihn's curve is liberal for small values of knuckle to crownradri r/L. Accordingly, the ASME Code, VIII-I, developed an empirical curvethat parallels both Hdhn's curve for large values of rfL and the ellipsoid curvefor small r/Z ratios, as shown in Fig. 9.5. The ASME curve can be expressedby the equation

DIsION O? TORMID HTADS AND IRANSITION SICTIONS

Thus the ASME equation for the torispherical heads is given by

PLMzSE - O.zP

where r : thickness of torispherical head

P = internal pressure

Z = spherical crown radius

M = stress intensity factor obtained from Eq. 9.9

S = allowable stress

g = joint efficiency

r = knuckle radius

n =)(z + (e.8)

(9.e)

In practical applications researchers noticed that Eq. 9.9 gave conservafiveresults for the majority of head designs but became unconsirvative for lareeratios of r/t. This ratio was not considered by the ASME in its derivation of E-q.9.8 because Fig. 9.5 was based on a constant value of r/r. Accordingly, researchwas conducted to evaluate the buckling behavior of the knuckle region for headswith large ratios of //t. The plastic analysis concept was used arid shown2 thatthefollowing equation can adequately predict the behavior of torispherical headswith large rf t rutios:

o"/FS = (0, *,,;);.,'(, -,,;)(i) - o 0006

Sglving. for _r

and fetting or/FS : S, the following approximate equarionobtained and used by the ASME Code, VIII-2:

nl= -t.zent - 4.ss246(; + 28.e3318G)

9.3 ASMI DISIGN IOUATIONS 253

(9.10)+ fo.oozw

- z.z4,,os!) + rs.oszr(r)] r" I| /r\+ 10.26879 x 10 a - 0.442621;l +L \U/

where t : required torispherical thickness

-L : spherical crown radius

L ..r

' "'-,(;i]l('";)]'

r = knuckle radius

D = diameter of shell to which head is attached

P : intemal pressure

S : allowable stress

A plot of Eq. 9.10 is shown in Fig. 9.6.

Example 9,2. What is the required thickness of a 3 : t head ifD = 144 in.,P= 100 psi, S: 17,500 psi, E: 1.0?

Solutian. From Eq. 9.6, with a/b = 3,

K: 0.166712 + (3.0f1

= 1.83

From Eq. 9.7,

(100x144)(1.83)

2x 17,500 x 1.0 - 0.2 x 100

= 0.75 in.

Example 9.3, What is the required thickness of a flanged and dished head ifD=24O in., L=2AO in., r: 15 in., S:17.500 psi, E= 1.0, andP : 50 psi?

Solution.

I

.17 l2: I Etl,plo'dar h€.d)

,005

Thickne.s of Formed

.01 .02 .03 .04 _05

Heods (Courre.y of lhe Ane.icon Sociery of /vtechonicolFigure 9.6 RequiredEnsineers.)

254

9.3 ASMT DESIGN EOUATIONS

l;rrrn Eq.9.tl.

l-M=zG+vr6): 1.75

From Eq. 9.9,

50x240x1.75t:-'-2xl7joox1.o-02x50: 0.60 in.

Because the thickness is obtained from Eq. 9.9 and because this thickness issmall compared with the diameter of the head, the requirements of Eq. 9. l0 mustbe checked.

r15D 240

P;J

: 0.0625

50

17,500

= 0.002857

!I

and from Eq. 9.10,

/r\ln l;l : -s.54851

\L/

;= o.oo38e

t = 0.93 in.

Hence, for this head a minimum thickness of 0.93 is to be used. I

9.3.1 Ellipsoidol ond Torisphericol Heods under Exlernol Pressure

For extemal pressure, the knuckle area is subjected to a tensile stress. Hence the

critical area that is necessary for consideration under extemal pressure is the

spherical region. Thus the ASME criteria for all ellipsoidal and torisphericalheads under extemal pressure are the same as those for spherical heads.

116 DIIIO}I O' IOTMID H!ADs AND TRANSITION SGCTIONS

9.4 ASME EQUATIONS FOR CONICAL HEAD DESIGN

From Example 6.7 it is seen that the hoop force Np is twice as large as thelongitudinal force N5 in conical heads subjected to internal pr"rr*". Th" ASMECode, VI[-I, uses Eq. 1 of Example 6.7 as the basis for establishing therequired thickness of a conical section subjected to intemal pressure. The Jqua_tion is given by

PD2 cos a(Str' _ 0.6p)

where P = internal pressure

(e.l l)

D = inside diameter of cone at the point of consideration measuredperpendicular to the longitudinal axis

,S = allowable stress

g = joint efficiency

a = one-half the included apex angle of the cone at the center line ofthe cone

"I:TOr" 9.4.

. What is the required thickness of a conical head attached to a

cyllnder whose inside diameter is 40.0 in. if the intemal pressure is 450 psi, theallowable stress is 20,000 psi, E = 1.0, and a = 20.0?

Solution. From Eq. 9.11,

(2X0.940X20,000 x 1.0 - 0.6 x 450)

= 0.49 in. I9.4. I ASME Simplificotion of Discontinuity Anolysis due fo Iniernolrressur€

The ASME Code, VtrI-1, uses the stress expressions obtained in Example 6.9f9r inlernal pressu€ as the basis for establisiing simplified l.ite.iu fo, dir"on_tinuity analysis at the cylinder-to-cone junction. At ihe large end of the cone(Fig,. 9.7) the discontinuity analysis results in the following i*o.*piessions fo,me longlrudlnal and hoop stresses in the shell:

,=!(o'+x{rz\

""=7(, - rrn

"\

T]

I'I

_t_

9.4 ASME EQUATIONS FOR CONICAT HEAD DESIGN 257

I--+. l+I

2raon6

+il rvl_t

Fig'rr€ 9.7 Discontinuiiy iorcas dus ro internol preslur€.

where, in this chapter, a is used rather than q' and X and Y are given by

X = 4.559U2 tan d

Y = 1.316(V - 2V) tan a

and V and V2 are given as in Example 6.9.The longitudinal stress in Eq. 9.12 is in tension for all values of a and does

not govem the design criteria. The f term in the circumferential stress expressionin Eq. 9.12 is positive for all practical applications. Hence the quantity a" variesfrom a maximum tensile value of Prft to a compressive value that depends onthe angle a. The ASME Code, VIII-I, limits the maximum compressive circum-ferential stress to a value of Pr/r. Using these criteria, the quintity yl|rt;nEq. 9.12 must be limited to a value of 2.0. Values in excess of 2.0 must besupported by a ring added at the junction. The area of the ring is given by theequatron

I\l I-nT,l?

___L

tr2 tan q (Y\EIt - 2)(e.13)2 Y\6rt

A plot of the quantity y shows that it can be approximated by the expression0.005a as shown in Fig. 9.8. Substituting this value into Eq. 9.13 gives

The ASME Code, VI[-l,1.5SE and thus

(9.14)t2tanall, ['

400 \(9.12)

limits the comnressive circumferential stress to

258 DTSION OF fORMTO I{TADS AND TRANSITION SECTIONS

t.o

oS

o.e

o.2

/X=O.C t2 6- \-7

'7Li =,r.5591I

'z Lan o\

4 --2 <=-l---<ly,o.brri!Y= 1.3t6<,V | -zvz tLonc,.

tO' 30' 40. SO.

o<

Figure 9.8 X ond Y volues for externot prossurc.

l.sSE : At

E [ssz,V7= ! P

Equation 9.14 thus becomes

Prl /^,=G('_

or for the large end of cone,

12 = radius of cylinder at large end of cone

J = allowable stress

t : joint efficiency

A = 326.6\/FEE

a = one-half the apex angle of cone

326.6\/F7SE\-- " 1t^"

. P,1 /. A\o':2sE\'-i) t"

where A, = required area of ring

P = intemal pressure

(9.15)

9,4 ASMT IQUATIONS FOR CONICAL HIAD DISIGN 259

liquation 9. l5 is used by thc ASMB Cirde , Vl ll- I , as thc bitsis litr chcckinglllc.ioint bctween the cylinder and the large end of cone due to internal prcssuro.

At thc small end ofthe cone (Fig. 9.7) the circumferential and longitudinal stress

cquations due to internal pressure are

Because both expressions include a negative term, the equation for or controlsbecause X is numerically larger than Z Limiting the maximum compressivestress to Pr/t, the term in parentheses in the equation for o1 is

< 1.5

and a stiffening ring is needed for values of XYr/t greater than 1.5. Therequired area of the stiffening ring is

Prl ran q l- t.s 'll_=-=-tl _.:l

zSE I xt/r,/tl

From Fig. 9.8, it is shown that the quantity X is approximated by 0.012a. Hencethe required area can be expressed as

. Prl tan af , 125 'l

2sE L' d,rtl

,: !(o '-',4)""=?('- ",4)

^': #(' - *) "" "

where A = *G

(9.16)

(9.r7)

By assuming that the maximum allowable longitudinal stress is limited to ,St,the expression

Pr.SE =;

can be substituted into Eq. 9.17 to give for the small end of cone,

r, = radius of cvlinder at the small end of cone

(9.18)

-*"--"W*-ffiUfY6ffiU'iii,aibt iio mmnror uapng

Exanple 9.5. Doslgn thc conc ehown in Fig. 9.9 and check the cone-to-shcllI junctions. L€t S = 20 ksi, E = 1.0, and p = 150 psi.

Solutian. From Eq. 9.11,

For small end

For the large end,

A = 326.6\m =28.28"

and from Eq. 9.15 the required area at the large end is

^=ffiffi(r-ff)<os,t= 0.64 in.2

A=89 = 7.71

100)

Fisurc 9.9 figur. 9.10 DLconrinulty for.e. dua lo cx|lrnol pr€!!ure,

9,.1 ASrvU IOUAT|ONS fOR CONICAI HIAD DlllON 261

and the required area

,=ffi(t-1ff)<o.sttt= 3.70 in.2

9.4.2 Conicol Shells under Exlernol Pressure

The goveming equation for the design of cones subjected to extemal pressure is

obtained from @. 6.43. Using a factor of safety 3.0, Eq. 6.43 becomes

P. 0.87(t,/D)25 (9.19)Es. L"/D,

This equation that expresses the cone in t€mrs of an equivalent cylinder ofthickness r" and length tr, is analogous to 84. 6.42 for cylindrical shells. Thus

the ASME Code, VI[-l, applies the same equations for the design of cylin&icalshells under extemal pressure for the design of cones with applicable values oft" urd L":

9.4.3 ASME Simplificotion of Discontinuity Anolysis due to ExternolPresgure

The discontinuity forces due to external lnessurea at the large end of cone are

shown in Fig. 9.10 and expressed as

I

{irtt*-_:I_

262 DI3ION OI IORMTD HIADS AND NANSTTION STCTIONS

u,__p_!t*r{^r,I I'''

'--&1'-"2)r \ t/where X : 9.34 Vz tan tt

Y = 2.57 (Vy - 2V2) tan aN, : axial compressive force = Pr2f2t + g

Q : axial force due to wind, dead load, etc.

lhe v-alues of X and I in Eq. 9.20 can be approximated by the expressions inFig. 9.11. A comparison of o; and or given by Eq. 9.20 indicates that themaximum compressive stress is given by ar; a conservative maximum allowablecompressive stress value is Pr2/f and thus

o.o27a E Pr''V;=N*-l

A stiffening ring is necessary if this quantity is exceeded. The required area ofthe ring is given by

_ (Prz/N,) - 1fO .027 a\/ r2/ t l

(e.20)

A,: (N,tanr(;;)lt

oEJla

x

z.ot,et.61.1t.2t.oo.8o.6o.+o.2o

.7x=bs4vzrLno< -./.

X:O.O27d rrttt'

'.t'

Y =o. to64 \- f_le. SZ.y - zvr,'Lonq^to. 20.

o(Figure 9.ll X ond Y voluer for externol pressure.

o'

pressure load at small end stiffening n"t: O!&i#

cos a

9.4 ASMI EQUATIONS TOR CONICAT HEAD DESIGN 263

lly limiting the axial stress to an allowable value of SE, the above equation canbc written for large end of the cone,

o'=*'#""1'-i(|*2.4):] (9.2r)

(9.22)

where A: 104

At the small end of the cone the stress is eiven bv

and the maximum compressive stress by 4. Limiting the allowable compressivestrcss to -(P/r/r), the first of Eq. 9.22 reduces to

(9.23)

_r,,_r,ill.r.- _P,,^t

or N, s 0, which indicates that at the small end of the cone the axial force N,must be resisted by a ring with an area for the small end of the cone

o _ N,rr tzn a" .sE

- Pr. "l/2o" = 1- r;N,t I'' -

-N-/ /^r:;l,t * "v;/

In addition to providing the required arca at a cone-to-shell junction, it isnecessary to design the ring at the junction to prevent buckling due to externalpressure. The procedure is similar to that for the design of stiffening rings incylindrical shells. A conservative approach used by the ASME Code, Vltr-l, indesigning cylindrical shells under extemal pressure assumes that intermediatestiffening rings support all the load applied to the shell. Using the same criteria,the load on the cone in Fig. 9.10 due to extemal pressure can be proportionedat the large and small end stiffening rings as follows:

pressure load at large end stiffening ring : ZnP (r2 I r)(r2 - r)3sina

PSE

2U DTSION Of TORMID HIAOS AND TRANSITION SECTIONS

'Ii)t l l()td itt lirrgc cnd duc ttl axial conrprcssitln, prcssurc on conc, an<I pressureon cylinder is

, = +tan q +, tun o + !. eliffi*or

where

t,_r2tana, Lt ri- rl

Total load at small end due to axial compression, pressure on cone, and pressureon cylinder is

^ Prt PL, 14 - r?/"= ^ tan c + -=- + =-P -r f, t?trr ctz I br2 t0ll d

F = P(M) +, tan 0

F = P(N) + f2 tar, d.

(9.24)

(9.2s)

where

rv=?tun o+L: +:3 lrlL Z O/2 l&Il (t

Equations 9.24 and,9.25 establish the maximum applied force at the cone_to_cylinder junction. The critical buckling stress of a iircular ring is

3Eolo" :

nr,

where A, is the total effective area given by

A' = + + Lj + e, for large end of cone

A' = + + Lj + e, for small end of cone

NOMENCLATURI

Ily using rr1 = E,A. thc above equation reduces to

Ar2 A,(e.26)

which is the required moment of inertia of a cone-to-shell stiffening ring.For design purposes, the value of A in Eq. 9.26 is obtained from a

stress-strain relationship. This is achieved by considering the stress in the ringas

Fr(9.27)

Applying a factor of safety of two in the foregoing equation and using theExtemal Pressure Charts in ASME (hat have a factor safety two), a designcriteria can be established as follows:

1. Calculate f. from Eqs. 9.24 or 9.25.2. Establish

3. Enter the Extemal Pressure Charts with a (factor B) and calculate thestrain A.

4. Use Eq. 9.26 to establish the minimum required moment of inertia. TheASME allows a 307o increase in value if the composite ring-shell mo-ment of inertia is considered and the equation then becomes

ADZA,

15.6

NOMENCTATURE

A : strain as obtained from Fig. 8.10

At : area of stiffening ring at cone-to-shell junction

,B = stress factor as obtained from Fig. 8.11

D : diameter

D1 : base diameter at small end of cone

26 OISION O' IORMID HTAO3 AND TRANSITION SICTIONS

l)r - base diumeter at large end of cone

6 = joint efficiency

8s = modulus of elasticity

, = head depth

I = moment of inertia

lK : factor for ellipsoidal heads as determined from Eq. 9.6

Z = spherical crown radius of flanged and dished heads

Z' = effective length of cylindrical shell

L" : effective length of conical section

= 70 + Dt/D:.l

M = factor for flanged and dished heads as obtained from Eq. 9.g

P = pressure

P" = allowable extemal pressure

It = inside radius

R, = outside radius

r = knuckle radius

rr : base radius at small end of cone

/2 = base radius at large end of cone

S = allowable stress

r = thickness

t" = effective thickness of conical section: tcos d

a = one-half the cone apex angle

4 : circumferential stress

(,r = longitudinal stress

REFERENCES

L Brownell, L. E., and E.H.Yorlll.g, Process Equipment Design, lolmWiley, New york, 1959.2. Shield, R. T., and D. C. Drucker, "Design of Thin-Walled Torispherical and Toriconical

Pressure-Vessel Head,s" in Pressure Vessels anl Pipirrg: Design atd Anatysis-A Decade ofProSleJr, Ameican Society of Mechaoical E4ineers, 1972.

Bf sf.tooRAPHY 267

.1. Bourdtnun, l{, C., "Strc$scs al Junction ol Conc and Cylindcr in Tanks with Cono Bottuns orEnds" in Prc.rsare ye.rszl aul Piping Design: Colkcted papers 1927-,/9Jg, Amcrican Socictyof Mechanical Engineers, 1960.

4. Jawad, M. H., "Design of Conical Shells under Extemal Loads,,' Jounnl of pressure VesselTechnology, pp. 230-238. Vol. t02, 1980.

BIBLIOGRAPHY

Fliigg€, W., Strcrrer ir, S,ells, Springer-Verlag, New york, 1960.

CHAPTER I OBLIND FIANGES, COVERPLATES, AND FLANGES

264

Iypicolflons€: r.ody for in.tollins. (Courlesy G+W Tdylor-Bonmy Div., lqylor Forse)

269

SIIND fLANOTS, COVTR PIATES, AND TLANGES

IO.I INTRODUCTION

One of the more common types of closures for pressure vessels is the unstayedflat head or cover. This may be either integrally formed with the shell or weldedto the shell, as shown in Fig. 10.1; or it may be attached by bolts or somequick-opening device as shown in Fig. 10.2. It may be circular, obround,square, rectangular, or some other shape. Those circular flat heads that are

(b)(a)

'tt-I

|-..............

T|-?\ I) tt II Fi-)-"--w

Int srol or wlld€d tlct h€ods.

IT(e)

Figoro l0,l

Fetaining

IO.I INTRODUCTION 271

I .--.i.nili--l lhl/il rhreadedi I ll'hl- "*\ --ltll-llll-J- - -t+-u-

(d)

Fisur€ 10,2 Bolt€d or quick-op€nins flol h€o&.

bolted into place utilizing a gasket are called blind flanges. Usually, the blindflange is bolted to a vessel flange with a gasket between two flanges as shownin Fig. 10.3. Although flat heads or blind flanges may be either circular ornoncircular, they usually have uniform thickness. In addition to the flat head orblind flange+ype closures, many large vessels use a circular, spherically dishedcover with a bolting flange, as shown in Fig. 10.4. In all cases, the bolts ofthehead attach either to a bolting flange on the end of the shell or to a thickenedshell.

In general, "failure" of a bolted flanged joint is due to excessive leakage atthe gasket, with very few failures in the metallic pressure boundary. Occa-sionally, problems encountered with bolts result from excessive tightening inreducing the leakage problem during hydrostatic testing of the assembly. Some-times excessive stresses in the bolts cause them to break or to stretch until theclosure leaks.

The basic equations used for the design of flat plates and blind flanges in theASME Code, VI[-1, are based on a flat plate with uniform thickness anduniform loading over the entire surface due to pressure. Depending upon the

l

I

Figur. 10.3 Elind flonsG-intesrol flonge conn.cnon.

h'.1__{|I

Kiucrh \lU ! R.d,u' \-\ca,rer \

Loos F1169. Typ.

Figure l0-4 Sphericolly dish€d cove|'s with bolting[nsino6r', ftom Fig. l-6 of rhe ASr{E Code, V I-i.)

272

Gastel

fTI

a

t-I

tbt

-l-1n c::

|-1/2A-''--.------.1I F-l/a la + 8l+I I ---?,,/-"),.#f,1t2 I

t2c(d)

flonses. (Courte.y Americon Socisry of Mechdn.col

r0.t tNTRoDUcTtoN 273

dc(ails of the corner construction shown in Fig. 10.5, various C-factors are usedthat require different minimum head thicknesses. The maximum deflection ofthe plate is assumed as not more tltan one-half the thickness and all the stresses

are keDt within the elastic limit.

icmkor!!dd . _- _^_

(rt lb ll lb.2t l.)

r* = ? rr nr.. nor r.$ rh.n 1.25 rr

c' 013 sk.rch.r{.llr) lsl ctrclr..cov.d,c- 033n, cmin. - o.2o,0,

"on-C',cur.r Cov.^, C -033

i., ir) ror

i-- L- _ tth.^..2t, i jrrr,rf l:l'^ ,.. .,.

' .WZ 'hnror75n ,,-,.,.i'i-'::lffi:.,,,M .' lt M,.,,.:r:- i_4;"1"6;,: |', 1 14/4,.1,, l; rn:-,^.'o.25,,'"' {ll $-",+#' L",Jr{- | - -VA " "''^ #l-r.,

c'o.r7o, c. o.r? c-033D :J^H:" * **' c.o3o

C.0.10 C m'n.. 0.20 C-o.20or0.13

ffiStt [.\ i| firj'I

@*4) llNniI d ll\l llf r{lf ll

rcKlt 1- h+

C. o.33

Fis',r€ 10.5 Un3toyod flor heods dnd covers. (Court6y Americon Soci€ry of r{€€honicol Ensineerc, tom Fis.

UG-3,1 of rhe ASME Codo, Vlll-I.)

c-0,33

C - O.2S

274 BTIND FI,ANOTS, COVTR PTATTS, AND fIANGES

IO.2 CIRCULAR FI.AT PLATES AND HEADS WITH UNIFORM

TOADING

When an exact solution involving a discontinuity analysis at the shell-to-headjuncture is not wanted, flat heads are generally calculated based on the assump-

tion that the edges are simply supported or fully fixed. The true condition lies

somewhere between.Exact equations for circular plates were developed in Section 7. l. In using

those equations as a basis, the equations below were developed using the head

diameter d instead of the radius term a. The following nomenclature was also

substituted in the equations of Section 7.1:

Poisson's ratio (p = 9.3;p = pressure loading Qsi)

E' : modulus of elasticity (psi)

d = diameter of head (in.)

T = nominal thickness of head (in.)

P : total presswe load (lb) = 6.7*t Ort

With the edges assumed to be simple-supported,Maximum stess is located at the center and equals

o^ = o.zs+fi (10.1)

Maximum deflection is located at the center and equals

w* = 0'0554#

Maximum rotation is located at the edge and equals

o* = o.s2s:L"'"-'E',Tj

With the edges assumed to be fully-fixpd,Maximum stress is radial and located at the edge:

o* = o.zto f;Maximum deflection is located at the center and equals

(10.2)

(10.3)

IO.2 CIRCUTAR TIAT PI.ATTS AND HEADS WITH UNITORM TOADING 275

w^ = 0.0B6#

Tangential stress at edge equals

o': O'Ilrcf;

Radial and tangential stress at the center equals

^ . -_ POI: OI = U'IJJ'i

( 10.5)

(10.6)

(10.7)

(10.8)

(10.e)

When the circular flat plate is loaded under uniform pressure p over the entiresurface, simplified equations of stress are

,: o.3ose(xf

/ )\2o = o.r88P(;/

for simply supported edge

for fixed edge

In actual designs, neither of these edge conditions is likely to be realized. Fullyfixed is very difficult to obtain in any constiruction.

Example 10.1. Determine the maximum stress in a flat head under internalpresswe of 1000 psi, diameter / = 48 in., and thickness r = 7 in. for bothsimple-supported and fixed-edge conditions. Assume carbon steel withfr = 0.3.

Solutian. Determine the total loading

P : 0.785 pd2 = 0.785(1000)(48),

P = 1.809.000 lb

For simple-supported edge use Eq. 10.1:

, : o.:l+!!.ffi : 14.550 psi(10.4)

For fixed edge use Eq. 10.4:

276 0UND FTANOIS, COVfR ptAT[S, AND ftANOIS

,, = 0.239 !l -1109-(xx)t--lj-: tlttzopsi I

Example 10.2. For the flat head in Example 10. l, determine the deflection atthe center for both edge conditions.

Solution, For simple-supported edge use Eq. 10.2:

rr R09,000x48fw = 0.0554*r.l x toltzt' = u'uzrz rn'

For fixed edge use Eq. 10.5:

, = 0.0136#" r0.)or24gx4l

= 0.0057 in.

Problems

10.1 What is the maximum shess in a simple-supporied flat head whenp : 56psi, d : 24 in., t : 0.75 in., and p = 0.3t

Answer: o^ : 15,840 psi

10.2 What is the maximum stress for the conditions in hoblem 10.1 usins thesimplified equation?

Answer: c,,," = 15,820 psi

IO.3 ASME CODE FORMUTA FOR CIRCULAR FLAT HEADS ANDCOVERS

In the ASME Code, Vm-1,1 and Section I,2 the minimum required thickness ofcircular, unstayed flat heads and covers without bolting are calculated by thelollowine:

I

t=d (10.10)

E = butt-weld joint efficiency for a joint within the head

S : allowable tensile stress (psi)

p : design pressure (psi)

C = 0.10 through 0.33 depending upon the construction details at the

tcplsrwnere

Answer: t6n = 5.792 n.

IO.3 ASME CODT FORMUI.A FOR CTRCUTAR FIAT HEADS AND COVTRS 277

head-to-shell juncture (see Fig. 10.5) and contains a factor toincrease effectively the allowable stress to 1.5 S because the stressis predominandy a bending stress

d = effective diameter of head (in.) (see Fig. 10.5)

t = minimum required thickness of flat head (in.)

Example 10.3. Determine the minimum required thickness of an integral flathead with intemal pressurep equal to 1000 psi, S = 15,000 psi, d = 48 in. withno corrosion, and no weld joints within the head (E = 1.0). The arrangementis the same as that given in Fig. 10.5, sketch , - 2, with m : l.O.

Solation. From Fig. 10.5, sketch b - 2, C = 0.33 m: 0.33(l) = 0.j3.From Eq. (10.10),

t=48 : 7.120 in. I

Example 10.4. Determine the minimum corner radius to make Example 10.1acceptable (valid) to be used.

Soltrtion. The cylindrical shell thickness r, must be calculated using Eq. UG-27(c)(1) of the ASME Code, VII-1:

PRl::=---------: 1000 x 24' sE - o.6P 15,000x1 - 0.6x1000

From Fig. 10.5, sketch b - 2, the rule is for

= 1.667 in.

,, > 1.5 in. r.j" : 0.25t, : 0.25(1.667) = 0.417 n. ACl lProblems

10.3 A flat head is constructed according to Fig. 10.5, sketch d. The diameterd = 12in., thickness r = l.25in.,E: l.0,andS = 15,000psi. Whatis the MAWP (maximum allowable working pressure)?

Ansyer.' MAWP = 1250 psi

10.4 A large flat head is made from pieces that are weld€d together and spotexamined so that ,E : 0.85. The corner details are similar to Fig. 10.5,skerch/ with ltt = 1.0. The diamet d = 60 in.,,S = 12,500 psi, andthe internal pnessure p = 300 psi. What is the minimum requiredthickness?

0.33 x 1000

15,000 x 1.0

278 0UND ftaNoEs, cov[R plarts, AND fl.ANGtS

IO.4 COMPARISON OF THEORY AND ASMT CODE FORMULA FORCIRCUTAR FLAT HEADS AND COVERS WITHOUT BOLTING

As previously mentioned, the ASME Code formula contains a factor of 1.5within the C factor to adjust for the permitted higher level of allowable stressbecause it is chiefly caused by primary bending stress. If the 1.5 value isremoved from the values of C, the range of C values in the code adjusts toC = 0.15 to C = 0.5. Rearranging Eqs. 10.8 and 10.9 into the same form asEq. 10.10, we see that C : 0.15 through C : 0.5 encompasses the two ex-hemes from fully fixed edges where C = 0.188 to simply supported edgeswhere C = 0.309.

The low value of C : 0.15 in the ASME Code is for a special head-to-shellconfiguration with an inside comer radius of at least three times the headthickness. The structural effect of this edge condition results in reducing theequivalent pressurized diameter on the circular flat head from the normal di-ameter d to a diameter of 0.893d such that the Eq. 10.9 becomes

o=orssp(o8e'il=r"(il (10.11)

IO.5 BOLTED FTANGED CONNECTIONS

The most usual type of joint for easy assembly and disassembly used in theprocess vessels and piping system is the bolted flanged connection. A convenientmethod to design and calculate flanges with ring-type gaskets that are within thebolt circle was first published by Taylor Forge in 1 937 .

3 These rules were furtherdevelopeda and incorporated into the ASME Code, VI[-l, some years later.These rules, which are still used to calculate this type of flange, are in Appendix2 Of thE ASME COdC, VIII.1.

Rules for calculating flat face flanges with metal-to-metal contact outside ofthe bolt circle are given in Appendix Y ofthe ASME Code, VI[-1. This designincorporates a self-energizing-type gasket such as the O-ring gasket. fhe origi-nal rules were restricted to analyzing identical pairs of flat face flanges. Currentrules have been improved to permit analysis of both identical and nonidenticalpairs of flanges.

Further development of design rules in the ASME Code, VI[-1,5 came withthe issuance of reverse flange rules that use a ring-type gasket with no additionalcontact of the faces. These rules were added to Appendix 2, ASME Code,VIII-I.

In addition to these rules for flange design in the code, many designs are usedin which rules are not in the ASME Code. One common type is the full-facegasket flange. There are many others that may be designed for ASME Codeapproval by meeting the requirements of U-2(g) of the ASME Code, VIII_I.

Before any flange design calculations are performed for a vessel to be ap_

I0.6 CONTACT FACING 279

provcd according to the code the dcsigner should recognizc th t s(nr1c calculltions can be avoided. If the flange is the type described in Appentlix 2 ol thcASME Code, the code permits using flanges with recognized standards thatcstablish items such as dimensional standards, materials, and pressure/temper-ature ratings. The code accepts flanges designed to ANSI 816.5 .,pipe Flangesand Flanged Fittings,"6 API 605 "Large Diameter Carbon Steel Flanges",T andANSI 816.24 "Bronze Flanges and Fittings, 150 and 300 lb,'.8 Several otherstandards are not included; however, when the flanges are selected by thismethod, no additional calculations are required to satisfy the ASME Code.

When calculations are necessary according to Appendix 2 of the ASMECode, VIII-I, for a nonstandard design or when it is desired to upgrade astandard flange, similar design calculations are required for blind flanges (circu-lar flat heads with bolts) and for regular bolted flanges. Although each item isdiscussed in geater detail in the following paragraphs, the basic steps in de-signing a flange are as follows:

l. Establish design pressure and design temperature.2. Select gasket material and dimensions and facing type. Calculate N

and b.3. Calculate loads for both gasket seating and operating conditions.4. Determine bolting sizes and gasket width check.5. Establish flange dimensions (usually using those from a standard flange).6. Using loads and dimensions, calculate moments for both qasket seatine

and operating conditions.7. Determine required thickness of flange.

10.6 CONTACT FACINGS

For ring+ype gasket design as given in the ASME Code, VI[-I, Appendix 2,several types of flange facings are used. Some of the more usual types are theraised face, the tongue-and-groove, and the lapjoint. When these types are used,the full seal loading is taken by the gasket because no other part of the face isin contact with the adjacent face.

In addition to the types of facing where the gasket must carry the seating load,one type of closure and facing depends upon the adjacent faces to be in contactwith each other, but it does not require a large seating load for initial sealing.This kind of closure is used on both the ring-type gasket design of Appendix 2and the flat face flanges with metal-to-metal contact outside the bolt circle asdescribed in Appendix Y ofthe ASME Code, VIII-1. This construction utilizesa self-energizing or pressure-actuated O-ring gasket that is internally pressurizedto seal the gasket and does not depend upon initial gasket seating by the boltsthat cause compression of the gasket (Fig. 10.6).

There are also special types of gaskets and facing designs that become self-

ETIND TTANOES, COVTR PIATIS, AND IIANOES

FACING DETAILSo. Sell

Energizing

O.Rin9

b.

/'' -)/ l\Cold tr.t.t Finbi

Met.llic

4\T-

Ring JointGroove

,,.-1.n o/y %F*\ b=, I \-1\__r/T;'l-\-l

Wirh SearingNubbin

Figur. 10.6 Typicol Gocing d.roil..

sealing frorn the gaskets rotation and deflection that are caused by contactloading ftom a retaining ring and head closure. Some of these are called the deltagasket as used in Bridgeman closures, the double cone gasket, and the wedgegasket. In all these cases, the initial gasket seating load is low. As the pressurein the vessel is increased, the gasket rotates and deflects into a special facing inwhich the sealing load increases as the pressure increases. Care must be takenwith this type of closure because the gasket often "seizes" and it may be difficult!o get the closure apart. In many instances, the gasket may be silver, gold, orplatinum plated to help prevent the "seizing."

Delta R ing

Ring & BevelCold WaterFinish

Lap Joint

10.7 GASKETS 281

r0.7 GASKTTS

A large variety of gaskets is needed in process equipment. The diverse pro-cesses, temperatures, pressures, and corrosion environment require gaskets withdifferent configurations, materials, and properties. Some of the frequently used

gaskets are:

l.t3.4.I

6.

7.

Rubber O-rings

Metallic O- and C-rings

Asbestos

Flat metal

Spiral wound

Jacketed

Metal ring8. High-pressure type

10.7. I Rubber O-Rings

These gaskets shown in Fig. 10.7a are used extensively in low-pressure applica-tions such as storage tanks and air receivers. They are normally confined in agroove to prevent exhusion and their maximum temperature limit is about250'F. Because the required seating stress is negligible, the number of boltsneeded in the flange is kept to a minimum. A groove finish of 32 rms is usuallyspecified.

10.7.2 Metollic O- ond C-Rings

The metallic O- and C-Rings in Fig. 10.7b and c have a wide range of applica-tions for both extemal and internal pressures. They have a good springbackcharacteristic and a low seating stress. The gaskets may be manufactured fromwidely selected matbrials compatible with the flange. This eliminates the prob-lem of thermal expansion between the gasket and the flange and increases theirapplicable temperatue range.

O- and C-gaskets seals along a contact line. Accordingly, a finish of about32 rms is needed in the flange seating surface to properly seal the gaskets. Incritical applications a silver plating is specified to help sealing.

The O-rings manufactured in ftree different styles shown in Fig. 10.7c are theunpressurized, pressurized, and vented types. The unpressurized one is used athigh temperatures whereby the increased pressure from the sealed gas cornpen-sates for the loss of strength. The vented ring gasket is used at high pressures

for better sealine.

EtIND TTANOIS, COVER PIATES, AND TI.ANOES

10.7.3 Compressed Asbestos Goskels

'fhese gaskets in Fig. lD.1d normally consist of 707o asbestos, 2O7o ntbberhinder, and l0% filler material and curative. They can be cut to fit various shapes

und configurations such as heat exchangers with pass partitions and oval andsquare openings. Thicknesses are normally fumished beween fr and i in' and

rcquire a seating surface finish of about 250 rms. Asbestos gaskets are normallyused for tempemtures up to 850T. A rule of thumb for determining the adequacyof Nbestos gaskets for a given temperature and pressure is to limit the productof temperature in T times the pressure in psi to about 300,000.

ITilll I

_)l(a) Rubber O-Ring

tressurized tessurized

I7I Htr)t )tO- Ring

(d) Asbestos (e) Flat Metal

figur. 10.7 lypo6 of oq.kcls.

r0.7 oAsKtTs

Fisu.. lO.7 TyF of solk6b (conrinu.d)

10.7.4 Flqt Melol Goskets

These gaskets (Fig. 10.7a) are made from a wide variety of materials that canbe cut from sheet metal to any desired configuration and width. Some frequentlyused gasket materials and their temperature limit are:

Material Max. Temp. ('F)

IpadAluminumBrass

Copper

212

400

500

600

(t) sp iru t

800

900

1000

1200

1200

1200

1500

1500

1800

*l Tt1l q".9,: need a high searing force for proper seating. Accordingly,tney are best-suited for high-pressure applications . The seating stirface must havea finish of about 63 rms.

S€rated gaskets require a smaller seating force than flat gaskets and thus areused in screwed flanges where friction forces are to be miiimized.

2t1 IUttlD ttaNots, covER ptaTts, AND FtaNors

Materiul Max. Temp. ('F)

Titanium and zirconiumCarbon steel

Monel400 series stainless sieelNickelInconel

300 series stainless steel

IncoloyHastelloy

Lens

Double Cone Bridgeman(i) High Pressure

Fisure 10.7 TyF! of golkei! (conrinud)

r0.7 oASKETS 285

10.7.5 Spirol-Wound Goskels

Spirul-wound gaskets (Fig. l0.1f) are very versatile and used in numerousnpplications. They are especially suited for cyclic conditions where the excellentnpringback makes them ideal for repetitive loading. They:ue manufactured toI rlcsired width by spiral winding a preshaped metal strip with a filler materialbctween the strips that consists of asbestos or teflon composites. Asbestos-filled

Iuskets are limited to temperatures of 850T, whereas teflon is limited to 500"F.For most applications spiral-wound gaskets are retained in a groove. In raised

luce flanges, an outer ring is used to prevent the gasket from extruding and thet:hevrons from excessive deformation. Sometimes an inner ring is also used torninimize erosion and to reduce temperature fluctuation in cyclic conditions.

Many factors affect the performance of spiral-wound gaskets such as tightnessof wraps, material of filler and strips, height of strips, diameter of opening, andsurface finish. The gasket seating surface finish is about 125 rms.

10.7.6 Jockeled Goskels

'these gaskets in Fig. 10.79 are normally used for pressures up to 500 psi in largediameter vessels where flange out-of-roundness and tregularities are large com-pared with small flanges . They may be purchased as plain or corrugated and theyseal at the inner and outer laps. The outer metal jacket are made from a widevariety of metals. The filler material is normally made of asbestos, teflon, ormetallic. The asbestos filler limits the temperature application to about 850'F,whereas the teflon is limited to 500'F. The metallic filler is used at the hishtemperatures. The seating surface finish is about 63 rms.

10.7.7 Metol Ring Goskets

The metal ring gaskets shown in Fig. l0.1h are used in high-pressure andtemperature applications. Their small cross-sectional area makes them ideal forcompact flanges. The required high seating stress has the same magnitude as thepressure stress. The rings are made from many varied materials and are some-times silver plated to improve sealing. The gasket groove finish is about 63 rms.

| 0.7.8 High-Pressure Goskets

I€ns, delta, double-cone, and Bridgeman configurations shown in Fig. 10.7icomprise the majority of the pressure applications where the seating saess isrequired to be low due to physical limitations of bolt spacing and flange width.They are used extensively in pressure vessels operating above 1000 psi and aremade of softer materials than the seating surfaces to prevent damage to theflanges or covers. In general, these gaskets are expensive to fabricate and

216 IIIND 'IANOIS,

COVIR PIATES, AND TTANOES

m1lchinc; rcquirc vcry tight k)lerunces; and need very smooth seating surfaces of16 rms or better.

. High-pressure gaskets have a large surface that is subjected to the vessor

ll,.e.rnall pr:r*t::.

+ccordingly, a free-body diagram is no-ully n"""rrury rodetermlne the additional forces transmitted to the flanges and boltjresulting frompressure on the gasket.

The individual design requirements for lens, delta, and double-cone gasketoare given in the next three sections.

10.7.9 Lens Ring Goskets

The lens ring- gaskets in Fig. 10.8 are normally used in small flanges. Thc

:f:r.-::"j^",:1T19j ttg rins. through points a to b, must be equal to or larger

f:_T::i1t-Tl "f p: flanc: !o pyent crushing of the gasket. Thus knowingIne nslde-dameler of the gasket and the required bolt area, the outside diametercan be calculated from

oo =llr,+ (D),]'/

where OD : outside diameter of ring gasket (in.)

ID = inside diameter of ring gasket (in.)

Aa = actual bolt area of flange

5-

fcl"r16

Figur. 10.8 t.n. so3k6t,

10.7 oASKETS 287

l'hc outside thickness of the gasket is established to allow tbr an 0.25 in.clctrirnce plus 0.0625 in. for a centering ring, ifrequired. The pitch diameter lbrgrskct seating reaction is established as

c=(rD)*1(oo-ro)3

whcre G = diameier of gasket reaction.'the spherical radius of the gasket surface is taken as

p=G/z" sin 0

where R : spherical radius of gasket surface (in.)

I : angle of friction (for mild steel, it is 20")

l.'rom the geometry, the inside thickness of the flange is calculated from

where t = inside thickness of gasket (in.).The width of gasket seating is normally

(At(1.s)(design bolt stress)

z(GX3)(yield shength of gasket material)

where N = gasket seating width (in.).Figure 10.8 shows the flange and cover surfaces at the vicinity of the gasket

machined to have a slope of 20'.

10.7.10 Delto Goskets

The delta ring gaskets in Fig. 10.9 are extensively used in the United Staies forhigh-pressure applications. These gaskets rely on the inside pressure to wedge

them in the gasket groove for sealing and thus do not requhe any initial seating

or bolting shess. The general dimensions that are shown in Fig' 10.9 apply to

rings of all diameters. The pitch diameter G is normally taken as

G=ID+0.125

and tlle gasket seating width N is usually equal to 0.125 in.

t=:-+)l' 16 -l /ODYI- \, i I

IIINO II.ANOIS, COVER PI.AT!S, AND FTANOES

Figur. lO.9 Dclto go!k6t.

10.7.1| Double-ConeGoskets

Double-cone gaskets are very popular in Europe and can be fabricated in variousslzes. A typical detail is shown in Fig. 10.10. The required cross-sectional areais given by

f t 1tl2OD = l:A, + (rD),

IL1r Iwhere OD : outside diameter of gasket (in.)

ID : inside diameter of gasket (in.)

A, = actual bolt area of flange (in.r)

10.7 oasKlts

The seating length N is determined from

and the pitch diamet€r G is expressed as

G=OD-0'5N

The height of the gasket is usually set so that the net pressure force does not

exceed the seating force. Thus,

Figur6 '10.10 Doublo-cono gosk6t.

(cos 60)(aXc)(3)(yield strength of gasket material)

tzxrr(|)<,- 60) = (P)(,r)

290 ITIND IIANOIS, COVTR PI.ATES, AND TIANOES

' - (rXN)(sin 6o)--- P

where I = seating stress of gasket material (psi)

P = internal design pressure (psi)

10.7.12 Gcsket Design

!a1f1t.fei1e,_r cnaraderistics depend upon the material and the design of the

gasKet rtself. -fhe gasket faclor m and, the minimum design seating sEess )l irre

P^l:':j:o,ro th: cTk"jjlry and the gasket marerial. flrhou;h-the rn and ytactors have been in the ASME Boiler and pressure Vessel CodJsince the 194iedition, they are suggested values only and are not mandatory. The originatiesting and development of the m and y factors are described ln an articii UyRossheim and Markle that does not give the underlying background for thespecific.values. V9ry

_fu* changes have been made to ttiese faciors since theywere originally published.

.A.s the

-result of many inquiries to the ASME Code Committee regarding the

validity of the m and J factors, a large-scale investigation has been undertakenby the hessure Vessel Research Committeero of the frelding Research Councit.As experimental tests progressed, it became obvious that thJrz and y factors are."J1t"9 t" many items not previously considered. There is a closJ corretationwith the amount of tightening of the bolts, the gasket type, and the material, forthel are all related to the leakage rate of the j-oint. "

.g.nc-e the_gasket tlpe and material have bCen selected, the effective gasket

width for cabulation may be determined. For solid flat metal and for thJ ring_type joints' the basic gasket seating width bs is found by the formuras rn columnI of Table 7-5.2 of the ASME C;de, VI[_l, whereas for all other types ofgaskets' ba is delermined by the formulas in column rI. The effective gasketseating width D is found by applying the following rules:

D=06 when bo = iin. b: O.slbo when De )|rn.With D deiermined the location of the line of gasket load reaction can bedetermined as well as the values of G and h6 for calculating flange moments.

.ln designing a flange, it is important to recognize that two'desiln conditions

cxrst-the gasket seating and the operating conditions. Gasket seattg conditioncxlsts when an initial load is applied by the bolts to seat the gasket at ambienttcmperature with no intemal pressure. The minimum initial biolt loaO is

W"e = tbGy (10.12)

r0.7 GASKETS 291

'l'he operating condition exists when the hydrostatic end fbrce from the internal

dcsign pressure tends to open the joint, for the gasket retains enough resiliencyl() keep the joint tight. Loadings and stresses are determined at design pressure

rund design temperature. The loading for the operating condition is

W.t: H + n,:f,C'zp + mGp(Zb) (10.13)

To avoid crushing the gasket in those bolted flanged connections where the

gasket is carrying all the loading, it is recommended that the initial loading does

not exceed the gasket seating stress y. Once the actual bolting areaA, is selecied,

a check may be made to determine the required minimum gasket width by the

fbllowine formula:

N^"=ffi (10.14)

Example 10.5. A vessel has the following design data: design pressure

p = 250/U.. psi; design temperature = 250'F; a spiral-wound metal, fiber-filledstainless steel gasket with an inside diameter of 13.75in. and widthN = l.0in.The gasket factors are m = 3.0 and y = 10,000. Bolts are SA-325 Grade I with& = Sa : 19,200 psi. Is the gasket width sufficiendy wide to keep from crush-

ing out?

Sohttian. Determine the effective gasket seating width as follows: N =1.0 in., b" : N/2 = 0.5 in., b = 0.5!bo = 0.3535 in., effective gasket di-ameter ls

G = 13.75 + (2 x 1) - (2 x 0.3535)

G = 15'M3 in.

Gasket loadings are

H : o.785G2p: 0.785(15.043F(25N1 = 444,1ss

H,= 2brGmp = 2(0.3535)r(15.043X3X2500) = 250,600

Wa : H * He = 444,1A0 + 250'600 : 694'700

W^z: rbGy = z(0 3535)(15.043X10'000) = 167,100

Since fi = Sa : 19,200 psi, W.r sets the bolting arca A^ as

A^=Y# = 36.182in.,

292 IUND ftANOrS, COVIR plarrs, AND FtaNots

AD = octuul bolt {reu = 36.ti in.2 lbr l6-2-in. diumeter bolts. Minimum gaskctwidth is

.. 36.8( 19.200)fl^" = zlq000).(15.0a, : 0 748 in versus I in' actual

One-inch wide gasket is sufficient to prevent crushing. fPmblem

10.5 A solid, flat, stainless steel gasket with 'l2

: 6.5 and y : 26,000 is usedin the vessel described in Example 10.5. The preliminary gasket insidediametef is 12 in. and the gasket width is 1.0 in. whai is the easketseatine load?

Answer: W,,2 = 384,000 lb

IO.8 BOTTING DESIGN

In designing bolting for flanges, the initial item is selecting the bolting matenal.It must be a malerial compatible with the flange material . ihat is , there musr norbe any chemical or galvanic action between the bolting and ffange material thatwould cause the bolts to seize in the tkeads. Under certiin circurn'stances, rt maybe necessary to plate the bolts or to make them from special material to prot€;tthem from the environment. Although it is not neceisary to select a 6oltingmaterial with a tensile shength close to that of the flange material, one shouldcarefully consider the effects of strain elongation and rehxation of boltingmaterials that have a high tensile strength and requte a smaller cross-secdona'iarea.- In addition, when high sFength material is used for the bolting, care mustbe taken not to rcduce the number of required bolts to such a small

-number that

excessive bolt spacing is developed. r I

Wlren the bo_lt spacing exceeds (U + t), secondary flange bending is devel-oped between the bolts to the extent that it affects the ;ormal flange bending . Toaccount for this effect, the flange bending moment M0 must be iricreased by theIactor

(10.l5)

where d = nominal diameter of bolts (in.)

t : flange thickness (in.)

To detennine the total required cross-sectional area of bolting, both the gasket

actual bolt spacing

10.8 BotT|NG DESTGN 293

scnting and operating conditions must be examined.The minimum required bolting area A, is th€ greater of

W.z

s, otW^r

Sa

where & : allowable bolting stress at room temperature (psi)

S1 = allowable bolting stress at design temperature (psi)

W.1 = operating load: H + H, (see Eq. 10.13)

1Y.r = gasket seating load

From this minimum required bolting a;ea A. the actual bolting area A, is

selected. In order to obtain a bolt loading for calculating moment for gasket

seating, the minimum required bolting area and the actual bolting area are

averaeed as follows:

w:0.5(A.+Ars" (10.16)

Certain times during the operation of a process vessel the bolts in a boltedflanged connection are subjected to actual stresses in excess of the allowabledesign stresses. This may be especially true during hydrostatic testing. Caremust be taken to enswe that during this testing, no permanent elongation ofthebolting has occurred. If so, the bolting may have to be replaced before the vesselis put into service. Realizing this is especially important if each of two suppliersprovides half of the bolted flanged assembly and one does not know what boltingis supplied.

Example 10.6. A vessel flange uses 16-2-in. diameter bolts. Flange sfresscalculations indicate that a flange thickness of t : 4.5 in. is adequate. The boltcircle diameter is C = 22.5 in. Will secondary bending stresses be developed?

Solutian. The maximum permissible bolt spacing without a penalty is(2d + t1 = (2 x 2 + 4.5; = 3.5 in. The actual bolt spacing is

rd _ n(22.5\1V -]6- = +Az n.

Because the actual spacing is less than the rnaximum spacing without a penalty,no secondary bending shesses are developed. I

Example 10.7. Suppose a vessel requires Z-2 j -in. diameter bolts on a flangethat is 5.5 in. thick. What is the madmum bolt circle that will not causesecondary bending shesses? The minimum bolt spacing for 2 j -in. diameter boltsis 5i in.

294 lutito ftaNolg, covER ptATEs, AND frANOrS

Solullon, Maximum spucing is

(2d + t) = (2 >< 2i + 5.5; = 19.5 tn.

Diameter of bolt circle is

lo'5 x 24 = 80.2 in.n

based on maximum spacing. I

Problems

10.6 Sixteen bolts at I l-in. diameter are io be located on a bolt circle ofG = 32 in. The flange is 2+ in. rhick. What is the factor rhat is due tosecondary fl ange bending?

Anstrer: l.l2

10.7 The minimum spacing for the wrench to fit I |-in. diameter bolts is 3 | in.Twelve bolts are to be used on a bolt cide of 15l-in. diameter. What isthe minimum flange thickness that does not cause secondary flange bend-iog?

Answer: 1.0 in.

IO.9 BTIND FTANGES

The minimum required thickness of a cicular, unstayed flat head or blind flangeattachedty bolts and utilizing a ring-type gasket that causes an edge moment isderived from the assumption that the flat plate is simply suppofted; the gasketload line G and is loaded by a gasket seating load or a combination of easketloading and a uniform pressure loading . The combination of these loadingJat thegasket and at the bolt circle causes an edge moment ofMg/zrG, as shown-in Fig.10.11. ff the edge moment ffi is assumed to be equal ta Wh", tlre theoreticalstress at the center of the flat plate is

" _3(3 + tt)p (GY + 1 /wrrc\"- :z F)-"G\-F) (10'17)

Sening trc = 0.3 and E = weld joint efficiency within the flat plate,

sE:(*rye\-r..(W) (10.18)

IO.9 BI.IND FTANGES

H

)Fisurg lo.l I loodings on blind flons€.

Solving Eq. 10.18 for r gives

(10. r9)

This is identical with the equation in the ASME Code, VI[-l, except that theconstants of 0.3 and 1.9 are used in the code instead of the exact constants of0.31 and 1.91 and the gasket load G is substituted for d. The general ASMECode equation for circular flat-bolted heads is

,:oJgll\, *t'n-LYl',

t=G (10.21)

For operating condition the intemal pressure p as well as the gasket loading areapplied. For this condition, Wr : H + Il, at operating temperature with anallowable tensile stress of S1. The equation for operating condition is

(10.20)

For the gasket seating condition, the internal pressure equals zero and the onlyload is the gasket seating load W" at ambient temperature with the allowabletensile stress of S.. The equation for gasket seating condition is

Figure 10.12 shows a sample calculation of a blind flange.

E-xample_ 10.E. Considering the pressure vessel described in Example 10.5,the vessel is to have one end closed by a blind flange. What is the minimumrequired thickness of the blind flange? Design data are the following:

296 ITIND ?LANOI3, COVTR PLATIS, AND FTANOTS

t=G

Design pressure p = 2500 psi.Design temperature = 250'F.Flange material is SA-105.Bolting material is SA-325 Gr. lNo corrosion exists.Allowable bolt stress at gasket seating19,200 psi.

Allowable flange sfress17,500 psi.

(t0.22')

and operating conditions =

at gasket seating and operating conditions =

Gasket is spiral-wound metal, fiber filled, stainless steel, 13.75 in. insidediameter times width N = I in.

Solution. Following the information calculated in Example 10.5, once theactual bolt area A, is found, the design loading for the gasket seating conditionW, can be determined as:

W: 0.5(A^ + Ar)S, : 0.5(36.2 + 36.8X19,200)

% = 700,800

The moment arm is determined from

hc = 0.5(C - G) = 0.5(22.5 - 15.043) = 3.729 in.

fJgt Pxample 10.5, the design loading for operating condition is lll.r =694,700.

. The minimum required thickness is determined as the greater thickness of thatdctcrmined for gasket seating load according to Eq. 10.i1 or for operating loadrrccording to Eq. 10.22.l.i)r gasket seating,

_ ll.gwhc .__.. I t.9(700.800x3.729), = uV s,uoi = ts.o+:{ffi :4.343in.

0.3p | .9W,,1ha;

s"E - -srEcl-

t0.9 EUND FTANOES

For operating condition,

t:GrF

t = ls.(Ml.v 17,500 x 1.0 (t7,500 x 1.0x15.043t

Therefore, the minimum required flange thickness is t : 5.329 in. I

Problcms

10.8 Suppose the flat head in Example 10.8 is made by butt-welding flat platestogether. The welds are spot examined so that E = 0.85. What is theminimum required head thickness?

Answer: t, n = 5.780 in.

x -crr/. - 41{,12t. -,51c - d - ,-'12

l. OO, l..iir dillina. C.|l.1-+6. 6 l4hlig

2. Xc .addj.el rfiktid. I '.{ft.d l, l.d.!

3, Ud{ or[.Rin rF..tjrd, l!.G .l $r'l 0..!.Br b. d.ri..d .nt d.. .r- '.i!lni lo'

Fisur€ 10.12 Elind flonge romple colcolorion sh€€|. (Courrery G+w roybFBonney Div., Toylor Forge.)

O.3p , |.9W.rhcsrE- - s"EGt

298 IIIND IIANOIS, COVER PLATES, AND TLANOES

10.9 Suppose the bolt circle diametsr is incrcused ro C = 24 in. What is thcminimum required head thickness considering both gasket seating andoperating conditions?

Answer: tnn^ = 5.671 in.

IO.IO BOTTED FTANGED CONNECTIONS WITH RING-TYPE GASKETS

The design_rules for bolled flanged connections with a ring+ype gasket that isentirely within a circle enclosed by the bolt holes and with no iontact outside of

, Wh€n it is necessary to design a bolted flanged connection because no stan_dard flange of the proper size is available, the standard pressure/lemDeraureratings are not adequate, or sperial design factors are to be used for the gasket,the procedure in Appendix 2 of the ASME code is used. The design of nanges,bolting, and gaskets by the ASME Code rules is essentially a ftal_and-enorprocedure where some dimensions are set and remain fixed, whereas otherdimensions such as the flange thickness are varied. Stresses in the flange and thehub are calculated. If any of these shesses exceeds the allowable tensile stresses,a "new" flange thickness is selected and the stresses are recalculated until theyare satisfactory. The trial-and-error method is essentially due to the compleitheory used by Waters, Rossheim, Wesshom, and Williams to solve the problemin the original development. The WRWW method, which was ultimatelv incor_porated into the ASME Code in 1940, is ar elastic analysis of the intiractionbetween the vessel or pipe, the hub, and the flange ring assembly. The shell andhub are resolved by a discontinuity analysis that was previously described inChapter 5 and the flange ring is considered as a flat plate with ihe center Dartrcmoved (Fig. I0.13). Interactions of rotations and deflections are oermiiteduntil lhe balance is obtained.

. The basic assumptions in the analysis are that the flange materials are elastic,that is, no creep or_ plastic yield at lower temperature ocJurs, the bolt loading iiassumed or determined from the gasket factors, and the moments due to loadinssare essentially constant across the width ofthe flange. In addition, rotation of tie

IO,IO sOTTED fI.ANGED CONNECTIONS WITH RING.TYPE GASKETS

I ) (l| ) (llr

(a) Moment Loadlng

(b) Direct Loadlng

t-t+ f+.llrr,-ri'r'roIllll.l-----v------J

'tt 0z(c) Combined Loadlng

Figure '10.13 Flongo loodin$ for alo:ric anolysii.

flat plate is assurned as linear with no dishing effect and superposition is accept-able.

The solution of the complex problem is simplified for code use by cwves,formulas, and tables that contain constants depending upon the geometry of theflange assembly. Formulas for various coefficients are given in the code thatpermits cornputer programming of the basic equations for rapid solution.

The code designates flanges with ring-type gaskets to be three types for

EIIND FTANOIS, COVER PLATES. AND FTANOTS

Bnalysis: int€grsl, loose, und optionul. lntegral means that the pipe, hub, andring are one continuous assembly from their original manufacture as eithcrforging or casting welded together by full penetration welds. lnose means noattachment of the assembly to the pipe or no ability of the juncture to carry sheafsand moments other than those required to seal against pressure. These types offlanges are called slip-on, lap joint, and threaded, and they may or may not havchubs. Optional means flange designs that, by construction, are integral, but thcanalysis is permitted by the simpler method for loose-type flanges. Examples ofthese various types are depicted in Fig. 10.14.

The calculation of a flange with a ring-typ€ gasket first involves selecting thematerial for the flange, bolts, and gasket in a manner very similar to the blindflange. Next, the facing and gasket details are set, the loads due to intemalpresswe are determined, and the required bolting area and bolt sizes are selected.The bolt circle is then decided; and the loads, moment ams, and moments dueto both gasket seating and operatihg conditions are determined as with the blindflange. By knowing these and the geometry used to determine K and other hubcoefficients, stress calculations are made for both conditions. The longitudinalhub stress, the radial flange sness, the tangential flange stess, and their variouscombinations are comparcd with allowable stresses.

The method of calculation is virtually identical for welding neck flanges andslip-on or lap-joint flanges except that the axial pressure load is applied at aslightly different location. For the ring flange design, the tangential flange stressis the only one calculated. The minimum required thickness can be directlydetermined from

(10.23)tM*V s-B

,..IExample 10.9, What is the minimum required thickness of a welding neckflange as shown in Fig. 10.lzla with the following design data? (Nore.' These dataare the same as those used for the blind flange in Example 10.8. In Fig. 10.15is a sample calculation of a welding neck flange.

Design pressure, p. = 2,500 psi.

Design temperature = 25O"F.

Bolt-up and gasket seating temperature = 70'F.Flange material is SA-105.Bolting material is SA-325 Grade 1.

Gasket details are spiral-wound metal, fiber filled, stainless steel, insidediameter is 13,75 in. and width is 1.0 in.

Solutinn

1. Allowable bolt shess at design and seating temperatufes = S, =19,200 psi.

I6

IO.IO BOTTTD TTANGED CONNECTIONS WITH RING.fiPE GASKETS 301

I

FE--d *--'t1+

(.)

tr=-tr_rel \__

(d)

IT l-T

9t 12 l9lt\A 9 j 12

(s)

Opii.nrl lYF. Fl.n96.Th6. M.y b. C.ldl.r..t.r Eirh.. Looro.or Inr.9..t.Typ.

(i) (i) (k) (1)

Figur6 lo.l,t Typer of flonger. (Courl,ely Americon So.ieiy of Mechonicol Engin€.rs, from Fig. 2-.1 of theASME Cod., Vlll-I.)

2. Allowable flange sfress at design and seating temporatures = S/ =17,500 psi.

3. Gasket dirnensions are

b.= N/2 = 0.5in. and b=0.5Vro=0.3535G = 13.75 + (2 x l) - (2 x 0.3535) = 15.M3 in.

{-..*,o

(h)

302 BTIND fI.ANOES, COVER PIATES, AND FLANGES

'r= 13.154

Io-:". I P;Pe s'zera't L. tza

B = to.7t "

c. 2t,5'

t6- 7"' eot::s -

Figure lO.l,C(d) Flons€ dim€Bions ior Ex. 10.9.

Determine bolt loadings and sizing of bolts with N = l; b = 0.3535;y= 10,000; m=3.0.

, : Xo,, = 2|o5.o43)'z(zsoo) : 444,323

H, = 2btrGmp = 2(0.3535)zr( 15.043X3.0X2500) :250,591 ,Wa = H + Ho = @44323) + (250,591) = 694,914

W^z = nbGy : z(0.3535)(15.043X10,000) : 167,060

A. = the greater of W f Sbh : 694,914) /(19,200) = 36.2 in.'z

or W,ef 56" = 067 ,0ffi) /(19,2C0) : 8.7 in.'z

Aa = actual bolt area = 36.8 in.'z 16 bolts at 2-in. diameter

W.: 0.5(A^ + Ar)Sr. = 0.5(36.2 + 36.8X19,200) : 700,800

Wo=Wa=694,914

Calculate total flange moment for the design condition.

Flange Loads

a, : XB'p :

!{ro.ts)'{zsn) = 226906

He = Hp:250,591

Hr : H - Ho : 1444,3231 - Q26906) = 2r7,4r7

Leaer Arms

hp=R+'0.591 = (2.5) + 0.5(3.375) = 4.1875

h= 6'at

I

I

4.

hT

I orrron coxorrronr

lltr

a.115 '

t0.75 "

l6 - 2"toont t r. a41'

Figuro lo.l5 weldins neck flonse sdmple colculotion sheet (Courtesv G+w Tovlor-Bonnev Div ' Tovlor

Fors..)

J7i.^l u."^L azl-1,l;ber fi//el,s u/eer tt"elrl t5 t.o. i t" ",tt

/e'

Ac =,5lc -6r = t.1L

rc =.5(c - cr - t71'65

.s(s, + srfq .5 {sr + srl =

.5lsi + s,lr.5lsx +5rl = 13. oO

= 1/3t.+t = Z.o

lf Lh i9oc., no..g ro + t .uttiptE .il .. lr .6or au.rld b,

Cdrvnd- D.+

Crxt.a-- tt'i..

-

303

Itll'lD flANOt!, COVIR p[Arr3, Al,tD frANOrs

nr; - O..J(C * 6) = 0..5(22.5 - 15.043) : j12BS}r=0.5(R*g1 * ft6):0.5(2.5 + 3.375 + 3.7285) = 4.9613

Flange Moments

Mo = Hp x hD = (226,910)(4.1875) = g5g,17g

Mo- Hox hc= Q5O,5g0)(3.7285) : g34,33g

Mr = Hr x h, : (Zt7 ,420)(4.3018) = t,O43,gg}Ma" = Mo + Mc + Mr = 2,928,490

Calculate total flange moment for bolt_up condition.

Flange Load

He = W: 700,800

Lever Arm

hc=0.5(C-G):3.7285

Flange Moment

Mon= He x ft6 = (700,800)(3.7285) : 2,612,930

7. Use the greater of Md. or M6"g1/S); Ms = 2,92g,490.E. q}11pe consrants from the ASME Code, VI[-I, Appendix 2: K:A/B = Q6.s)/(ro.7s) = 2.16s. ^F,-":"

i,e..-i 7 i, "s"Ju'i uro-,,r = 1.35 z = 1.3e y : 2.2e ii : z.ii."*""^'sr/ go = 3.375/1.0 :3.275

h=Vrry'=V@jr1D=3.27sh/lro = 6.2s/3.2t9 = 1.906

From Fig. 2-7.2, Section VIII-I, I. = 0.57.From Fig. 2-7.3, Section WII_I, y = 0.04.From Fig. 2-7.6, Section WII-I;y = 1.g.

e = F/ho= (.57) / (3.27s) = 0.1738

d = (u /v)hoeT = e.s1 / .o4e.279)(t)2 = 2g5.76

9. Calculate stresses. Assume a flange thickness t = 4.5 in.L = (te + 1)/T + t3/d = (1.320) + (0.443) = 1.763

6.

IO.IO BOLTID FTANGED CONNECTIONS WITH RING-fiPE GASKITS 305

Lon g itudinal H ub Stre s s

sx = fMn/ LglB : (t)(2,928,490) / (r.763)(3.37 ra(0.7 s)

sa: !3,s70 q9!

Radial Flange Stress

s^= Qte + r)Mo/LtzB = <z.uzaf,s:x:rrii;li:..763)(4.s),Oo.7s)s" = lLsg-Q-pg

Tangential F lang e St e s s

g7: (YMs/t2B) - ZSa

= (2.29)(2,928,490) / (4. r, 00.7 5) - ( 1. 39X 1 5,590)

Sr = 9140 psi

10. Allowable stresses

56 < 1.5Sy: (1.5X17,500) = 26,250 >-11,10 pC)

Sa < Sr: 17,500 > 15,590 psi

$ < Sy: 17,500 > 9140 psi I

Example 10. 10. What is the minimum required thickness of a ring flange withthe same design data as given in Exarnple 10.9? The inside diameter has beenincreased to fit over the outside of the shell to where,B = 12.75 in. The boltloadings and bolt size are the same as in Example 10.9. A sample calculationsheet is shown in Figure 10.16.

Solution

1. Calculate total flange moment for design condition.

Flange Loads

H, = !oB2 p = X02l s),(zsn) = 31e,2oo

He : Hp = 250,600

Hr= H - HD= 444,300 - 319,200 = 125,100

Lever Arms

ho = o.5(C - B) = 0.5(22.5 - 12.75) = 4.875 in.

306 BTIND TI.ANOES, COVIR PTATES, AND FI.ANOCS

hc = 0.5(C - G, = 0.5(22.5 - 15.043) = 3.729 in.hr = 0.5(ho + hd = g.514.gr1 + 3.729) = 4.302 n.

Flange Moments

Mo = Ho x h": (319,26)(4.575) = 1,556,000

Mc = He x h6 : Q5O,ffi)(3.729) = %a,5ggMr = Hr x h, = (125,1N)(a.302) = 53g.26iMa" = Mo + Mc + Mr= 3,029,000

Bolt-up moment is the same as in Example 10.9, Muu = 2,613,000.Shape constants arc K = A/B = 26.5/12.75 = 2.078From Fig. 2-7 .l of tlrc ASME Code, Ylfi-l, Y = 2.812.Required thickness is based on design condition as

, : M*Y _ (3.029,000x2.812)

': iE = -1r21x1)@.n : 5'8e8 in r

t3.

4.

I 2 cerrn I r^oE!"., r zfoo ftL ssir.l ..:."d raf.l,

{itc v f;rl€dr stainl?r, jf.elItl'lD J( l'dii<

G. tr lt +QtD4r .'r<.A,B.ql

Korrcl focezrooFiA- to5

5A'tz6 GP.lroao AxD aott cltcuultoNS

litllapsi rv.' - 5,cr E i6'Lo6o

^- -a:j w-,/r q wi/s\- 9.1 4 tC.x, z 25tc^r: ZE O,5 tc @1."9

117., a';, H -efr/! - 4t14,tt. w =,!l^. + Ats. - 100,€ooJ1t wn-,r.+n- 6,j1,1t{

l.vlt arra

150 ,ti = rolb = tt5t6,096q.-*.t-h- zto,tq I It E.5t< - ct = 3.1295 ,14 = rdg = 934,j21tt,-r -i/!- lZ5)132- :.51 lD + ri.l = {.iol6 u' -8fi - t1a.79tv. = 1,028,618

- loorSoo A. =.rtc - cl - 2,1285 N. - 2,ctz,1r1

tb

6 tn^tr cof.3r^Nri r:^/. rz.o'18 r-- L^z_b.h p..hr .ro.c, 2. + ,.,Lnirtr /rdi;;Eclvr .!d n h '.qe.ri6r 5'| \J r" +-;-

,-,[x=r* @'={ *i.mi

adrd.d- Dd.-6.1.J-}lvib.r-

Figure 10.16 Rins llonse lomple col.ulorion sheet. (Court€sy c+W Toytor-Bonney Div., Toytor Forge.)

IO.I I REVERSE FTANGES

l'robhm

10,10 Suppose a solid flat 2j chrome steel alloy gasket with 13.75 in. inside

diameter and width N = Lin., m = 6.0, and y = 21,800 is used withthe flange in Example 10.9. What are the gasket seating and the

operating loads?

Answersz design condition is I7,' = 734,599 tO

seating condition is W^, : 244,rOO ,O

IO.I I REVERSE FTANGES

Rules for the design of reverse flanges are given in Appendix 2 of the ASMECode, VI[-1.5 This type of flange is often used to form a reducing joint. The

solution of the reverse flange is similar to that for the raised-face standard flange

with the ring-type gasket within the bolt circle except for some minor differ-ences. Figure 10.17 shows some loads that are applied in the reverse direction'

Rwer$ flonge looding ond dimen3ions.Fisurs 10.17

308 IUND ftANoxs, covrR puTEs. aND rrANG[S

'lhis muy cuuse sorns ol the moments t0 be applietl in thc opposite direction liomthose loads on a regular flange. However, the analysis is the same after thc"new" total moment is determined. Again, the moments arc determined for boththe gasket seating condition and the operating condition. In Figure l0.lg is asample calculation sheet of a reverse flange.

Additionally, a new term aa is introduced to convert some terms from reqularflanges io reverse flanges; fra and K are redefined and based on the reverse fl-anseinside diameter; and a new equation is added to calculate the tansential flaniestress at the inside flange bore.

A special precaution is noted. When K < 2, results are faidy satisfactory;however, when I( > 2, the results become increasingly conservative. For thisreason the ASME Code procedure is limited to where I< s 2.

Derivation of the new equations for reverse flanges is similar to that for theregular flange except shears and moments are applied at the outer edge of the ringflange where discontinuities occur between the flange ana *re truU. WittrK = A/B', the conversion term ca is determined for converting T, Il, and y toT,, U,, alrd If, which is obtained as

"=*['.*.#*)Substituting this expression into the regular equation for tangential shess

(1,0.24)

(10.2s)

Example 10.11. A reverse flange is joined to a regular ring_type joint flangeto form a reducing connection . The total bolt-up momint ls coitroiting anA equitto Ms = 2,613,W. The flange bore d, = 13.25 in.; the outside diameterA = 26.5.n.; and the flange thickness r = 5 in. What is the tangential flangestress at the hub and at the inside bore?

Solutian, The tangential flange stress 51 at the hub is

" _Mol,, zl.,z \jtr= *,zl+ - i\l* r*7it, = #*#f r.ut - fl(r* ] " s " o.r:r)l : szzopsitj.zJ\zJ) L 4.14\ 3 --/]

'fhe tangential flange stress Ji at the flange bore is

s+ = kly _zK'?(_t + gofB'f L (K, - l)i l

I Drt|ox coxofioxt

5/rr/ "t"-"/- 'a.tat,fi l. r {,t/el, t& ;ahs 4a//r.75't0 ' / ' ltde- -

Kai'ed {""e-

IOAO I llwt lr|| - l Or{t|.lt.=J(C+i'-23-t)=

r.r -,r'r' - _ I,Orl,tq lfi*r- x.t - t81,1o- zz 1q f. y't f

* - Z,trL'?V

r(5r+ s.)rJ(s' + e!) = 816EO ?e\

= I't,|o'l rtlr.ir. rta. 9r ( ! a') =?

rr = r:-rj.jb.t r = 0.957

I=.ch/d - Z5r loatr.+ l),,P=

4sr +tn.t 3l +n= qt6rlr".9, rrg, sr t^r r') = ff

t =./3r+ | = l.

Fisurs 10.18 Rwerse flonse somple colculotion she€|. (Courtesy G+W Toylor-Bonney Div.. Toylor Forge.)

309

ll0 luND ftaNots, covln puTrs, AND ItANOES

_ z(+Xl +lx5 x.(,. _ 2,6t3.(xx)L oar3.2s(25) L-'" #*itti = I6'o5opsi r

Example 10.12. With the reverse flange given in Example 10.11, what is theminimum required thickness based on an allowable flange stress of 17,500 psi?

Solutia3. The tangential flange stress at the flange bore of 16,050 psi iscontrolling. Because / appears in several terms in a no;linear manner, the easiestway to select the proper thickness is by nial and error. For the initial trial, usea square relationship as follows:

16,050 17.500--tl-: l1z- or t:48in.

The term ,\ is recalculated as 3.855. Using this, the .ti = 16,900 psi. Bysuccessive recalculations, the approximately correct thickness is , : 4.6g in.which gives Si: 17,500 psi. IPmhlcm

10.11 Using the details of the flange described in Example 10.11. what is theminimum required thickness if the material of the flange is changed to onewith an allowable stress of 15,000 psi?

Answer t-i\ = 5 -26 in.

IO.I2 FULI-FACE GASKET FTANGE

919 gp" ot flange that is frequently used but no design methods exist in the

lfME "C:* is the flange using a full-face gasket, as shown in Fig. 10.19.

Figure.10.20_ is a sample calculation sheet. This type of flange is designedaccording to the provisions of U-2(g) of the ASME Code, Sectiorivll, Divisionl. This code paragraph permits using good engineering design for those con-structions where no rules exist in the code. Although the analysis is similar tothat used for a raised-face , ring-type flange , a countermoment is introduced fromthat part of the gasket that is outside of the bolt circle. In addition, the decreasein section strengtl at the bolt circle from the bolt holes must be considered whenthe radial stress at the bolt circle is determined.

lmplolng this type of gasket is usually limited to designs where a ..soft,,gasket (with a low m and 1, factor) is used and the design prlssure is low. Thisis necessary to keep the loads and bolt size within reaso-n ti fit *ithin th" fl*n"geometry even though the countermoment usually results in a low flange mi_ment and a minimum required flange thickness.

IO.I2 FUI.L.FACT GASKET FIANGE 3l I

Figura I gosket loodin$.

Several important design assumptions made in the analysis are uniform gasket

pressure over the entire gasket, inner edge of flange assembly unrestrained, and

no reduction in gasket pressure area due to bolt holes. Other restrictions and

limitations necessary for the raised-face, ring-type gasket flange, such as linear

rotation about the centroid of the ring, prevail.Assuming a uniform gasket pressure, determinations are made of the dis

l-l_tA I

0.19 full-foce

312 tNo ttANots, covrR p[ATrs, AND FtANOIS

tonce$ or momcnt olTns lrom the bolt circle to the ccntroid of the annulus fromthe bolt circle to the outside diameter and from the bolt circle to the insidcdiameter. In solving for the distances, the angle is assumed to be small and thcarc lengths are evaluated as sfiaight lines. From Roarkr2 and using his termi-nology, the basic equation is

Horvever, converting to the terminology used in the ASME Code and as shownin Fig. 10.19a, the equation becomes

a-

b=

Therefore, substituting Eqs. 10.26b,

A_C

,=i(T#)

,, (A-C'^)(U+Cl- 6(A + C)

2

"^err,J

*(#)hL

c, d, and e, into Eq.

(lO.26a)

(10.26b)

(10.26c)

(10.26d)

(rO.26e)

lO.26a,

(10.27)

Fiour€ 10.19(0) Full-foce gosket dim€nsions.

3r3

I oraa.. corDmoxt 2

F'tt - F'.e ,/1eturtL.f 'be.z - |ltt=7 qrlE

5 to9

sA-rzt cv.l ro^0 aro fori calcul lo|trlj= 6tor + ti'.r = l,tq,tqQ

^- =ebcL'of wd/s. d w-Jt= 17., n )c.1b

Itt,=26ra6r = | 18,q86 ^,= N.A \^.2 t(.- z"Q tJte

tlEoo ftl trtD = t/tt.r1'

= 4 t r' 5 t1 w=tl .+^,1t.= 706,t16lr )qzoo Fi E=cnr/.= 1t,168 tr'at=r./:. ttc,= 94a, 4q6

14 reL wr'=tt +lt.r+t|'r = 7At,1l3I ltv ar

5 it - r)trl. - 1,'t,o4\ r. - r+ .5r, - tl,t 1 Pt-,d6- tZt,r72,tt-E-ED- 1L,17-4 |'-Jtr+.,+tEl - { Zt, & - rrir - l1{,,rO{

V. - ?t1,h.6

6 = !!:!E!-*9=- = 2.9915' (. =bg.4Lgl= = r.oz?2'

,.o=\a -l|.= LlAtSoa *" = h.L- = o. 11 f6 a.=8.t-t'= 1C4 tozL

8 t:* 3nls3 CIICULllDOT-€4din, 6 x er,o sur rrcrort16!, |kb, ti = nJIEI = 1tt 71 ?tar.rErFEgr: Pb/f - t1,71o ?,1r-r. tla 9=-.t/f -u.- ?,95':-j.5(5r+ s.l'Jl3r +5'l - q' ToL,?r L

'-i - 0 t118t/iDrrt 5trE$ al tolt oto.t..-:;iffi = n,qL4 ?,. .-lh&.'-20116t\I'ti;-,.217

16- 2"uotst t1.tt1"

/ nr.3s torMuta tactott

. -r.+l r l.:5t calak+r = l.4IO

-afi - t. ool l

, -.tla - O,o.lo1r -t+a - r,o1',

-=rL,rt - H?.l6

t br rp.d.. o....' 2. + r.*htrt l;EI 2.+l

C-r-a- Od.-

o.d.a- L'r.+

Figurc lo.2o Fvll-Goce flonge lomple colculorion 3ho6t

3I4 II.IND ?U\NO!S, COVER PTATTS, AND TIANOTS

ln a similar munncr, the distance or moment &rnt k)w&rd the inside diameter h1;is determined as

C_B -0"=+('!:'\5 \O + c /. C_Ba:

2

, = .r(?*)

,= *(?;And substituting Eqs. 10.269, h, and i into Eq. 10.26f gives

c-B

and solving for ft6 gives

(r0.28)

9n9" t"_*9 gasket loadings and two moment arms are determined, theanalysis of the flange is the same as for other flanges. The method is equallyapplicable to integral flanges, loose flanges, reverse flanges, and any othei typeof flange. It is important to remember to use a..soft,'gaiket that keeps tfre tottloading within acceptable limits.

Example_10.13. A welding-neck flange with the same geometry as that inExample.10.9 except for the thickness is used with a full-facJ gasket. The designpressure is 320 psi and the "soft" gasket is vegetable fiber with m = 1.75 andy = 1100. What is the minimum required thickness?

Solution

l. Determine the lever arms of the inner and outer parts of the sasket:

h. = G - B)(28 + C\ _ (22.5 - t0.75)\2 x 10.75 + zz.5)'" 6(8 + C) 6( 10.75 + 22.5\

-

(r0.26f)

(to.26s)

(10.26h)

(10.26D

6(10.75 + 22.5)

ha = 2.5915 in.

IO,I2 FUIL-FACE GASKET TIANGI 3I5

.. (A - c)(u + c) (26.5 - 22.5t(2 x 26.5 + 22.5)

6(C + A) 6(22.5 + 26.5\

hb : 1.0272 in.

Determine the gasket dimensions:

G = C - 2hc:22.5 - 2 x 2.5915 = 17.317 in.

. c-B\ (22.5-10.75)t =- = 2.93?5 in.-44

y= l10O and m = 1.75

Determine the loads:

n = Tc.p = +,r7.317)r(3z}) : 75,3684'4

H o = ZbtGmp = 2(2'937 5) n(17 .317)(r.7 5)(320) = 178,986

/r,^\ .? sqr 5)HI : \rt)H' = @;^t'is,r8rl: 45r'55e

Wt= H + Hp+H; = 75'368 + 178,986 + 451'559 = 705'913

Hsn = blrGy = (2.9375)r(17 .317X1100) = 175 '79o

/ h^\ /, 5015\Hb'= \nHc'= @rfi5rw= 443'4e7

W,a = Hay + Hb, = 175,799 + 443,496 = 619,286

Determine bolting requirements: A' is the greater of Wa/S"ot W^z/S.

A^ = 36.76 in.'z based on ffiAt = 36.8 in.'z based on 16-2-in. diameter bolts

W = 0.5(A^ + A) : s.5136.76 + 36.8)(19'200) = 706'176

Determine flange moments at operating condition.

Flange Loads

H, = loB2p = f,go.t s1,1szo) = 29,oM

Hr = H - Ho = 75,368 - 29,444 : 46,324

316'IIND

ILANOIS, COVER PI.ATES, AND FI.ANOES

Lever Arms

[o=R * 0.59' = 2.5 + 0.5(3.375) = 4.t875 in.

lr = 0.5(ft i g * h) : 0.5(2.5 + 3.375 + 2.Sgt5\ = 4.2333 n.

Flange Moments

Mp : Hphp : (29,044)(4.187 5) = 121,622

M7 = H7h7 = @6,324)(4.2333) = 196JOq

M. = Mo + M, = 121,622 + 196,104 = 317,726

Determine flange moment at gasket seating condition.

Flange Load

He = W" - H = 706,176 - 75,368 = 630,808

Lever arm

h',_ = hahh - (2.5915X1.0272)"o - ho + hb- Q.915) + [onr) = 0'7356 in'

Fhnge moment

Mr: H6h'[ = (630,808X0.7356 = a6a,O22

All flange geometry constants are the same as in Example 10.9.Calculate flange stresses. Assume flange thickness t = 2.03 in. This isset directly from the radial flange stress at the bolt circte which is

sF=, M' - 6(4u,022)" t"(rc - Nd) (2.0j),(22.5t - 16 x 2)

SF = n,a@ psi < 17,500 psi allowable. stress

Inngitudinal Hub Stre ss

L=

Sr=

+a= 1.0021 +0.0407 = 1.0428

\4e,022)

6.

1

te+lT

L'?n (1.0428X3.375f(10.7s)

Sa = 3634 psi < 26,250 psi allowable stress

IO.I4 f LAT-TACE FTANGE WITH MEIAI-TO-MTTAI CONTACT

Radial Flange Stess

3t7

^ tlte + l\M" (t.4'104)t4M,O22)Lt2B (1.042Sx2.03)110.75)

Sn : 14,77O psi < 17,500 psi allowable stress

Tan Be ntial F lange Stre s s

VMs,: --+ - zs,t'B

Sr = 3457 psi <

Problcm

2.29(464,022)(2.03f (10.75) - (1.39)(14,77 0)

17,500 psi allowable stress t

10.12 Assume a flange with a flat-face gasket has an applied moment ofMo -- 464,000 with a bolt circle of C = 22.5 in. using 16-2-in. bolts.What is the required thickness of the flange if the allowable skess is15,000 psi?

IO.I3 FTANGE CATCULATION SHEETS

Answer: t,"q'd. = 2.19 in.

Calculation sheets are included for the following types of flange design:

Sheet 1. Welding neck flange with ring-type gasket

Sheet 2. Slip-on or lap-joint flange design with ring+ype gasket

Sheet 3. Ring flange with ring{ype gasket

Sheet 4. Reverse welding neck flange with ring-type gasket

Sheet 5. Slip-on flange with full-face gasket

Sheet 6. Welding neck flange with full-face gasket

IO.14 FLAT.FACE FTANGE WITH METAT.TO-METAL CONTACTOUTSIDE OF THE BOIT CIRCLE'",'O,''

Rules for the design of flat-face flanges with metal-to-metal contact outside ofthe bolt circle are given in Appendix Y of the ASME Code, VIII-I. The rulesare for circular, bolted flanged connections with identical and nonidentical pairsof flanges. The pairs of flanges that are in metal-to-metal contact across thewhole face and the gasket load to compress the gasket are small. (see Fig. 10.21)The rules also apply to identical pairs of flanges with a metal spacer added at theouter edge between them.

torrD afto aorY calcut '|or{S

It

ro/ro r llvar /lr|r

,!lsr + 5r,.! tsi + S,j =

aEotr.

SlSr + trlo'.5(t, + t'i =

lr b.l iFdne d.od.2. + r,'.hi?t /i"L,Da.rr .d i. h .br .qd'or b', v -;=;-

Csrea- O.r.

6.r., |{Jih.'-

Sh6.r l. Welding neck flcng€ with ring-ryp€ g6kei. (Co,.rrresy c+W Toylor-Bonnoy Div., Toytor Forg€.)

3t8 3r9

Shsor 2. Slip-on or lop-ioint flonse v.irh rins-typ€ sd3ker. ( Courre3y G+W Toylor-Bonney Div,, Toylor

Forge.)

rOAO l val ^rr - |rorlxl

-----J@!!

tl0 ll.lND .ANO!s, COV!n p[AtES, AND trANOls

ning flonga yrith ring-iypo golk€t. (Cou|.tosy G+W Toylor_Bonney Div., To),lor Fo;Shs€i 3.

The.basic development assumes that the flanges are in tangential contact at theouter.diameter or at some point between the b;ft circle ;; th;i,r'te. oiam"te,at a distance llc from the boit circle. Ttre gasket is assurneJi. ti-rJrilr"o,ng u"ageneraGs a negligibre load during operation and is located in rir,"iia tn" u"rr"twall. The major difference uetrieen ttris ryp",f d;" ;il'tr,"'Jrg_typ" o"_scribed in Section 10.9 is the additional pryi'nl ;fil;;;iliit; contact ofthe two flanges.To organize the calculations svsiemadcally, it is necessary to classify assem_blies and to categorize each individual nann".

-- "--veJqr !v !ru!.

C hs s ifcation o! As s e mb ly

Class l. This is a pair of flanges that are identical except for the gasketg'oove.

Class 2 . Ii consists of a pair of nonidentical flanges where the inside diameterof the reducing flange Jxceeds one_half the U.it "i."f" Ou."to*

Class 3- This is a flange combined with a flat head or a reducing flange wherethe inside diameter is small and ao", oot "*"""0-

on"_tlif" fieiort ctctediameter.

2olrr --- 1,-.:...:.---:-:-=-._.....Y-

rv.r - brc,la,;,.-=^-=Zi w-,tt,- s-,,.-

trll*' Lsc I -e,r/. c"I.

5

.Jlrb + rEl E

t-

k_/, trb rlttat lt,

6 rxpt coxrr^'m l-tL3,nfl;:r,,1*-;E;F-

7

,=J+l @1, 5:t

Co"ur.d- O.r.

G.*.d_ ili5.,--

I prrr 2

4 to^o eno rotr cercur.r6ii

^-,44 ||-1/s.' w-tts,-

r tavl| atrI

1."'.e

iD=J(C+r,-2$-tJ=ti=.5(c-G)

^,=.r{.-F-iiiiiiS,iiiidiii'fi iF,il,,i -""'!,iit,y,,,:,.^*,-,.". M.l

I "',i,:. rtttsS carculaiox-{rf, rolin, 6 xAtD |lut t^cro$r{e, xu5, 5r =ti./I!r'

tcie. tle., S'= tufr/t, -ZS,tO,6rt.+rt/B= z

1:/ r rsr + sr,or J{s{ + s,, =r".".a,.s,o,r'r=+ [v-,xit(Jj ""+[+:!Ilr#rl t-)-9 "?.::' ttr$t cllcutano|r- sdd..

.rub, g = Ircllo,r rr =+.a, r

! rod,crae.,sr = pn./lt l;-------=_+r.^o fl!., S'= ^cfr/r' -Z',to.6rt.+n/R= Ur=drU

llji .la, +sa- rrs, + s,r =r-, n,.. s" 1^l 11 = f, [v

_ i::" i i,' ] = 7 trrls3 Forr ui,a ;^cort

? x r_..r-T*-+

e-rrr:-i-------.']lY El k_r'=

| [.n'rV I

t--|r---T-| | / h ]_-,_ ,__<.'t r ttI I 4'"F-r-bJ +il--l,o l.-----r=

aadl. <t_A:Cmptrd- D.h.-*----

Chxr.d- N!ib.r=--

Sh€€l ,a. R€verc€ w€ldins n€* flonse wirh rins-fyp€ goskst. (Court$y c+W Toylor-Sonney Div., TdytorFors€.)

321

IOAO : lavn rrr - |tot,Nt

,r lA-Cll2A+Cl=""=-:rc+.i-

h,i= l'.1"., -

7 ttrg totnutt ttctotg

I Llr !Fd'! .s.i' 2. + I rlrirt

C..'a.itr Dd.-

O.d.a-

Sh.at 5, Slip-on flons6 with tulLlcce soska. (Court€sy cfw ToyloFBonney Div., Taytor Forss.)

322 323

I o8rox cDaa!|no{t 2. 3

4 loao ^to

lott carcuraroxtw.r= bror + Nir = Ar = rr..tr.t wd/s. d w.Jt =

IrliWn=fi+i!+fi'?=

I TIVlt T

tc-tlra+o=-=-=iiT6-

E "i* ttatS CArqrlAnOx-Ot rato, o x |lo nof tac'otlttr|lr,h=nJr6rr.aEt tr- ar

= P*/If f

r@. 6E sr = ..1/ I -l5l rJl!, + li'Jl$ +5rl '

t^Dur str!3s At |olr ot<ll u

7 nra$ tourula tacloti

It !.lr D.d'. .B.dt 2. + t dldrlr I t lr r.drr.inlb.e!fu !r! I 2.+l

C6er.{-- o.r.-

6rr.a- f&iL'-

Shcet 6.

321 II.INO FIANOIS, COVTR PTATES, AND FI"ANOTS

Figur6 10.21 Flor-foce flonge wirh mebl_to-metolconto.r ouisid. of tfie bolt circle,

Catagory of Flonges

9i*g.IV^ t. This is an integral flange or an optional flange calculated as anmtegral flange.

ca'gory 2.It consists of a loose-type flange with a hub that is considercd toadd shength.

:l"g.y 3. A.loose-type flange that is with or without a hub or an optionaltype calculated as a loose type where no credit is taken for the hub in any case.

Once the class and category are established, the analysis is similar to thati"9" fr:g Appendix 2 flaige except for tt

" udaitionJrl"oings iuus"c uy ttr.prying effe{t where the contact near the ourcr diameter occurs . iiis Jontact torceH3 and its_moment arm lk involve an interaction between tfr" Uoft

"iongution anCflange deflection and the moments Me and Ms.

The required bolt load for operating conditions is

W:H+Hc+Hc (r0.29)

IO.I5 SPHERICALTY DISHED COVERS

'Ihe ASME.Code contains special rules for designing spherically dished covers

I-tll I TntnC, "ng flange. The formulas given in the code are approximate

Decause tney do not take into account the discontinuity existing between the

t0.15 sPHERtcattY orsHED covtRs

dished head and the flange ring. The flange thickness is set by the combinationof the circumferential ring stress and the tangential bending stress. Figure 10.22shows the head geometry and the loading applied to the ring flange resultingliom the reaction from the internal pressure against the dished head. It is derivedthus. Using the geometry in Fig. 10.22, the following is set:

adjacent side :(10.30a)

cos B, = adjacent side _\/L'''' - ( _ \/4L'1Err-=Ehypotenuse L' 2L'

membrane force in head to due to oressure : F' =pL2T

(10.30b)

(10.30c)circurnferential ring stress = 5" = pR

t:pD

2t

A ( outeide dianeter)

Fisuru 10.22 Sphcricolly-dirhed covcr.

t26 .tND ttANO!!, COVIR puTrs, AND traNols

Substituting l, = tr', cos B,; D = B; and , = (/ _ B)/2 in Eq. 10.30c,

^ F'cos B, I' a-B)/22When the value of ./c, is substituted in Eq. 10.30d, the equation becomes

s" = *2 ,,"o" 9,,,212@-B)/2

Whdn the value of cos B1 from Eq. 10.30a is substituted,

" - pn t/+Ld-=E. 4T A_B

]anCeryia gtrep] ring due to Me is as follows: from Eq. 9 of 2_51 of the ASMECode, Section VIII, the equation rs

- vM"-' T2B (10.32)

butY = (A + B)/(A - B),so

"= c^;(*=)

(r0.30d)

(10.30e)

(10.31)

(10.33)

Combining the circumferential and the tangential stresses in the ring grves

s, = s" + r, = r4@-=F * ( u"\(e * n\4r e= n - - \f,ri1, _, ) (to.34\

I€t

r = PBldt=E -^^ , _ Mo(A + B)8S(A - 8) s8(A _ B)

(10.3s)

Then

s,=f<n * fiut (10.36)

IO,I5 SPHERICATTY DISHED COVERS

l)ividing by S and rearranging terms to form a quadratic,

T2-27(F)-J=02p -r 1/@ufi or T:F t-1/vta1

(10.37)

(10.38)

This equation is identical with the equation given in 1-6(9) of the ASME Code,v[-1.

Application of this equation is permitted for either internal or extemal pres-sure. The term p is the absolute value for either the internal pressure or theexternal pressure. The value for ffi is determined by combining the momentsfrom bolt loading and gasket loading with the moment caused by the pressureloading from the spherical head on the inside edge of the ring. When this totalmoment is determined, the absolute value is inserted for Mo in the equation.Figure 10.22 shows the loadings caused by the pressure. In Fig. 10.22, theloading shown is due to intemal pressure. However, if the loading were due toextemal pressure, all directions of loadings would be reversed, but the effectivestresses on the ring would be the same. If external pressure were applied to thehead, it would have to be examined for compressive buckling that might set thethickness.

Example 10.14. A spherically dished head is to be bolted to the welding neckflange described in Example 10.9. The dished head is to b€ attached at the upperinside comer with the outside surface even with the ring's outside (see Fig.10.23). What is the minimum required thickness of the flange ring when thespherical head is dished to a radius of I = 28?

Fisure 10.23 Dimgnsioni of sphericolly-di.hed heod in

Ex. 10.14.

329 tuND traNots, covtR phTrs, aND H.ANotS

Solutlon. From the geometry of Example 10.9,

A = 26.5; a = 10.75; L=28=2(10.75\:21.5

D€termine the minimum required head thickness as follows:

, =t# =*ffii = 2.560 in.; use2.625 in.

From geometry calculations

L'=L+l:r.r.t*?@L' = 22.813 in.

cos p1 =22.8t3)z - (10.75

2(22.813) =0.972; h=13.626"

Membrane force in the head due io pressure is

r':pL -(25mV2l'sl' -n- :2r,s" =lo'zolb

Horizontal force = F' cos p1 = GO,24O)(0.972\ = 9950 lb

Vertical force = F' sin Br = O0,'2,40)(.236) = ?A2O lb

Total horizontal force = a(10.75)(9950) : 336,000

Total vedical force : zr(10.75)(Z4ZO) : 91,76

Moment at gasket seating condition is

Load furn Moment

He = W. = 700,800 ho = 0.5(C - G) : 3.729 Mo = H6h6: 2,613,000

Moment at operating condition is

t-oad Arm

IO.I5 SPHERICAI.I"Y DISHED COVERS

Moment

IIp = 11,: 8l,700

He = Hp:250,600

Hr = 336,100 hr = 0.5(R'r g'l h6)

= 4.8O2

Hn = 336,1.0O -hn : -O.5(T - t)

ho = 0.5(C - 8) : 5.875 Mp : H6hp :

he = 0.5(C - G) : 3.729 Mc = H6h6 =

M7 = H7h7 :

Ma = Hnhn :

480,000

934,000

1,614,000

- 168,050r+441,300: -0'5T + 1.313

Mo: Mp + MG + Mr + M, : 3,469,60 - 168'050f

The minimum thickness at gasket seating condition is

(2,613,0W)1?!ll_!!JI)_ = ?,, RsnF=u and r: (nJ:00xmj5)(26i - l0J5): rt'orv

T: F + \/F' + J = *2-s5o = 5.732 in.

The minimum thickness at operating condition is

" _ (2s00x10.75)y'4f11.8 I tP-:-(l0l75f = o 54o' 8(17,500x26.5 - 10.75)

If we assume T = 5.75 and M0 : 2,503,000,

r :32.8so 'ffi;ffi = 31.467

T = 0.540 + + (3r.467): 6.175 in.

If we assume T = 6.25 and Mo = 2,419,000,

I=32.850"?*i:# =ro4tlT = 0.52+0 * y'1g5a0t + (30'4lD = 6.081 in.

The minimum thickness is approximately 6.108 in. Although exact thickness

can be determined , T = 6.25 in. is satisfactory. I

330 EuND frANOrs, covER ptAlrs, AND FtANGTS

l'robhms

10.13 A spherically dished flange with an outside diameter ofA = 36 in. andan inside diameter ofB : l8in. is subjected to a gasket seating momenrof 3,500,000 in.lb. The allowable tensile stress of the flangi materialis 15,000 psi. What is the minimum required thickness at the gasketseating condition?

Answer: t^n = 6.24 in.

10.14 For the flange in Problem 10. 13, what is the maximum allowable flansemoment if the allowable tensile stress is increased to 17,500 psi and tieflange thickness is set at 6.25 in. ?

Answer: M51o* = 4,|O2,NO in. lbs./in. circumference

NOMENCTATURE 33I

NOMENCTATURT

lndividual nomenclature is used throughout Chapter 10. It is usually noted nearto where used. The following list gives some of the general nomenclature.

A : outside diameter of flange (in.)

I = inside diameter of flange (in.)

C = diameter of bolt circle (in.)

E : weld joint efficiency

G : diameter of gasket reaction (in.)

1u' : gasket seating width (in.)

N-6 = minimum gasket width (in.)

P = total load 0b)

R = spherical radius of gasket surface (in.)

S = allowable tensile stress (psi)

S. : allowable bolt stress, room temperature (psi)

56 : allowable bolt stress, design temperature (psi)

design pressure (psi)

nominal thickness of head (in.)

effective diameler of flat head (in.)

minimum requircd thickness (in.)

seating stress of gasket material (psi)

= modulus of elasticity (psi)

= flange loading for design condition (lb)

= flange loading for bolt-up condition (lb)

= basic gasket seating width (in.)

= effective gasket seating width (in.)

= outside diameter of gasket (in.)

= inside diameter of gasket (in.)

= actual bolt area (in.'z)

: flange moment for design conditions (in.lb)

Yory =E,

W^t

Wz

bn

b

OD

ID

Ms

332 tuNO fl-ANOts, COVTR PLATES, AND lt-ANOts

o,

a"

lL

Mn

0*o,

',

flange momcnt for bolt-up condition (in.-lb)

maximum bending stress (psi)

m dmum deflection (in.)

maximum rotation (radians)

or : tangential stess (psi)

ai = radial stress (psi)

a6 = circumferential sness (psi)

poisson's ratio

REFERENCES

ASME Boiler atrd Pressure Veso€l Code, Section VIII, Division I, Pressure Vessels,ANSVASME BPV-Vn-I, Amedcan Society of Me.hanical Engine€$, New Yo*, 1983.

ASME Boiler ard Pressure Vessel Code, S@ton I, Pover Boilcrs, ANSVASME BPV-I,American Society of Mechadcal Engineers, New York, 1983.

Wate$, E. O., D. B. Wesstiom, D. B. Rossheim, and F. S. G. Williams, "Formulas forSft€sses in Bolted Flanged Connections," f/dns. ASME, vol. 59, 1931, pp. 16l-169.Wate$, E. O., D, B, Rossheim, D. B, Wesstrom, aDd F. S. G. Wilham's, Development ofGeneral Formulas for Bohed Flaryes, Taylor Forge atrd Pipe Works, Chicago, 1949.

Waters, E. O., and R. W. Schneider, "Derivation of ASME Code Formulas for lhe Designof Reverse Flanges," WeldittS Research Council, Bulletin 262, October 1980, pp. 2-9.ANSI Standard 816.5, "Pipe Flatrges and Flanged Fittings," American NatioMl Srzndardshstitute. New York.

API St ndard 605, lzrge Diametcr Carbon Stee, FrarSer, ANSVAPI Std. 605, AmericatrPetroleum ltrstitute, Washington, D,C.

ANSI Staftlard B16.24, 'Bronze Flanges ard Fittings, 150 aod 300 lb," America! NationalStandatds Institute, New Yo*.Rossheim, D, 8,, and A. R, C. Markl, "Gasket Irading Con$tants," Mechanical Ebgi ceri g, Vol. 65, September 1943, W. 647-648.

Raut, H. D., and G. F. Iron, "Report of Oasket Factor Tests," Welding Research Council,Bulleti! 233. New Yort. De.ember 1977,

Modern Flange Design, Bulleth 502, Tttt ed, , G aDd W Taylor-Bomey Division, Southfield,Mich.

Roark, R. J,, FormnJas for Stress and Strain,3d ed., McGraw-Hill, New York, 1954.

Water$, E. O,, 'Dedvatior of Code Fordulas for Part B Flanges," WeUinB ResearchCoun il, Bnlletin 166, October 1971, pp. n47.Schneider, R. W., and E. O. Waters, "The Backgrcund of ASME Code Cas€ 1828: ASimplified Metbod of Analyzing Part B Flanges," Tranr, ASME, Jounal of Presrure VesselTechnology, Vol. 100, No. 2, Mt! 1978, pp. 215-219.

Schneide., R. W, and E, O, Wate$, "The Application of ASME Code Case 1828," Irarr,ASME, lournal of Pressure Vewel Technology, Vol. 101, No. I, February 1979, pp.87-94,

12.

13.

11.

14,

S|EUOORAPHY 333

BIBTIOGRAPHY

Bhch, A. E., and A. Bazergui, "M€thods of Analysis of Bolted, FlaDged Connections-A Re-view," Ecolc Pol)4echnique, Monti€al, Camda, Jaluary 1981.

InterprEtive Study on the Design of Notr-Circular Flaoges and Flanges with Extemal Loads" (aprivate report to PVRC , May 23 , 1979 .)

Rrut, H. D., A. Bazeigui aod L. Marchand, 'Casket l€akage Trends" (A private report to PVRC),April 1981.

Rodsbaugh, E. C., atrd S. E. Moore, "Evaluation of the Bolting and Flarges of ANSI 816.5Flarged Joints-ASME Part A Design Rules," ORNUSub-2913-3, Oak Ridge National Labo-ratory, Oak Ridge, TeDn., Septenb€r 30, 1976.

Timo$henko, 5., Theory of Plates and S&ells, McGraw-Hill, New Yo*, 1940.

334 335

CHAPTER llOPENINGS, NOZZLES, AND

EXTERNAL LOADINGS

336 OPENINGS, NOZZIES, AND EXTERNAI' LOADINOS

I I .I GENERAL

All process vessels require openings to get the contents in and out For somc

vessels, where the conients miy be large or some of the intemal parts may need

frequent changing, access is made through large openings in which the entirc

heai or a secioriof the shell is removed. However, for most process vessels,

the contents enter and exit through openings in the heads and shell to which

nozzles and piping are attached. In addition to these openings others may be

required, suCh- as those for personnel entering the vessel through a manway

opining. Other openings rnay be necessary for inspecting the vessel ftom the

outside-thmugh a handhole opening , and still others may be required-for cleaning

or draining tie vessel. Thesl openings do not always have a nozzle located at

the openin!. Sometimes the closure may be a manway cover o-r handhole cover

that ii eithlr directly welded or attached to the vessel or a built-up pad area by

bolts.For some nozzles, additional loading to the iniernal or extemal pressure may

be innoduced from dead loads ftom equipment and piping and ftom thermal

expansion flexibility loadings from the piping and equipment motions' This

Ml

f-F^l

figura tl.l Applied Pras$rre ond externol loodings on noz-zle' FigurG ll.2

I I.I OENERAI,

additional loading may require compensation as well as what is necessary to

rcsist the internal and external pressure loadings, as shown in Fig. 1l.l'Openings and nozzles similar to those occurring in pressure vessels also occur

in piping. This is the case where a branch run is attached to the main run ofpiping. The branch-to-run intersection is subjected to the same pressure and

ihermal expansion loadings as those applied to a vessel nozzle. Although the

nozzles have a similar construction, usually an important difference lies in the

relationship between the ratio of the nozzle diameter to the vessel diameter and

the branch diameter to the run diameter. For pressure vessels, this ratio d/D ismuch less than for piping. In many piping systems, this ratio may be very close

to 1:1, as as shown in Fig. 11.2.In designing openings and nozzles for resisting loadings from internal and

extemal pressures and from external loadings, two types of stress conditions are

important. Fkst, the primary rnembrane stresses in the vessel or run pipe, that

is, the necessary stresses mahtaining static equilibrium must be kept within the

limits set by the allowable tensile stresses. Second, the peak stresses caused by

abrupt changes in the geometry at the nozzle-to-shell comer and cause stress

concentrations must be kept within acceptable limits. These peak stresses are

important in a fatigue evaluation where the design life of the nozzle and the other

Voriotion in d/D rotio of nozzles ond piping.

OPININOS, NOZZICS, AND EXTERNAI TOADINOS

vcss€l parts or piping system are established. A slight change in the details at thointersection may enable the vessel to operate through many more cycles ofpressure and temperature loadings.

Detailed rules for designing vessels and piping to accommodale the primarymembrane stresses and loadings from intemal and external pressures are givenin codes and regulations such as the ASME Boiler and Pressure Vessel Codeland the ASME Code for Pressure Piping B31.'z In addition, some design rulcsare given in the more advanced sections of these codes to permit consideringshess intensity factors (SIF) and stress concentration factors (SCF) in deter-mining peak stresses. The peak stresses are used to determine the design fatiguelife of the vessel. Other codes do not mention peak stresses or fatigue evaluationsand leave the latter as the designer's responsibility. At the present time, none ofthese codes contains detailed design rules for the consideration of extemalloadings from either dead loadings or piping expansion loadings.

I1,2 Stresses ond Loodings ot Openings

Both single and multiple openings require calculations that show that the stresses

and loadings in the shell and head are kept within acceptable limits. Singleopenings are calculated by the reinforcement method, whereas multiple open-ings are calculated by either the reinforcement method or the ligament efficiencymethod. In both cases, the primary stresses are effectively kept less than theallowable stress by replacing the area removed for openings.

For a single circular opening in a flat plate with infinite boundaries in twodirections (not through the thickness) that is subjected to applied forces andstresses along opposite edges of the plate, stresses are increased above thenominal applied stress in the unperforated plate. The stresses decrease awayfrom the opening until the nominal stress in the plate is obtained. The ratio ofthe stress at the examined point divided by the nominal stess is the stress

intensity factor.The shess intensity around an opening may be expressed either in general

terms of applied stresses and geometry or in location of the considered point. Thebasic equation at an opening may be represenied in terms of o and 0, with theangle of the considered point measured from the loading axis.r For the loadingshown in Fig. 11.3a, the or is axial and 0 = rl2 at the maximum sfiess loca-tion. ForFig. ll.3b,lhe q is axial and 0: rl2 atthe maximum stress locationand, o2:0.5or is at right angles to a1 and 0 : 0 at the maximum stress

location. For Fig. 11.3c, the a1 is axial and o2 : o1 is at right angles to o1.

Values of 0 are the same as for the cylinder. When two loadings or stresses are

involved, the effects at the maximum stress location are added. The basiccouation for direct s[ess is

" = Zl'. (;)'] - i [' -' '(;)']

*' '' (11 . l)

I I.2 STRESS€S AND TOADINGS AT OPTNINOS

Figure ll.3 Two-direction'lood combinorio's on flot Plote with circ"ldr oPening'

The basic equation for the maximum stress at the edge of the opening in terms

of the component of stresses in each direction is

(1r.2)o^o:3o1 - o2

where the values of or ?rd oz include plus and minus signs depending upon

whether the applied stress is tension or compression. The stress intensity factor

at the edge oi circular openings for various ratios of applied edge stresses is

siven in Table 11. l.The stress intensity factor for various combinations of stresses is maximum

340 OPININO5, NOZZLES, AND EXTERNAT IOADINOS

Tqble I I.l Streu InlenallyFoctors for Vorious Rolios ofApplied Stress

Stress Ratio

1:0(axialonly)2 : I (cylinder)

I : I (sphere)

at the edge of the opening and decreases away from the opening until the stressapproaches a nominal shess factor close to 1.0.

Using the following nomenclature, various formulas for different combina-tions of applied stresses are developed:

r = radius of circular opening in plate (in.)

r : distance from centerline to point of SIF (in.)

For applied stress ratio of 1 : 0-the condition of an axial tension loadonly-the basic equation for the stress intensity factor is obtained by solving Eq.1l.l with 0 = rl2 wherc cos 20 = -1, giving

3.00

2.50

2.00

"' = Zl'. (;)l - i l' -''0']t-'r

',:ilr. (;)'-,(;)l (1 1.3)

At the edge of the opening, the stress iniensity factor is determined from Eq.ll.2 assuming that o1 = or and or = 6'

o^:3or-0=3.00or (r 1.4)

Substituting various values of r for x in Eq. 1 I .3, stress intensity factors atvarious distances from the edge of the opening are

r 3.00sr21 1.15ar

3r l.Mo14r l.O4o1

For an applied stress ratio of 2 : l-the condition of a cylindrical shell under

l l.2 STRESSES AND IOAOINOS AT OPININOS 341

intemal pressure-the basic equation for the stress intensity factor is found by

combinine the effects of sfesses in two directions according to Eq. l I l:

",, : Zl'. G)l - ; t' . '(;)']

cos 20 where o = I* =; i' . (;)']

The general equation is the summation of the two stresses and cos 20 = + 1

", =Ilz* /r)'* r - 3fr)'+ r + (r)'- t- rf:)'l4L- \r/ \x/ \r/ u/ I

", =ilo. ,(')' . ,(;)']

cos 20 = -1where 0 = 0

(l l.s)

At the edge of the opening, the sffess intensity factor is determined from Eq'

1 1.2 assuming cr : or vnd 02 = 0.5or so that

c* = 3or -O.5or = 2.5oot (1 l.6)

Substituting various values of r for.r in Eq' 11.5, stress intensity factors at various

distances from the edge of the opening are

r 2 5oor

2r l.23at31 1.09or

4r l.05or

For an applied stress ratio of 1 : 1 in a spherical shell or hemispherical head

under internal pressure, the basic equation for the stress intensity factor is

resolved by combining effects of stesses in two directions accoding to Eq. 11.1

that gives

-i['-':0']*'ze

'=t*$' (l l.7)

At the edge of the opening, the stress intensity factor is determined from Eq' 1 I .2

assuming that at = or dllrd cz: 6t as given by

c*:3c1 - ar=200or (11.8)

Substituting vsrious vulucs ol'r li)r r in Lq. 11.7, stress intensity tactors atdislances lrom the edge ol'the opening are

r 2.O0oy

2r l.25or31 1.11o1

41 1.06s,

Exarnple 11,1 A vertical vessel under intemal pressure and dead load containsan opening that is qubjected to applied stresses. The dead load stress is equal tothe circumferential pressure shess. For this stress condition, what is the basicequation for the stress intensity factor at any location from the center of theopening?

Solutian. The applied stress in the circumferential direction is o1, whereas inthe longitudinal direction it is the longitudinal pressure stress minus the longi-tudinal dead load shess. This equals oy, = IO.5o1 and c2DL: -or and thesummation equals -0.5o;. Using Eq. 11.1, the basic equations are

/r\ 4l+31:l lcos 20 where 0 = "

\r/l 2

cos 20 = -l

G): *' 2o where o: trcos 20: *l

oPlNrNoS, l{onfis, AND ExftRNAr toAD|NOS

Exarnple 11.2 For the vessel described in Example 11.1. what is the max-imum stress at the edge of the opening according to Eq. I 1.2?

Soltttion. Assuming that or = o1 and o, = -0.5 rr, the equation for max-imum stress at the edge of the opening is given by Eq. 11.2 as

",=url,. (t'] -;['of /r\21 -loa: --l l+l-l l+ _ll+4L V/l 4L

,, : "ol,

+ rG)' +,.'(:) -' - (")'.' ., (r'],, ="olo. (;)' .'(')'] .

o,o^: 3oy - oz = 3ot - (-0.5ar) : 3.50ar I

I'roblems

I I .1 What is the stress intensity factors for a plate under a stress ratio of 2 : - Ifor the vessel given in Example 1 1.1 at the edge of the opening and at

distances of 2r,3r, and 4r?

I I.3 THEOTY OF REINfORCID OPENINOS

Answeri Incat\on SIF

r 3.50

21 1.2O

31 1.05

41 1.Oz

11.2 In the ASME Code, VIII-2, a local stress region is one in which the stress

intensity of 1.15, does not extend more than VRt In terms of the radius

of the oiening r, how far ftom the edge of the opening in-a hemispherical

head under iniernal pressure is required to have a stress of 1' 15. assuming

S^ = o1?

Answert Ttrc SIF becomes equal to 1.1S. at a distance of 2'66r from

the edge of the oPening.

11.3 As in Problem 11.2, what is the distance from the edge of the opening

along the longitudinal axis in a cylindrical shell under intemal pressure

if required to have a stress of 1. lS,?

Answerz T\e SIF becomes equal to l.lS. at a distance of 2'90r from

the edge of the opening along the longitudinal axis'

I I,3 THEORY OF REINFORCED OPENINGS

AsdescribedinSectionll'2,thereisanincreaseinbasicshessesatanopenlngin aflat plate or shell under edge loadings due to the discontinuous pathway for

the loms (anO stresses) to pasJ from one side of the opening to the other side'

When this happens, otler pathways have to be established- in.order to keep the

orimarv sresies at an acce-ptable level. The basic theory of reinforced openings

ir to suppty pathways with additional material in the region of the opening to

"u.ty tiJ ioud. Uy

-ttt" opening. In designing process equipment and other

pr"s*r" u".t"l., tiis pattrway iJ supptied by the thickening of the basic shell or

iozzle material and Uy adding material such as a pad, as shown in Fig' l1'4'placement of the additional material is important. It must be sufficiently near

the opening to be effective; and yet, it must be added with caution to prevent

OPININOS, NO2IIES, AND TXTENNAI I.OADINOS

Figt re I L,4 ,{€lhods of odding r6inforc€msnt moteriol. (Court*y Arnericon Socisry of it€chonicol Engine€rc,From Fis. UM6.l of the ASr'tE Code, Vlll-l)

another problem such as high thermal stresses. Investigations by the pVRCa andothers indicate that the placement or location of the reinforcement is important.On most pressure vessels, the reinforcement is added to the outside as shown inFig. 11.5. However, on some vessels the reinforcement is added on the insideas in Fig. 11.6; and on still others, some of the reinforcehent material may beadded !o both the outside and the inside as in Fig. 11.7. The best arrangementfor reinforcement is the so-called balanced reinforcement which consists ofabout 35-407o of the area on the inside and about ffi-C|Vo of the area on theoutsidg. On many designs, it is difficult to place any reinforcement on the inside

Tn

.r!,

\,i:\:;i*

T

figure I L5 Reiniorcemsnl od&d to outside of oPening'

lT

5

\tl.t

Figur€ I1.6 R€infor.€ment odded to intid. of op€ning'

345

t.l6 OPININOS, NOZZTES, AND IXTIRNAI. I.OADINOS

Tn

{.a\i,:'l\i-;.jl:i

T

Figore I I .7 R€inlorc6manr odded to bofi insid€ ond o'rrside.

either because it is not accessible or it interferes with flow or drainage. Thebalanced reinforcement is often used at manway and inspection openings whereno nozzle is attached.

For applications in design problems, where the reinforcement requirementsare established, the method of replacing areas is chosen rather than a method thatbalances loads or stresses . An area at the opening for carrying primary loads andshesses is removed. Thus this required area must be replaced by another areaadjacent to the opening that is not used for that purpose. It is desirable to replacethat area required for primary loads by an adjacent reinforcement area. Withinthe reinforcement limits, the reinforcement areas are assumed to have the sameload-carrying capabilities as the area removed for the opening. Consequently,when the reinforcement areas are equal to or exceed the required area, primarystresses have been restored to as near the unperforated plate as possible.

I I.4 REINFOR.CEMENT LIMITS

As described in Section I 1 . 2 , the stress intensity factor for an opening in a shellor head is highest at the edge of the opening and decreases away from theopening (based on a shell wall with constant thickness). When the effectivethickness is increased, as happens with added reinforcement material, the ever-age stresses are lowered. Limits of reinforcement are set parallel and perpendic-

I I.4 RTINFORCEMENT TIMITS

ular to the surt'ace of the shell. These are set at a point at which it is l'elt that the

added reinforcement within the limits is effectively helping to replace the metal

removed at the opening.Two formulas are used for setting the limits measured from. the opening

c9!!er!!e3!91g.1@g!qg".jg@f, with--th" t-c* offlre two inswEis used.

The first answer is equal to the diam6fer of the opening d. The second limit equaltothesumof ?l + T^+ O.5d. AsshowninFig. 11.8, the thickness ofthe nozzle

wall usually determines which of the two limits controls. At a distance d fromthe centerline without reinforcement added, Eq. 1 1.5 for a cylinder gives a SIFof 1.23or and Eq. 11.7 for a sphere gives a SIF of 1.25 q. With additionalreinforcement material, the nominal stress is reduced close to that in an un-

perforated plate.If z nozzle is attached at the opening, it also offers reinforcement arca

available for replacing that area removed from the vessel at the opening.lU-

@iruegAUC!.d9l4bl9-lgled on the wave

damping length of a beam on an elastic foundation. For a cylindrical shell, thislength is a function of l/B, where B for a poisson's ratio of 0.3 is equal to1.285 /Y rt .

The vertical limit was set in the ASME Code as 2.5 [ many years ago when

an assumption was made that r/r of 10 was to be usEif-This limit is about rightfor an internal pressure p of 1200 psi and S : 12,000 psi. The development of2.5 T" is

I

_l\/,- p r.285\/o.tv

= O.246r = 2.46t1.285

For code application, the number was rounded off to 2.5 f. With the wide range

of r/t ratios, which are currently used in process vessel construction, some ofthe codes are setting this reinforcement limit in the vertical direction by

L = O.7S !r-I" ( 11.e)

where L = reinforcement limit perpendicular to shell (in.)

/, = mean radius of nozzle opening in shell (in.)

4, : nominal thickness of nozzle (in.)

Each pressure vessel and piping code treats the calculation of the reinforce-ment area somewhat differently and establishes both parallel and perpendicularlimits in different ways. A discussion of the reinforcement requirements forseveral different codes follows.

Example 11.3 A cylindrical pressure vessel that is 60-in. ID by 6-in. thickcontains a nozzle that is 12-in. ID by 3-in. thick. What is the stress intensityfactor at the reinforcing limit that is parallel to the surface of the vessel?

Ti+Tn+o,td

348

Figur. I1.8 Reintorcomsnt limits porollsl to she'l sur{o.6.

I r.4 RqNtoRctMENT tlMlrs

Solution. The two horizontal limits are set by the larger of

d:12in.

0r

T, + T^ + O.5d = 6 + 3 + 0.5(12) : 15 in'

The limit is set by the 15 in. from the nozzle centerline.

The stress intensity factor is obtained by using Eq' 11.5 to give

",:Xlo. '(*)' . '(iJl = l rkn I

Problams

11.4 If the new reinforcement limit in the vertical direction were based on

r/r = 5 instead of r/r = l0 that was used to obtain the present limit of2.5T,, what is this multiplying factor in terms of T^ for r/t = 5?

Ans*ert The vertical limit is 1.74?1, based on the limit of r/t = 5'

11.5 For an allowable stress of 15,000 psi, what is the maximum design

pressure permitted for an rft = 5 based on the circumferential stress

formula given in the ASME Code, VtrI-l?

Answer: Based on the circumferential stress, the maximum design

pressure is 3260 Psi.

i t.l. t Reinforcemenl Rules for ASME, Section I

The rules for reinforced openings in ASME, Section l, Power Boikrs, permlt

using the replacement of both area and ligament efficiency provided certain

limits are met. Ligament rules may be used for repeating pattems of openings

provided the maximum diameter of any hole in the pattern does not exceed a

diameter determined from Fig. PG-32 of Section I. This figure is a plot of the

following equation with limiting values of K between 0.5 and 0.99:

a,,, =z.ts{WJr-x) (11.10)

with the limits of 0.5 < K < 0.99.

p = intemal design pressure of maximum allowable working pressure

(psi)

3IO OPTNINOS, NOZZTES, AND TXTERNAI. TOADINGS

4nn" : msximum alkrwable diameter of opening (in.)

D, = outside diameter of shell (in.)

fl = nominal thickness of shell (in.)

S = allowable tensile stress (psi)

For shelfs designed to PG-27.2.2.1 of Section I,

K: P?'- 1.6 s?i

For shells designed to PG-27 .2.2 of Section I,

(1 l. l1)

(t|.r2)

N o Reintorcement Re quhe d

The rules for openings in Section I contain proyisions for single openings whenno calculations are required to prove the adequacy of the shell. No calculationsare needed for a cylindrical shell when either of the following is met:

l. d /D < O.25 arrd d^* = 2-in. NPS.

2. d* = maximum diameter using Fig. PG-32.

For openings in forrned heads, no calculations are required to prove the ade-quacy if all of the following are met:

1. Actual center-to-cenier distance between openings is less than I where

A+B

r= PD"-- 1.82 St

L= 2(r + K)

r = --!2:--- 1.82 S?:

A and B : diameiers of adjacent openings (in.)

D, = outside diameter of formed head (in.)

{ = nominal thickness of formed head (in.)

Other terms are the same as for Eq. 1 1. 10.

2. The edge of one opening is no closer than I to the edge of the adjacentopening.

(11.13)

(1 1.14)

I I.4 RTINFORCEMENT TIMITS 35I

3. Except for hemispherical heads, formed heads have dlD = 0.25 Bndd^* : 2-in. NPS, which is the same as for cylindrical shells.

4. For hemispherical heads, the actual center-to-center distance in item I ismet. The value of K is one-half the value of 1( as determined bv Eo.11. 14.

5. For formed heads, d,- of Eq. 11.10 is met.

Size and. Shape of Openings

The shape of the opening when these rules are applicable is limited to circular,elliptical, or obround where the ratio of the large-to-small dimension is < 2.0.When the ratio is ) 2.0, special requirements may be necessary to resist anytwisting moment. For shapes other than those above, a special analysis or prooftest is required.

No limitations are set on the size of an opening by Section I rules. However,the rules in the text are limited to the following sizes:

l. For vessels 60 in. and less in diameler, the opening shall not exceed0.5 D or 20 in.

2. For vessels over 60 in. in diameter, the opening shall not exceed 0.33 Dor 40 in.

When these sizes are exceeded, suggested rules place the available reinforcingarea close to the opening.

Required Area of Reinforcemenl

The total cross-sectional area of reinforcement required for any plane through thecenter of an opening is given by

A = dt,F (r1.15)

where d = diameter of opening (in.)

t, = minimum required thickness of seamless shell (in.)

f = 1.0 except that Fig. PG-33 may be used for integrally reinforcedopenings, where permitted,

F=0.5(cos'z0+l) (11.16)

For torispherical heads when the opening and its reinforcement are within thespherical part t, is the minimum required thickness for a hemispherical headwhen the radius is equal to that of the spherical part of the torispherical head.For a 2 : I ellipsoidal head when the opening and its reinforcement are within

352 OPININOS, NOZTLES, AND EXTERNAT I.OADINGS

u circlc of 0.8 D, ,. is the nrinimum required thickness fbr a hemispherical heud

when the radius is equal to 0.9 D.

Limit of Reinforcement Parallel to Shell

The limit of reinforcement parallel to the shell measured on each side of the

opening centerline is the greater of (1) / or (2) T, + T, + 0.5d.

Limil of Reinforcement Perpendicular to Shell

The limit of reinforcement perpendicular to the shell measured either inward or

outward from the surface is the smaller of (l) 2.5 T, or (2) 2.5 T" + T".

Available Area of Reinforcement

If the thicknesses of the shell and nozzle are uniform and the reinforcement areadoes not extend beyond this uniform thickness, the following formulas may beused for determining the available area of reinforcement. However, if the open-ing and its reinforcement extend into areas with different nominal thicknessesand different minimum required thicknesses, these formulas are not applicable.

1. Area available in shell wall is the ereater of

At=(EL-Ft,)(zd-d)or

At=z(EtT'- Ft,)(T'+n)

2, Area available in nozzle walT is the smaller of

Az=2(T"-t,)(2.57,f

Az: 2(7" - t^)(2.57, + T)

(11. l7)

(11.18)

(11.19)

(r 1.20)

When two or more openings are spaced so that theil limits of reinforcementoverlap, the combined area is used and counted only once. The spacing betweenany two openings is to be not less than 1.33 4". For a series of openings in apattern, the area between any two openings equals 0.7f' of the area obtained bymultiplying the center-to-center distance by the required thickness as shown inFie. 11.9.

I I.4 REINFORCEMENT IIMITS 353

Figorc tt.9 binfo.cing requirom6nb for mlltiple oPnings. (courles), Americdn so.i6tv of r{e.honicdl

Ensin€€rs, Froh fig. PG-38 of the ASME Code, S€crion l)

Example 11.4 Figure 11.10 shows a 66-in. ID steam drum containing fivedifferent diameters and two types of nozzles. What are the nozzle reinforcementrequirements? The design data are

Design pressure = 2875 psi.

Design iemperature : saturation at design pressure approximately 689'F.

Materials 70,000 psi UfS drum plate.

Allowable stress at safiiration temperature : 16,800 psi.

Weld joint efficiency is E : 0.95.

Nozzles are 3+in., 4 in., 5 in., 63 in., nd 24-in. ID.

Solution.

1. Minimum required thickness of shell at E : 0.95 is

PR 2875 x 33' sE - 0.6p 16.800 x 0.95 - 0.6 x 2875 .- -

use 6i in. plate

a

bd

T"l

EX--

x x

2X

A B( D=LIMIT OF REINFORCEMENT

T^

ABCD-LlMll OF REINFORC EMENTFigurc 11.10 .,S.t-on,, ond ,,3€t-in,, nozzles.

2, Minimum required thickness of seamless shell is

PR 2875 x 33=61!rrn' sE - o.6P t6.800 - 0.6 x 2875

3. Minimun required thickness of nozzle is

Pr rR75 r'' sE - 0.6P 16,800 - 0.6 x 2875

4, Maxirnum diameter of a single uffeinforced opening is determined as

D, = 66 + 2(6.75y = 79.5 in.

K from Eq. tt.tZ ls ,*, ,r-32r-

: 287 5 x 79.5

l3 -ffifi;Js= r'tur

K'- = 0'99354

Area available is greater than atea requied and values of 4 that were assumed arc correct.

I I.4 RIINFORCEMENT I.IMITS

Using Eq. I L 10, the maximum diameter is

d-^:2.75 \rr3t675x(i-r0.9 = 4.814 in.

The only single openings are 5 in. and 6{ in.;others are not consideredsingle. Consequendy, all nozzles have the reinforcement area calculated(see Table 11.2).

All nozzles except the 24-in. ID nozzle are "set-on" type and calculatedtogether.

Reinforcement area required by Eq. 11.15 is

A, = dt,F : d(6.294)(1.0) = 6.294 d

8. Limit parallel to shell is greater of

d or T,+7,+ O.5d:6.75 + T.+ r

9. Limit normal to shell is smaller of

2.57, = 2.5(6.75) = 16.875 in. or 2.57,

10. Area available in shell wall is

A, = (2x - d)(r" - t) = (2x - d)(6.7s - 6.294)

A1 =(2X-dX0.456)

11. Area available in nozzle wall rs

Az:2Y(T^-t^)

Tqble ll.2 Reinforcement Cqlculqtions for 3| in.,4 in.,5 in., ond 6f in.Nozzles on o 66 in. /D Steom Drum

,X= parallel Y= normal

355

6.

7

6.'75 +d A, T" d T"+r 2.57, 2.5T, t,^ At Az A,

3.5 22.03 1.8'75 3.s 10.375 4.688 t6.87s 0.334 7.8'7 t4.44 22.314.0 25.18 2.125 4.O 10.875 s.313 16.875 0.381 8.09 18.53 26.625.0 31.47 2.5 5.0 11.75 6.25 t6.8'75 0.477 8.44 25.29 33.736.875 43.27 3.O 6.875 13.188 7.5 16.875 0.656 8.89 35.16 ,t4.05

116 OPININOS, NOIZI.Is. AND EXTERNAT IOADINOS

12. A 24-in. lD downcomer has "set-in" nozzle and calculates d/D =24/66 = 0.36 exceeds limit of 0.33D for vessels over 60 in. ID. Alter-nate rules are also recommended.

13, Minimum required thickness of a downcomer nozzle is

t- = 0.lm(12) : 2.289 in. use ?i : 5l in.

14, Reinforcement area required is

A, = 6.294(24) = 151.06 in.2

Limit parallel to shell is greater of

d: 24 in. or T,'t T, + O.sd = 6.75 + 5.25 + 12 = 24 n.

Limit normal to shell is smaller of

16.875 in. or 2.5/5.25't: 13.125 in.

Area available in shell wall is

At = (2 x U - 12)(0.456) = 5.47 in.2

Area available in r:r.rrzzle wall; wall extends inward for 6j in. as fullreinforcing limit oulward:

At = 2(t3.t25)(5.25 - 2.289) : 77.73 in.'2

Azz : 2(6.5)(5.25 - o) : 68.25 in.'?

Total area available is

At = Ar r A21 I A22 = 151.45 in.2 > A, of 151.06 in.2

Also, check 'tlose-in" limit. Determine limit parallel io shell as thegreater of

0.75d: 0.75(24) = 18 in. or T" + T^ + r = 24 in.

Because the parallel limit is the same for the "close-in" limit, the arearequired of 0.674, is also satisfied without further calculations. I

Exrmple 11.5. Determine the minimum required thickness of a 36-in. IDcylindrical shell based upon reinforcement requirements. The nozzles arethrough-welded as shown in Fig. 11.9c and have 2.25-in ID on a staggered

16.

t7.

18.

19.

20.

I I,4 REINIORCEMENT IIMITS 357

Dattem of three fows on 3-in. centers and 4 5-in. longitudinal spacing, as shown

in Fig. I l.l l. The design pressure is 500 psi at 700"F design temperature The

allowable stress is 16,600 psi There is no corosion.

Solution

l. The minimum required thickness of the cylindrical shell is

PR 500 x l8" sE - o.6P 16,600 - 0.6 x 500

2. The minimum required thickness of the nozzle is

: 0.552 in.

3.

500 x 1.125 : 0.035 in.sE - 0.6P 16,600 - 0.6 x 500

Deiermine the reinforcement limits based on Ts = 1.125 in' and f,. =0.188 in. Limit parallel to shell surface = X = 2.25 in. or (1.125 +1.125 + 0.188) = 2.438 in., use 2.438 in.The limit normal to shell surface : y = 2.57" = 2.812 in. ot 2 5n =0.469 in., use 0.469 in.

2.25" diameter

Fisur€ I l.l I Muliiple op€ninss in cylindricol lhell.

358 OPININOS, NOZILIS, AND EXTIRNAI. I.OADINGS

4. Exnminc thc longitudinal plane l-2. Actual spacing = 4.5 in. parallcllimits without overlap = 2x = 2(2.438) : 4.875 in., exceeds actualspacing. Therefore, limits ovedap and special rules apply.

A,= dt,F = (2.2s)(0.552)(1.0) = 1.242in.2

Ar = (?i - r)(spacins d) = (r.tZs - 0.552X4.5 - z.z5) =Ar = 1.289 in.z

Az = (7, - t^)(2Y) : (0.188 - 0.035)(2 x 0.469) : 0.143 in.,A' = A1 t Az = 1.432 in.z

Asa = (0.552)(4.5X0.7 x 1.0) = 1.739 in.2

A561s: (l.125)(4.5 - 2.25') = 2.531 n.2 > 1.739 in.2

5. Examine the diagonal plane, 2-3. With a row-to-row spacing of 3 in. anda longitudinal spacing of 4.5 in., the diagonal spacing is \,tr + ZB: 3.75in.,0 = tan-' 3/2.25.0 = 53.13.. With a spacing of 3.75in.,the parallel limit of 4.875 in. exceeds the actual spacing; therefore, thelimits overlap and the special rules apply. From Eq. 11.16,F:0.5(cos2 0 + l) = 0.68 for g = 53.t3'.

A, = dt,F : (2.2s)(0.ss2)(0.68) = 6.30r .n.,

Ar = (4 - r'r,)(spacing - d) = (1.125 - 0.68 x 0.552)

(3.75 - 2.25) = 1.124 in.2

A: = 0.143 in.'z

A, = A1 ! Az = 1.267 in.2 > 0.845 in.2

A14a: (0,552)(3.7 5)(O.7 x 0.68) : 0.985 in.,4567s = 0.125)(3.75 - 2.25) = 1.688.in., > 0.985 in.,

6. The assumed values of T, = 1.125 in. and ?i = 0.188 in. are satis-factory. I

Problzms

11.6 What is the minimum required wall thickness (rounded uo to the nextl/8 in.) of a t2 3/4-in. ID nozzle atrached to a 60-in. d by 3.75-in.thick drum? The allowable stess of both the shell and nozzle material is15.0 ksi. The nozzle is attached by a full penetration weld with comerfillet welds with a throat of 0.7 ?i. The design pressure if 1400 psi at room@mperature.

Answer: t-in = 2.50 in.

I I,4 REINFORCEMENT IIMITS 359

ll.7 For triangular anangement of openings shown in Fig. I l.l I with oPen-

ings that are 2.25-in.lD, what is the minimum side length of a spacingthat forms a series of equilateral triangles?

Ans#sr.' Minimum lensth of side is 4.631 in.

I1.4.2 Reinforcemenl Rules for ASME, Section Vlll, Division I

The rules for reinforced openings in ASME Section VIII, Division 1, Pressure

Vessek, arc similar to those for ASME Section L However, the rules forreinforcernent are given as the main choice, with ligament rules used only as an

alt€rnative for repeating pattems of openings. Rules iue contained in both the

iext and the appendices. They are given for both internal pressure and extemalpressure. The rules are essentially the same except that only 50% of the replace-

ment area is required for extemal pressure assuming that the minimum requiredthickness in each case is based on the appropriate formula and design rules forboth intemal and extemal pressures.

N o Reintorcement Re quiNed

Single openings in vessels that are not subjected to special applied loadings, suchas cyclic loading, do not require reinforcement calculations if the openings donot exceed the following size limits.

1. In plate thickness of $in. or less, d,,. = 3-in. NPS.

2. In plate thickness greater than;i"., d* = 2-in. NPS.

Size anl Shape of Openings

The nrles apply to openings that are circular, elliptical, or obround. The lattershapes often result from an opening in a curved surface or from a nonradialnozzle. However, other shapes arc permitted when considered according tou-2(e).

For openings in cylindrical shells, the rules in the text are limited to openingsof the following size limits:

1. For shells 60-in. and less in diameter, the opening is not to exceed 0'5 Dor 20 in.,

2. For shells over 60 in. in diameter, the opening is not to exceed 0.33 Dor 210-in.

When these size limits are exceeded, in addition to the rules in tle text, the rules

in Appendix l-7 are also to be met. These additional rules may require some

reinforcement to be placed closer to the opening than required by the rules in the

360 oP!N|NOS, NOZZUS, AND CXTEnNAL tOADINOS

text. l.'tlr opcnings in sphcrical shells and lbrmed heads, the text rures are merby considering the use of reverse curves and conical sections where possible.There are no specific limitations on size and shape of openings in spherical shellsand formed heads.

Required Area of Reintorceme nt

The total cross-sectional area of reinforcement required for any plane through thecenter of an opening is given by

A: dt,F (ll.2l)

where d = the diameter of the opening on the longitudinal plane of acylindrical shell or any plane of a spherical shell or formed head(in.)

F : conection factor for pressure stress on plane being examined withrespect to longitudinal axis, as shown in Fig. 11.12. This factor isapplicable only to nozzles with integral reinforcement,

t, = minimum requked thickness of a seamless shell based on thecircumferential stress (longitudinal plane) or of a seamless formedhead with the following additional provisions:

1 When the opening and its reinforcement are totally within the sphericalpart of a torispherical head, l. is determined using the hemispherical headformula with both E and M = 1.0 (see Fig. t1.l3a).

2. When the opening and its reinforcement are in a cone, r, is the requiredthickness of a seamless cone.

3. When the opening and its reinforcement are in an ellipsoidal head andwithin a circle that is equal to 8070 of the shell diameter, r, is determinedusing the hemispherical head formula for a seamless shell of radius K1 D ,where D is the shell diameter and K1 is obtained.from Table 1 1 . 3 and asshown in Fig. 11.130.

The value of t, obtained from any of the methods given above is used onlyto determine the required area of reinforcement. The value of t" used to set theminimum required thickness of the shell or head is based on the thicknessfbrmulas that consider all the design loadings and weld joint efficiencies.

Limit of Reintorcement Parallel ta Shell Surlace

When the size of tle opening is within the limits in the text, the limits ofreintbrcement parallel to the shell surface measured on each side of the center-line are the larger of (l) d or (2) T" + Tn + 0.5d.

I I ,4 RETNIORCEMINT TIMITS 361

o" r;o 2oo so" aoo loo coo loo loo eoP

Arrelr ol Pt.na wlih Loneitldltral AtbFisur€Il.l2 Charr tor der€rmins F. (Courreiy Amsricon Socistv of l'lechonicol Ensine€rs, From Fis UG-37

of d's ASME Code, Vlll-l .)

Limit o! Reinforcement Perpendicular to Shell

The limit of reinforcement p€rpendicular or normal to the shell measured either

inward or outward from the surface of the shell is the smaller of (1) 2.57' or (2'S

2.5n+ L.

Avaibble Area of Reinforcement

When the reinforcement limits do not extend outside the zone of nominal wall

thickness of the shell and nozzle, the area available for reinforcement may be

calculated by the following formulas:

t62 OPININO!, NO! T!3, AND TXTIRNAT TOADINGS

1. Area available in shell wall is the greater of

41 =(2"d._d)(ET,_Ft)or

tu = 2(7" + T^ + 0.5 d) - d(ET, - Ft,)

2. Area available in nozzle wall is the smaller of

A,=(sDQ.-h)

A2= 6n+2.5t"\(7.-t,)

(a) rimits for

0.8D = sneci '1

forisphericat Head

(b) Limits for Etlipsoidaf HeadFigt r€ I L I 3 Dorerminorion of sp€ciol limirs ior determing t, to u!€ in roinfior€€rnenr colculations.

(11.22)

(r1.23)

(r1.24)

(11.2s)

When the size of the opening exceeds the limits in which the rules in the textapply' the supplemental rules in r-7 are used in addition to the text rules. Theseadditional requirements follow.

I I.4 REINTORCTMENT TIMITS

Tqble I 1.3'

D /2h 3.0 2.2 2.O

0.99 0.90Kr l 36

2.8

1.27

2.6l.l8

2.4

1.08

D /2hKI

1.8

0.81

1.6

o.73

1.4

0.65

1.2 l 0

0.57 0.50

Required Area of Reinforcement

The total cross-sectional area of reinforcement required for any plane through thecenter of an opening using rules l-7 is given by

A : 0.67 dt,F (l l.26)

Limit of Reinforcement Parallel to SheA

The limit of reinforcement parallel to the shell measured on each side of theopening centerline is the greater of (1) 0.75 d or (2) T, + T, + 0.5d.

Limil of Reinforcement Perpendicular n Shell

The limit is set exactly the same way as for a nozzle that is within the rules ofthe text.

Wlen any two adjacent openings are spaced so that their reinforcementoverlaps, the combined area is used, but is evaluated only once in the combinedarea. The preferred spacing is at least 1.5d"", with 507o of the area requiredbetween the two openings.

Example 11.6. Determine the reinforcement requirements of an 8 in. 1Dnozzle that is centrally located in a 2 : 1 ellipsoidal head. The inside diameterof the head skirt is 41.75 in. The allowable shess of both the head and nozzlematerial is 17.5 ksi. The design pressure is 700 psi and the design temperatureis 500'F. There is no corrosion and the weld joint efficiency is E : 1.0. See Fig.11. 13. 1 for details of a nozzle.

+^' o.taq"- T^r t'6

Fisure I l.l3.l Deroils of nozzle in exomple I L6.

t6a oPtNtNo3, NOZZII3, AND tXTrRNAt" LOAD|NGS

Solullon

l. The minimum required thickness of aZ: I ellipsoidal head without anopening is determined from UG-32(d) of the A3ME Code, VIII-I as

= = - 7oo (41'751PD

= 0.838 in.;zSE - 0.2P 2(17,sn x 1.0) - 0.2(700)

use 1.0 in.

As noied in the definition of r, ro use with Eq. 11.20 and shown in Fig.11.13.1, when an opening and its reinforcement are located in an ellio_soidal head and within a circle equal to 80Zo of the shell diameter, ,. tobe used in reinforcement calculations is the thickness required for aseamless sphere of radius K1D, where D is the shell ID and E for a 2 : Iellipsoidal head is 0.9 frorn Table 11.3. For this head, the opening andreinforcement are within 0.8 D = 0.8(41.75) = 33.4 in.Using the spherical shell radius of R = K tD = 0.9(41.7 5) = 37 .57 5 n.in the hemispherical head formula gives

! = = PR

= - 7oo (37.575)PR

= u. /f) rn.' 2SE - 0.2p 2(17,500 x 1.0) _ 0.2(700)

4. The minimum required thickness of the nozzle is

t

3.

9,

,^= PR' ='- sE - 0_6P

7OO (4)(17,500 x 1.0) - 0.6(700)

= 0.164 in.

!.

6.

,|

E.

Limits paraUel to head surface = X = d or (T, + T, + r), whichever rslarger. X = 8 in. or (4 + | + t.125 = 6.125 in.); use 8 in.Lirnits perpendicular to head surface i s y = 2.57, or 2.52,, whichever rssmaller. y = 2.5(1) = 2.5 in. or 2.5(1.125) = 2.813 in.; use 2.5 in.Because the limit of 2X = 2(8): 16 in. is less ihan 33.4 in. of item 2above, the provision of the spherical head may be used.Reinforcement area required following Eq. 11.20 is

A, = d+F = 8(0.7s5)(1.0) = 6.040 in.,

Reinforcement area available in head according to Eq. 11.21 is

A1 = (ET, - Ft)(U - d) = (1.0 - 0.755X16 - 8) = 1.960 in.,

Reinforcement area available in nozzle according to Eq. 11.23 is

Az= 57"(7, - tJ = 5(1.0X1.r25 - 0.164) = 4.805 in.,

10.

Figure ll.l,1 12 x 16 Monwoy op€nins detoils for exompl€ l l 7.

I t.4 RHNFORCEMENT UMI1S 365

ll. Total reinforcement available from head and nozzle is

A,: A1 'r Az: 1.960 + 4'805 : 6 765 in''

Area provided = 6.765 in.z > area required = 6.040 in.'? If additional

area is needed, use fillet weld area. I

Example 11.7. Determine the reinforcement requirements for a l2-in.x 16-in. manway opening. The l2-in. dimension lies along the longitudinal axis

of the cylindrical shell. The manway cover seals against the outside surface ofthe opening so that the opening nozzle is under intemal pressure' The ID of the

shell is 41.875 in. Both the shell and manway material have an allowable tensile

stress of 17,500 psi. The design pressure is 700 psi at a design temperature of500'F. There is no corrosion and the joint efficiency is E = 1.0' Details are

shown in Fig. 11.14.

Solution

1. The minimum rcquired thickness of the shell is found from

700 (20.938)":'::- = 0.858 in.:

(17,500 x 1.0) - (0.6 x 700)

use 1

2. The minimum required thickness of the nozzle is obtained as

PRt=-=- sE - 0.6P

, : --!lo- =' sE - o.6P

700 (6)

(17,500 x 1.0) - (0.6 x 700)= 0.246 in.:

use 1 i in.

ln.

-a-

ABCD=Limit of Reinf

OPININOS, NOZZUS, AND TXTERNAI. I,OADINOS

Uxuntinution ol thc longitudinal planc where t' = | .0 is determined tiomFig. l l.12.Limits parallel to shell, whichever is larger:

d: 12 in.

T, + T, + 0.5d : 6+ 1 + 1* = S+in.

Limits perpendicular to shell, whichever is smaller:

2iT, = 2.5(r) = 2.5 in.

2+r": z.s(1.25) = 3.125 in.

Reinforcement area required by Eq. 11.20 is

A,: dt,F : 12 (0.858X1.0) = 10.296 in.2

Reinforcement area available in shell by Eq. 11.21 is

Ar = (ET" - Ft)(2x - d) = (r - 0.85sX24 - r2) = 1.7s4.o.2

Reinforcement area available jn nozzle wall by Eq. 11.23

At = 57,(7, - t,,) = S(t)(1.2s - 0.246) = 5.020 in.,Azz = 57,(7, - t^) = 5(1)(1.2s - 0) = 6.256 ;n.z

Total reinforcement area available:

At = At + A^ + An = 1.704 + 5.020 + 6.25O-_ 12.974 in.2

Area provided of 12.974 in.2 ) area required of 10.296 in.24. Examination of circumferential plane where F = 0.5 from Fig. 11.12

gives the reinforcement area required by Eq. 11.20 as

A,: dt,F = 16 (0.858)(0.s) : 6.864 in.2

If the arc length of l6.4in. is used,A, = 7.036 in., Either of these areasis less than A. in item 3 and does not control. The longitudinal planeconhols. I

Example 11.8. Determine the reinforcement requirements for a 6-in. ID noz_zle that is located at the junction of a cylindrical shell and a hemispherical head.The entire opening is in the cylindrical shell, but ttre reinforcemlni extends inDotn drectlons-some into the head and some into the shell. The ID of the shellis 41.875 in. The allowable shess of all material is 17.5 ksi. The design pressure

3.

I r.4 RflNFORCTMENT UMTTS 367

is 70O psi at a design temperature of 500'F. All joints are fully radiographed witht joint efficiency of E : 1.0. There is no corrosion.

Solutinn

1. The minimum required thickness of the cylindrical shell from Example11.7 is

t = 0.858 in.: usel in.