Structural Analysis and Design of Process Equipment
Structural Analysis and Design of Process Equipment
Maan H. Jawad
James R. Farr
Third Edition
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vii
Contents
Preface to the Third Edition xv
Preface to the Second Edition xvii
Preface to the First Edition xix
Acknowledgements xxi
Part I
1
1.1
1.2
1.3
1.4
1.5
1.6
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.12.1
2.12.2
2.12.3
2.12.4
2.12.5
Background and Basic Considerations
1
4
Use of Process Vessels and Equipment 4
History of Pressure Vessel Codes in the United States 4
Organization of the ASME Boiler and Pressure Vessel Code 6
Organization of the ANSI B31 Code for Pressure Piping 6
Some Other Pressure Vessel Codes and Standards in the United States 6
Worldwide Pressure Vessel Codes 7
References 7
Further Reading 7
History and Organization of Codes
Selection of Vessel, Specifications, Reports, and Allowable Stresses 10
Selection of Vessel 10
Which Pressure Vessel Code is Used 10
Design Specifications and Purchase Orders 10
Special Design Requirements 11
Design Reports and Calculations 11
Materials Specifications 11
Design Data for New Materials 11
Factors of Safety 12
Allowable Tensile Stresses in the ASME Code 12
Allowable External Pressure Stress and Axial Compressive Stress in the ASME Boiler
and Pressure Vessel Code 13
Allowable Stresses in the ASME Code for Pressure Piping 14
Allowable Stress in Other Codes of the World 14
European Union (EN) Countries 14
Japanese Code 15
People’s Republic of China 15
Indian Code 15
Australian Code 16
References 16
3
Strength Theories, Design Criteria, and Design Equations
3.1
3.2
Strength Theories 18
Design Criteria 18
18
viii
Contents
3.3
3.4
3.5
3.6
Design Equations 19
Stress–Strain Relationships 19
Strain–Deflection Equations 20
Force–Stress Expressions 22
References 23
Further Reading 23
4
26
Material Selection 26
Corrosion 26
Strength 26
Specified Minimum Yield Stress 27
Specified Minimum Tensile Stress 28
Creep Rate 28
Rupture Strength 28
Material Cost 30
Nonferrous Alloys 31
Aluminum Alloys 31
Annealing 31
Normalizing 31
Solution Heat Treating 31
Stabilizing 31
Strain Hardening 31
Thermal Treating 32
Copper and Copper Alloys 32
Nickel and High-Nickel Alloys 32
Titanium and Zirconium Alloys 33
Ferrous Alloys 34
Carbon Steels 34
Low-Alloy Steels 34
High-Alloy Steels 34
Martensitic Stainless Steels 34
Ferritic Stainless Steels 34
Austenitic Stainless Steels 34
Heat Treating of Steels 35
Normalizing 35
Annealing 35
Postweld Heat Treating 35
Quenching 35
Tempering 35
Brittle Fracture 35
Charpy V-Notch Test (C v ) 36
Drop-Weight Test (DWT) 37
Fracture Analysis Diagram (FAD) 37
Theory of Fracture Mechanics 39
Relationship Between K IC and C V 41
Hydrostatic Testing 42
Factors Influencing Brittle Fracture 42
ASME Pressure Vessel Criteria 43
Hydrogen Embrittlement 50
Nonmetallic Vessels 50
References 50
Further Reading 51
4.1
4.1.1
4.1.2
4.1.2.1
4.1.2.2
4.1.2.3
4.1.2.4
4.1.3
4.2
4.2.1
4.2.1.1
4.2.1.2
4.2.1.3
4.2.1.4
4.2.1.5
4.2.1.6
4.2.2
4.2.3
4.2.4
4.3
4.3.1
4.3.2
4.3.3
4.3.3.1
4.3.3.2
4.3.3.3
4.4
4.4.1
4.4.2
4.4.3
4.4.4
4.4.5
4.5
4.5.1
4.5.2
4.5.3
4.5.4
4.5.5
4.5.6
4.5.7
4.5.8
4.6
4.7
Materials of Construction
Contents
Part II
Analysis of Components 53
5
Stress in Cylindrical Shells 56
5.1
5.2
5.2.1
5.2.2
5.3
5.3.1
5.3.2
5.3.3
5.4
5.4.1
5.4.2
5.4.3
Stress Due to Internal Pressure 56
Discontinuity Analysis 60
Long Cylinders 61
Short Cylinders 66
Buckling of Cylindrical Shells 69
Uniform Pressure Applied to Sides Only 70
Uniform Pressure Applied to Sides and Ends 71
Pressure on Ends Only 72
Thermal Stress 72
Uniform Change in Temperature 75
Gradient in Axial Direction 76
Gradient in Radial Direction 77
Nomenclature 80
References 81
Further Reading 81
6
Analysis of Formed Heads and Transition Sections 84
6.1
6.1.1
6.1.2
6.1.3
6.1.4
6.2
6.3
6.4
6.4.1
6.4.2
6.4.3
6.5
Hemispherical Heads 84
Various Loading Conditions 86
Discontinuity Analysis 88
Thermal Stress 91
Buckling Strength 91
Ellipsoidal Heads 93
Torispherical Heads 95
Conical Heads 95
Unbalanced Forces at Cone-to-Cylinder Junction 96
Discontinuity Analysis 97
Cones Under External Pressure 98
Nomenclature 99
References 100
Further Reading 100
7
Stress in Flat Plates 102
7.1
7.2
7.3
7.4
Introduction 102
Circular Plates 102
Rectangular Plates 106
Circular Plates on Elastic Foundations 107
Nomenclature 109
Reference 109
Further Reading 109
Part III
Design of Components
111
8
Design of Cylindrical Shells 114
8.1
8.2
8.3
8.4
8.5
ASME Design Equations 114
Evaluation of Discontinuity Stresses 115
ASME Procedure[2] for External Pressure Design in VIII-1 121
Design of Stiffening Rings 126
Allowable Gaps in Stiffening Rings 129
ix
x
Contents
8.6
8.7
Out-of-Roundness of Cylindrical Shells Under External Pressure 129
Design for Axial Compression 132
Nomenclature 132
References 133
Further Reading 133
9
Design of Formed Heads and Transition Sections 136
9.1
9.1.1
9.1.2
9.1.3
9.1.4
9.1.5
9.2
9.3
9.4
9.4.1
9.4.1.1
9.4.2
9.4.2.1
Introduction 136
Flanged Heads 136
Hemispherical Heads 136
Elliptical and Torispherical (Flanged and Dished) Heads 136
Conical and Toriconical Heads 136
Miscellaneous Heads 136
ASME Design Equations for Hemispherical Heads 137
ASME Design Equations for Ellipsoidal, Flanged, and Dished Heads 139
ASME Design Equations for Conical Heads 143
Internal Pressure 143
Discontinuity Analysis for Internal Pressure 143
External Pressure 145
Discontinuity Analysis for External Pressure 145
Nomenclature 147
References 148
Further Reading 148
10
Blind Flanges, Cover Plates, and Flanges 150
10.1
10.2
10.3
10.4
10.5
10.6
10.7
10.7.1
10.7.2
10.7.3
10.7.4
10.7.5
10.7.6
10.7.7
10.7.8
10.7.9
10.7.10
10.7.11
10.7.12
10.8
10.9
10.10
10.11
10.12
10.13
10.14
10.14.1
10.14.2
Introduction 150
Circular Flat Plates and Heads with Uniform Loading 151
ASME Code Formula for Circular Flat Heads and Covers 153
Comparison of Theory and ASME Code Formula for Circular Flat Heads and Covers Without Bolting 154
Bolted Flanged Connections 154
Contact Facings 155
Gaskets 155
Rubber O-Rings 155
Metallic O- and C-Rings 155
Compressed Fiber Gaskets 158
Flat Metal Gaskets 158
Spiral-Wound Gaskets 158
Jacketed Gaskets 158
Metal Ring Gaskets 158
High-Pressure Gaskets 158
Lens Ring Gaskets 159
Delta Gaskets 159
Double-Cone Gaskets 159
Gasket Design 160
Bolting Design 161
Blind Flanges 163
Bolted Flanged Connections with Ring-Type Gaskets 164
Reverse Flanges 170
Full-Face Gasket Flange 171
Flange Calculation Sheets 176
Flat-Face Flange with Metal-to-Metal Contact Outside of the Bolt Circle 177
Classification of Assembly 177
Categories of Flanges 177
Contents
10.15
Spherically Dished Covers 177
Nomenclature 184
References 184
Further Reading 185
11
Openings, Nozzles, and External Loadings 188
11.1
11.2
11.3
11.4
11.4.1
11.4.1.1
11.4.1.2
11.4.2
11.4.3
11.4.3.1
11.4.4
11.4.4.1
11.4.5
11.4.5.1
11.5
11.6
11.7
11.7.1
11.7.2
11.7.2.1
General 188
Stresses and Loadings at Openings 188
Theory of Reinforced Openings 192
Reinforcement Limits 193
Reinforcement Rules for ASME Section I 195
No Reinforcement Required 195
Size and Shape of Openings 195
Reinforcement Rules for ASME Section VIII, Division 1 198
Reinforcement Rules for ASME, Section VIII, Division 2 204
Nomenclature 204
Reinforcement Rules for ANSI/ASME B31.1 210
No Reinforcement Calculations Required 210
Reinforcement Rules for ANSI/ASME B31.3 212
Nomenclature 213
Ligament Efficiency of Openings in Shells 215
Fatigue Evaluation of Nozzles Under Internal Pressure 217
External Loadings 218
Local Stresses in the Shell or Head 218
Stresses in the Nozzle 226
Nomenclature 227
References 230
Bibliography 231
12
234
Introduction 234
Skirt and Base-Ring Design 234
Anchor-Chair Design 239
Design of Support Legs 241
Lug-Supported Vessels 242
Ring Girders 243
Saddle Supports 245
Nomenclature 248
References 249
Further Reading 249
12.1
12.2
12.2.1
12.3
12.4
12.5
12.6
Vessel Supports
Part IV
Theory and Design of Special Equipment 251
13
Flat-Bottom Tanks 254
13.1
13.2
13.2.1
13.2.1.1
13.2.1.2
13.2.1.3
13.2.2
13.2.3
Introduction 254
API 650 Tanks 254
Roof Design 254
Dome Roofs 254
Conical Roofs 256
Small Internal Pressure 256
Shell Design 258
Annular Plates 261
xi
xii
Contents
13.3
13.3.1
13.3.1.1
13.3.1.2
13.3.1.3
13.3.1.4
13.3.2
13.4
13.4.1
13.5
API 620 Tanks 263
Allowable Stress Criteria 266
Compressive Stress in the Axial Direction with No Stress in the Circumferential Direction 266
Compressive Stress with Equal Magnitude in the Meridional and Circumferential Directions 266
Compressive Stress with Unequal Magnitude in the Meridional and Circumferential Directions 267
Compressive Stress in One Direction and Tensile Stress in the Other Direction 267
Compression Rings 267
Aluminum Tanks 270
Design Rules 270
AWWA Standard D100 271
References 273
Further Reading 273
14
Heat-Transfer Equipment 276
14.1
14.2
14.3
14.4
14.4.1
14.4.2
14.4.3
14.5
14.6
14.6.1
14.6.2
14.6.3
14.7
14.8
Types of Heat Exchangers 276
TEMA Design of Tubesheets in U-tube Exchangers 276
Theoretical Analysis of Tubesheets in U-tube Exchangers 280
ASME Equations for Tubesheets in U-tube Exchangers 283
Nomenclature 283
Preliminary Calculations 285
Design Equations 288
Theoretical Analysis of Fixed Tubesheets 291
ASME Equations for Fixed Tubesheets 293
Nomenclature 293
Preliminary Calculations 294
Design Equations 294
Expansion Joints 300
Tube-to-Tubesheet Junctions 303
References 305
Further Reading 305
15
Vessels for High Pressures 308
15.1
15.2
15.3
15.4
15.5
Basic Equations 308
Prestressing (Autofrettaging) of Solid-Wall Vessels
Layered Vessels 311
Prestressing of Layered Vessels 315
Wire-Wound Vessels 317
Nomenclature 317
References 318
Further Reading 318
16
Tall Vessels 320
16.1
16.2
16.2.1
16.2.2
16.3
16.3.1
16.3.2
16.4
16.5
16.6
Design Considerations 320
Earthquake Loading 320
Lateral Loads 320
Numerical Method for Calculating Natural Frequency 324
Wind Loading 331
External Forces from Wind Loading 332
Dynamic Analysis of Wind Loads 332
Vessel Under Internal Pressure Only 336
Vessel Under Internal Pressure and External Loading 338
Vessel Under External Pressure Only 340
309
Contents
16.7
Vessel Under External Pressure and External Loading 341
References 342
Bibliography 342
17
Vessels of Noncircular Cross Section 344
Types of Vessels 344
Rules in Codes 345
Openings in Vessels with Noncircular Cross Section 345
Ligament Efficiency for Constant-Diameter Openings 345
Ligament Efficiency for Multidiameter Openings Subject to Membrane Stress 349
Ligament Efficiency for Multidiameter Openings Subject to Bending Stress 350
Design Methods and Allowable Stresses 352
Basic Equations 353
Equations in the ASME Code, VIII-1 356
Design of Noncircular Vessels in Other Codes 360
Method of the British Code BS 1113 360
Method of the European Standards EN 12952 and EN 13445 360
Forces in Box Headers due to Internal Pressure 361
Square Headers 362
Rectangular Headers 362
References 364
Further Reading 364
17.1
17.2
17.3
17.4
17.5
17.6
17.7
17.8
17.9
17.10
17.10.1
17.10.2
17.11
17.11.1
17.11.2
18
18.1
18.2
18.3
18.3.1
18.3.2
18.4
18.5
18.6
18.6.1
18.7
18.7.1
Power Boilers 366
A
Guide to ASME Code
B
Sample of Heat-Exchanger Specification Sheet 383
C
Sample of API Specification Sheets 387
D
Sample of Pressure Vessel Design Data Sheets 393
E
Sample Materials for Process Equipment 407
F
Required Data for Material Approval in the ASME Code 411
G
Procedure for Providing Data for Code Charts for External-Pressure Design
H
Corrosion Charts
General 366
Materials 366
General Design Requirements 366
Allowable Stress Values 366
Cylinders under Internal Pressure 366
Formed Heads under Internal Pressure 368
Loadings on Structural Attachments 368
Watertube Boilers 369
Special Design Requirements and Rules 369
Firetube Boilers 373
Special Design Requirements and Rules 373
References 373
415
375
413
xiii
xiv
Contents
I
Various ASME Design Equations 431
J
Joint Efficiency Factors
K
Simplified Curves for External Loading on Cylindrical Shells 445
L
Conversion Tables 453
Index 455
433
xv
Preface to the Third Edition
The third edition includes revisions to various chapters
due to advancement in technology since the second
edition was written over 30 years ago. These advancements include earthquake and wind analysis, fracture
mechanics, and creep analysis of equipment operating in
high temperatures. Additional changes were also needed
due to the reduction of safety factors in various codes
and standards in the last three decades. These reductions
were due to improvements in material manufacturing,
more accurate analyses due to computerized technology, and better inspection methodology. Additional
structural analysis methods were added in few chapters
to assist the designer in solving complicated problems
not covered by the prevailing codes and standards.
These include a natural frequency analysis required
in earthquake evaluation for vessels with nonuniform
cross sections and analysis of vessels with rectangular
cross section having sides with different thicknesses and
moduli of elasticity.
Many of the chapters in the first and second editions
were written by the late James R. Farr. An effort was made
in this third edition to preserve these chapters in their
original format with only the necessary changes needed
to bring them up to date to the current technology and
standards.
The tendency of the newer editions of the codes
such as the American Society of Mechanical Engineers
(ASME) Boiler and Pressure Vessel Code is to replace
existing charts needed in the design of components
with equations that are more suitable for computerized programs. These equations are obtained in one
of two methods. The first is to go back to the origin
of a given chart. If the original chart was drawn from
equations, then these equations are now used in the new
code edition and the chart deleted. The format of these
equations, more often than not, leads to the original
derivation or the assumptions made in developing the
equations. The second method is to take the charts that
were drawn based on experience and/or experimental
data with no background equations and simulate these
charts with equations obtained from regression analysis. The resulting equations normally have no physical
significance even though the results obtained from them
are essentially the same as those obtained from the
original chart. Accordingly in this book, equations from
the first method were incorporated, as much as possible,
in the text since they can be traced back to their original
derivation. Equations from the second method were
not incorporated in order to minimize the confusion
regarding their original background.
Camas, WA, USA
January 2018
Maan H. Jawad
xvii
Preface to the Second Edition
The second edition includes a number of new topics not
included in the first edition, which are useful in designing pressure vessels. A new chapter has been added to
the design of the power boilers, which are an integral
part of a chemical plant or refinery. Some of the existing
chapters have been expanded to include new topics
such as toughness criteria, design of expansion joints,
tube-to-tubesheet parameters. In addition, portions of
three chapters and one appendix have been rewritten
to reflect current practice. The first such passage concerns the design of water tanks, where new equations are
added in accordance with the revised criteria given in the
American Water Works Association (AWWA) Standard.
The second concerns the design of tubesheets in U-tube
heat exchangers, where simplified equations are used in
lieu of the cumbersome charts shown in the first edition.
The third concerns the design of noncircular vessels,
where new equations are added to reflect new changes
made in the American Society of Mechanical Engineers
(ASME) Boiler and Pressure Vessel Code. Appendix J on
joint efficiencies has been rewritten to reflect the current
criteria of the ASME code, VIII-1.
We thank all of our colleagues for their numerous comments, which promoted us to revise the first edition. Special thanks are given to Mr E. L. Thomas, Jr., and Dr L. J.
Wolf for their help.
St Louis, MO, USA
Barberton, OH, USA
June 1988
Maan H. Jawad
James R. Farr
xix
Preface to the First Edition
We wrote this book to serve three purposes. The first
purpose is to provide structural and mechanical engineers associated with the petrochemical industry a
reference book for the analysis and design of process
equipment. The second is to give graduate engineering
students a concise introduction to the theory of plates
and shells and its industrial applications. The third is to
aid process engineers in understanding the background
of some of the design equations in the American Society
of Mechanical Engineers (ASME) Boiler and Pressure
Vessel Code, Section VIII.
The topics presented are separated into four parts.
Part 1 is intended to familiarize the designer with some
of the common “tools of the trade.” Chapter 1 details
the history of pressure vessels and various applicable
codes from around the world. Chapter 2 discusses
design specifications furnished in the purchasing process equipment as well as in various applicable codes.
Chapter 3 establishes the strength criteria used in different codes and the theoretical background needed in
developing design equations in subsequent chapters.
Chapter 4 includes different materials of construction
and toughness considerations.
Part 2 is divided in to three chapters outlining the basic
theory of plates and shells. Chapter 5 develops the membrane and bending theories of cylindrical shells. Chapter
6 discusses various approximate theories for analyzing
heads and transition sections, and Chapter 7 derives the
equations for circular and rectangular plates subjected
to various loading and support conditions. These three
chapters form the basis from which most of the design
equations are derived in the other chapters.
Part 3, which consists of five chapters, details the
design and analysis of components. Chapters 8 and 9
derive the design equations established by the ASME
Code, VII-1 and -2, for cylindrical shells as well as heads
and transition sections. Chapter 10 discusses gaskets,
bolts, and flange design. Chapter 11 presents openings
and their reinforcement; Chapter 12 develops design
equations for support systems.
Part4 outlines the design and analysis of some specialized process equipment. Chapter 13 describes the design
of flat-bottom tanks; Chapter 14 derives the equations for
analyzing heat-transfer equipment. Chapter 15 describes
the theory of thick cylindrical shells in high-pressure
applications. Chapter 16 discusses the stress analysis of
the tall vessels. Chapter 17 outlines the procedure of the
ASME Code, VIII-1, for designing rectangular pressure
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 the industry for that particular type
of equipment or component discussed in the pertinent chapter. Accordingly, a summary of nomenclature
appears at the end of most of the chapters in which
mathematical expressions are given.
In using this book as a textbook for plates and shells,
Chapters 3, 5, 6, and 7 form the basis for establishing
the basic theory. Instructors can select other chapters to
supplement the theory according to the background and
needs of the graduate engineer.
In deriving the background of some of the equations
given in the ASME Boiler and Pressure Vessel Code,
attention was focused on Section VIII, Divisions 1 and
2. Although these same equations do occur in the other
sections of the ASME Code, such as the Power and
Heating Coilers, no consideration is given in this book
regarding other sections unless specifically stated.
Saint Louis, MO, USA
Barberton, OH, USA
September 1983
Maan H. Jawad
James R. Farr
xxi
Acknowledgements
Thanks to the many people and organizations that helped
during the rewrite of the third edition. Special thanks are
given to the following people for helping with the international standards: Dave I. Anderson for the British code,
Anne Chaudouet for the French code, Susumu Terada
for the Japanese code, Jay Vattappilly for the Indian code,
and Jinyang Zheng for the Chinese code. Thanks are also
given to Basil Kattula for his help with the wind load and
earthquake requirements of ASCE 7-10.
The Nooter Corporation of St. Louis, Missouri, is
acknowledged for its continual support of the author
in publishing this book as well as participating in other
standards’ activity.
Special thanks is also extended to the editors and staff
of Wiley for doing an excellent job in editing as well
as updating the old charts, figures, and tables from the
Second edition to the Third edition.
1
Part I
Background and Basic Considerations
Old timers. Source: (Top) Courtesy Babcock & Wilcox Company; (bottom) Courtesy Nooter Corporation.
4
1
History and Organization of Codes
1.1 Use of Process Vessels
and Equipment
Throughout the world, the use of process equipment has
expanded considerably. In the petroleum industry, process vessels are used at all stages of oil processing. At the
beginning of the cycle, they are used to store crude oil.
Many different types of these vessels process the crude
oil into oil and gasoline for the consumer. The vessels
store petroleum at tank farms after processing and finally
serve to hold the gasoline in service stations for the consumer’s use. The use of process vessels in the chemical
business is equally extensive. Process vessels are used
everywhere.
Pressure 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 ocean; most are positioned on the ground or supported on platforms; and some actually are found in storage tanks and hydraulic units in aircraft.
The internal pressure to which the process equipment is designed is as varied as the size and shape.
Internal pressure may be as low as 1 in. water-gage
pressure or 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
American Society of Mechanical Engineers (ASME)
Boiler and Pressure Code, Section VIII, Division 1 [1],
specifies a range of internal pressure from 15 psi at
the bottom to no upper limit; however, at an internal pressure above 3000 psi, the ASME Code, VIII-1,
requires that special design considerations may be
necessary [1]. However, any pressure vessel that meets
all the requirements of the ASME Code, regardless
of the internal or external design pressure, may still
be accepted by the authorized inspector and stamped
by the manufacturer with the ASME Code symbol.
Some other pressure equipment, such as American
Petroleum Institute (API) [2] storage tanks, may be
designed for and contain internal pressure not more
than that generated by the static head of fluid contained
in the tank.
1.2 History of Pressure Vessel Codes
in the United States
Through the late 1800s and early 1900s, explosions in
boilers and pressure vessels were frequent. A firetube
boiler explosion on the Mississippi River steamboat
Sultana on April 27, 1865, resulted in sinking of the
boat within 20 minutes and the death of 1500 soldiers
who were going home after the Civil War. This type of
catastrophe continued unabated into the early 1900s.
In 1905, a destructive explosion of a firetube boiler in
a shoe factory in Brockton, Massachusetts (Figure 1.1)
killed 58 people, injured 117 others, and caused $400 000
in property damage. In 1906, another explosion in a shoe
factory in Lynn, Massachusetts, resulted in death, injury,
and extensive property damage. After this accident,
the Massachusetts governor directed the formation of
a Board of Boiler Rules. The first set of rules for the
design and construction of boilers was approved in
Massachusetts on August 30, 1907. This code was three
pages long!
In 1911, Colonel E. D. Meier, the president of the
ASME, established a committee to write a set of rules for
the design and construction of boilers and pressure vessels. On February 13, 1915, the first ASME Boiler Code
was issued. It was entitled “Boiler Construction Code,
1914 Edition.” This was the beginning of the various
sections of the ASME Boiler and Pressure Vessel Code,
which ultimately became Section I, Power Boilers [3].
The first ASME Code for pressure vessels was issued as
“Rules for the Construction of Unfired Pressure Vessels,”
Section VIII, 1925 edition. The rules applied to vessels
over 6 in. in diameter, volume over 1.5 ft [3], and pressure over 30 psi. In December 1931, a Joint API–ASME
Committee was formed to develop an unfired pressure
vessel code for the petroleum industry. The first edition
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
1.2 History of Pressure Vessel Codes in the United States
Figure 1.1 Firetube boiler explosion in shoe factory
in Brockton, Massachusetts in 1905. Source: Courtesy
Hartford Steam Boiler Inspection and Insurance Co.,
Hartford, Ct.
was issued in 1934. For the next 17 years, two separate
unfired pressure vessel codes existed. In 1951, the last
API–ASME Code was issued as a separate document
[4]. In 1952, the two codes were consolidated into one
code – the ASME Unfired Pressure Vessel Code, Section
VIII. This continued until the 1968 edition. At that time,
the original code became Section VIII, Division 1, Pressure Vessels, and another new part was issued, which was
Section VIII, Division 2, Alternative Rules for Pressure
Vessels.
The ANSI/ASME Boiler and Pressure Vessel Code
is issued by the ASME with approval by the American
National Standards Institute (ANSI) as an ANSI/ASME
document. One or more sections of the ANSI/ASME
Boiler and Pressure Vessel Code have been established
as the legal requirements in 47 of the 50 states in the
United States and in all the provinces of Canada. Also, in
many other countries of the world, the ASME Boiler and
Pressure Vessel Code is used to construct boilers and
pressure vessels.
In the United States, most piping systems are built
according to the ANSI/ASME Code for Pressure Piping B31. There are a number of different piping code
sections for different types of systems. The piping section
that is used for boilers in combination with Section I
of the ASME Boiler and Pressure Vessel Code is the
Code for Power Piping, B31.1 [5]. The piping section
that is often used with Section VIII, Division 1, is the
5
6
1 History and Organization of Codes
code for Chemical Plant and Petroleum Refinery Piping,
B31.3 [6].
1.3 Organization of the ASME Boiler
and Pressure Vessel Code
The ASME Boiler and Pressure Vessel Code is divided
into many sections, divisions, parts, and subparts. Some
of these sections relate to a specific kind of equipment
and application; others relate to specific materials and
methods for application and control of equipment; and
others relate to care and inspection of installed equipment. The following sections specifically relate to the
design and construction of boiler, pressure vessel, and
nuclear components:
Sections.
(I) Rules for Construction of Power Boilers
(II) Materials
Part A. Ferrous Material Specifications
Part B. Nonferrous Material Specifications
Part C. Specifications for Welding Rods,
Electrodes, and Filler Metals
Part D. Properties
(III) Rules for Construction of Nuclear Facility Components
Division 1.
Subsection NB. Class 1 Components.
Subsection NC. Class 2 Components.
Subsection ND. Class 3 Components.
Subsection NE. Class MC Components.
Subsection NF. Supports.
Subsection NG. Core Support Structures.
Division 5. High-Temperature Reactors.
(IV) Rules for Construction of Heating Boilers
(VIII) Rules for Construction of Pressure Vessels
Division 1.
Division 2. Alternative Rules.
Division 3. Alternative Rules for Construction of
High Pressure Vessels.
(X) Fiber-Reinforced Plastic Pressure Vessels
(XII) Rules for Construction and Continued Service of
Transport Tanks
A new edition of the ASME Boiler and Pressure Vessel
Code is issued every 2 years. A new edition incorporates all the changes made to the previous edition. The
new edition of the code becomes mandatory when it
appears.
Code Cases [7] are also issued periodically after
each code meeting. They contain permissive rules for
materials and special constructions that have not been
sufficiently developed to include them in the code itself.
Finally, there are Code Interpretations [8]. These are in
the form of questions and replies that further explain the
items in the code that have been misunderstood.
1.4 Organization of the ANSI B31 Code
for Pressure Piping
In the United States, the most frequently used design
rules for pressure piping are the ANSI B31 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:
B31.1 Power Piping
B31.3 Process Piping
B31.4 Pipeline Transportation Systems for Liquids and
Slurries
B31.5 Refrigeration Piping and Heat Transfer Components
B31.8 Gas Transmission and Distribution Piping Systems
B31.9 Building Services Piping
B31.12 Hydrogen Piping and Pipelines
The ANSI B31 Piping Code Committee prepares and
issues new editions and addenda with 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.
1.5 Some Other Pressure Vessel Codes
and Standards in the United States
In addition to the ANSI/ASME 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 as follows:
ANSI/API Standard 620. Design and Construction of
Large, Welded, Low-Pressure Storage Tanks, American
Petroleum Institute (API), Washington, D.C.
ANSI/API Standard 650. Welded Steel Tanks for
Fuel Storage, American Petroleum Institute, Washington, D.C.
ANSI-AWWA Standard D100. Welded Carbon Steel
Tanks for Water Storage, American Water Works Association (AWWA), Denver, Colorado.
UL 644. Standard for Container Assemblies for LP-Gas,
9th ed., Underwriters Laboratories, Northbrook, Illinois.
Standards of Tubular Exchanger Manufacturers Association, 9th ed., Tubular Exchanger Manufacturer’s Association, New York.
Further Reading
Standards of the Expansion Joint Manufacturers Association, 10th ed., Expansion Joint Manufacturer’s Association, New York.
A number of standards are available in the United
States for repairing and altering existing boilers and
pressure vessels. Frequently, the repairs and alterations
involve design considerations that are outside the scope
of ASME Sections I and VIII. Some of these standards
are as follows:
National Board Inspection Code. National Board of
Boiler and Pressure Vessel Inspectors, Ohio.
Fitness-for-Service. API 579–1/ASME FFS-1, American Society of Mechanical Engineers, New York.
Pressure Vessel Inspection Code. API-510, American
Petroleum Institute, Washington, D.C.
1.6 Worldwide Pressure Vessel Codes
In addition to the ASME Boiler and Pressure Vessel
Code, which is used worldwide, many other pressure
vessel codes have been legally adopted in various countries. Difficulty often occurs when vessels are designed
in one country, built in another country, and installed in
still another country. This is often the case.
The following list is a partial summary of some of the
various codes used in different countries:
Australia. Pressure Equipment: AS 1200. Standards
Association of Australia. Sydney, Australia.
China. Pressure Vessel Standard GB 150. China
National Institute of Standardization (CNIS). Beijing,
China.
European Union. Countries belonging to the European Union (EN) including France, Germany, Italy,
and the United Kingdom use the European Pressure
equipment Directive (PED) for the design of boilers
and pressure vessels. Hence, Standard EN 12953 is used
for boilers and Standard EN 13445 is used for pressure
vessels. Local codes are also used when specific rules
are not covered by these two standards. These include
CODAP in France, A. D. Merkblatter in Germany, and
BS 5500 in the United Kingdom.
Japan. In Japan, the Japanese Industrial Standard for
pressure vessels is JIS B 8265, 8266, and 8267. For boilers,
the standard is JIS B 8201.
More complete details, discussions of factors of safety,
and applications of the codes mentioned are given in
Section 2.12.
References
1 (2017). ASME Boiler and Pressure Vessel Code, Section
5 ASME Code for Pressure Piping B31 Power Piping,
VIII, Division 1, Rules for Construction of Pressure
Vessels. New York: American Society of Mechanical
Engineers.
2 API Standard 620 (2013). Design and Construction of
Large, Welded, Low-Pressure Storage Tanks,” ANSI/API
Std. 620. Washington, D.C.: American Petroleum
Institute.
3 (2017). ASME Boiler and Pressure Vessel Code, Section
I, Rules for Construction of Power Boilers. New York:
American Society of Mechanical Engineers.
4 API-ASME Code (1951). Unfired Pressure Vessels for
Petroleum Liquids and Gases, 5ee. New York: American Society of Mechanical Engineers and American
Petroleum Institute.
ANSI/ASME B31.1,. New York: American Society of
Mechanical Engineers.
6 ASME Code for Pressure Piping B31 Chemical Plant
and Petroleum Refinery Piping, ANSI/ASME B31.3.
New York: American Society of Mechanical Engineers.
7 ASME Boiler and Pressure Vessel Code, Code Cases,
Boilers and Pressure Vessels. New York: American
Society of Mechanical Engineers.
8 ASME Boiler and Pressure Vessel Code Interpretations
(issued periodically). New York: American Society of
Mechanical Engineers.
Further Reading
Steel Tanks for Liquid Storage. In: Steel Plate Engineering
Data, 1976the, vol. 1. Washington, D.C: American Iron
and Steel Institute.
7
Design standards.
10
2
Selection of Vessel, Specifications, Reports, and Allowable Stresses
2.1 Selection of Vessel
Although many factors contribute to the selection of
pressure vessels, the two basic requirements that affect
the selection are safety and economics. Many items
are considered, such as materials availability, corrosion
resistance, materials strength, types and magnitudes of
loadings, location of installation including wind loading
and earthquake loading, location of fabrication (shop or
field), position of vessel installation, and availability of
labor supply at the erection site.
With increasing use of special pressure vessels in the
petrochemical and other industries, the availability of
the proper materials is fast becoming a major problem.
The most usual material for vessels is carbon steel. Many
other specialized materials are also used for corrosion resistance or the ability to contain a fluid without
degradation of the material’s properties. Substitution
of materials is prevalent, and cladding and coatings
are used extensively. The design engineer must be in
communication with the process engineer in order that
all materials used will contribute to the overall integrity
of the vessel. For those vessels that require field assembly (in contrast to those that can be built in the shop),
proper quality assurance must be established for acceptable welding regardless of the adverse conditions under
which the vessel is made. Provisions must be established
for radiography, stress relieving, and other operations
required in the field.
For those vessels that will operate in climates where
low temperatures are encountered or that contain fluids
operating at low temperatures, special care must be taken
to ensure impact resistance of the materials at low temperatures. To obtain this property, the vessel may require
a special high-alloy steel, nonferrous material, or some
special heat treatment.
2.2 Which Pressure Vessel Code is Used
The first consideration must be whether or not there is
a pressure vessel law at the location of the installation.
If there is, the applicable codes are stated in the law. If
the jurisdiction has adopted the American Society of
Mechanical Engineer (ASME) Code, Section VIII, the
decision may be narrowed down to selecting whether
Division 1 or Division 2 is used.
There are many opinions regarding the use of Division 1 versus Division 2, but the “bottom line” is
economics. In the article “ASME Pressure-Vessel Code:
Which Division to Choose?,” [1] the authors have listed
a number of factors for consideration. Division 1 uses
approximate formulas, charts, and graphs in simple
calculations. Division 2, on the other hand, uses a complex method of formulas, charts, and design by analysis,
which must be described in a stress report. Sometimes,
so many additional requirements are added to the minimum specifications of a Division 1 vessel that it might be
more economical to supply a Division 2 vessel and take
advantage of the higher allowable stresses.
2.3 Design Specifications and Purchase
Orders
Currently, all ASME code sections, with the exception of
VIII-1, require user design specifications (also called user
design requirements) as part of the code requirements.
These codes require a User Design Specifications to be
prepared and certified by a registered professional engineer experienced in pressure vessel design. This certification by the professional engineer is given on the ASME
Manufacturer’s Data Report. The manufacture is responsible for retaining the user’s Design Report for a specified
number of years.
For the ASME Code, VIII-1, there is no specific statement that any design specifications are required. The
only indication of some sort of design specifications
is the list of minimum loadings in UG-22 that is considered for all construction. The Manufacturer’s Data
Report, Form U-1 for the ASME Code, VIII-1, requires
many items to be listed, which means that most of the
basic design information must be given in a design
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
2.7 Design Data for New Materials
specification or purchase order. Although some codes
help the purchaser regarding what data are needed for
inclusion in the design specifications, this is usually done
by mutual agreement between the purchaser and the
manufacturer.
For those process vessels that do not have a “suggested”
list of items in design requirements and specifications
as part of code requirements, it is necessary to establish
them in the purchase order or contract agreement. The
contract information is supplied by the purchaser or user
with the manufacturer’s advice about what is needed
and what shall be considered. Some design standards
help the user and manufacturer by offering fill-in forms
that specifically list the requirements for designing a
process vessel. Design specification forms for a heat
exchanger built according to the standards of the Tubular Exchanger Manufacturers Association [2] are given in
Appendix B, and those for an API Standard 650 Storage
Tank [3] are given in Appendix C. It is always necessary
to maintain a document containing design specifications
so that a permanent record is kept for reference. Often
on a large process vessel, some loadings from attached
or supported equipment are not known until after the
job has started.
incorporated in the stress analysis calculations. These
calculations are prepared and certified by a registered
professional engineer experienced in pressure vessel
design. As with the User’s Design Specification, the
Manufacturer’s Design Report is mandatory and the certification reported on the Manufacturer’s Data Report.
This is kept on file by the manufacturer for at least
3 years.
For vessels not requiring design reports, the manufacturer has those necessary calculations for satisfying
U-2(g) or other design formulas available for the Authorized Inspector’s review. The pressure vessel design
sheets should contain basic design and materials data
and at least the basic calculations of pressure parts as
given in the design formulas and procedures in the applicable code or standard. For a simple vessel, an example of
calculation sheets is given in Appendix D. This example
depicts only those calculations that are required for
the Authorized Inspector and for construction. Other
vessels may require much more extensive calculations depending upon the complexity and contractual
agreements.
2.6 Materials Specifications
2.4 Special Design Requirements
In addition to the standard information required on
all units, such as design pressure, design temperature,
geometry, and size, many other items of information
are necessary and must be recorded. The corrosion and
erosion amounts are to be given, and a suitable material
and method of protection are to be noted. The type of
fluid that will be contained (e.g., such as lethal) must be
noted because of the required specific design details.
Supported position, vertical or horizontal, and support
locations, as well as any local loads from supported
equipment and piping, must be listed. Site location is
given so that wind, snow, and earthquake requirements
can be determined. Impact loads and cyclic requirements
are also included.
For the ASME Code, VIII-2, a statement as to whether
or not a fatigue analysis is required according to 2.2.2.1
(f ) is given. If a fatigue analysis is required, the specific
cycles and loadings will be given. In addition, the design
specifications state whether or not certain loadings are
sustained or transient. The allowable stresses vary with
the type of loadings.
2.5 Design Reports and Calculations
The ASME Code, VIII-2, requires a formal design report
with the assumptions in the User’s Design Specification
All codes and standards have materials specifications and
requirements describing which materials are permissible. Those materials that are permitted with a specific
code are either listed or limited to the ones that have
allowable stress values given. Depending upon the code
or standard, permitted materials for a particular process
vessel are limited. For instance, only materials with an
SA or SB designation or in a Code Case can be used in
ASME Boiler and Pressure Vessel Code construction.
Most of the SA and SB specifications are the same as
an A or B specification in the ASTM Standards [4].
In specific instances, certain materials that have been
manufactured to some other specification, such as the
DIN Standard [5] may be recertified to an SA or SB
specification for an ASME certified vessel. Depending
upon the contract specifications, permissible materials
for construction 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 to
the Code Committee for evaluation. The ASME Code
Committee lists this information to develop allowable
stresses, strength data, and other required properties
11
12
2 Selection of Vessel, Specifications, Reports, and Allowable Stresses
for accepting a new material into the code. Each section
of the code contains an appendix listing these requirements such as the one for the ASME Code, VIII-1, in
Appendix F. The code also provides data to establish
external pressure charts for new materials; this is given
to those who want to establish new external pressure
charts. The required information is given in Appendix G.
It is the person’s responsibility requesting the addition to
supply all the data needed 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 based on complex theories and various
modes of failure, and the actual design formulas used for
setting the minimum required thicknesses and the stress
levels, a factor of safety (FS) is applied to various materials’ properties that are used to set the allowable stress
values. The factors of safety are directly related to the theories and modes of failure, the specific design criteria of
each code, and the extent to which various levels of actual
stresses are determined and evaluated.
3) 80% of the minimum stress to produce rupture at the
end of 100,000 h.
In the temperature range in which tensile strength or
yield strength sets the allowable stresses, higher allowable stresses are permitted for austenitic stainless steels
and nickel-alloy materials where greater deformation is
not objectionable. In this case, the criterion of 2/3 yield
strength at temperature may be increased to 90% yield
strength at temperature. However, the factor 2/3 specified minimum yield strength is still maintained.
For the ASME Code, VIII-1, bolting material whose
strength has been enhanced by heat treatment or strain
hardening is subject to the additional criteria of (i) 1/5
of the specified minimum tensile strength and (ii) 1/4 of
the specified minimum yield strength.
For the ASME Code, Section VIII-2, the factor used
to set the allowable stress values for all materials except
bolting is the least of:
1)
2)
3)
4)
1/2.4 of the specified minimum tensile strength.
1/2.4 of the tensile strength at temperature.
2/3 of the specified minimum yield strength.
2/3 of the yield strength at temperature (except as
noted earlier where 90% is used).
At temperatures in the creep and rupture strength
range, the factors are the least of:
2.9 Allowable Tensile Stresses in the
ASME Code
As previously discussed, the basis for setting the allowable stress values or the design stress intensity values
is directly related to many different factors depending
upon the section of the code used. The criteria for setting
allowable tensile stresses for each section of the ASME
Boiler and Pressure Vessel Code are as follows:
For Section I, Power Boilers, the ASME Code, VIII-1,
Pressure Vessels, and Section III, Division 1, Subsections
NC, ND, and NE, except for bolting whose strength has
been enhanced by heat treatment, the factors used to set
the allowable tensile stresses, at temperatures in the tensile strength and yield strength range, are the least of:
1)
2)
3)
4)
1/3.5 of the specified minimum tensile strength.
1/3.5 of the tensile strength at temperature.
2/3 of the specified minimum yield strength.
2/3 of the yield strength at temperature (except as
noted in the following where 90% is used).
At temperatures in the creep and rupture strength
range, the factors are the least of:
1) 100% of the average stress to produce a creep rate of
0.01 per 1000 h (1% in 105 h).
2) 67% of the average stress to produce rupture at the end
of 100,000 h.
1) 100% of the average stress to produce a creep rate of
0.01 per 1000 h (1% in 105 h).
2) 67% of the average stress to produce rupture at the end
of 100,000 h.
3) 80% of the minimum stress to produce rupture at the
end of 100,000 h.
For the ASME Code, Section III, Division 1, Subsection
NB and NC-3200 of Subsection NC, the factor used to set
the design stress intensity values for all materials except
bolting is the least of:
1)
2)
3)
4)
1/3 of the specified minimum tensile strength.
1/3 of the tensile strength at temperature.
2/3 of the specified minimum yield strength.
2/3 of the yield strength at temperature except as
noted in the following paragraph.
Higher design stress intensity values are permitted
for austenitic stainless steels and nickel-alloy materials
where greater deformation is not objectionable. In this
case, the criterion of 2/3 yield strength at temperature
may be increased to as high as 90% yield strength at
temperature or any value between 2/3 and 90% yield
strength at temperature depending upon the acceptable
amount of deformation. However, the factor of 2/3
specified minimum yield strength is still maintained.
There are two criteria for setting bolting design stress
intensity values in the ASME Code, VIII-2. For design by
2.10 Allowable External Pressure Stress and Axial Compressive Stress in the ASME Boiler and Pressure Vessel Code
materials’ properties are used to develop the materials
charts.
The allowable compressive stress in the ASME VIII-1,
III-NB, III-NC, and III-ND is as follows:
External Pressure in Cylindrical Shells
Appendix 3, the criteria are the same as for the ASME
Code, VIII-1, because these values are used for the
design of bolts for flanges. For design by Appendix 4 of
the ASME Code, VIII-2, and by Section III, Division 1,
Subsection NB and NC-3200 of Subsection NC, the
criteria for setting bolting design stress intensity values
are the lesser of the following: (i) 1/3 of the specified
minimum yield strength and (ii) 1/3 of the yield strength
at temperature.
For Section IV, Heating Boilers, the criterion for setting
allowable stresses is the least of:
1)
2)
3)
4)
1) A knock down factor of 1.0 is applied to the theoretical
external pressure buckling interaction chart [6].
2) A design factor of 3.0 is applied to the external pressure design equation as discussed in Chapter 8.
3) The allowable external pressure obtained from aforementioned 1 and 2 cannot exceed the allowable tensile
stress.
1/5 of the specified minimum tensile strength.
1/5 of the tensile strength at temperature.
2/3 of the specified minimum yield strength
2/3 of the yield strength at temperature.
External Pressure in Spherical Shells
1) A knock down factor of 1.25 is applied to the theoretical external pressure buckling equation [7].
2) A design factor of 4 is applied to the external pressure
chart correlating compressive stress with strain factor
A as discussed in Chapter 9.
3) The allowable compressive stress obtained from
aforementioned 1 and 2 cannot exceed the allowable
tensile stress.
The aforementioned multiplying factors are summarized in Table 2.1.
2.10 Allowable External Pressure Stress
and Axial Compressive Stress in the
ASME Boiler and Pressure Vessel Code
Axial Compression in Cylindrical Shells
Within the ASME Boiler Code, simplified methods are
given to determine the maximum allowable external
pressure and the maximum allowable axial compressive
stress on a cylindrical shell without having to resort
to complex analytical solutions. Various geometric
values are contained in the geometry chart, whereas
1) A knock down factor of 5 is applied to the theoretical
axial buckling equation [8] of a long cylinder.
2) A design factor of 2 is applied to the external pressure
chart correlating axial compression to strain factor A
as discussed in Chapter 8.
Table 2.1 Multiplying factors on materials’ properties to determine maximum allowable tensile-stress or design-stress intensity values for
the ASME Boiler and Pressure Vessel Code.
Code section
Note
Minimum
specified
tensile
strength
Tensile strength
at temperaturea
Minimum
specified
yield strength
Yield strength
at temperature
Creep rate
of 0.01% in
1000 h
Average
I, III-NC, III-ND,
III-NE, VIII-1, XII
III-B
VIII-2
Bolting
0.80
1/3.5
1/3.5
2/3
0.9
1.0
0.67
0.80
1
1/3.0
1/3.0
2/3
2/3
NA
NA
NA
2
1/3.0
1/3.0
2/3
0.9
NA
NA
NA
1
1/2.4
1/2.4
2/3
2/3
1.0
0.67
0.8
2
1/2.4
1/2.4
2/3
0.9
1.0
0.67
0.8
1
1/4
1/4
2/3
2/3
1.0
0.67
0.8
2
1/5
1/5
2/3
0.9
1.0
0.67
0.8
2 For austenitic SS and nickel alloys only
1.0
0.67
2
1 Minimum for all materials
2/3
Minimum
1/3.5
Notes
2/3
Average
1
a Values in this column are multiplied by 1.1.
1/3.5
Stress to rupture in
100,000 h
13
14
2 Selection of Vessel, Specifications, Reports, and Allowable Stresses
3) The allowable compressive stress obtained from
aforementioned 1 and 2 cannot exceed the allowable
tensile stress.
2.11 Allowable Stresses in the ASME
Code for Pressure Piping
The allowable stresses given in various sections of the
ASME B31 Code for Pressure Piping are similar to the
corresponding sections of the ASME Boiler and Pressure
Vessel Code; however, in some sections, the basis is
different. In the Code for Power Piping B31.1, the allowable tensile stresses are set by the same criteria as used
for ASME Code, Section I. In the Code for Chemical
Plant and Petroleum Refinery Piping B31.3, the allowable tensile stresses for other than bolting are set on a
similar basis as used for ASME Section VIII, Division
1, except a factor of 1/3 is substituted for 1/3.5 on the
tensile strength. The factor of 2/3 on yield strength is
used in both codes. This makes B31.3 in the tensile- and
yield-strength range half way between Divisions 1 and 2
and in the creep and rupture-strength range similar to
Divisions 1 and 2.
2.12 Allowable Stress in Other Codes
of the World
Throughout the world, various factors of safety are
applied to materials data to establish allowable stresses
for the design of boilers, pressure vessels, and piping.
For the temperature range below that temperature
where creep or rupture sets the allowable stresses,
the universal factor for setting allowable stresses is
based on yield strength. In some countries, a factor is
applied to sets of data that have been established from
many tests; in others, the data are determined by the
low yield point or the high yield point. In still other
countries, the actual data for the component being
designed are used, its yield strength being determined
by tests. The actual data on the part are then factored
into the design formulas. Not all countries choose to
use the ultimate tensile strength as a criterion for setting
allowable stresses. When they do, the factor of safety
between various countries is sometimes very different.
In order to show these differences, a discussion follows
regarding the allowable stress basis of several different
countries.
The terms, symbols, and definitions used are as
follows:
UTS = ultimate tensile strength (either specified
minimum or data at design temperature)
YS = yield strength (either specified minimum or
data at design temperature)
R = stress to cause rupture in 100,000 h
C = stress to cause total creep or creep rate in
100,000 h
na = not applicable
n = none or not used
2.12.1
European Union (EN) Countries
Member states of the European Union (EU) and other
countries that are part of the European Economic Area
(EEA), which includes France, Germany, Italy, and the
United Kingdom, use Standard EN 12952 for Water Tube
Boilers, EN 12953 for Shell Boilers and EN 13445 for
Unfired Pressure Vessels.
Rolled and forged steel (carbon and low alloy)
UTS
YS
R
C
Rm20
ReH tc
Rp0,2 tc
RmT tc
1.25
Water tube boilers
EN12952
2.4
1.5
1.5
Shell boilers
EN12953
2.4
n
1.5
n
Unfired pressure
vessels
EN13445
2.4
1.5
1.5a)
1.5 or
1.25b)
a) Replace with Rp1.0T tc /1.3 for creep design.
b) Use 1.25 if creep life monitoring system is installed.
Austenitic steels
A
Water tube
boilers
EN12952 A < 30
A ≥ 30
A ≥ 35
Shell boilers
Unfired
pressure
vessels
EN12953
EN13445 30 < A
≤ 35
A > 35
UTS
YS
R
C
Rm20
ReH tc
Rp0,2 tc
RmT tc
n
n
1.5
1.25
Rm20
ReH tc
Rp1.0 tc
RmT tc
n
n
1.5
1.25
Rm20
Rm tc
Rp1.0 tc
RmT tc
n
3
1.2
1.25
RmT tc
Rm20
ReH tc
Rp0,2 tc
n
n
n
n
Rm20
Rm tc
Rp1.0 tc
RmT tc
n
n
1.5a)
1.5 or
1.25b)
Rm20
Rm tc
Rp1.0 tc
RmT tc
3
1.2a)
1.5 or
1.25b)
a) Replace with Rp1.0T tc /1.3 for creep design.
b) Use 1.25 if creep life monitoring system is installed.
2.12 Allowable Stress in Other Codes of the World
YS = yield strength (either specified minimum or data
at design temperature)
R = 0.2% proof stress (ferritic) or 1.0% proof stress
(austenitic)
Factors for design by rules
UTS
YS
Rave
C
1.0
Carbon and low-alloy steel
2.7
1.5
1.5
C = stress to cause total creep or creep rate at
temperature and defined time
High-alloy steel
2.7
1.5
1.5
1.0
Titanium and titanium alloy
2.7
1.5
1.5
1.0
A = Elongation
Nickel and nickel alloy
2.7
1.5
1.5
1.0
n = none or not used
Aluminum and aluminum alloy
3.0
1.5
—
—
Copper and copper alloy
3.0
1.5
—
—
For precise definitions and use, refer to the individual
standards.
2.12.2
Factors for design by analysis
Japanese Code
UTS
The following codes are used in Japan for boilers and
pressure vessels:
Boilers
JIS B 8201
UTS
YS
R
C
4.0
1.5
1.5 av
1.0
Pressure vessels
4.0
1.5
JIS B 8267
3.5
1.5
JIS B 8266
3.0
1.5
1.5 av
1.0
1.25 min
1.5 av
1.0
1.25 min
Alternative
1.5 av
Carbon and low-alloy steel
2.4
1.5
High-alloy steel
2.4
1.5
UTS = ultimate tensile strength at room temperature
YS = yield strength at design temperature
R = stress to cause rupture in 100,000, 150,000, or
200,000 h according to design life
1.25 min
JIS B 8265
C = stress to cause total creep or creep rate in
100,000 h
For precise definitions and use, refer to the individual
standards.
1.0
1.25 min
UTS = ultimate tensile strength (either specified
minimum or data at design temperature)
2.12.4
Indian Code
The Indian Boiler Regulations Code has the following
safety factors:
YS = yield strength (either specified minimum or
data at design temperature)
R = stress to cause rupture in 100,000 h
C = stress to cause total creep or creep rate in
100,000 h
na = not applicable
For precise definitions and use, refer to the individual
standards.
2.12.3
YS
People’s Republic of China
The rules used in the People’s Republic of China for
design of pressure vessels are given in Supervision Regulation on Safety Technology for Stationary Pressure
Vessel and Supervision Regulation on Safety Technology
for Transportable Pressure Vessel. The allowable stresses
are set by applying the factors in the following tables:
TS
Et
SR
Sc
Boilers
For service temperature at or below 850 ∘ F 2.7 1.5 na na
For service temperature above 850 ∘ F
na 1.5 1.5 1.0
TS = minimum tensile strength of the material at
room temperature
Et = warm yield point (0.2% proof stress) at
temperature t
SR = the average stress at the service temperature to
produce rupture in 100,000 h and, in no case,
more than 1.33 times, the lowest stress to
produce the rupture at service temperature
Sc = the average stress at service temperature to
produce an elongation of 1% creep in 100,000 h
na = not applicable
15
16
2 Selection of Vessel, Specifications, Reports, and Allowable Stresses
For precise definitions and use, refer to the individual
standards.
2.12.5
Australian Code
The Australian unfired pressure vessel Standard AS 1210
has the following safety factors:
Material
type and
design
Material
temperature, ∘ F verification
Min. tensile
stress at
room
temp.
Min. yield
stressa)
at room
temp.
Min. yield
stressa)
at design
temp.
Min.
yield
stressa)
at room
temp.
Min.
yield
stressa)
at design
temp.
120 < T < 300b)
T > 300
2.35
1.5
2.35
1.5
T ≥ 120
hot tested
Not verified
nor hot
tested
Material
verification
Min.
tensile
stress at
room
temp.
Verified or
hot tested
2.5
1.35
Not verified
nor hot
tested
2.5
1.45
Nonferrous alloys
Carbon and low-alloy steels
T ≥ 120 Verified or
Material
type and
design
temperature,
∘F
Verified or
hot tested
3
1.5
Not verified
nor hot
tested
3
1.5
Verified or
hot tested
3
1.5
Not verified
nor hot
tested
3
120 < T < 210b)
120 < T < 300b)
T > 300 Verified or
2.35
1.5
2.35
1.6
hot tested
Not verified
nor hot
tested
High-alloy steels
T ≥ 120 Verified or
2.5
1.5
2.5
1.5
hot tested
Not verified
nor hot
tested
T > 210
1.5
1.7
a) Proof stress determined at 0.2% offset value for ferritic steels and
1% offset value for high-alloy steels.
b) Strength values are based on linear interpolation between the
120 ∘ F values and the 300 ∘ F (or 210 ∘ F) values.
References
1 Smolen, A.M. and Mase, I.R. (January 11, 1982).
ASME Pressure-Vessel Code: Which Division to
Choose? Chemical Engineering.
2 (2007). Standards of Tubular Exchanger Manufactures
Association, 9th ed. Terrytown, NY: Tubular Exchanger
Manufacturers Association.
3 ANSI/API Standard 650 (2014). Welded Tanks for
Oil Storage, 12th ed. Washington, D.C.: American
Petroleum Institute.
4 (2014). 2014 Annual Book of ASTM Standards.
Philadelphia, PA: American Society for Testing and
Materials.
5 DIN Standard (Deutsche Normen DIN), Deutschen
Normenausschu (DNA), Berlin, Germany.
6 Sturm, R. (1941). “A Study of the Collapsing Pressure
of Thin-Walled Cylinders.” Engineering Experiment
Station Bulletin 329. Urbana, IL: University of Illinois.
7 von Karman, T. and Tsien, H.-S. (1939). “The Buckling
of Spherical Shells by External Pressure.” Pressure
Vessel and Piping Design Collected Papers 1927–1959.
ASME, New York, NY.
8 Timoshenko, S. and Gere, J. (1961). Theory of Elastic
Stability. New York, NY: McGraw Hill.
τmax
ε 1 + 𝜇 ε2
σ1 = E
1–𝜇2
σx , τxy
σ2
ε2 + 𝜇 ε1
σ2 = E
1–𝜇2
σ1
τxy = 2(1+𝜇) τxy
E
σy , τxy
σy
τyx
τzy
σx
Mx =
𝜕2 w + 𝜇 𝜕 2 w
–E t3
𝜕y2
12(1–𝜇2) 𝜕x2
My =
–E t3 𝜕2w + 𝜇 𝜕2w
12(1–𝜇2) 𝜕y2
𝜕x2
τzx
σz
Mxy =
E t3
𝜕2 w
12(1+𝜇) 𝜕x𝜕y
Theories, criteria, and basic equations.
18
3
Strength Theories, Design Criteria, and Design Equations
3.1 Strength Theories
In the design of process vessels and pressure equipment,
subjected to tensile loads at temperatures where brittle fracture or creep is not a consideration, two basic
modes of failure may be assumed: elastic failure, based
on the theory of elasticity, and plastic 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 considered to have occurred when the
elastic limit of the material is reached. Beyond this limit,
excessive deformation or rupture is expected. These
limits are usually measured in terms of tensile strength,
yield strength, and (to some degree) rupture strength.
Of the many theories developed to predict elastic failure, the three most commonly used are the
maximum-principal-stress theory, the maximum-shearstress theory, and the distortion-energy theory. The
maximum-principal-stress theory considers failure to
have occurred when any one of the three principal
stresses has reached a value equal to the elastic limit
as determined from a uniaxial tension or compression
test. The maximum-shear-stress theory (also called the
Tresca criterion) considers failure to have occurred
when the maximum shear stress equals the shear stress
at the elastic limit as determined from a pure shear test.
The maximum shear stress is defined as one-half of the
algebraic difference between the largest 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 distortion
energy accumulated in the part under stress reaches the
elastic limit as determined by the distortion energy in a
uniaxial tension or compression test.
Engineers have known for some time that the
maximum-shear-stress theory and the distortion-energy
theory predict yielding and fatigue failure in ductile
materials better than does the maximum-stress theory [1]. However, the maximum-stress theory is easier
to apply, and with an adequate safety factor, it gives
satisfactory designs. But where a more exact analysis is
desired, the maximum-shear-stress theory is used.
Three basic theories of strength are used in the American Society of Mechanical Engineers (ASME) Boiler
and Pressure Vessel Code. The maximum-stress theory is used in Section I [2]; Section IV [3]; VIII-1; and
Section III, Division 1, Subsections NC [4], ND [5],
and NE [6]. The maximum-shear-stress theory is used
in Section III, Division 1, Subsection NB [7], and the
optional part of NC. The von Mises theory is used in the
ASME Code, VIII-2.
In the two sections of the ASME/ANSI Code for Pressure Piping B31 that are used primarily with the ASME
Boiler and Pressure Vessel Code, both ANSI B31.1 [8] and
B31.3 [9] use the maximum-stress theory. B31.3 is unique
in that it uses the maximum-stress theory but permits
allowable stresses to be established on the same basis as
the ASME Code, VIII-2, which requires the use of the
von Mises theory. The other sections of B31 also use the
maximum-stress theory. They require that, in addition
to the stresses caused by internal and external pressures,
those stresses caused by thermal expansion of the piping
be considered.
3.2 Design Criteria
The design criteria for both Sections I and IV basically
call for determining the minimum wall thickness that will
keep the basic circumferential stress below an allowable
stress level. Additional rules and charts are included for
determining the minimum thickness of various components. However, in general, a detailed stress analysis is
required only for special designs. Sections I and IV recognize that 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, VIII-1, and
Section III, Division 1, Subsections NC except NC-3200,
ND, and NE, are similar to those for Sections I and IV
except that the ASME Code, VIII-1, and Section III,
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
3.4 Stress–Strain Relationships
Table 3.1 Stress categories.
Allowable stress valuea)
Factor based on yield strength
General primary membrane (Pm )
S
Sy /1.5
Local primary membrane (PL )
1.5 S
Sy
Primary membrane plus primary bending (Pm + Pb )
1.5 S
Sy
Primary plus secondary (PL + Pb + Q)
3S
2 Sy
Equivalent stress category in VIII-2 or stress intensity category in III
a) The allowable stress in VIII-2 is designated as S, and the allowable stress in III is designated as Sm .
Division 1, Subsections NC, ND, and NE require calculations of cylindrical shell thicknesses based on both
the circumferential and the longitudinal directions. The
minimum required thickness may be set by stresses in
either direction. In addition, the ASME Code, VIII-1,
permits the combination of primary membrane stress
and primary bending stress to reach as high as 1.5S at
temperatures where tensile and yield strength control
and 1.25S at temperatures where creep and rupture
control, where S is the allowable tensile-stress value.
The design criteria for the ASME Code, VIII-2, provide
formulas and rules for the more common configurations
of shells, nozzles, and formed heads. Requirements
include detailed evaluations of actual stresses in complex geometries and with unusual loadings, especially
if cyclic loading occurs. These calculated stresses are
assigned various categories and subcategories that have
different allowable stress values as multiples of the basic
allowable stress intensity value. The various categories
and subcategories are as follows:
• Primary stresses, including general primary membrane stress, local primary membrane stress, and
primary bending stress
• Secondary stresses
• Peak stresses
Primary stress is caused by loadings that are necessary
to satisfy the laws of equilibrium between applied forces
and moments. Primary stresses are not self-limiting.
Secondary stress is developed by self-constraint of the
structure. Its basic characteristic is that it is self-limiting.
That is, rotation and deformation or deflection take place
until the forces and moments are balanced even though
some permanent geometric changes may have taken
place.
Lastly, peak stress is the highest stress condition in
a structure and is usually due to a stress concentration
caused by an abrupt change in geometry. This stress is
important in considering a fatigue failure because of
cyclic load application.
In general, thermal stresses are considered only in
the secondary and peak categories. Thermal stresses
that cause a distortion of the structure are categorized
as secondary stresses; thermal stresses caused by suppression of thermal expansion, but that 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. The primary
plus secondary limits are set to prevent plastic deformation leading to incremental collapse and to validate using
an elastic analysis to perform a fatigue analysis. Finally,
peak stress limits are set to prevent fatigue failure due to
cyclic loadings.
The basic equivalent stress limits for various categories
relating to an analysis according to the ASME Code,
VIII-2, and the basic stress intensity limits in Section III,
Division1, Subsection NB, and optional Part NC-3200 of
Subsection NC are shown in Table 3.1.
The design criteria for Section III, Division 1, Subsection NB, are very similar to those for the ASME Code,
VIII-2, except there is less use of design formulas, curves,
and tables and greater use of design by analysis. The categories of stresses and stress intensity limits are the same
in both sections.
3.3 Design Equations
Once the allowable stresses are set, the basic design
equations must be developed. The design of process
equipment is based on the assumption that the material
generally behaves elastically at the design pressure and
design temperature. Accordingly, most of the equations
are derived from the theory of elasticity and strength of
materials.
3.4 Stress–Strain Relationships
The stress–strain relationship at any point within a
homogeneous, isotropic, and linearly elastic body that is
subjected to a system of forces is obtained from the theory of elasticity. Referring to Figure 3.1, the stress–strain
19
20
3 Strength Theories, Design Criteria, and Design Equations
f2
𝜏xz =
f1
𝜀x , 𝜀y , 𝜀z = axial strain in the x, y, and z directions,
respectively
𝜎 x , 𝜎 y , 𝜎 z = axial stress in the x, y, and z directions,
respectively
f4
(a)
𝛾 xy , 𝛾 yz , 𝛾 xz = shearing strain in the x, y, and z directions,
respectively
dx
𝜏 xy , 𝜏 yz , 𝜏 xz = shearing stress in the x, y, and z directions,
respectively
σz
dy
τzx
τzy
E = modulus of elasticity of material (psi)
τxz
G = shear modulus of material (psi)
𝜇 = Poisson’s ratio.
σx
τyz
τyx
τxy
σy
In most pressure vessel applications, the values of 𝜎 z ,
𝜏 yz , and 𝜏 xz are small compared with 𝜎 x and 𝜎 y . Hence,
they are normally ignored, and the equations reduce to
z
x
y
(b)
Figure 3.1 Stress resultants at a point within a homogeneous,
isotropic, and linearly elastic body.
relationship is given by
1
𝜀x = [𝜎x − 𝜇(𝜎y + 𝜎z )]
E
1
𝜀y = [𝜎y − 𝜇(𝜎z + 𝜎x )]
E
1
𝜀z = [𝜎z − 𝜇(𝜎x + 𝜎y )]
E
𝜏xy
2(1 + 𝜇)
𝛾xy =
=
𝜏xy
G
E
2(1 + 𝜇)
𝛾yz =
𝜏yz
E
2(1 + 𝜇)
𝛾xz =
(3.1)
𝜏xz .
E
or, in a different form,
E
𝜎x =
[𝜀 (1 − 𝜇) + 𝜇(𝜀y + 𝜀z )]
(1 + 𝜇)(1 − 2𝜇) x
𝜎y =
E
[𝜀 (1 − 𝜇) + 𝜇(𝜀x + 𝜀z )]
(1 + 𝜇)(1 − 2𝜇) y
𝜎z =
E
[𝜀 (1 − 𝜇) + 𝜇(𝜀x + 𝜀y )]
(1 + 𝜇)(1 − 2𝜇) z
𝜏xy =
𝜏xy =
(3.2)
where
f3
dz
E 𝛾xz
,
2 (1 + 𝜇)
E𝛾xy
2(1 + 𝜇)
E𝛾yz
2(1 + 𝜇)
1
(𝜎 − 𝜇𝜎y )
E x
1
𝜀y = (𝜎y − 𝜇𝜎x )
E
−𝜇
𝜀z =
(𝜎 + 𝜎y )
E x
2(1 + 𝜇)
𝛾xy =
𝜏xy
E
or, in a different form,
𝜀x =
E
(𝜀 + 𝜇𝜀y )
1 − 𝜇2 x
E
𝜎y =
(𝜀 + 𝜇𝜀x )
1 − 𝜇2 y
𝜎z = 0
E
𝜏xy =
𝛾 .
2(1 + 𝜇) yz
(3.3)
𝜎x =
(3.4)
3.5 Strain–Deflection Equations
Figure 3.2 is cross section of a pressure vessel wall.
It undergoes an extension in the middle surface of 𝜀0
due to stretching plus extension due to bending. The
original length l1 at a distance z from the middle surface
is given by
(
)
z
.
l1 = ds 1 −
rx
The final length l2 after extension is
(
)
z
l2 = ds(1 + 𝜀0x ) 1 − ′ ,
rx
3.5 Strain–Deflection Equations
Hence, Eq. (3.5) may be written as
( 2
[
)]
𝜕 w
E
𝜕2w
+
𝜇𝜀
−
z
+
𝜇
𝜀
𝜎x =
0y
1 − 𝜇2 0x
𝜕x2
𝜕y2
( 2
[
)]
𝜕 w
E
𝜕2w
+
𝜇𝜀
−
z
+
𝜇
𝜎y =
𝜀
.
0x
1 − 𝜇2 0y
𝜕y2
𝜕x2
(3.7)
ds
1
The shearing strain–displacement relationship can be
obtained from Figure 3.3. The quantity 𝛾 xy is shown in
Figure 3.3a and can be expressed as
ЄoXds
𝛾xy = 𝛾0xy + 𝛼 + 𝛽,
2
rʹx
where 𝛾 0xy is the shearing stress due to in-place forces
and 𝛼 and 𝛽 are due to twisting moments. Also, from the
figure,
z
rx
(𝜕u∕𝜕y) dy 𝜕u
=
dy
𝜕y
(𝜕v∕𝜕x) dx 𝜕v
=
𝛽 ≈ sin 𝛽 ≈
dx
𝜕x
𝛼 ≈ sin 𝛼 ≈
and
𝛾xy = 𝛾0xy +
Figure 3.2 Cross section of a shell wall subjected to stretching
and bending loads.
v
Substituting the values of l1 and l2 into the aforementioned equation and omitting all small terms results in
(
)
1
1
𝜀x = 𝜀0x − z ′ −
= 𝜀0x − z𝜒x ,
rx rx
Substitution of the aforementioned two equations into
Eq. (3.4) gives
E
𝜎x =
[𝜀 + 𝜇𝜀0y − z(𝜒x + 𝜇𝜒y )]
1 − 𝜇2 0x
E
𝜎y =
[𝜀 + 𝜇𝜀0x − z(𝜒y + 𝜇𝜒x )].
(3.5)
1 − 𝜇2 0y
β
dy
α
u+
y
𝝏u
dy
𝝏y
(3.6)
d x + 𝝏v d x
𝝏x
(a)
x
–𝝏 w
𝝏x
Note that 𝜒 x is related to the deflection by the expression
d2 w∕dx2
.
𝜒x =
[1 + (dw∕dx)2 ]3∕2
However, because dw/dx is small compared with unity,
the aforementioned expression becomes
v + 𝝏v d x
𝝏x
dy +
where 𝜒 x is the change in curvature. Similarly,
(
)
1
1
𝜀y = 𝜀0y − z ′ −
= 𝜀0y − z𝜒y .
ry ry
d2 w
and 𝜒y =
.
dy2
x
u
l2 − l1
.
l1
d2 w
𝜒x =
dx2
(3.8)
dx
whereas the strain is given by
𝜀x =
𝜕u 𝜕v
+ .
𝜕y 𝜕x
z
(b)
Figure 3.3 Shear deformations of a unit cross section.
𝝏u
dy
𝝏y
21
22
3 Strength Theories, Design Criteria, and Design Equations
From Figure 3.3b, which represents the middle surface,
the rotation is given by −𝜕w/𝜕x. The minus sign indicates
counterclockwise rotation. As a result of this rotation,
any point at a distance z from the middle surface will have
a deflection of
𝜕w
u = −z .
𝜕x
Similarly,
𝜕w
v = −z .
𝜕y
Hence, Eq. (3.8) becomes
𝛾xy = 𝛾0xy − 2z
𝜕2w
𝜕x𝜕y
and Eq. (3.4) becomes
(
)
𝜕2w
𝜏xy = G 𝜏0xy − 2z
.
𝜕x𝜕y
(3.9)
The force–stress relationship for the cross section shown
in Figure 3.4a, b can be expressed as
(
)
t∕2
z
𝜎x 1 −
dz
Nx =
∫−t∕2
ry
x
y
t
Et
(𝜀 + 𝜇𝜀0y )
1 − 𝜇2 0x
Et
Ny =
(𝜀 + 𝜇𝜀0x )
1 − 𝜇2 0y
𝛾0xy Et
Nxy =
2(1 + 𝜇)
)
( 2
−Et 3
𝜕2w
𝜕 w
Mx =
+𝜇 2
12(1 − 𝜇2 ) 𝜕x2
𝜕y
)
(
3
2
−Et
𝜕2 w
𝜕 w
+𝜇 2
My =
12(1 − 𝜇2 ) 𝜕y2
𝜕x
Nx =
Mxy =
z
rx
z
(a)
My
ry
Qy
Qx
Nxy
Nyx
My
Ny
y
Mxy
Et 3 (1 − 𝜇) 𝜕 2 w
.
12(1 − 𝜇2 ) 𝜕x𝜕y
= 13,630 psi
(b)
Figure 3.4 Forces in a unit cross section.
(3.11)
Example 3.1
Stresses are to be determined at the inside corner of an
opening in a cylindrical shell by applying strain gages at
the location. The cylindrical shell is carbon steel with
E = 29.9 × 106 psi and 𝜇 = 0.3. The strain readings from
the three gages are 𝜀x = +360 × 10−6 , 𝜀y = +180 × 10−6 ,
and 𝜀z = −230 × 10−6 . What are the stresses in the three
principal directions at the opening?
Solution:
Using Eq. (3.2), the stresses are determined as
29.9
𝜎x =
[(360)(0.7) + 0.3(180 − 230)]
(1.3)(0.4)
Myx
(3.10)
In the majority of cases, the quantity z/r is small with
respect to unity and can thus be disregarded. Also, substituting Eqs. (3.7) and (3.9) into Eq. (3.10) gives
3.6 Force–Stress Expressions
Nx
(
)
z
𝜎y 1 −
dz
∫
rx
(
)
z
1−
dz
𝜏
Nxy =
∫ xy
ry
(
)
z
Nyx =
dz
𝜏yx 1 −
∫
rx
(
)
z
𝜎x z 1 −
dz
Mx =
∫
ry
(
)
z
𝜎 z 1−
dz
My =
∫ y
rx
)
(
z
dz
Mxy = − 𝜏xy z 1 −
∫
ry
)
(
z
dz.
𝜏 z 1−
Myx =
∫ yx
rx
Ny =
𝜎y =
29.9
[(180)(0.7) + 0.3(360 − 230)]
(1.3)(0.4)
= 9490 psi
Further Reading
𝜎z =
29.9
[(−230)(0.7) + 0.3(360 + 180)]
(1.3)(0.4)
the circumferential and longitudinal axes. The tube
is carbon steel with E = 29.9 × 106 psi, 𝜇 = 0.3, and
the stress at the surface in the circumferential direction is 17,500 psi. What are the strain gage readings
in the two directions?
= 60 psi.
Example 3.2
What are the stresses in the two principal directions of
the cylindrical shell with 𝜎 2 = 0?
Solution:
Using the simplified equations in Eq. (3.4), the stresses
are determined as
29.9
𝜎x =
(360 + 0.3 × 180) = 13,600 psi
0.91
𝜎y =
29.9
(180 + 0.3 × 360) = 9460 psi.
0.91
Problems
3.1
Strain gages are attached to the surface of a tube
subjected to internal pressure. The gages lie along
Answer:
𝜀x = +498 × 10−6 .
𝜀y = +117 × 10−6 .
3.2
In the tube of Problem 3.1, what is the strain in the z
direction? Using that answer and the other answers
in Problem 3.1, what are the calculated stresses in
the three directions?
Answer:
𝜎 x = 17,500 psi.
𝜎 y = 8750 psi.
𝜎 z = 0.
References
1 (1969). Criteria of the ASME Boiler and Pressure Vessel
2
3
4
5
Code for Design by Analysis in Sections III and VIII,
Division 2. New York: American Society of Mechanical
Engineers.
ASME Boiler and Pressure Vessel Code, Section
I Power Boilers. New York: American Society of
Mechanical Engineers.
ASME Boiler and pressure vessel code, section IV
Heating Boilers. New York: American Society of
Mechanical Engineers.
ASME Boiler and pressure vessel code, section III,
division 1, subsection NC Class 2 Components. New
York: American Society of Mechanical Engineers.
ASME Boiler and pressure vessel code, section III,
division 1, subsection ND Class 3 Components. New
York: American Society of Mechanical Engineers.
6 ASME Boiler and pressure vessel code, section
III, division 1, subsection NE Class MC Components. New York: American Society of Mechanical
Engineers.
7 ASME Boiler and pressure vessel code, section
III, division 1, subsection NB Class 1 Components. New York: American Society of Mechanical
Engineers.
8 ASME Code for Pressure Piping B31 Power Piping,
ANSI/ASME B31.1. New York: American Society of
Mechanical Engineers.
9 ASME Code for pressure piping B31 Chemical
Plant and Petroleum Refinery Piping, ANSI/ASME
B31.1. New York: American Society of Mechanical
Engineers.
Further Reading
Brownell, L.E. and Young, E.H. (1959). Process Equipment
Design. New York: Wiley.
Faupel, J.H. (1964). Engineering Design. New York:
Wiley.
Harvey, J.F. (1991). Theory and Design of Pressure Vessels.
2nd ed. Princeton, NJ: Van Nostrand-Reinhold.
Seely, F.B. and Smith, J.O. (1952). Advanced Mechanics of
Materials, 2nde. New York: Wiley.
23
Top: Metalograph of titanium weld. Bottom: Tantalum-clad vessel. Source: Courtesy of the Nooter Corp., St. Louis, MO.
26
4
Materials of Construction
4.1 Material Selection
The vast majority of vessels are constructed of ferrous
and nonferrous alloys. Ferrous alloys are defined as
those having more than 50% iron. They are used in
the ASME Code, VIII-1 and 2 and include carbon and
low-alloy steels, stainless steels, cast iron, wrought iron,
and quenched and tempered steels. Nonferrous alloys
include aluminum, copper, nickel, titanium, and zirconium. The American Society for Testing and Materials
(ASTM) designates all ferrous alloys by the letter A
and all nonferrous alloys by B. American Society of
Mechanical Engineers (ASME) uses the prefixes SA
and SB, respectively. In most cases, the ASME and
ASTM specifications are identical. However, vessels
built according to the ASME Code usually refer to the
ASME specifications.
Nonmetallic pressure vessels may also be constructed
according to the ASME Code. ASME Section X includes
fiber-reinforced plastic (FRP) vessels. Details of the construction are given in Section 4.7. Concrete vessels are
also being considered by the ASME. However, no specific
rules are available at this time.
Figure 4.1 shows an experimental steel–concrete composite pressure vessel developed and owned by Oak
Ridge National Laboratory in Oak Ridge, Tennessee.
The vessel is designed for 6250 psi and was successfully
hydrotested at 8900 psi. The design was performed by
Global Engineering & Technology in Camas, Washington. The inner steel vessel with 4 in. thick layered shell
was built by Kobelco in Japan, and the 12 in. thick outer
reinforced and prestressed concrete shell was built by
Thompson Pipe Group in Dallas, Texas.
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, which is defined as the deterioration of
metals by chemical action, is probably the single most
important consideration in selecting materials. A slight
change in the chemical composition of the environment
can significantly change the corrosive behavior of a
given metal. This is illustrated in Appendix H, which
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 4.2 shows an example of a heat exchanger used
in city water service. The corroded tubesheet is made of
carbon steel, and the uncorroded tubes are made of copper. Another example, shown in Figure 4.3, is a titanium
tubesheet exhibiting crevice corrosion.
In highly corrosive environments, every phase of the
pressure vessel fabrication process, such as burning,
forming, welding, stress relieving, and polishing, must
be evaluated for corrosion. 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 cleanness and finish of the inside surface of a
pressure vessel before its operation are very important in
preventing subsequent corrosion in service. Many users
require special cleaning of the inside surface prior to its
installation.
4.1.2
Strength
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 the yield
strength and ultimate strength are relatively low and the
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
4.1 Material Selection
Figure 4.1 Steel–concrete vessel. Source: Courtesy of Oak
Ridge National Lab.
Figure 4.2 Corroded carbon steel tubesheet. Source:
Courtesy of the Nooter Corp., St. Louis, MO.
creep and rupture behavior may control the allowable
stress values. In the ASME Code, VIII-1, the criteria
for allowable stress at elevated temperatures take into
account both the creep and rupture behaviors as discussed in Section 2.9. For applying the ASME criteria for
allowable stress as given there, the following procedures
are used.
4.1.2.1
Specified Minimum Yield Stress
In obtaining the minimum yield stress of a given material, the test data are plotted at various temperatures as
shown in Figure 4.5. A smooth trend curve is then drawn
through the averages of the data for individual test temperatures. The specified minimum-yield-stress curve is
obtained by applying to the yield trend curve by the ratio
27
28
4 Materials of Construction
Figure 4.3 Corroded titanium tubesheet. Source:
Courtesy of the Nooter Corporation, St. Louis, MO.
Figure 4.4 Crack in a Carpenter 20 tube weld. Source:
Courtesy of Nooter Corp., St. Louis, MO.
of the specified minimum value, as given in the material
specification, to the trend value at room temperature.
4.1.2.2
Specified Minimum Tensile Stress
The tensile trend curve is determined by the same
method as the yield trend curve including the ratio
factor. The specified minimum tensile stress is arbitrarily
taken as 110% of the tensile trend curve, as illustrated in
Example 4.1.
4.1.2.3
Creep Rate
In order to establish the creep rate of 1%/100,000 h, data
are plotted as shown in Figure 4.6. Interpolation and
extrapolation may be needed to establish the creep rate
for various temperature levels.
4.1.2.4
Rupture Strength
Test data are normally plotted as shown in Figure 4.7. In
some cases, the data need to be extended to 100,000 h,
so that extreme care must be taken to extrapolate
accurately.
Example 4.1
A user is requesting code approval for a new material that
has a minimum specified tensile stress of 120 ksi and a
minimum specified yield stress of 60 ksi at room temperature. Tensile and yield values for various heats and
temperatures are shown in Figure 4.5. Creep and rupture
data are given in Figures 4.6 and 4.7, respectively. What
are the allowable stress values at 300 and 1200∘ F based
on the criteria for VIII-I?
4.1 Material Selection
Figure 4.5 Tensile and yield strength.
150
140
130
120
Stress (ksi)
110
100
90
80
70
60
50
40
30
0
200
400
600
800
1000
1200
1400
1600
Temperature (°F)
Figure 4.6 Creep strength.
100.0
Stress (ksi)
10.0
1.0
Creep strength
0.1
0.00001
0.0001
0.001
0.01
% (h)
Solution:
Allowable stress at 300 ∘ F
1) From Figure 4.5, average tensile stress = 130 ksi.
Tensile stress reduced to minimum = 130 × 120/140
= 111 ksi.
Specified minimum tensile stress = 111 × 1.10
= 122 ksi.
Maximum stress to be used cannot exceed
120 ksi.
Allowable stress based on tensile stress = 120/3.5
= 34.3 ksi.
2) From Figure 4.5, average yield stress = 60 ksi.
Yield stress reduced to minimum = 60 × 60/75 =
48 ksi.
29
4 Materials of Construction
Figure 4.7 Rupture strength.
100.0
10.0
Stress (ksi)
30
1100° F
1150° F
1200° F
1300° F
1400° F
1500° F
1.0
Rupture strength
0.1
100
10 000
1000
100 000
Life (h)
Allowable stress based on yield stress = 48 × 23 =
32 ksi.
3) From Figures 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 is
34.3 ksi.
Allowable stress at 1200 ∘ F
1) From Figure 4.5, average tensile stress = 112 ksi.
Tensile stress reduced to minimum = 112 × 120/140
= 96 ksi.
Specified minimum tensile stress = 96 × 1.1 =
106 ksi, which is less than the maximum allowed
(120 ksi).
Allowable stress based on tensile stress = 106/3.5
= 30.3 ksi.
2) From Figure 4.5, average yield stress = 52 ksi.
Yield stress reduced to minimum = 52 × 60/75 =
42 ksi.
Allowable stress based on yield stress = 42 × 23 =
28 ksi.
3) From Figure 4.6, creep stress for 0.01% in
1000 h = 15 ksi.
Allowable stress based on creep = 15 ksi.
4) From Figure 4.7, stress to cause rupture at
105 h = 22 ksi.
Allowable stress based on rupture = 0.67 × 22 =
14.7 ksi.
5) Therefore, maximum allowable stress at 1200 ∘ F is
14.7 ksi.
4.1.3
Material Cost
Because costs of materials vary significantly, the designer
must evaluate material cost versus other factors such
as corrosion, expected life of equipment, availability
of material, replacement cost, and code restrictions on
fabrication and repairs. A summary of the cost of some
frequently used materials is given in Table 4.1. In view of
the large difference in costs, the designer should consider
all factors carefully.
Table 4.1 Approximate cost of materialsa) used in
pressure vessel construction.
Type
Cost ($/lb)
Aluminum
2.00
Carbon steel
0.80
Copper, brass
8.50
Hastelloysb)
24.00
Incoloy
10.00
Inconel
18.00
Low-alloy steel
Monel
Stainless steel
Tantalumc)
1.60
15.00
2.70
80.00
Titanium
35.00
Zirconium
55.00
a) As of January 2017.
b) The cost varies based on grade and product form.
c) The price fluctuates substantially from year to year.
4.2 Nonferrous Alloys
4.2 Nonferrous Alloys
The 2017 ASME Code, VIII-1, lists five nonferrous
alloys for code construction: aluminum, copper, nickel,
titanium, and zirconium. These alloys are normally used
in corrosive environment or at elevated temperatures
where ferrous alloys are unsuitable. Nonferrous alloys
are nonmagnetic except for commercially pure nickel,
which is slightly magnetic.
4.2.1
Aluminum Alloys
All aluminum alloys are categorized by ASME specification number, alloy designation, and temper designation
as shown in Tables 4.2 and 4.3. Some of the terms in the
tables are defined as follows:
4.2.1.1
4.2.1.2
Aluminum alloys have a unique combination of properties that make them useful in process equipment
applications. They are nonmagnetic, are light in
weight, have good formability, and have an excellent
weight–strength ratio. Aluminum surfaces exposed to
the atmosphere form an invisible oxide skin that protects
the metal from further oxidation. This characteristic
gives aluminum a high resistance to corrosion.
Aluminum alloys have a systematic numbering system
as shown in Table 4.2. The specification number also designates the various product forms. For example, SB-209
applies to plate products, and SB-210 applies to drawn
seamless tube products. The first digit of the alloy designation number indicates its major alloying element as
shown in Table 4.2.
Annealing
Heating the material to a given temperature and then
slowly cooling it down. The purpose is to soften the
material in order to remove cold-working stress.
Normalizing
Heating the material to a temperature slightly higher than
the annealing temperature and then cooling at a rate that
is faster than that of annealing.
4.2.1.3
Solution Heat Treating
Heat treating at a temperature high enough for the alloys
to be randomly dispersed.
4.2.1.4
Stabilizing
Low-temperature heating to stabilize the properties of an
alloy.
4.2.1.5
Strain Hardening
Modification of metal structure by cold working resulting
in an increase in strength with a loss in ductility.
Table 4.2 Aluminum alloy designation.
Example
Spec. no.
Alloy designation
Temper
SB-209
3003
↓↓↓
H114
↓↓
𝛼𝛽𝛾
𝛿𝜀
Condenser and heat
exchanger tube
SB-234
Castings
SB-26
Plates
SB-209
Drawn seamless tube
SB-210
SB-211
Seamless pipe and
seamless extruded tube
SB-241
Bolting
Rods, bars, and shapes
SB-221
Die and hand forgings
SB-247
𝛼 – major alloy elements
1) – 99% aluminum. Excellent corrosion resistance; high thermal and electrical conductivity;
low mechanical properties and excellent workability.
2) – Copper. In heat-treated condition, mechanical properties are similar to and sometimes
exceed those of mild steel.
3) – Manganese. Used for moderate-strength applications requiring good workability.
4) – Silicon. Not used in boiler code. Has architectural applications.
5) – Magnesium. Moderate-to-high strength non-heat-treatable alloy
6) – Magnesium and silicon. Main advantage is that it is heat-treatable.
7) – Zinc. Ultrahigh-strength properties. Not used in the boiler code.
𝛽 – modification of the aluminum alloy.
𝛾 – first two digits identify the aluminum alloy.
𝛿 – O: annealed; H: strain hardened; T: thermally treated; F: as fabricated; W: solution heat treated.
𝜀 – degree of annealing (see Table 4.3).
31
32
4 Materials of Construction
Table 4.3 Temper classification for aluminum alloys.
Example
H1 1 4
αβγ δ
α– strain hardened by cold working;
β– 1 strain hardened only;
2 strain hardened then partially annealed;
3 strain hardened then stabilized;
γ– amount of cold work with 8 representing full-hard condition;
δ– degree of control of temper or to identify a special set of mechanical properties.
Example
T6 5 1
αβ
α– thermally treated;
β– 2 annealed;
3 solution HT then cold worked;
4 solution HT then naturally aged to stable condition;
5 artificially aged only;
6 solution HT and then artificially aged;
7 solution HT then stabilized;
8 solution HT, cold worked, the artificially aged;
9 solution HT, artificially aged, then cold worked;
10 artificially aged then cold worked.
4.2.1.6
Thermal Treating
Temperature treatment of an alloy to produce a stable
temper.
4.2.2
Copper and 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 alloy designation system serves to identify the type
of material, as shown in Table 4.4. Alloys 101–199 are
high-grade copper with very few alloys added. Alloys
201–299 are brasses (mainly copper and zinc). Alloys
501–665 are bronzes, composed of copper and elements
other than zinc. Other properties of copper alloys are
also shown in the table.
Most copper alloys are distinguishable by their color,
except for Cu–Ni alloys, which tend to lose their color as
the amount of Ni is increased.
4.2.3
Nickel and High-Nickel Alloys
Nickel and high-nickel alloys have excellent corrosion and oxidation resistance, which makes them
ideal for high-temperature applications with corrosive
Table 4.4 Copper alloys.
Alloy designation of coppers
101–199
Coppers
201–299
Copper–zinc alloys (brass)
301–399
Copper–zinc–lead alloys (leaded
brass)
401–499
Copper–zinc–tin alloy (tin brass)
501–599
Copper–tin alloy (phosphor
bronze)
601–645
Copper–aluminum alloys
(aluminum bronze)
645–665
Copper–silicon (silicon bronze)
666–699
Miscellaneous copper alloys
701–730
Copper–nickel alloys
Cold-worked temper
designation description
Approximate % reduction by cold
working
Quarter hard
10.9
Half hard
20.7
Three-quarters hard
29.4
Hard
37.1
Extra hard
50.0
Spring
60.5
Extra spring
68.7
4.2 Nonferrous Alloys
Table 4.5 Commercial names for some nickel alloys.
Trade namea
(UNSN number)
Composition
Alloy
designation
ASME number
Plate and
sheets
Pipe and
tube
Tube
Rods,
bars shapes,
and forgings
Bolt
Nickel ⟨02200⟩
Ni
200
SB-162
SB-161
SB-163
SB-160
SB-160
Nickel ⟨02201⟩
Ni–low-C
201
SB-162
SB-161
SB-163
SB-160
Monel ⟨04400⟩
Ni–Cu
400
SB-127
SB-165
SB-163
SB-160
}
{
SB-164
Monel ⟨04405⟩
Ni–Cu
405
—
—
{
SB-167
—
SB-564
SB-163
SB-164
}
{
SB-166
SB-516
SB-564
SB-164
SB-164
SB-166
Inconel 600
⟨06600⟩
Ni-Cr-Fe
600
SB-168
Inconel 625
⟨06625⟩
Ni–Cr–Mo–Cb
625
SB-443
SB-444
—
SB-446
SB-446
Inconel 690
⟨06690⟩
Ni–Cr–Fe
690
SB-168
SB-167
SB-166
Incoloy 800 and
800H ⟨08800 and
08810⟩
Ni–Fe–Cr
{800 and
800H}
SB-409
Incoloy 825
⟨08825⟩
Ni–Fe–Cr–Mo–Cu
825
SB-424
SB-423
Hast. B-2 ⟨10665⟩
Ni–Mo
B-2
SB-333
SB-619
Hast. C-4 ⟨06445⟩
Ni–Mo–Cr
C-4
SB-575
SB-517
SB-163
SB-166
{
SB-407
SB-163
}
SB-408
SB-514
SB-515
SB-564
SB-163
SB-425
SB-425
SB-335
SB-335
SB-574
SB-574
SB-574
SB-574
SB-581
SB-581
SB-581
SB-581
SB-581
—
{
}
SB-622
{
SB-626
}
SB-622
{
SB-626
SB-619
Hast. C-276
⟨10276⟩
Ni–Mo–Cr
C-276
SB-575
SB-619
Hast. G ⟨06007⟩
Ni–Cr–Fe–Mo–Cu
G
SB-582
SB-619
SB-622
}
SB-626
}
{
SB-622
{
SB-408
SB-626
}
SB-622
Hast. G-2 ⟨06975⟩
Ni–Cr–Fe–Mo–Cu
G-2
SB-582
Hast. G-3
⟨069857⟩
Ni–Cr–Fe–Mo–Cu
G-3
SB-582
Carp. 20 ⟨08020⟩
Cr–Ni–Fe–Mo–Cu–Cb
20 Cb
SB-463
SB-464
SB-468
Rolled alloy 330
⟨08330⟩
Ni–Fe–Cr–Si
330
SB-536
SB-535
SB-710
SB-511
—
904L ⟨08904⟩
Ni–Fe–Cr–Mo–Cu–low-C
904L
SB-625
SB-677
SB-677
SB-649
—
SB-619
}
{
SB-622
SB-619
SB-626
SB-626
}
{
SB-462
SB-473
SB-462
SB-674
a) Hast. = Hastelloys; Carp. = Carpenter.
environments. Such products are normally called by their
commercial names rather than their ASME designation
number, as shown in Table 4.5.
4.2.4
Titanium and Zirconium Alloys
Titanium and zirconium alloys are used in the process
equipment subjected to severe environment. In the
ASME Code, VIII-1, alloyed titanium and unalloyed
titanium are listed for various grades. 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 of steel. Also, the coefficient
of thermal expansion of both is about half that of steel.
The density of zirconium is slightly less than that of steel,
whereas the density of titanium is about 0.58 times that
of steel.
33
34
4 Materials of Construction
4.3 Ferrous Alloys
Iron alloys with carbon content of less than 2% are known
as steels, and those with more than 2% are known as cast
iron. Steels are further divided into those with carbon
content of more than 0.8%, called hypereutectoid steels,
and those with carbon content of less than 0.8%, called
hypoeutectoid steels. Most steels used in pressure vessel applications have a carbon content of less than 0.4%.
Steels with carbon content of over 0.4% 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 complicated components and configurations. The ASME
Code, VIII-1, also imposes limitations on the pressure
and temperature ranges and the repair methods.
Steel alloys can be produced with a wide variety of
alloying elements. Some of the common elements and
their effect on steel products are shown in Table 4.6.
The ASME Code, VIII-1, divides steel alloys into the
following categories:
4.3.2
Low-Alloy Steels
These are essentially chromium (up to 10%), molybdenum, and nickel-alloy steels. These elements enhance the
steel for high-temperature applications and in hydrogen
service.
4.3.3
High-Alloy Steels
These are commonly referred to as stainless steels. They
have mainly chromium (over 10%), nickel, and molybdenum alloys. The three basic types of stainless steel used
in process equipment are as follows:
4.3.3.1
Martensitic Stainless Steels
This group includes type 410, which has a low chromium
content, slightly above 12%. They behave like steel and are
magnetic, heat-treatable, and difficult to fabricate.
4.3.3.2
Ferritic Stainless Steels
This group includes types 405 and 430. They are magnetic
but not heat-treatable.
4.3.1
Carbon Steels
These are widely used in pressure vessels. They have
mainly silicon and manganese as the main alloying elements and are limited to application temperatures below
about 1000 ∘ F.
4.3.3.3
Austenitic Stainless Steels
This group includes all 200 and 300 series, which are
chromium–nickel and chromium–nickel–manganese
steels. They are nonmagnetic and not heat-treatable.
Table 4.6 Effect of alloying elements in steel.
Element
Advantages
Element
Advantages
Aluminum
Restricts grain growth
Nickel
Strengthens annealed steels
Chromium
Increases resistance to corrosion and
oxidation
Increases hardenability
Toughens steels
Silicon
Improves oxidation resistance
Titanium
Prevents the formation of austenite in
high-chromium steels
Adds strength at high temperature
Manganese
Counteracts sulfur brittleness
Molybdenum
Increases hardenability
Increases hardenability. Strengthens steel
Prevents localized depletion of chromium in
stainless steel during long heating
Raises grain-coarsening temperature
Counteracts tendency toward temper
brittleness
Enhances corrosion resistance
Vanadium
Increases hardenability
Resists tempering
4.5 Brittle Fracture
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 Figure 4.8. Between room temperature and 1333 ∘ F,
the steel consists of ferrite and pearlite. Ferrite is a solid
solution of a small amount of carbon dissolved in iron.
Pearlite, which is shown in Figure 4.9, is a mixture of
ferrite and iron carbide. The carbide is very hard and
brittle.
In Figure 4.8, between lines A1 (lower critical temperature) and A3 (upper critical temperature), the carbide
dissolves more readily into the lattice, which is now
a mixture of ferrite and austenite. Austenite is a solid
solution of carbon and iron that is denser than ferrite.
Above line A3 , the lattice is uniform in its properties, with the austenite being the main component.
The minimum temperature for this austenite range is a
function of the carbon content of the steel, as shown in
Figure 4.8.
With this brief description, we can now discuss various
heat treatments of carbon steel.
4.4.1
This consists of heating the steel to about 100∘ F above
the upper critical line A3 and then cooling in still air. The
purpose is to homogenize the steel structure and produce
a steel harder than that in the annealed condition.
4.4.2
Temperature (°F)
1667
A3
Ferrite
+
Austenite
A1
4.4.4
0.4
Quenching
The rate of cooling of steel after heat treating is very
important in establishing the hardness of steel. Some
steels, such as SA-517, obtain most of their high strength
by quenching. The rate of cooling depends on many
factors, such as the quenching medium, its temperature,
and the size and mass of the part.
Ferrite
+
Pearlite
0.2
Postweld Heat Treating
This consists of heating to a temperature below the lower
critical temperature line A1 for the purpose of reducing
the fabrication and welding stress and softening the weld
heat-affected zones.
Austenite
0
Annealing
This consists of heating the steel to about 50 ∘ F above the
upper critical line A3 and then furnace-cooling slowly.
The purpose is to refine the grain and induce softness.
4.4.3
1333
Normalizing
0.6
0.8
4.4.5
Figure 4.8 Iron–iron carbide equilibrium diagram.
Iron carbide
Tempering
Quenched steels are very brittle. In order to increase
toughness, they are heat treated below A1 and then
cooled to produce the desired property of high strength
and good toughness.
Ferrite
4.5 Brittle Fracture
Annealed
pearlite
Figure 4.9 Pearlite structure.
Normalized
pearlite
Pressure vessel components constructed of ferrous alloys
occasionally fail during hydrotest, initial start-up, or normal operating temperature at a pressure well below the
designed value. Such failures generally occur at low
temperatures and could be minimized by incorporating
brittle-fracture considerations at the design stage. The
required degree of sophistication varies from simple traditional methods to the most complicated mathematical
analyses. Both extremes are useful to the pressure vessel
35
4 Materials of Construction
energy. The magnitude of the measured energy, shape of
the energy curve, and appearance of the cross section of
tested specimens are all significant in evaluating material
toughness.
The energy level at a given temperature varies with different steels, as shown in ASTM A-593. An energy level
of 15 ft-lb is considered adequate for A-283 steel at room
temperature. However, such a level is exceedingly low for
A-387 steels. Recognizing this fact is imperative in specifying energy requirements for various steels at different
temperatures.
The slope of the energy curve in Figure 4.11 gives the
rate of change of steel toughness with temperature. At
the bottom shelf of the curve, the steel is very brittle,
as indicated by the cleavage of the tested specimen.
designer, and their application depends on the amount
of information available and the required reliability of a
given component.
4.5.1
Charpy V-Notch Test (C v )
The C v test is the simplest and most popular method
qualitatively determining the fracture toughness of
low-carbon steels. The test procedure is detailed in
ASTM A-370 and consists of impact testing a notched
specimen (Figure 4.10a) taken from a specific location
of a product form. The specimen is struck with a falling
weight (Figure 4.10b), and the energy required to fracture
it at various temperatures is recorded. Figure 4.11 shows
two typical plots of the temperature versus absorbed
Figure 4.10 Charpy V-Notch specimen.
Striking edge
0.394 in.
0.394 in.
2.165 in. – 2.362 in.
1.575 in.
(a) Standard specimen
(b) Test arrangement
Notch
Shear area
(Dull)
Cleavage area
(Shiny)
(c) Percent of shear fracture
after testing.
Figure 4.11 C v energy transition curves.
Charpy V-Notch values
36
High shelf
energy
A
Low shelf
energy
NDT
Curve (A)
B
NDT
Curve (B)
Temperature
4.5 Brittle Fracture
Failure is normally abrupt. At the upper shelf, material
fails in shear, and the cross section shows a dull area.
Failure occurs after excessive yielding. Low-strength
steels show a sharp increase in toughness as the temperature increases, as shown in curve A of Figure 4.11.
Higher-strength steels show a slight increase, as shown
by curve B. This slight increase in toughness makes the
C v test impractical to use in high-strength steels.
The percentage of dull and bright areas in the cross
section of tested specimens at a given temperature is
a measure of the ductility at failure (Figure 4.10c). It is
also helpful in comparing the ductility of two steels at a
given temperature as well as determining the magnitude
of the test temperature with respect to the nil ductility
temperature.
The nil ductility transition (NDT) temperature shown
in Figure 4.11 is of importance when considering
low-strength steels. Below this temperature, the fracture
appearance of steel changes from part shear to complete
cleavage. Thus, vessels of low-strength steel must not
be operated below this temperature without a detailed
fracture evaluation.
The C v tests give a good qualitative indication of fracture trends. They do not, however, give any correlation
between energy and stress levels. Such information
is needed where a stress analysis is required. For
Figure 4.12 Fracture analysis diagram.
this reason, other methods were devised, such as the
drop-weight test (DWT) established by the US Naval
Research Laboratory.
4.5.2
Drop-Weight Test (DWT)
The DWT procedure is given in ASTM E-208 and
consists of welding a brittle bead on a test specimen.
The bead is then notched, and the specimen is impact
tested at various temperatures. The NDT temperature
is obtained when the specimen does not break upon
impact.
In testing the specimens, deflection can be limited so
that the stress at failure does not exceed the yield value.
Thus, a direct correlation is established between the NDT
temperature and yield stress. Such information is used in
constructing the fracture analysis diagram (FAD).
4.5.3
Fracture Analysis Diagram (FAD)
The FAD is one of the earliest applications of brittlefracture rules to fail-safe designs. The results obtained
from the curve are very conservative but require the
minimum of engineering analysis. A simplified version of the diagram for low-strength steels is shown in
Figure 4.12 and indicates the types of tests required to
construct the diagram.
σu
Small flaw
Initiation
Curve
σy
100%
Shear
A
Break
No break
(In drop. weight test)
Run no run
(In explosion test)
Cat curve
B
Temp.
“Flat”
fracture
(NDT+60°F)
(NDT+120°F)
FTE
FTP
NDT
Bulge
&
fracture
‘Bulge’
&
partial
fracture
“Bulge’’
&
shear
tears
37
4 Materials of Construction
Point A is obtained from the DWT, and it establishes
the location of the NDT temperature with respect to
yield stress. The crack-arrest temperature (CAT) curve,
developed by the Naval Research Laboratory, is obtained
by running explosive tests on sample plates at various
temperatures and observing the crack pattern. From
such tests, the fracture-tear elastic (FTE) point is determined as the temperature at which the crack pattern
changes from bulge and fracture to bulge and partial
fracture, as shown in Figure 4.12. The fracture-tear
plastic (FTP) point is obtained when the crack pattern
changes from bulge and partial fracture to bulge and
shear tears. The FTE point also locates the yield stress
with respect to temperature, whereas the FTP point
locates the ultimate stress.
Below point B in Figure 4.12, fracture does not
propagate regardless of the temperature as long as the
stress is below 5–8 ksi. Between points A and B, other
stress lines are drawn to correlate various stress levels.
These lines are obtained from the Robertson test, which
consists of impact testing a specimen that is stressed
to a certain level and heated from one side to create a
temperature gradient as shown in Figure 4.13.
Figure 4.14 shows the complete fracture analysis diagram [1]. The range of flow sizes at various stress levels
has been obtained from experiments as well as experience. The experiments consisted of using large spheres
of tough material and replacing portions of them with
a notched brittle material. The spheres were then pressurized to a given stress level at the NDT temperature
σ
Figure 4.13 Diagram of specimen used in the
Robertson crack-arrest test.
Notch
Crack
Impact
Arrest
point
ΔT
Hot
end
Cold
end
FTP
Tensile
stress
Yield
stress
(1")
4"
3 Y.S.
4
1 Y.S.
2
Nominal stress
38
1 Y.S.
4
0
8"
Increasing
flaw sizes
NDT
FTE
Fractures
do not
propagate
8"
12"
(Temperature limitations)
1'
2'
Cat curve
5-8KSI
(Stress limitation)
NDT
(NDT + 30°F) (NDT + 60°F)
Temp.
Figure 4.14 Generalized fracture analysis diagram.
(NDT + 120°F)
4.5 Brittle Fracture
of the brittle material. The size of the notch was varied
with different stress levels to obtain the range shown in
the figure.
In using Figure 4.14, the following limitations must be
considered:
1) It applies only to low-carbon steels.
2) It is valid only for thicknesses of less than 2 in. Larger
thicknesses require special evaluation; it has been
proposed that the FTE temperature for thicknesses
over 6 in. should be taken as NDT + 120∘ F rather
than NDT + 60 ∘ F. The FTP temperature should be
NDT + 210∘ F instead of NDT + 120 ∘ F. This indicates that for thick sections, Figure 4.14 is on the
unconservative side, and the safe operating temperature should be greater than those indicated by
the figure.
In using Figure 4.14, the designer must establish the
minimum operating temperature of the equipment.
Figure 4.15 can be used as a guide for determining
the minimum temperature at various locations in the
United States and Canada, unless more accurate data are
available.
Example 4.2
A low-carbon steel material with NDT temperature of
15 ∘ F is used in a pressure vessel. What is the minimum
safe operating temperature for such material?
Solution:
Because no stress level is given, the minimum stress
is assumed at yield. Entering Figure 4.14 at yield
stress, the CAT curve is intersected at the FTE point.
Moving vertically, a temperature of NDT + 60 ∘ F is
obtained. Thus, the minimum safe operating temperature
is 75 ∘ F.
If stress concentrations are assumed in the vessel and
the stress level is beyond yield 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.
Example 4.3
A low-carbon steel vessel with an NDT temperature of
−20 ∘ F is to have a start-up temperature of 0 ∘ F and a
stress level of one-half yield. Is the start-up temperature
safe?
Solution:
From the CAT curve in Figure 4.14, the minimum safe
temperature is at NDT + 30∘ or 10 ∘ F for a stress of
one-half yield. Thus, start-up temperature is on the
unsafe side because it is less than 10 ∘ F. If start-up temperature is critical, the stress will have to be decreased
or a better impact material selected.
4.5.4
Theory of Fracture Mechanics
Basically, the brittle-fracture theory assumes that stress
in the vicinity of a crack (Figure 4.16) due to a load applied
perpendicular to the direction of crack is given by the following expressions:
)(
)
K (
𝜃
𝜃
3𝜃
1 − sin sin
cos
𝜎x = √ I
2
2
2
2πr
)
(
)
(
K
𝜃
𝜃
3𝜃
1 + sin sin
cos
𝜎y = √ I
2
2
2
2πr
)
(
K
𝜃
𝜃
3𝜃
,
sin cos cos
𝜏xy = √ I
2
2
2
2πr
where
𝜎 x , 𝜎 y , 𝜏 xy = stress components at a point (ksi)
r, 𝜃 = polar coordinates from tip of crack
√
K I = fracture toughness factor (ksi in.)
The fracture toughness factor K I is a function of the
applied load as well as the configuration of the body and
crack. Thus, K I can be expressed by
KI = 𝜎F,
(4.1)
where F = crack shape factor.
Unstable crack propagation occurs when the value of
K I reaches a critical value K IC , which is a function of the
properties of the material. Temperature variation can
have a drastic effect on the value of K IC , as in the case
with low-strength carbon steels.
Some published K IC values are given in Table 4.7.
Experimental determination of the K IC factors is
described in ASTM E-399. The methodology of
E-399 is mainly applicable to high-strength steels.
For low-strength steel with substantial plastic deformations, the methodology of E-399 results in unpractical
specimen sizes. For such low-strength materials, the
J-integral method [7] described in ASTM E-1820 is used
where the specimen sizes are kept to a minimum.
Values for the crack shape factor F are normally
obtained from the theory of elasticity. Because of the
complexity of such analysis, only a few cases are suited
for practical use. Some of them are shown in Table 4.8.
Materials, in general, lose their toughness as the yield
strength increases. One measure of toughness is the ratio
K IC /𝜎 y . Ratios larger than 1.5 indicate tough materials,
whereas lower ratios indicate more brittle materials. A
study of K IC /𝜎 y and Eq. (4.1) reveals that the defect factor F has to be very small when 𝜎 y is high and K IC is low.
In other words, very small defects in high-strength materials can lead to catastrophic failures.
Fracture theory is one of the most accurate methods
presently available for evaluating the maximum tolerable defect size. Its main drawback is the difficult task of
39
Compiled from U.S. weather bureau and
meterological div. depl. of transport
of dominion of canada records uo to 1952.
Figure 4.15 Isothermal lines of lowest one-day mean temperature (∘ F). Source: Courtesy of the American Petroleum Institute.
4.5 Brittle Fracture
σ
Table 4.8 Shape factors for common configurations.
Case 1: Flow in a sheet of infinite width
σ
2a
y
Crack
tip
σ
Magnitude of
stress along
x axis
F=
√
Ref. [2, p. 49]
πa
Case 2: Internal circular flow in a sheet of finite width
Stress
σ
Stress
σ
σy
σx
r
θ
2a
σx
σy
W
x
σ
Figure 4.16 Elastic stress distribution near the tip of a crack.
Table 4.7 Some approximate K IC values.
√
KIC (ksi in.)
[
]1
W
πa ∕2
tan
πa
W
Case 3: Internal circular flow in a thick plate
√
a
F=2
,
π
where “a” is the radius of crack
F=
√
πa
Ref. [2, p. 50]
Ref. [3, p. 39]
−300 ∘ F
−200 ∘ F
−100 ∘ F
0 ∘F
A302, grade B
25
34
48
—
A517, grade F
34
44
77
—
A203, grade A normalized
38
50
—
—
42
83
—
—
Case 4: Internal elliptic flow in a thick plate
√
πa
,
Ref. [3, p. 39]
F=
3π π a2
+
8
8 c2
where “2a” is the minor axis and “2c” is the major
axis.
A533, grade B
35
40
46
78
Case 5: Single edge notch
HY-80
55
—
—
—
Material
A203, grade A quenched
and tempered
obtaining K IC for different materials. Economics might
dictate a simplified approach such as FAD or the ASME
criteria with a small permissible defect size rather than a
fracture-theory approach that might allow a larger tolerable defect.
4.5.5
Relationship Between K IC and C V
Determination of K IC values is tedious and expensive,
especially for low-strength steels. Various methods relating K IC to the relatively inexpensive C V test have been
suggested. One empirical method, proposed by Rolfe
and Barsom [3] consisted in preparing two equations for
correlation purposes. One equation relates the C V and
K IC values in the transition temperature region, whereas
the other equation is applicable in the upper shelf region.
Thus, for the transition region [6],
KIC at temp.(T − ΔT) = KId at temp.(T), (4.2)
σ
α
l
σ
√
F = C π•a
Ref. [4, p. 328]
( )2
( )3
√
a
a
a
− 28.32
C = 1.11 π + 0.22 + 10.28
𝓁
𝓁
𝓁
( )4
a
+42.27
𝓁
Case 6: Elliptical surface flow
√
1.12 πa
Ref. [[5], p. 315]
F = √(
)
3π π a2
− 0.212𝜎 2 ∕𝜎y2
+
8
8 c2
where “2c” is the crack length, “a” is the crack depth, 𝜎 is the
actual material stress, and 𝜎 y is the yield stress.
41
42
4 Materials of Construction
where
KId2 ∕E = 5CV
{
215 − 1.5𝜎y
ΔT =
0
raisers should be ground off to minimize their effect. Hot
forming substantially improves the situation because it
increases the NDT temperature and thus prevents brittle
fracture.
36 < 𝜎y ≤ 140 ksi
,
𝜎y > 140 ksi
whereas for the upper shelf range,
(
(
)
)
KIC 2
CV
= 4.0
− 0.1 ,
𝜎y
𝜎y
(4.3)
√
where C V is in ft-lb, 𝜎 y in ksi, and K IC in ksi in. Eqs. (4.2)
and (4.3) are for medium-strength steels such as A517 Gr.
F and A302 Gr. B.
4.5.6
Hydrostatic Testing
Hydrostatic testing of a pressure vessel is the best available method for determining the maximum tolerable
defect size. Thus, if a thick pressure vessel is hydrotested
at a pressure that is 50% greater than the design pressure,
the critical K I is given by Eq. (4.1) as
KIC = 𝜎F.
Assuming an internal defect represented by case 3 in
Table 4.8, the maximum K IC immediately after hydrotesting is
( √ )
a
KIC = 1.5Sm 2
.
π
Maximum defect size x at the design pressure is
given by
( √ )
( √ )
a
x
1.5Sm 2
= Sm 2
π
π
or
x = 2.25a.
Hence, a crack that is discovered after hydrotesting can
grow 2.25 times its original size before causing failure.
This fact illustrates the importance of hydrotesting. A
hydrostatic temperature that is the same as the lowest
operating temperature of the vessel should be used.
4.5.7
Factors Influencing Brittle Fracture
Many factors such as torch cutting, arc strikes, and cold
forming affect the brittle-fracture behavior of metals
and should be considered in fabricating pressure vessels.
Torch cutting or beveling of the plate edges may lead to
hard and brittle areas. In cases where this condition is
undesirable, the plate should be heated to minimize this
effect. Grinding the edges eliminates the hard surfaces.
Arc strikes can create failure by brittle fracture, especially if the strike is made over a repaired area. It is desirable to grind and repair all arc strikes before hydrotesting,
especially at low temperatures.
Cold forming of thick plates may lead to fracture in
areas with stress raisers or plate scratches. All stress
Example 4.4
A titanium pipe (ASTM B265 Gr. 5) with a 2.375 in.
outer diameter and a 0.154 in. wall thickness has an
actual stress√of 30 ksi, a yield stress of 120 ksi, and
KIC = 40 ksi in. at a given temperature. The pipe contains a flaw of depth 0.05 in. and length 0.25 in., which
is similar to case 6 in Table 4.8. What is the maximum
internal pressure the pipe can hold?
Solution:
From a conventional strength of material analysis, the
pressure required to yield the pipe is given in Figure 5.4 as
P=
𝜎(R2o − R2i )
R2o + R2i
=
120(1.1882 − 1.0342 )
1.1882 + 1.0342
= 16.5 ksi.
Using the fracture-toughness approach, the maximum
stress is
K
𝜎 = IC .
F
From case 6 in Table 4.8,
√
1.12 (π)(0.05)
F=√
(1.178 + 0.063)2 − 0.212(30∕120)2
= 0.359.
Hence, 𝜎 = 40/0.359 = 111.4 ksi, and
111.4(1.1882 − 1.0342 )
= 15.4 ksi.
1.1882 + 1.0342
Therefore, fracture-toughness criteria control the
design.
max P =
Example 4.5
An A302-B material with a yield stress of 50 ksi is to be
used in a pressure vessel. The C V value is 15 ft-lb, and an
examination of the percentage shear in the cross section
of tested specimens indicates a temperature in the transition range. Ultrasonic examination of the plate uncovered an elliptical defect under the surface that is 0.375 in.
long and 0.25 in. deep. How safe is the vessel if the operating stress is 2/3 yield? Let K IC = K Id and E = 30 × 103 ksi.
Solution:
√
√
From Eq. (4.2), KIC = 5 × 30 × 15 = 47 ksi in..
From Table 4.8, case 4,
√
π(0.125)
= 0.46.
F=
3π∕8 + (π∕8)(0.125∕0.1875)2
4.5 Brittle Fracture
Hence, from Eq. (4.1), 𝜎 cr = 47/0.46 = 102 ksi. Actual
stress = 23 × 50 = 33 ksi, which is less than the critical
brittle stress. Therefore, operation of the vessel is safe
unless the defect grows in size.
4.5.8
ASME Pressure Vessel Criteria
The ASME Code, VIII-1, uses a simplified approach
for preventing brittle fracture in pressure vessels constructed with carbon steel. The code uses exemption
curves (Figure 4.17) to determine the acceptable minimum temperature for a given material and thickness
where impact testing is not required. The figure is based
on experience as well as test data. The notes to the
figure list only a small number of specifications that are
assigned to the various exemption curves. Additional
specifications are to be added as more data become available. The figure requires impact testing for all welded
thicknesses over 4 in. and all nonwelded thicknesses over
6 in. Also, the minimum acceptable temperature without
impact testing is −55 ∘ F. The 0.394 in. cutoff limit on
the left-hand side corresponds to 1 cm, which is the
size of a Charpy V-notch specimen. In using the curves,
the designer must specify the minimum design metal
temperature at which the vessel is to operate. The ASME
Code does not provide minimum-temperature charts for
Figure 4.17 Impact-test exemption curves.
Assignment of materials to curves. Source:
Courtesy of ASME.
various locations. Figure 4.15 can be used as a guide to
minimum temperatures for vessels located in the United
States and Canada.
The exemption curves in Figure 4.17 can be constructed from theoretical brittle-fracture considerations.
This is done by expressing the relationship between 𝛽 IC
and K IC as [2]
)2
(
1 KIC
𝛽IC =
,
(4.4)
B 𝜎y
where
K IC = stress intensity factor
B = thickness
𝛽 IC = factor
𝜎 y = yield stress.
Tests have shown that a value 𝛽 IC = 1.5 represents a
leak-before-failure condition in the vessel. On the other
hand, a value 𝛽 IC = 0.4 represents the upper limit of
small-scale yielding where the linear theory of elastic
fracture mechanics is applicable. Using a conservative
𝛽 IC value of 1.5, Eq. (4.4) becomes
1.5B =
2
KIC
𝜎y2
.
(4.5)
Impact test exemption curves
160
140
120
Minimum design metal temperature (°F)
100
A [Note (1)]
B [Note (2)]
80
60
C [Note (3)]
40
D [Note (4)]
20
0
−20
−40
−55
−60
Impact texting required
−80
0.394
1
2
3
4
Governing thickness (in.)
[Limited to 4 in. for welded construction]
5
6
43
4 Materials of Construction
Tests correlating K IC to the temperature and yield
stress have shown that an approximate equation can be
written as
𝜎y
,
(4.6)
KIC =
C1 − C2 T
theoretical solution results in lower temperature values
compared to the conservative curve in the ASME code.
Similar equations can be developed for various materials
after establishing the values of C 1 and C 2 .
Notes to Figure 4.17:
where
1) Curve A applies to:
a) all carbon and all low-alloy steel plates, structural
shapes, and bars not listed in Curves B, C, and D.
b) SA-216 Grades WCB and WCC if normalized
and tempered or water-quenched and tempered;
SA-217 Grade WC6 if normalized and tempered
or water-quenched and tempered.
2) Curve B applies to:
a) see the following:
SA-216 Grade WCA if normalized and tempered
or water-quenched and tempered
SA-216 Grades WCB and WCC for thicknesses
not exceeding 2 in. (50 mm), if produced to
fine-grain practice and water-quenched and
tempered
SA-217 Grade WC9 if normalized and tempered
SA-285 Grades A and B
SA-414 Grade A
SA-515 Grade 60
SA-516 Grades 65 and 70 if not normalized
SA-612 if not normalized
SA-662 Grade B if not normalized
SA/EN 10028-2 Grades P235GH, P265GH,
P295GH, and P355GH as rolled
SA/AS 1548 Grades PT430NR and PT460NR;
b) except for cast steels, all materials of Curve A if produced to fine-grain practice and normalized, which
are not listed in Curves C and D;
c) all pipe, fittings, forgings, and tubing not listed for
Curves C and D;
d) parts permitted under UG-11 shall be included in
Curve B even when fabricated from plate that otherwise would be assigned to a different curve.
3) Curve C applies to:
a) see the following:
SA-182 Grades F21 and F22 if normalized and tempered
SA-302 Grades C and D
SA-336 F21 and F22 if normalized and tempered,
or liquid-quenched and tempered
SA-387 Grades 21 and 22 if normalized and tempered, or liquid-quenched and tempered
SA-516 Grades 55 and 60 if not normalized
SA-533 Grades B, C, and E
SA-662 Grade A;
b) all materials listed in 2(a) and 2(c) for Curve B if
produced to fine-grain practice and normalized,
normalized and tempered, or liquid-quenched and
C 1 , C 2 = constants
T = temperature.
Substituting Eq. (4.5) into Eq. (4.6) gives
T=
C1
1
−√
.
C2
1.5BC 2
(4.7)
Eq. (4.7) forms the basis for plotting various exemption
curves. The constants C 1 and C 2 have to be established
experimentally. For SA 302 grade B material, tests have
shown that the approximate values of C 1 and C 2 are
C1 = 1.15
C2 = 0.0208,
and Eq. (4.7) becomes
39.3
T = 55.3 − √ .
B
(4.8)
A plot of this equation is shown in Figure 4.18. Also
shown is a plot of curve B from Figure 4.17, which is
for SA 302 grade B material. Figure 4.18 shows that the
100
80
Minimum temperature (°F)
44
Figure 4.17
60
Eq. (4.8)
40
20
0
−20
1
2
3
4
5
Thickness (in.)
Figure 4.18 Thickness–temperature relationship for SA 302-B
material.
6
4.5 Brittle Fracture
For bolting, the following impact-test exemption temperature shall apply:
Specification number
Grade
Exemption temp. (∘ F)
SA-193
B5
−20
SA-193
B7
−40
SA-193
B7M
−50
SA-193
B16
−20
SA-307
B
−20
SA-320
L7, L43
Impact tested
SA-325
1
−20
SA-354
BB
−20
SA-354
BC
0
SA-354
BD
+20
SA-449
−20
The derivation of Eq. (4.8) extends the brittle-fracture
theory to materials with relatively low yield strength.
This produces a number of uncertainties, since the theory is intended for materials with high yield strength.
Also, the brittle-fracture theory disregards numerous
factors that affect the toughness of low-strength steels
such as fine-grain practice, normalizing, quenching,
and postweld heat treating. Because of these factors,
the shape and temperature exemption of the curves in
Figure 4.17 are based on the experience of the industry
with the various materials listed in the figure rather than
on theoretical equations.
When a material is required to be impact tested in
accordance with Figure 4.17, the specified minimum
energy level is obtained from Figure 4.19. This figure
takes into consideration the interaction between thickness, yield strength, and toughness levels. The energy
level given by the figure is obtained from experience with
various materials and thicknesses.
0.394 in.
50
Minimum
specified
yield
strength
40
65 ksi
Cv (ft–lb)
tempered as permitted in the material specification
and not listed for Curve D.
4) Curve D applies to:
SA-203
SA-508 Grade 1
SA-516 if normalized or quenched and tempered
SA-524 Classes 1 and 2
SA-537 Classes 1, 2, and 3
SA-612 if normalized
SA-662 if normalized
SA-738 Grade A
SA-738 Grade A with Cb and V deliberately added
in accordance with the provisions of the material
specification, not colder than −20 ∘ F (−29 ∘ C)
SA-738 Grade B not colder than −20 ∘ F (−29 ∘ C)
SA/AS 1548 Grades PT430N and PT460N.
30
55 ksi
50 ksi
45 ksi
20
38 ksi
and
lower
10
0
0
1
2
Thickness of material (in.)
3
Figure 4.19 Charpy V-notch test requirements.
Curves similar to those shown in Figure 4.19 for the
interaction between thickness, energy, and yield strength
can be expressed theoretically by using the equation [2]
2
KIC
∕E = 5CV .
(4.9)
Combining this equation with Eq. (4.5) gives
CV = 0.01B𝜎y2 .
This equation gives the required values that tend to be
too high for most available low-carbon steels. Accordingly, Figure 4.19 was generated from experience rather
than theoretical formulation.
In developing Figures 4.17 and 4.19, consideration was
given to the fact that the actual relationships between
C V , K IC , and 𝜎 y as given by Eqs. (4.2) and (4.9) are all
based on dynamic strain rates. The strain rate developed
in a Charpy V-notch test specimen is about 10 (in./in.)/s,
while that in a pressure vessel is significantly lower and
is of the magnitude of 10−3 (in./in.)/s. Hence, a given
energy level of a C V specimen at a given temperature can
be acceptable in a pressure vessel of low-carbon steel
with lower strain rate at a significantly lower temperature. Because of this, the ASME gives some allowance
for low-strength steel to be used at a design temperature
lower than the Charpy V-notch test temperature, as
shown in Table 4.9.
It has been shown [1] that vessels constructed of
low-carbon steel having a stress level below 6000 psi do
not fail in brittle fracture. Accordingly, the ASME Code
permits materials to be used at temperatures below
those given in Figure 4.15 if the stress in a component is
45
4 Materials of Construction
Table 4.9 Impact text temperature differential.
1.6
a
Minimum specified
yield strength (ksi)
Temperature
difference (∘ F)a)
≤40
10
≤55
5
>55
0
1.4
a/l = 0
β
l
0°
90°
1.2
t
a) Impact-test temperature may be higher than the minimum
design metal temperature by the amount shown.
Mb
1.0
decreased below the allowable stress level. The amount
of decrease in temperature as the stress level decreases
is based on many theoretical considerations. The first is
the intensity factor, which is defined as
√
(4.10)
KI = πa∕Q(Sm Mm + Sb Mb ),
a/l = 0.3
a/l = 0.1
0.8
a/l = 0.2
0.6
0.4
where
√
K I = stress intensity factor, ksi in.
0.2
a/l = 0.2
a/l = 0.3
Exact solution (β = 0°)
Estimate (β = 0°)
Estimate (β = 90°)
a/l = 0.4
a/l = 0.5
Sm , Sb = membrane and bending stress, respectively, ksi
Q = flaw shape factor
0.1
a = depth of a surface flaw or one-half the minor
diameter of an embedded flaw
0.2
0.3
a/t
0.4
0.5
0.6
Mm , Mb = correction factors for membrane and bending
stress.
Figure 4.21 Bending-stress correction factor. Source: Courtesy of
ASME.
The ASME code, VIII-1, assumes a maximum flaw
size of
a = t∕4 and l∕a = 6,
Based on these assumptions, the values of Mm and Mb
are obtained from Figures 4.20 and 4.21 and are given by
where
Mm = 1.18 and Mb = 0.80.
l
=
major diameter of flaw, in.
t
=
vessel wall thickness, in.
These values are based on a surface flaw and are higher
than those obtained from subsurface flaws. It is also
Figure 4.20 Membrane-stress correction factor.
Source: Courtesy of ASME.
2.0
a =0
l
1.9
a
1.8
1.7
a = 0.2
l
a = 0.05
l
a = 0.1
l
l
1.6
Mm
46
a = 0.25
l
a = 0.15
l
t
1.5
a = 0.3
l
1.4
1.3
1.2
0.35 < a < 0.5
l
1.1
0.1
0.2
0.3
0.4
0.5
a/t
0.6
0.7
0.8
4.5 Brittle Fracture
assumed that the total stress is expressed as
K I curve. One such curve is shown in Figure 4.23 for
SA-533 Gr. B class 1 and SA-508 class 2 and 3 steels. The
K I value at NDT + 60 ∘ F is used as a basis for establishing
the temperature reduction shown in Figure 4.23. The
temperature reduction is given in Table 4.11 and is
plotted in Figure 4.24. This figure is used by the ASME
for determining the reduction in minimum allowable
temperature for various stress levels without impact
testing.
Exact values of the stress intensity factor K I near a flaw
in a stress field with a gradient distribution can be calculated by the rules given in many codes and standards.
The American Petroleum Institute [8] API-579 Standard
gives the detailed methodology in Annex C for calculating K I . Similar solutions are given in Article A-3000 of
the ASME [9] Section XI “Rules for Inservice Inspection
of Nuclear Power Plant Components.”
Sm + Sb ≤ Sy .
Letting Sm = 23 Sy and Sb = 13 Sy , Eq. (4.10) reduces to
√
KI = 1.867Sy a∕Q.
(4.11)
K I in Eq. (4.11) is a function of Sy , Q, and a. The value
of Sy is assumed to be 40 ksi, which is an average value
for the majority of materials used in pressure vessel construction. The value of Q is obtained from Figure 4.22
and is a function of the stress level. The actual value of
a in Eq. (4.11) depends on the thickness of the vessel
wall. Table 4.10 lists values of K I as a function of stress
level using Eq. (4.11), different wall thicknesses, and
Figure 4.22. The table also gives the ratio of the K I values
at reduced stress levels to that at the full stress level of
40 ksi.
The K I ratios calculated in Table 4.10 are used to
obtain temperature-reduction values from an actual
Figure 4.22 Shape factors. Source: Courtesy of ASME.
(Sm + Sb)/Sy
0.4
1.0
0.8
0.3
a/l
0.5
0.3
0
0.2
a
l
l
0.1
2a
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
Q
Table 4.10 K IC values and ratios.
S/Sy = 1.0
0.8
0.5
0.4
Q = 1.045
1.105
1.18
1.20
Thickness (in.)
a (in.)
K c (1)
K IC (2)
(2)/(1)
K IC (3)
(3)/(1)
K IC (4)
(4)/(1)
6
1.5
89
70
0.79
42
0.47
33
0.37
4
1.00
73
57
0.78
34
0.47
27
0.37
2
0.50
52
40
0.77
24
0.46
19
0.37
1
0.25
37
28
0.76
17
0.46
14
0.38
0.5
0.125
26
20
0.77
12
0.46
10
0.38
Average
where Sy = yield stress, ksi.
0.77
0.47
0.37
47
4 Materials of Construction
Figure 4.23 K IC test data. Source: Courtesy of ASME.
200
180
KIC (ksi– in.)
160
140
120
100
80
60
40
20
NDT – 50
NDT
NDT + 50 NDT + 100 NDT + 150
Temperature (°F)
Figure 4.24 Temperature reduction.
Source: Courtesy of ASME.
1.00
0.90
Ratio of actual stress to allowable stress
48
0.80
0.70
0.60
0.50
0.40
0.35
0.30
0.20
0.10
0.00
0
20
40
60
80
100
Reduction in minimum temperature (°F)
120
140
4.5 Brittle Fracture
Table 4.11 Temperature-reduction values.
K IC ratio
average
from Table 4.10
K IC from
Figure 4.23
√
(ksi in.)
1.0
114a)
0.77
88
0.47
54
0.37
42
Temperature
Temperature
difference (∘ F)
NDT + 60 ∘ F
NDT + 41 ∘ F
19
NDT + 0 ∘ F
NDT – 46 ∘ F
0
60
106
a) This value is obtained from entering Figure 4.23 at NDT + 60 ∘ F.
Figure 4.25 The Nelson chart. Source: Courtesy of the
American Petroleum Institute.
1400
Stainless steel
1200
6Cr–1/2 Mo steel
Temperature (°F)
1000 1Cr–1/2 Mo steel
800
3Cr–1/2 Mo steel
2Cr–1/2 Mo steel
600
1/2 Mo steel
400
Carbon steel
200
1000
2000
3000
Hydrogen partial pressure (p.s.i)
4000
49
50
4 Materials of Construction
4.6 Hydrogen Embrittlement
Essentially, the two different ways in which hydrogen can
embrittle steels are as follows:
1) Hydrogen decarburization. In this case, hydrogen penetrates the steel and combines with the carbides in the
structure (Figure 4.8) to form methane gas. This gas
accumulates in the space of the original carbide and
builds up pressure that leads to cracking. This process
normally accelerates with an increase in temperature
and in operating pressure. One method of minimizing
hydrogen attack is by using Cr–Mo steels. Here the
carbides are in solution with the Cr or Mo and do not
readily combine with the hydrogen. The type of steel
to be used in a given combination of temperature and
pressure is normally determined by the Nelson chart
in Figure 4.25.
2) Hydrogen attack. Researchers have observed that
hydrogen attacks certain regions of a pressure vessel
at temperatures below 200 ∘ F when they have high
hardness zones in the range of 200 Brinell and higher.
The exact mechanism is not known exactly, but it
is believed that the hydrogen is attracted to hard
regions with higher-stressed zones. Accordingly,
many users require soft heat-affected zones with a
Brinell hardness below 200 to avoid hydrogen attack
at low temperatures.
4.7 Nonmetallic Vessels
Rules for 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, and filament-winding
processes.
In the contact-molding 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 cylinder and head preforms
against an outer heated mold. The vessels constructed by
this process are also limited to 150 psi pressure.
In the centrifugal-casting process, the cylindrical
sections are formed from chopped fiberglass strands and
a resin system in a mandrel, which is spun to produce
a suitable laminate and heated to cure the resin system.
Pressure vessels constructed by this method are also
limited to 150 psi design pressure.
In the filament-winding 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 3000 psi for filament-wound vessels with uncut filaments.
FRP vessels normally operate at low temperatures. The
ASME Code, Section X, limits the temperature between
a minimum of −65 ∘ F and a maximum of 150 ∘ F. Also,
because the modulus of elasticity is about 1 × 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 accepted as adequate, a prototype must be cycled
100,000 times between zero and design pressure and
then burst at a pressure not less than six times the design
pressure.
Recently, alternative rules were added to the ASME
Code, Section X, for the construction of vessels based
on theoretical analysis rather than the proof-testing
method.
References
1 Pellini, W.S. (1972). Principles of fracture safe
design—Part 1. In: Pressure Vessels and Piping: Design
and Analysis, vol. 1. New York: American Society of
Mechanical Engineers.
2 Tetelman, A.S. and McEvily, A.J. Jr. (1967). Fracture of
Structural Materials. New York: Wiley.
3 Rolfe, S.T. and Barsom, J.M. (1977). Fracture and
Fatigue Control in Structures. Englewood Cliffs, NJ:
Prentice-Hall.
4 Riccardella, P.C. and Mager, T.R. (1972). Fatigue
crack growth in pressurized water reactor vessels.
In: Pressure Vessels and Piping: Design and Analysis,
vol. 1. New York: American Society of Mechanical
Engineers.
5 Wessel, E.T. and Mager, T.R. (1972). Fracture mechanics technology as applied to thick-walled nuclear
pressure vessels. In: Pressure Vessels and Piping: Design
and Analysis, vol. 1. New York: American Society of
Mechanical Engineers.
6 Roberts, R. and Newton, C. (1981). Interpretive Report
on Small-Scale Test Correlations with KIC Data,”
Bulletin 265. New York: Welding Research Council.
Further Reading
7 Dowling, N. (2007). Mechanical Behavior of Materials.
9 ASME Section XI (2017). Rules for Inservice Inspection
New Jersey: Prentice-Hall.
8 API-579/ASME FFS-1 (2016). Fitness-For-Service. New
York, NY: ASME.
of Nuclear Power Plant Components. New York: American Society of Mechanical Engineers.
Further Reading
1 Aluminum Standards and Data, Aluminum Associa-
5 Tada, H., Paris, P.C., and Irwin, G.R. (2000). The
tion, Washington, D.C., 1979.
2 Anderson, T.L. (2005). Fracture Mechanics. Boca
Raton: CRC.
3 Avner, S.H. (1964). Introduction to Physical Metallurgy. New York: McGraw-Hill.
4 Nichols, R.W. (1971). Pressure Vessel Engineering
Technology. England: Applied Science Publishers.
Stress Analysis of Cracks Handbook. New York: ASME
Press.
6 Thielsch, H. (1965). Defects and Failures in Pressure
Vessels and Piping, R. E. New York: Krieger.
51
53
Part II
Analysis of Components
A multitude of cylindrical shells in a chemical plant. Source: Courtesy of E.I. du Pont de Nemours and Co.
56
5
Stress in Cylindrical Shells
5.1 Stress Due to Internal Pressure
The classic equation for determining stress in a thin
cylindrical shell subjected to pressure is obtained from
Figure 5.1. Summation of forces perpendicular to plane
ABCD gives
PL•2r = 2𝜎𝜃 Lt
or
𝜎𝜃 =
Pr
,
t
(5.1)
where:
P = pressure
L = length of the cylinder
𝜎 𝜃 = hoop stress
Equations (5.1) and (5.4) give accurate results when
r/t > 10. As r/t decreases, however, a more accurate
expression is needed because the stress distribution
through the thickness is not uniform. Recourse is then
made to the thick-shell theory [1] first developed by
Lame. The derived equations are based on the forces
and stresses shown in Figure 5.3. The theory assumes
that all shearing stresses are zero due to symmetry and
that a plane section that is normal to the longitudinal
axis before pressure is applied remains plane after pressurization. In other words, 𝜀l is constant at any cross
section.
A relationship between 𝜎 r and 𝜎 𝜃 can be obtained by
taking a free-body diagram of the ring dr as shown in
Figure 5.3b. Summing forces in the vertical direction and
neglecting higher-order terms, we then have
r = inside radius
𝜎𝜃 − 𝜎r = r
t = thickness
The strain 𝜀𝜃 is defined as
E
[𝜖 (1 − 𝜇) + 𝜇(𝜖y + 𝜖l )]
(1 + 𝜇)(1 − 2𝜇) x
E
𝜎y =
[𝜖 (1 − 𝜇) + 𝜇(𝜖x + 𝜖l )]
(1 + 𝜇)(1 − 2𝜇) y
(5.6)
𝜎l = 0.
𝜎x =
and from Figure 5.2,
2π(r + w) − 2πr
2πr
𝜖𝜃 =
w
.
r
or
(5.2)
Also,
dw
(5.3)
dr
The radial deflection of a cylindrical shell subjected to
internal pressure is obtained by substituting the quantity
𝜀𝜃 = 𝜎 𝜃 /E into Eq. (5.2). Hence, for thin cylinders,
𝜖r =
Pr2
w=
,
Et
where,
w = radial deflection
E = modulus of elasticity
(5.5)
A second relationship is obtained from Eq. (3.2), which
is written as
final length − original length
𝜖𝜃 =
,
original length
𝜖𝜃 =
d𝜎r
.
dr
Substituting Eqs. (5.2) and (5.3) into the first two
expressions of Eq. (5.6) and substituting the result into
Eq. (5.5) results in
d2 w 1 dw w
+
− 2 = 0.
dr2
r dr
r
A solution of this equation is
w = Ar +
(5.4)
B
,
r
(5.7)
where A and B are constants of integration and are determined by first substituting Eq. (5.7) into the first one of
Eq. (5.6) and then applying the boundary conditions
𝜎r = −Pi
at r = ri
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
5.1 Stress Due to Internal Pressure
Figure 5.1 Free-body diagram of a cylindrical
shell subjected to internal pressure.
P
σθ
C
σθ
D
B
L
A
r
t
thick cylinders as
r
𝜎𝜃 =
Pi ri2 − Po ro2 + (Pi − Po )(ri2 ro2 ∕r2 )
ro2 − ri2
𝜎r = −
r+w
P
Po ro2
−
Pi ri2
+ (Pi − Po )(ri2 ro2 ∕r2 )
ro2 − ri2
,
(5.9)
where:
𝜎 r = radial stress
𝜎 𝜃 = hoop stress
Pi = internal pressure
Po = external pressure
ri = inside radius
Figure 5.2 Thin cylindrical shell.
ro = outside radius
Po
r = radius at any point
2(r+ dr)(σθ + dσr)
σr + dσr
dr
dr
σ
θ
σθ
σr
r
(2r)(σr)
r
r1
ro
Pi
σθ
σθ
(a)
(b)
The distribution given by Eq. (5.9) of the stresses
through the thickness of a cylinder due to internal
and external pressures is shown in Figures 5.4 and 5.5,
respectively.
A comparison between Eqs. (5.1) and (5.9) is shown in
Figure 5.6. The figure illustrates the adequacy of Eq. (5.1)
for ro /ri ratios of less than or equal to 1.1 (or, conversely,
ri /t ≥ 10).
The longitudinal stress (Figure 5.7) in a thick cylinder is
obtained by summing forces in the longitudinal direction
to give
𝜎l =
Figure 5.3 Cross section of a thick cylindrical shell.
Pi ri2 − Po ro2
ro2 − ri2
.
(5.10)
With 𝜎 l known, Eq. (5.8) for the deflection of a cylinder
can be expressed as
and
at r = ro .
𝜎r = −Po
w=
Eq. (5.7) then becomes
w=
Er(ro2 − ri2 )
.
(5.11)
1
− ri2 )
Er(ro2
2
×[r
r2 (Pi ri2 − Po ro2 )(1 − 𝜇) + (Pi − Po )ri2 ro2 (1 + 𝜇)
(1 − 𝜇)(Pi ri2 − Po ro2 ) + ri2 ro2 (1 + 𝜇)(Pi − Po )].
(5.8)
Once w is obtained, the values of 𝜎 𝜃 and 𝜎 r are determined from Eqs. (5.2), (5.3), and (5.6) and expressed for
The deflection pattern for external and internal pressures is shown in Figure 5.8.
Example 5.1
The inside radius of a hydraulic cylinder is 12.0 in. What is
the required thickness if P = 7500 psi and 𝜎 𝜃 = 20,000 psi?
57
58
5 Stress in Cylindrical Shells
σθ
Pi
Eq. (5.1)
Eq. (5.1)
–
σθ
Po
Po
+
Pi
σr
Pi
σθ =
Piri2
ro2 – ri2
ro2 + r2
σr =
–Piri2
ro2 – ri2
ro2 – r2
σθ =
r2
2ri2
σθ
=
Pi ro2 – ri2
7500 × 12
Pr
=
t=
𝜎𝜃
20,000
= 4.50 in.
From Lamé’s equation (see Figure 5.4),
ro2 + ri2
𝜎𝜃
= 2
P
ro − ri2
t = r1
𝜎𝜃 + P
− r1
𝜎𝜃 − P
At outer surface
Hence, the error of using Eq. (5.1) in this case is 22%.
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
and corresponding deflection if the allowable stress is
15,000 psi, 𝜇 = 0.3, and E = 30 × 106 psi?
t=
50 × 72
= 0.24 in.
15,000
σθ = –2Poro
ro2 – ri2
σθ =
–Po(ro2 + ri2)
ro2 – ri2
σr
= –1.0
Po
Figure 5.5 Stress distribution in a thick cylinder due to external
pressure.
From Eq. (5.4),
50 × 722
= 0.0360 in.
30 × 106 × 0.24
Using the thick-shell theory, we obtain the required
thickness from Figure 5.4 as
(√
)
𝜎𝜃 + P
t = r1
−1
𝜎𝜃 − P
)
(√
15,000 + 50
−1
= 72
15,000 − 50
w=
= 0.24 in.
= 5.80 in.
Solution:
From the membrane Eq. (5.1),
r2
σr
=0
Po
Figure 5.4 Stress distribution in a thick cylinder due to internal
pressure.
√
r2 – ri2
2
At Inner Surface
σr
=0
Pi
or
–Poro2
ro2 – ri2
r2
Po
σθ ro2 + ri2
=
Pi ro2 – ri2
Solution:
From the membrane Eq. (5.1),
σr =
σr
σr
= –1
Pi
At outer surface
r2 + ri2
r2
–
At inner surface
–Poro2
ro2 – ri2
and from Figure 5.8,
)
(
(50)(72) 72.242 + 722
+ 0.3
w=
30 × 106 72.242 − 722
= 0.0361 in.
Examples 5.1 and 5.2 indicate that Eq. (5.1) is adequate
when the ratio r1 /t is greater than 10.
Problems
5.1
A cylinder with an inside diameter of 24 in. is subjected to an internal pressure of 10,000 psi. Using
5.1 Stress Due to Internal Pressure
Figure 5.6 Comparison of formulas for hoop stress
in a cylindrical shell.
20.0
σθ
P
10.0
Eq. (5.9)
Eq. (8.1)
1.0
Eq. (5.1)
0.10
1.0
1.5
2.0
2.5
3.0
ro
ri
3.5
4.0
Internal pressure
Pi
w=
Pi.ri2[r2(1–μ) + ro2(1+μ)]
E.r [ro2 – ri2)
Max.w at inner surface
σ
wE/Pi
wi =
Po
α=
Piri
(α+μ)
E
ro2 + ri2
r o2 – r i2
P1
Figure 5.7 Longitudinal pressure and stress.
an allowable stress of 25,000 psi, determine the
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
internal pressure that can be applied if the maximum allowable stress is 20,000 psi.
Answer:
Pi = 7690 psi.
P
External pressure
w=
–Po.ro2[r 2(1–μ) + ri 2(1+μ)]
E.r [ro2 – ri2)
Max.w at inner surface wi =
–Po . ro2 . ri(2)
E( ro2 – ri2)
Figure 5.8 Radial deflection due to internal and external pressure.
59
60
5 Stress in Cylindrical Shells
5.3
A cylinder is subjected to an external pressure of
5000 psi and an internal pressure of 2000 psi. If
r1 = 15 in. and r2 = 19 in., what are the maximum
circumferential, longitudinal, and radial stresses?
At what location do they occur?
Mθ
Qx
Nx
x
y
z
Answer:
𝜎 𝜃 =−17,900 psi at inside surface
𝜎 r =−5000 psi at outside surface
𝜎 l =−9960 psi uniform through thickness
Mx + dMx dx
dx
Nx + dNx dx
dx
Qx + dQx dx
dx
p
Mx
Nθ
Mθ
Nθ
r
dx
dθ
(a)
5.2 Discontinuity Analysis
Nθdθdx
All the previous equations were based on the assumption
that the cylinder is free to deform under pressure. In
practical applications, the cylinder is attached to end
closures that restrain its deformation. Other items such
as stiffening rings and internal bulkheads affect the cylinder deformation and introduce local stresses. These local
stresses can be evaluated by a discontinuity-type analysis
using the general bending theory of thin cylindrical
shells. The theory assumes that the loads are symmetric
around the circumference and that the thickness of
the shell is small compared with its radius. It is also
assumed that the in-plane shearing forces and moments
are zero. The problem then reduces to that of solving
the forces shown in Figure 5.9. The relationship between
these forces can be obtained from statics. Hence, from
Figure 5.9,
∑
Fx = 0
or
dNx
dx r d𝜃 = 0,
dx
which indicates that N x must be a constant.
Let
dθ
(b)
Figure 5.9 Forces in a unit length of a cylindrical shell.
Deleting Qx from Eqs. (5.12) and (5.13) gives
N𝜃 d2 Mx
= P.
(5.14)
+
r
dx2
This equation has two unknowns, N 𝜃 and Mx . Both
unknowns can be expressed in terms of the deflection w.
The relationship between Mx and w is given by Eq. (3.11)
as
( 2
)
𝜕 w
𝜕2w
+
𝜇
Mx = −D
𝜕x2
𝜕𝜃 2
( 2
)
𝜕 w
𝜕2w
M𝜃 = −D
+
𝜇
,
𝜕𝜃 2
𝜕x2
D=
Fz = 0
or
dQx N𝜃
+
= P.
dx
r
Similarly,
∑
My = 0
(5.12)
Et 3
.
12(1 − 𝜇2 )
(5.15)
Because the rate of change of deflection in the 𝜃 direction is zero due to symmetry, the aforementioned two
equations reduce to
( 2 )
dw
(5.16)
Mx = −D
dx2
and
(
M𝜃 = −𝜇D
or
dMx
− Qx = 0.
dx
Nθdx
where
Nx = 0.
Also,
∑
Nθdx
d2 w
dx2
)
or
(5.13)
M𝜃 = 𝜇Mx .
(5.17)
5.2 Discontinuity Analysis
The expression for N 𝜃 is derived from the axial and
hoop strains. On referring to Figure 5.9, the axial strain
is given by
Q0
M0
du
(5.18a)
dx
and the hoop strain as
w
(5.18b)
𝜖𝜃 = −
r
Substituting Eqs. (5.18a) and (5.18b) into Eq. (3.11)
gives
𝜖x =
Nx =
Et
(𝜖 + 𝜇𝜖𝜃 ) = 0
1 − 𝜇2 x
or
( )
du
w
.
=𝜇
dx
r
Also,
M0
x
Et
N𝜃 =
(𝜖 + 𝜇𝜖x )
1 − 𝜇2 𝜃
Figure 5.10 Edge force and moment in a cylindrical shell.
or
−Etw
.
(5.19)
r
Substituting Eqs. (5.16) and (5.19) into Eq. (5.14) yields
( 4 )
dw
Etw
+D
= −P.
r2
dx4
N𝜃 =
Defining
3(1 − 𝜇2 )
Et
,
=
4r2 D
r2 t2
the differential equation becomes
𝛽4 =
−P(x)
d4 w
+ 4𝛽 4 w =
,
dx4
D
where P is a function of x.
5.2.1
(5.20)
(5.21)
Long Cylinders
One of the most practical applications of Eq. (5.21) is for
long cylinders subjected to end shears and moments as
shown in Figure 5.10. The force and deformation distribution at any point x along the cylinder due to Q0 and M0
can be obtained from Eq. (5.21) with P = 0. Hence,
4
dw
+ 4𝛽 4 w = 0.
dx4
A solution of this equation can be expressed as
(C3 cos 𝛽x + C4 sin 𝛽x) − f (x).
(5.22b)
The constants C 3 and C 4 can be evaluated from the
boundary conditions
( 2 )|
dw |
Mx |x=0 = M0 = −D
|
dx2 ||x=0
and
(
)
dM ||
Qx |x=0 = Q0 = −D
| .
dx ||x=0
Hence,
−1
C3 = 3 (Q0 + 𝛽M0 )
2𝛽 D
M
C4 = 20 .
2𝛽 D
Eq. (5.22b) then becomes
w=
e−𝛽x
[𝛽M0 (sin 𝛽x − cos 𝛽x) − Q0 cos 𝛽x].
2𝛽 3 D
By defining
B𝛽x = e−𝛽x (cos 𝛽x − sin 𝛽x)
C𝛽x = e−𝛽x cos 𝛽x
w = e (C1 cos 𝛽x + C2 sin 𝛽x)
+e
w = e−𝛽x (C3 cos 𝛽x + C4 sin 𝛽x).
A𝛽x = e−𝛽x (cos 𝛽x + sin 𝛽x)
𝛽x
−𝛽x
infinity. Thus, the constants C 1 and C 2 must be set to
zero, and the solution becomes
(5.22a)
By observation, we can conclude that the deflection
due to Q0 and M0 approaches zero as x approaches
D𝛽x = e−𝛽x sin 𝛽x,
(5.23)
the deflection and its derivatives can be expressed as
1
w = 3 (𝛽M0 B𝛽x + Q0 C𝛽x )
2𝛽 D
61
62
5 Stress in Cylindrical Shells
−1
dw
= 2 (2𝛽M0 C𝛽x + Q0 A𝛽x )
dx
2𝛽 D
determine the distance x at which the moment is about
7% of the original applied moment M0 .
1
d2 w
=
(2𝛽M0 A𝛽x + 2Q0 D𝛽x )
dx2
2𝛽D
d3 w −1
=
(5.24)
(2𝛽M0 D𝛽x − Q0 B𝛽x ).
dx3
D
Values of A𝛽x , B𝛽x , C 𝛽x , and D𝛽x are given in
Table 5.1.
Using the terminology of Eqs. (5.23) and (5.24), the
expressions for Mx and Qx are represented by
( 2 )
dw
1
=
(2𝛽M0 A𝛽x + 2Q0 D𝛽x )
Mx = −D
dx2
2𝛽
Qx = (2𝛽M0 D𝛽x − Q0 B𝛽x ).
(5.25)
The relationship between Mx , M𝜃 , Q, N 𝜃 , w, and 𝜃 for
various boundary conditions is shown in Table 5.2.
Example 5.3
A long cylindrical shell is subjected to an end moment
M0 . Plot the value of Mx from 𝛽x = 0 to 𝛽x = 4.0. Also,
Solution:
From Eq. (5.25), Mx = M0 A𝛽x .
The values of A𝛽x are obtained from Table 5.1, and a
plot of Mx is shown in Figure 5.11. From Table 5.1, the
values of 𝛽x at which Mx is equal to 7% of M0 is about
2.00:
𝛽x = 2.00
or
x=
and
2
𝛽
√
x = 1.56 rt.
(1)
√
The significance of the quantity 1.56 rt is apparent
from Figure 5.11. It shows that a moment applied at the
end dissipates very
√ rapidly and decreases by as much as
94% at x = 1.56 rt. This indicates that any other force
Table 5.1 Values of functions A𝛽x , B𝛽x , C 𝛽x , D𝛽x .
𝜷x
A𝜷x
B𝜷x
C 𝜷x
D𝜷x
0
1.0000
1.0000
1.0000
0.0000
0.05
0.9976
0.9025
0.9500
0.0475
0.10
0.9907
0.8100
0.9003
0.0903
0.15
0.9797
0.7224
0.8510
0.1286
0.20
0.9651
0.6398
0.8024
0.1627
0.30
0.9267
0.4888
0.7077
0.2189
0.40
0.8784
0.3564
0.6174
0.2610
0.50
0.8231
0.2415
0.5323
0.2908
0.55
0.7934
0.1903
0.4919
0.3016
0.60
0.7628
0.1431
0.4530
0.3099
0.80
0.6354
−0.0093
0.3131
0.3223
1.00
0.5083
−0.1108
0.1988
0.3096
1.20
0.3899
−0.1716
0.1091
0.2807
1.40
0.2849
−0.2011
0.0419
0.2430
1.60
0.1959
−0.2077
−0.0059
0.2018
1.80
0.1234
−0.1985
−0.0376
0.1610
2.00
0.0667
−0.1794
−0.0563
0.1231
2.50
−0.0166
−0.1149
−0.0658
0.0491
3.00
−0.0423
−0.0563
−0.0493
0.0070
3.5
−0.0389
−0.0177
−0.0283
−0.0106
4.0
−0.0258
0.0019
−0.0120
−0.0139
5.0
−0.0045
0.0084
0.0019
−0.0065
6.0
0.0017
0.0031
0.0024
−0.0007
7.0
0.0013
0.0001
0.0007
0.0006
5.2 Discontinuity Analysis
Table 5.2 Various discontinuity functions.
o
M0
Functionsa)
o
O
Q0
θ0
Δ0
Edge functions
w
−M0
2𝛽 2 D
Q0
2𝛽 3 D
Δ0
0
𝜃
M0
𝛽D
−Q0
2𝛽 2 D
0
𝜃0
Mx
M0
0
2𝛽 2 DΔ0
2𝛽D𝜃 0
N𝜃
−2M0 𝛽 2 r
2𝛽rQ0
EtΔ0
r
0
Q0
0
Q0
4𝛽 3 DΔ0
2𝛽 2 D𝜃 0
w
−M0
B
2𝛽 2 D 𝛽x
Q0
C
2𝛽 3 D 𝛽x
Δ0 (2C 𝛽x − B𝛽x )
𝜃0
(C − B𝛽x )
𝛽 𝛽x
𝜃
M0
C
𝛽D 𝛽x
−Q0
A
2𝛽 2 D 𝛽x
2𝛽Δ0 (A𝛽x − C 𝛽x )
𝜃 0 (A𝛽x − 2C 𝛽x )
Mx
M0 A𝛽x
Q0
D
𝛽 𝛽x
2𝛽 2 DΔ0 (A𝛽x − C 𝛽x )
2𝛽D𝜃 0 (D𝛽x − A𝛽x )
N𝜃
−2M0 𝛽 2 rB𝛽x
2𝛽rQ0 C 𝛽x
Et
Δ (2C𝛽x − B𝛽x )
r 0
Et𝜃0
(C𝛽x − B𝛽x )
r𝛽
Qx
−2𝛽M0 D𝛽x
Q0 B𝛽x
4𝛽 3 DΔ0 (B𝛽x − D𝛽x )
2𝛽 2 D𝜃 0 (2D𝛽x + B𝛽x )
General functions
a) Clockwise moments and rotation are positive at point 0. Outward forces and deflections are positive at point 0. M0 = 𝜇Mx .
applied at that distance x can be analyzed without regard
to the applied moment M0 .
To find the maximum moment Mx , the aforementioned equation can be differentiated with respect to x
and equated to zero. Hence,
Example 5.4
A long cylinder is subjected to end shear Q0 . Plot
the value of Mx as a function of Q0 from 𝛽x = 0 to
𝛽x = 4.0, and derive the location of the maximum value
of Mx .
Solution:
From Eq. (5.25),
Mx =
Q0
D
𝛽 𝛽x
Referring to Table 5.1 for values of D𝛽x , a plot of
Mx /(Q0 /𝛽) can be constructed as shown in Figure 5.11.
Q0 −𝛽x
e sin 𝛽x
𝛽
dMx
Q
= 0 = 0 (−𝛽e−𝛽x sin 𝛽x + 𝛽e−𝛽x cos 𝛽x)
dx
𝛽
Mx =
or
𝛽x =
π
4
and maximum moment is given by
0.322Q0
M=
𝛽
63
64
5 Stress in Cylindrical Shells
b) The discontinuity stress in the shell and ring if the ring
has a thickness of 0.375 in. and a depth of 4.0 in.
Mx
M0
1.0
0.8
0.6
0.4
0.2
0
–0.2
–0.4
1.0
0.8
Mx 0.6
Q0 /β 0.4
0.2
0
–0.2
–0.4
Solution:
A free-body diagram of the shell-to-ring junction is
shown in Figure 5.13. Because the ring is assumed to
have infinite rigidity, the deflection due to pressure must
be brought back to zero by a force Q0 . Also, because the
slope at the shell-to-ring junction is zero (due to symmetry), a moment M0 must be applied at the junction
to reduce the slope created by force Q0 to zero. From
Figure 5.13,
βx
1.0
2.0
3.0
4.0
5.0
deflection due to P − deflection due to Q0
+deflection due to M0 = 0.
1.0
2.0
3.0
4.0
5.0
The deflection due to P is obtained from Eq. (1) in
Example 5.5, whereas the deflections due to M0 and Q0
are obtained from Eq. (5.24). Hence,
(
Q
M
𝜇)
Pr2
1−
− 30 + 20 = 0.
Et
2
2𝛽 D 2𝛽 D
From Eq. (5.15),
βx
Figure 5.11 Longitudinal moment distribution.
D = 0.00143E,
Example 5.5
Determine the expression for the deflection of a long
cylinder with end closures due to internal pressure P.
Solution:
For internal pressure P, the axial force is N x = Pr/2 and
the hoop force is N 𝜃 = Pr. Also from Eqs. (3.11) and
(5.18), assuming outward deflection as positive,
)
(
w
Pr
Et
𝜖
+
𝜇
=
x
2
1 − 𝜇2
r
and
Pr =
Et
1 − 𝜇2
(
)
w
+ 𝜇𝜖x .
r
Pr 1 − 2𝜇
Et 2
w=
(
𝜇)
Pr2
1−
.
Et
2
and
𝛽 = 0.3636.
Hence, the deflection compatibility equation becomes
M0 − 2.750Q0 = −321.39.
(1)
The second compatibility equation gives
rotation due to Q0 − rotation due to M0 = 0
or
Q0 − 2𝛽M0 = 0.
(2)
Solving Eqs. (1) and (2) gives
M0 = 321.4 in.-lb∕in.
Solving for 𝜀x and w gives
𝜖x =
and from Eq. (5.20),
The maximum longitudinal stress is given by
Pr 6M0
+ 2
𝜎x =
2t
t
= 40,900 psi.
(1)
Example 5.6
A stiffening ring is placed around a cylinder at a distance
from the ends as shown in Figure 5.12. The cylinder has
a radius of 50.0 in. and a thickness of 0.25 in. and is subjected to an internal pressure of 100 psi.
Assuming E = 30 × 106 psi and 𝜇 = 0.3, find
a) The discontinuity stress in the shell with the ring
assumed to have infinite rigidity.
The maximum hoop moment is given by Eq. (5.17) as
M𝜃 = 96.4 in.-lb∕in.
The hoop force N 𝜃 is given by Eq. (5.19) as
Etw
.
r
But because w = 0, N 𝜃 is equal to zero and the maximum hoop stress is
6M
𝜎𝜃 = 2 𝜃
t
= 9300 psi.
N𝜃 =
5.2 Discontinuity Analysis
Figure 5.12 Cylindrical shell with stiffening ring.
4ʺ × 3/8ʺ ring
t = 0.25ʺ
d
b
50ʺ
Figure 5.13 Sign convention at point 0:
clockwise 𝜃 and M0 are +; outward w and Q0
are −.
M0 M
0
Q0
Pr(r + d∕2)
dE
𝜃 = 0.
w=
Due to Q0 ,
2Q0 r(r + 2)
w=
b dE
𝜃 = 0.
Q0
𝜃 = 0.
The deflection compatibility is
[wp − wQ0 + wM0 ]cyl = [wp + w2Q0 ]ring
(
Q
M
P (r + d∕2)
𝜇)
Pr2
1−
− 30 + 20 = r
Et
2
2𝛽 D 2𝛽 D
dE
2Q (r + d∕2)
+ 0
bdE
or
Due to M0 ,
The ring deformations are expressed as follows:
Due to P,
Q 0 Q0
w=0
w=
M
w = 20
2𝛽 D
M
𝜃 = 0.
𝛽D
0
Due to M0 ,
Due to Q0 ,
Q0
2𝛽 3 D
−Q
𝜃 = 20 .
2𝛽 D
M0
0
r
Solution (b):
The shell deformations are expressed as follows:
Due to P,
(
𝜇)
Pr2
1−
w=
Et
2
𝜃 = 0.
M0
M0 − 4.06Q0 = −296.8.
(3)
Similarly,
[𝜃p + 𝜃M0 − 𝜃Q0 ]cyl = [𝜃p + 𝜃M0 − 𝜃Q0 ]ring
or
M0
Q0
−
=0
𝛽D 2𝛽 2 D
and
2𝛽M0 − Q0 = 0.
Solving Eqs. (3) and (4) yields
M0 = 152.0 in.-lb∕in.
Q0 = 110.6 lb∕in.
(4)
65
66
5 Stress in Cylindrical Shells
The hoop stress in the ring is
5.5
Pr 2Q0 r
+
d
bd
= 1250 + 7370
= 8620 psi.
𝜎r =
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 maximum discontinuity stress?
Answer:
The maximum longitudinal stress in the cylinder is
Pr 6M0
+ 2
2t
t
100 × 50 6 × 152
=
×
2 × 0.25
(0.25)2
= 24,600 psi.
0.966Pr
Pr
+ √
2t t 3(1 − 𝜇2 )
= 18,740 psi
𝜎=
𝜎x =
The hoop force at the discontinuity is
Etw
,
r
N𝜃 =
where
w = wp − wQ0 + wM0
(
M
Q
𝜇)
Pr2
=
1−
− 30 + 20
Et
2
2𝛽 D 2𝛽 D
447,500
=
E
or
E(0.25)(447,500∕E)
50
= 2238 lb∕in.,
N𝜃 =
and the hoop stress at the discontinuity is
N𝜃 6M𝜃
+ 2
t
t
2238 6(0.3 × 152)
=
+
0.25
0.252
= 13,300 psi,
𝜎𝜃 =
whereas away from discontinuity,
𝜎𝜃 =
Pr
= 20,000 psi.
t
Problems
5.4
A long cylindrical shell is welded to rigid bulkheads
such that the deflection and rotation due to applied
pressure are zero. If r = 36 in., t = 0.5 in., P = 240 psi,
𝜇 = 0.3, and E = 29 × 106 psi, what is the maximum
longitudinal stress?
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 1.56 rt. 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 C 1 and C 2 in Eq. (5.22a) must be considered. Consequently, Eqs. (5.23) and (5.24) have to be modified to
include the effect of all four constants. Eq. (5.22a) may be
rewritten in a different form as
w = A1 sin 𝛽x sinh 𝛽x + A2 sin 𝛽x cosh 𝛽x
+A1 cos 𝛽x sinh 𝛽x + A4 cos 𝛽x cosh 𝛽x, (5.26)
and a solution is obtained for various boundary conditions.
The most frequent 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 alone or in conjunction with
Table 5.2.
Example 5.7
Derive N 𝜃 for the case of applied bending moment M0 at
edge x = 0 for a short cylinder of length l.
Solution:
The four boundary conditions are as follows:
At x = 0,
( 2 )
dw
−D
= M0
dx2
( 3 )
dw
−D
= 0.
dx3
At x = l,
(
Answer:
−D
Pr
3Pr
𝜎=
+ √
2t t 3(1 − 𝜇2 )
= 40,015 psi
(
−D
d2 w
dx3
d3 w
dx3
)
=0
)
= 0.
Table 5.3 Various functions of short cylinders.
l
x
x
Function
w
𝜃
Mx
N𝜃
Qx
[
]
M0 −C2
C
V + 3 V − V8
2𝛽 2 D C1 7 C1 2
[
]
2C3
M0 C2
V +
V − V2
2𝛽D C1 1
C1 7
[
]
C
C2
M0
V − 3 V − V7
C1 8 C1 1
[
]
C
C
2B2 rM0 − 2 V7 + 3 V2 − V8
C1
C1
(
) ]
[
2C3
C2
−𝛽M0
V2 −
+ 1 V8
C1
C1
l
θ0
Q0
M0
[
]
Q0
C
C
C4
V − 5V − 6V
2𝛽 3 D C1 7 C1 5 C1 6
[
]
−Q0 C4
C
C
V + 5V + 6V
2𝛽 2 D C1 1 C1 4 C1 3
[
]
Q0
C
C
C
− 4 V8 − 5 V6 + 6 V5
𝛽
C1
C1
C1
[
]
C
C
C4
V7 − 5 V5 − 6 V6
2𝛽rQ0
C1
C1
C1
[
]
C5
C6
C4
V2 +
V3 −
V4
Q0
C1
C1
C1
Δo
[
]
[
]
C6
C
C
C
C
Δ0 V7 − 3 V1 − 2 V8
V − 5V − 4V
C1 5 C1 6 C1 8
C1
C1
[
]
[
]
C6
C
C
C
C
V − 5V − 4V
𝜃0
𝛽Δ0 −V1 + 2 3 V8 − 2 V2
C1 4 C1 3 C1 2
C1
C1
[
]
[
]
C6
C
C
C
C
2𝛽D𝜃0
V + 5 V − 4 V 2𝛽 2 DΔ0 −V8 + 3 V2 − 2 V7
C1 6 C1 5 C1 7
C1
C1
[
[
]
]
Et𝜃0 C6
EtΔ0
C
C
C
C
V6 + 5 V5 − 4 V7
V7 + 3 V1 − 2 V8
𝛽
C1
C1
C1
r
C1
C1
[
]
[
]
C6
C5
C3
C4
C
2
3
V3 +
V4 +
V1 −2𝛽 DΔ0 −V2 + 2 V7 + 2 V1
2𝛽 D𝜃0
C1
C1
C1
C1
C1
𝜃0
𝛽
Constants
Variables
C 1 = sinh2 𝛽l − sin2 𝛽l
V 1 = cosh 𝛽x sin 𝛽x − sinh 𝛽x cos 𝛽x
C 2 = sinh2 𝛽l + sin2 𝛽l
V 2 = cosh 𝛽x sin 𝛽x + sinh 𝛽x cos 𝛽x
C 3 = sinh 𝛽l cosh 𝛽l + sin 𝛽l cos 𝛽l
V 3 = cosh 𝛽x cos 𝛽x − sinh 𝛽x sin 𝛽x
C 4 = sinh 𝛽l cosh 𝛽l − sin 𝛽l cos 𝛽l
V 4 = cosh 𝛽x cos 𝛽x + sinh 𝛽x sin 𝛽x
C 5 = sin2 𝛽l
V 5 = cosh 𝛽x sin 𝛽x
C 6 = sinh2 𝛽l
V 6 = sinh 𝛽x cos 𝛽x
V 7 = cosh 𝛽x cos 𝛽x
V 8 = sinh 𝛽x sin 𝛽x
68
5 Stress in Cylindrical Shells
From Eq. (5.26), the second derivative is given by
d2 w
= 2𝛽 2 (A1 cos 𝛽x cosh 𝛽x
dx2
+ A2 cos 𝛽x sinh 𝛽x − A3 sin 𝛽x cosh 𝛽x
− A4 sin 𝛽x sinh 𝛽x),
(1)
whereas the third derivative is expressed as
From Eq. (5.19),
Etw
N𝜃 =
r
Et
= (A1 sin 𝛽x sinh 𝛽x + A2 sin 𝛽x cosh 𝛽x
r
+ A3 cos 𝛽x sinh 𝛽x + A4 cos 𝛽x cosh 𝛽x). (3)
Using the values of A1 , A2 , A3 , A4 thus obtained and the
terminology of Table 5.3, Eq. (3) reduces to
d3 w
= 2𝛽 2 [A1 (cos 𝛽x sinh 𝛽x − sin 𝛽x cosh 𝛽x)
dx3
+ A2 (cos 𝛽x cosh 𝛽x − sin 𝛽x sinh 𝛽x)
N𝜃 =
− A3 (sin 𝛽x sinh 𝛽x + cos 𝛽x cosh 𝛽x)
− A4 (sin 𝛽x cosh 𝛽x + cos 𝛽x sinh 𝛽x)].
(2)
Substituting Eq. (1) into the first boundary condition
gives
A1 =
−M0
.
2D𝛽 2
or
Substituting Eq. (2) into the second boundary condition gives
A2 = A3 ,
and from the third and fourth boundary conditions, the
relationships
(
)
M0
sin 𝛽l cos 𝛽l + sinh 𝛽l cosh 𝛽l
A3 =
2D𝛽 2
sinh2 𝛽l − sin2 𝛽l
and
A4 =
−M0
2D𝛽 2
(
sin2 𝛽l + sinh2 𝛽l
)
sinh2 𝛽l − sin2 𝛽l
are obtained.
Et M0
r 2D𝛽 2
[
C
− sin 𝛽x sinh 𝛽x + 3 (sin 𝛽x cosh 𝛽x
C1
]
C2
cos 𝛽x cosh 𝛽x
+ cos 𝛽x sinh 𝛽x) −
C1
(
)
C
C
N𝜃 = 2rM0 𝛽 2 −V8 + 3 V2 − 2 V7 .
C1
C1
Example 5.8
Determine the maximum stress at point A of the thin
cylinder shown in Figure 5.14a. Let 𝜇 = 0.3.
Solution:
A free-body diagram of junction A is shown in
Figure 5.14b. The deflection at point A in the thick
cylinder due to P is obtained from Table 5.3 by letting 𝛽x
equal to 𝛽l. Hence,
(
)
C5
C6
C4
−P
V7 −
V −
V .
wp = 3
C1 5 C1 6
2𝛽1 D1 C1
Figure 5.14 Discontinuity forces.
1 = 1.0ʺ
P
A
t1 = 0.50ʺ
t2 = 0.25ʺ
r=6
(a)
P
M0
A
M0
A
1
2
Q0
Q0
x2
x1
(b)
5.3 Buckling of Cylindrical Shells
For 𝛽 1 = 0.7421, D1 = 0.011 45E, and 𝛽x = 𝛽l, the
following values are obtained:
C1 = 0.2028
C2 = 1.1164
C3 = 1.5444
C4 = 0.5481
C5 = 0.4568
C6 = 0.6596
V1 = 0.2721
V3 = 0.4006
V4 = 1.4984
V5 = 0.8707
V6 = 0.5986
V7 = 0.9495.
which reduces to
(2)
12.58M0 + Q0 = −3.48P.
Solving Eqs. (1) and (2) yields
M0 = −0.27P
Q0 = −0.08P.
Hence, the maximum axial stress is
6M
6(0.27P)
𝜎 = 20 =
= 25.9P psi.
t
0.252
Thus, the expression for wp due to P is given by
The circumferential bending moment is given by
144.02P
wp =
.
E
The deflection compatibility equation at point A is
M𝜃 = 𝜇Mx = 0.08P.
The circumferential force N 𝜃 is given by Eq. (5.19) as.
Etw
r
Et 2
=
(w − wM0 )
r ( Q0
)
Et 2 −0.08P −0.27P
=
−
R
2𝛽22 D2
2𝛽23 D2
𝛿1 |x=0 = 𝛿2 |x=0
N𝜃 =
or from Tables 5.2 and 5.3 with x = 0,
( )
(
)
M
Q
C4
−C2
144.02P
− 20
− 30
E
C1
2𝛽1 D1
2𝛽1 D1 C1
Q
M
= 30 − 2 0
2𝛽2 D2 2𝛽2 D2
P
(−23.86 + 84.53) = 2.53P
24
N
6M
𝜎𝜃 = 𝜃 + 2 𝜃
t
t
2.53P 6(0.08P)
=
+
0.25
0.252
= 17.80P.
=
and with 𝛽 2 = 1.0495 and D2 = 0.00145E, the equation
becomes
144.02P − 288.82Q0 + 436.59M0
= 298.30Q0 − 313.07M0
or
5.206M0 − 4.077Q0 = −P.
(1)
The rotation at point A due to P is obtained from
Table 5.3 as
(
)
C
C
C4
P
V1 + 5 V4 + 6 V3
𝜃p = 2
C1
C1
2𝛽1 D1 C1
Problems
5.6
Find the discontinuity stress in the figure shown due
to an internal pressure of 375 psi. Let E = 29 × 106 psi
and 𝜇 = 0.3.
or
429.33P
.
E
The rotation compatibility equation at point A is
𝜃p =
3.0ʺ
t = 1/2ʺ
t = 1ʺ
𝜃1 |x=0 = 𝜃2 |x=0 .
t = 3/4ʺ
r = 40ʺ
Hence,
(
)
M0
2C3
Q
429.33P
+
+ 20
E
2𝛽1 D1 C1
2𝛽1 D1
(
)
C5 C6
Q
−M0
×
+
+ 20
=
C1 C1
𝛽2 D2 2𝛽2 D2
5.3 Buckling of Cylindrical Shells
429.33P
+ 896.45M0 + 436.59Q0
E
= −657.13M0 + 313.07Q0 ,
Most cylindrical shells are subjected to various compressive forces such as dead weight, wind loads, earthquakes,
and vacuum. The behavior of cylindrical shells under
or
69
5 Stress in Cylindrical Shells
where f (x, 𝜃) is a function cf. x and 𝜃, which expresses
the variation of N 𝜃 from the average value. When the
deflection w of the shell is very small, f (x, 𝜃) is also very
small. Similarly, the axial and shear stresses are expressed
by
these compressive forces is different from those under
internal pressure. In most instances, the difference is due
to the buckling phenomena that render cylindrical shells
weaker in compression than in tension.
The derivation of the equations for the axial and lateral
buckling of cylindrical shells is given by many authors
such as Timoshenko [2], Gerard [3], Donnell [4], and
Sturm [5]. The derivations differ slightly from each other
based on the assumptions made regarding the deformation shape and the effect of plasticity on buckling. The
methodology used by Sturm is used by ASME [6] since it
can easily take into effect two factors needed in pressure
vessel design. These are buckling at elevated temperature
and inelastic strains.
Sturm used the system of forces shown in Figure 5.15
to establish the buckling characteristics of cylindrical
shells subjected to external compressive forces. From
the figure, he derived a relationship between strains and
deflections. Using this relationship and Eq. (3.11), he
obtained a system of equations that relates forces and
moments to deflections. These equations, together with
the equilibrium equations determined from Figure 5.15,
result in the four basic differential equations for the
buckling of cylindrical shells.
Nx = 0 + g(x, 𝜃)
N𝜃x = 0 + h(x, 𝜃)
Nx𝜃 = 0 + j(x, 𝜃).
Substituting these expressions into the four basic differential equations for the buckling of cylindrical shells,
and using boundary conditions for simply supported
ends, the solution for the elastic buckling of a simply
supported cylindrical shell due to uniform pressure
applied to sides only is given by
(
t
Pcr = KE
D0
)3
,
(5.27a)
where
Pcr = buckling pressure
E = modulus of elasticity
t = thickness
5.3.1
D0 = outside diameter
Uniform Pressure Applied to Sides Only
For this case, the hoop force is
(
K = K1 + K2
N𝜃 = −Pr + f (x, 𝜃),
Mx
Mxθ
dx
70
Nθ
Mθx
Mθ
𝜕Mxθ
dx
𝜕x
𝜕Nxθ
Nxθ +
dx
𝜕x
Mxθ +
Nθz
𝜕Nxz
dx
Nxz +
𝜕x
𝜕Mx
dx
Mx +
𝜕x
𝜕Nx
dx
Nx +
𝜕x
r
dθ
)2
Figure 5.15 Effective length of T-stiffener.
Nx
Nxθ
Nxz
Nθx
D0
t
𝜕Mθx
dθ
𝜕θ
𝜕Mθ
Mθ +
dθ
𝜕θ
𝜕Nθ
dθ
Nθ +
𝜕θ
𝜕Nθz
Nθz +
dθ
𝜕θ
𝜕Nθx
Nθx +
dθ
𝜕θ
Mθx +
5.3 Buckling of Cylindrical Shells
Figure 5.16 Buckling modes of a cylindrical
shell [5].
=
= πR
N
(a)
πR
N
Final shape
(d)
Center
Edges simple supported
symmetrical about center line line
2
{N 2 [N 2 𝜆2 − 𝜇(𝜆 − 1) − 1]
3F(1 − 𝜇2 )
200
}
𝛼+1+𝜇 2
{N [1 + (𝜆 − 1)(2 − 𝜇)] − 1}
−
𝛼𝜆
2
,
𝛼2F
2 2
πr
+1
N 2 L2
2 2
N L
𝛼= 2 2 +1
πr
𝜇 = Poisson’s ratio
𝜆=
F = N2 − 1 +
Edges fixed
symmetrical about center line
8
7
40
where N = number of lobes as defined in Figure 5.16
(e)
N = 12
11
10
9
100
80
60
Values of K
K2 =
πR
N
(c)
(b)
Original shape
K1 =
=
6
5
20 Oo
= 20
t
50
10
100
8
200
6
500
4
1000
4
3
2
2
𝜇
1
−
𝛼 2 𝜆𝛼
I
r2 (1 − 𝜇2 )t𝜆𝛼
{N 2 [1 + (𝜆 − 1)(2 − 𝜇)] − 1}
{(
)
P r
×
1 − cr
Et
1
0.4 0.6
1
4
2
−
6
10
Values of
20
40 60 100 200
L
r
Figure 5.17 Collapse coefficients of round cylinders with
pressures on sides only, edges simply supported; 𝜇 = 0.3 [5].
}
× [𝛼(1 − 𝜇2 ) + (1 + 𝜇)2 ] + 𝛼 + 1 + 𝜇
I = moment of inertia =
t3
.
12
A plot of the K value in Eq. (5.27a) based on the first
two terms of the expression for F is shown in Figure 5.17.
5.3.2
Substituting those expressions into the four basic differential equations for the buckling of cylindrical shells,
and using boundary conditions for simply supported
ends, the solution for the elastic buckling of a simply
supported cylindrical shell due to uniform pressure
applied to sides and ends is given by
Uniform Pressure Applied to Sides and Ends
The values of N 𝜃 , N x𝜃 , and N 𝜃x are the same as those for
pressure applied to sides only. The value of N x is given by
−Pr
Nx =
+ g(x, 𝜃).
2
Pcr = [Pcr for side pressure only]
or
(
1
Pcr = KE
D0
F
F+
(π2 r2 ∕2L2 )
)3
.
(5.27b)
71
5 Stress in Cylindrical Shells
200
Solution:
L
= 4.0
r
D
= 160.
t
From Figure 5.18, K = 16, and from Eq. (5.27b),
)
(
0.375 3
Pcr = (16)(29,000,000)
60
Pcr = 113 psi.
N = 12
11
10
9
100
80
60
8
7
40
Values of K
72
6
5
20 Oo
= 20
t
50
10
100
8
200
6
500
4
1000
4
3
2
2
5.4 Thermal Stress
1
0.4 0.6
1
2
4
6
10
20
40 60 100 200
L
Values of
r
Figure 5.18 Collapse coefficients of round cylinders with
pressures on sides and ends, edges simply supported; 𝜇 = 0.3 [5].
(
where K = K3 + K4
K3 = K1
K4 = K2
D0
t
)2
,
F
F+
(π2 r2 ∕2L2 )
F
.
F + (π2 r2 ∕2L2 )
𝛿T = 𝛿F ,
A plot of the K value in Eq. (5.27b) using the first two
terms of the expression for F is shown in Figure 5.18.
5.3.3
If a cylinder is subjected to a uniform change in temperature and is allowed to deform freely, no thermal stress is
produced. Any restraint that prevents free deformation
produces thermal stress. The amount of restraint affects
the stress level. Figure 5.19 illustrates a bar restrained
in one and then in two directions. In Figure 5.19a, the
bar is fixed in the x direction only and is subjected to an
increase in temperature. To determine the thermal stress,
the restraint is first removed, and the bar is allowed to
deform due to change in temperature. A force F is then
applied to produce an equal but opposite deformation.
The thermal stress in the bar can be calculated from the
compatibility equation
where
𝛿 T = deformation due to temperature
𝛿 F = deformation due to force F.
Pressure on Ends Only
Substituting the values of 𝛿 F and 𝛿 T in the compatibility
equation gives
For this case,
𝜎 = −𝛼ΔTE
N𝜃 = 0 + f (x, 𝜃)
Nx = −P + g(x, 𝜃)
(5.29a)
where
N𝜃x = 0 + h(x, 𝜃)
𝜎 = stress (positive values indicate tension)
Nx𝜃 = 0 + j(x, 𝜃),
𝛼 = coefficient of thermal expansion
and the four differential equations are solved for the value
of Pcr . For small values of t/r, the critical compressive longitudinal stress 𝜎 l can be expressed as
( )
P
t
.
𝜎t = cr = 0.60E
t
r
for a uniaxial case,
(5.28)
Example 5.9
A cylindrical shell with r = 30 in. is simply supported at
the ends. If L = 10 ft and t = 0.375 in., find the critical
buckling pressure for a uniform applied pressure to sides
and ends. Let E = 29 × 106 psi.
ΔT = 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 Figure 5.19b, the deformations due to F x and F y are calculated in the same manner as the uniaxial deformation.
The two compatibility equations then become
𝛿T + 𝛿Fx − 𝜇𝛿Fy = 0
𝛿T − 𝜇𝛿Fx + 𝛿Fy = 0,
5.4 Thermal Stress
+ Δτ
Figure 5.19 Thermal expansion of
infinitesimal element.
=
–
L
L
δτ = α . Δτ . L
F
L
δτ
δF
δF σ . L/E
(a)
Fy
+Δτ
=
L
–
Fx
Fx
δτ
L
δF
δτ
–
μδF
x
α = Coeff. of expansion
Δτ = Change in temp. (positive when temp. increases)
μ = Poisson’s ratio
δF
y
x
Fy
μδF
y
(b)
where
𝜇 = Poisson’s ratio
Weld A
𝛿Fx = deformation due to force F x
𝛿Fy = deformation due to force F y .
Solving the
equations gives
aforementioned
𝜎x = 𝜎y = −
𝛼ΔTE
1−𝜇
Rod
two
simultaneous
for a biaxial case.
(5.29b)
A comparison between Eqs. (5.29a) and (5.29b) indicates that a higher stress level is obtained when the
number of restraints is increased. Hence, for a bar with
𝜇 = 0.3, a stress increase of 43% results when the number
of restraints increases from one to two. Another interesting feature of Eqs. (5.29a) and (5.29b) is that the thermal
stress is independent of thickness and length.
Eq. (5.29b) can also be obtained from the theory of elasticity. If 𝜀 = 𝛼ΔT is substituted into the first two expressions of Eq. (3.3), the following results:
−1
(𝜎 − 𝜇𝜎y )
E x
−1
𝛼ΔT =
(𝜎 − 𝜇𝜎x ),
E y
from which the expression
80ʺ
Figure 5.20 Rod in a cylindrical shell.
In a triaxial case, the thermal stress can be determined
easily from the theory of elasticity. Substituting 𝜀 = 𝛼ΔT
in the first three expressions of Eq. (3.1) results in
𝜎x = 𝜎y = 𝜎z =
−𝛼ΔTE
for a triaxial case.
1 − 2𝜇
(5.29c)
𝛼ΔT =
𝜎x = 𝜎y =
is obtained.
−𝛼ΔTE
1−𝜇
Example 5.10
An internal stainless-steel rod is welded to the inside of a
carbon-steel vessel as shown in Figure 5.20. If the coefficient of thermal expansion is 9.5 × 10−6 in./in.-∘ F for the
rod and 6.7 × 10−6 in./in.-∘ F for the vessel, what is the
stress in weld A due to a temperature increase of 400 ∘ F?
Use E = 28 × 106 psi.
73
74
5 Stress in Cylindrical Shells
Equation (a) can be written as
deflection of tray due to temperature + F3
t
= deflection of shell due to
1
Tray
T=1
MO
3
F1
F3
MO
40ʺ
temperature + F1 − M0
Fr
(𝛼ss )(ΔT)(r) + 3 (1 − 𝜇)
ET
M
F
= (𝛼cs )(ΔT)(r) + 31 − 20 .
2𝛽 D 2𝛽 D
F2
Eqs. (b)–(d) can be written as
M
F1
− 0 =0
2
2𝛽 D 𝛽D
2
(a)
(b)
(1)
(2)
Figure 5.21 Tray in a cylindrical shell.
Solution:
Weld A is essentially subjected to a uniaxial stress. Hence,
max𝜎 = EΔ𝛼ΔT
M
F2
− 0 =0
2𝛽 2 D 𝛽D
(3)
F1 + F2 + F3 = 0.
(4)
From Eq. (2),
= (28 × 106 )(9.5 − 6.7)(10−6 )(400)
= 31,400 psi.
F1 = 2𝛽M0 .
From Eq. (3),
Example 5.11
An internal stainless tray is welded to the inside of
a carbon-steel vessel as shown in Figure 5.21a. If the
coefficient of thermal expansion is 9.5 × 10−6 in./in.-∘ F
for the tray and 6.7 × 10−6 in./in.-∘ F for the vessel, what
is the stress in the weld due to a temperature increase of
400 ∘ F? Use E = 28 × 106 psi and 𝜇 = 0.3.
Solution:
A conservative answer can be obtained by assuming the
tray attachment to be rigid. Because the weld is subjected
to both hoop and axial stresses, it can be treated as a biaxial condition. Hence, from Eq. (5.29b),
−(28 × 106 )(9.5 − 6.7)(10−6 )(400)
1 − 0.3
= −44,800 psi.
𝜎=
If a more accurate result is desired, then a discontinuitytype analysis can be performed. Referring to Figure 5.21b,
and taking account of symmetry, the equations of compatibility and equilibrium can be written as
𝛿3 = 𝛿1
(a)
𝜃1 = 0
(b)
𝜃2 = 0
∑
F = 0.
(c)
(d)
From the aforementioned four equations, the four
unknowns F 1 , F 2 , F 3 , and M0 can be obtained.
F2 = 2𝛽M0 .
From Eq. (4),
F3 = −4𝛽M0 ,
and Eq. (1) becomes
[
]
r(1 − 𝜇)
1
+ 3
= (𝛼cs − 𝛼ss )(Δt)(r).
F3
ET
8𝛽 D
(5)
Assuming the thickness of the cylinder is t = 0.1875 in.
and using the other given values, the value of F 3 from
Eq. (5) is
F3 = −618 lb∕in. or
𝜎 = 618 psi compression in weld.
The value 618 psi is significantly lower than the conservative value 44,800 psi obtained from Eq. (5.29b), because
of the flexibility of the cylinder. If the thickness of the
cylinder is t = 3.0 in., then Eq. (5) gives
F3 = 21,200 lb∕in.
or
𝜎 = 21,200 psi compression in weld.
This value indicates that as the cylinder gets thicker, the
stress approaches that of Eq. (5.29b). In fact, if the cylinder is taken as infinitely rigid, then Eq. (1) becomes
F3 r
(1 − 𝜇) = (𝛼cs )(ΔT)(r),
Et
and the equation yields F 3 = −44,800 lb/in. or 𝜎 = 44,800
psi compression in the weld, which is the same as that
obtained from Eq. (5.29b).
(𝛼ss )(ΔT)(r) +
5.4 Thermal Stress
5.4.1
Uniform Change in Temperature
A uniform 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 of the various components. The following examples illustrate 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 Figure 5.22a. What is the maximum
thermal stress if 𝛼 = 6.5 × 10−6 in./in.-∘ F, E = 30 × 106 psi,
and 𝜇 = 0.3?
Solution:
A solution can be obtained by taking a free-body diagram
at the gas–liquid boundary as shown in Figure 5.22b.
Compatibility at the interface requires that the deflection
in (1) equals the deflection in (2). Hence, from Eq. (5.24),
H
M
(𝛼)(ΔT1 )(r) − 30 + 20
2𝛽 D 2𝛽 D
H
M
= (𝛼)(ΔT2 )(r) + 30 + 20 ,
2𝛽 D 2𝛽 D
from which
3
H0 = (𝛼)(ΔT1 − ΔT2 )(r)(𝛽 )(D)
= (6.5 × 10−6 )(240 − 30)(6)(1.4843)3 (5366)
The circumferential force in the pipe due to H 0 is
obtained from Eqs. (5.19) and (5.24):
N𝜃 =
The maximum value of C 𝛽x is obtained from Table 5.1
as 1.0. Hence,
N𝜃 = 2565 lb∕in.
Mx = 0
and
2565
= 20,500 psi.
0.125
The maximum bending moment due to H 0 was derived
in Example 5.4 as
max𝜎 =
Mx =
1
Gas
Liquid
at 40 °F
r = 6ʺ
MO
MO
t=
1ʺ
8
2
HO
HO
0.322H0
𝛽
at 𝛽x =
π
.
4
The stress due to the bending moment at 𝛽x = 𝜋/4 is
(
)
6M
6 0.322H0
= 12,000 psi.
𝜎x = 2 = 2
t
t
𝛽
The deflection due to H 0 at 𝛽x = 𝜋/4 is obtained from
Eqs. (5.23) and (5.24) as
= 144 lb∕in.
Gas
at 250 °F
at interface.
Also,
w=
M0 can be obtained from the second compatibility
equation, whereby the slope in (1) at the interface is
equal to the slope in (2):
H
M
H
M0
− 20 = − 0 − 20 ,
𝛽D 2𝛽 D
𝛽D 2𝛽 D
and M0 = 0.
Et H0
C .
r 2𝛽 3 D 𝛽x
0.322H0
.
2𝛽 3 D
Hence,
N𝜃 = 827 lb∕in.,
and the circumferential stress is
827
𝜎𝜃 =
+ 0.3 × 12,000 = 10,200 psi.
0.125
Thus, the maximum stress occurs at the interface with
magnitude of 20,500 psi.
Example 5.13
Determine the bending stress in a cylinder fixed at one
edge (Figure 5.23a) due to a uniform rise in temperature
of 200 ∘ F, if
∘
𝛼 = 605 × 10−6 in.∕in.- F
E = 30 × 106 psi
𝜇 = 0.3.
Liquid
Solution:
The radial deflection of the cylinder if the ends are free is
(a)
(b)
Figure 5.22 Pipe with abrupt change in temperature.
w =(𝛼)(ΔT)(r)
=0.039 in.
75
76
5 Stress in Cylindrical Shells
and
−Pr
(5.30a)
= −E𝛼Tx
t
Because the cylinder does not have any applied loads
on it, the external force Px used to reduce the deflection to zero must be eliminated by applying an equal
and opposite force in the cylinder. Hence, Eq. (5.21)
becomes
Et𝛼Tx
d4 w
+ 3𝛽 4 w =
(5.30b)
4
dx
rD
The total stress in the cylinder is determined from Eqs.
(5.30a) and (5.30b).
𝜎𝜃 =
HO
MO
30ʺ
0.25ʺ
(a)
(b)
Figure 5.23 Cylindrical shell fixed at end.
From Figure 5.23a and 5.23b, the rotation at the end is
zero because the cylinder is fixed. Hence,
Q
M0
= 20
𝛽D 2𝛽 D
or
Example 5.14
A vessel that operates at 800 ∘ F is supported by an
insulated skirt. The thermal distribution in the skirt is
shown in Figure 5.24a. If the top and bottom of the skirt
are assumed to be fixed with respect to rotation, what
is the maximum stress due to temperature gradient?
𝛼 = 7 × 10−6 in./in.-∘ F, 𝜇 = 0.3, E = 30 × 106 psi.
Q0 = 2𝛽M0 .
In the second compatibility equation, the deflection
due to temperature plus moment plus shear is equal to
zero:
M
Q
0.039 + 20 − 30 = 0
2𝛽 D 2𝛽 D
800 °F
Tt = 800 °F
and
M0 = (0.039)(2𝛽 2 D)
= (0.039)(2)(0.4694)2 (42,930)
= 738 in.-lb∕in.
6M
𝜎 = 2 = 70,800 psi.
t
5.4.2
Gradient in Axial Direction
The stress in a cylinder due to a thermal gradient T x in the
axial direction can be obtained by first subdividing the
cylinder into infinitesimal rings of length dx. Hence, the
radial thermal expansion due to T x in each ring is given
by 𝛼T x r. This expansion can be eliminated by applying an
external force Px such that
t = 1ʹʹ
3ʹ
Tb = 200 °F
= 10ʹ
(a)
MO
x
Ty =
Tb – Tt
x = –5x
deflection due to Px = deflection due to Tx
or
Px r2
= 𝛼rT x .
Et
Hence,
Et𝛼Tx
Px =
r
(b)
Figure 5.24 Thermal gradient in a vessel skirt.
5.4 Thermal Stress
Solution:
The equation for a linear temperature gradient is
T − Tt
Ty = Tt + b
x.
l
The temperature change can be expressed as
T − Tt
T= b
x,
l
and the circumferential stress due to ring action obtained
from Eq. (5.30a) is
(
)
Tb − Tt
𝜎𝜃 = −E𝛼
x.
(1)
l
Eq. (5.30b) gives
Et𝛼
d4 w
+ 4𝛽 4 w =
dx4
rD
(
Tb − Tt
l
)
x
A particular solution takes the form
(
)
Et𝛼 Tb − Tt
w = c1
x + c2 ,
rD
l
which upon substituting into the differential equation
gives
1
c1 = 4 and c2 = 0,
4𝛽
and w reduces to
(
)
Tb − Tt
w = r𝛼
x.
l
From Eq. (5.19),
(
)
Tb − Tt
N𝜃
𝜎𝜃 =
= E𝛼
x.
t
l
(2)
Adding Eqs. (1) and (2) results in
𝜎𝜃 = 0,
which means that for a linear distribution, the thermal
stress along the skirt is zero.
The slope due to axial gradient is given by
(
)
Tb − Tt
dw
𝜃=
= r𝛼
.
dx
l
Because the ends are fixed against rotation, a moment
must be applied at the ends to reduce 𝜃 to zero as shown
in Figure 5.24b. From Eq. (5.24),
(
)
Tb − Tt
−M0
r𝛼
=
l
𝛽D
or
( )
𝛼r
(Tb − Tt ).
M0 = −𝛽D
l
Since 𝛽 = 0.2142, D = 2.747 × 106 ,
M0 = 742 in.-lb∕in.
and
𝜎 = 4450 psi.
5.4.3
Gradient in Radial Direction
The thermal stress in a cylinder due to a temperature gradient in the radial direction can be obtained from the
theory of elasticity. Hence, Eq. (3.1) can be written as
1
(5.31)
𝜖r = [𝜎r − 𝜇(𝜎𝜃 + 𝜎z )] + 𝛼T
E
𝜖𝜃 =
1
[𝜎 − 𝜇(𝜎z + 𝜎r )] + 𝛼T
E 𝜃
(5.32)
1
(5.33)
[𝜎 − 𝜇(𝜎r + 𝜎𝜃 )] + 𝛼T,
E z
where 𝛼T is the strain due to temperature change. If the
temperature is symmetric with respect to 𝜃, all shearing
stresses are zero, and the radial and hoop strains can be
expressed as
dw
w
𝜖r =
𝜖𝜃 =
𝜖z = 0
dr
r
or
(
)
d𝜖𝜃
𝜖r = 𝜖𝜃 + r
.
(5.34)
dr
Substituting Eqs. (5.31) and (5.32) into (5.34) gives
(
)
)
(
d𝜎r
d𝜎r
𝜇
𝜎r =𝜎𝜃 + r
− 𝜎𝜃
−
𝜎r + r
dr
1−𝜇
dr
(
)
(
)
E
dT
+
𝛼(1 + 𝜇)r
.
(5.35)
2
1−𝜇
dr
In Section 5.1, it was shown that the equilibrium of an
element in a cylinder (Figure 5.3) can be expressed by the
equation
(
)
d𝜎r
𝜎𝜃 − 𝜎r = r
.
(5.36)
dr
Solving Eqs. (5.35) and (5.36) for 𝜎 r gives
]
(
[
)
dT
−E
d 1 d 2
𝛼(1 + 𝜇)r
(r 𝜎r ) =
.
r
dr r dr
1 − 𝜇2
dr
(5.37)
𝜖z =
Solving the aforementioned differential equation and
applying the boundary conditions
𝜎r |r=rt = 0
𝜎r |r=ro = 0
gives
E𝛼 1 + 𝜇
𝜎r =
1 − 𝜇2 r2
(
r2 − ri2
ro2 − ri2 ∫ri
ro
)
r
Tr dr−
∫ri
Tr dr .
From Eq. (5.36),
E𝛼 1 + 𝜇
𝜎𝜃 =
1 − 𝜇2 r2
(
)
r
r2 + ri2 ro
2
Tr dr +
Tr dr − Tr ,
∫ri
ro2 − r2 ∫ri
i
(5.38)
77
78
5 Stress in Cylindrical Shells
and from Eq. (5.31) for a cylinder unrestrained in the z
direction,
)
(
ro
E𝛼
2
𝜎z =
Tr dr − T .
1 − 𝜇 ro2 − ri2 ∫ri
0.5
σ Eα Ti
(1 – u)
σz ,σθ
From these equations, some cases can be derived.
Case 1. Linear Thermal Distribution.
For thin vessels, a steady-state condition produces linear thermal distribution through the thickness that can
be expressed as
T = Ti
0.0
σr
–0.5
ro − r
,
ro − ri
r1
where T i = inside wall temperature relative to outside
wall temperature. Substituting T into Eq. (5.38) gives
E𝛼Ti
𝜎r = 2
r (1 − 𝜇)
[
(r2 −ri2 )(2ri +ro )
6(ri + ro )
−
2(r3 − ri3 ) − 3ro (r2 − ri2 )
]
6(ri − ro )
E𝛼Ti
𝜎𝜃 = 2
r (1 − 𝜇)
[
]
(r2 +ri2 )(2ri +ro ) 2(2r3 +ri3 )−3ro (r2 + ri2 )
−
6(ri + ro )
6(ri − ro )
]
[
ro − r
2ri − ro
E𝛼Ti
.
(5.39)
−
𝜎z =
(1 − 𝜇) 3(ri + ro ) ro − ri
Figure 5.25 is a typical plot of 𝜎 r , 𝜎 𝜃 , and 𝜎 z . The plot indicates that 𝜎 r is small compared with 𝜎 𝜃 and 𝜎 z . For all
practical purposes, 𝜎 𝜃 and 𝜎 z are equal. The maximum
values of 𝜎 𝜃 and 𝜎 z occur at the inside and outside surfaces. From Eq. (5.39),
[
]
⎧ −E𝛼Ti 2ro + ri
for inside surface
⎪
⎪ 1 − 𝜇 3(ro + ri )
𝜎𝜃 =𝜎z = ⎨
.
[
]
⎪ E𝛼Ti ro + 2ri
for outside surface
⎪ 1 − 𝜇 3(r + r )
o
i
⎩
(5.40)
For thin-walled vessels, Eq. (5.40) reduces to
⎧ −E𝛼T
i
⎪
⎪ 2(1 − 𝜇)
𝜎𝜃 = 𝜎z = ⎨
⎪ E𝛼Ti
⎪ 2(1 − 𝜇)
⎩
for inside surface
.
for outside surface
(5.41)
Case 2. Logarithmic Thermal Distribution.
In thick vessels, a steady-state thermal condition gives
rise to a logarithmic temperature distribution that can be
ro
Figure 5.25 Linear thermal stress distribution in a vessel shell.
expressed as
T = Ti
(
ln ro − ln r
ln ro − ln ri
)
.
Substitution of this expression in Eq. (5.38) results in
𝜎r =
𝜎𝜃 =
𝜎z =
−E𝛼Ti
2(1 − 𝜇) ln(ro ∕ri )
[
(
) ( )]
ri2
ro2
r
ro
1 − 2 ln o
× ln + 2
r
r
ri
ro − ri2
E𝛼Ti
2(1 − 𝜇) ln(ro ∕ri )
[
(
) ( )]
ri2
r
ro2
ro
1 + 2 ln o
× 1 − ln − 2
2
r
r
ri
ro − ri
E𝛼Ti
2(1 − 𝜇) ln(ro ∕ri )
[
( )]
2ri2
r
ro
ln o
.
× 1 − 2 ln − 2
2
r
ri
ro − ri
(5.42)
Again disregarding 𝜎 r as being small compared with 𝜎 𝜃
and 𝜎 z , Eq. (5.42) have a maximum value of
[
( )]
⎧
r
2ro2
E𝛼Ti
⎪
ln o
1− 2
2
ri
ro − ri
⎪ 2(1 − 𝜇) ln(ro ∕ri )
⎪
⎪for inside surface
⎪
𝜎𝜃 = 𝜎z ⎨
.
[
( )]
⎪
2ri2
ro
E𝛼Ti
ln
1− 2
⎪
ri
ro − ri2
⎪ 2(1 − 𝜇) ln(ro ∕ri )
⎪
⎪for outside surface
⎩
(5.43)
5.4 Thermal Stress
And for thin-walled cylinders, Eq. (5.43) reduces to
⎧ −E𝜎T
i
⎪
2(1
−
𝜇)
⎪
𝜎𝜃 = 𝜎z = ⎨
⎪ E𝛼Ti
⎪ 2(1 − 𝜇)
⎩
700 °F
for inside surface
,
for outside surface
which are the same as those for the linear case.
Case 3. Complex Thermal Distribution.
In many instances, such as transient and upset conditions, the temperature distribution through the wall of a
vessel cannot be represented by a mathematical expression. In this case, a graphical solution can be obtained
for the thermal stress. From Eq. (5.38),
]
[
ro
r
2
E𝛼 1 + (ri ∕r)
1
Tr dr + 2 Tr dr − T .
𝜎𝜃 =
1 − 𝜇 ro2 − r2 ∫ri
r ∫ri
i
For a cylinder where the thickness is small compared
to the radius, the first expression in the brackets can be
expressed as
1 + (ri ∕r)2
r
2π ∫r o Tr dr
ro
i
Tr dr =
ro2 − ri2 ∫ri
π(ro2 − ri2 )
= mean value of the temperature
distribution through the wall
= Tm .
(5.44)
r
r
2π ∫0 Tr dr
1
Tr dr =
r2 ∫ri
2πr2
= one-half the mean value of the
temperature distribution from the
axis of the vessel to r.
(5.45)
However, because the temperature distribution from the
axis to ri is zero, this latter expression for all practical purposes can be neglected. Hence, 𝜎 𝜃 can be expressed as
E𝛼
(T − T),
1−𝜇 m
ro = 13ʺ
Figure 5.26 Linear temperature gradient in a vessel shell.
in.-∘ F, and 𝜇 = 0.28, determine the maximum thermal
stress (a) using Eq. (5.40) and (b) using Eq. (5.41).
Solution:
(a): T i = 400–700 = −300 ∘ F. Hence, at the inside surface,
−(27 × 106 )(9.5 × 10−6 )(−300)
(1 − 0.28)
(
)
2(13) + 10
×
3(13 + 10)
= 55,800 psi
𝜎=
and at the outer surface,
The second expression can be expressed as
𝜎𝜃 =
r1= 10ʺ
400 °F
(5.46)
where.
T m = mean value of temperature distribution through
the wall
T = temperature at desired location.
From Eq. (5.38), it can be seen that 𝜎 z can also be
expressed by Eq. (5.46).
Example 5.15
A thin cylindrical vessel is heated by a jacket from
the outside such that the temperature distribution is as
shown in Figure 5.26. If E = 27 × 106 psi, 𝛼 = 9.5 × 10−6 in./
(27 × 106 )(9.5 × 10−6 )(−300)
(1 − 0.28)
(
)
13 + 2 × 10
×
3(13 + 10)
= −51,000 psi
𝜎=
(b): At the inner surface,
(−27 × 106 )(9.5 × 10−6 )(−300)
2(1 − 0.28)
= 53,400 psi,
𝜎=
and at the outer surface, 𝜎 = −53,400 psi.
Example 5.16
A pressure vessel operating at 300 ∘ F is subjected to
a short excursion temperature of 600 ∘ F. At a given
time, the temperature distribution in the wall is
shown in Figure 5.27. Find the maximum thermal
stress at that instance. Let 𝜇 = 0.3, E = 30 × 106 psi, and
𝛼 = 6.0 × 10−6 in./in.-∘ F.
Solution:
This problem can be visualized as a biaxial case where
the inside surface heats quickly while the rest of the wall
remains at 300 ∘ F. Using Eq. (5.29b) results in
(6 × 10−6 )(600 − 300)(30 × 106 )
1 − 0.3
= − 77,100 psi,
𝜎=
79
80
5 Stress in Cylindrical Shells
It is of interest to note that the high stress occurs at the
surface only. Thus at one-tenth of the thickness inside the
surface, the stress is
(30 × 106 )(6 × 10−6 )
(356 − 460)
1 − 0.3
= −26,700 psi.
600 °F
𝜎=
300 °F
r1 = 40ʺ
The high stress at the inside surface indicates that local
yielding will occur.
0
Nomenclature
Et 3
12(1 − 𝜇2 )
Do = outside diameter of cylinder
D=
ro = 42ʺ
E = modulus of elasticity
Figure 5.27 Nonlinear temperature gradient in a vessel shell.
K = constant
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
Figure 5.27 and tabulated as follows:
L = length of cylinder
Mx = axial bending moment
M𝜃 = hoop bending moment
P = pressure
Pcr = buckling pressure
Locations as ratios
of thickness
Temperature (∘ F)
Area
Pi = internal pressure
Po = external pressure
0
600
0.1
460
53.0
0.2
400
43.0
ri = inside radius
0.3
370
38.5
ro = outside radius
0.4
340
35.5
0.5
320
33.0
0.6
310
31.5
0.7
305
30.8
0.8
300
30.3
0.9
300
30.0
1.0
300
30.0
∑
355.6
Qx = shearing force
And T m ≈ 356 ∘ F.
From Eq. (5.46), at the inside surface,
(30 × 106 )(6.0 × 10−6 )
(356 − 600)
1 − 0.3
= −62,700 psi,
𝜎=
and at the outside surface
(30 × 106 )(6.0 × 10−6 )
(356 − 300)
1 − 0.3
= 14,400 psi.
𝜎=
r = radius of cylinder
t = thickness of cylinder
T = temperature
ΔT = temperature change
w = deflection
𝛼 = coefficient of thermal expansion
√
3(1 − 𝜇2 )
4
𝛽=
r2 t 2
𝛿 = deflection due to temperature change
𝜀z = longitudinal strain
𝜀r = radial strain
𝜀𝜃 = hoop strain
𝜃 = rotation
𝜇 = Poisson’s ratio
𝜎 = stress
𝜎 z = longitudinal stress
𝜎 r = radial stress
𝜎 𝜃 = hoop stress
Further Reading
References
1 Popov, E. and Balan, T. (1998). Engineering Mechanics
5 Sturm, R. (1941). A Study of the Collapsing Pressure of
of Solids. Upper Saddle River, NJ: Prentice Hall.
2 Timoshenko, S. and Gere, J. (2009). Theory of Elastic
Stability. Dover Publications.
3 Gerard, G. (1962). Introduction to Structural Stability
Theory. McGraw Hill.
4 Donnell, L. (1934). A new theory for the buckling of
thin cylinders under axial compression and bending”.
Transactions of the ASME. 56: 795–806.
Thin-Walled Cylinders, vol. XXXIX, 12. University of
Illinois Bulletin.
6 The American Society of Mechanical Engineers American Society of Mechanical Engineers boiler and pressure
Vessel Code, Section VIII, Pressure Vessels, Division 1.
New York: ANSI/ASME BPV-VIII-1.
Further Reading
American Society of Mechanical Engineers Pressure Vessel
and Piping Design – Collected Papers 1927–1959,
vol. 1960. New York: ASME.
Baker, E.H. et al. (1968). Shell Analysis Manual.
Washington, DC: NASA.
Burgreeen, D. (1971). Elements of Thermal Stress Analysis.
New York: C.P. Press.
Flugge, W. (1973). Stresses in Shells. New York: Springer
New York.
Gibson, J.E. (1965). Linear Elastic Theory of Thin Shells.
New York: Pergamon Press.
Harvey, J.F. (1991). Theory and Design of Pressure Vessels.
New York: Van Norstand Reinhold.
Jawad, M.H. (2018). Stress in ASME Pressure Vessels,
Boilers, and Nuclear Components. New York: John Wiley
Publishing.
Jones, D. (1965). Thermal Stress Analyses. New York:
Pergamon Press.
Timoshenko, S. and Woisowskey-Krieger, S. (1959). Theory
of Plates and Shells. New York: McGraw Hill.
81
Conical head consisting of two eccentric cones. Source: Courtesy of the Nooter Corporation, St. Louis, MO.
84
6
Analysis of Formed Heads and Transition Sections
6.1 Hemispherical Heads
The required thickness of hemispherical heads is determined from a free-body diagram as shown in Figure 6.1.
Hence,
σ
p
Pπr2 = 2πr𝜎t
r
t
or
𝜎=
Pr
,
2t
(6.1)
where
𝜎 = membrane stress
Figure 6.1 Cross-section of a spherical shell.
P = pressure
and
r = radius
dw
.
dr
Hence, Eqs. (6.2) and (6.3) become
t = thickness
This equation, which assumes uniform stress distribution through the thickness, is adequate for thin heads.
As the thickness becomes comparable 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,
𝜎𝜃 = 𝜎𝜙 .
From Eq. (3.1),
1
𝜀r = (𝜎r − 2𝜇𝜎𝜙 ),
(6.2)
E
1
𝜀𝜙 = [(1 − 𝜇)𝜎𝜙 − 𝜇𝜎r ].
(6.3)
E
The strain–displacement relationship is the same as
that derived from cylindrical shells:
w
𝜀𝜙 =
r
𝜀r =
(1 − 𝜇)
d
d
(r𝜎 ) − 𝜇 (r𝜎r ) − 𝜎r + 2𝜇𝜎𝜙 = 0.
dr 𝜙
dr
(6.4)
Figure 6.2 shows an infinitesimal segment of a spherical
head. Summation of forces in the radial direction gives
( )
1 d
r𝜎𝜙 = −
(6.5)
(r2 𝜎r ).
2 dr
Solving Eqs. (6.4) and (6.5), we obtain
( )[
]
d
1 d 3
r
𝜎
)
= 0.
(r
r
dr
r2 dr
Its solution is expressed as
A B
+ .
3 r3
The boundary conditions are given by
𝜎r =
𝜎r = −Pi
at r = ri
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
(6.6)
6.1 Hemispherical Heads
σr + dσr
dr
σθ
σϕ
σϕ
10.0
σθ
σ
P
1.0
Eq. (6.7a)
σr
r
Eq. (9.1)
Eq. (6.1)
0.10
1.0
2.0
ro
3.0
4.0
rl
Figure 6.2 Biaxial stress in a unit thin spherical shell.
Figure 6.3 Comparison of stress calculated by various equations.
and
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).
𝜎r = −Po
at r = ro .
Solving the boundary conditions for A and B and substituting into Eqs. (6.5) and (6.6), we obtain
(
)
(
)
ri3
ri3 Pi
ro3
ro3 Po
1− 3 ,
𝜎r = 3
1− 3 − 3
r
r
ro − ri3
ro − ri3
(
)
3
3
r Pi
r
𝜎𝜙 = 𝜎𝜃 = 3 i 3 1 + o3
2r
ro − ri
(
)
3
3
ri
ro Po
− 3
1+ 3 .
(6.7a)
2r
ro − ri3
Equation (6.7a) can be simplified for the following cases:
Solution:
From Eq. (6.1),
Pr
2𝜎
(4000)(15)
=
2(23,000)
= 1.30 in.
t=
From Eq. (6.7),
Case 1. Internal pressure only
max 𝜎r = −Pi
[
max 𝜎𝜙 = 𝜎𝜃 =
2ri3 + ro3
2(ro3
−
ri3 )
𝜎=
at r = ri ,
]
Pi
at r = ri .
or
(6.7b)
max 𝜎𝜙 = 𝜎𝜃 = −
=
at r = ro ,
3
2
ro3 Po
3
ro − ri3
at r = ri .
3
1+
ro3
]
2ri3
2ri3 Pi + 2𝜎ri3
2𝜎 − Pi
2(15)3 (4000) + 2(23,000)(15)3
2(23,000) − 400
= 16.31
(6.7c)
A comparison between Eqs. (6.1) and (6.7a) is shown
in Figure 6.3.
3
√
Case 2. External pressure only
max 𝜎r = −Po
ro3 − ri3
√
ro =
[
ri3 Pi
or
t = 1.31 in.
85
86
6 Analysis of Formed Heads and Transition Sections
6.1.1
Various Loading 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, which assumes that
the loads are carried by membrane action rather than
bending moments.
Referring to Figure 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 is found and by giving
the value of a second coordinate that varies along the
r
z
dz
dϕ
B
ϕ r
2
+r
dθ
Nϕ
Nϕθ
Nθ
Pr
Pθ
Nθϕ
Pϕ
dϕ
Nϕθ +
дNϕθ
r1
r2
(
𝜕N𝜃𝜙
𝜕𝜃
)
− r1 N𝜃 cos 𝜙 + P𝜙 rr1
Nϕ +
ϕ
дNϕ
дϕ
dϕ
Nθϕ +
дNθ
dθ
дθ
Nθ +
дNθ
dθ
дθ
(b)
Figure 6.4 Forces in a unit spherical shell.
= 0.
Summation of forces in the direction of parallel circles
gives
)
(
𝜕N𝜃
𝜕
(rN 𝜙𝜃 ) − r1
− r1 N𝜃𝜙 cos 𝜙 + P𝜃 r1 r = 0.
𝜕𝜙
𝜕𝜃
(6.9)
For the majority of pressure vessel applications, the
loads are symmetric with respect to the axis of revolution. Hence, all derivatives with respect to 𝜃 in Eqs. (6.8)
and (6.9) can be deleted. Shearing stresses due to torsion
are small compared with other stresses. Thus, Eq. (6.9)
can be omitted completely.
The last equation of equilibrium is obtained by summing the forces in Figure 6.4b perpendicular to the middle surface:
(a)
r
Figure 6.4b is a free-body diagram of a section of a surface of revolution. Summing forces parallel to the tangent at the meridian and simplifying by deleting terms
of higher order, we obtain
(6.8)
dr = ds cos ϕ
o
ds = r1 d𝜙.
ds = r1 d ϕ
ds
r1 dr
r = r2 sin 𝜙,
𝜕
(rN 𝜙 ) − r1
𝜕𝜙
r = r2 sin ϕ
ϕ
A
r
meridian and is constant on a circle around the shell’s
axis. Because these circles are parallel to one another,
they are called the parallel circles.
The definitions of r, r1 , r2 , and 𝜙 are shown in
Figure 6.4a. The radius r1 is measured from point 0,
which is the center of curvature of the meridian; r2 is
measured from the z axis and is normal to the meridian.
The parallel circle is defined by r. From Figure 6.4a,
dϕ
dϕ
N𝜃 N𝜙
+
= Pr .
r2
r1
(6.10)
Substituting Eq. (6.10) into Eq. (6.8) gives
N𝜙 =
1
r2 sin2 𝜙
]
[
×
r1 r2 (Pr cos 𝜙 − P𝜙 sin 𝜙) sin 𝜙 d𝜙 + C .
∫
(6.11)
The right-hand side of Eq. (6.11) is equal to the sum of
all the N 𝜙 forces around a circle of angle 𝜙. Therefore,
we can solve N 𝜙 at any given location 𝜙 by summing all
6.1 Hemispherical Heads
rO
Equation (6.15) can be solved for the deflections once
N 𝜃 and N 𝜙 are established from Eq. (6.12). Table 6.1
shows the solution of Eqs. (6.12) and (6.15) for various
loading conditions.
w
v
Example 6.2
Determine the forces in a spherical shell due to
snow load.
r1
Solution:
From Figure 6.6 and Table 6.1,
Pr = −P0 cos2 𝜙,
w+
dw
dϕ
dϕ
v+
dv
dϕ
dϕ
ϕ
dϕ
Figure 6.5 Deformation in a unit thin spherical shell.
forces in the 𝜙 direction. Once N 𝜙 is obtained, N 𝜃 can be
determined from Eq. (6.10).
For a spherical shell, r1 = r2 = r. Hence, Eqs. (6.10) and
(6.11) can be simplified as follows:
N𝜙 + N𝜃 = Pr r
r
N𝜙 =
sin2 𝜙
[
]
×
(P cos 𝜙 − P𝜙 sin 𝜙) sin 𝜙 d𝜙 + C .
∫ r
(6.12)
The displacement for various loading conditions is
derived from Figure 6.5. The total change in length AB is
dv
ΔL =
(d𝜙 − w d𝜙).
d𝜙
The strain is therefore
(
)
1 dv
w
𝜀𝜙 =
−
.
(6.13)
r1 d𝜙 r1
The change in r0 is given by
w
Δr0 = v cos 𝜙 −
,
sin 𝜙
and the strain is expressed as
1
(6.14)
𝜀𝜃 = (v cos 𝜙 − w sin 𝜙).
r0
Substituting r0 = r2 sin 𝜙 and Eq. (3.3) into expressions
(6.13) and (6.14) gives
v = sin 𝜙
[
]
N𝜙 (r1 + 𝜇r2 ) − N𝜃 (r2 + 𝜇r1 )
1
×
d𝜙 + C ,
Et ∫
sin 𝜙
r2
(6.15)
w = v cot 𝜃 − (N𝜃 − 𝜇N𝜙 ).
Et
P = P0 cos 𝜙 sin 𝜙.
From Eq. (6.12),
r
N𝜙 =
sin2 𝜙
[
×
(−P0 cos3 𝜙 − P0 sin2 𝜙 cos 𝜙)
∫
]
× sin 𝜙 d𝜙 + C
[
−P0 r
=
(cos2 𝜙 + sin2 𝜙)
sin2 𝜙 ∫
]
× sin 𝜙 cos 𝜙 d𝜙 + C
]
−P0 r [ 1 2 𝜙
𝜙|
+
C
sin
o
sin2 𝜙 2
−P0 r P0 rC
N𝜙 =
−
.
2
sin 𝜃
As 𝜙 approaches zero, the second expression of N 𝜙
approaches infinity unless C is set to zero.
=
P = P0 cos ϕ
r
ϕ
Figure 6.6 Snow load on a spherical head.
87
88
6 Analysis of Formed Heads and Transition Sections
Table 6.1 Membrane forces and deflections in spherical shells.
Pcos ϕ
p
P
ϕ
r
Live load
Dead load
Pr
−P cos 𝜙
P𝜙
P sin 𝜙
(
−Pr cos 𝜙 −
N𝜃
N𝜙
𝛿
𝜃
1
1 + cos 𝜙
)
−Pr
1 + cos 𝜙
]
[
1+𝜇
Pr2
(1
−
cos
𝜙)
sin 𝜙 − cos 𝜙 +
Et
sin2 𝜙
−Pr
(2 + 𝜇) sin 𝜙
Et
Uniform pressure
−P cos2 𝜙
P
P cos 𝜙 sin 𝜙
0
−Pr cos 2𝜙
2
−Pr
2
[
]
1+𝜇
Pr2
sin 𝜙 −cos2 𝜙 +
Et
2
−Pr
(3 + 𝜇) sin 𝜙 cos 𝜙
Et
Pr
2
Pr
2
Pr2
(1 − 𝜇) sin 𝜙
2Et
0
F
h
ϕo
Hydrostatic pressure
Lantern landing
Pr
−𝛾[h + r(1 − cos 𝜙)]
0
P𝜙
0
N𝜃
−𝛾r2
6
0
+F sin 𝜙0
N𝜙
𝛿
𝜃
(
)
h 4cos2 𝜙 − 6
−
r
1 + cos 𝜙
(
)
2
−𝛾r
2cos2 𝜙
h
−1 + 3 −
6
r
1 + cos 𝜙
]
[ (
)
3
−𝛾r
2(1 + 𝜇)
h
(cos
𝜙
−
1)
sin 𝜙 3 1 +
(1 − 𝜇) − 6 cos 𝜙 −
6Et
r
sin2 𝜙
2
𝛾r
sin 𝜙
Et
−1 + 3
Thus, for C = 0,
r
N𝜙 = −P0 .
2
From Eq. (6.12),
P0 r
2
)
(
1
N𝜃 = −P0 r cos2 𝜙 −
2
−P0 r
=
cos 2𝜙.
2
N𝜃 = −P0 r cos2 𝜙 +
or
6.1.2
sin2 𝜙
−F sin 𝜙0
sin2 𝜙
sin 𝜙0
−Fr
(1 + 𝜇)
Et
sin 𝜙
0
Discontinuity Analysis
The membrane analysis discussed in the previous section
fails to give adequate results when the loads are localized or when the hemispherical section is attached to
another shell that acts differently under certain loads.
In these cases, the bending moments must be considered in the analysis. For a given loading condition,
the membrane and the bending moments can be considered as shown in Figure 6.7. Proceeding as before,
where both the free-body forces and the compatibility
6.1 Hemispherical Heads
Accordingly, the equations reduce to
r
Nϕ
(6.19)
d2 Q
= Et𝜃.
d𝜙2
(6.20)
and
Nθ
Eliminating 𝜃 from Eqs. (6.19) and (6.20) gives
Nθ
r2
ϕ
−Qr2
d2 𝜃
=
2
d𝜙
D
Nϕ +
r1
дNϕ
дϕ
d4 Q
+ 4𝜆4 Q = 0,
d𝜙4
dϕ
(6.21)
where
( )2
r
.
t
The solution of Eq. (6.21) can be expressed as
𝜆4 = 3(1 − 𝜇2 )
Membrane forces
(a)
Q = e𝜆𝜙 (c1 cos 𝜆𝜙 + c2 sin 𝜆𝜙)
r
r
+ e−𝜆𝜙 (c3 cos 𝜆𝜙 + c4 sin 𝜆𝜙).
Q
Once the value of Q is determined for a given loading and boundary conditions, the other quantities can be
obtained from
Mθ
Mϕ
Mθ
r2
Mϕ +
r1
dϕ
Q+
дMϕ
дϕ
N𝜙 = −Q cot 𝜙,
−dQ
N𝜃 =
,
d𝜙
( )
−D d𝜃
M𝜙 =
,
r
d𝜙
M𝜃 = 𝜇M𝜙 ,
dϕ
дQ
dϕ
дϕ
Bending forces
(b)
Figure 6.7 (a) Membrane and (b) bending forces in a unit thin
spherical shell.
equations are taken into consideration, a complicated
set of simultaneous differential equations result. The
solution of these equations is impractical without some
simplifications. By assuming symmetric Pr forces only,
the differential equations for a spherical shell reduce to
Qr
d𝜃
d2 𝜃
+
cot 𝜙 − 𝜃(cot2 𝜙 + 𝜇) = −
2
d𝜙
d𝜙
D
2
(6.16)
and
d2 Q
dQ
+ cot 𝜙
− Q(cot2 𝜙 − 𝜇) = Et𝜃,
2
d𝜙
d𝜙
(6.17)
where 𝜃 is the angle of rotation and is given by
𝜃=
v 1 dw
+
.
r r d𝜙
(6.22)
𝜃 = slope
)
(
1 d2 Q
=
,
Et d𝜙2
w = radial deflection
r
(6.23)
= (N𝜃 − 𝜇N𝜙 ).
Et
The solution of Eq. (6.23) for various common loading
conditions is given in Table 6.2.
Example 6.3
Calculate the head discontinuity forces of the head-toshell junction shown in Figure 6.8a. Let P = 300 psi and
𝜇 = 0.3. Use thin-shell theory.
Solution:
From Figure 6.8b and Table 6.1, the deflection in the head
due to pressure is
Pr2
(1 − 𝜇) sin 𝜙
2Et
(300)(50)2
(1 − 0.3)
=
2E(0.50)
525,000
=
E
𝛿p =
(6.18)
By a rigorous analysis, Gibson [1] has shown that
in Eqs. (6.16) and (6.17), only the higher-order terms
are significant in usual pressure vessel applications.
89
90
6 Analysis of Formed Heads and Transition Sections
Table 6.2 Approximate force and deflection functions for spherical segments.
Mo
ϕ
Ho
ϕ0
Y
Q
√
− 2e−𝜆𝛾 sin 𝜙0 cos(𝜆𝛾 + π∕4)H0
N𝜃
2𝜆e−𝜆𝛾 sin 𝜙0 (cos𝜆𝛾)H 0
N𝜙
√
2e−𝜆𝛾 sin 𝜙0 cot 𝜙 cos(𝜆𝛾 + π∕4)H0
M𝜃
𝜇M𝜙
M𝜙
𝛿
𝛼
Mo
Ho
2𝜆 −𝜆𝛾
e sin(𝜆𝛾)M0
r
√ 𝜆2
2 2 e−𝜆𝛾 cos(𝜆𝛾 + π∕4)M0
r
2𝜆 −𝜆𝛾
− e cot 𝜙 sin(𝜆𝛾)M0
r
r −𝜆𝛾
e sin 𝜙0 sin(𝜆𝛾)H0
𝜆
√
H0 −𝜆𝛾
{re sin 𝜙0 [2𝜆 sin 𝜙 cos 𝜆𝛾 − 2𝜇 cos 𝜙 cos(𝜆𝛾 + π∕4)]}
Et
√
H0
[−2 2𝜆2 e−𝜆𝛾 sin 𝜙0 sin(𝜆𝛾 + π∕4)]
Et
𝜇M𝜙
√
2e−𝜆𝛾 sin(𝜆𝛾 + π∕4)M0
√
M0
{2𝜆e−𝜆𝛾 [ 2𝜆 sin 𝜙 cos(𝜆𝛾 + π∕4) + 𝜇 cos 𝜙 sin 𝜆𝛾]}
Et
)
(
M0 −4𝜆3 −𝜆𝛾
e cos 𝜆𝛾
Et
r
and
t = ½”
√
𝜆=
r = 50”
4
3(1 − 0.32 )
(
50
0.5
)2
= 12.854.
r = 50”
t = 1”
(a)
MO
HO
HO
MO
r = 50”
r = 1”
From Table 6.2,
2570.8
𝛿Ho =
Ho ,
E
−660.9
𝛼Ho =
Ho ,
E
660.9
𝛿Mo =
Mo ,
E
−339.8
𝛼Mo =
Mo .
E
Similarly, the deflection in the shell due to pressure is
obtained from Eq. (1) of Example 5.5 as
(
𝜇)
Pr2
1−
𝛿=
Et
2
(300)(50)2
=
(1 − 0.15)
E(1.0)
637,500
=
,
√ E
𝛽=
(b)
Figure 6.8 Discontinuity forces in a spherical
head-to-cylindrical-shell junction.
4
3(1 − 0.32 )
(50)2 (1.0)2
= 0.1818,
E(1.0)3
D=
= 0.0916E,
12(1 − 0.32 )
and from Table 6.2,
−908.43
𝛿Ho =
Ho ,
E
6.1 Hemispherical Heads
−165.15
Ho ,
E
165.15
𝛿Mo =
Mo ,
E
60.05
𝜃Mo =
Mo ,
E
total deflection of head
can be expressed as
(
)
r
r3 − ri3 r0 2
2E𝛼
2
Tr dr −
Tr dr ,
𝜎r =
∫ri
(1 − 𝜇)r3 ro3 − ri3 ∫ri
(6.24a)
E𝛼
𝜎𝜙 = 𝜎𝜃 =
(1 − 𝜇)r3
(
)
r
2r3 + ri3 r0 2
2
3
×
Tr dr +
Tr dr − r T .
∫ri
r3 − r3 ∫ri
𝜃Ho =
= total deflection of shell
525,000 2570.8
660.9
+
Ho +
Mo
E
E
E
637,500 908.43
165.15
=
−
Ho +
Mo
E
E
E
0
(6.24b)
or
8Ho + Mo = −995.17.
(1)
Similarly,
rotation of head = rotation of shell
339.8
165.15
−660.9
Ho −
Mo = −
Ho
E
E
E
60.05
+
Mo
E
or
Ho = −0.818Mo ,
Example 6.4
Determine the circumferential thermal stress on the
inside surface of a hemispherical head subjected to
an inside temperature of 600 ∘ F and varying linearly
to a temperature of 400 ∘ F at the outside surface.
Let r1 = 30 in., r2 = 40 in., E = 30 × 106 psi, 𝛼 = 7.0 ×
10−6 in./in.-∘ F, and 𝜇 = 0.3.
Solution:
The temperature distribution across the thickness can be
expressed as
T = 600 −
and from Eq. (1),
Mo = −48.5 lb-in.∕in.
T = 1200 − 20r.
Ho = 39.3 lb∕in.
From Table 6.2, N 𝜃 at the discontinuity is
2𝜆2
Pr
Mo +
N𝜃 = 2𝜆Ho +
r
2
2(12.854)2 (−48.5)
= 2(12.854)(39.3) +
50
300(50)
+
,
2
N𝜃 = 1010.3 + −320.54 + 7500,
The first integral in Eq. (6.24b) gives
r0
∫ri
𝜎𝜃 =
pr
= 7500 lb∕in.,
2
M𝜙 = 48.5 lb-in.∕in.,
and
(1200r2 − 20r3 )dr
(30 × 106 )(7 × 10−6 )
(1 − 0.3)(30)3
{
2(30)3 + (30)3
×
(6,050,000)
(40)3 − (30)3
}
3
− (30) [1200 − 20(r)]
= −32,800 psi.
M𝜃 = 14.6 lb-in∕in.
The derivation of the thermal stress in a spherical
segment due to a radial distribution of temperature
can be derived similarly to cylindrical shells (see
Section 5.4.3). The meridional and circumferential
stresses in a sphere are the same due to symmetry and
∫30
The second integral is zero because the limits of integration at the inner surface are both ri . Hence, Eq. (6.24b)
gives
N𝜙 =
Thermal Stress
40
Tr2 dr =
= 6,050,000.
N𝜃 = 8190 lb∕in.,
6.1.3
200
(r − ri )
r0 − ri
or
and
and
i
6.1.4
Buckling Strength
The buckling equations developed by Von Karman and
Tsien [2] are the basis of the design equations developed
by the ASME. Von Karman’s equations, which are substantiated by tests, give a more accurate prediction of
the buckling strength of spherical sections than those
91
92
6 Analysis of Formed Heads and Transition Sections
Z
𝛽
×
P
∫0
𝛽
ϕ
rO
ZO
[(
+P
θ
∫0
d𝜃
−1
d𝜙
(
)2
+
𝜙2 (𝜃 − 𝜙)d𝜙.
𝜃
−1
𝜙
)2 ]
𝜙 d𝜙
(6.29)
The solution of Eq. (6.29) is obtained by the Rayleigh–Ritz
method by finding an expression for the deflection that
satisfies the boundary condition
{
0 at 𝜙 = 0,
𝜃=
𝛽 at 𝜙 = 𝛽.
r
ϕ
β
Such an expression can be written as
(
)]
[
𝜙2
,
𝜃 = 𝜙 1 − C1 1 − 2
𝛽
Figure 6.9 Buckling shape of a spherical segment.
(6.30)
where 𝜃 is the slope and is related to the deflection by
developed earlier by Flugge, Timoshenko, and others.
Von Karman and Tsien took the out-of-roundness
imperfections into consideration. They also used the
energy equations as a basis for derivation. Referring to
Figure 6.9, it can be shown that the strain energy due to
the extension of the sphere is given by
)2
𝛽(
( )
cos 𝜙
3 t
π
− 1 sin 𝜙 d𝜙.
U1 = Er
∫0
r
cos 𝜃
(6.25)
Similarly, the strain energy due to bending is
expressed as
𝛽
( )3
π
t
U2 = Er3
sin 𝜙
r 12 ∫0
[(
)2 (
)2 ]
sin 𝜃
cos 𝜃 d𝜃
d𝜙.
−1 +
−1
×
cos 𝜙 d𝜙
sin 𝜙
(6.26)
The potential energy of the external pressure P is
given by
U3 = Pr3 π
𝛽
∫0
P sin2 𝜙(tan 𝜃 − tan 𝜙) cos 𝜙 d𝜙.
(6.27)
The total energy of the system is the sum of Eqs.
(6.25)–(6.27):
U = U1 + U2 + U3 .
(6.28)
Equation (6.28) can be simplified by assuming 𝛽 to be
small. By neglecting terms of higher order and expanding
the sine and cosine functions in a power series, Eq. (6.28)
becomes
E(t∕r) 𝛽 2
U
=
(𝜃 − 𝜙2 )𝜙 d𝜙
πr3
4 ∫0
E(t∕r)3
+
12
𝛿=r
𝛽
∫0
(𝜙 − 𝜃)d𝜙.
(6.31)
From Eqs. (6.30) and (6.31), the value of C 1 can be determined to be
4𝛿
(6.32)
C1 = 2 .
r𝛽
Substituting Eq. (6.30) into Eq. (6.29), the energy expression becomes
)
(
3
4
)
(
C
C
Et
U
B6 C12 − 1 + 1
=
πr3
60r
2
14
Et 3 2 2 P𝛽 4
𝛽 C1 −
C.
18r3
12 1
This expression can be minimized by taking its derivative
with respect to C 1 and equating the derivative to zero.
This gives
( )
4 𝛿
𝜎r
=
Et
105
[ t
)
]
( ) t∕r (
(t∕r)2
𝛿2
𝛿
× 21 − 63
+ 48 2 + 70
,
t 𝛽2
t
𝛽4
(6.33)
+
where
Pr
,
2t
and 𝛿 is obtained from Eq. (6.32).
A plot of Eq. (6.33) is shown in Figure 6.10. The minimum value of Eq. (6.33) can be found by taking the
derivative with respect to 𝛽 and equating the result
to zero:
3
(𝛿∕t)2
( )1+
4 𝛿
𝜎r
280
,
(6.34)
=
24
Et
5 t
1 + (𝛿∕t)2
35
which is shown as a dashed line in Figure 6.10. This figure
illustrates the effect of 𝛿/t on the buckling strength of
𝜎=
6.2 Ellipsoidal Heads
Figure 6.10 Minimum envelope of the buckling
strength of a spherical shell.
1.2
1.1
1.0
β2R = 30
t
0.9
20
0.8
15
0.7
σR
Et
10
0.6
5
0.5
7
0.4
0.3
0.2
Envelope
0.1
0.0
0
2
4
6
8
14
12
14
16
18
δ
t
spherical sections. The minimum value of the buckling
strength is obtained from the figure as
𝜎r
𝛿
= 0.24 at = 9.35.
Et
t
The value 0.24 can be reduced if the strain energy due
to membrane stress before buckling is considered. If Eq.
(6.25) is modified to include this strain energy and if
the revised expression is substituted into Eq. (6.29), the
differentiation results in an expression whose minimum
value is
𝜎r
= 0.183.
Et
Experimentally, the minimum value obtained is of the
order of
𝜎cr r
= 0.125.
(6.35)
Et
Example 6.5
What is the required thickness of a hemispherical head
subjected to an external pressure of 15 psi? Let r = 96 in.,
E = 27 × 106 psi, and the factor of safety (FS) = 10.
Solution:
From Eq. (6.35), with 𝜎 cr = (FS)𝜎 and 𝜎 = Pr/2t,
( )
( )
r
Pr
(FS)
= 0.125
2t
Et
or
√
Pr2 (FS)
t=
0.25E
√
=
(15)(96)2 (10)
0.25(27 × 106 )
= 0.45 in.
6.2 Ellipsoidal Heads
The governing equations for the design of ellipsoidal and
torispherical heads are obtained from Eqs. (6.10) and
(6.11). For internal pressure, Pr = P, P𝜙 = 0, and the two
equations give [3]
Pr
N𝜙 = 2 ,
2
2r − r2
.
(6.36)
N𝜃 = +Pr2 1
2r1
We can write Eq. (6.36) in terms of the major and minor
radii a and b. Using the notations of Figure 6.11, we then
obtain
a2 b 2
r1 =
2
2
(a sin 𝜙 + b2 cos2 𝜙)3∕2
and
a2
r2 =
.
2
(a2 sin 𝜙 + b2 cos2 𝜙)1∕2
Expressions (6.36) then become
N𝜙 =
Pa2
1
,
2 (a2 sin2 𝜙 + b2 cos2 𝜙)1∕2
N𝜃 =
2
Pa2 b2 − (a2 − b2 )sin 𝜙
.
2
2
2b (a2 sin 𝜙 + b2 cos2 𝜙)1∕2
(6.37)
93
94
6 Analysis of Formed Heads and Transition Sections
ro
3
2
o
ho
r1
b
γ
5 a/b
4
3
2
1
0
2
Nθ
pa
r2
ϕ
3
−2
a
4
5
−4
−5
3
5
Figure 6.11 Ellipsoidal head.
2
The radial deflection w and meridional deflection u due
to internal pressure are given by
{√
(a∕b)2 − 1
Pa2
w=
F sin 𝛾
2Et
2(a∕b)
}
( )2
a
2
[(2 − 𝜇)G − 1] ,
+
b
√
Pa2 (a∕b)2 − 1
F cos 𝛾,
u=
2Et
2(a∕b)
where
{[ ( )2
]
a
F=
2
− (1 − 2𝜇)
b
(
)
√( )
a
a 2
× ln
− 1 sin 𝛾 G
+
b
b
[
]
}
( ) √( )2
a
a
2
− (1 − 2𝜇)
− 1 G sin 𝛾 ,
b
b
1
.
G= √
1 + [(a∕b)2 − 1]cos2 𝛾
At any given point on the ellipse given by x0 and y0 , the
angle 𝜙 can be obtained from
x0
.
sin 𝜙 = √
(a∕b)4 y20 + x20
A plot of Eq. (6.37) in Figure 6.12 shows that for
ellipsoidal heads with a/b ratios over 1.4, the hoop stress
at 𝜙 = 90∘ is compressive. The curves indicate that this
compressive force increases as the head gets shallower.
Design of heads based on these high compressive membrane forces tends to give ultra-conservative answers.
This is because discontinuity forces tend to lower the
maximum compressive stress, so that taking them into
Nϕ
pa
a/b
4
3
1
2
1
0.0
0.0
0.2
0.4
0.6
0.8
1.0
r0/a
Figure 6.12 Stress distribution due to internal pressure in
ellipsoidal heads.
account results in more realistic designs. The ASME
Code uses such an approach in the design of elliptic and
torispherical heads.
Example 6.6
A 2 : 1 ellipsoidal head is subjected to an internal pressure
of 100 psi. If a = 48 in. and t = 0.5 in., determine the hoop
and meridional stress at 𝜙 = 90∘ .
Solution:
With 𝜙 = 90∘ , Eq. (6.37) becomes
pa
N𝜙 =
2
(100)(48)
=
= 2400 lb∕in.
2
and
2400
= 4800 psi.
𝜎𝜙 =
0.5
Similarly,
pa
N𝜃 = 2 (2b2 − a2 )
2b
(100)(48)
=
(2 × 242 − 482 )
2(24)2
= −4800 lb∕in.
6.4 Conical Heads
where
and
𝜎𝜃 =
−4800
= −9600 psi.
0.5
A𝛽s = e−𝛽s (cos 𝛽s + sin 𝛽s)
B𝛽s = e−𝛽s (cos 𝛽s − sin 𝛽s)
C 𝛽s = e−𝛽s cos 𝛽s
6.3 Torispherical Heads
In formulating the discontinuity equations for torispherical heads at the cylinder junction, two assumptions
must be made. First, the ratio a/t must be over 30.
Second, all deflections dissipate rapidly away from the
junction. With these two assumptions (known as Geckeler’s approximations), the discontinuity analysis of a
torispherical head near a cylinder junction is similar
to that for a cylindrical shell. Hence, the governing
equations are (see Section 5.2.1)
d4 w
+ 4𝛽 4 w = 0,
dx4
where
√
𝛽=
4
6.4 Conical Heads
The stress distribution in a conical head can be obtained
from Eqs. (6.10) and (6.11). From Figure 6.14 with
𝜙 constant,
r = s sin 𝛼 ′ ,
(6.38)
r1 = ∞,
r2 = s tan 𝛼 ′ .
Redefining N 𝜙 as N s and P𝜙 as Ps , Eq. (6.10) becomes
3(1 − 𝜇2 )
.
r22 t 2
N𝜃 = Pr s tan 𝛼 ′
Equation (6.38) is similar to Eq. (5.21) for cylindrical
shells except that in Eq. (6.38), the quantity 𝛽 is a function of r2 that is variable along the meridian. This requires
numerical integration of all moment, force, deflection,
and slope expressions at angles less than 𝜙 = 90∘ .
If a discontinuity force is applied at the edge as shown
in Figure 6.13, Eq. (6.38) yields the following values [4]:
1
w = 3 (C𝛽s Q0 − 𝛽0 B𝛽s M0 ),
2𝛽0 D
Et
a2
N𝜃 = w = 2𝛽0 (C𝛽s Q0 − 𝛽0 B𝛽s M0 ),
r
r2
√2
a3
Q=
(B𝛽s Q0 + 2𝛽0 D𝛽s M0 ),
r23
a
M𝜙 =
(−D𝛽s Q0 + A𝛽s M0 ),
𝛽 0 r2
M𝜃 = 𝜇M𝜙 ,
D𝛽s = e−𝛽s sin 𝛽s
√
3(1 − 𝜇2 )
4
𝛽0 =
a2 t 2
𝜇 = Poisson’s ratio
and Eq. (6.11) becomes
Ns =
−1
(P − Pr tan 𝛼 ′ )s ds.
s ∫ s
The forces and deflections obtained from Eq. (6.40) due
to some typical loading conditions are shown in Table 6.3.
Example 6.7
A conical shell with 𝛼 ′ = 45∘ and base diameter of 8 ft is
subjected to an internal pressure P. Find the expressions
for N 𝜃 and N s .
αʹ
s
(6.39)
ℓ
ℓ′
r
ϕ
Figure 6.13 Edge forces in ellipsoidal heads.
(6.40)
Figure 6.14 Conical heads.
r2
95
96
6 Analysis of Formed Heads and Transition Sections
Table 6.3 Membrane forces and deflections in conical shells [3].
Po
Po
S
S
α′o
α′o
Internal pressure
Live load
Dead load
Pr
−P0 cos 𝛼0′
−P0 cos2 𝛼0′
P0
Ps
P0 sin 𝛼 0
P0 cos 𝛼0′ sin 𝛼0′
o
−P0 Scos2 𝛼0′
sin 𝛼0′
2
1 P0 S
−
S 2 sin 𝛼0′
2
−P0 S cos3 𝛼0′
sin 𝛼0′
P0 S cot 𝛼0′
N𝜃
NS
(
𝜇)
−S
P0 cot 𝛼0′ cos2 𝛼0′ −
Et
2
]
P0 S cos 𝛼0′ [
1
(2 + 𝜇)cos2 𝛼0′ − − 𝜇
2 ′
2
Etsin 𝛼0
𝛿
𝜃
P0 S
cot 𝛼0′
2
(
P S2
𝜇)
− 0 cos 𝛼0′ cot 𝛼0′ cos2 𝛼0′ −
Et
2
]
P0 S 2 ′ [
1
cot 𝛼0 (2 + 𝜇)cos2 𝛼0′ − 𝜇 −
Et
2
−
Solution:
For internal pressure, N s = 0 and N 𝜃 = P. Therefore, from
Eq. (6.40),
PS
cot 𝛼0′
2
(
P0 S 2
𝜇)
cos 𝛼0′ cot 𝛼 ′ 1 −
Et
2
−
3 PS 2 ′
cot 𝛼0
2 Et
Summation of vertical forces at point 0 gives
2πRV = PπR2
or
N𝜃 = Ps tan 𝛼 ′
PR
.
2
Since V is the resultant of components N s and H, it
follows that H is an inward force with magnitude
V =
or
N𝜃 =
Pr
.
cos 𝛼 ′
(1)
Also,
P
s2
Ns = tan 𝛼 ′ + C
s
2
at
s = 0,
Ns = 0.
PR tan 𝛼 ′
.
2
This force H must be resisted by ring action at the junction. The required area of the ring is given by
H=
Hence,
C=0
Ps tan 𝛼 ′
Ns =
2
and
A=
HR
𝜎
A=
PR2 tan 𝛼 ′
,
2𝜎
or
(6.41)
where
Pr
Ns =
.
2 cos 𝛼 ′
A = required ring area
P = internal pressure
6.4.1 Unbalanced Forces at Cone-to-Cylinder
Junction
The junction of cones to cylinders must always be considered as part of the cone design because of the large
stresses that occur there. Refer to Figure 6.15. The force
N s at point 0 was found in Example 6.7 to be
Ns =
PR
.
2 cos 𝛼 ′
R = radius at the base of cone
𝛼 ′ = one-half of the apex angle
𝜎 = 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 in the ring be 10,000 psi.
6.4 Conical Heads
Solution:
From Eq. (6.41),
20 × 482 × 1.00
2 × 10,000
= 2.30 in.2 .
A=
o
p
6.4.2
v
NS
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, approximate solutions for various edge
loading conditions can be expressed in a simple form as
shown in Table 6.4. In this table,
√
3(1 − 𝜇2 )
l
𝛽=
.
sin 𝜙
r22 t 2
v
V
V
H
H
Discontinuity Analysis
Example 6.9
Calculate the maximum longitudinal and circumferential
stresses in the cylinder shown in Figure 6.16 due to internal pressure P.
NS
o
Solution:
From Figure 6.16,
(1)
f + F = H,
(b)
where from Section 6.4.1,
Figure 6.15 Discontinuity forces at a conical
head-to-cylindrical-shell junction.
H=
Pr tan 𝛼 ′
.
2
Table 6.4 Forces and deflections in conical shells due to edge loads.
ℓ
xℓ
Ho
Q
N𝜃
N𝜙
M𝜙
r2
ϕ
Ho
√
− 2e−𝛽x sin 𝜙 cos(𝛽x + π∕4)H0
Mo
Mo
−2𝛽e−𝛽x sin 𝜙
sin(𝛽x)M0
l
√
2 2r2 𝛽 2 −𝛽x 2
e sin 𝜙 cos(𝛽x + π∕4)M0
l2
−2𝛽 −𝛽x
e cos 𝜙 sin(𝛽x)M0
l
√
− 2e−𝛽x sin(𝛽x + π∕4)M0
−2r2 𝛽 −𝛽x 2
e sin 𝜙 cos(𝛽x)H0
l
(
)
√
π
− 2e−𝛽x cos 𝜙 cos 𝛽x +
H0
4
l −𝛽x
e sin 𝛽x H0
𝛽
M𝜃
𝜇M𝜙
𝛿
H0 l3 e−𝛽x
𝜇l cot 𝜙
cos 𝛽x − √
cos(𝛽x + π∕4)
2D𝛽 3 sin 𝜙
2r2 𝛽 sin 𝜙
]
[√
−l2 e−𝛽x
1 cos 𝜙 sin 𝛽x
M0
2 cos(𝛽x + π∕2) + 𝜇
2
2
2D𝛽 sin 𝜙
r2 𝛽sin 𝜙
𝜃
−l2 e−𝛽x H0
sin(𝛽x + π∕4)
√
2D𝛽 2 sin 𝜙
le−𝛽x
cos 𝛽x M0
D𝛽 sin 𝜙
[
]
𝜇M𝜙
97
98
6 Analysis of Formed Heads and Transition Sections
f
M
6.4.3
H
t
f
F
r
M
F
αʹ
Figure 6.16 Discontinuity forces at a head-to-ring-to-shell
junction.
The deflection compatibility between the cylinder and
cone is given by
deflection of cylinder at junction due to M and f
= deflection of cone at junction
due to M and F.
(2)
Similarly,
The solution for the buckling of a conical section subjected to external hydrostatic pressure is normally
obtained by energy methods. The resultant equation is
very cumbersome due to the iterative process needed
for the solution [5]. Experimental research comparing
the buckling equations of conical and cylindrical shells
has shown that the buckling of a conical shell is similar
to the buckling of cylindrical shells with a length equal
to the slant length of the cone and a radius equal to the
average radius of curvature of the cone. Research has
also shown that the quantity 1 − D1 /D2 has a significant
influence on the buckling of a cone. Accordingly, the
buckling equation for a cone can be expressed as
(
)
D1
,
Pcr = pf 1 −
D2
where p is the modified buckling equation of a cylindrical
shell and f (1 − D1 /D2 ) is a function of D1 /D2 .
A simplified equation for the buckling of a cylindrical
shell [6] is
Pcr =
rotation of cylinder at junction due to M and f
= rotation of cone at junction
due to M and F.
Cones Under External Pressure
(3)
Using Tables 5.2 and 6.4 and solving Eqs. (1)–(3) results
in the following expressions:
f = HV 1 ,
F = H(1 − V1 ),
( )
2
V2 ,
M=H
𝛽
where
Pr tan 𝛼 ′
,
2
cos2 𝛼 ′ (3 + cos2 𝛼 ′ )
V1 =
1 + cos2 𝛼 ′ (6 + cos2 𝛼 ′ )
V1
V2 =
,
3 + cos2 𝛼 ′
√
3(1 − 𝜇2 )
4
𝛽=
.
r2 t2
The maximum longitudinal stress due to M and pressure is expressed as
√
)
(
Pr
r
𝜎x =
tan 𝛼 ′ ,
0.5 + 4.559V2
t
t
H=
whereas the maximum circumferential stress due to M
and pressure is given by
√
]
[
Pr
r
𝜎𝜃 =
(V1 − 2V2 ) tan 𝛼 ′ .
1.0 − 1.316
t
t
(t∕2r)2.5
2.42E
.
(1 − 𝜇2 )3∕4 [L∕2r − 0.45(t∕2r)1∕2 ]
(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 (Figure 6.17) may be written as
)
2.5 (
D1
2.42E (t∕2𝜌av )
Pcr =
f
1
−
D2
(1 − 𝜇2 )3∕4 l′ ∕2𝜌av
or
(
)
Pcr
D1
2.6(cos 𝛼 ′ )2.5 (t)2.5
f 1−
.
=
E
D2
l[(D1 + D2 )∕2]1.5
The magnitude of the function f (1 − D1 /D2 ) can be
determined theoretically. Based on this plus the “scatter”
αʹ
D1
ℓʹ
ℓ
ρ
D2
Figure 6.17 Dimensions of a conical section.
6.5 Nomenclature
band of experimental data, a value of 1.0 was used for the
function at D1 /D2 of 1.0 (cylinder). The function changes
linearly to a value of 0.8 for D1 /D2 of zero (full cone).
Thus, the buckling equation becomes
)2.5
(
⎤
⎡
t cos 𝛼 ′
2.6
⎥ [ 21.5 (0.8 + 0.2D ∕D ) ]
⎢
D2
Pcr ⎢
1
2
⎥
.
=
⎥
⎢
E
L∕D2
1 + D1 ∕D2
⎥
⎢
⎦
⎣
The second bracketed expression can be approximated by
the quantity
Solution:
𝛼 ′ = 30.96∘ and cos 𝛼 ′ = 0.858.
(
)
60
30
1+
= 24.375 in.
Le =
2
90
From Eq. (6.43),
(
)
(Pe )(FS)(Le ∕D2 ) 0.4
te = D2
2.6(E)
)0.4
(
(15)(4)(24.375∕96)
= 96
2.6(30 × 106 )
= 0.20 in.
and
2
.
1 + D1 ∕D2
t=
Using an FS, the allowable external pressure on a cone is
given by
Pa
2.6(te ∕D2 )2.5
=
,
E
(FS)(Le ∕D2 )
(6.43)
te
= 0.23 in.
cos 𝛼 ′
6.5 Nomenclature
where
a = major radius of ellipse
Pa = allowable external pressure
b = minor radius of ellipse
D = Et 3 /12(1 − 𝜇2 )
E = modulus of elasticity
t e = tcos 𝛼 ′
D1 = diameter at the small end of cone
Le = (l/2)(1 + D1 /D2 )
D2 = diameter at the large end of cone
D1 = diameter at the small end of cone
E = modulus of elasticity
D2 = diameter at the large end of cone
L = axial length of cylinder
Le = (l1 /2)(1 + D1 /D2 )
Example 6.10
Design the cone shown in Figure 6.18 for an external
pressure of 15 psi. Let FS = 4.0 and E = 30 × 166 psi.
l = axial length of cone
l′ = slant length of cone
M𝜃 = bending moment in hoop direction
M𝜙 = bending moment in meridional direction
N s = axial force in cone
N 𝜃 = force in hoop direction
D2 = 96ʺ
N 𝜙 = force in meridional direction
P = pressure
ℓ = 30ʺ
Pa = allowable external pressure
Pi = internal pressure
P0 = external pressure
D1 = 60ʺ
Pr = radial pressure
Ps = axial pressure in cone
Q = shearing force in head
r = radius
r1 = radius of curvature as defined in Figure 6.4
Figure 6.18 Conical transition section.
r2 = radius of curvature measured from the axis of
symmetry
99
100
6 Analysis of Formed Heads and Transition Sections
ri = inside radius
ro = outside radius
s = distance along the slant length of cone, measured
from the apex
T = temperature
t = thickness
t e = t cos 𝛼 ′
𝛾 = 𝜋/2 − 𝜙
𝛿 = deflection measured perpendicular to the axis of
symmetry
𝜃 = rotation
√
𝜆 = 4 3(1 − 𝜇2 )(r∕t)2
𝜇 = Poisson’s ratio
𝜎 = stress
v = axial deformation
𝜎 cr = critical buckling stress
w = radial deformation
𝜎 r = radial stress
𝛼 = coefficient of thermal expansion
𝜎 s = longitudinal stress in cone
𝛼 ′ = one-half of the apex angle of a cone
⎧√
2
⎪ 4 3(1 − 𝜇 ) for cylinders
2
2
⎪
r t
𝛽 = ⎨√
3(1
− 𝜇2 )
⎪4
for cones
⎪
r22 t 2
⎩
𝜎 𝜃 = hoop stress
𝜎 𝜙 = meridional stress
𝜙 = angle defined in Figure 6.4
References
1 Gibson, J.E. (1965). Linear Elastic Theory of Thin
Shells. New York: Pergamon Press.
2 Von Karman, T. and Tsien, H.-S. (1960). The buckling
of spherical shells by external pressure. In: Pressure
Vessel and Piping Design: Collected Papers 1927–1959.
New York: American Society of Mechanical Engineers.
3 Baker, E.H. et al. (1968). Shell Analysis Manual, NASA
CR-912. Washington, DC: National Aeronautics and
Space Administration.
4 Coates, W.M. (1960). The state of stress in full heads
of pressure vessels. In: Pressure Vessel and Piping
Further Reading
1 Flugge, W. (1967). Stresses in Shells. New York:
Springer-Verlag.
2 Timoshenko, S. and Woinowsky-Krieger, S. (1959).
Theory of Plates and Shells. New York: McGraw-Hill.
Design: Collected Papers 1927–1959. New York:
American Society of Mechanical Engineers.
5 Jawad, M.H. (1979). Design of conical shells
under external loads. J. Pressure Vessel Technol.
102: 230.
6 Raetz, R.V. 1959. An experimental investigation of
the strength of small-scale conical reducer sections
between cylindrical shells under external hydrostatic
pressure. U.S. Department of the Navy, David Taylor
Model Basin, Report 1187.
Various flat plates. Source: Courtesy of the Nooter Corporation, St. Louis, MO.
102
7
Stress in Flat Plates
7.1 Introduction
A
Flat plates are very common in process equipment.
Circular plates are used in such areas as nozzle covers,
bulkheads, and tubesheets, whereas rectangular plates
are used as segmental trays, as baffles, and in rectangular
pressure vessels. This chapter presents a brief description
of the theoretical background of circular and rectangular
plates.
The theory of symmetric bending of laterally loaded
plates [1] is generally based on the following assumptions:
1) The thickness of the plate is significantly smaller than
its least lateral dimension.
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
deformation remain perpendicular to the deformed
middle surface.
5) These lines are inextensible.
6) These lines remain straight lines.
These assumptions form the basis for developing the
bending theory of plates and apply to plates where buckling is not a consideration.
7.2 Circular Plates
The relationship between the radius of curvature and the
deflection of a circular plate is obtained from Eq. (3.6) as
d2 w
1
=
,
rx
dx2
or in the terminology of Figure 7.1,
1
d2 w −d𝜙
=
=
.
rn
dx2
dr
(7.1)
The second radius of curvature is also obtained from
Figure 7.1. Line AB is the radius of curvature rt of all
ϕ
rt
r
B
r
ϕ
dr
Figure 7.1 Deflection of a circular plate.
points at a distance r forming a cone:
r
sin 𝜙 ≈ 𝜙 = .
rt
Using the sign convention that clockwise angles and
moments are positive and downward deflections are positive, the relationship becomes
𝜙 1 dw
1
=− =
.
(7.2)
rt
r
r dr
The moment–curvature relationship is based on Eq.
(3.11) and is given by
(
)
1
1
Mr = −D
+𝜇
.
rn
rt
Substituting Eqs. (7.1) and (7.2) into this expression
gives
( 2
)
d w 𝜇 dw
Mr = −D
+
(7.3a)
dr2
r dr
(
)
𝜙
1 dw
=D
+𝜇
.
(7.3b)
r dr
r
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
7.2 Circular Plates
Similarly,
(
1 dw
d2 w
+𝜇 2
r dr
dr
(
)
𝜙
d𝜙
= −D
+𝜇
.
r
dr
Substituting Eqs. (7.3a), (7.3b), and (7.4) into Eq. (7.6)
gives
)
Mt = −D
(7.4)
For a uniformly loaded plate, the forces acting on an
element are shown in Figure 7.2a. Taking moments about
a–a gives
(
)
dMr
(Mr rd𝜃) − Mr +
dr (r + dr)d𝜃
dr
(
)
d𝜃
dr
+ 2 Mt dr
− Qr d𝜃
2
2
(
)
dQ
dr
− Q+
dr (r + dr)d𝜃
= 0.
(7.5)
dr
2
The quantity Mt dr(d𝜃/2) is the component of Mt perpendicular to axis a–a as shown in Figure 7.2b.
Disregarding higher-order terms, Eq. (7.5) can be
reduced to
dMr
Mr +
(7.6)
r − Mt + Qr = 0.
dr
(Q+ dQ dr)
dr
(Mr+
dMr
dr)
dr
Mt
a
dθ
dr
a
(7.7a)
[
(
)]
Q
dw
d 1 d
r
= .
dr r dr
dr
D
(7.7b)
Similarly, substituting Eqs. (7.3a), (7.3b), and (7.4) into
Eq. (7.6) gives
d2 𝜙 1 d𝜙 𝜙
−Q
+
− 2 =
dr2
r dr
r
D
or
(7.8)
[
]
Q
d 1 d
(r𝜙) = − .
dr r dr
D
Eqs. (7.7a), (7.7b), and (7.8) are the basic differential
equations for the bending of circular plates due to symmetric loading. Eqs. (7.7a) and (7.7b) can also be written
in terms of the local load as
{
[
(
)]}
q
d 1 d
dw
1 d
r
r
= .
(7.9)
r dr
dr r dr
dr
D
Once w is determined from Eqs. (7.7a) and (7.7b), then
the moments are obtained from Eqs. (7.3a), (7.3b), and
(7.4). The shearing force is determined from Eq. (7.6) and
is expressed as
)
( 3
1 dw
d w 1 d2 w
+
− 2
.
(7.10)
Q=D
dr3
r dr2
r dr
Example 7.1
Derive the moment expression for a uniformly loaded,
simply supported circular plate of radius a. For 𝜇 = 0.3,
plot the moment diagram and determine the maximum
deflection, rotation, and stress values.
Mr
Mt
or
1 dw Q
d3 w 1 d2 w
+
− 2
=
3
2
dr
r dr
r dr
D
Q
Solution:
From Figure 7.3, the shear Q at any radius r is given by
2𝜋rQ = 𝜋r2 P, or
Pr
Q=
.
2
P
dθ
dθ
2
r
a
Mt
(b)
Figure 7.2 Forces in a unit circular plate.
Figure 7.3 Uniformly loaded simply supported plate.
103
104
7 Stress in Flat Plates
Therefore, from Eqs. (7.7a) and (7.7b),
[
(
)]
d 1 d
dw
Pr
.
r
=
dr r dr
dr
2D
4.0
Integrating both sides gives
(
)
d
dw
Pr3
+ C1 r.
r
=
dr
dr
4D
A second integration gives the expression for the slope:
Cr C
dw
Pr3
𝜃=
=
+ 1 + 2,
(1)
dr
16D
2
r
and the third integration gives the deflection w as
16M
a2 P
C r2
Pr4
(2)
+ 1 + C2 ln r + C3 .
64D
4
At the center of the plate, r = 0 and the slope is zero due
to symmetry. Hence, from Eq. (1), C 2 = 0.
At r = a, the moment Mr = 0 and Eqs. (7.3a) and (7.3b)
give
(
)
Pa2 3 + 𝜇
C1 = −
.
8D 1 + 𝜇
dw
Pr
𝜃=
=
dr
16D
(
)
3+𝜇 2
•a
.
r −
1+𝜇
2
Mr
1.0
–1.0
–2.0
0
0.2
0.4
r/a
0.6
0.8
1.0
and with 𝜇 = 0.30
(4)
The maximum deflection occurs in the middle where
r = 0. Hence,
(
)
Pa4 5 + 𝜇
wmax =
64D 1 + 𝜇
(
)
Pa4 5 + 𝜇 12(1 − 𝜇2 )
=
64 1 + 𝜇
Et 3
and with 𝜇 = 0.30
0.696Pa4
Et 3
The maximum rotation occurs at the edge where r = a.
Hence,
−Pa3
−3 3 1 − 𝜇
𝜃max =
=
Pa
8D(1 + 𝜇)
2
Et 3
wmax =
2.0
Figure 7.4 Moment distribution for simply supported plate.
Hence, the deflection as expressed by Eq. (2) becomes
(
)
Pr4
r2 3 + 𝜇 Pa2
w=
−
64D
4 1 + 𝜇 8D
)
(
Pa4 6 + 2𝜇
−1
+
64D 1 + 𝜇
or
(
)
5+𝜇 2
P
•a − r 2
w=
(3)
(a2 − r2 )
64D
1+𝜇
and
Mt
0.0
w=
At r = a, the deflection is zero, and Eq. (2) gives
)
(
Pa4 6 + 2𝜇
C3 =
−1
64D 1 + 𝜇
3.0
Pa3
.
Et 3
The moment expression is obtained by substituting
Eq. (3) into Eqs. (7.3a), (7.3b), and (7.4). Hence,
𝜃max = −1.05
P
(5)
(3 + 𝜇)(a2 − r2 )
16
P 2
(6)
[a (3 + 𝜇) − r2 (1 + 3𝜇)]
Mt =
16
A plot of Eqs. (5) and (6) for 𝜇 = 0.3 is shown in
Figure 7.4. The plot indicates that the maximum moment
occurs in the center and is given by
Mr =
Mmax =
3.3Pa2
16
or
𝜎max =
6M
1.24Pa2
=
.
2
t
t2
Example 7.2
Derive the moment expression for a uniformly loaded circular plate of radius a that is fixed at the edge. For 𝜇 = 0.3,
plot the moment diagram and determine the maximum
deflection and stress values.
Solution:
From Example 7.1,
w=
C r2
Pr4
+ 1 + C2 ln r + C3
64D
4
(1)
7.2 Circular Plates
and
and
Cr C
Pr
+ 1 + 2.
(2)
16D
2
r
At the center of the plate, r = 0 and the slope is zero due
to symmetry. Hence, from Eq. (2), C 2 = 0.
At r = a, the slope is zero and from Eq. (2),
𝜃=
3
−Pa2
.
8D
Also at r = a, the deflection is zero and from Eq. (1),
P 2
(5)
[a (1 + 𝜇) − r2 (1 + 3𝜇)].
16
A plot of Eqs. (4) and (5) for 𝜇 = 0.3 is shown in
Figure 7.5. The plot indicates that the maximum moment
occurs at the edge and is given by
Mt =
Mmax =
C1 =
and
6M
−0.75Pa2
=
.
2
t
t2
𝜎max =
Pa4
C3 =
.
64D
Hence, Eq. (1) becomes
Pr4
Pa2 r2
Pa4
−
+
64D
32D
64D
P
2
2 2
(3)
(a − r ) .
=
64D
The maximum value of deflection occurs at r = 0:
Pa4
Pa4 12(1 − 𝜇2 )
w=
,
=
64D
64
Et 3
and for 𝜇 = 0.3,
( 4)
Pa
wmax = 0.171
.
Et 3
The moment expression is obtained by substituting
Eq. (3) into Eqs. (7.3a), (7.3b), and (7.4). Hence,
P 2
Mr =
(4)
[a (1 + 𝜇) − r2 (3 + 𝜇)]
16
−Pa2
8
Problems
w=
4.0
3.0
7.1
Determine the maximum deflection, slope, and
bending moment for a simply supported plate
subjected to edge moment M0 .
Answer:
a2 M 0
at center
2D(1 + 𝜇)
aM0
at edge
max 𝜃 =
D(1 + 𝜇)
max Mt = M0 throughout plate
max w =
max Mr = M0 throughout plate
7.2
A circular plate is fixed at the edge and is at an ambient temperature of 70 ∘ F. What is the maximum
stress if the top surface is heated to a temperature
of 170 ∘ F and the bottom surface is cooled to a
temperature of −30 ∘ F? Let 𝛼 = 9 × 10−6 in./in.-∘ F,
t = 0.5 in., a = 60 in., E = 30 × 106 psi, 𝜇 = 0.3.
Answer:
𝛼 ΔTE
1−𝜇
= 38,600 psi
𝜎=
2.0
16M
1.0
a2 P
Mt
Mr
P
0.0
–1.0
–2.0
0
b
a
0.2
0.4
r/a
0.6
Figure 7.5 Moment distribution for fixed plate.
0.8
1.0
Figure 7.6 Uniformly loaded simply supported plate with a
central hole.
105
106
7 Stress in Flat Plates
7.3
Determine the maximum moment in the circular plate shown in Figure 7.6 if a = 4 in., b = 2 in.,
𝜇 = 0.3, and P = 100 psi.
Summation of forces in the z direction gives
(
)
𝜕Qx
dx dy
q(x, y)dx dy − Qx dy + Qx +
𝜕x
(
)
𝜕Qy
−Qy dx + Qy +
dy dx = 0.
𝜕y
Answer:
Mt = 384.6 in.-lb./in.
This equation can be reduced to
𝜕Qx 𝜕Qy
q(x, y) +
+
= 0.
𝜕x
𝜕y
7.3 Rectangular Plates
Summing moments around the x axis and deleting all
quantities of higher order gives
𝜕My 𝜕Mxy
(7.12)
−
− Qy = 0
𝜕y
𝜕x
or
𝜕Qy
𝜕 2 My 𝜕 2 Mxy
−
=
.
(7.13)
𝜕y
𝜕y2
𝜕x𝜕y
In developing the differential equation for circular plates,
the shearing stress was ignored because the load was
symmetric with respect to 𝜃. In rectangular plates under
uniform loads, the shearing stress interacts with the
normal stresses in the x and y directions and thus cannot
be ignored. This results in a more complicated differential equation than that for circular plates. In addition,
the solution of the differential equation of rectangular
plates is more elaborate and involves the use of Fourier
series. Because of this, only the case of a simply supported rectangular plate loaded throughout its surface is
discussed here. Nonsymmetric loadings and boundary
conditions other than simply supported result in quite
complicated solutions that are beyond the scope of this
book. The examples given in this section are intended
to give the reader an idea of the general behavior of
rectangular plates and the difference between them and
circular plates.
Consider an infinitesimal section of a rectangular plate.
The forces acting on it will be as shown in Figure 7.7.
Similarly, summing moments around the y axis and
deleting all quantities of higher order gives
𝜕Mx 𝜕Myx
(7.14)
+
− Qx = 0
𝜕x
𝜕y
or
𝜕Qx
𝜕 2 Mx 𝜕 2 Myx
+
=
.
(7.15)
𝜕x
𝜕x2
𝜕y𝜕x
Substituting Eqs. (7.13) and (7.15) into Eq. (7.11) and
using Mxy = − Myx from Eq. (3.10) gives
q(x, y) +
Z
Qy
Mx
Mxy
dy
X
Mx +
q(x,y)
Mx
Mxy
Qx +
My +
Y
𝜕Mx
dx
𝜕x
Mxy +
𝜕My
dy
𝜕y
Qy +
𝜕 2 Mx 2𝜕 2 Mxy 𝜕 2 My
−
= 0.
+
𝜕x2
𝜕x𝜕y
𝜕y2
(7.16)
Figure 7.7 Forces in a unit rectangular plate.
dx
Qx
(7.11)
𝜕Qy
dy
𝜕y
Myx +
𝜕Qx
dx
𝜕x
𝜕Myx
dy
𝜕y
𝜕Mxy
dx
𝜕x
7.4 Circular Plates on Elastic Foundations
The differential equation relating deflections and
applied loads is obtained by substituting Eq. (3.11) into
Eq. (7.16):
2𝜕 4 w
𝜕 4 w q(x, y)
𝜕4w
+
+
=
,
(7.17)
𝜕x4
𝜕x2 𝜕y2
𝜕y4
D
which is the differential equation of the bending of a rectangular plate subjected to lateral loads.
For any given loading and boundary conditions, the
deflection w can be obtained from Eq. (7.17). The bending moments can then be determined from Eq. (3.11)
as
( 2
)
𝜕 w
𝜕2w
+𝜇 2
(7.18a)
Mx = −D
𝜕x2
𝜕y
( 2
)
𝜕 w
𝜕2w
+𝜇 2
My = −D
(7.18b)
2
𝜕y
𝜕x
𝜕2w
,
(7.18c)
Mxy = D
𝜕x𝜕y
and the shearing forces Qx and Qy are determined from
Eqs. (7.12), (7.14), and (3.11) as
)
(
𝜕 𝜕2w 𝜕2w
Qx = −D
+
(7.19a)
𝜕x 𝜕x2
𝜕y2
)
(
𝜕 𝜕2w 𝜕2w
+
Qy = −D
.
(7.19b)
𝜕y 𝜕x2
𝜕y2
Example 7.3
Determine the maximum moment in a simply supported
rectangular plate of length a and width b if the applied
load is expressed as
πy
πx
q = q0 sin
sin .
a
b
Let a = 60 in., b = 25 in., q0 = 3 psi, and 𝜇 = 0.3.
Solution:
Assume w to be of the form
πy
πx
w = C sin
sin .
a
b
This expression satisfies the boundary conditions
of w = 0 and M = 0 at all four edges. Substituting this
expression into Eq. (7.17) gives
q0
,
C=
Dπ4 (1∕a2 + 1∕b2 )2
and the deflection expression becomes
q0
πy
πx
w=
sin
sin .
Dπ4 (1∕a2 + 1∕b2 )2
a
b
Substituting this expression into Eq. (7.18) gives
(
q0
πy
𝜇)
πx
1
Mx = 2
sin
+
sin
2
2
2
2
2
π (1∕a + 1∕b ) a
b
a
b
(𝜇
)
q0
πy
πx
1
sin
+
My = 2
sin
π (1∕a2 + 1∕b2 )2 a2 b2
a
b
q0 (1 − 𝜇)
πy
πx
Mxy = 2
cos
cos .
π (1∕a2 + 1∕b2 )2 ab
a
b
The maximum values of Mx and My occur when x = a/2
and y = b/2. A comparison of the factors in parentheses
in the expressions for Mx and My indicates that My will
always give a larger value of Mx for the given values of a
and b. Accordingly, the maximum value of M is given by
(𝜇
)
q0
1
Mmax = 2
+
π (1∕a2 + 1∕b2 )2 a2 b2
(
)
0.3
3.0
1
Mmax =
+
π2 (1∕602 + 1∕252 )2 602 252
= 145.1 in.-lb∕in.
The maximum value of Mxy occurs when x = 0 and
y = 0. Hence, the maximum value of Mxy is given by.
3.0(0.7)
+ 1∕252 )2 (60)(25)
= 40.2 in.-lb∕in.
Mxy =
π2 (1∕602
7.4 Circular Plates on Elastic
Foundations
Many tubesheets of heat exchangers are designed as
plates on elastic foundations as discussed in Chapter 14.
The solution of the differential equation for a plate on
elastic foundation involves Bessel functions. The four
Bessel functions used in this section are
∞
∑
(−1)j x4j
Z1 (x) = ber(x) =
4j
2
j=0 2 [(2j)!]
(x∕2)4 (x∕2)8 (x∕2)12
+
−
+···
2!2
4!2
6!2
∞
∑
(−1)j x4j+2
Z2 (x) = −bei(x) =
4j+2
[(2j + 1)!]2
j=0 2
=1−
(x∕2)2 (x∕2)6 (x∕2)10
+
−
+···
1!2
3!2
5!2
[
]
Z (x) 2
𝛾x
2
V1 + ln Z2 (x)
Z3 (x) = − Kei(x) = 1
−
π
2
π
2
]
Z2 (x) 2 [
𝛾x
−2
V2 + ln Z1 (x) ,
Z4 (x) =
Ker(x) =
+
π
2
π
2
where
( )2 𝜙(3) ( )6 𝜙(5) ( )10
x
x
x
− 2
+ 2
−···
V1 =
2
3!
2
5!
2
𝜙(2) ( x )4 𝜙(4) ( x )8 𝜙(6) ( x )12
− 2
+ 2
+···
V2 = 2
2!
2
4!
2
6!
2
1 1 1
1
𝜙(n) = 1 + + + + · · · +
2 3 4
n
𝛾 = 0.577216.
=−
The limits of the Z functions as x → 0 and as x → ∞ are
given in Table 7.1. The table also shows the limits for the
first derivatives of Z1 through Z4 .
107
108
7 Stress in Flat Plates
Table 7.1 Limits of Z functions.
Function
Limit as x → 0
Limit as x → ∞
Z 1 (x)
1.0
−x2
4
0.5
2
ln
π
−x3
16
−x
2
x
ln
π
2
πx
𝜁cos 𝜅
Z 2 (x)
Z 3 (x)
Z 4 (x)
dZ1 (x)
dx
dZ2 (x)
dx
dZ3 (x)
dx
dZ4 (x)
dx
and (7.7b) as
{
[
(
)]}
q−p
d 1 d
dw
1 d
.
r
r
=
r dr
dr r dr
dr
D
The solution of this equation is expressed as
−𝜁 sin 𝜅
−𝜈cos 𝜓
𝜁
√ (cos 𝜅 − sin 𝜅)
2
−𝜁
√ (cos 𝜅 + sin 𝜅)
2
𝜈
√ (cos 𝜓 − sin 𝜓)
2
𝜈
√ (cos 𝜓 + sin 𝜓)
2√
2
−x
exp √
𝜈=
πx
2
𝛾x
2
x
1
exp √
𝜁= √
2πx
2
x
π
𝜅= √ −
8 8
1
x
𝜁=√
exp √
2πx
2
x
π
𝜅=√ −
8 8
w = C1 Z1 (𝛼r) + C2 Z2 (𝛼r) + C3 Z3 (𝛼r) + C4 Z4 (𝛼r),
𝜈 sin 𝜓
𝛾x
2
π
x
𝜓= √ +
8
2
√
−x
2
exp √
πx
2
x
π
𝜓=√ +
2 8
𝜈=
The relations between the various derivatives of the Z
functions are as follows:
d2 Z1 (x)
1 dZ1 (x)
= Z2 (x) −
dx2
x dx
d2 Z2 (x)
1 dZ2 (x)
= −Z1 (x) −
2
dx
x dx
d2 Z3 (x)
1 dZ3 (x)
= Z4 (x) −
dx2
x dx
2
d Z4 (x)
1 dZ4 (x)
= −Z3 (x) −
.
2
dx
x dx
The force exerted by an elastic foundation on a circular
plate due to deflection of the foundation is expressed as
p = Ko w,
where
where
𝛼=
K o = stiffness of foundation
= (modulus of elasticity of foundation)(area of
foundation)/depth of foundation
The differential equation for a circular plate on an elastic foundation can be obtained by modifying Eqs. (7.7a)
√
4
Ko ∕D
Z1 − Z4 = Bessel functions
C 1 − C 4 = constants of integration.
Example 7.4
Determine the maximum deflection of a circular plate on
an elastic 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, Z1
and Z 2 also approach infinity. Therefore, C 1 and C 2 must
be set to zero. Thus,
w = C3 Z3 (𝛼r) + C4 Z4 (𝛼r)
and
dw
= C3 𝛼Z3′ (𝛼r) + C4 𝛼Z4′ (𝛼r).
dr
As r approaches zero, 𝜃 must be zero due to symmetry. But from Table 7.1, Z4′ approaches infinity as r
approaches zero. Hence, C 4 must be set to zero.
Thus,
𝜃=
w = C3 Z3 (𝛼r)
dw
= C3 𝛼Z3′ (𝛼r)
dr
[
]
1 ′
d2 w
2
Z
=
C
𝛼
(𝛼r)
−
(𝛼r)
Z
3
4
dr2
𝛼r 3
[
]
1 ′′
d3 w
2
′
𝛼Z
=
C
𝛼
(𝛼r)
−
(𝛼r)
.
Z
3
4
dr3
r 3
Substituting these derivatives into Eq. (7.10) and equating this to F gives
p = foundation load
w = deflection of foundation
(7.20)
C3 =
F
4𝛼 2 D
and
w=
F
Z (𝛼r)
4𝛼 2 D 3
and
wmax =
F
.
8𝛼 2 D
Further Reading
Nomenclature
D=
Et 3
12(1 − 𝜇2 )
E = modulus of elasticity
K o = stiffness of foundation
= (modulus of elasticity of foundation)(area of
foundation)/(depth of foundation)
Mr = radial moment in circular plates
Mt = tangential moment in circular plates
Mx = moment in x direction of rectangular plates
My = moment in y direction of rectangular plates
P = applied pressure
Q = shearing force
q = applied load
r = radius
T = temperature
t = thickness
w = deflection
Z 1 to Z4 = Bessel functions
√
𝛼 = 4 Ko ∕D
𝜇 = Poisson’s ratio
Mxy = shearing moment
Reference
1 Tiomoshenko, S. and Woinowsky-Krieger, S. (1959).
Theory of Plates and Shells. New York: McGraw-Hill.
Further Reading
Birman, V. (2011). Plate Structures. New York: Springer.
Hetenyi, M. (1964). Beams on Elastic Foundation. Ann
Arbor, Michigan: University of Michigan Press.
Jawad, M.H. (2018). Stress in ASME Pressure Vessels,
Boilers, and Nuclear Components. New York: Wiley.
McFarland, D.E., Smith, B.L., and Bernhart, W.D. (1972).
Analysis of Plates. New York: Spartan Books.
Szilard, R. (1974). Theory and Analysis of Plates.
Englewood Cliffs, N.J: Prentice-Hall.
109
111
Part III
Design of Components
An aluminum tower used by a fertilizer manufacturer. Source: Courtesy of the Nooter Corporation, St. Louis, MO.
114
8
Design of Cylindrical Shells
Cylindrical vessels are very frequently used in the petrochemical industry. They are easy to fabricate and install
and economical to maintain. The required thickness
is generally controlled by internal pressure, although
in some instances, applied loads and external pressure
have control. Other factors such as thermal stress and
discontinuity forces may also influence the required
thickness.
8.1 ASME Design Equations
A simplified equation was developed by the American
Society of Mechanical Engineers (ASME) Code, VIII-1,
for determining the required thickness of a cylinder subjected to internal pressure. It is a simplification of Eq.
(5.9) and gives accurate results over a wide range of r2 /r1 .
This equation is expressed as
t=
PR
,
SE − 0.6P
(8.1)
In Section VIII, Division 2, of the ASME Code, the
equation for the required thickness is based on plastic
analysis and is given by
S=
P
ln(t∕R + 1)
or
t = R(eP∕S − 1).
(8.2)
The required thickness of the cylindrical shell is normally calculated by Eq. (8.2). However, Lame’s formulation given by Eq. (5.9) is used for detailed elastic stress
analysis.
Example 8.1
A pressure vessel with an inside diameter of 50.0 in. is
subjected to an internal pressure of 100 psi. Using an
allowable stress of 17,500 psi, find the required thickness
according to Section VIII, Division 1. Assume that all circumferential and longitudinal seams are double-welded
butt joints and are spot radiographed.
where
t = required thickness
P = internal pressure
R = inside radius
S = allowable stress
E = joint efficiency factor
A comparison of Eqs. (8.1) and (5.9) is shown in
Figure 5.6. It indicates the wide range of applicability of
Eq. (8.1). The ASME Code, VIII-1, has, however, limited
the use of Eq. (8.1) to t less than or equal to R/2 and
pressure less or equal to 0.385SE. Various forms of Eq.
(8.1) are shown in Appendix I together with an alternate
equation that expresses the thickness in terms of R0
rather than R.
The factor E in Eq. (8.1) is an efficiency factor, and its
magnitude depends on the extent of radiography performed at the various seams of the cylinder. Appendix J
illustrates the effect of radiography of various seams on
the values of E as established by the ASME Code, VIII-1.
Solution:
From Appendix J, a value of E = 0.85 is obtained. From
Eq. (8.1),
t=
100 × 25
PR
=
SE − 0.6P
17,500 × .85 − 0.6 × 100
t = 0.17 in.
Example 8.2
A seamless cylindrical shell with an outside diameter of
30.0 in. is butt-welded to seamless ellipsoidal heads. The
circumferential seams are not X-rayed. Find the required
shell thickness if the allowable stress is 15,000 psi and the
internal design pressure is 250 psi. Use Section VIII, Division 1 rules.
Solution:
From Appendix J, with a value of E = 1.0 (seamless shell),
the allowable circumferential stress must be reduced to
85%, since the circumferential seams are not X-rayed.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
8.2 Evaluation of Discontinuity Stresses
From Appendix I, the required thickness equation in
terms of outside radius is given by
t=
PR
SE + 0.4P
or
250 × 15
t=
(15,000 × 0.85)(1.0) + (0.4 × 250)
t = 0.29 in.
Problems
8.1
An ASME pressure vessel with an inside diameter
of 4 ft has a seamless shell. The head-to-shell seams
are spot radiographed. Find the required thickness
if the allowable stress is 20,000 psi and the design
pressure is 2900 psi.
Answer:
t = 3.81 in.
8.2
Structural
discontinuity
A source of stress or strain
intensification which affects
a relatively large portion of
a structure and has a
significant effect on the
overall stress or strain
pattern. Examples of gross
structural discontinuities are
head-to-shell and flange-toshell junctions, nozzles, and
juctions between shells of
different diameters or
thickness
8.2 Evaluation of Discontinuity Stresses
In Chapter 5, we showed how stresses are evaluated at
different locations due to thermal and mechanical conditions. The magnitude of these stresses must be kept below
a given allowable stress. This allowable stress is established in the ASME Code, VIII-2. The designer has to
establish first whether the stress is at a local or a gross
structural discontinuity, as defined in Figure 8.1. Next,
the stress is categorized as a primary, secondary, or a peak
stress as shown in Figures 8.2 and 8.3. The two categories
of thermal stress are described in Figure 8.4. Once the
stress categories are established, the stresses at a vessel’s
different locations can be classified as in Table 8.1.
Section VIII-2 has specific procedure for setting limits
on various combinations of the stresses Pm , PL , Pb ,
and Q. The maximum values of these combinations are
called equivalent stress values as shown in Table 8.2.
A source of stress or strain
intensification which affects
a relatively small volume of
material and does not have a
significant effect on the
overall stress or strain pattern
or on the structure as a
whole
Figure 8.1 Structural discontinuities. Source: Courtesy of the
American Society of Mechanical Engineers.
What is the maximum allowable pressure that can
be applied to a cylinder shell with an outside diameter of 6 ft, thickness of 1.25 in, and an allowable
stress of 17,500 psi? Let E = 0.85.
Answer:
P = 524 psi.
Local structural
discontinuity
Gross structural
discontinuity
Stress
categories
Primary
Secondary
Peak
See Figure 8.3
A stress developed by the constraint
of a structure. Secondary stress is
self-limiting. Local yielding and
minor distortions can satisfy the
conditions which cause the stress to
occcur and failure from one application
of the stress is not to be expected.
Examples of secondary stress are general
thermal stress and bending stress at a
gross structural discontinuity
Peak stress does not cause any noticeable distortion
and is objectionable only as a possible source of a
fatigue crack or a brittle fracture. Examples of peak
stress are: (i) thermal stress in austenitic steel cladding
of carbon steel vessels, (ii) thermal stress in the wall of
a vessel caused by a rapid change in temperature of the
contained fluid, (iii) the stress at a local structural discontinuity
Figure 8.2 Stress categories. Source: Courtesy of the American
Society of Mechanical Engineers.
115
116
8 Design of Cylindrical Shells
Figure 8.3 Primary stress categories. Source:
Courtesy of the American Society of Mechanical
Engineers.
Primary stress
A stress developed by the imposed loading
which is necessary to satisfy the laws of
equilibrium. The basic characteristic of
a primary stress is that it is not self-limiting.
Primary stress which considerably
exceed the yield strength will result in failure
or at least in gross distortion. A thermal
stress is not classified as a primary stress
Membrane
Bending
An example is the bending
stress in the central
portion of a flat head due
to pressure
General
Local
A general primary membrane
stress is one which is so
distributed in the structure
that no re-distribution of load
occurs as a result of yielding.
An example is the stress in a
cylinder due to internal pressure
An example of a local primary
membrane stress is the membrane
stress in a shell produced by
external load, and moment at a
permanent support or at a nozzle
connection
Figure 8.4 Thermal stress categories. Source:
Courtesy of the American Society of Mechanical
Engineers.
Thermal stress
A self-balancing stress produced by a non-uniform
distribution of temperature or by differing thermal
coefficients of expansion. Thermal stress is developed
in a solid body whenever a volume of material is
prevented from assuming the size and shape that it
normally should under a change in temperature
General
General thermal stress is classified
as secondary stress. Example of
general thermal stress are:
(1) stress produced by an axial
temperature distribution in
a cylindrical shell
(2) stress produced by the
temperature difference
between a nozzle and the
shell to which it is attached
(3) the equivalent linear stress
produced by the radial temperature
distribution in a cylindrical shell
Local
Local thermal stress which
is associated with almost
complete suppression of the
differential expansion and
thus produces no significant
distortion. Examples of local
thermal stresses are:
(1) stress in a small hot spot
in a vessel wall
(2) the difference between the
actual stress and the
equivalent linear stress
(3) the thermal stress in a
cladding material
8.2 Evaluation of Discontinuity Stresses
Table 8.1 Classification of stresses.
Vessel component
Cylindrical
or spherical
shell
Any shell or
head
Location
Shell plate remote
from discontinuities
Perforated
head or
shell
Internal pressure
General membrane
Pm
Axial thermal gradient
Gradient through plate thickness
Membrane
Q
Q
Bending
Q
Membrane
Bending
PL
Q
Any section across
entire vessel
External load or moment,
or internal pressure
General membrane averaged across full
section. Stress component
perpendicular to cross section
Pm
External load or moment
Bending across full section. Stress
component perpendicular to cross
section
Pm
External load moment, or
internal pressure
Local membrane
PL
Bending
Q
Peak (fillet or corner)
F
Temp. diff. between shell
and head
Membrane
Q
Bending
Q
Crown
Internal pressure
Membrane
Pm
Knuckle or junction
to shell
Internal pressure
Bending
Membrane
Pb
PL
Bending
Q
Center region
Internal pressure
Membrane
Pm
Bending
Pb
Junction to shell
Internal pressure
Membrane
PL
Bending
Q
Membrane (average through cross
section)
Bending (average through width of
ligament, but gradient through plate)
Pm
Typical ligament in
a uniform pattern
Isolated or atypical
ligament
Nozzle
Classification
Internal pressure
Any location
Flat head
Type of stress
Junction with head
or flange
Near nozzle or
other opening
Dished head
or conical
head
Origin of stress
Pressure
Pressure
Pb
Peak
F
Membrane
Q
Bending
F
Peak
F
General membrane (average across full
section). Stress component
perpendicular to section
Pm
Cross section
perpendicular to
nozzle axis
Internal pressure or
external load or moment
External load or moment
Bending across nozzle section
Pm
Nozzle wall
Internal pressure
General membrane
Pm
Local membrane
PL
Bending
Q
Peak
F
117
118
8 Design of Cylindrical Shells
Table 8.1 (Continued)
Vessel component
Location
Origin of stress
Type of stress
Differential expansion
Classification
Membrane
Q
Bending
Q
Peak
F
Membrane
Bending
F
F
Cladding
Any
Differential expansion
Any
Any
Radial temperature
distribution
Equivalent linear stress
Q
Nonlinear portion of stress distribution
F
Any
Stress concentration (notch effect)
F
Any
Any
Source: Reprinted from Ref. [1].
Table 8.2 Stress categories and limits of equivalent stress.
Stress
Category
Description
(For
examples,
see Table
8.1)
Symbol
Secondary
Membrane
plus Bending
Primary
Local
Membrane
General
Membrane
Average primary
stress across
solid section.
Excludes discontinuities and
concentrations.
Produced only
by mechanical
loads
Average stress
across any
solid section.
Considers discontinuities but
not concentrations.
Produced only
by mechanical
loads.
Component of
primary stress
proportional to
distance from
centroid of solid
section. Excludes
discontinuities and
concentrations.
Produced only
by mechanical
loads.
PL
Pb
Pm
Pm
Bending
Peak
Self-equilibrating
1. Increment
added to
stress necessary
primary or
to satisfy continsecondary
uity of structure.
stress by a
Occurs at strucconcentration
tural discontinuities.
(notch).
Can be caused by
mechanical load or 2. Certain thermal
by differential
stresses which
thermal expansion.
may cause
Excludes local
fatigue but not
stress
distortion of
concentrations.
vessel shape.
Q
F
S
PL + Pb + Q
PL
SPS
SPL
Use design loads
Use operating loads
PL + Pb
SPL = 1.5S; SPS = 3S; Sa = alternating stress used in fatigue analysis.
Source: ASME.
SPL
PL + Pb + Q + F
Sa
8.2 Evaluation of Discontinuity Stresses
The procedure for calculating equivalent stress values
from stress values is given by the following steps.
1) At a given time during the operating cycle, separate
the calculated stress at a given point in the vessel
into Pm , (PL + Pb ), and (PL + Pb + Q). The quantity
PL is either general or local primary membrane
stress.
2) Each of these bracketed quantities is actually six
quantities acting as normal and shear stresses on an
infinitesimal cube at the point selected.
3) Repeat steps 1 and 2 at a different time frame in the
cycle.
4) Find the algebraic sum of the stresses from steps 1
through 3. This sum is the stress range of the quantities Pm , (PL + Pb ), and (PL + Pb + Q).
5) From step 4, determine the three principal stresses
S1 , S2 , and S3 for each of the stress categories Pm ,
(PL + Pb ), and (PL + Pb + Q).
6) From step 5, determine the equivalent stress, Se ,
for each of the stress categories Pm , (PL + Pb ), and
(PL + Pb + Q) in accordance with the strain energy
(von Mises) equation
Se = 0.707[(S1 − S2 )2 + (S2 − S3 )2
+[(S3 − S1 )2 ]1∕2 .
(8.3)
7) The equivalent stress values in step 6 are analyzed in
accordance with Table 8.2.
Example 8.3
Calculate the stress at points A, B, and C of the vessel shown in Figure 8.5. Let R = 60 in., t s = 2.0625 in.,
t h = 1.0313 in., S = 15,000 psi, 𝜇 = 0.3, E0 = 30 × 106 psi.
The pressure cycle ranges from 0 to 500 psi. Use thin-shell
theory since R/t > 10.
Solution:
From Figures 8.6 and 8.7 and Eq. (5.1), the membrane and
secondary stresses, in psi, at points A and B are shown as
follows:
Stress location
Point A
Point B
Circumferential membrane stress (Pmc )
14,550a)
14,550
Longitudinal membrane stress (PmL )
7 275b)
7 275
Radial membrane stress (Pmr )
−250c)
−250
Circumferential secondary stress (Qc )
0
0
Longitudinal secondary stress (QL )
0
0
Radial secondary stress (Qr )
−250d)
+250
Circumferential membrane + secondary
stress (Pmc + Qc )
14,550
14,550
Longitudinal membrane + secondary
stress (PmL + QL )
7 275
7 275
Radial membrane + secondary stress
(Pmr + Qr )
−500
0
a) S = PR/t.
b) S = Pr/2t.
c) The radial stress distribution through the thickness is equal to −P at
the inside surface and zero at the outside surface. Hence, the membrane stress is the average of these two values.
d) The secondary stress is the difference between the total stress and
the average stress at the location.
The equivalent stresses, in psi, at points A and B are
obtained from Eq. (8.3) and are shown as follows:
Stress location
Point A
Point B
Equivalent membrane stress, Pm
12,820
12,820
Equivalent membrane plus secondary
stress, (Pm + Q)
13,035
12,600
Figure 8.5 Pressure vessel.
ts = 2 1 in.
16
A
R = 60 in.
B
60 in. rad.
C
th = 2 1 in.
32
Gross structural
discontinuity
Local structural
discontinuity
119
120
8 Design of Cylindrical Shells
σθ
σr
E0 ts3
= 0.8035E0 .
12(1 − 𝜇2 )
The first compatibility equation is given by
Ds =
deflection of shell = deflection of head
σ
σ
or
wp + wNx e + wMs − wQ |shell = wp + wQ
+ wMh |head ,
where for the shell,
PR2
741 820
wp =
(1 − 𝜇∕2) =
,
E0 ts
E0
σθ
Figure 8.6 Stress orientation at outer surface.
wNx e =
Nx e
366 454
=
,
2𝛽 2 Ds
E0
wMs =
47.3822Ms
,
E0
σθ
wQ =
σ
σ
and for the head,
PR2
621 430
wp =
(1 − 𝜇) =
,
2E0 th
E0
wQ =
σr
σθ
Q
413.4569Q
=
,
3
2𝛽 Ds
E0
wMh =
2rQh 𝜆 1155.5665Q
=
,
E0 th
E0
188.022Ms 1,454,162
2Mh 𝜆2
=
+
.
E0 th
E0
E0
Figure 8.7 Stress orientation at inner surface.
The maximum equivalent membrane stresses, Pm , are
within the limit S as shown in Table 8.2.
The maximum equivalent primary plus secondary
stresses (Pm + Q) are within the limit 3S as shown in
Table 8.2.
Point C. The discontinuity forces at point C are shown
in Figure 8.8. From this figure,
( )
PR ts
500 × 60 × 2.0625
=
Nx e =
2
4
2×4
= 7734 in.-lb∕in.
Also,
Mh = Ms + Nx e = Ms + 7734,
2.0625
rs = 60 +
= 61.0313 in.,
2
1.0313
rh = 60 +
= 60.5157 in.,
2
√
3(1 − 𝜇2 )
= 0.1146,
rs2 ts2
√
( )2
r
4
𝜆 = 3(1 − 𝜇2 ) h
= 9.8465,
th
𝛽=
4
e
Nx
Q Nx-e M
s
Mh
Nx
Nx
Q
Figure 8.8 Discontinuity forces at the head-to-cylindrical-shell
junction.
(1)
8.3 ASME Procedure [2] for External Pressure Design in VIII-1
Substituting these values into Eq. (1) gives
Ms + 10.8719 Q = −6878.
512.53 262.55
(2)
512.53
The second compatibility equation is obtained from
15 000
rotation in shell = rotation in head
15 000
or
𝜃Nx e + 𝜃Ms − 𝜃Q |shell = −𝜃Mh − 𝜃Q |head ,
(3)
262.55
where for the shell,
Ne
83,991
𝜃Nx e = x =
,
𝛽Ds
E0
10.86Ms
,
𝜃Ms =
E0
Q
47.3822Q
𝜃Q = 2 =
,
2𝛽 D
E0
Figure 8.9 Shear and moment values at the
head-to-cylindrical-shell junction.
From Tables 8.1 and 8.2, the maximum allowable local
membrane stress is equal to
1.5 S = 22,500 > 20,950 psi acceptable
and for the head,
61.1861Ms 473,213
4𝜆3 Mh
=
+
,
𝜃Mh =
E0 th rh
E0
E0
2𝜆2 Q 188.022Q
𝜃Q =
=
.
E0 th
E0
2) Local membrane plus secondary stress (PL + Q):
𝜎𝜃 = 23,430 − 80 = 23,350 psi,
𝜎t = 7275 − 264 = 7011 psi,
𝜎r = −250 − 250 = −500 psi,
maximum stress difference = 21,120 psi.
Substituting these values into Eq. (3) gives
Ms + 1.9521Q = −7733.99.
(4)
Solving Eqs. (2) and (4) gives
Q = 95.97 lb∕in.,
From Tables 8.1 and 8.2, the maximum allowable local
membrane plus secondary stress is equal to
3 S = 45,000 psi > 21,120 acceptable
Ms = −7921.3 in.-lb∕in.,
Mh = −187.3 in.-lb∕in.,
693 266
total w =
.
E
The actual forces are shown in Figure 8.9. We have
Etw
hoop stress at point C = 0 s = 23,430 psi,
rs
PR
axial stress at point C =
= 7275 psi,
2ts
6M
axial bending stress at point C = 2
ts
= 264 psi,
circumferential bending stress at point C
= (0.3)(264) = 80 psi.
These stresses are divided into two categories in accordance with Table 8.2:
1) Local membrane stress (PL ):
𝜎𝜃 = 23,430 psi,
𝜎t = 7275 psi,
𝜎r = −250 psi,
maximum stress difference = 20,950 psi.
8.3 ASME Procedure [2] for External
Pressure Design in VIII-1
A comparison of Figures 5.17 and 5.18 indicates that the
buckling due to pressure applied to sides and ends is
more critical than the pressure applied to the ends only.
Accordingly, the ASME Code, VIII-1, uses Eq. (5.27b) as
the basis for the design. This equation is modified to take
nonelastic buckling into consideration and expresses the
basic relationship as
( )3
Pr r
t
.
= KE0
𝜎𝜃 =
t
t
Do
Defining A = 𝜖 cr , and using a knock-down factor of 1.0,
( )2
𝜎
K t
A= 𝜃 =
.
(8.4)
E0
2 Do
Equation (8.4) is plotted in Figure 8.10. Hence, for any
given value of L, Do , and t, a value of A can be determined
from Figure 8.10.
The allowable compressive stress in the elastic region
can be determined from the equation
𝜎=
AE0
,
DF
121
Do / r = 4
Do / r = 5
Do / r = 6
Do / r = 8
Do / r = 10
Do / r = 15
Do / r = 20
Do / r = 25
Do / r = 30
Do / r = 40
Do / r = 60
Do / r = 50
Do / r = 80
Do / r = 100
Do / r = 120
30.0
Do / r = 150
40.0
35.0
Do / r = 200
50.0
Do / r = 250
8 Design of Cylindrical Shells
20.0
18.0
16.0
14.0
12.0
10.0
9.0
8.0
7.0
6.0
5.0
4.0
3.5
3.0
5
=6
/t
Do
2.0
1.8
1.6
1.4
/t =
Do
2.5
0
0
30
00
00
00
00
40
=8
1.0
/t =
= 6 /t = 5 /t =
t
t
=
/
o
/
D
/t
Do
Do
Do
Do
Do
Length ÷ outside diameter = L/Do
25.0
8
10
=
/t
t=
D o D o/
1.2
1.0
.90
.80
.70
0
0
5
0
0
5
0
=2 t=2 =3 t=4 =5 =6
=1
80
/t
/t
/
/
/t
=
/t /t
Do
Do
Do Do
Do
/t
0
Do Do
D o 10
=
/t
5
D o 12
=
/ t 50
Do =1
/t
D o 200
=
/t
D o 250
=
/ t 00
Do 3
=
/t
0
Do
40
=
/t
D o 500
=
0
/t
D o 60
=
/t
Do
00
=8 0
/t
0
D o = 1,
/t
Do
.60
.50
.40
.35
Length ÷ outside diameter = L/Do
122
.30
.25
.20
.10
.16
.14
.12
.10
.090
.080
.070
.060
.050
0.00001
2
3
4 5 6 7 89
0.0001
2
3
4 5 6 7 89
0.001
Factor A
2
3
4 5 6 7 89
0.01
2
3 4 5 6 78 9
0.1
Figure 8.10 Geometric chart for cylindrical vessels under external or compressive loadings – for all materials. Source: Reprinted from
Ref. [3].
8.3 ASME Procedure [2] for External Pressure Design in VIII-1
where DF is the design factor and E0 is the modulus of
elasticity. In terms of pressure,
2AE0
P=
.
(Do ∕t)(DF)
In the inelastic region, ASME uses a chart expressing
axial compression stress versus factor A. This relationship is obtained by calculating the tangent modulus Et
at a point on a stress–strain curve and then calculating
strain factor A at a specific stress value using the tangent
modulus. The chart is plotted on a log–log paper, and the
B stress is traditionally taken as one-half of the compressive stress. One such chart for carbon steel is shown in
Figure 8.1. The allowable compressive stress is given by
𝜎=
(factor B from chart)
(factor of safety of chart)
ASME uses a design factor (DF) of 3.0 for buckling of
cylindrical shells subjected to lateral and end external
pressures. Hence, for the elastic region (Do /t ≥ 10),
2AE0
P=
,
(8.6)
3(Do ∕t)
and for the elastic or the inelastic region (Do /t ≥ 10),
4B
,
P=
3(Do ∕t)
where
A = factor determined from Figure 8.10
B = stress determined from Figure 8.11
Do = outside diameter of cylinder
E0 = modulus of elasticity
(8.5)
DF
or
P = allowable external pressure
t
2B
.
𝜎=
DF
For the allowable pressure, if needed, we have
4tB
2t𝜎
=
P=
Do
(DF)Do
or
B
.
P=4
(Do ∕t)DF
= thickness of cylinder
For Do /t values less than 10, ASME uses a variable DF
that ranges from 3.0 for values of Do /t = 10 to a factor
of 2.0 for values of Do /t = 4.0. This reduction occurs
because for very thick cylinders, buckling ceases to be
a consideration and the allowable values in tension and
compression are about the same. Hence, for Do /t < 10,
the allowable value of P is the lower of the quantities P1
and P2 given as follows.
25 000
up to 300 F
700 F
20 000
18 000
16 000
800 F
14 000
500 F
900 F
12 000
9 000
8 000
7 000
Factor B
10 000
6 000
5 000
E = 29.0 × 106
E = 27.0 × 106
E = 24.5 × 106
E = 22.8 × 106
E = 20.8 × 106
2
0.00001
3
4
5 6 78 9
0.0001
4 000
3 500
3 000
2
3
4
5 6 78 9
0.001
2
3
4
5 6 78 9
0.01
2
3
4
2 500
5 6 789
0.1
Factor A
Figure 8.11 Chart for determining shell thickness of cylindrical and spherical vessels under external pressure when constructed of carbon
or low-alloy steels (specified yield strength 30 000–38 000 psi inclusive) and types 405 and 410 stainless steels. Source: Reprinted from
Ref. [3].
123
124
8 Design of Cylindrical Shells
For Do /t < 10,
(
)
2.167
P1 =
− 0.0833 B,
Do ∕t
(
)
2𝜎
1
P2 =
1−
,
Do ∕t
Do ∕t
Then,
Do
= 384,
t
(8.7)
where 𝜎 = 2 times the allowable stress in tension or 0.9
times the yield stress of the material, whichever is less.
Note also that for values of Do /t < 4, the following
equation can be used:
1.1
.
(Do ∕t)2
A=
(8.8)
The ASME procedure for the design of cylindrical
shells under external pressure is complicated because
of the various parameters that must be considered. A
summary of the procedure is shown in Figure 8.12 as an
aid to the designer.
Example 8.4
The length of a cylindrical shell is 15 ft, outside diameter
is 10 ft, and is constructed of carbon steel with minimum
yield strength of 36,000 psi. The shell is subjected to an
external pressure of 10 psi. Find (a) the required thickness
using ASME DF and (b) the required thickness using a DF
of 2.0.
t=
2(0.00014) × 29 × 106 3.0
×
3(384)
2.0
=
= 10.6 psi acceptable
Returning to Figure 8.11 with A = 0.00014, it is seen
that the value from the chart indicates an elastic behavior.
Use t =
5
in.
16
Example 8.5
A cylindrical shell with length 18 ft and outside diameter
6 ft is constructed of carbon steel with a yield stress of
38,000 psi. Determine the thickness needed to resist an
external pressure of 300 psi.
Solution:
Let t = 1.25 in.
L
= 1.25.
Do
(2)(0.00018)(29 × 106 )
= 10.9 psi acceptable
(3)(320)
A check is needed to ascertain that the buckling is
in the elastic rather than the inelastic region. From
Figure 8.11 with A = 0.0018, a value of B = 2600 psi is
obtained in the elastic region of the curve. Hence, the
aforementioned solution of 10.9 psi is adequate.
3
in.
8
(b) For a DF 2.0, assume
t = 0.3125 in.
2 × 0.00095 × 29 × 106
= 320 psi.
3 × 57.6
Now check for the inelastic region. From Figure 8.11,
factor B = 12,000 psi in the plastic region. Hence, the first
of Eq. (8.6) cannot be used. From the second of Eq. (8.6),
Pa =
From Figure 8.10, factor A = 0.00018. From Figure 8.11,
the modulus of elasticity at room temperature is
29,000,000 psi. Hence, from Eq. (8.6),
Use t =
Do
= 57.6.
t
From Figure 8.10, the factor A = 0.00095. From Eq. (8.6)
with E0 = 29 × 106 psi,
3
in.
8
Then,
Do
= 320,
t
P=
From Figure 8.10, the factor A = 0.00014, and from Eq.
(8.6),
2AE0 ASME design factor
•
P=
3(Do ∕t) specified design factor
L
= 3.0,
Do
Solution:
(a) Assume
L
= 1.25.
Do
4 × 12,000
= 278 psi inadequate.
3 × 57.6
Try t = 1.375 in. Then,
P=
Do
= 52.4 and A = 0.0011.
t
From Figure 8.11,
B = 12,400 psi,
4 × 12,400
P=
= 316 psi acceptable
3 × 52.4
Use t = 1.375 in.
Figure 8.11 and Eqs. (8.6) and (8.7) are applicable only
when Do /t is less than or equal to 1000. For ratios above
1000, many engineers use the following equation, which
8.3 ASME Procedure [2] for External Pressure Design in VIII-1
Calculated Do / t
Yes
Calculate
A = 1.1
Do /t 2
Is Do /t less than 4?
No
Assume L & calculated L/Do
Is A greater
than 0.1?
No
Yes
Is L/Do greater than 50?
No
Yes
Is L/Do less than 0.05?
Yes
Use A
= 0.1
Use this
value of A
Use L/Do = 50
Use L/Do = 0.05
No
Enter Figure 8.10
with Do / t and L/D
Calculated A and use
Enter materials chart with A
Yes
Is A off scale to right?
Yes
Calculate
2AB
P2 =
3 Do /t
Extend appropriate
temperature line
horizontally and
read B
No
Is A off scale to left?
No
Read B at appropriate
temperature line
Yes
Is Do /t ≥ 10?
No
Calculate
4B
P2 =
3 Do /t
Determine 2St where
St is allowable tensile
stress from ASME II-D
Determine 0.9Sy where Sy is
yeild strength from ASME II-D
Is 2St < 0.9Sy?
No
Set S = 0.9Sy
Yes
Set S = 2St
Calculate
Pa2 =
2s
Do /t
1–
1
Do /t
Calculate
Pa1 =
Yes
Pa = Pa1
2.167
– 0.0833 B
Do/t
Is Pa1 < Po2?
No
Pa = P2
Figure 8.12 The ASME VIII-1 method for determining the maximum allowable external pressure on cylinders.
125
126
8 Design of Cylindrical Shells
was developed by the US Experimental Model Basin:
Pcr =
factor of 1.0. Determine the required stiffener spacing from Figures 8.10 and 8.11.
2.42E0
(t∕Do )2.5
.
(1 − 𝜇2 )3∕4 [L∕Do − 0.45(t∕Do )1∕2 ]
Answer:
L = 16.2 in.
A minimum DF of 3.0 is to be applied to this equation
in order to obtain the allowable external pressure.
8.6
Problems
8.3
The thickness of a 13 ft diameter reactor is 5.50 in.,
and its effective length is 18 ft. If the design temperature is 900 ∘ F, what is the maximum allowable
external pressure?
Answer:
T = 900 ∘ F.
8.4 Design of Stiffening Rings
Answer:
P = 300 psi.
8.4
In deriving Eq. (5.27) for the maximum strength of a
cylindrical shell under external pressure, it was assumed
that the ends of the shell were simply supported. For
this to be true, stiffening rings, flanges, and so on
(Figure 8.13) are needed as lines of supports. These
supports are assumed to carry all the load that the shell
carries due to external pressure. From Figures 8.13 and
8.14, the total force in the stiffener is
A vessel has a 15 ft outside diameter and effective
length of 6 ft. If it is subjected to 15 psi external pressure, what is the required thickness (to
the nearest 1/16 in.) if the design temperature is
300 ∘ F?
Answer:
7
in.
t=
16
8.5
h/3
A distillation tower is subjected to a vacuum
of 15 psi. If Do = 9 ft, t = 0.75 in., and stiffener
spacing = 8 ft, what is the maximum permissible
temperature?
PDo L = 2F
or
PDo L
.
2
The stress in the shaded area of Figure 8.14 is
PDo
F
𝜎=
=
.
L(t + As ∕L) 2(t + As ∕L)
A jacketed pressure vessel with an internal diameter
of 12 ft is subjected to an internal pressure of 400 psi
and a jacket pressure of 200 psi. The shell thickness
is controlled by the internal pressure using an allowable tensile stress of 15,000 psi at 800 ∘ F with an E
F=
t
Moment axis of ring
h/3
L
L
L
L
L
L
L
L
L
L
Do
t
Do
h/3
h/3
h = depth of head
L
Do
L
t
L
h/3
h = depth
of head
L
Do
Figure 8.13 Diagrammatic representation of variables for design of cylindrical vessels subjected to external pressure. Source: Reprinted
from Ref. [2].
8.4 Design of Stiffening Rings
0.55 DOt
0.55 DOt
L
Stiffener
Not to overlap area
from adjacent rings
Figure 8.15 Effective length of T-stiffener. Source: Reprinted from
Ref. [2].
Figure 8.14 Effective length of the stiffener.
In this equation, it is assumed that the area As of stiffening ring is “smeared” over the total length L.
Using the terminology of Figure 8.11 and Eq. (8.5),
𝜎=
2B
DF
or
𝜎(DF)
,
2
PDo
DF
B=
.
2 2(t + As ∕L)
B=
3 PDo
.
4 t + As ∕L
(8.9)
The stiffening ring must also be checked against buckling. The classical expression for the buckling of a ring
due to external pressure is
F=
12 E0 I
D2o
or
𝜎=
12E0 I
F
.
= 2
area Do (t + As ∕L)L
The expression for strain is
12I
.
A= 2
Do (t + As ∕L)L
And the expression for I becomes
D2o L(t + As ∕L)A
.
(8.10)
12
This equation can be used in conjunction with Eq.
(8.9) and Figure 8.11. In doing so, a trial As is normally
selected, and B is calculated from Eq. (8.9). Using the
value of B, which already incorporates a factor of safety
I=
D2o L(t + As ∕L)A
.
(8.11a)
14
The required I s obtained from Eq. (8.11a) must be equal
to or lower than the available moment of inertia, I, of
the stiffening ring. This inertia is calculated without considering the contribution of the adjacent cylinder. If the
composite moment of inertia of the ring and the effective
cylinder is considered, then a penalty of 28% is applied to
Eq. (8.11a) and a new expression for the required moment
of inertia of the composite section, Is′ , given by
Is =
With a DF of 3.0, this expression becomes
B=
of 3.0, a value of A is obtained from Figure 8.11. With
this A, the required moment of inertia is calculated from
Eq. (8.10).
Because the stability of the stiffening ring is essential
in calculating the shell stability, a higher factor of safety
is used by ASME in the stiffening-ring calculations than
in the shell calculations. With a factor of safety of 3.5, Eq.
(8.10) for the required moment of inertia, I s , becomes
Is′ =
D2o L(t + As ∕L)
10.9
(8.11b)
must be used. The available I′ of the composite section
is calculated from the available ring and shell areas as
shown in Figure 8.15 for Is′ ≤ I ′ .
Example 8.6
A long cylindrical shell is constructed of carbon steel with
a yield stress of 38,000 psi and a radius of 36 in. If the stiffeners are spaced at a 4 ft interval, calculate the required
shell thickness and the size of stiffening rings for an external pressure of 10 psi at 100 ∘ F.
Solution:
Let
t = 0.1875 in.,
L
= 0.67,
Do
Do
= 384.
t
127
128
8 Design of Cylindrical Shells
Gap (not to exceed 8 times
the thickness of the shell plate)
See UG-29(b)
This section shall have moment of inertia
required for ring unless requirements of
UG-29 (b)(2) are met
Gap
E
Butt weld
Shell
A
Web of stiffener
Flange of
stiffener
Butt weld
Gap in ring
for orainage
F
This section shall have moment of
inertia required for ring
J
Strut
member
D
K
Section J–K
Length of any gap in unsupported shell not to
exceed length of arc shown in Figure. UG-29.2
B
Butt weld
in ring
Unstiffened cylinder
Type of construction
when gap is greater
than length of arc
shown in Figure UG-29.2
C
This section shall have moment of
inertia required for ring
At least 120°
Support
K
Figure 8.16 Various arrangements of stiffening rings for cylindrical vessels subjected to external pressure. Source: Reprinted from Ref. [2].
8.6 Out-of-Roundness of Cylindrical Shells Under External Pressure
From Figure 8.10,
shell. Referring to Figure 5.16, the maximum gap length
can be expressed as
A = 0.00028.
B = 4000 psi.
Hence,
P=
(
)
4 4000
= 10.4 psi.
3 384
Use a shell thickness of 3/16 in., and try a 3 × 3 ×
angle stiffening ring:
3
16
As = 1.09 in.2 and I = 0.9622 in.4 ,
(10)(72)
3
B=
= 2570 psi.
4 0.1875 + 1.09∕48
Example 8.7
What is the maximum gap allowed in a stiffening ring of
a shell with Do = 7.0 ft, L = 15.0 ft, and t = 1.0 in.?
From Figure 8.11, A = 0.00018, so
722 × 48(0.1875 + 1.09∕48)(0.00018)
14
= 0.67 in.4 acceptable
I=
Solution:
8.5 Allowable Gaps in Stiffening Rings
Gaps in stiffening rings are normally provided to allow
for drainage of vessel contents or permit piping and other
internals to extend through the ring. Examples of various
gap arrangements are shown in Figure 8.16. The maximum allowable gap can be calculated by treating the distance between points a and b in Figure 8.17 as a simply
supported column of length l. The maximum buckling
load that can be applied to this column is given by
π2 E0 I
.
(8.12a)
l2
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 strength
of the shell approaches that of a simply supported flat
plate. The minimum critical buckling strength of the
simply supported plate abcd (Figure 8.17) loaded in the
circumferential direction is given by
Fcr =
Fcr =
4π2 E0 I
4π2 E0 I
.
≈
l2 (1 − 𝜇2 )
l2
πDo
l
=
.
(8.13)
2
4N
ASME has developed curves that are based on Eq.
(8.13). These curves are shown in Figure 8.18. A comparison, however, between Eq. (8.13) and Figure 8.18
indicates some differences. These differences are due to
the fact that Figure 5.18, which is used with Eq. (8.13),
is plotted using, in Eq. (5.27a), the first two terms of
the expression for F, whereas Figure 8.18 uses all terms.
Therefore, the results shown in Figure 8.18 are more
accurate.
G=
From Figure 8.11,
(8.12b)
By comparing Eqs. (8.12a) and (8.12b), it can be concluded that in order for the column ab to be as strong
as the plate, its length must be about one-half that of
the plate. Therefore, the maximum gap length must be
one-half that of a buckling lobe length of a cylindrical
Do
L
= 84,
= 4.29
t
r
or from Figure 5.18, N = 4, and from Eq. (8.13),
G=
πDo
= 0.20 Do
(4)(4)
or
G = 16.8 in.
8.6 Out-of-Roundness of Cylindrical
Shells Under External Pressure
In the fabrication of cylindrical shells, slight out-ofroundness invariably results. This is due to forming,
welding, or postweld heat-treating operations. Normally,
internal pressures tend to minimize out-of-roundness,
whereas external pressures tend to increase it. Because
of that, and to prevent failure, extra precautions must be
taken in fabricating shells that are subjected to external
pressures.
In Figure 5.16, it is assumed that the shell is approximated by a series of columns connected end to end. The
length of each column is one-half of a lobe length, or
πDo
.
2N
The slenderness of each column is expressed by the
ratio l/r, where r is the radius of gyration. Since r is equal
l=
129
8 Design of Cylindrical Shells
Figure 8.17 Length of gap in a cylindrical
shell.
d
c
b
a
=
0.
Ar
03
c
=
0
Ar
0
D
c = .0
o
3
5
0
Ar
Ar .04 D
c=
c= 0 o
Ar
0.0
0.0 Do
c=
55
45
0
Ar
D
.
0
c=
D
o
65
o
Ar
D
c = 0.07
o
5
0
Ar
D
.08
c=
5D o
0.1
o
0D
Ar
c=
o
0.1
Ar
25
c=
D
0
o
.
15
Arc
0
=0
D
Arc
.17
o
5D
=0
Arc
.20
o
0D
=0
.25
Arc
o
0
=0
D
.30
o
0D
Arc
o
=0
.39
0D
o
2000
Ar
c
1000
800
Outside diameter ÷ thickness, Do/t
130
600
500
400
300
200
100
80
60
50
40
30
20
10
0.01
0.02
0.04 0.06
0.10
0.2
0.4 0.6
1.0
2
Design length ÷ outside diameter, L/D0
3
8 5 6
8 10
20
Figure 8.18 Maximum arc of shell left unsupported because of gap in the stiffening ring of cylindrical shell under external pressure.
Source: Reprinted from Ref. [2].
8.6 Out-of-Roundness of Cylindrical Shells Under External Pressure
√
to t∕ 12 in a shell wall, the equation becomes
slightly. Thus, Eq. (8.14) needs modification to take
into account the two extreme cases. Disregarding the
increase in e/t as L/Do decreases, an empirical equation
of the form
C1
e
=
+ C2 N
t
N(t∕Do )
is found satisfactory. On using the values of C 1 and C 2 as
obtained from tests, the equation becomes
0.018
e
=
+ 0.015N.
(8.15)
t
N(t∕Do )
Figure 8.19 is a plot of this equation.
5.44
l
=
.
r
N(t∕Do )
The eccentricity of each column is expressed by e. If it is
assumed that the eccentricity ratio e/r affects the strength
of a column in the same way as for the shell, it can be
concluded that
l
e
∝ for columns
r
r
and
e
5.44
for cylinders
√ ∝
N(t∕D
o)
t∕ 12
Example 8.8
Calculate the maximum out-of-roundness allowed in a
cylinder subjected to external pressure with Do = 5.0 ft,
L = 14.0 ft, and t = 0.75 in. Compare the result with that
obtained from Figure 8.19.
or
C1
e
=
,
t
N(t∕Do )
(8.14)
where C 1 = 1.57.
Experiments have shown that for constant t/Do
ratio, the value of e/t increases with an increase in
L/Do ratio. This, however, applies only in the case of
intermediate-length shells. For this range, Eq. (8.14)
was found to give adequate results. As the shell gets
longer, tests have shown that an increase in L/Do has no
influence on e/t. On the other hand, tests have shown
that as the length decreases, the value of e/t increases
Solution:
Do
L
= 5.60,
= 80.
Ro
t
From Figure 5.18, N = 3. From Eq. (8.14),
]
[
0.018
+ 0.015(3) (0.75),
e=
3(0.75∕60)
e = 0.39 in.
1000
900
800
700
600
500
Outside diameter ÷ thickness, Do/t
400
e=
300
1.0
t
0.8
t
0.6
t
0.5
t
e=
200
e=
150
e=
100
90
80
70
60
e=
e=
50
e=
40
e=
30
25
0.05
0.10
0.2
0.4
t
0.3
t
0.2
5t
0.2
0t
0.3 0.4 0.5 0.6 0.8 1.0
2
Design length ÷ outside diameter, L/D0
3
4
5 6 7 8 9 10
Figure 8.19 Maximum permissible deviation from a circular form e for vessels under external pressure. Source: Reprinted from Ref. [2].
131
132
8 Design of Cylindrical Shells
From Figure 8.18,
From Figure 8.11,
E0 = 29 × 106 psi.
e = 0.53t
or
e = 0.40 in.
For the elastic region,
1
(0.0625 × 29 × 106 )
𝜎t =
144
= 12,600 psi.
8.7 Design for Axial Compression
For axial compression, Eq. (5.28) may be written as
𝜎
𝜀cr = cr
E0
or
P
0.6
𝜀cr = cr =
.
E0 t
Ro ∕t
The allowable strain can be expressed as
0.6
A=
.
(KD)(Ro ∕t)
A large knock-down factor (KD) is applied to this
equation based on experimental data and to account for
such items as geometric imperfections. ASME uses a KD
factor of 5, and the equation becomes
0.125
.
(8.16)
A=
(Ro ∕t)
In the elastic range,
𝜎cr = 𝜀cr E0 .
For design purposes, a DF of 2.0 is incorporated into
this equation to take into account such items as inaccuracy in determining the modulus of elasticity and variation in material properties. Hence, the equation becomes
B = S∕2 = AE0 ∕2.
In the inelastic range, the elastic modulus, E0 , must
be replaced by the tangent modulus, Et . This is accomplished by constructing an external pressure chart where
Et is used to correlate factor A to a stress B = S/2 as
shown in Figure 8.11. Hence,
For the plastic region,
0.125
= 0.00087
A=
144
and
B = 11,000 psi (use).
Nomenclature
A
= strain obtained from external pressure charts
As = area of the stiffening ring
B
= stress magnitude in a cylinder due to external
pressure
Do = outside diameter of the cylinder
DF = design factor
E
= joint efficiency
E0 = modulus of elasticity
F
= peak stress as defined in Table 8.2
I
= moment of inertia of the stiffening ring
I′
= combined moment of inertia of the stiffening
ring and adjacent shell
KD = knock-down factor
L
= effective length of the shell
P
= internal or external pressure
Pb = primary bending stress as defined in Table 8.2
PL = primary local membrane stress as defined in
Table 8.2
B = 0.5AE
Pm = primary general membrane stress as defined in
Table 8.2
B = 0.55.
Q
= secondary membrane plus bending stress as
defined in Table 8.2
R
= inside radius of cylinder
or
Example 8.9
A cylindrical tower is constructed of stainless steel 410
material. Its radius is 6 ft, and its thickness is 0.5 in.
Determine the maximum allowable compressive stress
at room temperature.
Ro = outside radius of cylinder
S
= allowable tensile stress in the ASME Code,
VIII-1
Solution:
Sa
= alternating stress in the ASME Code, VIII-2
Ro
72
=
= 144.
t
0.5
Sm = allowable tensile stress in the ASME Code,
VIII-2
t
= thickness of the cylindrical shell
Further Reading
References
1 ASME Boiler and Pressure Vessel Code, Section VIII,
3 ASME Boiler and Pressure Vessel Code, Section II,
Division 2Alternative Rules—Pressure Vessels. New
York: American Society of Mechanical Engineers.
2 ASME Boiler and Pressure Vessel Code, Section VIII,
Division 1Pressure Vessels. New York: American Society of Mechanical Engineers.
Part D, Material Properties. New York: American
Society of Mechanical Engineers.
Further Reading
Windenburg, D.F. (1960). Vessels under external pressure.
In: Pressure Vessel and Piping Design: Collected Papers
1927–1959. New York: American Society of Mechanical
Engineers.
Holt, M. (1960). A procedure for determining the allowable
out-of-roundness for vessels under external pressure. In:
Pressure Vessel and Piping Design: Collected Papers
1927–1959, 7. New York: American Society of
Mechanical Engineers.
133
Inside surface of a vessel head Source: Courtesy of the Nooter Corp., St. Louis, MO.
136
9
Design of Formed Heads and Transition Sections
9.1 Introduction
A large variety of end closures and transition sections are
available to the design engineer. Using one configuration
versus another depends on many factors such as method
of forming, material cost, and space restrictions. Some
frequently used heads are as follows:
9.1.1
Flanged Heads
These heads (Figure 9.1a) are normally found in vessels
operating at low pressures such as gasoline tanks, and
boilers. They are also used in high-pressure applications
where the diameter is small. Various details for their
design and construction are given by the American
Society of Mechanical Engineers (ASME) Code, VIII-1.
9.1.2
Hemispherical Heads
Generally, the required thickness of hemispherical heads
due to a given temperature and pressure is one-half
that of cylindrical shells with equivalent diameter and
material. Hemiheads (Figure 9.1b) are very economical
when constructed of expensive alloys such as nickel and
titanium – either solid or clad. In carbon steel, hemiheads are not as economical as flanged and dished heads
because of the high cost of fabrication. Hemiheads are
normally fabricated from segmental “gore” sections or
by spinning or pressing. Segmental gore hemiheads are
economical in thin, large-diameter equipment or thick,
small-diameter reactors. Because hemispherical heads
are thinner than the cylindrical shells to which they
are attached, the transition area between the head and
shell must be contoured so as to minimize the effect of
discontinuity stress. Figure 9.2 illustrates the transition
requirements in the ASME Code, VIII.
9.1.3 Elliptical and Torispherical (Flanged
and Dished) Heads
These heads are very popular in pressure vessels
(Figure 9.1c and d). Their thickness is usually the same
as the cylinder to which they are attached. This reduces
considerably the weld build-up shown in Figure 9.2.
Thus, because the required thickness in areas away from
the knuckle region is less than the furnished thickness,
the excess can be advantageously used in reinforcing
nozzles in these areas. Many mills can furnish such
heads in various diameters and thicknesses that are
competitive in price.
In a true elliptical head, the radii of curvature vary
between adjacent points along a meridian. To simplify
the calculations and fabrication, the ASME Code established the following various approximations. A 2 : 1
elliptical head can be assumed to consist of a spherically
dished head with a radius of 90% and a knuckle radius of
17% of the shell diameter to which they are attached, as
shown in Figure 9.3. The smallest knuckle radius allowed
for a flanged and dished head is 6% of the shell diameter,
and the smallest spherical radius is 100% of the shell
diameter.
9.1.4
Conical and Toriconical Heads
These heads, shown in Figure 9.1e and f, are used in hoppers and towers as bottom end closures or as transition
sections between cylinders with different diameters. The
cone-to-cylinder junction must be considered as part of
the cone design due to the high unbalanced forces at the
junction. Because of these high forces, the ASME Code,
VIII-1, limits the apex angle to a maximum of 30∘ when
the cone is subjected to internal pressure. Above 30∘ , a
discontinuity analysis is done or a toriconical head used
to avoid the unbalanced forces at the junction.
9.1.5
Miscellaneous Heads
Many chemical processes require unusual vessel configurations. The heads of such vessels can have an infinite
number of contours. One such contour is shown in
Figure 9.1g. The design of these heads is very complicated, and there are no simple methods of analysis.
Experience, proof testing, and sophisticated analyses are
generally used to determine the required thicknesses.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
9.2 ASME Design Equations for Hemispherical Heads
Figure 9.1 Commonly used formed
closure heads.
t
t
R
sf
sf
ID
ID
(b) Hemispherical
(a) Flanged
t
t
h
h
sf
ID
L
(d) Flanged & dished
(Torispherical)
sf
(c) Elliptical
α
t
t
α
r
ID
ID
(e) Conical
(f) Toriconical
sf
R
t
sf
ID
(g) Miscellaneous
9.2 ASME Design Equations
for Hemispherical Heads
The ASME Code, VIII-1, has combined Eqs. (6.1) and
(6.7) for internal pressure into one simplified equation:
PR
,
(9.1)
t=
2SE − 0.2P
where
t = required thickness
P = internal pressure
R = inside radius
S = allowable stress
E = joint efficiency.
This equation with E = 1.0 is plotted in Figure 6.3, and
it approximates the more complicated Eq. (6.7) over a
large range of ro /ri . Other 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 𝜀cr = A, r = Ro , and
modulus of elasticity as E0 , Eq. (6.35) can be written as
𝜎cr
0.125
=
.
E0
Ro ∕t
This equation includes a knock-down factor of 1.25.
Or
𝜀cr =
0.125
,
Ro ∕t
(9.2a)
137
9 Design of Formed Heads and Transition Sections
th
l>3y
l>3y
Tangent line
y
Thinner part
th
Thinner part
<1/2(ts–th)
Figure 9.2 Head-to-shell junction.
Source: Courtesy of the American
Society of Mechanical Engineers.
<1/2(ts–th)
y
Length of required
taper, l, may include
the width of the weld.
ts
ts
(a)
(b)
th
th
y
Tangent line
y
<1/2(ts–th)
l>3y
<1/2(ts–th)
ts
Thinner part
l>3y
Thinner part
138
ts
(c)
(d)
Figure 9.3 Head contours approximating various elliptical
shapes.
R = 0.9D
r = 0.17D
D
SF
(a) ASME 2:1 head
R=D
r = 0.06D
D
(b) ASME flanged and dished head
SF
9.3 ASME Design Equations for Ellipsoidal, Flanged, and Dished Heads
where, 𝜀cr = A for design purposes.
From Figure 8.11, with A = 0.001, plastic behavior prevails, and Eq. (9.2c) must be used. Hence, from Figure
8.11,
𝜀cr = critical strain
Ro = outside radius
B = 11,000 psi
t = thickness.
and
The critical stress in a spherical section is given by
𝜎cr =
Pcr Ro
2t
Pcr =
2𝜎cr
2𝜀 E
= cr 0 ,
Ro ∕t
Ro ∕t
or
(9.3)
and the allowable pressure is expressed as
P=
2AE0
.
(FS)Ro ∕t
(9.4)
Using factor of safety (FS) = 4.0 and substituting Eq.
(9.2a) into Eq. (9.4) gives
0.0625E0
,
(9.2b)
P=
(Ro ∕t)2
where
P = allowable external pressure
E0 = modulus of elasticity.
The ASME procedure for determining the allowable
external pressure for a spherical section is to determine
first the A value from Eq. (9.2a). The allowable pressure
can then be obtained by referring to a stress–strain chart
similar to the one shown in Figure 8.11. If A falls in
the elastic region, then P is calculated from Eq. (9.2b).
If A falls in the plastic region, however, a value of B is
determined first from the chart. The allowable pressure
is then calculated from Eq. (9.3) as
2𝜎cr
.
(9.5)
Pall =
(FS)Ro ∕t
Substituting B = 𝜎 cr /2 and FS = 4 into Eq. (9.5) gives
B
P=
,
(9.2c)
Ro ∕t
where B = factor determined from Figure 8.11.
Eqs. (9.2) form the ASME basis for determining the
allowable external pressure for spherical sections.
Example 9.1
Using the ASME criteria, determine the allowable external pressure on a spherical shell with Ro = 60 in. and
t = 0.5 in. Use the 300 ∘ F line shown in Figure 8.11.
Solution:
From Eq. (9.2a),
A=
0.125
= 0.001.
60∕0.5
P=
11,000
= 91 psi.
60∕0.5
9.3 ASME Design Equations
for Ellipsoidal, Flanged, and Dished
Heads
The general solution of Eq. (6.39) is very cumbersome
because r2 is a variable function. However, the forces and
bending moments obtained from Eq. (6.39) are important
because they can be added to the membrane forces of Eq.
(6.37), significantly reducing the total stress at the vicinity
of the junction. Therefore, this equation can be advantageously utilized by the designer in reducing the required
head thickness. The ASME used this fact in developing
design parameters for ellipsoidal and torispherical heads.
A study was made [1] with ellipsoidal heads to determine the effect of the ratio a/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 stress in the head changes with a
change of a/b. For heads shallower than a/b of 2.5, the
maximum stress is in the hoop direction at the outer
surface of the knuckle region and is in compression as
shown in Figure 9.4. For ratios of a/b between 2.5 and
1.2, the maximum stress occurs at the junction and is
a hoop tensile stress. The stress magnitude for various
ratios of a/b is shown in Figure 9.4. A simplified equation
used by the ASME Code, VIII-1, approximates the theoretical stress ratios shown in Figure 9.4 for values of a/b
between 2.6, which is the maximum allowed by the code,
and 1.0 for a spherical head:
[
( )2 ]
D
1
.
(9.6)
2+
K=
6
2h
A plot of this equation is also shown in Figure 9.4. The
ASME Code uses the K values given by Eq. (9.6) to determine the required stress ratios needed in obtaining the
thickness of ellipsoidal heads. This is accomplished by
multiplying the calculated thickness of a cylindrical shell
with diameter D by K. Hence, for ellipsoidal heads,
PDK
,
(9.7)
t=
2SE − 0.2P
where
t = thickness of the ellipsoidal head
P = internal pressure
139
9 Design of Formed Heads and Transition Sections
3.5
3.0
Maximum stress in head
Hoop stress in shell
140
2.5
Figure 9.4 Maximum stress in ellipsoidal
heads. Source: Brownell and Young (1959) [1].
Reproduced with permission of John Wiley &
Sons.
Theoretical values
Approximate values given by Eq. (9.6)
tensile hoop stress at
Junction
tensile meridional stress at
Inside surface of knuckle area
2.0
1.5
Compressive hoop stress
at outside surface
of knuckle area
1.0
1
a
2 +( )2
6
b
0.5
0
1.0
1.5
2.0
a
b
2.5
3.0
D = inside diameter of shell to which the head is
attached
3.5
M = stress intensity factor obtained from Eq. (9.8)
S = allowable stress
K = stress intensity factor obtained from Eq. (9.6)
E = joint efficiency
S = allowable stress
r = inside knuckle radius.
E = joint efficiency.
For torispherical heads, tests conducted by Höhn [1]
and others have shown that the stress at the knuckle
area due to internal pressure reaches the yield value
long before the spherical region does. Höhn plotted
an empirical equation that correlates well with the
available test data, as shown in Figure 9.5. To evaluate
Höhn’s empirical curve, Figure 9.5 shows another curve
that indicates the stress in an equivalent ellipsoidal head
whose thickness is equal to the shell thickness. This curve
indicates that Höhn’s curve is liberal for small values
of knuckle-to-crown radii r/L. Accordingly, the ASME
Code, VIII-1, developed an empirical curve that parallels
Höhn’s curve for large values of r/L and the ellipsoidal
curve for small r/L ratios, as shown in Figure 9.5. The
ASME curve can be expressed by the equation
(
√ )
L
1
3+
.
(9.8)
M=
4
r
Thus, the ASME equation for the torispherical heads is
given by
t=
PLM
,
2SE − 0.2P
where
t = thickness of torispherical head
P = internal pressure
L = inside spherical crown radius
(9.9)
In practical applications, researchers noticed that Eq.
(9.9) gave conservative results for the majority of head
designs but became unconservative for large ratios of r/t.
This ratio was not considered by the ASME in its derivation of Eq. (9.8), because Figure 9.5 was based on a constant value of r/t. Accordingly, research was conducted to
evaluate the buckling behavior of the knuckle region for
heads with large ratios of r/t. Plastic analysis was used,
and it was shown [2] that the following equation can adequately predict the behavior of torispherical heads with
large r/t ratios:
)
(
r t
P
= 0.33 + 5.5
𝜎y ∕FS
D L
) ( )2
(
r
t
+28 1 − 2.2
− 0.0006.
D
L
Solving for t and letting 𝜎 y /FS = S, the following
approximate equation is obtained and used by the ASME
Code, VIII-2:
( )2
( )
t
r
r
+ 28.93318
ln = −1.26177 − 4.55246
L
D
D
[
( )2 ]
( )
r
r
+ 15.68299
+ 0.66299−2.24709
D
D
( )
r ⎤
−4
⎡
P ⎢0.26879 × 10( )− 0.44262 D ⎥
× ln +
2
⎥
S ⎢ +1.88783 r
⎦
⎣
D
)2
(
P
,
(9.10)
× ln
S
9.3 ASME Design Equations for Ellipsoidal, Flanged, and Dished Heads
3.5
Höhn’s empirical function
for F. & D. heads
(20r/L) +3
(20r/L) +1
2.5
2.0
1.5
Eq. (9.8)
1.0
Minimum ratio
of r/L permitted
by ASME code
Ratio of maximum stress to crown stress
3.0
0.5
0
6.0
0
4.0
3.0
2.4
Elliptical heads (computed
maximum stresses for head
thickness equal to shell
thickness)
Ratio of major axis to minor axis of ellipse
2.2
2.0
1.8
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.20
Ratio of minimum knuckle radius to crown radius. r/L
Figure 9.5 Stress in flanged and dished heads. Source: Brownell and Young (1959) [1]. Reproduced with permission of John Wiley & Sons.
where
t = required torispherical thickness
L = inside spherical-crown radius
r = inside knuckle radius
D = inside diameter of shell to which the head is
attached
P = internal pressure
S = allowable stress.
A plot of Eq. (9.10) is shown in Figure 9.6.
Example 9.2
What is the required thickness of a 3 : 1 head if
D = 144 in., P = 100 psi, S = 17,500 psi, E = 1.0?
Solution:
From Eq. (9.6), with a/b = 3,
K = 0.1667[2 + (3.0)2 ]
= 1.83.
From Eq. (9.7),
(100)(144)(1.83)
2 × 17 500 × 1.0 − 0.2 × 100
= 0.75 in.
t=
Example 9.3
What is the required thickness of a flanged and dished
head if D = 240 in., L = 240 in., r = 15 in., S = 17,500 psi,
E = 1.0, and P = 50 psi?
Solution:
L
= 16
r
From Eq. (9.8),
√
1
M = (3 + 16)
4
= 1.75.
From Eq. (9.9),
50 × 240 × 1.75
t=
2 × 17,500 × 1.0 − 0.2 × 50
= 0.60 in.
Because the thickness if obtained from Eq. (9.9) and
because this thickness is small compared with the diameter of the head, the requirements of Eq. (9.10) must be
checked:
15
r
=
D 240
= 0.0625
50
P
=
S
17,500
= 0.002857,
141
9 Design of Formed Heads and Transition Sections
0.10
0.05
r/D
0.17 (2:1 Eilipsoidal head)
0.10
P/S
142
0.20
0.15
0.06
0.01
t
h
L
0.005
r
D/2
C
L
0.001
0.002
0.005
0.01
NOTE: For 2:1 Eilipsoidal heads use L = 0.9D
to calculate t/L
0.02
0.03
0.04 0.05
t/L
Figure 9.6 Required thickness of formed heads. Source: Courtesy of the American Society of Mechanical Engineers.
and from Eq. (9.10),
( )
t
= −5.54851
ln
L
t
= 0.00389
L
t = 0.93 in.
Hence, for this head, a minimum thickness of 0.93 is to
be used.
For external pressure, the knuckle area is subjected
to a tensile stress. Hence, the critical area that has to
be considered under external pressure is the spherical
9.4 ASME Design Equations for Conical Heads
Solution:
From Eq. (9.11),
region. Thus, the ASME criteria for all ellipsoidal and
torispherical heads under external pressure are the same
as those for spherical heads.
(450)(40.0)
(2)(0.940)(20,000 × 1.0 − 0.6 × 450)
= 0.49 in.
t=
9.4 ASME Design Equations for Conical
Heads
9.4.1
9.4.1.1
Internal Pressure
From Example 6.7, it is seen that the hoop force N 𝜃 is
twice as large as the longitudinal force N S in conical
heads subjected to internal pressure. The ASME Code,
VIII-1, uses Eq. (1) of Example 6.7 as the basis for
establishing the required thickness of a conical section
subjected to internal pressure. The equation is given by
t=
PD
,
2 cos 𝛼(SE − 0.6P)
(9.11)
where
where, in this chapter, 𝛼 is used rather than 𝛼 ′ ; X and Y
are given by
P = internal pressure
X = 4.559V2 tan 𝛼
D = inside diameter of the cone at the point
considered, measured perpendicular to the
longitudinal axis
Y = 1.316(V1 − 2V2 ) tan 𝛼,
and V 1 and V 2 are as in Example 6.9.
The longitudinal stress in Eq. (9.12) is in tension for
all values of 𝛼 and does not govern the design criteria.
The Y term in the circumferential stress expression in
Eq. (9.12) is positive for all practical applications. Hence,
the quantity 𝜎 c varies from a maximum tensile value of
Pr/t to a compressive value that depends on the angle 𝛼.
The ASME Code, VIII-1, limits the maximum compressive circumferential stress
√ to a value of Pr/t. Using these
criteria, the quantity Y r2 ∕t in Eq. (9.12) must be limited
to a value of 2.0. Values in excess of 2.0 must be supported
S = allowable stress
E = joint efficiency
𝛼 = one-half of the included apex angle of the
cone at the center line of the cone.
Example 9.4
What is the required thickness of a conical head attached
to a cylinder whose inside diameter is 40.0 in. if the internal pressure is 450 psi, the allowable stress is 20,000 psi,
E = 1.0, and 𝛼 = 20.0?
Figure 9.7 Discontinuity forces due to
internal pressure.
Discontinuity Analysis for Internal Pressure
The ASME Code, VIII-1, uses the stress expressions
obtained in Example 6.9 for internal pressure as the basis
[3] for establishing simplified criteria for discontinuity
analysis at the cylinder-to-cone junction. At the large end
of the cone (Figure 9.7), the discontinuity analysis results
in the following two expressions for the longitudinal and
hoop stresses in the shell:
(
√ )
Pr
r
𝜎l = t 2 0.5 + X t2
(
√ )
(9.12)
Pr
r
𝜎c = t 2 1 − Y t2 ,
M
P• r2
2
f
M
f
F
P• r tan ∝
2
∝
F
f
F
P• r1
2
r2
F
f
M
r1
143
9 Design of Formed Heads and Transition Sections
√
Δ = 326.6 P∕SE
1.0
𝛼 = one-half of the apex angle of the cone.
0.8
X & Y Values
144
0.6
X = 0.012 α
0.4
X = 4.559V2 tan α
0.2
0
Y = 0.005 α
Y = 1.316 (V1 – 2V2) tan α
0°
10°
20°
30°
α
40°
50°
60°
Figure 9.8 X and Y values for internal pressure.
by a ring added at the junction. The area of the ring is
given by the equation
)
( √
tr2 tan 𝛼 Y r2 ∕t − 2
At =
.
(9.13)
√
2
Y r ∕t
2
A plot of the quantity Y shows that it can be approximated by the expression 0.005𝛼 as shown in Figure 9.8.
Substituting this value into Eq. (9.13) gives
)
(
tr2 tan 𝛼
400
At =
.
(9.14)
1− √
2
𝛼 r ∕t
2
The ASME Code, VIII-1, limits the compressive
circumferential stress to 1.5SE, and thus,
1.5SE =
or
√
r2
=
t
Pr2
t
√
Eq. (9.15) is used by the ASME Code, VIII-1, as the
basis for checking the joint between the cylinder and the
large end of the cone due to internal pressure. At the small
end of the cone (Figure 9.7), the circumferential and longitudinal stress equations due to internal pressure are
(
√ )
r1
Pr1
𝜎l =
0.5 − X
t
t
(
√ )
r1
Pr
.
(9.16)
𝜎c = 1 1 − Y
t
t
Because both expressions include a negative term, the
equation for 𝜎 l controls because X is numerically larger
than Y . Limiting the maximum compressive stress to
Pr/t, the factor in parentheses in the equation for 𝜎 l is
√
r1
X
≤ 1.5,
t
√
and a stiffening ring is needed for values of X r1 ∕t
greater than 1.5. The required area of the stiffening ring
is
]
[
Pr12 tan 𝛼
1.5
.
At =
1− √
2SE
X r ∕t
1
From Figure 9.8, it is shown that the quantity X is
approximated by 0.012𝛼. Hence, the required area can
be expressed as
]
[
Pr12 tan 𝛼
125
At =
.
(9.17)
1− √
2SE
𝛼 r ∕t
1
Assuming that the maximum allowable longitudinal
stress is limited to SE, the expression
1.5SE
.
P
Eq. (9.14) thus becomes
(
)
√
Pr22
326.6 P∕SE
1−
tan 𝛼
At =
3SE
𝛼
or for the large end of the cone,
)
Pr2 (
Δ
At = 2 1 −
tan 𝛼,
2SE
𝛼
where
Pr1
2t
can be substituted into Eq. (9.17) to give for the small end
of the cone
)
Pr2 (
Δ
tan 𝛼,
(9.18)
At = 1 1 −
2SE
𝛼
where
√
P
Δ = 89 SE
SE =
(9.15)
At = required area of the ring
r1 = radius of the cylinder at the small end of the cone.
P = internal pressure
r2 = radius of the cylinder at large end of the cone
S = allowable stress
E = joint efficiency
Example 9.5
Design the cone shown in Figure 9.9, and check the
cone-to-shell junctions. Let S = 20 ksi, E = 1.0, and
P = 150 psi.
9.4 ASME Design Equations for Conical Heads
For the small end,
√
150
= 7.71,
Δ = 89
20,000
D = 8’
and the required area is.
)
(150)(48)2 (
7.71
1−
(0.577)
A=
(2)(20,000)
30
= 3.70 in.2
°
30
9.4.2
External Pressure
The governing equation for the design of cones subjected
to external pressure is obtained from Eq. (6.43). Using a
design factor of 3.0, Eq. (6.43) becomes
D = 12’
Pa
0.87(te ∕D2 )2.5
=
.
E0
Le ∕D2
Figure 9.9 Conical transition section.
This equation that expresses the cone in terms of an
equivalent cylinder of thickness t e and length Le is analogous to Eq. (6.42) for cylindrical shells. Thus, the ASME
Code, VIII-1, applies the same equations for the design of
cylindrical shells under external pressure for the design of
cones with applicable values of t e and Le .
Solution:
From Eq. (9.11),
(150)(144)
2(0.866)(20,000 × 1.0 − 0.6 × 100)
= 0.62 in.
t=
For the large end,
√
Δ = 326.6
9.4.2.1
∘
150
= 28.28 ,
20,000
Nx
Q
1∕2
r
Pr2
+ Y 23∕2 Nx
t
t
Nx (
r )
1+X 2 ,
𝜎l = −
t
t
𝜎c = −
Q1
Nx
Mx
Discontinuity Analysis for External Pressure
The discontinuity forces due to external pressure [4] at
the large end of the cone are shown in Figure 9.10 and
expressed as
and from Eq. (9.15), the required area at the large end is
)
(150)(72)2 (
28.28
A=
1−
(0.577)
(2)(20,000)
30
= 0.64 in.2
P
P
Nx
Mx
(9.19)
α
Mx
r2
Nx
Nx
Mx
Nx
Q1
r1
Q
Figure 9.10 Discontinuity forces due to external pressure.
P
(9.20)
145
9 Design of Formed Heads and Transition Sections
where
(9.22) reduces to
1∕2
r
Pr
−Pr1
− Y 13∕2 Nx ≤ − 1
t
t
t
or N x ≤ 0, which indicates that at the small end of the
cone, the axial force N x must be resisted by a ring with
area
N r tan 𝛼
A= x 1
.
(9.23)
SE
In addition to providing the required area at a coneto-shell junction, it is necessary to design the ring at the
junction to prevent buckling due to external pressure.
The procedure is similar to that for the design of stiffening rings in cylindrical shells. A conservative approach
used by the ASME Code, VIII-1, in designing cylindrical
shells under external pressure assumes that intermediate
stiffening rings support all the load applied to the shell.
Using the same criteria, the load on the cone shown in
Figure 9.10 due to external pressure can be proportioned
at the large- and small-end stiffening rings as follows:
X = 9.34V 2 tan 𝛼
Y = 2.57(V 1 − 2 V 2 )tan𝛼
N x = axial compressive force = Pr2 /2 t + Q
Q = axial force due to wind, dead load, and so on.
The values of X and Y in Eq. (9.20) can be approximated
by the expressions in Figure 9.11. A comparison of 𝜎 c and
𝜎 l given by Eq. (9.20) indicates that the maximum compressive stress is given by 𝜎 l ; a conservative maximum
allowable compressive stress value is Pr2 /t, and thus,
√
r2
Pr
0.027𝛼
≤ 2 − 1.
t
Nx
A stiffening ring is necessary if this quantity is
exceeded. The required area of the ring is given by
[
]
(r )
(Pr2 ∕Nx ) − 1
2
1−
At = (Nx tan 𝛼)
.
√
SE
0.027𝛼 r ∕t
pressure load at large-end stiffening ring
2πP(r2 + r1 )(r2 − r1 )
=
cos 𝛼
2 sin 𝛼
pressure load at small-end stiffening ring
πP(r2 + r1 )(r2 − r1 )
.
=
3 sin 𝛼
The total load at the large end due to axial compression,
pressure on the cone, and pressure on the cylinder is
PL
−Pr2
F=
tan 𝛼 + f1 tan 𝛼 + 1
2
2
P(r2 + r1 )(r2 − r1 ) cos 𝛼
+
3r2 sin 𝛼
or
2
By limiting the axial stress to an allowable value of SE,
the aforementioned equation can be written for the large
end of the cone,
) ]
[
(
Nx r2 tan 𝛼
1 Pr2 − Nx Δ
At =
1−
, (9.21)
SE
4
Nx
𝛼
√
P
.
where Δ = 104 SE
At the small end of the cone, the stress is given by
1∕2
r
−Pr1
𝜎c =
−Y 1 N
t ( t 3∕2 x )
√
−Nx
r1
𝜎l =
1+X
,
t
t
F = P(M) + f1 tan 𝛼,
(9.22)
where
and the maximum compressive stress by 𝜎 c . Limiting the
allowable compressive stress to −(Pr1 /t), the first of Eq.
M=−
1.8
1.6
1.4
X = 9.34V2 tan α
1.2
1.0
X = 0.027α
0.8
0.6
0.4
Y = 0.010 α
0.2
0
r2 − r12
r2 tan 𝛼 Ll
+ + 2
.
2
2
3r2 tan 𝛼
Figure 9.11 X and Y values for external pressure.
2.0
X & Y Values
146
Y = 2.57 (V1 – 2V2) tan α
0°
10°
20°
30°
α
40°
50°
60°
(9.24)
Nomenclature
The total load at the small end due to axial compression, pressure on the cone, and pressure on the cylinder
is
r2 − r12
PL
Pr
F = 1 tan 𝛼 + s + 2
P + f2 tan 𝛼
2
2
6r2 tan 𝛼
or
F = P(N) + f2 tan 𝛼,
(9.25)
3E0 I
,
At r2
E0 = modulus of elasticity
h = head depth
= moment of inertia
K = factor for ellipsoidal heads as determined from
Eq. (9.6)
L l t s Lc t c
+
+ As for large end of cone
2
2
Lt
Lt
At = s s + c c + As for small end of cone.
2
2
On using 𝜎 c = E0 A, the aforementioned equation
reduces to
At =
2
Ar At
,
(9.26)
3
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 ring as
I=
(9.27)
Applying a design factor of 2 in the foregoing equation
and using the external-pressure charts in the ASME Code
(which have a factor of safety of 2), a design criterion can
be established as follows:
1) Calculate F from Eqs. (9.24) or (9.25).
2) Establish
2Fr
.
𝜎=
At
3) Enter the external-pressure charts with 𝜎 (factor B),
and calculate the strain A.
4) Use Eq. (9.26) to establish the minimum required
moment of inertia. The ASME allows a 30% increase
in value if the composite ring–shell moment of inertia
is considered, and the equation then becomes
AD2 At
.
15.6
E = joint efficiency
I
where At is the total effective area given by
I=
B = stress factor as obtained from Figure 8.11
D2 = base diameter at the large end of the cone
Eqs. (9.24) and (9.25) establish the maximum applied
force at the cone-to-cylinder junction. The critical buckling stress of a circular ring is
Fr
.
At
At = area of the stiffening ring at the cone-to-shell
junction
D1 = base diameter at the small end of the cone
r2 + r12
L
r
N = 1 tan 𝛼 + s + 2
.
2
2
6r2 tan 𝛼
𝜎=
A = strain as obtained from Figure 8.10
D = diameter
where
𝜎c =
Nomenclature
L = spherical crown radius of flanged and dished
heads
L′ = effective length of the cylindrical shell
Lc = length of cone
Le = effective length of conical section
(
)
′
D
= L2 1 + D1
2
Ll = length of large cylinder
Ls = length of small cylinder
M = factor for flanged and dished heads as obtained
from Eq. (9.8)
P = pressure
Pa = allowable external pressure
R = inside radius
Ro = outside radius
r
= knuckle radius
r1 = base radius at the small end of the cone
r2 = base radius at the large end of the cone
S
= allowable stress
t
= thickness
t e = effective thickness of conical section
= t cos 𝛼
t l = thickness of large cylinder
t s = thickness of small cylinder
𝛼 = one-half of the cone apex angle
𝜎 c = circumferential stress
𝜎 l = longitudinal stress
147
148
9 Design of Formed Heads and Transition Sections
References
1 Brownell, L.E. and Young, E.H. (1959). Process Equip-
ment Design. New York: Wiley.
2 Shield, R.T. and Drucker, D.C. (1972). Design
of thin-walled torispherical and toriconical
pressure-vessel heads. In: Pressure Vessels and Piping:
Design and Analysis—A Decade of Progress. American
Society of Mechanical Engineers.
3 Boardman, H.C. (1960). Stresses at junction of cone
and cylinder in tanks with cone bottoms or ends. In:
Further Reading
Flügge, W. (1960). Stresses in Shells. New York:
Springer-Verlag.
Pressure Vessel and Piping Design: Collected Papers
1927–1959. American Society of Mechanical Engineers.
4 Jawad, M.H. (1980). Design of conical shells under
external loads. Journal of Pressure Vessel Technology
102: 230–238.
Typical flanges ready for installing. Source: Courtesy G + W Taylor – Bonney Div., Taylor Forge.
150
10
Blind Flanges, Cover Plates, and Flanges
10.1 Introduction
One of the more common types of closures for pressure vessels is the unstayed flat head or cover. This may
be either integrally formed with the shell or welded to
the shell, as shown in Figure 10.1; or, it may be attached
by bolts or some quick-opening device as shown in
Figure 10.2. It may be circular, obround, square, rectangular, or some other shape. Those circular flat heads that
are bolted into place utilizing a gasket are called blind
flanges. Usually, the blind flange is bolted to a vessel
flange with a gasket between two flanges as shown in
Figure 10.3. Although flat heads or blind flanges may be
either circular or noncircular, they usually have uniform
t
t
(a)
(b)
Retaining
ring
t
t
(c)
(d)
t
ts
ts
Threaded
ring
(e)
Figure 10.2 Bolted or quick-opening flat heads.
£
t
t
(a)
(b)
ts
t
ts
t
(c)
(d)
ts
ts
t
t
(e)
Figure 10.1 Integral or welded flat heads.
Figure 10.3 Blind-flange–integral-flange connection.
(f)
thickness. In addition to the flat-head or blind-flange closures, many large vessels use a circular, spherically dished
cover with a bolting flange, as shown in Figure 10.4. In
all cases, the bolts of the head attach either to a bolting
flange on the end of the shell or to a thickened shell.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
10.2 Circular Flat Plates and Heads with Uniform Loading
Figure 10.4 Spherically dished
covers with bolting flanges. Source:
Courtesy of American Society of
Mechanical Engineers; from Figure
1-6 of the ASME Code, VIII-1.
t
t
Kunckle
radius
L
Kunckle
radius
t
L
T°
Gasket
Gasket
loose flange type
integral flange type
(a)
1/2 A
L
1/2 B
1/2 C
Ring
gasket
shown
T° = T > r
(c)
1/2 A
1/2 C
1/4 (A + B)
t
t
βL
L
T
T
1/2 B
Ring
gasket
shown
(b)
In general, failure of a bolted flanged joint is due
to excessive leakage at the gasket; there are very few
failures at the metallic pressure boundary. Occasionally,
problems encountered with bolts result from excessive tightening in reducing the leakage problem during
hydrostatic testing of the assembly. Sometimes, excessive
stresses in the bolts cause them to break or to stretch
until the closure leaks.
The basic equations used for the design of flat plates
and blind flanges in the ASME Code, VIII-1, are based
on a flat plate with uniform thickness and uniform loading over the entire surface due to pressure. Depending
upon the details of the corner construction shown in
Figure 10.5, various C factors are used that require
different minimum head thicknesses. The maximum
deflection of the plate is assumed to be not more than
one-half of the thickness, and all the stresses are kept
within the elastic limit.
10.2 Circular Flat Plates and Heads
with Uniform Loading
When an exact solution involving a discontinuity analysis at the shell-to-head juncture is not wanted, stresses
on the flat heads are generally calculated based on the
assumption that the edges are simply supported or fully
fixed. The actual condition lies somewhere between.
Exact equations for circular plates were developed
in Chapter 7.2. In using those equations as a basis, the
following equations were developed using the head
diameter d instead of the radius term a. The following
L
hr
Centroid
t
HD
Hr
Use any suitable
type of gasket
1/2 C
(d)
1/2 B
nomenclature was also substituted in the equations in
Chapter 7.2:
𝜇 = Poisson’s ratio = 0.3
P = pressure loading (psi)
E′ = modulus of elasticity (psi)
d = diameter of head (in.)
T = nominal thickness of head (in.)
Pt = total pressure load (lb) = 0.785pd2 .
With the edges assumed to be simply supported:
Maximum stress is located at the center and equals
Pt
.
(10.1)
T2
Maximum deflection is located at the center and
equals
𝜎max = 0.394
Pt d2
.
(10.2)
E′ T 3
Maximum rotation is located at the edge and equals
wmax = 0.0554
Pt d
.
(10.3)
E′ T 3
With the edges assumed to be fully fixed:
Maximum stress is radial and located at the edge:
𝜃max = 0.167
Pt
.
(10.4)
T2
Maximum deflection is located at the center and
equals
𝜎max = 0.239
wmax = 0.0136
Pt d2
.
E′ T 3
(10.5)
151
152
10 Blind Flanges, Cover Plates, and Flanges
Center of weld
L
tf min. = 2ts
Tangent
t
ts
t
line
s
ts
aper
r = 3t
min.
t
d
tmin = 0.375 in.
(a)
t
ts
t
d
(b-2)
d
t
(e)
than 1.25 rr
t
(b-1)
C = 0.13
(d)
d
C = 0.33 m
C min. = 0.20
0.7 ts
t = 1/4 t
min.
r = 3t
min.
C = 0.17
t
tmin. = 0.25 ts for
ts > 1.1/2 in.
t
t
d
Retaining ring
t
Threaded ring
t
C = 0.25
d
C = 0.75
d
C = 0.3
(k)
d
t
C = 0.30
(o)
ts
ts
t
t
C = 0.3
C = 0.30
(n)
C = 0.30
(m)
d
(j)
d
t
d
hG
hG
(i)
t
C = 0.30
C = 0.20 or 0.13
tw = 2 ts, min. not less that 1.25 ts
but need not be greater that t
Projection
0.7 ts
ts
beyond weld
is optional
0.7 ts
Bevel optional
Continuation
ts
d
45° max.
d
t of shell
t
optional
Sketches (e) (f) (g) Circular covers, C = 0.33 m, C min. = 0.20
Non-circular covers, C = 0.33
(g)
(f)
C = 0.33 m
C min. = 0.20
(h)
r = 3t
min.
t
0.7 ts
ts
C = 0.33
d
Tangent
line
(c)
0.7 ts
d
l
t
d
but need not be greater
than 3/4 in.
ts not less
ts
ts
t or ts < 1.1/2 in.
d
C = 0.17 or
C = 0.10
Center of lap
ts
d
t
C = 0.33
(r)
30° min.
45° max.
min. ts = t or ts
whichever
is greater
Seat weld
t
or
d
C = 0.33
% t min.
0.8 ts
min.
ts
(s)
Figure 10.5 Unstayed flat heads and covers. Source: Courtesy of American Society of Mechanical Engineers; from Figure UG-34 of the
ASME Code, VIII-1.
10.3 ASME Code Formula for Circular Flat Heads and Covers
Tangential stress at the edge equals
P
(10.6)
𝜎t = 0.0716 t2 .
T
Radial and tangential stress at the center equals
P
(10.7)
𝜎t = 𝜎r = 0.155 t2 .
T
When the circular flat plate is loaded under uniform
pressure P over the entire surface, the simplified
equations of stress are
{
( )2
d
𝜎 = 0.309P
for simply supported edge ,
T
{
𝜎=
( )2
d
0.188P
T
For a fixed edge, use Eq. (10.5):
w = 0.0136
Problems
10.1
(10.9)
In actual designs, neither of these edge conditions is
likely to be realized. Fully fixed edge is very difficult to
obtain in any construction.
Example 10.1
Determine the maximum stress in a flat head under
internal pressure of 1000 psi, diameter d = 48 in., and
thickness t = 7 in., for both simply supported and
fixed-edge conditions. Assume carbon steel with
𝜇 = 0.3.
Solution:
Determine the total loading:
Pt = 0.785Pd2 = 0.785(1000)(48)2
Pt = 1,809,000 lb.
For a simply supported edge, use Eq. (10.1):
(1,809,000)
𝜎 = 0.394
= 14,550 psi.
(7)2
For a fixed edge, use Eq. (10.4):
1,809,000
𝜎 = 0.239
= 8820 psi.
(7)2
What is the maximum stress in a simply supported
flat head when P = 50 psi, d = 24 in., t = 0.75 in.,
and 𝜇 = 0.3?
Answer:
𝜎 max = 15,840 psi
(10.8)
for fixed edge .
(1,809,000)(48)2
= 0.0057 in.
(29 × 106 )(7)3
10.2
What is the maximum stress for the conditions
mentioned in Problem 10.1 using the simplified
equation?
Answer:
𝜎 max = 15,820 psi
10.3 ASME Code Formula for Circular
Flat Heads and Covers
In the ASME Code, VIII-1 [1], and Section I [2], the minimum required thicknesses of circular, unstayed flat heads
and covers without bolting are calculated as follows:
√
CP
,
(10.10)
t=d
SE
where
E = butt-weld joint efficiency for a joint within the
head
S = allowable tensile stress (psi)
P = design pressure (psi)
Example 10.2
For the flat head mentioned in Example 10.1, determine the deflection at the center for both edge conditions.
Solution:
For a simply supported edge, use Eq. (10.2):
w = 0.0554
(1,809,000)(48)2
= 0.0232 in.
(29 × 106 )(7)3
C = 0.10 through 0.33, depending upon the
construction details at the head-to-shell juncture
(see Figure 10.5) and containing a factor to
effectively increase the allowable stress to 1.5S
because the stress is predominantly a bending
stress
d = effective diameter of head (in.) (see Figure 10.5)
t = minimum required thickness of flat head (in.)
153
154
10 Blind Flanges, Cover Plates, and Flanges
Example 10.3
Determine the minimum required thickness of an
integral flat head with interval pressure P = 1000 psi,
S = 15,000 psi, d = 48 in., with no corrosion and no weld
joints within the head (E = 1.0). The arrangement is
the same as that given in Figure 10.5, sketch b-2, with
m = 1.0.
Solution:
From Figure 10.5, sketch b-2, C = 0.33, m = 0.33(1) = 0.33.
From Eq. (10.10),
√
0.33 × 1000
= 7.120 in.
t = 48
15,000 × 1.0
Example 10.4
Determine the minimum corner radius to make the solution to Example 10.1 acceptable (valid).
Solution:
The cylindrical-shell thickness t s must be calculated
using Eq. UG-27(c)(1) of the ASME Code, VIII-1:
1000 × 24
PR
=
SE − 0.6P
15,000 × 1 − 0.6 × 1000
= 1.667 in.
10.4 Comparison of Theory and ASME
Code Formula for Circular Flat Heads
and Covers Without Bolting
As previously mentioned, the ASME Code formula contains a factor of 1.5 within the C factor to adjust for the
permitted higher level of allowable stress because it is
chiefly caused by primary bending stress. If that factor
is removed, the range of C values in the code becomes
0.15–0.5. Rearranging Eqs. (10.8) and (10.9) into the same
form as Eq. (10.10), we see that this range encompasses
the two extremes from fully fixed edges, where C = 0.188,
to simply supported edges, where C = 0.309.
The low value of C = 0.15 in the ASME Code is for a
special head-to-shell configuration with an inner corner radius of at least three times the head thickness.
The structural effect of this edge condition results in
reduction of the equivalent pressurized diameter on
the circular flat head from the normal diameter d to a
diameter of 0.893d so that Eq. (10.9) becomes
( )2
)2
(
d
d
= 0.15P
.
(10.11)
𝜎 = 0.188P 0.893
t
t
t=
From Figure 10.5, sketch b-2, the rule is
ts > 1.5 in.
rmin = 0.25ts = 0.25(1.667)
= 0.417 in.
Problems
10.3
A flat head is constructed according to Figure 10.5,
sketch d. The diameter d = 12 in., thickness
t = 1.25 in., E = 1.0, and S = 15,000 psi. What
is the MAWP (maximum allowable working
pressure)?
Answer:
MAWP = 1250 psi
10.4
A large flat head is made from pieces that are
welded together and spot-examined, so that
E = 0.85. The corner details are similar to those in
Figure 10.5, sketch f , with m = 1.0. The diameter
d = 60 in., S = 12,500 psi, and the internal pressure P = 300 psi. What is the minimum required
thickness?
Answer:
t min = 5.792 in.
10.5 Bolted Flanged Connections
The most usual type of joint for easy assembly and disassembly used in the process vessels and piping systems is
the bolted flanged connection. A convenient method to
design and calculate flanges with ring-type gaskets that
are within the bolt circle was first published by Taylor
Forge in 1937 [3]. These rules were further developed
[4] and incorporated into the ASME Code, VIII-1, some
years later. These rules, which are still used to calculate
this type of flange, are given in Appendix 2 of the ASME
Code, VIII-1.
Rules for calculating flat-face flanges with metal-tometal contact outside the bolt circle are given in
Appendix Y of the ASME Code, VIII-1. This design
incorporates a self-energizing-type gasket such as the
O-ring gasket. The original rules were restricted to analyzing identical pairs of flat-face flanges. Current rules
have been improved to permit analysis of both identical
and nonidentical pairs of flanges.
Further development of design rules in the ASME
Code, VIII-1 [5], came with the issuance of reverse
flange rules that use a ring-type gasket with no additional contact of the faces. These rules were added to
Appendix 2, ASME Code, VIII-1.
In addition to these rules for flange design in the code,
many designs are used for which there are no rules in
the ASME Code. One common type is the full-face gasket flange. There are many others that may be designed
10.7 Gaskets
for ASME Code approval by meeting the requirements
of U-2(g) of the ASME Code, VIII-1.
Before any flange design calculations are performed
for a vessel to be approved according to the code, the
designer should recognize that some calculations can be
avoided. If the flange is the type described in Appendix 2
of the ASME Code, the code permits using flanges with
recognized standards that establish items such as dimensional standards, materials, and pressure/temperature
ratings. The code accepts flanges designed according
to ANSI B16.5 “Pipe Flanges and Flanged Fitings,” [6]
API 605 “Large Diameter Carbon Steel Flanges” [7],
and ANSI B16.24 “Bronze Flanges and Fittings, 150 and
300 lb” [8]. Several other standards are not included;
however, when the flanges are selected by this method,
no additional calculations are required to satisfy the
ASME Code.
When calculations are necessary according to
Appendix 2 of the ASME Code, VIII-1, for a nonstandard design or when it is desired to upgrade a
standard flange, similar design calculations are required
for blind flanges (circular flat heads with bolts) and for
regular bolted flanges. Although each item is discussed
in greater detail in the following paragraphs, the basic
steps in designing a flange are as follows:
1) Establish the design pressure and design temperature.
2) Select the gasket material and dimensions and facing
type. Calculate N and b.
3) Calculate the loads for both gasket seating and operating conditions.
4) Determine the bolting sizes and gasket width check.
5) Establish the flange dimensions (usually using those
from a standard flange).
6) Using loads and dimensions, calculate the moments
for both gasket seating and operating conditions.
7) Determine the required thickness of flange.
10.6 Contact Facings
For ring-type gasket design as given in the ASME Code,
VIII-1, Appendix 2, several types of flange facings are
used. Some of the more usual types are the raised face,
the tongue-and-groove, and the lap joint. When these
types are used, the full-seal loading is taken by the gasket,
because no other part of the face is in 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 requires the adjacent faces to be in contact with
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 described in Appendix 2 and the
flat-face flanges with metal-to-metal contact outside of
the bolt circle as described in Appendix Y of the ASME
Code, VIII-1. This construction utilizes a self-energizing
or pressure-actuated O-ring gasket that is internally
pressurized to seal the gasket and does not depend upon
initial gasket seating by the bolts that cause compression
of the gasket (Figure 10.6).
There are also special types of gaskets and facing
designs that become self-sealing from the gasket rotation and deflection that are caused by contact loading
from a retaining ring and head closure. Some of these
are the delta gasket as used in Bridgeman closures,
the double-cone gasket, and the wedge gasket. In all
these cases, the initial gasket seating load is low. As the
pressure in the vessel is increased, the gasket rotates and
deflects into a special facing in which the sealing load
increases as the pressure increases. Care must be taken
with this type of closure because the gasket often seizes,
and it may be difficult to get the closure apart. In many
instances, the gasket may be silver, gold, or platinum
plated to help prevent the seizing.
10.7 Gaskets
A large variety of gaskets are needed in process equipment. The diverse processes, temperatures, pressures,
and corrosion environment require gaskets with different configurations, materials, and properties. Some of
the frequently used gaskets are as follows:
1)
2)
3)
4)
5)
6)
7)
8)
Rubber O-rings
Metallic O- and C-rings
Fiber
Flat metal
Spiral-wound
Jacketed
Metal ring
High-pressure type.
10.7.1
Rubber O-Rings
These gaskets, shown in Figure 10.7a, are used extensively in low-pressure applications such as storage tanks
and air receivers. They are normally confined in a groove
to prevent extrusion, and their maximum temperature
limit is about 250 ∘ F. Because the required seating stress
is negligible, the number of bolts needed in the flange is
kept to a minimum. A groove finish of 32 rms is usually
specified.
10.7.2
Metallic O- and C-Rings
The metallic O- and C-rings (Figures 10.7b,c) have a
wide range of applications for both external and internal
155
156
10 Blind Flanges, Cover Plates, and Flanges
(a)
Raised face
Facing details
(e) Self-energizing 1°
64 Rellet
I.D.
O.ring
Groove
smooth finish
(b)
Male and
female
Groovecold water finish
O°−5°
–O°
Metallic
Elastomer
Delta ring
I.D.
(c)
Tongue and
groove
I.D.
Spherical radius
Lens type
1°
20° ± 2
Ring and bevel
cold water
finish
(d)
Lap joint
(g)
Ring joint
(f)
Tongue and
groove
I.D.
D
D
I.D.
D
With seating
nubbin
Figure 10.6 Typical facing details.
pressures. They have a good springback characteristic
and a low seating stress. The gaskets may be manufactured from widely selected materials compatible with the
flange. This eliminates the problem of thermal expansion
between the gasket and the flange and increases their
applicable temperature range.
O- and C-gaskets seal along a contact line. Accordingly,
a finish of about 32 rms is needed in the flange seating
surface to properly seal the gaskets. In critical applications, a silver plating is specified to help sealing.
Metal O-rings are manufactured in three different
styles, shown in Figure 10.7c: the unpressurized, pressurized, and vented types. The unpressurized ring is used
at high temperatures where the increased pressure from
the sealed gas compensates for the loss of strength. The
vented ring is used at high pressures for better sealing.
10.7 Gaskets
For external press
For internal press
(a) Rubber O-Ring
With outer
ring
Plain
(b) C-Ring
With inner
ring
(f) Spiral wound
Plain
Unpressurized
Pressurized
Vent
Corrugated
Vented
(g) Jacketed
(c) Metal O-ring
Serrated
Circular
Oval
Plain
(d) Fiber
(e) Flat Metal
(h) Metal ring
Lens
Delta
Double cone
Bridgeman
(i) High-pressure gaskets
Figure 10.7 Types of gaskets.
157
158
10 Blind Flanges, Cover Plates, and Flanges
10.7.3
Compressed Fiber Gaskets
10.7.5
These gaskets (Figure 10.7d) normally consist of 70%
fiber, 20% rubber binder, and 10% filler material and
curative. They can be cut to fit various shapes and configurations such as heat exchangers with pass partitions
and oval and square openings. Thicknesses are normally
furnished between 1/64 and 1/4 in. and require a seating surface finish of about 250 rms. Fiber gaskets are
normally used for temperatures up to 850 ∘ F. A rule of
thumb for determining the adequacy of fiber gaskets for
a given temperature and pressure is to limit the product
of temperature in ∘ F times the pressure in psi to about
300,000.
10.7.4
Flat Metal Gaskets
These gaskets (Figure 10.7e) are made from a wide
variety of materials that can be cut from sheet metal to
any desired configuration and width. Some frequently
used gasket materials and their temperature limits are as
follows:
Maximum
Material
temperature (∘ F)
Lead
212
Aluminum
400
Brass
500
Copper
600
Titanium and zirconium
800
Carbon steel
900
Monel
1000
400 series stainless steel
1200
Nickel
1200
Inconel
1200
300 series stainless steel
1500
Incoloy
1500
Hastelloy
1800
Spiral-wound gaskets (Figure 10.7f ) are very versatile
and used in numerous applications. They are especially
suited for cyclic conditions where the excellent springback makes them ideal for repetitive loading. They are
manufactured to a desired width by spiral winding a
preshaped metal strip with a filler material between the
strips that consists of composites.
For most applications, spiral-wound gaskets are
retained in a groove. In raised-face flanges, an outer ring
is used to prevent the gasket from extruding and the
chevrons from excessive deformation. Sometimes, an
inner ring is also used to minimize erosion and to reduce
temperature fluctuation in cyclic conditions.
Many factors affect the performance of spiral-wound
gaskets such as tightness of wraps, material of filler
and strips, height of strips, diameter of opening, and
surface finish. The gasket seating surface finish is about
125 rms.
10.7.6
Jacketed Gaskets
These gaskets (Figure 10.7g) are normally used for pressures up to 500 psi in large-diameter vessels where flange
out-of-roundness and irregularities are large compared
with those in small flanges. They may be purchased
as plain or corrugated, and they seal at the inner and
outer laps. The outer metal jackets are made from a wide
variety of metals. The filler material is normally made of
composites or metal. The seating surface finish is about
63 rms.
10.7.7
Metal Ring Gaskets
The metal ring gaskets (Figure 10.7h) are used in
high-pressure and high-temperature applications. Their
small cross-sectional area makes them ideal for compact
flanges. The required high seating stress has the same
magnitude as the pressure stress. The rings are made
from many materials and are sometimes silver-plated
to improve sealing. The gasket groove finish is about
63 rms.
10.7.8
Flat metal gaskets need a high seating force for
proper seating. Accordingly, they are best suited for
high-pressure applications. The seating surface must
have a finish of about 63 rms.
Serrated gaskets require a smaller seating force compared to flat gaskets and thus are used in screwed flanges
where friction forces are to be minimized.
Spiral-Wound Gaskets
High-Pressure Gaskets
Lens, delta, double-cone, and Bridgeman configurations
(Figure 10.7i) are used in the majority of the pressure
applications where the seating stress is required 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 are made of softer materials compared to the seating surfaces to prevent damage
10.7 Gaskets
to the flanges or covers. In general, these gaskets are
expensive to fabricate and machine, require very tight
tolerances, and need very smooth seating surfaces of
16 rms or better.
High-pressure gaskets have a large surface that is
subjected to the vessel internal pressure. Accordingly, a
free-body diagram is normally necessary to determine
the additional forces transmitted to the flanges and bolts
resulting from pressure on the gasket.
The individual design requirements for lens, delta, and
double-cone gaskets are given in the next three subsections.
The outside thickness of the gasket is established to
allow for a 0.25 in. clearance plus 0.0625 in. for a centering ring, if required. The pitch diameter for gasket
seating reaction is established as
1
G = (ID) + (OD − ID),
3
where
G = diameter of gasket reaction
The spherical radius of the gasket surface is taken as
R=
G∕2
,
sin 𝜃
where
10.7.9
Lens Ring Gaskets
R = spherical radius of gasket surface (in.)
The lens ring gaskets shown in Figure 10.8 are normally
used in small flanges. The cross-sectional area of the ring,
through points a to b, must be equal to or larger than the
bolt area of the flange to prevent crushing of the gasket.
Thus, knowing the inside diameter of the gasket and the
required bolt area, the outside diameter can be calculated
from
[
]1∕2
4
OD = Ab + (ID)2
,
π
where
𝜃 = angle of friction (for mild steel, it is 20∘ )
From the geometry, the inside thickness of the gasket
is calculated from
]
[√
√
)2
( )2
(
5
ID
OD
t=
,
R2 −
− R2 −
+2
16
2
2
where t = inside thickness of gasket (in.).
The width of gasket seating is normally
N=
OD = outside diameter of ring gasket (in.)
ID = inside diameter of ring gasket (in.)
Ab = summation of actual bolt areas in flange (in.2 )
where N = gasket seating width (in.).
Figure 10.8 shows the flange and cover surfaces in the
vicinity of a gasket machined to have a slope of 20∘ .
10.7.10
20°
G
a
b
5/16 in.
t
I.D.
(Ab )(1.5)(design bolt stress)
,
π(G)(3)(yield strength of gasket material)
Delta Gaskets
The delta ring gaskets shown in Figure 10.9 are extensively used for high-pressure applications in the United
States. These gaskets rely on the inner pressure to wedge
them in the gasket groove for sealing and thus do not
require any initial seating or bolting stress. The general
dimensions that are shown in Figure 10.9 apply to rings of
all diameters. The pitch diameter G is normally taken as
G = ID + 0.125,
1/16 in.
and the gasket seating width N is usually equal to 0.125 in.
O.D.
R
10.7.11
θ°
Figure 10.8 Lens gasket.
Double-Cone Gaskets
Double-cone gaskets are very popular in Europe and can
be fabricated in various sizes. A typical detail is shown in
Figure 10.10. The required ouside diameter of the gasket
is given by
]1∕2
[
4
,
OD = Ab + (ID)2
π
159
10 Blind Flanges, Cover Plates, and Flanges
G
I.D.
r = 1/16 in.
45°
0.75 in.
160
G
45°
I.D.
h
a
60°
b
O.D.
15°
O.D.
49°
60°
Figure 10.9 Delta gasket.
where
Figure 10.10 Double-cone gasket.
OD = outside diameter of gasket (in.)
ID = inside diameter of gasket (in.)
Ab = summation of actual bolt area in flange (in.2 )
The seating length N is determined from
N=
(Ab )(1.5)(design bolt stress)
,
∘
(cos 60 )(π)(G)(3)
(yield strength of gasket material)
and the pitch diameter 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,
( )
∘
N
(2)(Y )
(sin 60 ) = (P)(h)
2
or
∘
(Y )(N)(sin 60 )
h=
,
P
where
Y = seating stress of gasket material (psi)
P = internal design pressure (psi)
10.7.12
Gasket Design
Gasket design characteristics depend upon the material
and the design of the gasket itself. The gasket factor m and
the minimum design seating stress y are both related to
the gasket type and the gasket material. Although m and
y have been included in the ASME Boiler and Pressure
Vessel Code since the 1942 edition, they are suggested
values only and are not mandatory. The original testing
and development of m and y are described in an article
by Rossheim and Markl [9] that does not give the reasons
for the specific values. Very few changes have been made
in these factors since they were originally published.
As a result of many inquiries to the ASME Code
Committee regarding the validity of the m and y factors,
a large-scale investigation has been undertaken by the
Pressure Vessel Research Committee [10] of the Welding
Research Council. As experimental tests progressed, it
became obvious that the m and y factors are related to
many items not previously considered. There is a close
correlation with the amount of tightening of the bolts,
the gasket type, and the material, for they are all related
to the leakage rate of the joint.
10.8 Bolting Design
Once the gasket type and material have been selected,
the effective gasket width for calculation may be determined. For solid flat metal and for the ring-type joints,
the basic gasket seating width b0 is found by the formulas in column I of Table 2-5.2 of the ASME Code, VIII-1,
whereas for all other types of gaskets, b0 is determined
by the formulas in column II. The effective gasket seating
width b is found by applying the following rules:
Solution:
Determine the effective gasket seating √
width as follows:
N = 1.0 in., b0 = N/2 = 0.5 in., b = 0.5 b0 = 0.3535 in.,
effective gasket diameter
G = 13.75 + (2 × 1) − (2 × 0.3535)
G = 15.043 in.
The gasket loadings are as follows:
H = 0.785G2 P = 0.785(15.043)2 (2500)
= 444,100
1
when b0 ≤ in. and
4
√
1
b = 0.5 b0 when b0 > in.
4
b = b0
Hp = 2bπGmP = 2(0.3535)π(15.043)(3)(2500)
= 250,600
With b determined, the location of the line of gasket load
reaction, as well as the values of G and hG , can be determined for calculating the flange moments.
In designing a flange, it is important to recognize that
two design conditions exist – the gasket seating and the
operating conditions. The gasket seating condition exists
when an initial load is applied by the bolts to seat the
gasket at ambient temperature with no internal pressure.
The minimum initial bolt load is
Wm2 = πbGy.
(10.12)
The operating condition exists when the hydrostatic
end force from the internal design pressure tends to pen
the joint, for the gasket retains enough resilience to keep
the joint tight. Loadings and stress are determined at
design pressure and design temperature. The loading for
the operating condition is
Wm1
π
= H + Hp = G2 P + πmGP(2b).
4
(10.13)
To avoid crushing of 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
area Ab is selected, a check may be made to determine
the required minimum gasket width by the following
formula:
Nmin =
Ab Sa
.
2yπG
(10.14)
Example 10.5
A vessel has the following design data: design pressure P = 2500 psi; design temperature = 250 ∘ F; a
spiral-wound metal, fiber-filled stainless-steel gasket
with inside diameter 13.75 in. and width N = 1.0 in.
The gasket factors are m = 3.0 and y = 10,000. Bolts are
SA-325 Grade 1 with Sa = Sb = 19,200 psi. Is the gasket
sufficiently wide to keep from crushing out?
Wm1 = H + Hp = 444,100 + 250,600 = 694,700
Wm2 = πbGy = π(0.3535)(15.043)(10,000)
= 167,100.
Since Sa = Sb = 19,200 psi, W m1 sets the bolting area Am
as
694,700
Am =
= 36.182 in.2
19,200
Ab = actual bolt area = 36.8 in.2 for sixteen 2-in. diameter bolts. The minimum gasket width is
36.8(19,200)
2(10,000)π(15.043)
= 0.748 in. versus 1 in.actual.
Nmin =
A 1 in. wide gasket is sufficient to prevent crushing.
Problem
10.5
A solid, flat, stainless-steel gasket with m = 6.5
and y = 26,000 is used in the vessel described
in Example 10.5. The preliminary gasket inside
diameter is 12 in., and the gasket width is 1.0 in.
What is the gasket seating load?
Answer:
W m2 = 384,000 lb.
10.8 Bolting Design
In designing bolting for flanges, one first selects the
bolting material. It must be compatible with the flange
material. That is, there must not be any chemical or
galvanic action between the bolting and flange material
that would cause the bolts to seize in the threads. Under
certain circumstances, it may be necessary to plate the
bolts or to make them from special material to protect
them from the environment. Although it is not necessary to select a bolting material with a tensile strength
161
162
10 Blind Flanges, Cover Plates, and Flanges
close to that of the flange material, one should carefully
consider the effects of strain elongation and relaxation
of bolting materials that have a high tensile strength and
require a smaller cross-sectional area. In addition, when
high-strength material is used for the bolting, care must
be taken not to reduce the number of bolts so far that
the bolt spacing becomes excessive [11].
When the bolt spacing exceeds 2d + t, secondary flange
bending occurs between the bolts to the extent that it
affects the normal flange bending. To take account of this
effect, the flange bending moment M0 must be increased
by the factor
√
actual bolt spacing
,
(10.15)
2d + t
where
Example 10.6
A vessel flange uses 16 2 in. diameter bolts. Flange stress
calculations indicate that a flange thickness of t = 4.5 in.
is adequate. The bolt circle diameter is C = 22.5 in. Will
secondary bending stresses be developed?
d = nominal diameter of bolts (in.)
Example 10.7
Suppose that a vessel requires 24 2 - 1/2 in. diameter bolts
on a flange that is 5.5 in. thick. What is the maximum bolt
circle that will not cause secondary bending stresses? The
minimum bolt spacing for 2 1/2 in. diameter bolts is 5
1/4 in.
t = flange thickness (in.)
To determine the total required cross-sectional area of
bolting, both the gasket seating and operating conditions
must be examined.
The minimum required bolting area Am is the greater
of
Wm2
Wm1
or
,
Sa
Sb
Solution:
The maximum permissible bolt spacing without a penalty
is 2d + t = 2 × 2 + 4.5 = 8.5 in. The actual bolt spacing is
π(22.5)
πC
=
= 4.42 in.
N
16
Because the actual spacing is less than the maximum
spacing without a penalty, no secondary bending stresses
are developed.
Solution:
The maximum spacing is
1
2d + t = 2 × 2 + 5.5 = 10.5 in.
2
where
Sa = allowable bolting stress at room temperature
(psi)
Sb = allowable bolting stress at design temperature
(psi)
The diameter of the bolt circle is
10.5 × 24
= 80.2 in.
π
based on maximum spacing.
W m1 = operating load = H + H p (see Eq. (10.13))
W m2 = gasket seating load
From this minimum required bolting area Am , the
actual bolting area Ab is selected. In order to obtain
a bolt loading for calculating the moment for gasket
seating, the minimum required bolting area and the
actual bolting area are averaged as follows:
W = 0.5(Am + Ab )Sa .
Problems
10.6
Sixteen bolts of 1 1/4 in. diameter are to be located
on a bolt circle of G = 32 in. The flange is 2 1/2 in.
thick. What is the factor that is due to secondary
flange bending?
(10.16)
At certain times during the operation of a process
vessel, the bolts in a bolted flanged connection are subjected to actual stresses in excess of the allowable design
stresses. This may be especially true during hydrostatic
testing. Care must be taken to ensure that during this
testing, no permanent elongation of the bolting has
occurred. If so, the bolting may have to be replaced
before the vessel is put into service. Realizing this is
especially important if each of two suppliers provides
half of the bolted flanged assembly, and one does not
know what bolting is supplied.
Answer:
1.12
10.7
The minimum spacing for the wrench to fit 1
1/2 in. diameter bolts is 3 1/4 in. Twelve bolts are
to be used on a bolt circle of 15 1/4 in. diameter.
What is the minimum flange thickness that does
not cause secondary flange bending?
Answer:
1.0 in.
10.9 Blind Flanges
W
t
hG
W
P
G=d
P
For the gasket seating condition, the internal pressure
equals zero, and the only load is the gasket seating
load W a at ambient temperature with the allowable
tensile stress of Sa . The equation for the gasket seating
condition is
√
1.9Wa hG
t=G
.
(10.21)
Sa EG3
For the operating condition, the internal pressure P, as
well as the gasket loading, is applied. For this condition,
W m1 = H + H p at operating temperature with an allowable tensile stress of Sb . The equation for the operating
condition is
√
0.3P 1.9Wm1 hG
t=G
.
(10.22)
+
Sb E
Sb EG3
Hp
M0
Figure 10.12 shows a sample calculation for a blind
flange.
Figure 10.11 Loadings on blind flange.
10.9 Blind Flanges
The minimum required thickness of a circular, unstayed
flat-head or blind flange attached by bolts and utilizing a ring-type gasket that causes an edge moment is
derived from the assumption that the flat plate is simply
supported at the gasket load line G and is loaded by a
gasket seating load or a combination of gasket loading
and uniform pressure loading. The combination of these
loadings at the gasket and at the bolt circle causes an
edge moment of M0 /𝜋G, as shown in Figure 10.11. If the
edge moment M0 is assumed to be equal to WhG , the
theoretical stress at the center of the flat plate is
)
(
WhG
3(3 + 𝜇)P ( G )2
6
+
S=
.
(10.17)
32
t
πG
t2
Setting 𝜇 = 0.3 and E = butt-weld joint efficiency
within the flat plate,
(
)
( )2
WhG
G
SE = 0.31P
+ 1.91
.
(10.18)
t
Gt 2
Solving Eq. (10.18) for t gives
√
0.31P 1.91WhG
t=G
.
+
SE
SEG3
(10.19)
This is identical with the equation in the ASME Code,
VIII-1, except that the constants 0.3 and 1.9 are used in
the code instead of the exact constants 0.31 and 1.91, and
the gasket load G is substituted for d. The general ASME
Code equation for circular flat-bolted heads is
√
0.3P 1.9WhG
.
(10.20)
+
t=G
SE
SEd3
Example 10.8
Considering the pressure vessel described in Example 10.5,
the vessel is to have one end closed by a blind flange.
What is the minimum required thickness of the blind
flange? Design data are as follows:
Design pressure P = 2500 psi
Design temperature = 250 ∘ F
Flange material is SA-105
Bolting material is SA-325 Gr. 1
No corrosion exists.
Allowable bolt stress at gasket seating and operating conditions = 19,200 psi
Allowable flange stress at gasket seating and operating
conditions = 17,500 psi
Gasket is spiral-wound metal, fiber-filled, stainless steel,
with inside diameter 13.75 in. and width N = 1 in.
Solution:
Following the information calculated in Example 10.5,
once the actual bolt area Ab is found, the design loading
for the gasket seating condition W a can be determined
as
Wa = 0.5(Am + Ab )Sa = 0.5(36.2 + 36.8)(19,200)
Wa = 700,800.
The moment arm is determined from
hG = 0.5(C − G) = 0.5(22.5 − 15.043) = 3.729 in.
From Example 10.5, the design loading for operating
condition is W m1 = 694,700.
163
164
10 Blind Flanges, Cover Plates, and Flanges
Figure 10.12 Blind-flange sample calculation sheet. Source: Courtesy of G + W Taylor – Bonney Div., Taylor Forge.
The minimum required thickness is determined as the
greater thickness of that determined for the gasket seating load according to Eq. (10.21) and for the operating
load according to Eq. (10.22). For gasket seating,
√
1.9Wa hG
t=G
Sa EG3
√
1.9(700,800)(3.729)
= 15.043
= 4.343 in.
(17,500 × 1)(15.043)3
For the operating condition,
√
0.3P 1.9Wm1 hG
t=G
+
Sb E
Sb EG3
√
1.9(694,700)(3.729)
0.3(2500)
= 15.043
+
17,500 × 1.0 (17,500 × 1.0)(15.043)3
= 5.329 in.
Therefore, the minimum required flange thickness is
t = 5.329 in.
Problems
10.8
Suppose that the flat head mentioned in
Example 10.8 is made by butt-welding flat plates
together. The welds are spot-examined, so that
E = 0.85. What is the minimum required head
thickness?
Answer:
t min = 5.780 in.
10.9
Suppose that the bolt circle diameter is increased
to C = 24 in. What is the minimum required head
thickness, considering both gasket seating and
operating conditions?
Answer:
t min = 5.671 in.
10.10 Bolted Flanged Connections
with Ring-Type Gaskets
The design rules for bolted flanged connections with a
ring-type gasket that is entirely within a circle enclosed
by the bolt holes and with no contact outside of the bolt
circle are given in the ASME Code, VIII-1. According to
the code rules, an acceptable design of a bolted flanged
connection may be accomplished by two means: standard rated flanges and flange design calculations. The
easiest way to select a suitable flange is to choose one that
10.10 Bolted Flanged Connections with Ring-Type Gaskets
has been dimensionally standardized, uses standardized
materials, and has pressure–temperature ratings previously determined. The ASME Code presently permits
the use of standard flanges as follows: ANSI B16.5, “Pipe
Flanges …,” [6] API 605, “Large Diameter Carbon Steel
Flanges,” [7] and ANSI B16.24, “Bronze Flanges . . . .” [8]
When a standard flange is selected from one of these
specifications, regardless of the m and y factors used
for the gasket, no additional calculations are required.
Essentially, any variations in flange stresses are incorporated by using some of the factors of safety in the
allowable tensile stress.
When it is necessary to design a bolted flanged connection because no standard flange of the proper size is
available, the standard pressure–temperature ratings are
not adequate, or special design factors are to be used for
the gasket, the procedure in Appendix 2 of the ASME
code is used. The design of flanges, bolting, and gaskets
by the ASME Code rules is essentially a trial-and-error
procedure where some dimensions are set and remain
fixed, whereas other dimensions such as the flange thickness are varied. Stresses in the flange and the hub are
calculated. If any of these stresses exceeds the allowable
tensile stresses, a “new” flange thickness is selected, and
the stresses are recalculated until they are satisfactory.
The trial-and-error method is essentially due to the complex theory used by Waters, Rossheim, Wesstrom, and
Williams to solve the problem in the original development. The WRWW method, which was ultimately incorporated into the ASME Code in 1940, is an elastic analysis
of the interaction between the vessel or pipe, the hub, and
the flange ring assembly. The shell and hub are resolved
by a discontinuity analysis that was previously described
in Chapter 5, and the flange ring is considered as a flat
plate with the center part removed (Figure 10.13). Interactions of rotations and deflections are permitted until
the balance is obtained.
The basic assumptions in the analysis are that the
flange materials are elastic (i.e. no creep or plastic
yield at lower temperature occurs), the bolt loading is
assumed or determined from the gasket factors, and the
moments due to loadings are essentially constant across
the width of the flange. In addition, rotation of the flat
Figure 10.13 Flange loadings for elastic
analysis.
W
W
(a) Moment loading
HD
HD
HD
HD
HD
HD
π Bp
π Bp
π Bp
(b) Direct loading
W
HG
HD
HD
HD HD
π Bp
π Bp
HT
HD
π Bp
165
166
10 Blind Flanges, Cover Plates, and Flanges
plate is assumed to be linear, with no dishing effect, and
superposition is acceptable.
Calculating the stress in a flange using theoretical
equations is a complex problem. This problem is simplified for code use by curves, formulas, and tables that
contain constants depending upon the geometry of the
flange assembly. Formulas for various coefficients are
given in the code that permits computer programming
of the basic equations for rapid solution.
The code classifies flanges with ring-type gaskets into
three types for analysis: integral, loose, and optional.
Integral means that the pipe, hub, and ring are one
continuous assembly from their original manufacture: either forgings or castings welded together by
full-penetration welds. Loose means no attachment of
the assembly to the pipe: no ability of the juncture to
carry shears and moments other than those required
to seal against pressure. Such flanges are classified as
slip-on, lap joint, and threaded, and they may or may
not have hubs. Optional means flange designs that,
by construction, are integral, but for which the analysis is permitted by the simpler method for loose-type
flanges. Examples of these various types are depicted in
Figure 10.14.
The calculation of a flange with a ring-type gasket
involves first selecting the material for the flange, bolts,
and gasket in a manner very similar to the blind flange.
Next, the facing and gasket details are set, the loads due
to internal pressure are determined, and the required
bolting area and bolt sizes are selected. The bolt circle
is then decided, and the loads, moment arms, and
moments due to both gasket seating and operating
Loose-type fianges
t
Gasket
h
>
t1 = tn
W
A
g1
hG or hT
HT
r hD
C
HD
G
A
Full-penetration weld
single or double
Gasket
g0
B
(a)
hG
W
r
hT
G HT
tn
g1
HG
hD
g0
C
HD
B
(c)
(b)
(d)
h
t
Gasket
t1
(e)
Integral-type fianges
Gasket
t
hG
w
r R
hD C
hT
HT
A
HD
G HG
B
g1 = g0
g1/2
hG
W
r
hT
HT
h > 1.5 g0
t
Gasket
G H
G
(f)
HD
R
hD
g1
B
Gasket
Slope
1:3 (max.)
B1/2
t
h
w
A h
G
hT
HG
C
HT
g0
(g)
G
R h C
D
HD
g0
B
g1
g1/2
(h)
Optional-type fianges
these may be calculated as either loose. or integral·typo
(i)
(j)
(k)
(l)
Figure 10.14 Types of flanges. Source: Courtesy of American Society of Mechanical Engineers, from Figure 2-4 of the ASME Code, VIII-1.
10.10 Bolted Flanged Connections with Ring-Type Gaskets
conditions are determined as with the blind flange. By
knowing these and the geometry used to determine K
and other hub coefficients, stress calculations are made
for both conditions. The longitudinal hub stress, the
radial flange stress, the tangential flange stress, and
their various combinations are compared with allowable
stresses.
The method of calculation is virtually identical for
welding neck flanges and slip-on or lap-joint flanges
except that the axial pressure load is applied at a slightly
different location. For the ring flange design, the tangential flange stress is the only one calculated. The minimum
required thickness can be directly determined from
√
M0 y
.
(10.23)
t=
Sa B
Additional requirements for the deflection of the
flange between the bolts are given in the ASME VIII-1
and VIII-2 codes. Limiting the deflection between the
bolts prevents the flange from leakage due to internal
pressure and prevents the bolts from being spaced too
far apart.
Example 10.9
What is the minimum required thickness of a welding
neck flange as shown in Figure 10.15 with the following
design data? (Note: These data are the same as those
used for the blind flange in Example 10.8. In Figure 10.16
is a sample calculation for a welding neck flange.)
Design pressure P = 2500 psi
Design temperature = 250 ∘ 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; inside diameter is 13.75 in., and width is
1.0 in.
1) Allowable bolt stress at design and seating temperatures = Sb = 19,200 psi
2) Allowable flange stress at design and seating temperatures = Sf = 17,500 psi
3) Gasket dimensions are as follows:
√
b0 = N∕2 = 0.5 in. and b = 0.5 b0 = 0.3535
G = 13.75 + (2 × 1) − (2 × 0.3535) = 15.043 in.
4) Determine the bolt loadings and sizing of bolts with
N = 1, b = 0.3535, y = 10,000, m = 3.0:
π
π
H = G2 P = (15.043)2 (2500) = 444,323
4
4
Hp = 2bπGmP = 2(0.3535)
π(15.043)(3.0)(2500) = 250,591
Wm1 = H + Hp = (444,323) + (250,591)
= 694,914
Wm2 = πbGy = π(0.3535)(15.043)(10,000)
= 167,060
Am = the greater of Wm1 ∕Sbh = (694,914)∕
(19,200) = 36.2 in.2 or
Wm2 ∕Sbc = (167,060)∕(19,200) = 8.7 in.2
Ab = actual bolt area = 36.8 in.2
16 bolds at 2 in.diameter
Wa = 0.5(Am + Ab )Sbc = 0.5(36.2 + 36.8)
(19,200) = 700,800
Wb = Wm1 = 694,914.
5) Calculate the total flange moment for the design condition. Flange loads:
π
π
HD = B2 p = (10.75)2 (2500) = 226,906
4
4
HG = Hp = 250,591
HT = H − HD = (444,322) − (226,906)
= 217,417.
Solution:
Figure 10.15 Flange dimensions mentioned
in Example 10.9.
g0 = 1''
g1 = 3.375''
h = 6.25''
Nominal pipe size
12'' Sc h.120
E = 2''
R = 2.5''
B = 10.75''
W
HD
hD
t = 4.5''
C = 22.5''
hT
16 – 2''· bolts
hC
HC
HT
G = 15.043''
A = 26.5''
167
168
10 Blind Flanges, Cover Plates, and Flanges
Figure 10.16 Welding-neck-flange sample calculation sheet. Source: Courtesy of G + W Taylor – Bonney Div., Taylor Forge.
Lever arms:
Flange moments:
hD = R + 0.5g1 = (2.5) + 0.5(3.375) = 4.1875
hG = 0.5(C − G) = 0.5(22.5 − 15.043) = 3.7285
MD = HD × hD = (226,910)(4.1875) = 950,170
MG = HG × hG = (250,590)(3.7285) = 934,330
hT = 0.5(R + g1 + hG ) = 0.5(2.5 + 3.375 + 3.7285)
= 4.8018.
MT = HT × hT = (217,417)(4.8018) = 1,043,990
Mde = MD + MG + MT = 2,928,490.
10.10 Bolted Flanged Connections with Ring-Type Gaskets
6) Calculate the total flange moment for bolt-up condition. Flange load:
10) Allowable stresses:
SH ≤ 1.5Sf ∶
Lever arm:
hG = 0.5(C − G) = 3.7285.
Mbu = HG × hG = (700,800)(3.7285) = 2,612,930.
7) Use the greater of Mde or Mbu (Sh /Sc ); M0 = 2,928,490.
8) Shape constants from the ASME Code, VIII-1,
Appendix 2 : K = A/B = 26.5/10.75 = 2.465. From
Figure 2-7.1, Section VIII-1,
T = 1.35, Z = 1.39, Y = 2.29, U = 2.51
g1 ∕g0 = 3.375∕1.0 = 3.375
√
√
h0 = Bg 0 = (10.75)(1.0) = 3.279
h∕h0 = 6.25∕3.279 = 1.906.
From Figure 2-7.2, Section VIII-1, F = 0.57.
From Figure 2-7.3, Section VIII-1, V = 0.04.
From Figure 2-7.6, Section VIII-1, f = 1.0
(1.5)(17,500) = 26,250
> 13,570 psi
SR ≤ Sf ∶ 17,500 > 15,590 psi
ST ≤ Sf ∶ 17,500 > 9140 psi.
HG = Wa = 700,800.
Flange moment:
169
Example 10.10
What is the minimum required thickness of a ring flange
with the same design data as given in Example 10.9? The
inside diameter has been increased to fit over the outside
of the shell to where B = 12.75 in. The bolt loadings and
bolt size are the same as mentioned in Example 10.9. A
sample calculation sheet is shown in Figure 10.17.
Solution:
1) Calculate the total flange moment for design condition. Flange loads:
π
π
HD = B2 P = (12.75)2 (2500) = 319,200
4
4
HD = Hp = 250,600
HT = H − HD = 444,300 − 319,200 = 125,100.
Lever arms:
hD = 0.5(C − B) = 0.5(22.5 − 12.75) = 4.875 in.
e = F∕h0 = 0.57∕3.279 = 0.1738
U
2.51
d = h0 g02 =
(3.279)(1)2 = 205.76.
V
0.04
9) Calculate the stresses. Assume a flange thickness
t = 4.5 in.
hG = 0.5(C − G) = 0.5(22.5 − 15.043) = 3.729 in.
hT = 0.5(hD + hG ) = 0.5(4.875 + 3.729) = 4.302 in.
Flange moments:
MD = HD × hD = (319,200)(4.875) = 1,556,000
3
L=
MG = HG × hG = (250,600)(3.729) = 934,500
MT = HT × hT = (125,100)(4.302) = 538,200
te + 1 t
+ = 1.320 + 0.443 = 1.763.
T
d
Longitudinal hub stress:
SH = fM0 ∕Lg 21 B = (1)(2,928,490)∕
×(1.763)(3.375)2 (10.75)
SH = 13,570 psi.
Radial flange stress:
(
)
4
SR =
te + 1 M0 ∕Lt 2 B
3
= (2.0428)(2,928,490)∕(1.763)(4.5)2 (10.75)
Mdc = MD + MG + MT = 3,029,000.
2) Bolt-up moment is the same as mentioned in
Example 10.9, MBU = 2,613,000.
3) Shape constants are K = A/B = 26.5/12.75 = 2.078.
From Figure 2-7.1 of the ASME Code, VIII-1,
Y = 2.812.
4) Required thickness based on design condition is
√
√
Mdc Y
(3,029,000)(2.812)
=
= 6.179 in.
t=
Sf B
(17,500)(12.75)
= 15,590 psi.
Tangential flange stress:
YM0
− ZSR
t2 B
(2.29)(2,928,490)
=
− (1.39)(15,590)
(4.5)2 (10.75)
= 9140 psi.
ST =
Problem
10.10
Suppose that a solid flat 2 1/4 chrome steel alloy
gasket with 13.75 in. inside diameter and width
N = 1/2 in., m = 6.0, and y = 21,800 is used with
the flange mentioned in Example 10.9. What are
the gasket seating and the operating loads?
170
10 Blind Flanges, Cover Plates, and Flanges
Figure 10.17 Ring-flange sample calculation sheet. Source: Courtesy of G + W Taylor – Bonney Div., Taylor Forge.
Answer:
Design condition is W m1 = 734,500 lb.
Seating condition is W m2 = 244,000 lb.
10.11 Reverse Flanges
Rules for the design of reverse flanges are given in
Appendix 2 of the ASME Code, VIII-1 [5]. This type of
flange is often used to form a reducing joint. The solution
for the reverse flange is similar to that for the raised-face
standard flange with the ring-type gasket within the bolt
circle, with some minor differences. Figure 10.18 shows
some loads that are applied in the reverse direction.
This may cause some of the moments to be applied in
the opposite direction from those loads on a regular
flange. However, the analysis is the same after the “new”
total moment is determined. Again, the moments are
determined for both the gasket seating condition and
the operating condition. Figure 10.19 shows a sample
calculation sheet for a reverse flange.
Additionally, a new term 𝛼 R is introduced to convert
some terms from regular flanges to reverse flanges; h0
and K are redefined and based on the reverse-flange
inside diameter; and a new equation is added to calculate
the tangential flange stress at the inside flange bore.
A special precaution is noted. When K ≤ 2, the results
are fairly satisfactory; however, when K > 2, they become
increasingly conservative. For this reason, the ASME
Code procedure is limited to K ≤ 2.
Derivation of the new equations for reverse flanges is
similar to that for the regular flanges except that shears
and moments are applied at the outer edge of the ring
flange where discontinuities occur between the flange
and the hub. With K = A/B′ , the conversion term 𝛼 R is
determined for converting T, U, and Y to T r , U r , and Y r ,
which is obtained as
[
]
3(K + 1)(1 − 𝜇)
1
𝛼r = 2 1 +
.
(10.24)
K
Y
Substituting this expression into the regular equation
for tangential stress, we have
[
]
M0
2K 2 (1 + 0.67te)
.
(10.25)
𝜎T = 2 ′ Y −
t B
𝜆(K 2 − 1)
Example 10.11
A reverse flange is joined to a regular ring-type joint
flange to form a reducing connection. The total bolt-up
moment is controlling and equal to M0 = 2,613,000.
The flange bore B′ = 13.25 in., the outside diameter
A = 26.5 in., and the flange thickness t = 5 in. What is the
tangential flange stress at the hub and at the inside bore?
10.12 Full-Face Gasket Flange
Figure 10.18 Reverse-flange loading and
dimensions.
HG
W
B′
hG
G
hT
hD
HD
HT
HD
g1
B
g0
A
Solution:
The tangential flange stress ST at the hub is
(
)]
M [
2
Z
1 + te
ST = ′ G2 Yr −
Bt
𝜆
3
[
(
)]
2,613,000
1.67
2
1.241 −
1 + × 5 × 0.131
ST =
13.25(25)
4.14
3
= 5220 psi.
The tangential flange stress ST′ at the flange bore is
(
)
⎡
⎤
2
2
1
+
te
2K
M
⎢
⎥
3
G
ST′ = ′ 2 ⎢Y −
Bt ⎢
(K 2 − 1)𝜆 ⎥⎥
⎣
⎦
)
(
⎡
⎤
2
×
5
×
0.131
2(4)
1
+
⎥
3
2,613,000 ⎢
′
ST =
2.96 −
⎢
⎥
13.25(25) ⎢
(3)(4.14)
⎥
⎣
⎦
= 16,050 psi.
Example 10.12
With the reverse flange given in Example 10.11, what is
the minimum required thickness based on an allowable
flange stress of 17,500 psi?
Solution:
The tangential flange stress at the flange bore of
16,050 psi is controlling. Because t appears in several terms in a nonlinear manner, the easiest way to
select the proper thickness is by trial and error. For the
initial trial, use a square relationship as follows:
16,050 17,500
=
t2
(5)2
or
t = 4.8 in.
The value of 𝜆 is recalculated as 3.855. Using this,
the ST′ = 16,900 psi. By successive recalculations, the
thickness is approximately t = 4.68 in., which gives
ST′ = 17,500 psi.
Problem
10.11
Using the details of the flange described in
Example 10.11, what is the minimum required
thickness if the material of the flange is changed
to one with an allowable stress of 15,000 psi?
Answer:
t min = 5.26 in.
10.12 Full-Face Gasket Flange
One type of flange that is frequently used but for which
no design methods exist in the ASME Code is the
flange using a full-face gasket, as shown in Figure 10.20.
171
172
10 Blind Flanges, Cover Plates, and Flanges
Figure 10.19 Reverse-flange sample calculation sheet. Source: Courtesy of G + W Taylor – Bonney Div., Taylor Forge.
Figure 10.21 shows a sample calculation sheet. This
type of flange is designed according to the provisions of
U-2(g) of the ASME Code, Section VIII, Division 1. This
code permits using good engineering design for those
constructions where no rules exist in the code. Although
the analysis is similar to that used for a raised-face,
ring-type flange, a countermoment is introduced from
that part of the gasket that is outside of the bolt circle.
In addition, the decrease in section strength at the bolt
circle from the bolt holes must be considered when the
radial stress at the bolt circle is determined.
This type of gasket is usually limited to designs where
a “soft” gasket (with low m and y factors) is used and the
design pressure is low. This is necessary to keep the loads
and bolt size such as to fit within the flange geometry,
even though the countermoment usually results in a
low flange moment and a minimum required flange
thickness.
10.12 Full-Face Gasket Flange
becomes
A−C
,
2(
)
2𝛼
,
b = πA
( 360 )
2𝛼
c = πC
,
360
y = h′G .
d=
G
h G'
G
hG
A
HG'
hG'
hG
HG
HT
G
hT
C
hD
HD
W
B
Figure 10.20 Full-face gasket loadings.
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 distances or moment arms from the bolt
circle to the centroid of the annulus from the bolt circle
to the outside diameter and from the bolt circle to the
inside diameter. In solving for the distances, the angle is
assumed to be small, and the arc lengths are evaluated as
straight lines. From Roark [12] and using his terminology,
the basic equation is
(
)
d 2b + c
.
(10.26a)
y=
3 b+c
However, converting to the terminology used in the
ASME Code and as shown in Figure 10.22, the equation
(10.26b)
(10.26c)
(10.26d)
(10.26e)
Therefore, substituting Eqs. (10.26b)–(10.26e) into Eq.
(10.26a),
[
]
A − C 2πA(2𝛼∕360) + πC(2𝛼∕360)
h′G =
2 × 3 πA(2𝛼∕360) + πC(2𝛼∕360)
(A − C)(2A + C)
h′G =
.
(10.27)
6(A + C)
In a similar manner, the distance or moment arm
toward the inside diameter hG is determined as
(
)
d 2b + c
C−B
− hG =
,
(10.26f)
2
3 b+c
C−B
,
(10.26g)
d=
2(
)
2𝛼
,
(10.26h)
b = πC
360
(
)
2𝛼
c = πB
.
(10.26i)
360
Substituting Eqs. (10.26g)–(10.26i) into Eq. (10.26f)
gives
[
]
C −B
C − B 2πC(2𝛼∕360) + πB(2𝛼∕360)
−hG =
,
2
2 × 3 πC(2𝛼∕360) + πB(2𝛼∕360)
and solving for hG gives
hG =
(C − B)(2B + C)
.
6(B + C)
(10.28)
Once the two gasket loadings and two moment arms
are determined, the analysis of the flange is the same
as for other flanges. The method is equally applicable
to integral flanges, loose flanges, reverse flanges, and
any other type of flange. It is important to remember
to use a “soft” gasket that keeps the bolt loading within
acceptable limits.
Example 10.13
A welding-neck flange with the same geometry as that
mentioned in Example 10.9 except for the thickness
is used with a full-face gasket. The design pressure is
320 psi, and the “soft” gasket is vegetable fiber with
m = 1.75 and y = 1100. What is the minimum required
thickness?
173
174
10 Blind Flanges, Cover Plates, and Flanges
Figure 10.21 Full-face-flange sample calculation sheet.
Solution:
1) Determine the lever arms of the inner and outer parts
of the gasket:
hG =
(22.5 − 10.75)(2 × 10.75 + 22.5)
6(10.75 + 22.5)
hG = 2.5915 in.
(A − C)(2A + C)
h′G =
6(C + A)
=
(C − B)(2B + C)
6(B + C)
RR
10.12 Full-Face Gasket Flange
B
hG' hG
C
A
Figure 10.22 Full-face gasket dimensions.
(26.5 − 22.5)(2 × 26.5 + 22.5)
6(22.5 + 26.5)
h′G = 1.0272 in.
=
2) Determine the gasket dimensions:
G = C − 2hG = 22.5 − 2 × 2.5915 = 17.317 in.
(C − B) (22.5 − 10.75)
b=
=
= 2.9375 in.
4
4
y = 1100 and m = 1.75.
175
5) Determine the flange moments at operating condition. Flange loads:
π
π
HD = B2 P = (10.75)2 (320) = 29,044
4
4
HT = H − HD = 75,368 − 29,044 = 46,324.
Lever arms:
hD = R + 0.5g1 = 2.5 + 0.5(3.375) = 4.1875 in.
hT = 0.5(R + g1 + hG )
= 0.5(2.5 + 3.375 + 2.5915)
= 4.2333 in.
Flange moments:
MD = HD hD = (29,044)(4.1875) = 121,622
MT = HT hT = (46,324)(4.2333) = 196,104
M0 = MD + MT = 121,622 + 196,104
= 317,726.
6) Determine the flange moment at gasket seating condition. Flange load:
HG = Wa − H = 706,176 − 75,368 = 630,808.
Lever arm:
hG h′G
(2.5915)(1.0272)
h′′G =
=
(2.5915) + (1.0272)
hG + h′G
= 0.7356 in.
Flange moment:
Mg = HG h′′G = (630,808)(0.7356) = 464,022.
All flange geometry constants are the same as those
mentioned in Example 10.9.
7) Calculate the flange stresses. Assume flange thickness
3) Determine the loads:
t = 2.03 in. This is set directly from the radial flange
π
π
H = G2 P = (17.317)2 (320) = 75,368
stress at the bolt circle, which is
4
4
6Mg
Hp = 2bπGmP = 2(2.9375)
6(464,022)
=
SRbc = 2
2
t (πC − Nd) (2.03) (22.5π − 16 × 2)
×π(17.317)(1.75)(320) = 178,986
( )
)
(
bc
h
=
17,464
psi < 17,500 psi (allowable stress).
S
2.5915
G
R
(178,986) = 451,559
Hp′ =
Hp =
Longitudinal hub stress:
1.0272
h′
G
Wm1 = H + Hp + Hp′ = 75,368 + 178,986
+ 451,559 = 705,913
HGy = bπGy = (2.9375)π(17.317)(1100)
= 175,790
( )
hg
(2.5915)(175,790)
′
HGy =
HGy =
(1.0272)
h′G
= 443,497
′
Wm2 = HGy + HGy
=175,790 + 443,496=619,286.
4) Determine bolting requirements. Am is the greater of
W m1 /Sa or W m2 /Sa :
Am = 36.76 in.2 based on Wm1
Ab = 36.8 in.2 based on 16-2 in. diameter bolts
Wa = 0.5(Am + Ab )Sa = 0.5(36.76 + 36.8)
×(19,200) = 706,176.
te + 1 t 3
+ = 1.0021 + 0.0407 = 1.0428
T
d
fMg
1(464,022)
SH = 2 =
Lg1 B (1.0428)(3.375)2 (10.75)
L=
SH = 3634 psi < 26,250 psi (allowable stress).
Radial flange stress:
(
)
4
te
+
1
Mg
3
(1.4704)(464,022)
SR =
=
(1.0428)(2.03)2 (10.72)
Lt 2 B
SR = 14,770 psi < 17,500 psi (allowable stress).
Tangential flange stress:
YMg
− ZSR
t2 B
2.29(464,022)
=
− (1.39)(14,770)
(2.03)2 (10.75)
ST = 3457 psi < 17,500 psi (allowable stress).
ST =
176
10 Blind Flanges, Cover Plates, and Flanges
Problem
10.13 Flange Calculation Sheets
10.12
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-type gasket.
Sheet 3. Ring flange with ring-type gasket.
Sheet 4. Reverse welding-neck flange with ring-type
gasket.
Assume that a flange with a flat-face gasket has an
applied moment of M0 = 464,000 with a bolt circle of C = 22.5 in. using 162 in. bolts. What is the
required thickness of the flange if the allowable
stress is 15,000 psi?
Answer:
t reqd = 2.19 in.
Sheet 1 Welding-neck flange with ring-type gasket. Source: Courtesy G + W Taylor – Bonney Div., Taylor Forge.
10.15 Spherically Dished Covers
Sheet 5. Slip-on flange with full-face gasket.
Sheet 6. Welding-neck flange with full-face gasket.
10.14 Flat-Face Flange
with Metal-to-Metal Contact Outside
of the Bolt Circle
Rules for the design of flat-face flanges with metal-tometal contact outside of the bolt circle are given in
Appendix Y of the ASME Code, VIII-1 [13–15]. The
rules are for circular, bolted flanged connections with
identical and nonidentical pairs of flanges. The pairs
of flanges that are in metal-to-metal contact across the
whole face and the gasket load to compress the gasket are
small (see Figure 10.23). The rules also apply to identical
pairs of flanges with a metal spacer added at the outer
edge between them.
The basic development assumes that the flanges are
in tangential contact at the outside diameter or at some
point between the bolt circle and the outside diameter at
a distance hC from the bolt circle. The gasket is assumed
to be self-sealing, generates a negligible load during
operation, and is located in line with the vessel wall. The
major difference between this type of flange and the ring
type described in Section 10.9 is the additional prying
effect at the point of contact of the two flanges.
To organize the calculations systematically, it is necessary to classify assemblies and to categorize each individual flange.
10.14.1
Classification of Assembly
Class 1. A pair of flanges that are identical except for the
gasket groove.
Class 2. A pair of nonidentical flanges where the inside
diameter of the reducing flange exceeds one-half of the
bolt-circle diameter.
Class 3. A flange combined with a flat-head or a reducing flange where the inside diameter is small and does not
exceed one-half of the bolt-circle diameter.
10.14.2
Categories of Flanges
Category 1. An integral flange or an optional flange calculated as an integral flange.
Category 2. A loose-type flange with a hub that is considered to add strength.
Category 3. A loose-type flange that is with or without a
hub, or an optional type 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 that made for an Appendix 2 flange except
for the additional loadings caused by the prying effect
where the contact near the outside diameter occurs. This
contact force H C and its moment arm hC involve an interaction between the bolt elongation and flange deflection
and the moments MP and MS .
The required bolt load and operating condition is
Wm1 = H + HC + HG .
(10.29)
10.15 Spherically Dished Covers
The ASME Code contains special rules for designing
spherically dished covers with a bolting ring flange. The
formulas given in the code are approximate because
they do not take into account the discontinuity existing
between the dished head and the flange ring. The flange
thickness is set by the combination of the circumferential
ring stress and the tangential bending stress. Figure 10.24
shows the head geometry and the loading applied to the
ring flange resulting from the reaction from the internal
pressure against the dished head. It is derived thus.
Using the geometry shown in Figure 10.24, the following
is set:
Adjacent side
√
( )2
B
2
adjacent side
cos 𝛽1 =
hypotenuse
√
√
L′ 2 − (B∕2)2
4L′ 2 − B2
=
, (10.30a)
=
′
L
2L′
=
Figure 10.23 Flat-face flange with metal-to-metal contact outside
of the bolt circle.
L′ 2 −
177
178
10 Blind Flanges, Cover Plates, and Flanges
Sheet 2 Slip-on or lap-joint flange with ring-type gasket. Source: Courtesy G + W Taylor – Bonney Div., Taylor Forge.
Membrane force in head to due
to pressure
PL′
,
2
Circumferential ring stress
(horizontal force)(D)
.
= Sc =
2TT ′
= F′ =
Substituting horizontal force = F′ cos 𝛽 1 ; D = B; and
T = (A − B)/2 in Eq. (10.30c),
(F ′ cos 𝛽1 ) B
•
Sc =
(10.30d)
(A − B)∕2 2T
When the value for F′ is substituted in Eq. (10.30d), the
equation becomes
PL′ B cos 𝛽1
.
(10.30e)
Sc =
2T 2 (A − B)∕2
′
(10.30b)
(10.30c)
10.15 Spherically Dished Covers
Sheet 3 Ring flange with ring-type gasket. Source: Courtesy G + W Taylor – Bonney Div., Taylor Forge.
When the value for cos 𝛽 1 from Eq. (10.30a) is substituted,
√
PB 4L′ 2 − B2
Sc =
.
(10.31)
4T
A−B
The tangential stress in the ring due to M0 is as follows:
from Eq. (9) of (2-51) of the ASME Code, Section VIII,
the equation is
ST =
St =
2S
S
F + 2 J.
T
T
T 2 − 2T(F) = 0
or
(10.33)
Combining the circumferential and the tangential
stresses in the ring gives
√
PB 4L′ 2 − B2
St = Sc + ST =
A−B
)4T
(
M0 ( A + B )
.
(10.34)
+
T 2B
A−B
Let
and J =
M0 (A + B)
.
SB(A − B)
(10.35)
√
4F 2 + 4J
2
√
T = F ± F 2 + J.
2F ±
(10.32)
but Y = (A + B)/(A − B), so
(
)
M0 ( A + B )
.
ST =
T 2B
A−B
(10.36)
Dividing by S and rearranging terms to form a
quadratic,
T=
YM0
,
T 2B
√
PB 4L′ 2 − B2
F=
8S(A − B)
Then,
(10.37)
(10.38)
This equation is identical with the equation given in
1-6(g) of the ASME Code, VIII-1.
Application of this equation is permitted for either
internal or external pressure. The quantity P is the absolute value for either the internal pressure or the external
pressure. The value for M0 is determined by combining
the moments from bolt loading and gasket loading with
the moment caused by the pressure loading from the
spherical head on the inside edge of the ring. When
this total moment is determined, the absolute value is
inserted for M0 in the equation. Figure 10.24 shows the
loadings caused by the pressure. The loading shown in
Figure 10.24 is due to internal pressure. However, if the
loading were due to external pressure, all directions of
loadings would be reversed, but the effective stresses
on the ring would the same. If external pressure were
179
180
10 Blind Flanges, Cover Plates, and Flanges
Sheet 4 Reverse welding-neck flange with ring-type gasket. Source: Courtesy G + W Taylor – Bonney Div., Taylor Forge.
applied to the head, it would have to be examined for
compressive buckling that might set the thickness.
Example 10.14
A spherically dished head is to be bolted to the
welding-neck flange described in Example 10.9. The
dished head is to be attached at the upper inner corner
with the outer surface even with the ring’s outer surface
(see Figure 10.25). What is the minimum required thickness of the flange ring when the spherical head is dished
to a radius of L = 2B?
Solution:
From the geometry mentioned in Example 10.9,
A = 26.5, B = 10.75, L = 2B = 2(10.75) = 21.5.
10.15 Spherically Dished Covers
Sheet 5 Slip-on flange with full face gasket. Source: Courtesy G + W Taylor – Bonney Div., Taylor Forge.
Determine the minimum required head thickness as
follows:
5PL 5(2500)(21.5)
t=
=
= 2.560 in. use 2.625 in.
6S
6(17,500)
From geometry calculations,
t
2.625
= 21.5 +
2
2
L′ = 22.813 in.
L′ = L +
181
182
10 Blind Flanges, Cover Plates, and Flanges
Sheet 6 Welding-neck flange with full-face gasket.
√
4(22.813)2 − (10.75)2
cos 𝛽1 =
= 0.972,
2(22.813)
𝛽1 = 13.626∘ .
The membrane force in the head due to pressure is
F′ =
(2500)(22.813)
PL′
=
= 28,520 lb∕in.
2
2
Horizontal force = F ′ cos 𝛽1
= (28,520)(0.9720) = 27,720 lb∕in.
Vertical force = F ′ sin 𝛽1 = (28,520)(0.2356)
= 6720 lb∕in.
Total horizontal force = π(10.75)(27,720)
= 936,150 lb.
10.15 Spherically Dished Covers
β1
T'
The moment at the operating condition is as follows:
t
Load
B/2
(Inner radius)
F'
F'
β1
L'
T
A-B
L
Moment
HD = Hv
= 226,900
hD 0.5(C − B)
= 5.875
MD = H D hD
= 1,333,000
HG = Hp
= 250,600
hG = 0.5(C − G)
= 3.729
MG = hG hG
= 934,000
H T = 217,400
hT = 0.5(R + g 1
+ hG ) = 4.802
MT = H T hT
= 1,044,000
H H = 936,150
−hH = −0.5(T − t)
= −0.5 T + 1.313
MH = H H hH
= −468,100 T
+ 1,229,200
B (Inner diameter)
2
Arm
β1
A (Outer diameter)
M0 = MD + MG + MT + MH = 4,540,200–468,100 T
The minimum thickness at the gasket seating condition
is
Figure 10.24 Spherically dished cover.
F = 0 and J =
(2,613,000)(26.5 + 10.75)
(17,500)(10.75)(26.5 − 10.75)
= 32.850
√
√
T = F ± F 2 + J = 32.850 = 5.732 in.
Hv
W
The minimum thickness at the operating condition is
√
(2500)(10.75) 4(22.813)2 − (10.75)2
F=
8(17,500)(26.5 − 10.75)
= 0.540.
ha
HH
hD
T = 5.75''
B = 10.75''
C = 22.5''
If we assume T = 5.75 and M0 = 1,849,100,
1,849,100
J = 32.850 ×
= 23.246
2,613,000
√
T = 0.540 ± (0.540)2 + 23.246 = 5.392 in.
A = 26.5''
G
hG
hT
HG
HT
Figure 10.25 Dimensions of spherically dished head mentioned
in Example 10.14.
The minimum thickness of 5.75 in. based on the gasket seating condition is also satisfactory for the operating
condition.
Problems
10.13
A spherically dished flange with an outside diameter A = 36 in. and an inside diameter B = 18 in.
is subjected to a gasket seating moment of
3,500,000 in.-lb. The allowable tensile stress of
the flange material is 15,000 psi. What is the minimum required thickness at the gasket seating
condition?
Answer:
t min = 6.24 in.
10.14
For the flange mentioned in Problem 10.13, what
is the maximum allowable flange moment if the
allowable tensile stress is increased to 17,500 psi
and the flange thickness is set at 6.25 in.?
Total vertical force = π(10.75)(6270)
= 226,900 lb.
The moment at the gasket seating condition is as follows:
Load
HG = W a
= 700,800
Arm
hG = 0.5(C − G)
= 3.729
Moment
MG = H G hG
= 2,613,000
183
184
10 Blind Flanges, Cover Plates, and Flanges
Answer:
N = gasket seating width (in.)
Mallow = 4,102,000 in.-lb/in. (circumference).
N min = minimum gasket width (in.)
OD = outside diameter of gasket (in.)
Nomenclature
P = design pressure (psi)
Pt = total load (lb)
Individual nomenclature is used throughout Chapter 10.
R = spherical radius of gasket surface (in.)
It is usually noted near to where it is used. The following
S = allowable tensile stress (psi)
list gives some of the general nomenclature.
Sa = allowable bolt stress, room temperature
(psi)
A
=
outside diameter of flange (in.)
Ab
=
actual bolt area (in.2 )
B
=
inside diameter of flange (in.)
Sb = allowable bolt stress, design temperature
(psi)
b
=
effective gasket seating width (in.)
T = nominal thickness of head (in.)
b0
=
basic gasket seating width (in.)
C
=
diameter of bolt circle (in.)
W m1 = flange loading for design condition (lb)
d
=
effective diameter of flat head (in.)
W m2 = flange loading for bolt-up condition (lb)
E
=
weld joint efficiency
E′
=
modulus of elasticity (psi)
𝜎 max = maximum bending stress (psi)
G
=
diameter of gasket reaction (in.)
wmax = maximum deflection (in.)
ID
=
inside diameter of gasket (in.)
𝜃 max = maximum rotation (radians)
M0
=
flange moment for design conditions
(in.-lb)
𝜎 t = 𝜎 T = tangential stress (psi)
flange moment for bolt-up condition
(in.-lb)
𝜎 c = 𝜎 C = circumferential stress (psi)
Mbu
=
t = minimum required thickness (in.)
Y or y = seating stress of gasket material (psi)
𝜎 r = 𝜎 R = radial stress (psi)
𝜇 = Poisson’s ratio
References
1 ASME Boiler and Pressure Vessel Code, Section VIII,
2
3
4
5
Division 1, Pressure Vessels. New York: American
Society of Mechanical Engineers.
ASME Boiler and Pressure Vessel Code, Section
I, Power Boilers. New York: American Society of
Mechanical Engineers.
Waters, E.O., Wesstrom, D.B., Rossheim, D.B., and
Williams, F.S.G. (1937). Formulas for stresses in
bolted flanged connections. Transaction of the ASME
59: 161–169.
Waters, E.O., Rossheim, D.B., Wesstrom, D.B., and
Williams, F.S.G. (1949). Development of general formulas for bolted flanges. Chicago: Taylor Forge and
Pipe Works.
Waters, E.O. and Schneider R.W. (1980). Derivation
of ASME Code Formulas for the Design of Reverse
Flanges, Welding Research Council, Bulletin 262,
pp. 2–9.
6 ANSI Standard B16.5 Pipe Flanges and Flanged
7
8
9
10
11
12
Fittings. New York: American National Standards
Institute.
API Standard 605 Large Diameter Carbon Steel
Flanges, ANSI/API Std. 605. Washington, DC: American Petroleum Institute.
ANSI Standard B16.24 Bronze Flanges and Fittings,
150 and 300 lb. New York: American National Standards Institute.
Rossheim, D.B. and Markl, A.R.C. (1943). Gasket loading constants. Mechanical Engineering 65:
647–648.
Raut, H.D. and Leon G.F. (1977). Report of Gasket
Factor Tests, Welding Research Council, Bulletin 233,
New York.
Modern Flange Design, Bulletin 502, 7th ed., G + W
Taylor—Bonney Division, Southfield, Mich.
Young, W., Budynas, R., and Sadegh, A. (2012).
Roark’s Formulas for Stress and Strain. New York:
McGraw Hill.
Further Reading
13 Waters, E.O. (1971). Derivation of Code Formulas for
15 Schneider, R.W. and Waters, E.O. (1979). The appli-
Part B Flanges, Welding Research Council, Bulletin
166, 1971, pp. 27–37.
14 Schneider, R.W. and Waters, E.O. (1978). The Background of ASME code case 1828: a simplified method
of analyzing part B flanges. Transaction of the
ASME Journal of Pressure Vessel Technology 100
(2): 215–219.
cation of ASME code case 1828. Transaction of the
ASME Journal of Pressure Vessel Technology 101 (1):
87–94.
Further Reading
Blach, A.E. and Bazergui, A. (1981). Methods of Analysis of
Bolted, Flanged Connections—A Review. Ecole
Polytechnique, Montreal, Canada.
Interpretive Study on the Design of Non-circular Flanges
and Flanges with External Loads (1979). Private report
to PVRC, May 23, 1979.
Raut, H. D., A. Bazergui, and L. Marchand, “Gasket
Leakage Trends” (private report to PVRC), April 1981.
Rodabaugh, E.C. and Moore, S.E. (1976). Evaluation of the
Bolting and Flanges of ANSI B16.5 Flanged
Joints—ASME Part A Design Rules, ORNL/Sub-2913-3.
Oak Ridge, Tennessee: Oak Ridge National Laboratory.
Timoshenko, S. and Woinowsky-Krieger, S. (1959). Theory
of Plates and Shells. New York: McGraw hill.
185
Vessel with multiple nozzles.
188
11
Openings, Nozzles, and External Loadings
11.1 General
All process vessels require openings to get the contents
in and out. For some vessels, where the contents may be
large or some of the internal parts may need frequent
changing, access is through large openings in which the
entire head or a section of 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 manways for personnel.
Other openings may be necessary for inspecting the vessel from the outside through a handhole opening, and still
others may be required for cleaning or draining the vessel. These openings do not always have a nozzle located
at the opening. Sometimes, the closure may be a manway
cover or handhole cover that is either welded to the vessel
or to a built-up pad area or attached to it by bolts.
For some nozzles, additional loading to the internal or
external pressure may be introduced by dead loads from
the equipment and piping and by thermal-expansion flexibility loadings from the piping and equipment motions.
This additional loading may require compensation as well
as what is necessary to resist the internal and external
pressure loadings, as shown in Figure 11.1.
Openings and nozzles similar to those occurring in
pressure vessels also form in piping. This is the case
where a branch run is attached to the main run of the
piping. The branch-to-run intersection is subjected to
the same pressure and thermal-expansion loadings as
those applied to a vessel nozzle. Although the nozzles
have a similar construction, there is usually an important difference in the relationship between the ratio of
the nozzle diameter to the vessel diameter and that of
the branch diameter to the run diameter. For pressure
vessels, this ratio d/D is much less than that for piping, where it may be very close to 1 : 1, as shown in
Figure 11.2.
In designing openings and nozzles for resisting loadings from internal and external pressures and from
external loadings, two types of stress conditions are
important. First, the primary membrane stresses in
the vessel or run pipe, that is, the necessary stresses
maintaining 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 corner 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 vessel parts or piping system is established.
A slight change in the details at the intersection may
enable the vessel to operate through many more cycles
of pressure and temperature loadings.
Detailed rules for designing vessels and piping to
accommodate the primary membrane stresses and loadings from internal and external pressures are given in
codes and regulations such as the ASME Boiler and Pressure Vessel Code [1] and the ASME Code for Pressure
Piping B31 [2]. In addition, some design rules are given
in the more advanced sections of these codes to permit
considering stress intensity factors and stress concentration factors (SCFs) in determining peak stresses. The
peak stresses are used to determine the design fatigue
life of the vessel. Other codes do not mention peak
stresses or fatigue evaluations and leave the latter as
the designer’s responsibility. At the present time, none
of these codes contains detailed design rules for the
consideration of external loadings – either dead loadings
or piping expansion loadings.
11.2 Stresses and Loadings at 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. Single openings
are calculated by the reinforcement method, whereas
multiple openings are calculated by either the reinforcement method or the ligament efficiency method. In both
cases, the primary stresses are effectively kept less than
the allowable stress by replacing the area removed for
openings.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
11.2 Stresses and Loadings at Openings
Mb
Figure 11.1 Applied pressure and external loadings on nozzle.
Fa
Fh
d
P
P
D
P
Figure 11.2 Variation in d/D ratio of nozzles
and piping.
d
d
d/D = 0.15
d/D = 0.33
D
d
d
d/D = 0.5
d/D = 0.25
D
d
d/D = 1.0
D
For a single circular opening in a flat plate with infinite boundaries in two directions (not through the thickness) that is subjected to applied forces and stresses along
opposite edges of the plate, stresses are increased above
the nominal applied stress in the unperforated plate. The
stresses decrease away from the opening until the nominal stress in the plate is attained. The ratio of the stress
at the examined point to the nominal stress is the stress
intensity factor.
The stress intensity around an opening may be
expressed either in general terms of applied stresses
and geometry or with respect to the considered point.
The basic equation at an opening may be written in
terms of 𝜎 and 𝜃, with the angle of the considered point
measured from the loading axis [3]. For the loading
shown in Figure 11.3a, 𝜎 1 is axial and 𝜃 = 𝜋/2 at the
maximum stress location. For the loading shown in
Figure 11.3b, 𝜎 1 is axial, 𝜃 = 𝜋/2 at the maximum stress
location, 𝜎 2 = 0.5𝜎 1 is at right angles to 𝜎 1 , and 𝜃 = 0 at
the maximum stress location. For the loading shown in
Figure 11.3c, the 𝜎 1 is axial, and 𝜎 2 = 𝜎 1 is at right angles
to 𝜎 1 . The values of 𝜃 are the same as for the cylinder.
189
190
11 Openings, Nozzles, and External Loadings
σ1
Table 11.1 Stress intensity factors
for various ratios of applied stress.
3.00σ1
1.07σ1 1.15σ1
r
2r
3r
1 : 0 (axial only)
σ1
2.50σ1
1.09σ1 1.23σ1
r
2r
3r
SIF
1 : 0 (axial only)
3.00
2 : 1 (cylinder)
2.50
1 : 1 (sphere)
2.00
(a)
σ1
0.5σ1
Stress ratio
0.5σ1
2 : 1 (cylinder)
(b)
or compression. The stress intensity factor at the edge
of circular openings for various ratios of applied edge
stresses is given in Table 11.1.
The stress intensity factor for various combinations
of stresses is maximum at the edge of the opening
and decreases away from the opening until the stress
approaches a nominal stress factor close to 1.0.
Using the following nomenclature, various formulas for different combinations of applied stresses are
developed:
r = radius of circular opening in plate (in.)
σ1
x = distance from centerline to point of SIF (in.).
σ1
For applied stress ratio of 1 : 0 – the condition of an
axial tension load only – the basic equation for the stress
intensity factor is obtained by solving Eq. (11.1) with
𝜃 = 𝜋/2, where cos 2𝜃 = −1, giving
[
[
( )2 ]
( )4 ]
𝜎
𝜎
r
r
𝜎1 =
−
(−1)
1+
1+3
2
x
2
x
[
( )4 ]
( )2
𝜎
r
r
+3
.
(11.3)
𝜎1 =
2+
2
x
x
2.00σ1
1.11σ1 1.25σ1
σ1
r
2r
3r
σ1
1 : 1 (sphere)
(c)
σ1
Figure 11.3 (a–c) Two-direction load combinations on flat plate
with circular opening.
At the edge of the opening, the stress intensity factor
is determined from Eq. (11.2) assuming that 𝜎 1 = 𝜎 1 and
𝜎 2 = 0:
𝜎max = 3𝜎1 − 0 = 3.00𝜎1 .
When two loadings or stresses are involved, the effects
at the maximum-stress location are added. The basic
equation for direct stress is
[
[
( )2 ]
( )4 ]
𝜎
𝜎
r
r
𝜎T =
−
cos 2𝜃.
1+
1+3
2
x
2
x
(11.1)
The basic equation for the maximum stress at the edge
of the opening in terms of the component of stresses in
each direction is
𝜎max = 3𝜎1 − 𝜎2 ,
(11.4)
Substituting various values of r for x in Eq. (11.3), stress
intensity factors at various distances from the edge of the
opening are
x
𝝈
r
3.00𝜎 1
2r
1.15𝜎 1
3r
1.07𝜎 1
4r
1.04𝜎 1
(11.2)
where the values of 𝜎 1 and 𝜎 2 are positive or negative
depending upon whether the applied stress is tension
For an applied stress ratio of 2 : 1 – the condition of
a cylindrical shell under internal pressure – the basic
11.2 Stresses and Loadings at Openings
equation for the stress intensity factor is found by combining the effects of stresses in two directions according
to Eq. (11.1):
[
[
( )2 ]
( )4 ]
𝜎
r
r
𝜎
1+
1+3
−
cos 2𝜃,
𝜎t1 =
2
x
2
x
where 𝜃 =
𝜋
2
and cos 2𝜃 = –1,
[
[
( )2 ]
( )4 ]
𝜎
𝜎
r
r
𝛼t2 =
−
cos 2𝜃,
1+
1+3
4
x
4
x
where 𝜃 = 0 and cos 2𝜃 = +1.
The general equation is the summation of the two
stresses, and thus,
[
( )4
( )2
( )2
𝜎
r
r
r
𝜎1 =
+1−3
+1+
2+
4
x
x
x
( )4 ]
r
−1 − 3
x
[
( )4 ]
( )2
𝜎
r
r
𝜎1 =
+3
.
(11.5)
4+3
4
x
x
At the edge of the opening, the stress intensity factor
is determined from Eq. (11.2) assuming 𝜎 1 = 𝜎 1 and
𝜎 2 = 0.5𝜎 1 , so that
𝜎max = 3𝜎1 − 0.5𝜎1 = 2.50𝜎1 .
(11.6)
Substituting various values of r for x in Eq. (11.5), stress
intensity factors at various distances from the edge of the
opening are as follows:
x
2.50𝜎 1
2r
1.23𝜎 1
3r
1.09𝜎 1
4r
1.05𝜎 1
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 found by combining effects of stresses in two directions according to
Eq. (11.1):
( )2
r
.
(11.7)
𝜎1 = 1 +
x
At the edge of the opening, the stress intensity factor
is determined from Eq. (11.2) assuming that 𝜎 1 = 𝜎 1 and
𝜎 2 = 𝜎 1 as given by
𝜎max = 3𝜎1 − 𝜎1 = 2.00𝜎1 .
r
2.00𝜎 1
2r
1.25𝜎 1
3r
1.11𝜎 1
4r
1.06𝜎 1
Example 11.1
A vertical vessel under internal pressure and dead load
contains an opening that is subjected to applied stresses.
The dead-load stress is equal to the circumferential pressure stress. For this stress condition, what is the basic
equation for the stress intensity factor at any distance
from the center of the opening?
Solution:
The applied stress in the circumferential direction is 𝜎 1 ,
whereas in the longitudinal direction, it is the longitudinal pressure stress minus the longitudinal dead-load
stress. This equals 𝜎 2P = −0.5𝜎 1 and 𝜎 2DL = −𝜎 1 , and the
summation equals −0.5𝜎 1 . Using Eq. (11.1), the basic
equations are
[
[
( )2 ]
( )4 ]
𝜎
𝜎
r
r
𝜎t1 =
−
cos 2𝜃,
1+
1+3
2
x
2
x
where 𝜃 =
𝜋
2
and cos 2𝜃 = −1,
[
[
( )2 ]
( )4 ]
𝜎
𝜎
r
r
+
cos 2𝜃,
1+
1+
𝜎t2 = −
4
x
4
x
where 𝜃 = 0 and cos 2𝜃 = +1,
[
( )4
( )2
( )2
𝜎
r
r
r
+2+6
−1−
𝜎1 =
2+2
4
x
x
x
( )4 ]
r
+1 + 3
x
[
( )4 ]
( )2
𝜎
r
r
𝜎1 =
+9
,
4+
4
x
x
𝝈
r
𝝈
x
(11.8)
Substituting various values of r for x in Eq. (11.7), stress
intensity factors at distances from the edge of the opening
are as follows:
Example 11.2
For the vessel described in Example 11.1, what is the
maximum stress at the edge of the opening according to
Eq. (11.2)?
Solution:
Assuming that 𝜎 1 = 𝜎 1 and 𝜎 2 = −0.5𝜎 1 , the equation for
the maximum stress at the edge of the opening is given
by Eq. (11.2) as
𝜎max = 3𝜎1 − 𝜎2 = 3𝜎1 − (−0.5𝜎1 ) = 3.50𝜎1 .
Problems
11.1
What is the stress intensity factor for a plate
under a stress ratio of 2 : −1 for the vessel given
191
192
11 Openings, Nozzles, and External Loadings
in Example 11.1 at the edge of the opening and at
distances of 2r, 3r, and 4r?
Answer:
1
t
Location
tn
tn
3
tn
t
SIF
r
3.50
2r
1.20
3r
1.05
4r
1.02
tn
t
t
11.2
In the ASME Code, VIII-2, a local stress region is
one in which the stress
√ intensity of 1.1Sm does not
extend more than Rt. In terms of the radius r of
the opening, how far from the edge of the opening
in a hemispherical head under internal pressure
must one be to have a stress of 1.1Sm , assuming
Sm = 𝜎 1 ?
tn
tn
d
Answer:
The SIF becomes equal to 1.1Sm 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 internal pressure at which
one will have a stress of 1.1Sm ?
Answer:
The SIF becomes equal to 1.1Sm at a distance of
2.90r from the edge of the opening along the longitudinal axis.
11.3 Theory of Reinforced Openings
As described in Section 11.2, there is an increase in basic
stresses at an opening in a flat plate or shell under edge
loadings due to the discontinuous pathway for the loads
(and stresses) to pass from one side of the opening to the
other side. When this happens, other pathways have to
be established in order to keep the primary stresses at
an acceptable level. The basic theory of reinforced openings is to supply pathways with additional material in the
region of the opening to carry the loads by the opening.
In designing the process equipment and other pressure
vessels, this pathway is supplied by thickening the basic
shell or nozzle material and by adding material such as a
pad, as shown in Figure 11.4.
Placement of the additional material is important. It
must be sufficiently near the opening to be effective,
Weld to pad
t
t
Figure 11.4 Methods of adding reinforcement material. Source:
Courtesy of American Society of Mechanical Engineers, from
Figure UW-16.1 of the ASME Code, VIII-1.
and yet it must be added with caution to prevent other
problems such as high thermal stresses. Investigations
by the PVRC [4] and others indicate that the placement
of the reinforcement is important. On most pressure
vessels, the reinforcement is added to the outside as
shown in Figure 11.5. However, on some vessels, the
reinforcement is added on the inside as in Figure 11.6,
and on still others, some of the reinforcement material
may be added to both the outside and the inside as shown
in Figure 11.7. The best arrangement is the so-called balanced reinforcement, which consists of about 35–40% of
the area on the inside and about 60–65% on the outside.
On many designs, however, it is difficult to place any
reinforcement on the inside, either because it is not
accessible or because it interferes with flow or drainage.
Balanced reinforcement is often used at manway and
inspection openings where no nozzle is attached.
For applications in design problems, where the reinforcement requirements are established, the method
of replacing areas is chosen rather than a method that
balances loads or stresses. An area at the opening for
11.4 Reinforcement Limits
Tn
Tn
Ts
Ts
Figure 11.5 Reinforcement added to outside of opening.
Figure 11.7 Reinforcement added to both inside and outside.
11.4 Reinforcement Limits
Tn
Ts
Figure 11.6 Reinforcement added to inside of opening.
carrying primary loads and stresses is removed. Thus,
this required area must be replaced by another area adjacent to the opening that is not used for that purpose. It is
desirable to replace that area required for primary loads
by an adjacent reinforcement area. Within the reinforcement limits, the reinforcement areas are assumed to have
the same load-carrying capabilities as the area removed
for the opening. Consequently, when the reinforcement
areas are equal to or exceed the required area, primary
stresses have been restored to as near the values for the
unperforated plate as possible.
As described in Section 11.2, the stress intensity factor
for an opening in a shell or head is the highest at the
edge of the opening and decreases away from the opening
(based on a shell wall with constant thickness). When the
effective thickness is increased, as happens with added
reinforcement material, the average stresses are lowered.
Limits of reinforcement are set parallel and perpendicular to the surface of the shell. These are set at a point at
which it is felt 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 centerline along the surface of the
shell, and the larger of the two answers is used. The first
formula is simply the diameter of the opening, d. The
second is T s + T n + 0.5d. As shown in Figure 11.8, the
thickness of the nozzle wall usually determines which
of the two limits controls. At a distance d from the
center line without reinforcement added, Eq. (11.5) for a
cylinder gives a SIF of 1.23𝜎 1 , and Eq. (11.7) for a sphere
gives a SIF of 1.25𝜎 1 . With additional reinforcement
material, the nominal stress is reduced close to that in an
unperforated plate.
If a nozzle is attached at the opening, it also offers reinforcement area available for replacing the area removed
from the vessel at the opening. Although the vertical limits are different in various codes, all are based on the wave
damping length of a beam on an elastic foundation. For a
cylindrical shell, this length is a function of √
1/𝛽, where 𝛽
for a Poisson’s ratio of 0.3 is equal to 1.285∕ rt.
193
194
11 Openings, Nozzles, and External Loadings
codes are setting this reinforcement limit in the vertical
direction by
√
(11.9)
L = 0.78 rm Tn ,
Tn
where
L = reinforcement limit perpendicular to shell (in.)
rm = mean radius of nozzle opening in shell (in.)
T n = nominal thickness of nozzle (in.).
Ts
d
d
Ts + Tn + 0.5d
Each pressure vessel and piping code treats the calculation of the reinforcement area somewhat differently and
establishes both parallel and perpendicular limits in different ways. A discussion of the reinforcement requirements for several different codes follows.
Example 11.3
A cylindrical pressure vessel that is 60 in. ID by 6 in. thick
contains a nozzle that is 12 in. ID by 3 in. thick. What is
the stress intensity factor at the reinforcing limit that is
parallel to the surface of the vessel?
tp
Solution:
The two horizontal limits are set by the larger of
d = 12 in.
or
Ts + Tn + 0.5d = 6 + 3 + 0.5(12) = 15 in.
Tn
d
Ts
d
Ts + Tn + 0.5d
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
[
( )4 ]
( )2
𝜎
6
6
𝜎1 =
+3
= 1.14𝜎1 .
4+3
4
15
15
Problems
11.4
Figure 11.8 Reinforcement limits parallel to shell surface.
The vertical limit was set in the ASME Code as 2.5T n
many years ago when an assumption was made that
r/t = 10 was to be used. This limit is about right for an
internal pressure p of 1200 psi and S = 12,000 psi. The
derivation of 2.5T n is
√
√
rt
0.1r2
1
L= =
=
= 0.246r = 2.46t
𝛽
1.285
1.285
For code application, the number was rounded off to
2.5T. With the wide range of r/t ratios, which are currently used in process-vessel construction, some of the
If the reinforcement limit in the vertical direction
were based on r/t = 5 instead of r/t = 10 that was
used to obtain the present limit of 2.5T n , what
would the multiplier of T n be for r/t = 5?
Answer:
The vertical limit would be 1.74T n 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 r/t = 5,
based on the circumferential stress formula given
in the ASME Code, VIII-1?
Answer:
Based on the circumferential stress, the maximum
design pressure is 3260 psi.
11.4 Reinforcement Limits
11.4.1
Reinforcement Rules for ASME Section I
The rules for reinforced openings in ASME Section I,
Power Boilers, permit using the replacement of both area
and ligament efficiency, provided certain limits are met.
Ligament rules may be used for repeating patterns of
openings, provided the maximum diameter of any hole
in the pattern does not exceed a diameter determined
from the equation
√
(11.10)
dmax = 2.75 3 Do Ts (1 − K)
within the limits 0.5 ≤ K ≤ 0.99. Here and in the
following,
P = internal maximum allowable working pressure
(psi)
dmax = maximum allowable diameter of opening (in.)
Do = outside diameter of shell (in.)
T s = nominal thickness of shell (in.)
S = allowable tensile stress (psi)
K=
11.4.1.1
PDo
.
1.82ST s
(11.11)
No Reinforcement Required
The rules for openings in ASME Section I contain provisions for single openings when no calculations are
required to prove the adequacy of the shell. No calculations are needed for a cylindrical shell when either of the
following is met:
d∕D ≤ 0.25 and dmax = 2 in.NPS
dmax = maximum diameter using Figure PG-32.
(11.12)
For openings in formed heads, no calculations are
required to prove the adequacy if all of the following
are met:
1) Actual center-to-center distance between openings is
equal to or greater than
L=
A+B
2(1 − K)
2) The edge of one opening is no closer than T s to the
edge of the adjacent opening.
3) Except for hemispherical heads, formed heads have
d/D ≤ 0.25 and dmax = 2 in. NPS, which is the same
as for cylindrical shells.
4) For hemispherical heads, the actual center-to-center
distance in item 1 is met. The value of K is one-half of
the value determined by Eq. (11.14).
5) For formed heads, dmax of Eq. (11.10) is met.
11.4.1.2
Size and Shape of Openings
The shape of the opening when these rules are applicable
is preferably circular, elliptical, or obround where the
ratio of the large to the small dimension is ≤ 2.0. When
the ratio is > 2.0, special requirements may be necessary to resist any twisting moment. For shapes other
than those mentioned, a special analysis or proof test 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:
1) For vessels 60 in. and less in diameter, the opening
shall not exceed 0.5D or 20 in.
2) For vessels over 60 in. in diameter, the opening shall
not exceed 0.33D or 40 in.
When these sizes are exceeded, suggested rules place
the available reinforcing area close to the opening.
Required Area of Reinforcement The total cross-sectional
area of reinforcement required for any plane through the
center of an opening is given by
(11.15)
A = dt r F,
where
d = diameter of opening (in.)
t r = minimum required thickness of seamless
shell (in.)
F = 1.0 except that Figure PG-33 [1] may be used for
integrally reinforced openings, where permitted
(11.13)
F = 0.5(cos2 𝜃 + 1) when Figure PG-33
with
K=
PDo
,
1.82ST s
(11.14)
where
A and B = diameters of adjacent openings (in.)
Do = outside diameter of formed head (in.)
T s = nominal thickness of formed head (in.)
and the other terms are the same as for Eq. (11.10).
is permitted.
(11.16)
For torispherical heads when the opening and its reinforcement are within the spherical part, t r is the minimum required thickness for a hemispherical head when
the radius is equal to that of the spherical part of the torispherical head. For a 2 : 1 ellipsoidal head when the opening and its reinforcement are within a circle of diameter
0.8D, t r is the minimum required thickness for a hemispherical head when the radius is equal to 0.9D.
195
11 Openings, Nozzles, and External Loadings
Limit of Reinforcement Parallel to Shell The limit of rein-
forcement parallel to the shell measured on each side
of the opening centerline is the greater of (1) d or (2)
T s + T n + 0.5d.
Ts
tr
1
2
4
3
Limit of Reinforcement Perpendicular to Shell The limit
8
of reinforcement perpendicular to the shell measured
either inward or outward from the surface is the smaller
of (1) 2.5T s or (2) 2.5T n + T e .
5
Ts
shell and nozzle are uniform and the reinforcement area
does not extend beyond this uniform thickness, the following formulas may be used for determining the available area of reinforcement. However, if the opening and
its reinforcement extend into areas with different nominal thicknesses and different minimum required thicknesses, these formulas are not applicable.
A1 = 2(E1 Ts − Ftr )(Ts + Tn ).
(11.18)
or
tr
6
1
2
4
3
8
7
(b)
5
Ts
1) Area available in the shell wall is the greater of
(11.17)
7
(a)
Available Area of Reinforcement If the thicknesses of the
A1 = (E1 Ts − Ftr )(2d − d)
6
5
tr
6
1
2
4
3
8
7
(c)
Figure 11.9 (a–c) Reinforcement requirements for multiple
openings. Source: Courtesy of American Society of Mechanical
Engineers, from Figure PG-38 of the ASME Code, Section I.
2) Area available in the nozzle wall is the smaller of
(11.19)
B
A
(11.20)
When two or more openings are spaced so that their
limits of reinforcement overlap, the combined area
is used and counted only once. The spacing between
any two openings has to be not less than 1.33dav . For
a series of openings in a pattern, the area between
any two openings shall equal at least 0.7F of the area
obtained by multiplying the center-to-center distance
by the required thickness as shown in Figure 11.9.
r
Ts
X
X
2X
D
ABCD = Limit of reinforcement
A
Design pressure = 2875 psi
Design temperature = saturation at design pressure
approximately 689∘ F
Materials: 70,000 psi UTS drum plate
Allowable stress at saturation temperature = 16,800 psi
Weld joint efficiency is E = 1.0.
Nozzles are 3 1/2 in., 4 in., 5 in., 6 7/8 in., and 24 in. ID.
d
trn
Tn
C
B
r
tr
Ts
tL
Example 11.4
Figure 11.10 shows a 66 in. ID steam drum containing five
different diameters and two types of nozzles. What are
the nozzle reinforcement requirements? The design data
are as follows:
trn
Tn
tr
R
A2 = 2(Tn − tn )(2.5Tn + Te ).
d
Y
or
Y
A2 = 2(Tn − tn )(2.5Ts )
R
196
X
X
2X
D
C
ABCD = Limit of reinforcement
Figure 11.10 “Set-on” and “set-in” nozzles.
11.4 Reinforcement Limits
Solution:
9) Area available in the shell wall is
A1 = (2X − d)(Ts − tr )
1) Minimum required thickness of shell at E = 1.00 is
PR
SE − 0.6P
2875 × 33
=
16,800 × 1.0 − 0.6 × 2875
= (2X − d)(6.75 − 6.294)
t=
A1 = (2X − d)(0.456).
10) Area available in the nozzle wall is
A2 = 2Y (Tn − trn ).
= 6.294 in.
Use 6 3/4 in. plate.
2) Minimum required thickness of nozzle is
Pr
2875r
=
= 0.190r.
SE − 0.6P
16,800 − 0.6 × 2875
3) Maximum diameter of a single unreinforced opening
is determined as
trn =
11) A 24 in. ID downcomer has “set-in” nozzle whose
calculated d/D = 24/66 = 0.36 exceeds the limit of
0.33D for vessels over 60 in. ID. Alternate rules are
also recommended.
12) Minimum required thickness of a downcomer nozzle is
trn = 0.190(12) = 2.289 in.
Use Tn = 5 14 in.
13) Reinforcement area required is
Do = 66 + 2(6.75) = 79.5 in.
K from Eq. (11.12) is
PDo
2875 × 79.5
=
= 1.107
1.82ST s
1.82 × 16,800 × 6.75
Ar = 6.294(24) = 151.06 in.2
14) Limit parallel to the shell is the greater of
d = 24 in. or Ts + Tn + 0.5d
Kmax = 0.99.
Using Eq. (11.10), the maximum diameter is
√
dmax = 2.75 3 79.5 × 6.75 × (1 − 0.99)
= 6.75 + 5.25 + 12 = 24 in.
15) Limit normal to the shell is the smaller of
16.875 in. or 2.5(5.25) = 13.125 in.
= 4.814 in.
4) The only single openings are 5 in. and 6 7/8 in.;
others are not considered single. Consequently, all
nozzles have the reinforcement area calculated (see
Table 11.2).
5) All nozzles except the 24 in. ID nozzle are “set-on”
type and calculated together.
6) Reinforcement area required, as given by Eq. (11.15),
is
Ar = dt r F = d(6.294)(1.0) = 6.294d.
16) Area available in the shell wall is
A1 = (2 × 24 − 12)(0.456) = 5.47 in.2
17) Area available in the nozzle wall (wall extends inward
for 6 1/2 in.. as full reinforcing), limit outward:
A21 = 2(13.125)(5.25 − 2.289) = 77.73 in.2
A22 = 2(6.5)(5.25 − 0) = 68.25 in.2
18) Total area available is
At = A1 + A21 + A22 = 151.45 in.2 ≥ Ar
7) Limit parallel to the shell is the greater of
d or Ts + Tn + 0.5d = 6.75 + Tn + r.
8) Limit normal to the shell is the smaller of
= 151.06 in.2
19) Also, check “close-in” limit. Determine the limit parallel to the shell as the greater of
2.5Ts = 2.5(6.75) = 16.875 in.or 2.5Tn .
0.75d = 0.75(24) = 18 in. or Ts + Tn + r = 24 in.
Table 11.2 Reinforcement calculations for 3 1/2 in., 4 in., 5 in., and 6 5/8 in. nozzles on a 66 in. ID steam druma).
d
Ar
Tn
d
6.75 + T n + r
2.5T n
2.5T s
trn
A1
A2
At
3.5
22.03
1.875
3.5
10.375
4.688
16.875
0.334
7.87
14.44
22.31
4.0
25.18
2.125
4.0
10.875
5.313
16.875
0.381
8.09
18.53
26.62
5.0
31.47
2.5
5.0
11.75
6.25
16.875
0.477
8.44
25.29
33.73
6.875
43.27
3.0
6.875
13.188
7.5
16.875
0.656
8.89
35.16
44.05
a) X = parallel, Y = normal. Area available is greater than the area required, and values of T n that were assumed are correct.
197
198
11 Openings, Nozzles, and External Loadings
Because the parallel limit is the same for the “close-in”
limit, the area required of 0.67Ar is also satisfied without
further calculations.
Example 11.5
Determine the minimum required thickness of a 36 in.
ID cylindrical shell based upon reinforcement requirements. The nozzles are through-welded as shown in
Figure 11.9c and have 2.25 in. ID in a staggered pattern
of three rows on 3 in. centers and 4.5 in. longitudinal
spacing, as shown in Figure 11.11. The design pressure is
500 psi at 700∘ F design temperature. The allowable stress
is 16,600 psi. There is no corrosion and E = 1.0.
𝜃 = tan−1 (3/2.25) = 53.13∘ . With a spacing of 3.75 in.,
the parallel limit of 4.875 in. exceeds the actual spacing; therefore, the limits overlap and the special rules
apply. From Eq. (11.16), F = 0.5(cos2 𝜃 + 1) = 0.68 for
𝜃 = 53.13∘ .
Ar = dt r F = (2.25)(0.552)(0.68)
= 0.845 in.2
A1 = (Ts − Ftr )(spacing − d)
= (1.125 − 0.68 × 0.552)
× (3.75 − 2.25) = 1.124 in.2
A2 = 0.143 in.2
At = A1 + A2 = 1.267 in.2 > 0.845 in.2
Solution:
1) The minimum required thickness of the cylindrical
shell is
PR
500 × 18
=
= 0.552 in.
tr =
SE − 0.6P
16,600 − 0.6 × 500
2) The minimum required thickness of the nozzle is
Pr
500 × 1.125
=
= 0.035 in.
SE − 0.6P
16,600 − 0.6 × 500
3) Determine the reinforcement limits based on
T s = 1.125 in. and T n = 0.188 in. Limit parallel to
the shell surface = X = 2.25 in. or (1.125 + 1.125 +
0.188) = 2.438 in.; use 2.438 in. Limit normal to
the shell surface = Y = 2.5T s = 2.812 in. or 2.5T n =
0.469 in.; use 0.469 in.
4) Examine the longitudinal plane 1–2. Actual spacing = 4.5 in. Parallel limits without overlap = 2X =
2(2.438) = 4.875 in.; exceeds actual spacing. Therefore, limits overlap and special rules apply:
A1234 = (0.552)(3.75)(0.7 × 0.68) = 0.985 in.2
A5678 = (1.125)(3.75 − 2.25)
= 1.688 in.2 > 0.985 in.2
6) The assumed values of T s = 1.125 in. and T n =
0.188 in. are satisfactory.
trn =
Problems
11.6
Ar = dt r F = (2.25)(0.552)(1.0)
Answer:
t min = 2.50 in.
= 1.242 in.2
A1 = (Ts − tr )(spacing d)
= (1.125 − 0.552)(4.5 − 2.25)
A1 = 1.289 in.2
A2 = (Tn − trn )(2Y )
= (0.188 − 0.035)(2 × 0.469)
= 0.143 in.2
What is the minimum required wall thickness
(rounded up to the next 1/2 in.) of a 12 3/4 in.
ID nozzle attached to a 60 in. ID by 3.75 in. thick
drum? The allowable stress of both the shell and
nozzle material is 15.0 ksi. The nozzle is attached
by a full-penetration weld with corner fillet welds
with a throat of 0.7T n . The design pressure is
1400 psi at room temperature.
11.7
For the triangular arrangement of openings shown
in Figure 11.11 with openings that are 2.25 in. ID,
what is the minimum side length of a spacing that
forms a series of equilateral triangles?
Answer:
Minimum length of side is 4.631 in.
At = A1 + A2 = 1.432 in.2
A1234 = (0.552)(4.5)(0.7 × 1.0)
= 1.739 in.
2
A5678 = (1.125)(4.5 − 2.25)
= 2.531 in. > 1.739 in.
2
2
5) Examine the diagonal plane 2–3. With a row-to-row
spacing of 3 in. and a longitudinal
√ spacing of 4.5 in.,
32 − 2.252 = 3.75 in.,
the diagonal spacing is
11.4.2 Reinforcement Rules for ASME Section VIII,
Division 1
The rules for reinforced openings in ASME Section VIII,
Division 1, Pressure Vessels, are similar to those for ASME
Section I. However, the rules for reinforcement are given
as the main choice, with ligament rules used only as an
alternative for repeating patterns of openings. Rules are
contained in both the text and the appendices. They are
11.4 Reinforcement Limits
Figure 11.11 Multiple openings in cylindrical
shell.
2.25′′
2.25′′
2.25′′
Longi tudinal
2
1
3.
75
′′
3′′
2.25′′ diameter
of all openings
3′′
axis
3
4.5′′
given for both internal pressure and external pressure.
The rules are essentially the same except that only 50%
of the replacement area is required for external pressure,
assuming that the minimum required thickness in each
case is based on the appropriate formula and design rules
for both internal and external pressures.
No Reinforcement Required Single openings in vessels that
are not subjected to special applied loadings, such as
cyclic loading, do not require reinforcement calculations
if the openings do not exceed the following size limits:
1) In plate thickness of 3/8 in. or less, dmax = 3 in. NPS.
2) In plate thickness greater than 3/8 in., dmax = 2 in.
NPS.
Size and Shape of Openings The rules apply to openings
that are preferably circular, elliptical, or obround. The latter shapes often result from an opening in a curved surface or from a nonradial nozzle. However, other shapes
are permitted when considered according to U-2(g).
For openings in cylindrical shells, the rules in the text
are limited to openings of the following size limits:
1) For shells 60 in. and less in diameter, the opening shall
not exceed 0.5D or 20 in.
2) For shells over 60 in. in diameter, the opening shall not
exceed 0.33D or 40 in.
When these size limits are exceeded, in addition to the
rules in the text, the rules in Appendix 1-7 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 text.
4.5′′
An alternate set of rules are listed in Appendix 1-10 of
VIII-1 for the design of nozzle reinforcement. These rules
differ slightly from the rules given in paragraph UG-37 of
VIII-1. The difference is in establishing the limit of reinforcement along the shell and nozzle. The limit along the
shell in UG-37 is essentially set to an arbitrary value of d/2
on each side of the nozzle center line where d is the nozzle diameter. The limits in Appendix 1-10 are based on
the attenuation property of the shell, which is a function
of the radius and thickness as detailed in Section 5.2.11.
There are no specific limitations on the size and shape
of openings in spherical shells and formed heads.
Required Area of Reinforcement The total cross-sectional
area of reinforcement required for any plane through the
center of an opening is given by
A = dt r F,
(11.21)
where d = the diameter of the opening on the longitudinal plane of a cylindrical shell or any plane of a spherical
shell or formed head (in.)
F = correction factor for pressure stress on plane being
examined with respect to longitudinal axis, as shown in
Figure 11.12. This factor is applicable only to nozzles with
integral reinforcement.
t r = minimum required thickness of a seamless shell
based on the circumferential stress (longitudinal plane)
or of a seamless formed head with the following additional provisions:
1) When the opening and its reinforcement are totally
within the spherical part of a torispherical head, t r
199
11 Openings, Nozzles, and External Loadings
limits of reinforcement parallel to the shell surface
measured on each side of the center line are the larger of
(1) d or (2) T s + T n + 0.5d.
1.00
0.95
Limit of Reinforcement Perpendicular to Shell The limit
of reinforcement perpendicular or normal to the shell
measured either inward or outward from the surface of
the shell is the smaller of (1) 2.5T s or (2) 2.5T n + t e .
0.90
0.85
Strength of Reinforcement When the reinforcement has a
lower (tensile) strength compared to the vessel wall, the
area provided by the reinforcement shall be decreased by
the ratio of the specified minimum tensile strengths. No
credit is given for tensile strength higher than that of the
vessel wall.
0.80
Value of F
200
0.75
0.70
Available Area of Reinforcement When the reinforcement
0.65
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:
0.60
1) Area available in the shell wall is the greater of
A1 = (2d − d)(Et s − Ftr )
0.55
0.50
0°
(11.22)
or
A1 = 2(Ts + Tn + 0.5d) − d(ET s − Ftr ).
10°
20° 30° 40° 50° 60° 70° 80°
Angle of plane with longitudinal axis
(11.23)
90°
Figure 11.12 Chart for determining F. Source: Courtesy of
American Society of Mechanical Engineers; from Figure UG-37 of
the ASME Code, VIII-1.
is determined using the hemispherical head formula
with both E and M = 1.0 (see Figure 11.13a).
2) When the opening and its reinforcement are in a cone,
t r is the required thickness of a seamless cone.
3) When the opening and its reinforcement are in an
ellipsoidal head and within a circle of diameter equal
to 80% of the shell diameter, t r is determined using
the hemispherical head formula for a seamless shell
of radius K 1 D, where D is the shell diameter and
K 1 is obtained from Table 11.3 and as shown in
Figure 11.13b.
The value of t r obtained from any of the given methods
is used only to determine the required area of reinforcement. The value of t r used to set the minimum required
thickness of the shell or head is based on the thickness
formulas that consider all the design loadings and weld
joint efficiencies.
2) Area available in the nozzle wall is the smaller of
A2 = (5Ts )(Tn − tn )
(11.24)
A2 = (5Tn + 2.5te )(Tn − tn ).
(11.25)
or
When the size of the opening exceeds the limits in
which the rules in the text apply, the supplemental rules
in Appendix 1–7 of VIII-I are used in addition to the text
rules. These additional requirements are as follows.
Required Area of Reinforcement The total cross-sectional
area of reinforcement required for any plane through
the center of an opening using Appendix 1–7 of VIII-I is
given by
A = 0.67dt r F.
(11.26)
Limit of Reinforcement Parallel to Shell The limit of rein-
forcement parallel to the shell measured on each side of
the opening center line is the greater of (1) 0.75d or (2)
T s + T n + 0.5d.
Limit of Reinforcement Perpendicular to Shell The limit is set
Limit of Reinforcement Parallel to Shell Surface When the
size of the opening is within the limits in the text, the
exactly the same way as for a nozzle, that is, within the
rules of the text.
11.4 Reinforcement Limits
Figure 11.13 Determination of special limits for
determining tr , to use in reinforcement calculations.
(a) Limits for torispherical head and (b) limits for
ellipsoidal head.
Spherical part = special limit for tr
(a)
0.8D = special limit for tr
h
D
(b)
Table 11.3 Factor K 1 for elliptical heads.
8′′ ID
2 1/2 T = 2.5′′
D/2h
K1
D/2h
K1
3.0
1.36
1.8
0.81
2.8
1.27
1.6
0.73
2.6
1.18
1.4
0.65
2.4
1.08
1.2
0.57
2.2
0.99
0
0.50
2.0
0.90
tn = 0.164′′
Tn = 1 1/8′′
2d = 16
0.8d = 33.4′′
tr = 0.755′′
T = 1′′
t = 0.838′′
41 5/8′′ ± 1/8′′
Figure 11.14 Details of nozzle in Example 11.7.
When any two adjacent openings are spaced so that
their reinforcement overlaps, the combined area is used,
but is evaluated only once in the combined area. The
preferred spacing is at least 1.5d, with 50% of the area
required between the two openings.
Example 11.6
Determine the reinforcement requirements of an 8 in.
ID nozzle that is centrally located in a 2 : 1 ellipsoidal
head. The inside diameter of the head skirt is 41.75 in.
The allowable stress of both the head and nozzle material
is 17.5 ksi. The design pressure is 700 psi, and the design
temperature is 500∘ F. There is no corrosion, and the weld
joint efficiency is E = 1.0. See Figure 11.14 for details of a
nozzle.
Solution:
1) The minimum required thickness of a 2 : 1 ellipsoidal head without an opening is determined from
UG-32(d) of the ASME Code, VIII-1, as
PD
2SE − 0.2P
700(41.75)
=
= 0.838 in.;
2(17,500 × 1.0) − 0.2(700)
tr =
use 1.0 in.
2) As noted in the definition of t r to use with Eq. (11.20)
and shown in Figure 11.14, when an opening and its
reinforcement are located in an ellipsoidal head and
within a circle of diameter equal to 80% of the shell
diameter, t r to be used in reinforcement calculations
is the thickness required for a seamless sphere of
radius K 1 D, where D is the shell ID and K 1 for a 2 : 1
ellipsoidal head is 0.9 as shown in Table 11.3. For
this head, the opening and reinforcement are within
0.8D = 0.8(41.75) = 33.4 in.
3) Using the spherical-shell radius of R = K 1 D =
0.9(41.75) = 37.575 in. in the hemispherical head
201
11 Openings, Nozzles, and External Loadings
formula gives
11) The total reinforcement available from head and
nozzle is
PR
2SE − 0.2P
700(37.575)
=
2(17,500 × 1.0) − 0.2(700)
= 0.755 in.
tr =
At = A1 + A2 = 1.960 + 4.805 = 6.765 in.2
Area provided = 6.765 in.2 > area required = 6.040 in.2 If
additional area is needed, use fillet-weld area.
4) The minimum required thickness of the nozzle is
PRn
SE − 0.6P
700(4)
=
(17,500 × 1.0) − 0.6(700)
= 0.164 in.
trn =
5) The limit parallel to the head surface is X = d
or T s + T n + r, whichever is greater. X = 8 in. or
(4 + 1 + 1.125 = 6.125 in.); use 8 in.
6) The limit perpendicular to the head surface is
Y = 2.5T s or 2.5T n , whichever is smaller. Y =
2.5(1) = 2.5 in. or 2.5(1.125) = 2.813 in.; use 2.5 in.
7) Because the limit of 2X = 2(8) = 16 in. is less than
33.4 in. of item 2, the provision for the spherical
head may be used.
8) The reinforcement area required following Eq.
(11.20) is
Ar = dt r F = 8(0.755)(1.0) = 6.040 in.2
9) The reinforcement area available in the head according to Eq. (11.21) is
A1 = (ET s − Ftr )(2d − d)
= (1.0 − 0.755)(16 − 8) = 1.960 in.2
10) The reinforcement area available in the nozzle
according to Eq. (11.23) is
A2 = 5Ts (Tn − trn )
= 5(1.0)(1.125 − 0.164) = 4.805 in.2
Example 11.7
Determine the reinforcement requirements for a 12 in. ×
16 in. manway opening. The 12 in. dimension lies along
the longitudinal axis of the cylindrical shell. The manway cover seals against the outer surface of the opening, so that the opening nozzle is under internal 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 of
500∘ F. There is no corrosion, and the joint efficiency is
E = 1.0. Details are shown in Figure 11.15.
Solution:
1) The minimum required thickness of the shell is
found from
PR
t=
SE − 0.6P
700(20.938)
=
(17,500 × 1.0) − (0.6 × 700)
= 0.858 in.
Use 1 in.
2) The minimum required thickness of the nozzle is
obtained as
PRn
t=
SE − 0.6P
700(6)
=
(17,500 × 1.0) − (0.6 × 700)
= 0.246 in.
Use 1 in.
Figure 11.15 Manway-opening details in Example 11.7.
OD
Tn = 1ʺ
d = 12ʺ
21/2ʺ
trn = 0.246ʺ
B
A
Y=
tr = 0.858ʺ
14ʺ
C
D
X = 12ʺ
X = 12ʺ
2x = 24ʺ
ABCD = Limit of reinforcement
21/2ʺ
Y=
Ts = 1ʺ
R= 20.9375ʺ
202
11.4 Reinforcement Limits
3) Examine the longitudinal plane where F = 1.0 is determined from Figure 11.12. Limits parallel to the shell,
whichever is greater:
d = 12 in.
at a design temperature of 500∘ F. All joints are fully
radiographed with a joint efficiency of E = 1.0. There is
no corrosion.
Solution:
Ts + Tn + 0.5d = 6 + 1 + 1 = 8 in.
Limits perpendicular to the shell, whichever is
smaller:
1
2 Ts = 2.5(1) = 2.5 in.
2
1
2 Tn = 2.5(1) = 2.5 in.
2
Reinforcement area required, as given by Eq. (11.20):
Ar = dt r F = 12(0.858)(1.0) = 10.296 in.2
Reinforcement area available in the shell, as given by
Eq. (11.21):
A1 = (ET s − Ftr )(2X − d)
= (1 − 0.858)(24 − 12)
= 1.704 in.2
Reinforcement area available in the nozzle wall, as
given by Eq. (11.23):
1) The minimum required thickness of the cylindrical
shell from Example 11.7 is
t = 0.858 in.
Use 1 in.
2) The minimum required thickness of the head is
PL
th =
2SE − 0.2P
700 × 20.938
=
2 × 17,500 × 1 − 0.2 × 700
= 0.420 in.;
use 1 in.
3) The minimum required thickness of the nozzle is
Pr
700 × 3
tn =
=
SE − 0.6P
17,500 × 1 − 0.6 × 700
= 0.123 in.
Use 3/4 in.
4) The limit parallel to vessel surface is the larger of
A21 = 5Ts (Tn − trn ) = 5(1)(1 − 0.246)
d = 6 in.
= 3.770 in.2
or
A22 = 5Ts (Tn − trn ) = 5(1)(1 − 0) = 5.000 in.2
Ts + Tn + 0.5d = 1 + 0.75 + 3 = 4.75 in.
Total reinforcement area available:
Use 6 in.
5) The limit perpendicular to vessel surface is the
smaller of
At = A1 + A21 + A22
= 1.704 + 3.770 + 5.000 = 10.474 in.2
Area
provided = 10.474 in.2 > area
required =
10.296 in.2
4) Examination of the circumferential plane, where
F = 0.5 from Figure 11.12, gives the reinforcement
area required, given by Eq. (11.20), as
Ar = dt r F = 16(0.858)(0.5)
= 6.864 in.2
2.5Ts = 2.5(1) = 2.5 in.
or
2.5Tn = 2.5(0.75) = 1.875 in.
Use 1.875 in.
6) The reinforcement area required, as given by
Eq. (11.20), is
Ar = dt r F = 6(0.858)(1.0) = 5.148 in.2
2
If the arc length of 16.4 in. is used, Ar = 7.036 in.
Either of these areas is less than Ar in item 3 and does
not control. The longitudinal plane controls.
Example 11.8
Determine the reinforcement requirements for a 6 in.
ID nozzle that is located at the junction of a cylindrical
shell and a hemispherical head. The entire opening is in
the cylindrical shell, but the reinforcement extends in
both directions – some into the head and some into the
shell. The ID of the shell is 41.875 in. The allowable stress
of all material is 17.5 ksi. The design pressure is 700 psi
7) The reinforcement area available in the shell and
head, as given by Eq. (11.21), is
A11 = (ET s − Ftr )
= (1 − 0.858)(6 − 3) = 0.852 in.2
A12 = (ET h − Fth )
= (1 − 0.420)(6 − 3) = 1.740 in.2
8) The reinforcement area available in the nozzle wall,
as given by Eq. (11.23), is
A2 = 5Tn (Tn − trn )
= 5(0.75)(0.75 − 0.123) = 2.351 in.2
203
204
11 Openings, Nozzles, and External Loadings
9) The total reinforcement available in the shell, head,
and nozzle is
At = A11 + A12 + A2 = 4.943 in.2
(not enough)
10) The reinforcement area available in the attachment
fillet weld assuming leg dimension of 0.5 in. is
2
A3 = (0.5) = 0.250 in.
2
11) The total area including the attachment fillet welds is
At = 5.193 in.2 > Ar = 5.148 in.2
Problems
11.8
What is the minimum required thickness of the
nozzle wall (rounded up to the next 1/8 in.) of
an opening whose reinforcement is based on
an available area in both the shell and the nozzle? The opening has a 15 in. diameter and is
located in a cylindrical shell of 22 in. diameter.
The design pressure is 450 psi, the design temperature is 450∘ F, and the allowable tensile stress is
S = 15,000 psi. There is no corrosion.
Answer:
Required nozzle wall thickness is 5/8 in.
LH = effective length of the nozzle wall outside the
vessel
LI = effective length of the nozzle wall inside the
vessel
LR = effective length of the nozzle wall inside the
vessel
Lpr1 = nozzle projection from the outside of the
vessel wall
Lpr2 = nozzle projection from the inside of the vessel
wall
PL = maximum local primary stress at the nozzle
intersection
Pmax = maximum allowable pressure at the
nozzle–shell connection
Reff = Di /2
Rn = nozzle inside radius
S = allowable stress of the shell
Sn = allowable stress of the nozzle
Sp = allowable stress of the pad
𝜎 avg = average primary membrane stress
𝜎 circ = general primary membrane stress
t = nominal thickness of the shell
11.9
Assume that the nozzle mentioned in Example
11.6 is not centrally located in the 2 : 1 ellipsoidal
head. Instead, some of the reinforcement area
extends into the knuckle region. Is the available
reinforcement area sufficient for this condition? If
not, how thick does the nozzle have to be?
Answer:
Required nozzle thickness is T n = 1.25 in.
11.4.3 Reinforcement Rules for ASME, Section VIII,
Division 2
The rules for reinforced openings in Section VIII, Division 2, differ substantially from those for Division 1. The
basis of the rules In Division 1 is to replace the metal
taken out of the opening with equivalent metal. The
rules in Division 2 are for checking the stress levels [5]
in the reinforced area. The rules for reinforced openings
in cylindrical shells are summarized as follows. Similar
rules are given in VIII-2 for nozzles in heads.
11.4.3.1
Nomenclature
AT = total area of reinforcement
Di = inside diameter of cylindrical shell
t e = nominal thickness of the reinforcing pad
t n = nominal thickness of the nozzle wall
W = effective width of the reinforcing pad
The nozzle design in VIII-2 consists of first assuming
a given reinforcement as shown in Figures 11.16 and
11.17. This reinforcement is then analyzed for adequate
stress level. The analysis is essentially in three steps. The
first is to find the limits of reinforcement, the second
is to determine the available area within the limits of
reinforcement, and the third is to calculate the stress
level and compare it to an allowable stress. The following
analysis for radial nozzles in cylindrical shells uses the
terminology in the aforementioned Nomenclature as
well as in Figures 11.16 and 11.17.
Step 1: Determine the limits of reinforcement along
the cylindrical shell
a) Integral nozzle
1) Set-in nozzle
LR is the smaller of
(Reff t)0.5
or
2Rn .
2) Set-on nozzle
LR is the smaller of
(Reff t)0.5 + tn
or
2Rn + tn .
11.4 Reinforcement Limits
Figure 11.16 Nomenclature for
nozzle openings. Source: Courtesy
of ASME.
Nomenclature for reinforced openings
CL
tn
CL
fN
LPr1
Lpr1
Leg41
Rn
Leg42
tw2
tw1
t
R
LH
w
LH
te
tw1
L1 L
pr2
Leg43
fs
fN
fs
LR
LR
CL
fY
fY
For set-in nozzles
For set-on nozzles
= A1 = Area contributed by shell
= A2 = Area contributed by nozzle projecting outward
= A3 = Area contributed by nozzle projecting inward
= A41 = Area contributed by outward weld
= A42 = Area contributed by pad to vessel weld
= A43 = Area contributed by inward weld
= A5 = Area contributed by reinforcing pad
AT = Total area contributed
(Rn tn )0.5 + 2.5t
b) Nozzle with reinforcing pads
1) Set-in nozzle
(Rn tn )
LR is the smaller of
or
2) Set-on nozzle
LH is the smaller of
2) Set-on nozzle
(Rn tn )0.5 + 1.5t
LR is the smaller of
or
[(Reff + t)(t + te )]0.5 + tn
or
2Rn + tn .
Step 2: Determine the limits of reinforcement along
the nozzle wall projecting outside the cylindrical shell
1) Set-in nozzle
LH is the smaller of
or
or
9t + 8te .
0.5
2Rn .
(Reff t)0.5 + w + tn
or
+ t + te
Lpr1 + t
(Reff t)0.5 + w or
[(Reff + t)(t + te )]
0.5
(Rn tn )
0.5
Lpr1
or
+ te
or
or
8(t + te ).
Step 3: Determine the limits of reinforcement along
the nozzle wall projecting inside the cylindrical shell
LI is the smaller of
(Rn tn )0.5
or
205
206
11 Openings, Nozzles, and External Loadings
Figure 11.17 Nomenclature for
variable thickness openings.
Source: Courtesy of ASME.
Nomenclature for variable thinkness openings
CL
tn2
CL
fN
tn
fN
Lpr1
LPr1
Lpr4
L41
Rn
Lpr3
LH
tw1
t
R
LH
Leg43
fs
tw1
L1 L
pr2
fs
LR
LR
CL
fY
fY
For set-in nozzles
For set-on nozzles
= A1 = Area contributed by shell
= A2 = Area contributed by nozzle projecting outward
= A3 = Area contributed by nozzle projecting inward
= A41 = Area contributed by outward weld
= A42 = Area contributed by pad to vessel weld
= A43 = Area contributed by pad to inward weld
= A5 = Area contributed by reinforcing pad
AT = Total area contributed
Lpr2
or
8(t + te ).
Step 4: Determine the available area of reinforcement
Area A1
A1 = the larger of
A1 = (t LR )
2Rn + tn
(Di + teff )teff
= t + (A5 )(frp )∕LR
𝜆=
or
A1 = (t LR )(𝜆∕5)0.85 ,
teff
teff = t + (A5 )(frp )∕(LR − tn )
for set-in nozzles
for set-on nozzles
frp = the smaller of
frp = 1 or
frp = Sp ∕S.
where
𝜆 = the smaller of
𝜆 = 12
or
Area A2
The nomenclature for the A2 reinforcement is given in
Figures 11.18 and 11.19.
A2 is obtained from the matrix shown as follows,
11.4 Reinforcement Limits
LH ≤ Lx3
tn = tn2
tn > tn2
Yes
A2 = t n L H
Lx3 < LH ≤ Lx4
LH > Lx4
A2 = A2b + A2b
A2 = A2a + A2c
Figure 11.18 A2 with variable thickness for set-in
nozzles. Source: Courtesy of ASME.
Metal area definition for A2 with variable thickness of set-in nozzles
tn2
tnx
tn2
A2c
A2b
LH
Lpr4
A
Lpr3 2a
LH
Lx4
A2a
LH
A2a
Lx3
t
tn
(a) LH≤Lx3
Figure 11.19 A2 with variable thickness for
set-on nozzles. Source: Courtesy of ASME.
tn
tn
(b) Lx3<LH≤Lx4
(c) LH>Lx4
Metal area definition for A2 with variable thickness of set-on nozzles
tnx
tn2
tn2
A2c
A2b
LH
Lpr4
A2a
Lpr3
LH
tn
(a) LH≤Lx3
Lx4
A2a
LH
A2a
Lx3
tn
(b) Lx3<LH≤Lx4
tn
(c) LH>Lx4
207
208
11 Openings, Nozzles, and External Loadings
Table 11.4 Length values.
Table 11.5 Values of f N , f s , and f Y .
Length
For set-in nozzle
For set-on nozzle
force
Set-in nozzle
Lx3 =
Lpr3 + t
Lpr3
fN
PRxn LH
PRxn (LH + t)
Lx4 =
Lpr4 + t
Lpr4
fs
PRxs (LR + t n )
PRxs LR
fY
PRxs Rnc
PRxs Rnc
where
Step 6: Determine the local primary stress
(f + Fs + fY )
𝜎avg = N
AT
A2a = tn Lx3
A2b is the smaller of
A2b = 0.39(tn + tnx )[Rn (tn + tnx )∕2]0.5
or
and
A2b = 0.5(tn + tnx )(LH − Lx3 )
𝜎circ =
A2c is the smaller of
A2c = 0.78tn2 (Rn tn2 )0.5
PRxs
teff
(11.27)
PL is the maximum of
or
A2c = 0.5(tn − tn2 )(Lpr4 − Lpr3 ) + tn2 (LH − Lx4 )
[
]
(tn − tn2 ) (Lx4 − LH )
.
tnx = tn2 1 +
tn2
(Lpr4 − Lpr3 )
PL = 2𝜎avg − 𝜎circ
or
PL = 𝜎circ .
Step 7: Determine the allowable stress value
PL ≤ Sallow
Lx3 and Lx4 are shown in Table 11.4.
Area A3
Sallow = 1.5SE for internal pressure
A3 = tn LI
Sallow = calculated allowable stress for
Area A4
external pressure.
A41 = 0.5L41 2
Step 8: Determine the pressure at the nozzle junction
A42 = 0.5L42 2
Pmax is the smaller of
Sallow
Pmax =
(2Ap ∕AT ) − (Rxs ∕teff )
A43 = 0.5L43 2
Area A5
(11.28)
or
A5 is the smaller of
A5 = A5a
Set-on nozzle
Pmax = S(t∕Rxs ),
or,
where
A5 = A5b
Ap = (fN + Fs + fY )∕P.
where
A5a = Wt e
A5b = LR te
A5b = (LR − tn )te
for set-in nozzle
for set-on nozzle.
Step 5: Determine the forces in the nozzle area
The forces f N , f s , and f Y are obtained from Table 11.5,
where
tn
Rxn =
ln(1 + tn ∕Rn )
teff
.
Rxs =
ln(1 + teff ∕Reff )
Example 11.9
Determine the reinforcement requirements for the 12 in.
ID nozzle shown in Figure 11.20. The ID of the shell is
40 in. The allowable stress of the shell material is 20.5 ksi
and of the nozzle material is 19.4 ksi. The design pressure
is 700 psi. There is no corrosion allowance. Disregard the
small area of the fillet welds.
Solution:
Check thickness of the shell wall from Eq. (8.2)
t = (40∕2)[exp(700∕20,500) − 1]
= 0.70 in. < 1.0 in. acceptable
11.4 Reinforcement Limits
or
14ʺ
8(t + te ) = 8(1.0 + 0.0) = 8.0 in.
12ʺ
5ʺ
1ʺ
Use LI = 2.45 in. for limit.
Step 4: Determine the available area of reinforcement.
Area A1
3ʺ
frp = the smaller of
Di = 40ʺ
frp = 1.0
or
frp = Sp ∕S = 0.0∕20,500 = 0.0.
Figure 11.20 Radial nozzle in a cylindrical shell.
Check thickness of the nozzle wall from Eq. (8.2)
Use f rp = 0.0.
For set-in nozzles,
t = (12∕2)(exp(700∕19,400) − 1)
= 0.22 in. < 1.0 in. acceptable
Step 1: Determine the limits of reinforcement along
the cylindrical shell.
a) Integral nozzle
1) Set-in nozzle
LR is the smaller of
(Reff t)0.5 = [(40∕2)(1.0)]0.5 = 4.47 in.
teff = t + (A5 )(frp )∕LR = 1.0 + 0.0 = 1.0
𝜆 = the smaller of
𝜆 = 12
or
𝜆=
2Rn + tn
2(6.0) + 1.0
=
= 0.317
(Di + teff )teff
(40.0 + 1.0)(1.0)
A1 = the larger of
A1 = the larger of
or
A1 = (t LR ) = (1.0)(4.47) = 4.47 in.2
2Rn = 2(12∕2) = 12 in.
or
Use LR = 4.47 in.
Step 2: Determine the limits of reinforcement along
the nozzle wall projecting outside the cylindrical shell.
1) Set-in nozzle
LH is the smaller of (Rn tn )0.5 + 2.5t
= [(6)(1.0)]0.5 + 2.5(1.0) = 4.95 in.
A1 = (t LR )(𝜆∕5)0.85 = (1.0)(4.47)(0.317∕5)0.85
= 0.44 in.2
Use A1 = 4.47 in.2
Area A2
tn2 = tn = 1.0
LH = 3.45 < 5.0
or
(Rn tn )
0.5
+ t + te = [(6)(1.0)]
0.5
+ 1.0 + 0.0
= 3.45 in.
Hence, A2 = tn LH = (1.0)(3.45) = 3.45 in.2
Area A3
A3 = tn LI = (1.0)(3.0) = 3.0 in.2
or
Lpr1 + t = 5.0 + 1.0 = 6.0 in.
Area A4
A4 = 0.0
or
9t + 8te = 9(1.0) + 0.0 = 9.0 in.
Use LH = 3.45 in. for limit.
Step 3: Determine the limits of reinforcement along
the nozzle wall projecting inside the cylindrical shell.
LI is the smaller of
LI is the smaller of
(Rn tn )0.5 = [(6)(1.0)]0.5 = 2.45 in.
or
Lpr2 = 3.0 in.
Area A5
A5 = 0.0
AT = A1 + A2 + A3 + A4 + A5 = 10.62 in.2
Step 5: Determine the forces in the nozzle area.
tn
1.0
=
= 6.487
Rxn =
ln(1 + tn ∕Rn ) ln(1 + 1.0∕6.0)
teff
1.0
=
Rxs =
ln(1 + teff ∕Reff ) ln(1 + 1.0∕20.0)
= 20.496
209
210
11 Openings, Nozzles, and External Loadings
f N = PRxn LH = (700)(6.487)(3.45) = 4544 lbs.
f s = PRxs (LR + tn ) = (700)(20.496)(4.47) = 64,132 lbs.
f Y = PRxs Rnc = (700)(20.496)(6.0) = 86,080 lbs.
Step 6: Determine the local primary stress.
(f + Fs + fY ) 1,54,756
=
𝜎avg = N
= 14,570 psi
AT
10.62
PRxs
(700)(20.496)
=
= 14,347 psi.
𝜎circ =
teff
1.0
PL is the maximum of
PL = 2𝜎avg − 𝜎circ = 2(14,570) − 14,347
= 14,793 psi
or
PL = 𝜎circ = 14,347 psi.
Hence, PL = 14,793 psi.
Step 7: Determine the allowable stress value.
Sallow = 1.5(20,500)(1.0) = 30,750 psi
PL = 14,793 psi ≤ 20,750 psi
acceptable
Step 8: Determine the pressure at the nozzle junction
Ap = (fN + fs + fY )∕P = 154,756∕700 = 221.1 in2 .
Pmax is the smaller of
Sallow
Pmax =
(2Ap ∕AT ) − (Rxs ∕teff )
30,750
=
[(2(221.1)∕10.62) − (20.496∕1.0)]
= 1454 psi
Reinforcement Rules for ANSI/ASME B31.1
The rules for welded, reinforced connections according
to ANSI/ASME B31.1, Power Piping, are similar to the
rules for reinforced openings in the ASME Code, I and
VIII-1. The following requirements give the basic considerations.
11.4.4.1
No Reinforcement Calculations Required
No calculations for reinforcement are required when the
following limits are met:
1) Connections made from fittings for which a standard
pressure–temperature rating are established.
2) dmax = 2 in. NPS with t min at least that for Schedule
160 pipe.
3) d/D ≤ 0.25.
4) Standard fittings of extra-heavy or Class 3000 rating.
Limitations. Angle between the branch and run or
nozzle and shell is between 45 and 90∘ .
Notation and Definitions. (See Figure 11.21).
A = additional thickness (in.)
𝛼 = angle between nozzle and shell (degrees)
Doh = outside diameter of run or header (in.)
Dob = outside diameter of branch or nozzle (in.)
d1 = [Dob − 2(T b − A)]/sin 𝛼
d2 = the greater of d1 or [(T b − A) + (T h − A) + d1 /2]
but not greater than Dob
L4 = the smaller of [2.5(T b − A) + t r ] or [2.5(T h − A)]
or
Pmax = S(t∕Rxs ) = (20,500)(1.0∕10.62)
= 1930 psi.
Hence, Pmax = 1454 psi > 700 psi acceptable.
Problems
11.10
11.4.4
A 10 in. ID nozzle is attached by a full-penetration weld and corner fillet weld to a 48 in. inside
diameter as shown in Figure 11.16b. The shell
material is SA-266 Class 1 carbon steel, and the
nozzle material is SA-182 F304 stainless steel.
The design pressure is 1250 psi at a design temperature of 500∘ F. What is the required thickness
of the shell, nozzle, and pad (if required) to
satisfy the reinforcement requirements?
Answer:
shell, T s = 1.93 in.
nozzle, T n = 0.375 in.
pad, t e = 2.125 in.
t r = thickness of pad reinforcement
Required Area of Reinforcement The total cross-sectional
area of reinforcement required for any plane through the
center of an opening is given by
A7 = (tmh − A)d1 (2 − sin 𝛼),
(11.29)
which for 𝛼 = 90∘ is
A7 = (tmh − A)d1 .
(11.30)
Available Area of Reinforcement The total area available for
reinforcement is the sum A1 + A2 + A3 + A4 + A5 , where
each area is determined as follows:
A1 = (2d2 − d1 )(Th − tmh )
2L (T − tmb )
A2 = 4 b
sin 𝛼
(11.31)
(11.32)
A3 = area of fillet welds
A4 = area of reinforcing rings, pads, and so on.
A5 = area of saddles.
11.4 Reinforcement Limits
Dob
Tb
Additional
thickness A
tmb
Excess wall
in branch
Reinforcement
zone
Reinforcement
zone
Mill tolerance
Ring or pad,
A4 [Notes (1)
and (2)]
Branch without
reinforcement
(for saddle
see detail on
Excess wall
next page)
in header
A3
A2
Branch pipe
or nozzle
A2
A3
L4
Nominal
thickness
A3
tr
Th
A6
A1
Nominal
thickness
tmh
Mill tolerance
d1
d2
d2
A1
Header or
run pipe
α
Additional
thickness, A
[see para.
104.1.2(A.6)]
C
L Run or header
Branch
Doh
C
L
Example A
Explanation of areas:
Area A1 — available reinforcement area (excess wall ) in header
Area A4 — metal in ring, pad, or integral reinforcement
Area A2 — available reinforcement area (excess wall ) in branch
Area A5 — metal in saddle parallel to run (see detail)
Area A3 — available reinforcement area filledt weld metal
Area A6 — pressure design area (expected
at the end of service life)
Reinforcement
zone
A2
Branch pipe
or nozzle
A3
Saddle, A5
[Note (3)]
A3
90°
tr
Tb
d1
Reinforcement
area
Excess wall
in header
A1
d1
Reinforcement
area
tr
tr
60°
Th
Header or
run pipe
Tb
Th
(a)
60°
(b)
Example B
Detail
for example A
Figure 11.21 Dimensions and notations for nozzle reinforcement in ASME B31.1. Source: Courtesy of ASME.
211
212
11 Openings, Nozzles, and External Loadings
Reinforcement Zone The limits of reinforcement are
formed by a parallelogram with sides d2 on each side of
the nozzle center line and an altitude L4 perpendicular
to the shell surface.
GENERAL NOTES:
2) Determine the minimum required thickness of the
branch pipe as follows:
Pr
2500 × 4.0
=
tmb =
SE − 0.6P
14,500 × 1 − 0.6 × 2500
= 0.769 in.
a) This figure illustrates the nomenclature of para.
104.3.1(D).
b) Required reinforcement area = A7 = A6 (2 − sin 𝛼) =
(t mh − A)dt (2 − sin 𝛼).
c) Available reinforcement areas = A1 + A2 + A3 + A4 +
A5 (as applicable).
d) Available reinforcement areas are greater than the
required reinforcement area.
Use T b = 2.0 in.
3) The area required for reinforcement according to
Eq. (11.29) for d1 = 8/sin 75∘ = 8.282 in. is
A = 1.07(2.308)(8.282)(2 − sin 75∘ ) = 21.150 in.2
7
4) The horizontal limits of reinforcement are the
larger of
d1 = 8.282 in.
= 2.5 + 2.0 + 4.141 = 8.641 in.
Notes:
1) When a ring or pad is added as reinforcement
(Example A), the value of reinforcement area may
be taken in the same manner in which excess header
metal is considered, provided the weld completely
fuses the branch pipe, header pipe, and ring or pad.
Typical acceptable methods of welding that meet the
aforementioned requirement are shown in Figure
127.4.8(D), sketches (c) and (d) of B31.1.
2) Width to height of rings and pads shall be reasonably
proportioned, preferably at a ratio as close to 4 : 1 as
the available horizontal space within the limits of the
reinforcing zone along the run and the outside diameter of the branch will permit, but in no case may the
ratio be less than 1 : 1.
3) Reinforcement saddles are limited to use on 90∘
branches (Example A Detail).
Use 8.641 in.
5) The perpendicular limit of reinforcement is as follows: Assume that a 0.875 in. thick pad is added
and attached by full-penetration welds that are
examined. Then,
tr = 0.875 in.
L = 2.5Tb + tr = 2.5(2) + 0.875 = 5.875 in.
6) The area available for reinforcement is
A1 = (2 × 8.641 − 8.282)(2.5 − 2.308)
= 1.728 in.2
2(5.875)(2 − 0.769)
A2 =
= 14.974 in.2
sin 75∘
A3 = (0.5)2 = 0.250 in.2
A4 = (2 × 8.5 − 8 − 4)(0.875) = 4.375 in.2
At = A1 + A2 + A3 + A4 = 21.327 in.2 > A7
Multiple Openings The following should be applied:
1) Count overlapping areas only once.
2) Try to limit the center-line spacing to 1.5dav , with at
least 50% of the area between openings.
Example 11.10
A steam pipe has a 24 in. inside diameter with a design
pressure of 2500 psi and an allowable stress of 14,500 psi
(y = 0.4) at the design temperature. A branch with an
inner diameter of 8 in. connects at an angle of 𝛼 = 75∘ .
The branch is attached by a full-penetration weld that is
radiographed, so that E = 1.0. Determine the thickness
and reinforcement requirements.
= 21.150 in.2
Shell requirement: 24 in. ID × 2.5 in. thick
Nozzle requirement: 8 in. ID × 2.0 in. thick
Fillet-weld requirements: 2 with 0.5 in. legs
Pad requirement: 17 in. OD ring × 0.875 in. thick
Nozzle is attached to shell and pad by full-penetration
welds.
Problem
11.11
Solution:
1) Determine the minimum required thickness of the
run pipe as follows:
PR
2500 × 12
=
tmh =
SE − 0.6P
14,500 × 1 − 0.6 × 2500
= 2.308 in.
Use T h = 2.5 in.
or Tb + Th + d1 ∕2
If the nozzle were attached at 𝛼 = 90∘ , what thickness would be required for the pad, if any?
Answer:
3.4 in. thick pad
11.4.5
Reinforcement Rules for ANSI/ASME B31.3
The reinforcement requirements for ANSI/ASME B31.3,
Chemical Plant and Petroleum Refinery Piping, are
11.4 Reinforcement Limits
Db
Tb
Reinforcement
zone
c
tb
Reinforcement
areas
L4
Branch
pipe
A1
Da
Tb
Branch
pipe
A3
A4
Tr
T h Th
Normal
thickness
Mill tolerance
Reinforcement
areas
A4
A2
Reinforcement
zone
Multiply this area
by 12 – Sin β1 to get
required area
th
c
Mill tolerance
A2
Normal
Run
pipe
thickness
d1
A2
d2
C
L Pipe
Run pipe
d2
β
Figure 11.22 Nomenclature and dimensions of ANSI/ASME B31.3 piping code. Source: Courtesy of American Society of Mechanical
Engineers: from Figure 304.3.3 of ASME/ANSI B31.3.
similar to the requirements for ANSI/ASME B31.1 and
for Section VIII, Division 1. Rules are given for branch
connections or nozzles, which are attached to run piping
or headers. Differing from other reinforcement calculations, the minimum required thickness of the branch
piping and the run piping is measured from the outside
of the piping. The area available for reinforcement is the
remainder of the piping’s nominal thickness as shown in
Figure 11.22.
Limitations of Geometry The angle between the nozzle
and header is restricted to those intersections where the
acute intersection angle 𝛽 is equal to 45∘ or more.
Limitation When No Reinforcement Calculations Are Required
1) Standard fittings for which pressure–temperature ratings have been determined
2) Standard fittings not exceeding 2 in. NPS that have
d/D ≤ 0.25 and a pressure rating of 2000 lb or more
3) Integrally reinforced connections that have been
proved adequate by tests, calculations, and use
11.4.5.1
Nomenclature
d1 = opening size in run or header (in.)
d2 = horizontal limit on one side measured from center
line of opening (in.)
L4 = vertical limit perpendicular to header surface (in.)
𝛽 = acute angle at intersection (degrees)
t h = required thickness of header (in.)
T h = nominal thickness of header (in.)
t b = required thickness of branch (in.)
T b = nominal thickness of branch (in.)
Db = outside diameter of branch (in.)
Dh = outside diameter of header (in.)
Required Area of Reinforcement, A1 For internal pressure,
A1 = th d1 (2 − sin 𝛽).
(11.33)
For external pressure,
th d1 (2 − sin 𝛽)
.
(11.34)
2
Horizontal Limits. The horizontal limit on each side
of the center line of the nozzle is the larger of
A1 =
d1
or
Th + Tb + 0.5d1
but not more than Dh .
Vertical Limits. The vertical limit measured from the
surface of the shell is the smaller of
2.5Th
or
2.5Tb + Tr ,
where
Tr = pad thickness (in.)
Areas Available for Reinforcement: A2 , A3 , and A4 The excess
thickness in header or run, A2 , is
A2 = (2d2 − d1 )(Th − th ).
(11.35)
213
214
11 Openings, Nozzles, and External Loadings
The excess thickness in nozzle or branch, A3 , is
A3 =
2L4 (Tb − tb )
.
sin 𝛽
Determine the vertical limits of reinforcement from the
following, whichever is smaller:
(11.36)
In other metals available with limits A4 , if reinforcement metal is weaker than vessel metal, the area available
for reinforcement is reduced by SR /SV .
Reinforcement Zone. Excess area within the following
is considered acceptable: 2d2 × L4 , where L4 is measured
perpendicular to the shell surface.
Multiple Openings. The following cautionary rules
are to be followed:
1) Center-to-center distance of at least 1.5dav
2) At least 50% of the total required area between the
openings
L4 = 2.5(0.519) = 1.298 in.
or
L4 = 2.5(0.438) + 0 = 1.095 in.
Use L4 = 1.095 in.
Determine the reinforcing area required according to
Eq. (11.33) as follows:
A = (0.450)(7.125)(2 − sin 90∘ ) = 3.206 in.2
1
Determine the areas of reinforcement as follows.
Excess in run (header):
A2 = (7.125)(0.519 − 0.450) = 0.492 in.2
Excess in branch (nozzle):
Example 11.11
An 8 in. NPS Schedule 80 branch (nozzle) is attached at
right angles to a 20-in. NPS Schedule 40 run (header)
with a full-penetration weld with fillet-weld cover. The
allowable stress is 13.1 ksi. The design pressure is 600 psi
at a design temperature of 900∘ F. Determine the reinforcing requirements and pad sizer, if required.
Solution:
Determine the actual and minimum required sizes at the
branch–run intersection as follows:
Th = (0.593)(0.875) = 0.519 in.
A3 = 2(1.095)(0.438 − 0.194) = 0.534 in.2
In fillet welds:
( ) ( )2
1
1
A4 = 2
= 0.063 in.2
2
4
Excess area in A2 + A3 + A4 = 1.089 in.2 This is less than
A1 ; consequently, a pad shall be provided. Determine the
thickness of pad based on the pad extending to the horizontal limits of reinforcement.
(7.125)H = 3.206 − 1.089 H = 0.297 in.
Assuming Tr =
width.
PD
600 × 20
=
2(SE + PY ) 2(13,100 + 0.4 × 600)
= 0.450 in.
Pd
600 × 8.625
tb =
=
2(SE + PY ) 2(13,100 + 0.4 × 600)
= 0.194 in.
The fillet-weld size is the smaller of 0.7Tb or 1/4 in.:
0.7(0.438) = 0.306 in.
Use 1/4 in.
Determine the horizontal limits of reinforcement from
the following, whichever is the greater:
8 − 2 × 438
= 7.125-in. opening size
sin 90∘
d2 = (0.519) + (0.438) + 0.5(7.125) = 4.519 in.
d1 =
or
d2 = d1 = 7.125 in.
Use d2 = 7.125 in.
determine the minimum pad
(2W − 7.125)(0.3125) = 2.117
Tb = (0.500)(0.875) = 0.438 in.
th =
5
in.,
16
W = 6.95 in. Use 14 in. ×
5
16
in. pad.
A5 = (14 − 7.125)(0.3125) = 2.148 in.2
The available area of reinforcement is A2 + A3 + A4 +
A5 = 3.237 in.2 This is greater than A1 and therefore is satisfactory.
It may be possible to obtain more refinement and a
thinner or narrower plate by reevaluating the vertical
limits by setting T r = 0.312 in. and by including the outer
fillet welds if they lie within the horizontal reinforcement
limits. This recalculation may reduce the pad thickness
and/or the pad width. However, a 14 in. by 5/16 in. pad
is satisfactory.
Problem
11.12
For the construction in Example 11.11, what is
the maximum allowable working pressure when
the allowable stress is increased to 18.8 ksi?
Answer:
MAWP = 880 psi
11.5 Ligament Efficiency of Openings in Shells
11.5 Ligament Efficiency of Openings
in Shells
In addition to the method of reinforced openings for
compensating for metal removed at openings in shells,
there is the method of ligament efficiency. This method
considers the load-carrying ability of the area between
two points in relationship to the load-carrying ability of the ligament remaining when the two points
become the centers of two openings. In the ASME
Code, only the shell plate is considered; however, Lloyd’s
Rules [6] permit some help from integrally attached
nozzles.
The basic method of diagonal ligament efficiency for
application in the ASME Code was developed in 1915 by
Black and Jones of the Babcock & Wilcox Company and
was published in 1920 in the Marine Engineers Handbook
[7]. In 1975, a limit design analysis was used to examine
the stresses in a perforated cylindrical shell [8]. This
limit analysis was further developed for ASME Code
application and used to update the original code rules.
The rules and curves are still given in several sections
of the ASME Code and several foreign codes that determine the ligament efficiency used in the cylindrical-shell
formulas.
The ligament-efficiency curves apply only to cylindrical pressure vessels where the circumferential tension
(stress) has twice the intensity of the longitudinal tension
(stress). Once this was established, Rankine’s ellipse of
stress was used to determine the intensity of tension
and of shear on any diagonal ligament. This is shown
in Figure 11.23. The total tension and the total shear
are obtained by multiplying the intensity of tension and
of shear, respectively, by the diagonal pitch between
openings and by the shell thickness, as expressed by the
following equations:
cos2 𝜃 + 1
2
(11.37)
sin 𝜃 cos 𝜃
intensity of shear on any plane =
2
(11.38)
cos2 𝜃 + 1 ′
total tension =
(11.39)
(p Ts )
2
sin 𝜃 cos 𝜃 ′
total shear =
(11.40)
(p Ts ).
2
The stress factor for tension for any section of the ligament is obtained by dividing the total tension by the
cross-sectional area of the ligament. The stress factor for
bending for any section of the ligament is obtained by
dividing the bending moment of the section by the modulus of the section. The bending moment is the total shear
multiplied by the distance between the section considered and the point of contraflexure, which is on the plane
passing through the centers of the openings. This distance is Y as shown in Figure 11.23. Thus,
√
length of ligament = L = p′ − d2 − 4Y 2 (11.41)
intensity of tension on any plane =
bending moment = (total shear)(Y )
2
section modulus =
′
Ts L
T (p −
= s
6
√
(11.42)
d2 − 4Y 2 )2
.
6
(11.43)
The stress factor for tension is determined as follows:
p′ Ts
cos2 𝜃 + 1
(11.44)
St =
√
2
(p′ − d2 − 4Y 2 )T
s
The stress factor for bending is determined as follows:
(p′ Ts )Y
sin 𝜃 cos 𝜃
Sb =
(11.45)
√
2
T (p′ − d2 − 4Y 2 )∕6
s
Figure 11.23 Diagonal ligaments.
d
ere
sid
n
o
gs
gc
nin
ein
e
b
p
n
o
of
ctio
Se
ine
l
r
nte
Ce
pʹ
L
d
Y
θ
Shell axis
s
215
216
11 Openings, Nozzles, and External Loadings
Table 11.6 Sample calculation of maximum ligament factor.
Y
Length
and area
Total
tension
Section
modulus
Total
shear
Bending
moment
Tension
factor
Bending
factor
Total
factor
0
3.32
6.12
1.84
1.72
—
1.84
—
1.84
0.50
3.45
6.12
1.98
1.72
0.86
1.77
0.43
2.20
0.75
3.62
6.12
2.18
1.72
1.29
1.69
0.59
2.28
0.90
3.75
6.12
2.34
1.72
1.55
1.63
0.66
2.29
1.00
3.86
6.12
2.48
1.72
1.72
1.59
0.69
2.28
1.50
4.68
6.12
3.65
1.72
2.58
1.31
0.71
2.02
2.00
7.32
6.12
8.93
1.72
3.44
0.84
0.39
1.23
The total stress factor for both tension and bending is
the sum of the stress factors for tension and for bending.
When the curve was originally developed for the ASME
Code, the maximum total stress factor was found by
trial by calculating the stress factor at several sections
between the sections through the centers of the two
openings to the plane that is tangent to the edge of the
openings. Table 11.6 shows a sample of this calculation
to determine the maximum factor for a particular angle
with the longitudinal-axis 𝜃 values equal to p′ /d and a
shell plate thickness T s .
Under the sponsorship of the Pressure Vessel Research
Committee, an extensive limit design analysis of perforated cylindrical shells with uniform patterns of openings
was completed. This analysis was used to determine the
upper and lower bounds of the limit pressure. A 2 : 1
stress–field ratio was considered, and the shell plate
curvature was not included. From this analysis, the basic
lower-bound equation was developed into
ligament efficiency
Example 11.12
Given
d = 4 in.
s = 6 in.
𝜃 = 35∘
Ts = 1 in.
calculate the ligament efficiency
Solution:
p′
s
6
7.32
=
= 7.32 in.
=
= 1.83
cos 𝜃
cos 𝜃
d
4.0
The lowest factor is used to calculate the minimum efficiency for the angle 𝜃 being examined:
p′ =
efficiency =
1
= 0.437 = 43.7%
2.29
Example 11.13
Determine the minimum required thickness of the shell
given in Example 11.5 using the ligament-efficiency rules.
Solution:
√
1 + cos2 𝜃 − (d∕P) 1 + 3 cos2 𝜃
=2
.
1 + 3 cos2 𝜃
(11.46)
For application in the ASME Code, the equation was
rearranged so that the diagonal efficiency term was
expressed as p′ /d, a number equal to or greater than 1.0,
and the efficiency was expressed as a percentage. The
equation for code use is
sec 𝜃 √
3 + sec2 𝜃
sec2 𝜃 + 1 − ′
p ∕d
E% =
.
(11.47)
0.015 + 0.005 sec2 𝜃
The minimum ligament efficiency was calculated by
examining various planes between openings at different
distances of Y as shown in Figure 11.23. Example 11.12
shows calculations based on the original ASME Code
work. It has to be repeated for different values of p′ /d
and for various angles 𝜃.
1) Determine the longitudinal efficiency based on the
longitudinal spacing of 4.5 in.:
E=
p′ − d
4.5 − 2.25
=
= 0.500
p′
4.5
2) Determine the equivalent longitudinal efficiency from
the diagonal efficiency using Eq. (11.47) as follows:
p′ = 3.75 𝜃 = 53.13∘ d = 2.25
p′
= 1.67 sec 𝜃 = 1.667
d
∘
∘
[sec2 53.13
√ + 1 − (sec 53.13 )∕
1.67 3 + sec2 53.13∘ ]
E% =
0.015 + 0.005 sec2 53.13∘
E% =
2
[(1.667)√
+ 1 − (1.667∕1.67)∕
3 + (1.667)2 ]
0.015 + 0.005(1.667)2
= 47.59%
11.6 Fatigue Evaluation of Nozzles Under Internal Pressure
3) Determine the minimum required thickness using the
equation of UG-27(c)(1) from the ASME Code, VIII-1,
as follows:
PR
t=
SE − 0.6P
500 × 18
=
16,600 × 0.476 − 0.6 × 500
= 1.184 in.
This thickness of t = 1.184 in. is based on the shell
thickness only, with no contribution from the nozzle.
Problems
11.13
What must the angle 𝜃 be between two openings
for the longitudinal and diagonal efficiencies to
be equal when the longitudinal spacing is 4.5 in.
and the opening diameter is 2.25 in.?
Answer:
Angle with longitudinal axis is 𝜃 = 54.1∘ .
11.14
The stress-index method permits easy calculation of
peak stresses at the nozzle–shell or nozzle–head intersection without resorting to any complex analysis. The
method gives conservative results, and if the exact multipliers for a specific geometry are known, they should be
used. Essentially, the nominal stress in the shell or head
is multiplied by the stress indices, and the peak stresses
are obtained.
The peak stresses are determined from the following
equations where the stress index I (Table 11.7) is multiplied by the nominal stress.
For spherical shells and formed heads:
PD
(11.48)
𝜎 = I m.
4Ts
For cylindrical shells:
PD
𝜎 = I m,
2Ts
(11.49)
where
R = inside radius of shell or head (in.)
r = inside radius of nozzle (in.)
In Problem 11.13, what is the circumferential
spacing for the same configuration?
T s = nominal thickness of shell or head (in.)
Dm = mean diameter of shell or head (in.) = 2R + T s
P = internal design pressure or pressure range (psi)
Answer:
The circumferential spacing is 3.108 in.
I = stress index for various locations (see Table 11.5)
𝜎 n = normal stress in the plane being examined (psi)
𝜎 t = tangential stress in the plane being examined
(psi)
11.6 Fatigue Evaluation of Nozzles
Under Internal Pressure
When a fatigue evaluation is required, it is necessary
to determine the peak stresses around the openings.
The current methods are the stress-index method,
experimental tests and measurements, and a theoretical
analysis procedure such as a finite-element analysis. The
stress-index method is the easiest method and is allowed
by the ASME Code, III-1, and VIII-2. It was developed
after reviewing a large amount of experimental and
analytical data determined in a program conducted by
the Pressure Vessel Research Committee.
𝜎 r = radial stress in the plane being examined (psi).
Example 11.14
A cylindrical shell that is 36 in. ID by 2.5 in. thick contains a perpendicular nozzle that is 4 in. ID by 0.75 in.
thick. The design pressure is 1900 psi at a design temperature of 450∘ F. The vessel is subjected to cyclic operation,
and a fatigue analysis is required. The peak stresses and
stress-concentration factors are not known for the specific geometry to be used. What method can be used to
evaluate the peak stresses for a fatigue analysis?
Table 11.7 Stress index I.
Spherical shells
and formed Heads
Cylindrical shells
Longitudinal plane
Inside
Outside
Transverse plane
Stress
Inside
Outside
Inside
Outside
𝜎n
2.0
2.0
3.1
1.2
1.0
2.1
𝜎t
−0.2
2.0
−0.2
1.0
−0.2
2.6
𝜎r
−4T s /Dm
0
−2T s /Dm
0
−2T s /Dm
0
217
218
11 Openings, Nozzles, and External Loadings
Solution:
At the intersection of the nozzle to the shell, peak stresses
are obtained according to Eq. (11.49). The nominal stress
is determined as
PDm
1900(36 + 2.5)
𝜎=
=
= 14,630 psi.
2Ts
2(2.5)
The peak stresses are determined as follows, using the
factors for the stress index I from Table 11.7:
Longitudinal plane
Transverse plane
Stress (psi)
Inside
Outside
Inside
Outside
Hoop stress
+45,350
+17,560
+14,630
+30,720
Longitudinal stress
−2 930
+14,630
−2 930
+38,040
Radial stress
−1 900
0
−1 900
0
These values are the peak stresses due to internal pressure only and must be combined with other peak stresses
occurring at the same location on the shell for a fatigue
evaluation.
Problem
11.15
A reactor vessel has 5 ft. 0 in. inner diameter
with hemispherical heads. The design pressure
is 450 psi at 650∘ F. The allowable stress of the
vessel is 17,500 psi. The head and shell are made
of minimum-thickness material rounded up to
the next 1/4 in. The vessel is operated under a
cycling condition, so that a fatigue analysis is
necessary. It is necessary to place an 8 in. ID
by 1 in. thick nozzle in the vessel. Is the peak
stress less in the head or in the shell at the nozzle
junction, and what are the values of the peak
stresses at the maximum location in the head
and shell? Thin-walled equations are used.
Answer:
𝜎 peak = +48,510 psi in shell
𝜎 peak = +27,220 psi in head
11.7 External Loadings
When external loadings are applied to nozzles or
branch piping, local stresses are generated at the
nozzle–shell intersection. Several types of loading
may be applied, such as sustained loadings, transient
loadings, and thermal-expansion flexibility loadings.
Sustained loadings, such as dead loads, are continuously
applied and combined with internal pressure. Transient
loadings, such as earthquake and wind loadings, pressure
fluctuations, and water-hammer loadings, are applied
for a short period of time. Thermal-expansion loadings
are caused by axial growth of piping due to heating.
When external loadings are applied to nozzles, stresses
are generated in both the nozzle and the shell or head.
The stresses in the nozzle are both general membrane
stresses, acting upon the entire nozzle cross section, and
local membrane stresses, acting through the nozzle cross
section, and local membrane stresses, acting through the
nozzle wall thickness. However, analysis procedures are
available only for the general stresses unless one resorts
to some procedure such as a finite-element analysis.
These general procedures usually involve a stress intensification factor (SIF) that predicts the local stresses in
the nozzle. The stresses generated in the shell or head
adjacent to the nozzle are local stresses. A procedure for
determining them is given in detail in Welding Research
Council Bulletin No. 107 [9].
11.7.1
Local Stresses in the Shell or Head
Although a considerable amount of theoretical development work on local stresses in shells from external loadings was conducted and reported by P. P. Bijlaard [10–19]
in the early 1950s, it was not until the Welding Research
Council Bulletin No. 107 (WRC 107) was issued that all
the miscellaneous information from Bijlaard and others
was put into a concise form for easy use. The range of
usage is restricted by limitations on various parameters,
but it is far better than anything available before. Currently, experimental and theoretical work is being conducted to extend the useful range.
External loadings considered by WRC are the longitudinal moment, transverse moment, torsional moment,
and axial force. Stresses at various locations on the inside
and outside surfaces are obtained by combining the
stresses from various effects. This involves considerable
“bookkeeping” that WRC 107 developed. Once the
stresses are obtained according to WRC, they must be
combined with internal-pressure stresses to determine
the overall stresses.
Bijlaard’s original problem was finding the effects of
structural supports on a cylindrical shell. This initial
work considered the radial loads and moments over a
flexible, rectangular loading surface. The initial treatment of nozzles was an approximation based on a rigid
attachment without the effects of nozzle wall flexibility.
Bijlaard extended this work to spherical shells, using a
shallow-shell theory, and considered both solid (rigid)
attachments and nozzles with flexibility parameters.
Usage of the rules is generally limited to the same
dimensional limits given in item 5 of the alternative rules
for nozzle design of Article 11.4.3.
11.7 External Loadings
The curves in WRC 107 are related to certain parameters at the intersection. The two important parameters
are the shell parameter and the attachment parameter.
The different applications to spherical and cylindrical
shells are as follows.
Spherical Shells
The shell parameter is
r
U=√ 0 .
Rm T
(11.50)
For a square attachment, the shell parameter is
U=
c1
,
√
0.875 Rm T
(11.51)
where
r0 = outside radius of cylindrical attachment (in.)
Rm = mean radius of spherical shell (in.)
T = thickness of spherical shell (in.)
c1 = half side dimension of square or rectangular
attachment (in.)
In the attachment parameter for all solid attachments,
no parameter is needed.
For a hollow cylinder (nozzle), we define
r
Υ= m
(11.52)
t
T
(11.53)
𝜌= .
t
For a hollow square,
rm
Υ=
0.875t
T
𝜌= .
t
(11.54)
(11.55)
Cylindrical Shells The shell parameter is
𝛾=
Rm
.
T
(11.56)
The attachment parameter for both solid and hollow is
for a cylinder
𝛽=
0.875ro
,
Rm
(11.57)
where
ro = outside radius of cylindrical attachment (in.)
Rm = mean radius of cylindrical shell (in.)
T = thickness of cylindrical shell (in.)
c1 = half side dimension of square or rectangular
attachment (in.).
For a square,
c
𝛽 = 1 where c1 = c2 .
Rm
(11.58)
For a rectangle,
c
𝛽1 = 1
Rm
c
𝛽2 = 2 .
Rm
If 𝛽 1 /𝛽 2 > 1,
[
)
]
(
√
1 𝛽1
𝛽 = 1−
− 1 (1 − K1 )
𝛽1 𝛽2 .
3 𝛽2
If 𝛽 1 /𝛽 2 < 1,
[
]
)
(
√
𝛽1
4
𝛽 = 1−
𝛽1 𝛽2 .
(1 − K2 )
1−
3
𝛽2
(11.59)
(11.60)
Using these parameters and the curves given in WRC
107, stresses may be calculated at the inner and outer
surfaces due to the various loadings. The proper values
must be read carefully because it may be necessary to
interpolate not only from line to line but from curve
to curve. The values on adjacent charts do not always
increase or decrease in a consistent direction. Computation sheets are given for: in Figure 11.24, solid
attachment to a spherical shell; in Figure 11.25, hollow
attachment to a spherical shell; and in Figure 11.26,
all attachments to a cylindrical shell.
In addition to the limitations on the geometry in the
analysis in WRC, there are other limitations. This analysis determines only the stresses in the shell or head due
to the external loadings, and thus, those from internal
pressure must be added to them. Because no nozzle
stresses are determined by this method, they must be
determined by a separate analysis. Engineers believe
in fact that when the external loadings are applied to a
relatively thin-walled nozzle, the highest stresses may
be in the nozzle. For thick-walled nozzles, it appears
that deformation is similar to that of a solid attachment,
and maximum stresses will occur in the shell or head
adjacent to the nozzle. Bijlaard’s method indicated that
for a longitudinal moment, the maximum stress occurs
on the longitudinal axis. However, experimental results
obtained in PVRC tests indicate that for larger nozzles
with a d/D = 0.5 or larger, the maximum stress may lie
somewhat off the longitudinal axis. Thus, adjustments
have been made to some of the curves in WRC 107. In
spite of these shortcomings, a reasonable estimate of
the stresses due to the external loadings is obtained by
following WRC 107.
Recently, in considering certain PVRC work to extend
the useful range of WRC 107, J. L. Mershon concluded
that within the range of its applicability, the curves for
loadings on a cylindrical shell could be reduced, for all
219
220
11 Openings, Nozzles, and External Loadings
Figure 11.24 Computation sheet for rigid attachment to spherical shell. Source: Courtesy of Welding Research Council, WRC Bulletin 107,
August 1965.
11.7 External Loadings
Figure 11.25 Computation sheet for hollow attachment to spherical shell. Source: Courtesy of Welding Research Council, WRC Bulletin
107, August 1965.
221
222
11 Openings, Nozzles, and External Loadings
Figure 11.26 Computation sheet for attachments to cylindrical shell. Source: Courtesy of Welding Research Council, WRC Bulletin 107,
August 1965.
practical purposes, to an easier-to-use set of curves given
in Appendix K. This set of simplified curves practically
eliminate the need to interpolate between various curves
in WRC 107 to determine the factors used to calculate
the stresses. When the simplified curves are used, it
will still be necessary to combine the internal pressure
stresses and to develop a method of “bookkeeping”
for the signs of the various stresses due to different
loadings.
The sign convention used with the Mershon method is
identical to that of WRC 107, as shown in Figure 11.26.
The figure shows that stresses may be obtained at the
same locations. The relationship of the curves given in
WRC and the Mershon curves given in Appendix K is as
follows:
11.7 External Loadings
Appendix K figures
WRC 107 figures
K.1
1A, 2A
K.2
3A, 4A
K.3
1B, 1B-1, 2B, 2B-1
K.4
3B, 4B
K.5
1C, 1C-1
K.6
2C, 2C-1
K.7
3C(1), 4C(1)
K.8
3C(2), 4C(2)
For the simplified method, only one parameter is
required in using the curves – the shell-opening parameter 𝜆, which is determined as follows:
d
𝜆= √ o ,
Dm T
where
do = outside diameter of attachment (in.)
Dm = mean diameter of shell (in.)
T = nominal thickness of shell (in.).
Because the ASME Code, VIII-1, has neither an acceptance criterion nor a method to classify stresses, the
designer has to establish a method that is acceptable
to the Authorized Inspector. For guidance, the method
in the ASME Code, VIII-2, may be followed by considering the differences in stress theory and allowable stresses
between the methods in VIII-1 and VIII-2. This method
permits the designer to assign stresses into such categories as primary stresses, secondary stresses, and peak
stresses, depending upon what loadings are included.
Example 11.15
A cylindrical shell that is 84 in. ID by 1.0 in. nominal
thickness contains a nozzle that is 8 in. ID by 1.0 in.
nominal thickness. The design pressure is 400 psi, and
the allowable stress of the material is 17.5 ksi. The nozzle
is subjected to an inward radial loading of 12,000 lb
and an applied moment in the longitudinal direction of
150,000 in.-lb. What are the combined stresses on the
longitudinal axis due to these two external loadings using
the Mershon method and the curves in Appendix K? The
vessel is not subjected to cyclic loading, and therefore,
no stress concentration factors need be considered.
Solution:
1) The shell parameter is
d
10
= 1.08.
𝜆= √ o = √
Dm T
(85)(1)
2) Using this parameter, the constants from the radial
loading on the longitudinal axis are as follows:
From Figure K.5, Mx /P = 0.127.
From Figure K.6, M𝜙 /P = 0.086.
From Figure K.8, N x T/P = 0.160.
From Figure K.8, N 𝜙 T/P = 0.176.
3) Using these constants, the stresses due to the radial
loading are determined as follows:
[
]
(6)(12,000)
Mx from P(bending) = 0.127
(1)2
= 9140 psi
]
[
(6)(12,000)
M𝜙 from P(bending) = 0.086
(1)2
= 6190 psi
[
]
(12,000)
Nx from P(membrane) = 0.160
(1)2
= 1920 psi
[
]
(12,000)
N𝜙 from P(membrane) = 0.176
(1)2
= 2110 psi
4) Using the parameter in item 1, the constants from the
applied longitudinal moment on the longitudinal axis
are determined as follows:
do
= 0.170
ML
d
From Figure K.3, M𝜙 o = 0.104
ML
do T
From Figure K.4, Nx
= 0.076
ML
d T
From Figure K.4, N𝜙 o = 0.260
ML
From Figure K.3, Mx
5) Using these constants, the stresses due to the longitudinal moment are determined as follows:
Mx from ML (bending)
]
[
(6)(1,50,000)
= 15,300 psi
= 0.170
(10)(1)2
M𝜙 from ML (bending)
[
]
(6)(1,50,000)
= 0.104
= 9360 psi
(10)(1)2
[
]
(1,50,000)
Nx from ML (membrane) = 0.076
(10)(1)2
= 1140 psi
[
]
(1,50,000)
N𝜑 from ML (membrane) = 0.260
(10)(1)2
= 3900 psi
223
224
11 Openings, Nozzles, and External Loadings
Table 11.8 Summary of membrane stresses at various locations.
𝝈𝝓
AU
𝝈x
AL
BU
BL
AU
AL
BU
BL
Due to P
−2110
−2110
−2110
−2110
−1920
−1920
−1920
−1920
Due to ML
−3900
−3900
+3900
+3900
−1140
+1140
1140
+1140
Total
−6010
−6010
+1790
+1790
−3060
−3060
−780
−780
Table 11.9 Summary of bending stresses at various locations.
𝝈𝝓
AU
AL
𝝈x
BU
BL
AU
AL
BU
BL
Due to P
−6 190
+6 190
−6 190
+6 190
−9 140
+9 140
−9 140
+9 140
Due to ML
−9 360
+9 360
+9 360
−9 360
−15,300
+15,300
+15,300
−15,300
−15,550
+15,550
+3 170
−3 170
−24,440
+24,440
+6 160
−6 160
Total
Table 11.10 Summary of membrane plus bending stresses at various locations.
𝝈𝝓
AU
Membrane
AL
𝝈x
BU
BL
AU
AL
BU
BL
−6 010
−6 010
+1 790
+1 790
−3 060
−3 060
−780
−780
Bending
−15,550
+15,550
+3 170
−3 170
−24,440
+24,440
+6 160
−6 160
Total
−21,560
+9 540
+4 960
−1 380
−27,500
+21,380
+5 380
−6 940
6) Summaries of membrane stresses, bending stresses,
and combined stresses at various locations for external loadings are given in Tables 11.8–11.10.
7) In addition to the stresses from the external loadings,
the stresses from internal pressure and other loadings
must be combined. For this example, assume that
the following multipliers of the circumferential stress
were determined:
Longitudinal plane
Circumferential plane
Membrane Bending
Membrane Bending
𝜎 n (= 𝜎 𝜙 )
1.40
∓0.20
𝜎 n (= 𝜎 x )
0.70
±0.35
𝜎 t (= 𝜎 x )
0.80
±0.10
𝜎 t (= 𝜎 𝜑 )
0.85
±0.55
Then,
S=
PDm
400 × 85
=
= 17,000 psi.
2T
2×1
8) The total combined stresses from internal pressure
and external loadings are given in Table 11.11.
Example 11.16
For the cylindrical shell given in Example 11.15, determine the stresses due to internal pressure and applied
external loading by the method in WRC Bulletin 107.
Solution:
1) For the WRC method, the following shell–nozzle
parameters are required:
r
5
= 0.103
𝛽 = 0.875 o = 0.875 ×
Rm
42.5
R
42.5
𝛾= m =
= 42.5
T
1
2) The following constants are determined from various
figures in WRC, and the stresses determined as follows:
N𝜙
12,000
= 7.3 ×
From Figure 4C ∶
P∕Rm
42.5 × 1
= 2060 psi
M𝜙
6 × 12,000
From Figure 2C-1 ∶
= 0.088 ×
P
(1)2
= 6340 psi
11.7 External Loadings
Table 11.11 Summary of stresses due to internal pressure and external loadings.
𝝈𝝓
Membrane
𝝈x
AU
AL
BU
BL
AU
AL
BU
BL
+23,800
+23,800
+23,800
+23,800
+13,600
+13,600
+13,600
+13,600
Bending
−3 400
+3 400
−3 400
+3 400
+1 700
−1 700
−1 700
−1 700
P + ML
−21,560
+9 540
+4 960
−1 380
−27,500
+21,380
+5 380
−6 940
−1 160
+36,740
+25,360
+25,820
−12,200
+33,280
+20,680
+4 960
Total
From Figure 3B ∶
N𝜙
= 4.5
ML ∕R2m 𝛽
150,000
×
= 3630 psi
(42.5)2 (0.103)(1)
M𝜑
= 0.045
From Figure 1B ∶
ML ∕Rm 𝛽
6 × 150,000
×
= 9250 psi
(42.5)(0.103)(1)2
Nx
= 6.0
From Figure 3C ∶
P∕Rm
12,000
×
= 1690 psi
42.5 × 1
M
From Figure 1C-1 ∶ x = 0.125
P
6 × 12,000
= 9000 psi
×
(1)2
Nx
= 1.3
From Figure 4B ∶
ML ∕R2m 𝛽
150,000
×
= 1050 psi
(42.5)2 (0.103)(1)
Mx
From Figure 2B ∶
= 0.072
ML ∕Rm 𝛽
6 × 150,000
×
= 14,800 psi
(42.5)(0.103)(1)2
𝝈x
AU
AL
BU
BL
Pressure membrane
+13,600
+13,600
+13,600
+13,600
Pressure bending
+1 700
−1 700
+1 700
−1 700
P membrane
−2 060
−2 060
−2 060
−2 060
P bending
−6 340
+6 340
−6 340
+6 340
+1 050
ML membrane
−1 050
−1 050
+1 050
ML bending
−14,800
+14,800
+14,800
−14,800
Totals
−11,240
+32,960
+20,460
+5 460
Problems
11.16
For the same vessel described in Example
11.15, what are the stresses on the transverse
plane when the applied moment is changed
from a longitudinal moment to a transverse
moment Mc = 150,000 in.-lb and the radial loading remains at 12,000 lb.2 Use the method in
Appendix F.
Answer:
𝜎 𝜙 : C U = −8180 psi
C L = +30,460 psi
3) Using the internal pressure stresses determined
for Example 11.15 and combining them with these
stresses gives:
DU = +33,280 psi
DL = −6200 psi
𝜎 x : C U = −3550 psi
C L = +19,970 psi
𝝈𝝓
Pressure membrane
AU
AL
BU
BL
+23,800
+23,800
+23,800
+23,800
DU = +23,030 psi
DL = +470 psi
Pressure bending
−3 400
+3 400
−3 400
+3 400
P membrane
−1 690
−1 690
−1 690
−1 690
P bending
−9 000
+9 000
−9 000
+9 000
Answer:
ML membrane
−3 630
−3 630
+3 630
+3 630
𝜎 𝜙 :C U = −7730 psi
ML bending
−9 250
+9 250
+9 250
−9 250
Totals
−880
+37,100
+24,880
+25,860
11.17
What are the same results using WRC 107?
C L = +31,150 psi
225
226
11 Openings, Nozzles, and External Loadings
When Sh > SL , the difference may be added to the term
0.25Sh in Eq. (11.57). This gives
DU = +34,670 psi
DL = −7050 psi
𝜎 x : C U = −2620 psi
SA = f [1.25Sc + 0.25Sh + (Sh − SL )]
or
C L = +19,240 psi
DU = +21,820 psi
SA = f [1.25(Sc + Sh ) − SL ],
DL = +920 psi
11.7.2
(11.65)
which may be used in place of Eq. (11.64).
Stresses in the Nozzle
The general membrane stresses in the nozzle are calculated using the basic equation
T
P M
𝜎t = ± ± c ± c .
(11.61)
A
I
J
However, in an attempt to make some correction for
local effects, the bending moments are adjusted by a SIF.
For piping thermal-expansion flexibility stresses in both
the ANSI B31.1 and ANSI B31.3 Codes, the procedure is
as follows:
The stress range SE is calculated by
√
SE = Sb2 + 4St2 ,
(11.62)
Example 11.17
A 12 in. NPS Schedule 160 branch and run pipe
are attached to one another. The design pressure is
2200 psi. The allowable stress at ambient temperature
is Sc = 17.5 ksi, and at design temperature, it is Sh =
12.0 ksi. In addition to the internal pressure, the
branch is subjected to externally applied forces and
moments from thermal expansion of connecting piping:
Mi = 600,000 in.-lb, Mo = 900,000 in.-lb, Mt = 750,000
in.-lb, and F axial = 900 lb. The nozzle is designed
for 20,000 cycles. Using the design procedure of
the ASME-ANSI B31.1 Code (see Figure 11.27 and
Tables 11.12 and 11.13), what is the total applied stress
and what is the allowable stress?
Solution:
where
St = Mt /2Z (psi)
Mt = torsional moment (in.-lb)
Z = section modulus of nozzle (in.3 ),
and Sb , the resultant bending moment, is
√
(ii Mi )2 + (io Mo )2
Sb =
,
Z
where
(11.63)
ii = in-plane SIF from Table 11.12
io = out-plane SIF from Table 11.12
Mi = in-plane bending moment (in.-lb)
Mo = out-plane bending moment (in.-lb).
The allowable stress range SA is
SA = f (1.25Sc + 0.25Sh ),
(11.64)
where
Sc = allowable stress at ambient (cold) temperature
(psi)
Sh = allowable stress at design temperature (psi)
f = reduction factor from Table 11.11 based on the
number of cycles.
The design is acceptable when SE ≤ SA .
Longitudinal stresses SL due to sustained loadings,
such as pressure and dead loading, shall not exceed Sh .
1) Properties of 12 in. NPS Schedule 160 are Do = 12.75
in.; inner area = 80.5 in.2 ; metal area = 47.14 in.2 ;
Z = 122.6 in.3 ; t m = 1.312 in.
2) Data at the juncture obtained from Table 11.6:
T
1.312
h= h =
= 0.229
Rm
5.719
0.9
io = 2∕3 = 2.40
h
1
ii = 0.75io + = 0.75(2.40) + 0.25 = 2.05
4
3) Determine the torsional stress:
M
750,000
St = t =
= 3060 psi
2Z
2(122.6)
4) Determine the bending stress:
√
(2.05 × 600,000)2
+(2.40 × 900,000)2
Sb =
122.6
= 20,270 psi
5) Determine the stress range:
√
√
SE = Sb2 + 4St2 = (20,270)2 + 4(3060)2
= 21,170 psi
6) Determine sustained longitudinal stress:
SL = (2200)
80.5
= 3760 psi
47.14
11.7 External Loadings
Figure 11.27 Stress intensification, flexibility, and
correction factors.
100
Stress intensification factor i and flexibility factor k
80
Flexibility factor
for elbows k = 1.65/h
Flexibility factor
for miters k = 1.52/h5/6
Stress intensification
factor i = 0.9/h2/3
Stress intensification
factor l = 0.75/h2/3
60
40
30
20
15
10
8
6
4
3
2
1.5
Chart A
1.00
2
0.8
1.0
0.5
0.4
0.3
0.2
0.15
0.10
0.25
0.06
2 ends flanged c1 = h1/3
0.04
1 end flanged c1 = h1/6
0.03
0.50
0.375
1.5
Chart B
0.75
0.02
Correction factor C1
1
Characteristic h
7) Determine the allowable stress range SA :
f = 0.8 for 20,000 cycles
SA = 0.8[1.25(17,500 + 12,000) − 3760]
= 26,490 psi
SE < SA ; design is acceptable.
11.7.2.1
Nomenclature
Special nomenclature is used throughout this chapter
and usually noted close to where it is used. The following
gives some general nomenclature:
P = internal design pressure or maximum allowable
working pressure (psi)
F a = externally applied axial force (lb)
Problem
11.18
An 8 in. NPS Schedule 160 branch pipe is
attached to a 16 in. NPS Schedule 160 run
pipe. The design pressure is 2000 psi, the
allowable stress cold is Sc = 17.5 ksi, and the
allowable stress at design temperature is 12.0 ksi.
The maximum allowable torsional moment
is 450,000 in.-lb. The pipe is designed for
10,000 cycles. The maximum allowable bending
moments are set equal. If rounded up to the next
even 100 in.-lb, what is the value of Mo and Mi ?
Answer:
Mo = Mi = 331,400 in.-lb
F h = externally applied horizontal force (lb)
Mb = externally applied bending moment (in.-lb)
𝜎 T = total local stress at opening (psi)
S = allowable tensile stress (psi)
D = inside diameter of the shell (in.)
d = inside diameter of the nozzle (in.)
r = inside radius of the opening (in.)
x = distance from the center of the opening to the
point being examined (in.)
T s = nominal thickness of the shell (in.)
T n = nominal thickness of the nozzle (in.)
t r = minimum required thickness of the shell (in.)
t rn = minimum required thickness of the nozzle (in.)
227
228
11 Openings, Nozzles, and External Loadings
Table 11.12 Flexibility factor k and stress intensification factor.
Stress intensification
factora),b)
Flexibility
factor
k
Description
Out-plane
io
In-plane
ii
Flexibility
characteristic
h
Sketch
T
Welding elbowa),c),d),e),f ) or
pipe bend
1.65
h
0.75
h2∕3
0.9
h2∕3
r2
TR1
(r2 )2
R1 = Bend
radius
T
Closely spaced miter
benda),c),d),f )
s < r2 (1 + tan 𝜃)
Single miter benda,c,f ) or
widely spaced miter bend
s ≥ r2 (1 + tan 𝜃)
1.52
h5∕6
0.9
h2∕3
0.9
h2∕3
Cot 𝜃 TS
2 (r2 )2
S
r2
θ
1.52
h5∕6
0.9
h2∕3
0.9
h2∕3
S cot θ
2
R1 =
T
S
1 + Cot𝜃 T
2
r2
r2
r2(1 + cot θ)
R1 =
θ
2
T
Welding teea,c,e,g,h per ANSI
1
B16.9 with rx ≥ Db
8
1
1
3
i +
4 o 4
0.9
h2∕3
3.1
T
r2
Tc
r2
rx
T c ≥ 1.5T
Reinforced fabricateda,c,h,i,j
tee with pad or saddle
1
3
1
i +
4 o 4
0.9
h2∕3
)5∕2
(
1
T + Tr
2
T
2∕3
r2
Tr
r2
Tr T
Saddle
Pad
r2
Unreinforceda,c,h,j fabricated
tee
1
0.9
h2∕3
1
3
i +
4 o 4
T
r2
T
T
Extrudeda,c,h welding tee
T c < 1.5T
rx ≥ 0.05Db
1
0.9
h2∕3
1
3
i +
4 o 4
(
1+
rx
r2
)
T
r2
r2
Tc
rx
T
a,c,g,h
Welded-in
insert
contour
1
0.9
h2∕3
1
3
i +
4 o 4
T
3.1
r2
r2
T
0.9
h2∕3
0.9
h2∕3
Brancha,c,j,k welded-on fitting
(integrally reinforced)
1
Butt-welded joint, reducer,
or weld-neck flange
1
1.0
Double-welded slip-on flange
1
1.2
3.3
T
r2
r2
11.7 External Loadings
Table 11.12 (Continued)
Stress intensification
factora),b)
Flexibility
factor
k
Description
Out-plane
io
In-plane
ii
Fillet-welded joint or
socket-weld flange
1
1.3l
Lap-joint flange (with ANSI
B16.9 lap joint stub)
1
1.6
Screwed pipe joint, or
screwed flange
1
2.3
Corrugated straight pipe or
corrugated or creased bendm
5
2.5
Flexibility
characteristic
h
Sketch
a) The flexibility factor, k, in the Table applies to bending in any plane. The flexibility factors, k, and stress intensification factors, i, shall apply
over the effective arc length (shown by heavy centerlines in the illustrations) for curved and miter bends, and to the intersection point for tees.
b) A single intensification factor equal to 0.9/h2/3 may be used for both ii and io if desired.
c) The values of k and i can be read directly from Chart A by entering with the characteristic h computed from the formulas given above.
Nomenclature is as follows:
Dd = outside diameter of branch
R1 = bend radius of welding elbow or pipe bend
rx = radius of curvature of external contoured portion of outlet, measured in the plane containing the axes of the header and branch
r2 = mean radius of matching pipe
s = miter spacing at centerline
T = for elbows and miter bends, the nominal wall thickness of the fitting
= for tees, the nominal wall thickness of the matching pipe
T c = crotch thickness of branch connections measured at the center of the crotch where shown in the illustrations
T r = pad or saddle thickness
𝜃 = one-half angle between adjacent miter axes
d) Where flanges are attached to one or both ends, the values of k and i in the table shall be corrected by the factors C 1 , which can be read
directly from Chart B, entering with the computed h.
e) The designer is cautioned that cast buttwelded fittings may have considerably heavier walls than that of the pipe with which they are used.
Large errors may be introduced unless the effect of these greater thicknesses is considered.
f ) In large diameter thin-wall elbows and bends, pressure can significantly affect the magnitude of k and i. To correct values form Table, divide k
by
( )( ) ( )
Pj
r2 7∕3 R1 1∕3
1 +6
Ej
r2
T
Divide i by
1 + 3.25
(
Pj
Ej
)(
r2
T
)5∕2 (
R1
r2
)2∕3
For consistency, use kPa and mm for SI units, and psi and in. for U.S. customary notation
g) If rx ≥ (1/8) Db and T c ≥ 1.5 T, a flexibility characteristic of 4.4 T/r2 may be used.
h) Stress intensification factors for branch connections are based on tests with at least two diameters of straight run pipe on each side of the
branch centerline. More closely loaded branches may require special consideration.
i) When T r is > 1 1/2 T, use h = 4 T/r2
j) The out-of-plane stress intensification factor (SIF) for a reducing branch connection with branch-to-run diameter ratio of 0.5 < d/D < 1.0 may
be nonconservative. A smooth concave weld contour has been shown to reduce the SIF. Selection of the appropriate SIF is the designer’s
responsibility.
k) The designer must be satisfied that this fabrication has pressure rating equivalent to straight pipe.
l) For welds to socket welded fittings, the stress intensification factor is based on the assumption that the pipe and fitting are matched in
accordance with ASME B16.11 and a fillet weld is made between the pipe and fitting as shown in Figure 328.5.2C of ASME B31.3. For welds to
socket welded flanges, the stress intensification factor is based on the weld geometry shown in Figure 328.5.2B, illustration (3) of ASME B.31.3
and has been shown to envelope the results of the pipe to socket welded fitting tests. Blending the toe of the fillet weld smoothly into the pipe
wall, as shown in the concave fillet welds in Figure 328.5.2A of ASME B31.3 has been shown to improve the fatigue performance of the weld.
m) Factors shown apply to bending. Flexibility factor for torsion equals 0.90.
229
230
11 Openings, Nozzles, and External Loadings
Table 11.13 Stress-range reduction factors f .
No. of cycles, N
Factor f
700 and less
1.0
Over 7 000–14 000
0.9
Over 14 000–22 000
0.8
Over 22 000–45 000
0.7
Over 45 000–100 000
0.6
Over 100 000
0.5
Source: Courtesy of American Society of Mechanical
Engineers.
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Pressure Vessels with Internal Pressure Loading. Phase
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Columbus, OH.
Rodabaugh, E.C. (1971). Review of Service Experience and
Test Data on Openings in Pressure Vessels with
Non-integral Reinforcing. WRC Bulletin No. 166,
Welding Research Council, New York.
Rodabaugh, E.C. and Gwaltney, E.C. (1971). Additional
Data on Elastic Stresses in Nozzles in Pressure Vessels
with Internal Pressure Loading. Phase Report 117-2,
Battelle-Columbus Laboratory, Columbus, OH.
Rodabaugh, E.C. and Gwaltney, R.C. (1976). Elastic
Stresses at Reinforced Nozzles in Spherical Shells with
Pressure and Moment Loading. Phase Report 117-9R,
Battelle-Columbus Laboratory, Columbus, OH.
Rodabaugh, E.C. and Moore, S.E. (1978). Evaluation of the
Plastic Characteristics of Piping Products in Relation to
231
232
11 Openings, Nozzles, and External Loadings
ASME Code Criteria. NUREG/CR-0261
ORNL/Sub-2913/8, Oak Ridge National Laboratory, Oak
Ridge, TN.
Schroeder, J., Srinivasaiah, K.R., and Graham, P. (1974).
Analysis of Test Data on Branch Connections Exposed
to Internal Pressure and/or External Couples. WRC
Bulletin No. 200, Welding Research Council, New York.
Schroeder, J. and Tugcu, P. (1978). Plastic Stability of Pipes
and Tees Exposed to External Couples. WRC Bulletin
No. 238, Welding Research Council, New York.
Sellers, F. (1969). A Note on the Correlation of Photoelastic
and Steel Model Data for Nozzle Connections in
Cylindrical Shells. WRC Bulletin No. 139, Welding
Research Council, New York.
Taylor, C.E. and Lind, N.C. (1966). Photoelastic Study of
the Stresses Near Openings in Pressure Vessels. WRC
Bulletin No. 113, Welding Research Council, New York.
Tso, F.K.W., Bryson, J.W., Weed, R.A., and Moore, S.E.
(1977). Stress analysis of cylindrical pressure vessels with
closely spaced nozzles by the finite element method. In:
Stress Analysis of Vessels with Two Closely Spaced
Nozzles under Internal Pressure, vol. 1, ORNL/
NUREG-18/VI. Oak Ridge, TN: Oak Ridge National
Laboratory.
Different vessel supports. Source: Courtesy of the Nooter Corporation: St. Louis, MO.
234
12
Vessel Supports
12.1 Introduction
Process equipment is normally supported by one of the
following methods:
1)
2)
3)
4)
5)
Skirts
Support legs
Support lugs
Ring girders
Saddles
Most vertical vessels are supported by skirts, as shown
in Figure 12.1a. Skirts are economical because they generally transfer the loads from the vessel by shear or direct
bearing action. They also transfer the loads to the foundation through anchor bolts and bearing plates.
Leg-supported vessels are normally lightweight, and
the legs provide easy access to the bottom of the vessel.
An economic design is shown in Figure 12.1b, where
the legs attach directly to the vessel and the loads are
transferred by shear action.
Figure 12.1c shows an alternative design where the legs
are attached to lugs that in turn are welded to the vessel. The bending stiffness of the shell and its ability to
resist the moments adequately must be considered. The
cross-bracing of the legs may be needed to minimize lateral and torsional movements.
Vessels supported by ring girders (Figure 12.1d) are
usually placed within a structural frame. The ring girder
has the advantage of supporting torsional and bending
moments resulting from the transfer of loads from the
vessel wall to the supports.
Horizontal vessels (Figure 12.1e) are normally supported by saddles. Stiffening rings may be required if the
shell is too thin to transfer the loads to the saddles. The
problem of thermal expansion must also be considered.
(a) Skirt
(b) Leg
(d) Ring girder
(c) Lug
(e) Saddles
Figure 12.1 Vessel supports.
wind and earthquake forces (see Chapter 16). The stress
in the skirt is then determined from
−W Mc
𝜎=
±
.
(12.1)
A
I
In most practical applications, the ratio R/t > 10.
Hence, the area A and the moment of inertia I of the
skirt are expressed as
A = 2πRt
I = πR3 t,
12.2 Skirt and Base-Ring Design
Design of the skirt involves first determining the dead
weight W of the vessel and bending moment M due to
and the equation for the stress in a skirt becomes
−W
M
𝜎=
± 2 ,
(12.2)
2πRt πR t
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
12.2 Skirt and Base-Ring Design
where
W = weight of vessel
of contents W 2 = 1440 kips, wind-bending moment
M = 1500 ft.-kips, and temperature = 300 ∘ F. Assume
A307 bolts, and use Figure 8.11 for the external-pressure
chart.
M = moment due to wind or earthquake forces
Solution:
𝜎 = axial stress in skirt
R = radius of skirt
Skirt design.
t = thickness of skirt.
Because the compressive stress is larger than the tensile stress, it usually controls the skirt design and is kept
below the skirt’s allowable axial compressive stress as
given by Eq. (8.15).
After the thickness t of the skirt is determined, the next
step is designing the anchor bolts. For a given number
of bolts, N, the total bolt area can be expressed as NA,
where A is the area of one bolt. The moment of inertia of
bolts about the vessel’s neutral axis is I = NAR2 /2. Thus,
Eq. (12.1) is
P=
−W 2M
±
,
N
NR
(12.3)
Let t = 0.4375 in. From Eq. (12.2),
160 + 1440
𝜎=−
2π(84 − 0.375∕2)(0.4375)
1500 × 12
−
π(83.813)2 (0.4375)
= −8.81 ksi.
From Eq. (8.15),
0.125
A=
Ro ∕t
= 0.00065.
Hence, from Figure 8.11, B = 9400 psi (acceptable).
Bolt design.
where
N = number of bolts
Let N = 12 bolts. From Eq. (12.3),
160 2(1500)(12)
Load∕bolt = −
+
12
12(84)
= 22.4 kips.
R = radius of bolt circle
From Table 12.1,
P = load/bolt
W = weight of vessel
M = bending moment.
22.4
20
= 1.12 in.2
area required =
The maximum load per bolt is based on the allowable
stress and corresponding area given in Table 12.1. The
allowable stress depends on the type of bolt furnished.
Table 12.2 shows various properties and required dimensions for bolts with different diameters.
From Table 12.2, use 1 38 -in. diameter bolts:
area = 1.155 in.2 > 1.12
Use 7/16 in. skirt with 12
Example 12.1
Determine the required skirt thickness and the number of bolts needed in a vessel with an outer radius
R = 7.0 ft. Let empty weight W 1 = 160 kips, weight
Table 12.1 Allowable stress for some bolts.a)
Bolt type
Allowable tensile
stress (ksi)
A307
20
A325
44
A449
40b)
A490
54
a) These values are approximate. Actual
values may be obtained from sources such
as AISC and ASME II-D.
b) For bolts equal to or smaller than 1 in. in
diameter.
(acceptable).
1 38 -in.
diameter bolts.
Having established the number and size of bolts,
the next step is to calculate the interaction between the
base plate, anchor bolts, and supporting structure. If the
supporting structure is a steel frame or foundation, then
Eq. (12.3) is all that is needed for designing the anchor
bolts. On the other hand, if the foundation is deep and/or
resting on a group of piles, it can be assumed that the
interaction between the bolts, base plate, and concrete is
similar to that for a reinforced-concrete beam. Referring
to Figure 12.2, the following assumptions are made:
1) The contribution of the bolts on the compression side
is negligible.
2) The bolts on the tension side are assumed to act as
a continuous ring of width t s calculated from the
equation
A
(12.4)
ts = s .
πd
235
236
12 Vessel Supports
Table 12.2 Bolt dimensions and clearance bolting data.
Threads
Bolt
size
Nut dimensions
No. of
threads
Root area
(in.2 )
Across
flats
Across
corners
Bolt
spacing B
Radial
distance R
Edge
distance E
Wrench
diameter a
1
2
5
8
3
4
7
8
13
0.126
0.969
1 14
0.202
13
16
15
16
1 18
1 14
1 38
1 12
1 34
1 78
5
8
3
4
13
16
15
16
1
1 16
1 18
1 14
1 38
1 12
1 58
1 34
1 78
1 12
11
7
8
1
1 16
1 14
7
1 16
1 58
1 13
16
10
0.302
9
0.419
1
8
0.551
1 18
1.175
3.035
1 34
1
2 16
2 14
2 12
2 13
16
1
3 16
3 14
3 12
3 34
3.242
4
3.449
4 14
4 34
5 14
5 34
6 14
6 34
7 18
7 58
8 18
1.383
1.589
1.796
8
0.728
1 14
1 38
1 12
1 58
1 34
1 78
8
0.929
2
2.209
8
1.155
2.416
8
1.405
2
8
3
2 16
2 38
9
2 16
2 34
2 15
16
3 18
3 12
3 78
4 14
4 58
2 14
2 12
2 34
8
1.608
8
1.980
8
2.304
2.652
8
3.423
8
4.292
8
5.259
2.002
2.622
2.828
3.862
4.275
4.688
3
8
6.324
3 14
8
7.487
5
5.515
8
8.749
5 38
5.928
3 12
3 34
8
10.108
4
8
11.566
5.102
5 34
6 18
1 12
6.341
6.755
2
2 18
2 14
2 38
2 12
2 34
1
3 16
3 38
3 58
3 78
4 18
4 38
4 58
1
2 16
2
2 14
2 38
2 58
2 78
2 38
2 58
2 78
3 14
3 12
3 34
4
4 14
4 12
4 34
5 14
5 78
6 12
7
7 12
3
3 14
8
3 12
8 12
3 58
a
1 34
9
B
E
R
d
3) The allowable stress of steel, f s , is taken from
Table 12.1.
4) Concrete on the compression side is assumed to have a
width t c that is the same as the width of the base plate.
5) The allowable compressive stress of concrete, f c , is
taken from Table 12.3.
6) The ratio of the modulus of elasticity of steel to that of
concrete is defined as n:
Es
Ec
f ∕ε
fε
= s s = s c.
fc ∕εc
f c εs
n=
In an elastic analysis, the strains in the concrete
and steel at any location are the same. Hence, 𝜀c = 𝜀s
and
f
n = s or fs = nf c .
fc
Also, from Figure 12.2c, using similar triangles,
nf
fs
= c
d − kd
kd
or
k=
1
.
1 + fs ∕nf c
(12.5)
12.2 Skirt and Base-Ring Design
Table 12.3 Properties of concrete.
fc
jd
fs
c
t
(c)
l1
N. A.
ts
tc
γ
t
c
l3
l2
Modulus of
elasticity (psi)
√
Ec = 57 000 fc′
n = E s /E c a)
2500
1125
2 850 000
11
3000
1350
3,120 000
10
3500
1575
3 370 000
9
4000
1800
3 610 000
8
Compressive
stress (psi) fc′
W M
(a)
Allowable
compressive stress
(psi) fc = 0.45fc′
kd
d
(b)
Figure 12.2 Skirt base ring on concrete foundation.
From these six assumptions and Figure 12.2, the following relationships are obtained:
l1
d∕2 − kd
= sin 𝛾 =
= 1 − 2k
(12.6)
r
d∕2
The total force T of the tensile area of the reinforcement can be determined by summing forces on the tensile side of the neutral axis, which gives [1]
( )
d
T = fs ts
2
{
[(
)
]}
2
𝜋
+ 𝛾 sin 𝛾 + cos 𝛾
1 + sin 𝛾
2
or
( )
d
(12.7)
K1 .
T = f s ts
2
The distance between T and the neutral axis, denoted
by l2 , is
⎡ (π∕2 + 𝛾)sin2 𝛾 + 1.5(sin 𝛾 cos 𝛾) ⎤
⎢
⎥
+0.5(π∕2 + 𝛾)
d
⎥ . (12.8)
l2 = ⎢
⎥
2⎢
(π∕2 + 𝛾) sin 𝛾 + cos 𝛾
⎢
⎥
⎣
⎦
a) Es = 30 × 106 psi.
Similarly, the total force C of the compressive area of
the concrete is given by
( )
d
C = (tc + nt s )
2
2[cos 𝛾 + 𝛾 sin 𝛾 − (π∕2) sin 𝛾]
(12.9)
fc
1 − sin 𝛾
or
( )
d
C = (tc + nt s )
fK.
2 c 2
The distance l3 between C and the neutral axis is
)
(
⎡
⎤
1
2
− 1.5(sin 𝛾 cos 𝛾) ⎥
𝛾
+
(π∕2
−
𝛾)
sin
2
d⎢
l3 = ⎢
⎥.
2⎢
cos 𝛾 − (π∕2 − 𝛾) sin 𝛾
⎥
⎣
⎦
The relationship between external forces M and W and
the internal forces T and C are derived from Figure 12.2c:
∑
Mc = 0
or
M − W (l1 + l3 ) − T(l2 + l3 ) = 0
and
T=
M − W (l1 + l3 )
.
l2 + l3
(12.10)
Similarly,
∑
Fv = 0
and
C = T + W.
(12.11)
The values of 𝛾, l1 , l2 , l3 , K 1 , and K 2 are given in
Table 12.4 for various values of k.
Example 12.2
In Example 12.1, it was found that 12 1 38 -in. in. A307
anchor bolts were needed for a vessel with an outer
radius R = 7 ft., W 1 = 160 kips, M = 1500 ft.-kips, and a
skirt thickness of 0.4375 in. If fc1 = 3000 psi, determine
the actual stress in the concrete and bolts.
237
238
12 Vessel Supports
Table 12.4 Various parameters as a function of k.
𝜸
k
2l1 /d
2l2 /d
2l3 /d
K1
K2
0.01
78.52
0.98
1.489
0.016
3.113
0.267
0.02
73.74
0.96
1.477
0.032
3.085
0.378
0.03
70.05
0.94
1.465
0.048
3.059
0.463
0.04
66.93
0.92
1.452
0.064
3.033
0.535
0.05
64.16
0.90
1.439
0.080
3.008
0.599
0.06
61.64
0.88
1.426
0.096
2.983
0.657
0.08
57.14
0.84
1.400
0.128
2.935
0.760
0.10
53.13
0.80
1.373
0.160
2.887
0.852
0.15
44.43
0.70
1.304
0.239
2.772
1.049
0.20
36.87
0.60
1.233
0.318
2.661
1.218
0.25
30.00
0.50
1.161
0.397
2.551
1.370
0.30
23.58
0.40
1.087
0.475
2.442
1.509
0.35
17.46
0.30
1.013
0.553
2.333
1.640
0.40
11.54
0.20
0.938
0.631
2.224
1.765
0.45
5.74
0.10
0.862
0.709
2.113
1.884
0.50
0.00
0.00
0.785
0.785
2.000
2.000
0.55
−5.74
−0.10
0.709
0.862
1.884
2.113
0.60
−11.54
−0.20
0.631
0.938
1.765
2.224
0.65
−17.46
−0.30
0.553
1.013
1.640
2.333
0.70
−23.58
−0.40
0.475
1.087
1.509
2.442
0.75
−30.00
−0.50
0.397
1.161
1.370
2.551
0.80
−36.87
−0.60
0.318
1.233
1.218
2.661
0.85
−44.43
−0.70
0.239
1.304
1.049
2.772
Solution:
By referring to Figure 12.3 and Table 12.2, for 1 38 -in. bolts,
the bolt circle can be calculated as
d = 2(84 + 0.25 + 1.875) = 172.25 in.
Also,
tc = 2(0.25 + 1.875 + 1.375) + 0.4375 = 7.438 in.
From Example 12.1,
22.4
fs =
= 19.39 ksi.
1.155
From Eq. (12.2),
12(1.155)
ts =
π(172.25)
= 0.0256 in.
From Table 12.3,
fc = 1350 psi
n = 10,
and Eq. (12.5) gives k = 0.41.
From Table 12.4 with k = 0.41,
2l1
= 0.180
d
2l2
= 0.923
d
R = 84.0"
1/4
3/8
1/4
1
7"
8
1
3"
8
d = 172.25"
tc = 7.438"
Figure 12.3 Base ring to skirt attachment.
2l3
= 0.647
d
K1 = 2.202
K2 = 1.789.
The magnitude of T is obtained from Eq. (12.10) as
T=
1,500,000 × 12 − 160,000(0.180 + 0.647)
(172.25)∕2
(0.923 + 0.647)(172.25)∕2
= 48,840 lb.
The value of f s is determined from Eq. (12.7) as
48,840
fs =
(0.0256)(172.25∕2)(2.202)
= 10,060 psi.
From Eq. (12.11),
C = 48,840 + 160,000
= 208,840 lb.
Eq. 12.9 gives
208,840
[0.0256 + (10)(7.438)](172.25∕2)(1.789)
= 18 psi.
fc =
The calculated values of fs1 and f c result in a k value of
1
k=
1 + 10,060∕(10)(18)
= 0.02,
which is considerably lower than the assumed value of
k = 0.41. Hence, another trial is needed with a k value
of 0.02. After recalculating the values of T, f s , C, and
f c , a new value of k is obtained and compared with the
assumed one. If both values are approximately the same,
the analysis is completed. If they are not, a new analysis
12.2 Skirt and Base-Ring Design
is performed. Thus, in this example after a few trials, for
k = 0.075, the following values were obtained:
2l1
= 0.85
d
2l2
= 1.407
d
2l3
= 0.120
d
K1 = 2.947
K2 = 0.734
T1 =
15 × 105 × 12 − 16 × 104 (0.85 + 0.120)
(172.25)∕2
l
t
fc
tc
(1.407 + 0.120)(172.25)∕2
= 35,230
Figure 12.4 Foundation pressure on base ring.
35,230
(0.0256)(172.25∕2)(2.947)
= 5420 psi
l
fs =
Pinned
d
a
and
1
= 0.076,
1 + 5420∕(10)(42)
which is approximately the same as the assumed value.
Hence, f s = 5420 psi and f c = 42 psi is the answer.
k=
12.2.1
Free
fc = 42 psi
Pinned
C = 195,230
Pinned
Partial view
of
base ring
b
(a)
(b)
Anchor-Chair Design
The base ring is designed both for the effect of the
concrete-bearing load on the side of the foundation
under compression and for the bolt force on the other
side of the foundation in tension. On the compressive
side, the base ring can be taken as a cantilever beam
subjected to f c as shown in Figure 12.4. The required
thickness is obtained from
√
6M
6M
𝜎= 2
or t =
.
t
𝜎
Substituting for M the value
fc l 2
,
2
the expression for t becomes
√
3fc l2
,
t=
𝜎
where
F
MP
MP
MP
a
Unit
Deflection
b
(c)
Figure 12.5 Base ring at anchor bolt vicinity.
M=
(12.12)
t = required base-ring thickness on the compressive
side of the neutral axis
f c = actual stress in concrete
l = cantilever length of base ring as defined in
Figure 12.4
𝜎 = allowable bending stress of base ring.
On the tensile side, the thickness of the base ring is
controlled by the amount of bolt force and dimensions
shown in Figure 12.5. The exact analysis for determining
the maximum bending moment in the base ring is rather
complicated because of the nature of the boundary conditions and the hole. However, an approximate solution
can be obtained by assuming the ring to act as a plate simply supported on three sides and free on the fourth side.
Using the yield-line theory [2]
external work = internal work
1
1
+ Mp (a − d)
F(1) = 2Mp (b − d)
a∕2
l
239
12 Vessel Supports
or
F
Mp =
.
2[2b∕a + a∕2l − d(2∕a + 1∕2l)]
Using a load factor of 1.7 and a factor of 1.15 to allow
for the yield-line corner effect [2] the equation
4M
Sy = 2
t
can be solved for the required thickness
√
3.91F
t=
, (12.13)
Sy [2b∕a + a∕2l − d(2∕a + 1∕2l)]
where.
t = required base ring thickness on the tensile side of
the neutral axis
F = bolt load
Sy = yield stress of base ring
and a, b, d, and l are as defined in Figure 12.5b.
The load in the shell is transferred to the anchor bolts
through the gussets. An approximate free-body diagram
of the forces is shown in Figure 12.6. Vertical forces
are transferred as shown in Figure 12.6a. The resulting
unbalanced bending moment in the gussets resulting
from the vertical forces requires equal and opposite horizontal forces as shown in Figure 12.6b. These horizontal
forces induce local stresses in the shell that are calculated
from the equation
1.5Fb
,
(12.14)
𝜎=
πt 2 h
where
𝜎 = allowable stress in shell
t = thickness of shell
F = bolt load.
Example 12.3
Design the base ring shown in Figure 12.7a. The stress
in the bolts is 17,500 psi, height of gussets 12 in., and
concrete-bearing stress 100 psi. The allowable stress for
base ring is 20,000 psi, and the yield stress is 36,000 psi.
Solution:
The required base-ring thickness due to the concretebearing stress is obtained from Eq. (12.12) as
√
3 × 100 × 62
t=
20,000
= 0.73 in.
From Table 12.2, the net area of 1 12 -in. bolts is
1.405 in.2 . Thus,
force F in bolts = 17,500 × 1.405
= 24,600 lb.
From this table, the clearance for the wrench diameter
is 3.75 in. Allowing for gusset fillet welds, the distance
3"
4
F
F
h = 12"
8'−0"O.D.
F
7'−6"
(12)1
8'−6"
F
1"
Bolts
2
9'−0"O.D.
(a)
(a)
l = 3"
H
H
H=
b
(F) (b/2)
2
h
3
(b)
d
1.625"
h
a = 5"
240
b = 6"
(b)
Figure 12.6 Forces in gusset-skirt-base ring area. (a) Vertical
forces. (b) Horizontal forces.
Figure 12.7 Details of vessel support.
12.3 Design of Support Legs
between gussets is as shown in Figure 12.7b. From
Eq. (12.13),
√
√
√
3.91 × 24,600
t=√
[
]
√
(2 × 6)∕5
+ 5∕(2 × 3)]
√
[
√ 36 000
2
+ 1∕(2 × 3)
−1.625
5
= 1.07 in.
The stress in the shell is obtained from Eq. (12.14) as
1.5 × 24,600 × 6
𝜎=
π × 0.752 × 12
= 10,400 psi.
B
A
A
B
W
Section A.A
V
A
M
A
W
This stress is combined with the axial stress, and the
total must be less than three times the allowable stress.
M
12.3 Design of Support Legs
Support legs are designed to take into consideration axial
loads, bending moments, and shear forces in the vessel.
Referring to Figure 12.8, we see that at cross section A–A,
all forces are expressed in terms of M, V , and W . The
axial force W is carried uniformly by all columns. The
bending moment M is carried by the columns away from
the neutral axis, and the shearing forces V are carried
by the columns closest to the neutral axis, as shown in
Figure 12.8.
Column A in Figure 12.8 is designed by using Eq. (12.3)
given by
P=
−W 2M
±
N
NR
where
V
Figure 12.8 Leg supported vessel.
Solution:
The axial force in columns A and B due to W is
W
240
F=
=
= 30 kips.
N
8
The axial force in column A due to M is
2M
F=
NR
2 × 2000
=
= 100 kips.
8×5
Thus,
total axial load in column A = −30 − 100
P = load per column
W = weight of vessel
N = number of columns
R = radius of column circle
M = moment due to wind or earthquake loads.
Column B in Figure 12.8 is designed to carry shear in
accordance with the equation
VQ
.
It
The shearing force H at the top of column B causes
bending moment in the column if no cross-bracing is
used. With cross-bracing, the force H is resolved into
axial forces as shown in Example 12.4.
H=
Example 12.4
Determine the forces in columns A and B of the vessel
shown in Figure 12.9.
= −130 kips
total axial load in column B = −30 kips.
The shearing stresses transferred to column A are zero.
Those transferred to column B are determined from
VQ
H=
.
It
The moment of inertia I of the whole cross section in
Figure 12.9c is given by 𝜋r3 t. The magnitude Q of the
cross-hatched area in this figure is given by
( )
2
Q = (πrt) r
π
= 2r2 t.
The force H is then given by
V (2r2 t)
V
2V
H=
=
=
(πr3 t)(2t) πrt
A
2 × 50
0.2653
H=
=
psi.
2π(60)t
t
241
12 Vessel Supports
D = 10'
and
total force in column A = −130 kips
A
W = 240 K
total force in column B = −30 − 69.61
= −99.61 kips
V = 50 K
total force in bracing = 70.91 kips.
M = 2000 K-ft.
B
20'−0"
242
(b)
(a)
2
πr
∝ = 180N – 360 = 67.5°
2N
r
X
U
B
H
(d)
(c)
If the cross-bracing is eliminated in Example 12.4, the
shear force tends to cause a bending moment in column
B. Assuming the bottom end of the columns pinned, the
horizontal force causes a bending moment at the top of
the column of magnitude 12.5 × 20 ft. = 250 k-ft. Thus,
without a bracing system, column B must be designed to
withstand a compressive force of 30 kips plus a bending
moment of 250 k-ft. rather than a compressive force of
99.61 kips with a bracing system.
Note that the absence of cross-bracing causes the tops
of the columns to sway laterally because of reduced rigidity. This can also cause excessive vibration or deformation
of the vessel.
X = 12.5
12.4 Lug-Supported Vessels
E
F
β
(e)
Figure 12.9 Forces in a leg supported vessel.
The main design consideration regarding lug-supported
vessels is the stress magnitude in the shell. Bijlaard’s
method is usually followed in such a design [3]. It consists of determining the stress in the shell in the vicinity
of a support lug of height 2C 2 and width 2C 1 , as shown
The horizontal force in column B is
)
(
2πr
t
H = (0.2653)
N
= 12.50 kips.
This force H is normally resolved into two components as shown in Figure 12.9d. U is a radial force on
the shell, and X is a horizontal force in the plane of the
cross-bracing. We have
U = H cot 𝛼 = 12.5 × 0.414 = 5.18 kips
H
12.5
X=
=
= 13.53 kips.
sin 𝛼
0.924
The force X introduces additional compressive force in
column B as shown in Figure 12.9e. The distance between
columns is
2πr
l=
= 3.93 ft.
N
The approximate height of the columns is 20 ft.
Hence, the angle 𝛽 is about 11∘ , and the axial force F in
column B is
X
= 69.61 kips
F=
tan 𝛽
so that the force
X
E=
= 70.91 kips
sin 𝛽
2C2
Rm
e
F
T
ϕ
Figure 12.10 Lug supports.
2C1
12.5 Ring Girders
in Figure 12.10. The bending moment in the shell due to
support eccentricity is given by
W
r
(12.15)
Mo = Fe,
θ
and the maximum stress in the shell is calculated from
Ref. [3]. Both membrane and bending stresses are
calculated. Details of the required calculations are well
established in Ref. [3]. Further treatment of this topic is
unnecessary in this book.
Ts
Ms
Vs
12.5 Ring Girders
Figure 12.11 Forces in ring girder support.
Ring girders (Figure 12.1d) are common in elevated vessels supported by a structural frame. An exact analysis of
the stresses in a ring girder due to various loading conditions is very complicated. For a uniform load, the stresses
and forces can be determined easily with the following
assumptions:
Mm , V m , T m = midspan moment, shear, and torsion,
respectively
K 3 –K 5 = constants obtained from Table 12.5
w = uniform load
r = radius
1) Supports are equally spaced.
2) Vertical deflection at supports is zero.
3) Slope of ring girder at supports is zero due to symmetry of loads and supports.
4) Torsion force at supports is zero. This assumes twisting of the girder due to flexibility of shell.
The maximum torsional moment occurs at the angles
shown in Table 12.5 and is given by
Tmax = K6 wr2 .
The moment, shear, and torsion expressions for any
given location between supports are obtained from
Based on these assumptions, the moments, shears,
and torsion at the supports and in between supports are
given by
Ms = K3 wr
Vs = K4 wr
Ts = 0
2
Mm = K5 wr
M𝜃 = Vs r sin 𝜃 + Ms cos 𝜃 − wr2 (1 − cos 𝜃)
V𝜃 = Vs − wr𝜃
T𝜃 = Vs r(1 − cos 𝜃) + Ms sin 𝜃 − wr2 (𝜃 − sin 𝜃),
(12.18)
2
Vm = 0
Tm = 0,
where
(12.16)
M𝜃 , V 𝜃 , T 𝜃 = moment, shear, and torsion at any location
where
𝜃 = angle defined in Figure 12.11.
Ms , V s , T s = support moment, shear, and torsion,
respectively. Positive direction is shown
in Figure 12.11
In deriving Eqs. (12.16), it is assumed that the loads
and the reactions act through the neutral axis of the
Table 12.5 Ring-girder coefficients.
Angle between
supports
(degrees)
K3
K4
K5
K6
Angle of maximum
torsion from
support (∘ )
2
180
−1.0000
1.5707
−0.5707
−3.307 × 10−1
39.55
3
120
−0.3954
1.0471
−0.2091
−8.278 × 10−2
25.80
4
90
−0.2146
0.7853
−0.1107
−3.313 × 10−2
19.21
5
72
−0.1351
0.6283
−0.0690
−1.654 × 10−2
15.30
6
60
−0.0931
0.5235
−0.0471
−9.471 × 10−3
12.74
−3
Number
of
supports
8
45
−0.0519
0.3926
−0.0262
−3.940 × 10
9.53
10
36
−0.0331
0.3141
−0.0166
−2.007 × 10−3
7.62
12
30
−0.0229
0.2617
−0.0115
−1.154 × 10−3
6.34
16
22.5
−0.0128
0.1963
−0.0065
−3.722 × 10−3
4.72
−0.0042
−4
3.79
20
(12.17)
18
−0.0082
0.1570
−2.469 × 10
243
244
12 Vessel Supports
t
where
𝜎 = stress
b
l = .78 r.t
y
w
x
b = width of flange
h
m
d = distance between flanges
d
e'
h = flange thickness
x
Shear center
I x = moment of inertia of girder
r = radius of vessel
l = .78 r.t
b
h
t = shell thickness
w = uniform applied load.
r
At the supports, the reaction eccentricity tends to
produce compressive forces in the top flange and tensile
forces in the bottom one as shown in Figure 12.13a. The
top and bottom flanges can be assumed to transfer the
loads as shown in Figure 12.13b. The forces are derived as
(
)
Hr
2
cos 𝜃
𝛼
−
0<𝜃<
Ml =
2 sin(𝛼∕2) 𝛼
2
H cos 𝜃
Fl =
2 sin(𝛼∕2)
−H sin 𝜃
Vl =
.
(12.20)
2 sin(𝛼∕2)
(a)
m
M
M
(b)
Figure 12.12 Cross section of ring girder support.
t
girder. In pressure vessels, the loads are transferred to
the ring girder through the shell. If the ring girder is
taken as a channel section as in Figure 12.13a, then the
loads in the shell cause a bending moment in the girder
because they are not applied through the shear (flexural) center. This moment, shown in Figure 12.12a, has
the magnitude
m = −we ,
′
h
l = 1.56 r.t + h
H
Shear
center
H=
d
e'
W (e + e')
N
d
H
where e’ is the shear center moment arm, which can be
expressed as
e′ =
e
W
N
b2 d 2 h
.
4Ix
The uniform bending moment m causes tension hoop
stress above the x axis and compression hoop stress
below the x axis as shown in Figure 12.12b.
The moment and corresponding stress can be
expressed as
α
θ
(a)
H
2 2
M = −mr =
𝜎=
mry
Ix
−wrb d h
4Ix
or
wb2 d2 hry
,
𝜎=
4Ix2
Vl
Fl
Ml
(12.19a)
(b)
(12.19b)
Figure 12.13 Effective length of ring girder support.
(c)
12.6 Saddle Supports
At the supports, 𝜃 = 0 and
(
)
Hr
2
1
Ml =
−
2 sin(𝛼∕2) 𝛼
H
1
Fl =
2 sin(𝛼∕2)
Vl = 0,
Table 12.6 Moments and forces at two locations around the ring
girder.
Example 12.5
The ring girder shown in Figure 12.14 is supported at
eight points. If W = 200 kips, find the forces in the ring
at the supports and at the point of maximum torsional
moment.
Ms , Mm
−82.65 k-in.
0 k-in.
V s, V m
12.50 k
7.20 k
T s, T m
0
−6.27 k-in.
Eqs. (12.19a) and (12.19b)
−60.05 k-in.
−60.05 k-in.
Ml b)
+19.35 k-in.
−38.45 k-in.
Fl
15.20 k
14.04 k
Vl
0k
−5.82 k
M
Eq. (12.20)a)
a) These equations apply at points B and D; points A and C have
opposite signs.
∘
W e + e′
200 4 + 1.58
360
b) H =
=
= 11.63; 𝛼 =
= 45 .
N d
8
12
8
Solution:
200
W
=
= 0.637 k-in.
w=
πd
(π)(100)
From Table 12.5, with N = 8,
b = 6"
1"
2
At point of
maximum torsion
Eq. (12.16)
and in between the supports, 𝜃 = 𝛼/2 and
(
)
𝛼 2
Hr
cot −
Ml =
2
2 𝛼
H
𝛼
Fl = cot
2
2
−H
Vl =
.
2
The positive directions of Ml , F l , and V l are shown in
Figure 12.13c.
t=
At support
K3 = −0.0519 K4 = 0.3926
1"
K5 = −0.0262 K6 = −3.940 × 10−3 ,
d = 12"
maximum torsion occurs at 9.53∘ from support.
The forces given by Eqs. (12.16), (12.19a), (12.19b), and
(12.20) are determined in Table 12.6 and illustrated in
Figure 12.15 at the supports and at the point of maximum
torsion.
h = 1"
6"
A
.78 r.t
C
19.80"
O.D. = 100" e = 4"
W
N
(a)
A
x
12.6 Saddle Supports
x
D
y
B
C
lx = 687.4 in.4
e' = 1.58 in.
y
D
B
(c)
Figure 12.14 Dimensions of ring girder.
(b)
Horizontal vessels supported by two saddles (Figure 12.1e)
act as simply supported beams. For thin vessels with
dished heads (Figure 12.16a), the equivalent beam length
is taken as L + 4H/3, where L is the tangent-to-tangent
length of the vessel and H is the depth of the heads.
The vertical load on each head is given by V = 2Hw/3
and is assumed to act at the center of gravity of the
head. The horizontal pressure on the heads due to liquid heads is resisted by a horizontal force F acting as
shown in Figure 12.16b. It is interesting to note that for
hemispherical heads where H is equal to r, the bending
moment at the head-to-shell junction due to the force F
and the vertical force V is zero. The bending moment at
any point in the vessel is obtained from statics as shown
in Figure 12.17b.
245
246
12 Vessel Supports
Figure 12.15 Forces in ring girder.
Ms = 82.65 K-in
M = 60.05 K-in
6.27 K-in
M = 60.05 K-in
5.82 K
0K
38.45 K-in
19.35 K-in
14.04 K
Vs = 12.5 K
15.20 K
Forces at supports
Forces at point of maximum
torsional moment
A
β
A
L
H
H
(a)
W=
2
V = 3 Hw
2 Hw
3
r
4
F = r.w
7.2 K
3H
8
W
L+ 4 H
3
β
6
θ
2
r
F
Figure 12.17 Cross section of cylindrical shell at saddle location.
W
2
W
2
Mm
(b)
Ms
Ms
Figure 12.16 Saddle supports.
The section modulus of the shell between the saddles
is I/c and is expressed as 𝜋r2 t. At the saddles, the effective section modulus is reduced by the deformation of
the shell, which renders the full cross section less effective. Research has shown [4] that the length of the effective cross section of the shell is equal to the arc length
of the contact angle of the saddle plus one-sixth of the
unstiffened shell, as shown in Figure 12.17. The section
modulus of the arc length that is in tension is expressed as
[
]
Δ + sin Δ cos Δ − 2(sin2 Δ)∕Δ
2
Z=r t
. (12.21)
(sin Δ)∕Δ − cos Δ
Thus, the maximum longitudinal stress values can be
expressed as
⎧C M for midspan between supports
⎪ 1 m
𝜎l = ⎨C2 Ms for unstiffened shells at saddles,
⎪C1 Ms for stiffened shells at saddles
⎩
(12.22)
12.6 Saddle Supports
where
saddle is distributed along an arc length of
)
(
𝛽
𝜃
+
,
l = 2r
2 20
𝜎 1 = longitudinal bending stress in shell (ksi)
W = weight of vessel plus its contents (kips)
as shown in Figure 12.18. The shearing stress can then be
calculated as
L = length of vessel between tangent lines (in.)
r = radius of vessel (in.)
𝜎s = C3 V ,
t = thickness of shell (in.)
where
sin 𝜙
⎧
⎪ rt(π − 𝛼 + sin 𝛼 cos 𝛼)
⎪
⎪
C3 = ⎨
⎪ sin 𝜙2 ( 𝛼 − sin 𝛼 cos 𝛼 )
⎪ πrt
π − 𝛼 + sin 𝛼 cos 𝛼
⎪
⎩
1
πr2 t[
]
(sin Δ)∕Δ − cos Δ
1
C2 = 2
r t Δ + sin Δ cos Δ − 2(sin2 Δ)∕Δ
𝛽
Δ=𝜃+ .
6
C1 =
The shear stress in the shell between the saddles is
computed by assuming a sinusoidal distribution of the
shear forces where the maximum value is at the equator,
given by
𝜎s =
V
sin 𝜑,
πrt
(12.23)
where 𝜑 is measured as shown in Figure 12.18.
The shearing stress in the saddle area is influenced by
the deformation of the unstiffened shell above the saddle.
Experimental research has shown that the shear near the
ϕ
α
𝜙2 = angle that varies between π − 𝛼 and π
𝜎 s = shear stress.
Eq. (12.24) is also used to check the stress in the head.
In this case, the value of t in the expression for C 3 is taken
as the thickness of the head rather than the shell.
The circumferential stress in the shell in the saddle area
is calculated by assuming the shell above the saddle to
act as a fixed arch subjected to shearing stress as illustrated in Figure 12.19. Using the theory of indeterminate
structures, the moment at any point along the arch can
be expressed as
A
θ
2
σt
r
θ
2
β
20
Figure 12.18 Shear distribution.
near heads
𝜙 = angle as measured in Figure 12.18
β
Shear Distribution
Away From Saddle
for saddles
𝛼 = 𝜃/2 + 𝛽/20
ϕ
Shear Distribution
At Saddle
away from
,
heads
r = radius of vessel
MA
β
for saddles
where
Pt
ϕ2
(12.24)
W
2
Figure 12.19 Shear forces.
Q sin ϕ
πr
247
12 Vessel Supports
)
[
(
sin2 𝛽 − 58 𝛽 sin 2𝛽
Wr
cos 𝜙
M𝜙 =
+ 14 𝛽 2 cos2 𝛽
4πC4
(
)
1
1
+ 𝜙 sin 𝜑 𝛽 2 + sin 2𝛽 − sin2 𝛽
2
4
1
+ 𝛽 cos 𝛽(2𝛽 + sin 2𝛽)
4 (
)]
1
1
1
𝛽 + sin 2𝛽 + 𝛽 cos 2𝛽 ,
− sin 𝛽
2
8
4
where
180
170
160
150
θ (°)
248
C5
140
130
120
1
1
C4 = sin2 𝛽 − 𝛽 2 − sin 2𝛽.
2
4
The maximum value of M𝜑 given by this equation
occurs at 𝜙 = 𝛽. Hence, the maximum circumferential
bending moment in the shell can be expressed as [4]
W
(12.25)
rC ,
2 5
where C 5 is plotted in Figure 12.20.
Experimental work has shown that the width of the
shell that is effective in resisting the moment in Eq.
(12.25) can be taken as four times the radius or one-half
of the length of the shell, whichever is smaller.
It has also been shown [4] that Eq. (12.25) is valid when
A/r is equal to or greater than 1.0. For A/r values of 0.5
or less, it is suggested that the value of C 5 be reduced by
a factor of 4 due to the stiffening effect of the head. For
in-between values of A/r, a reduction factor
)
(
A
3A
+ 0.5 0.5 < < 1.0
RF =
2r
r
can be used.
The maximum circumferential force P at the horn of
the saddle is determined from
W
P𝛽 =
(12.26)
C.
2 6
A plot of the quantity C 6 is shown in Figure 12.20.
When the stress in the shell as calculated from Eq.
(12.25) and (12.26) is excessive, stiffening rings are
used at the vicinity of the saddles to carry the bending
moment.
M𝛽 =
C6
110
100
90
0
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.2
C5 and C6
Figure 12.20 C 5 and C 6 as functions of the saddle angle 𝜃.
Source: Brownell and Young (1959) [1]. Reproduced with
permission of John Wiley & Sons.
f s = allowable tensile stress of steel bolts
k = constant given by Eq. (12.5)
K 1 , K 2 = constants given in Table 12.4
K 3 − K 6 = constants given in Table 12.5
l = length
l1 − l3 = lengths as specified by Figure 12.2
M = bending moment
M𝜃 = bending moment in a ring girder
N = number of bolts
n = f s /f c
R = radius
T = tensile force on foundation
T 𝜃 = torsion moment in a ring girder
t = thickness
t s = equivalent thickness of anchor bolts
Nomenclature
V = shearing force
C = compressive force on concrete foundation
V 𝜃 = shearing force in a ring girder
F = bolt load
W = weight
f c = allowable compressive stress of concrete
𝜎 = stress
Further Reading
References
1 Brownell, L.E. and Young, E.H. (1959). Process
Equipment Design. New York: Wiley.
2 Wood, R.H. (1961). Plastic and Elastic Design of Slabs
and Plates. New York: Ronald Press.
3 Wichman, K.R., Hopper, A.G., and Mershon,
J.L. (1965). Local Stresses in Spherical and
Cylindrical Shells due to External Loadings,”
Further Reading
(2013). AISC Steel Construction Manual, 14e. Chicago, IL:
American Institute of Steel Construction.
Young, W., Budynas, R., and Sadegh, A. (2012). Roark’s
Formulas for Stress and Strain. New York: McGraw-Hill.
Useful Information on the Design of Plate Structures, Steel
Plate Engineering Data, vol. 2. New York: American Iron
and Steel Institute.
WRC Bulletin 107. New York: Welding Research
Council.
4 Zick, L.P. Stresses in large horizontal cylindrical pressure vessels on two saddle supports. In: Pressure
Vessel and Piping Design, Collected Papers 1927–1959,
vol. 1960. New York: American Society of Mechanical
Engineers.
249
251
Part IV
Theory and Design of Special Equipment
Oil storage tank. Source: Courtesy of Nooter Corporation, St. Louis, MO.
254
13
Flat-Bottom Tanks
13.1 Introduction
where
Flat-bottom tanks are normally constructed according to
one of the following three standards:
1) API 650. Welded tanks for oil storage.
2) API 620. Design and construction of large welded
low-pressure storage tanks.
3) AWWA D 100. Welded carbon-steel tanks for water
storage.
Table 13.1 shows a general comparison between the
requirements of the various standards. The values in the
table serve as a general comparison; however, specific
requirements and limitations are obtained from the
standards themselves.
E = modulus of elasticity (psi)
P = applied load (psi)
rr = roof radius
t = roof thickness
The required thickness is obtained by assuming that
the basic pressure consists of an approximate live load
of 25 psf, which includes snow load, and an approximate
dead load of a roof thickness of 0.5 in. as allowed by API.
Hence,
P = 25 psf live load + 20.4 psf dead load
= 45.4 ≈ 45 psf
= 0.315 psi.
13.2 API 650 Tanks
The requirements of API 650 [1] are for flat-bottom tanks
containing liquids with little or no surface pressure. The
design criteria are based on simplified equations with a
minimum amount of analysis.
13.2.1
Roof Design
Flat-bottom tanks with large diameter and fixed roof
normally are designed with column-supported roofs. As
the diameter gets smaller, self-supporting roofs become
more economical. Dome and cone roofs are the most
popular types.
13.2.1.1
Dome Roofs
The following equation for designing self-supporting
dome roofs is obtained from Eq. (9.2b), which is based
on a factor of safety (FS) of 4:
P=
0.0625 E
,
(rr ∕t)2
(13.1)
Letting E = 29 × 106 psi, and expressing rr in ft and t in
inches, Eq. (13.1) is
r
t= r .
200
API modifies this equation to take into consideration
the actual loads used. Hence, this equation becomes
r
(13.2)
t = r (T∕45)0.5 ,
200
where T is the actual applied load combination.
Equation (13.2) gives the required thickness of a selfsupporting dome roof.
The roof-to-shell junction has a stiffening ring to provide for the discontinuity forces shown in Figure 13.1.
Force H is expressed as
H = N𝜙 cos 𝜃
Pr
(13.3)
= r cos 𝜃.
2
Defining sin 𝜃 = 0.5D/rr , the required area at the dome
roof-to-shell junction needed to resist the horizontal tensile force H is
A = (H)(0.5D)∕S,
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
(13.4)
13.2 API 650 Tanks
Table 13.1 Various standards requirements for flat-bottom tanks.
API 650
Basic
Annex A small tanks
Annex F internal P
Annex AL aluminum
Maximum internal pressure
Minimum temperature (∘ F)
atm
atm
2.5 psi
atm
NS
−20
NS
NS
Maximum temperature (∘ F)
200
200
200
400
Maximum shell thickness (in.)
1.75
0.50
1.75
1.75
Minimum shell thickness
D < 50 ft
3/16 in.
a)
50 ft < D < 120 ft
1/4 in.
b)
120 ft < D < 200 ft
5/16 in.
D > 200 ft
3/8 in.
Minimum roof thickness
3/16 in.
Minimum bottom-plate thickness
0.25 in. + CAc)
Minimum top angle
D < 35 ft
2 in. × 2 in. × 3/16 in.
2.5 in. × 2.5 in. × 1/4 in.
35 ft < D < 60 ft
2 in. × 2 in. × 1/4 in.
2.5 in. × 2.5 in. × 5/16 in.d)
D > 60 ft
3 in. × 3 in. × 3/8 in.
3 in. × 3 in. × 3/8 in.e)
API 620
Basic
AWWA
Annex R
Annex Q
Maximum internal pressure
Minimum temperature (∘ F)
15 psi
7 psi
7 psi
atm
−50
−60
−325
−40
Maximum temperature (∘ F)
250
40
250
RT
Maximum shell thickness
NS
NS
NS
2 in.f )
Minimum shell thickness
D < 25 ft
3/16 in.
3/16 in.g)
25 ft < D < 60 ft
1/4 in.
1/4 in.h)
60 ft < D < 100 ft
5/16 in.
5/16 in.i)
D > 100 ft
3/8 in.
3/8 in.j)
Minimum roof thickness
NS
0.1875 in.k)
Minimum bottom-plate thickness
NS
0.1875 in.
Minimum top angle
NS
Similar to API 650 for carbon steel
NS = not specified, CA = corrosion allowance, RT = room temperature.
a) For D < 20 ft.
b) For 20 ft < D < 120 ft.
c) t = 0.236 in. (6 mm) for carbon-steel tanks.
d) For 35 ft < D < 61 ft.
e) For D > 61 ft.
f ) For thicknesses over 2 in., additional requirements must be met.
g) For D < 50 ft.
h) 50 ft < D < 120 ft.
i) 120 ft < D < 200 ft.
j) D > 200 ft.
k) For cone roofs, the plate thickness may be 7 gage steel.
255
256
13 Flat-Bottom Tanks
P
t
0.020
θ
0.019
0.018
Roof
D
0.017
0.016
Eq.(13.6)
0.015
Nϕ v
θ
rr
v
0.014
0.013
t/D 0.012
H
FS = 4.0
FS = 3.0
0.011
0.010
FS = 2.0
0.009
Shell
0.008
0.007
D/2
1
400 sin θ
0.006
θ
0.005
0.004
5
10
15
20
θ
25
30
35
Figure 13.2 Value of t/D versus slope angle 𝜃.
Figure 13.1 Roof-to-shell junction.
where S = 0.6F y
A = PD2 ∕(8S tan 𝜃)
(13.5)
t = required thickness of cone roof (in.)
for the required area at dome-to-shell junction.
13.2.1.2
D = diameter of tank (ft)
𝜃 = angle between cone and horizontal base (∘ )
Conical Roofs
The required thickness of self-supporting conical roofs is
based on Eq. (6.43) and is
Pa
2.60(t sin 𝜃∕D)2.5
=
E
FS(L∕2D)
(
)
10.4
t sin 𝜃 2.5
=
.
FS tan 𝜃
D
PD
4 sin 𝜃
(13.8)
and
D (FS tan 𝜃)0.4
.
(13.6)
sin 𝜃 326.54
Figure 13.2 shows a plot of this equation for various factors of safety. A more simplified equation used by API is
t=
D
,
400 sin 𝜃
which closely matches Eq. (13.6) with an FS of 2.0. API
modifies this equation to take into consideration the
actual loads used. Hence, this equation becomes
t=
D
(T∕45)0.5 ,
400 sin 𝜃
where T is the actual applied load combination.
The required area at the cone roof-to-shell junction is
obtained from Eq. (13.4). In this case, H for a dead load
is given by
H=
Substituting E = 29 × 106 psi and Pa = 0.315 psi in this
equation and expressing D in ft and t in inches results in
t=
Equation (13.7) gives the required thickness of a selfsupporting conical roof,
where
(13.7)
P
(13.9)
8S tan 𝜃
for the required area at the cone-to-shell junction, where
A = D2
A = required area (in.2 )
D = diameter of tank (ft)
𝜃 = angle between cone surface and horizontal base (∘ )
13.2.1.3
Small Internal Pressure
For tanks with small internal pressure, the maximum
pressure is limited to that which does not cause uplift of
the tank in the empty condition. Hence, from Figure 13.3,
the upward force due to pressure is equal to the downward force resulting from the weight of the roof plus
shell plus wind load.
13.2 API 650 Tanks
th
Redefining D in feet, M in ft-lb, and P in inches of water,
this equation becomes
P = 0.245Wroof ∕D2 + 0.163Wshell ∕D2
+ 0.294M∕D3 .
P
θ
H
V
Nϕ
V
(13.10)
This equation is in API for internal pressure when the
tank is empty.
The required compression area at the roof-to-shell
junction must be checked due to internal pressure. From
Figure 13.3, the vertical force V is given by
πD(V ) + Wroof − PD2 ∕4 = 0.
D/2
Or,
(13.11a)
V = PD∕4 − Wroof ∕πD,
W
where
V = vertical force (lb/in.)
D = diameter of tank (in.)
Figure 13.3 Roof-to-shell forces due to internal pressure.
The equation considering the weight of the roof and
shell is
PπD2 ∕4 = Wroof + Wshell
W roof = weight of roof (lb)
From Figure 13.3,
H = V ∕ tan 𝜃.
The required compression area at the junction is
or
P = 1.273 Wroof ∕D2 + 1.273 Wshell ∕D2 ,
where
D = diameter of tank (in.)
P = internal pressure (psi)
W roof = weight of roof (lb)
W shell = weight of shell (lb)
The maximum downward force due to wind bending
near the bottom of the tank is
Fw = 4M∕πD2 .
Force in the shell due to pressure = PD/4.
Hence, P = 16M/πD3 = 5.093 M/D3 .
The total pressure due to roof weight, shell weight, and
effect of bending moment is
P = 1.273Wroof ∕D2 + 1.273Wshell ∕D2
+ 5.093M∕D3 .
API adjusts this equation to take into consideration
various uncertainties and variations in the construction and loading. Accordingly, the second term on the
right-hand side of this equation is divided by 1.5, and the
third term is multiplied by 0.3. The equation becomes
P = 1.273Wroof ∕D2 + 0.849Wshell ∕D2
+ 1.528M∕D3 .
(13.11b)
(13.11c)
A = HD∕2S,
where
A = compression roof-to-shell area (in.2 )
S = allowable stress of the compression area (psi)
Hence, combining Eqs. (13.11a)–(13.11c) gives
D2 (P∕8 − Wroof ∕2πD2 )
.
(13.11d)
S tan 𝜃
Redefining P in inches of water, D in feet, and letting
S = 0.625F y , where F y is the yield stress, Eq. (13.11d)
becomes
D2 (P − 0.245Wroof ∕D2 )
A=
.
(13.11e)
0.962Fy tan 𝜃
A=
This equation can also be written in terms of maximum
pressure as
0.962AF y tan 𝜃 + 0.245Wroof
.
(13.12)
D2
API 650 assumes failure to occur when the stress in
the junction area reaches the yield stress. Accordingly,
Eq. (13.11e) must be multiplied by a factor of 1.6 to
obtain yield stress since it was originally derived based
on S = 0.625F y . Hence, Eq. (13.11e) can be written as
P=
1.539AF y tan 𝜃
D2
= 1.6P −
0.147Wroof
.
D2
257
258
13 Flat-Bottom Tanks
The left-hand side of this equation is defined by API as
the equivalent maximum failure pressure P′ . Thus,
P′ = 1.6P −
0.147Wroof
,
D2
(13.13)
where
D = diameter of tank (ft)
P = design pressure (inches of water)
P′ = failure pressure
W roof = weight of roof (lb)
A
H
API 650 also includes an equation for frangible joints.
The roof-to-shell joint is defined as frangible when the
area at the joint is based on an equivalent pressure whose
total force is equal to the weight of the shell at the joint.
Hence,
1ʹ
X
D/2
P = 4Ws ∕πD2 ,
where
B
1ʹ
P = equivalent internal pressure
X
W s = weight of shell at the junction
D = diameter of the tank
The value of horizontal force H from Figure 13.3 is
H = PD∕4 tan 𝜃
Defining 𝛾 = 62.4 pcf and adding the corrosion
allowance to this expression gives
H = Ws ∕(πD tan 𝜃).
The equation for the area at the joint required to support the unbalanced force H around the circumference is
given by
t=
2.6D(H − 1)G
+ CA,
S
(13.15)
where
CA = corrosion allowance (in.)
A = DH∕2Fy
A = Ws ∕(2πFy tan 𝜃).
Figure 13.4 Shell with two different course thicknesses.
(13.14)
D = diameter of tank (ft)
G = specific gravity of liquid
13.2.2
Shell Design
API 650 includes two methods for the design of shells
[2]. The first, called the one-foot method, consists of
calculating the required thickness of shell course A
shown in Figure 13.4 based on the hydrostatic pressure
at 1 ft above point X, which is the circumferential seam
between courses A and B. This method considers that
the bottom plate on course B stiffens the next course at
point X, and the maximum stress occurs at a location
higher than X. This location is arbitrarily set at 1 ft.
At point X, the hoop stress is given by
S=
PD
2t
or
G𝛾(H − 1)D
.
t=
2S
H = liquid height (ft)
S = allowable stress (psi)
t = required thickness (in.)
The second method, the variable-point method, is an
extension of the one-foot method in that it calculates
a more exact location of the maximum stress near the
junction of the bottom or shell courses with differing
thickness. In this case, the bottom course is assumed to
be hinged at its junction with the bottom plate. Hence,
the deflection due to internal pressure at the junction is
equal to the deflection due to an applied shearing force
as shown in Figure 13.5. From Section 5.2.1,
𝛿 v = 𝛿p ,
PR2
V
=
,
2𝛽 3 D
Et
13.2 API 650 Tanks
The thickness of the second course is determined from
the following equations:
h
R
2
δ = PR
Et
X
O
V
if √rt1 ≤ 1.375,
⎧t1
1
⎪t + (t − t )
2a
]
⎪ 2a [ 1
h
⎪
h1
if 1.375 < √ 1 < 2.625
t2 = ⎨ × 2.1 −
,
√
rt 1
1.25 rt 1
⎪
⎪
h
if √ 1 ≥ 2.625,
⎪t2a
⎩
rt 1
(13.17)
where
t 1 = thickness of first course (in.)
t 2 = thickness of second course (in.)
Figure 13.5 Shell-to-flat-bottom junction.
t 2a = thickness of second course
calculated from the equation for
upper course (in.)
or
2𝛽 3 DPR2
.
Et
The hoop stress at any point along the cylinder close to
the junction is given by
V =
where C 𝛽x is given by Eq. (5.23). Taking the derivative of
this equation with respect to x and equating it to zero
gives the point of maximum N 𝜃 . This occurs at
and
r = radius of shell (in.)
Design of the upper courses is based on the equation
2.6D(H − x∕12)G
+ CA,
(13.18)
S
where x is a variable design point that is a function of the
thicknesses of layers, tank radius, and liquid height.
Referring to Figure 13.6, x is the minimum value of x1 ,
x2 , and x3 obtained from the following equations:
√
x1 = 0.61 rt u + 0.32Chu ,
t=
N𝜃 = PR(1 − C𝛽x ),
x=
h1 = height of first course (in.)
3π
4𝛽
x2 = Chu ,
√
x3 = 1.22 rt u ,
)
(
3π
N𝜃 = PR 1 − e−3π∕4 cos
4
= 1.06PR.
where
Hence,
t = N𝜃 ∕S
1.06
=
PR
S
or, using the terminology of API 650,
)
(
2.6HDG
.
t = 1.06
S
Tests have shown that this equation is too conservative
because the maximum stress can be many feet away from
the junction where the pressure is reduced and the stiffness of the second course becomes significant. Accordingly, the equation for the design of the bottom course is
modified to read
(
)
√
)
(
0.463D HG
2.6HDG
t1 = 1.06 −
+ CA.
H
S
S
(13.16)
√
K(K − 1)
C=
√
1+K K
K = t L /t u
hu = height of upper shell (in.)
t L = thickness of lower shell (in.)
t u = thickness of upper shell (in.)
After establishing the shell thickness due to hydrostatic
pressure, stability under wind loads must be checked. The
applied wind pressure is normally approximated as
Pw = 18(V ∕120)2 ,
where Pw (psf)
= 0.125(V ∕120) ,
2
where
Pw = wind pressure
V = wind velocity, MPH
where Pw (psi),
(d)
259
260
13 Flat-Bottom Tanks
tu
Initial
location
of tank
shell
3
Variable
design point
hu
X3 =
1.22
rtu
0.32 Chu
Maximum
deflection
1
X1
0.61
rtu
Min. height of X2
when tL =1.0; C = 0 = X2
tu
0.61 rtu
0.32C
γhu r 2
Etu
Girth joint
2
C
γr 2
Etu
γhu r 2
Etu
tL
Unrestrained
radial
growth
γhu r 2
Etu
Because the wind pressure distribution may cause vacuum on part of the shell, the shell is designed to withstand
the vacuum pressure described by the aforementioned
equation. A simplified expression of the buckling of a thin
cylindrical shell is given by Eq. (6.42) as
(
)
(t∕D)2.5
2.42E
,
(e)
Pcr =
(1 − 𝜇2 )0.75 H∕D − 0.45(t∕D)0.5
where
D = diameter of shell (in.)
where
D = diameter of tank (ft)
H = distance between shell stiffeners (ft)
t = thickness of tank (in.)
V = wind velocity (MPH)
The required section modulus of the stiffening ring
necessary for resisting the lateral pressure is obtained
from the following classical buckling equation of a ring:
H = effective height of shell (in.)
Fcr = 24EI∕D3
E = modulus of elasticity (psi)
or
𝜇 = Poisson’s ratio
Fcr =
Pcr = buckling pressure (psi)
For thin cylinders, the quantity 0.45(t/D)0.5 can be set
to zero. Also, substituting E = 30,000,000 psi and 𝜇 = 0.3,
Eq. (e) can be solved for H as
H≈
Figure 13.6 Elastic movement of shell courses of
girth joint [3].
77.9 × 106 t(t∕D)1.5
.
Pcr
(f )
Combining Eqs. (d) and (e) and using an FS of 2.0 gives
where
E = modulus of elasticity (psi)
D = diameter of tank (in.)
F = force on tank (lb/in. of circumference)
F cr = buckling force (lb/in. of circumference)
FS = factor of safety
H = 312 × 106 t(t∕D)1.5 (120∕V )2 .
I = moment of inertia of stiffening ring (in.4 )
API expresses D and H in feet rather than inches.
Accordingly, this equation becomes approximately
H ≈ 600,000 t(t∕D)1.5 (120∕V )2 ,
24EI
,
(FS)D3
(13.19)
The force F can be expressed as
F = PH,
13.2 API 650 Tanks
where
H = distance between shell stiffeners (in.)
D/2
P = wind pressure (psi)
Hence,
I=
PHD3 (FS)
.
24E
w = p.G.H
From the strength of materials,
tb
Z = I∕c,
where
L
c = one-half of the depth of the bar stiffening ring (in.)
(a)
Z = section modulus (in.3 )
w
The expression for Z becomes
PHD2 (FS)
Z=
.
24E(c∕D)
tb
API assumes that the ratio of the outstanding leg of a
stiffener (2c) to the diameter of the tank is not less than
0.015. Hence, c/D = 0.0075. Substituting Eq. (13.11a) for
P, 2.5 for FS, E = 30,000,000 psi, and expressing D and H
in feet rather than inches gives
HD2 (V ∕120)2
Z=
.
10,000
M
(13.20)
(13.21)
L
(b)
Figure 13.7 Tank bottom plate with annular outer plate.
where
F y = yield stress of annular base plate (psi)
G = specific gravity of liquid (>1.0)
13.2.3
H = height of liquid in tank (ft)
Annular Plates
The required thickness of the bottom plate in an API 650
tank is given in Table 13.1 [3]. At the shell-to-bottom
plate junction, the API standard requires a butt-welded
annular plate whose thickness varies between 0.25 and
0.75 in. and is a function of the stress and thickness of
the first shell course. The width of the annular plate
must be adequate to support the column of water on top
of it in case of a foundation settlement. By referring to
Figure 13.7,
M=
wL2
.
2
L = width of annular plate (in.)
𝜌 = density of water (psi/ft of height)
t b = thickness of annular base plate (in.)
Example 13.1
The steel tank shown in Figure 13.8a contains a liquid
up to the roof-to-shell junction level. Design the various
tank components if G = 1.1, CA = 0.0, allowable stress in
shell S = 15,000 psi, live load plus dead load P = 45.0 psf,
wind velocity V = 100 MPH, yield stress of stiffener at
roof-to-shell junction F y = 27,500 psi. Use the one-foot
method for the shell design.
Using plastic analysis,
𝜎y =
4M
.
tb2
Combining these two equations, letting w = 𝜌GH, and
using an FS of 2.0 gives the following equation used by
API
L = 2tb (Fy ∕2𝜌GH)0.5 ,
(13.22)
Solution:
For the roof design, Eq. (13.2) gives
r
t = r (T∕45)0.5
200
80
t=
(45.0∕45)0.5
200
t = 0.40 in.
Use t = 7/16 in. for the dome roof.
261
262
13 Flat-Bottom Tanks
rr = 80ʹ
the roof-to-shell junction from Eq. (13.5) is
A = PD2 ∕(8S tan 𝜃),
10ʹ
10ʹ
80ʹ
where
S = 0.6Fy = 0.6(27,500) = 16,500 psi,
∘
𝜃 = 26.57
from Figure 13.7.
Hence,
(a)
A = (45)(80)2 ∕[8(16,500)(0.5)]
= 4.36 in.2
t=
Angle 4 × 4 × 5
1
8
4
1
4
1ʺ
t=
4
7ʺ
16
t=
Use 4 × 4 × 5/8 in. angle with A = 4.61 in.2 . For the bottom plate, use t = 1/4 in. according to Table 13.1.
Assume that the annular plate is 1/4 in. thick. Then the
width of the annular plate from Eq. (13.22) is
L = 2 tb (Fy ∕2𝜌GH)0.5
5ʺ
16
L = 2(0.25)(27 500∕[2(0.433)(1.1)(20)]0.5 )
L = 19.92 in.
1ʺ
4
1ʺ
4
24ʺ
Use a 24 in. wide annular plate.
The aforementioned details of construction are shown
in Figure 13.8b.
(b)
Figure 13.8 Tank details.
For the shell design, the required thickness for the bottom course is given by Eq. (13.15) as
2.6(80)(20 − 1)(1.1)
+ 0.0
15,000
T = 0.29 in.
T=
Example 13.2
In Example 13.1, determine (a) the maximum allowable
internal pressure and the maximum failure pressure,
(b) the required roof-to-shell area if a frangible joint
is required, and (c) the thickness of the shell using the
design conditions of Example 13.1 and the variable-point
method.
Solution:
Use t = 5/16 in. for the bottom course. For the top
course
2.6(80)(10 − 2)(1.1)
T=
+ 0.0
15,000
T = 0.14 in.
Use t = 1/4 in. for the top course according to
Table 13.1.
The required intermediate stiffener spacing is obtained
from Eq. (13.19) as
H = 600,000t(t∕D)1.5 (120∕V )2 .
Using a conservative value of t = 0.25 in.,
H = 600,000(0.25)(0.25∕80)1.5 (120∕100)2
= 37.7 ft.
Because this is larger than the height of the tank, no
intermediate stiffeners are needed. The required area of
a) The maximum pressure that does not cause uplift of
the shell is obtained from Eq. (13.10):
Weight of shell
= (40.82)(π)(80)(10)(5∕16 + 1∕4)
= 57,800 lbs
Weight of roof(see Appendix I)
= (40.82)(0.8418)(802 )(7∕16)
= 96,200 lbs
Bending moment = 0
From Eq.(13.10),
P = [(0.245)(96,200)
+ (0.163)(57,800) + 0]∕802
= 5.16 in.of water
= 0.91 psi.
13.3 API 620 Tanks
The maximum pressure that does not cause excessive
stress at the roof-to-shell junction is given by Eq. (13.12):
P = [(0.962)(4.36)(27,500)(0.577)
+ (0.245)(96,200)]∕802
= 14.08 in.of water
= 0.51 psi.
Hence, the maximum allowable internal pressure is
5.16 in. of water.
The maximum failure pressure P′ is given by Eq.
(13.13):
P′ = (1.6)(5.16) − (0.147)(96,200)∕802
= 6.05 in. of water.
b) The frangible joint area given by Eq. (13.14) is
√
x3 = 1.22 (40 × 12)(0.14)
= 10 in.
Hence, x = 10 controls and
2.6D(H − x∕12)G
t2 =
S
2.6(80)(10 − 10∕12)(1.1)
=
15,500
= 0.14 in.
Because this value is the same as the assumed one,
the analysis is complete, and no additional iteration is
needed. Hence, use
5
t1 = 16
in. for the bottom course.
t2 = 14 in. for the top course as governed by Table 13.1.
A = 57,800∕[(2π)(27,500)(0.577)]
= 0.58 in.2
c) The thickness of the bottom course is calculated from
Eq. (13.16) as
)
(
√
0.463 × 80 20 × 1.1
t = 1.06 −
20
15,000
(
)
2.6 × 20 × 80 × 1.1
×
15,000
= 0.30 in.
For the top course, the quantity
h1
10 × 12
=√
= 10
√
rt 1
(40 × 12)(0.3)
indicates that t 2 = t 2a as given by Eq. (13.17). Equation
(13.18) is based on an iterative process that is initiated
by assuming a value of t 2 , which can be obtained from
the approximate equation
2.6(H − 1)DG
t2 =
15,000
2.6 × 9 × 80 × 1.1
=
15,000
= 0.14 in.
From Eq. (13.18),
t
0.30
K= L =
= 2.14
tu
0.14
C = 0.40
√
x1 = 0.61 (40 × 12)(0.14)
+ 0.32(0.40)(10 × 12)
= 20.36 in.
x2 = 0.40(10 × 12)
= 48.00 in.
13.3 API 620 Tanks
API 620 tanks [4] tend to be more complicated in geometry and are generally subjected to higher internal pressure
than API 650 tanks. Accordingly, the requirements of API
620 differ significantly from those of API 650 because the
thickness of the components is obtained from stress analysis that considers the biaxial stress state, rather than a set
of simplified formulas.
The stress analysis procedure in API 620 is based
on Eqs. (6.10) and (6.11). Equation (6.11) for N 𝜙 can
be determined for any shell configuration by using the
summation of forces obtained from a free-body diagram.
The advantage of a free-body diagram is that forces
other than pressure can be accounted for without going
through an integration process. Once N 𝜙 is determined,
the value of N 𝜃 is obtained from Eq. (6.10) as
N𝜃 N𝜙
+
= P,
R2
R1
(13.23)
where N 𝜙 is the sum of forces at a given cross section.
Example 13.3 illustrates the application of Eq. (13.23)
to API 620 tanks.
Example 13.3
The tower shown in Figure 13.9 is filled with a liquid
whose specific gravity is 1.0 up to point a. Above point
a, the tower is subjected to a gas pressure of 5 psi. Determine the forces in the various components of the tower,
disregarding the dead weight of the tower.
Solution:
Roof Forces
The maximum force in the roof is obtained from
Figure 13.10a. Below section a–a, a 5 psi pressure is
263
264
13 Flat-Bottom Tanks
section b–b is
40ʹ
W = 62.4(π)(20)2 (35)
= 2,744,500 lb.
P = 5 psi
a
The total pressure at b–b is 5 + (62.4/144)(35):
P = 20.17 psi.
R = 48ʹ
35ʹ
γ = 62.4 pcf
The sum of the forces at b–b is equal to zero. Hence,
2,744,500 − (20.17)(π)(240)2 + V (π)(480) = 0
V = 600 lb∕in.
and
b
N𝜙 = 600 lb∕in.
10ʹ
c
In a cylindrical shell, R1 = ∞ and R2 = R. Hence, Eq.
(13.23) becomes
N𝜃 = PR = (20.17)(240)
25ʹ
= 4841 lb∕in.
Conical Transition
d
20ʹ
Figure 13.9 Vertical storage tank.
At section b–b, the force V in the 40-ft shell must equal
the force V in the cone, due to continuity, as shown in
Figure 13.10b:
V = 600 lb∕in.
needed to balance the pressure above section a–a. Force
N 𝜙 in the roof has a vertical component V around the
perimeter of the roof. Summation of forces in the vertical
direction gives the following equation:
P
=0
4
PD
480
V =
=5×
4
4
= 600 lb∕in.
πDV − πD2
Hence,
V
600
=
N𝜙 =
sin 𝜃
0.42
= 1440 lb∕in.
and from Eq. (13.23) with R1 = R2 = 48 ft,
PR 5 × 576
=
2
2
= 1440 lb∕in.
N𝜃 =
and the unbalanced force H a = 1309 lb/in. (inward).
40-ft Shell
The maximum force in the shell is at section b–b as
shown in Figure 13.10b. The total weight of liquid at
and
600
0.707
= 849 lb∕in.
N𝜙 =
In a conical shell, R1 = ∞ and R2 = R/sin 𝜃. Hence, Eq.
(13.23) becomes
(20.17)
PR
= 240
N𝜃 =
sin 𝜃
0.707
= 6847 lb∕in.
The horizontal force at point b is H b = 600 lb/in.
(inward).
Figure 13.10c shows the forces at point c. The weight of
the liquid in the conical section is
π𝛾H 2
(R1 + R1 R2 + R22 )
W=
3
π × 62.4 × 10 2
=
(10 + 10 × 20 + 202 )
3
= 457,400 lb.
The total liquid weight is
W = 2,744,500 + 457,400 = 3,201,900 lb.
The pressure at section c–c is 5 + (62.4/144)(45):
P = 24.5 psi.
13.3 API 620 Tanks
480ʺ
W
5 psi
a
θ = 24.62°
H
a
5 psi
V
Nϕ
V
(a)
c
H
p
W
V
R = 240ʺ
Nϕ V
p
c
b
c
c
b
V
V
p
(c)
V
V
p
Nϕ
θ = 45°
b
H
W
b
d
d
(b)
p
V
(d)
Figure 13.10 Free-body diagrams.
Summing forces at section c–c gives
(24.5)(π)(120)2 − 3,201,900 − (V )(π)(240) = 0
V = −2777 lb∕in.
The negative sign indicates that the vertical component
of N 𝜙 is opposite to that assumed in Figure 13.10c and is
in compression rather than tension. This is caused by the
column of liquid above the cone, whose weight is greater
than the net pressure force at section c–c. Thus,
−2777
N𝜙 =
0.707
= −3927 lb∕in.
(compressive),
RP
24.5
= 120 ×
N𝜃 =
sin 𝜃
0.707
= 4158 lb∕in.,
Hc = 2777 lb∕in.
(inwards).
20-ft Shell
At section c–c, the value of V in the 20-ft shell is the
same as in the cone, due to continuity. Thus,
N𝜙 = V = −2777 lb∕in.
N𝜃 = PR = (24.5)(120)
= 2940 lb∕in.
At section d–d, the liquid weight is given by
W = 3,201,900 + (62.4)(𝜋)(10)2 (25)
= 3,692,000 lb,
and the pressure is calculated as
)
(
62.4
(70)
P =5+
144
= 35.3 psi.
265
13 Flat-Bottom Tanks
13.3.1.1 Compressive Stress in the Axial Direction
with No Stress in the Circumferential Direction
From Figure 13.10d, the summation of forces about
d–d is
The rules for this case are based on the axial buckling of a cylindrical shell as given by Eq. (5.28). With
E = 30 000 000 psi and an FS of 10, this equation becomes
( )
t
.
(13.25)
𝜎 = 1.8 × 106
R
3,692,000 − 35.3(π)(120)2 + V (π)(240) = 0
N𝜙 = V = −2777 lb∕in.,
which is the same as that at point c, and
N𝜃 = PR = (35.3)(120)
To prevent the stress in Eq. (13.25) from exceeding
the allowable tensile stress of the material, an arbitrary value of 15,000 psi is established as the upper
limit of the allowable compressive stress. This is shown
in Figure 13.11 as line OABC, where AB is a transition line between Eq. (13.25) and the upper limit of
15,000 psi.
= 4236 lb∕in.
13.3.1
Allowable Stress Criteria
The required thickness of API 620 components in tension
is determined from the larger of the values obtained from
these two expressions:
t=
t=
13.3.1.2 Compressive Stress with Equal Magnitude
in the Meridional and Circumferential Directions
N𝜙
SE
N𝜃
,
SE
The governing equation is obtained from Eq. (6.35) for
the buckling of a spherical shell with an FS of 4. Using
E = 30 000 000 psi, the equation becomes
( )
t
,
𝜎 = 937,500
R
(13.24)
where
t = required thickness of component (in.)
which is approximated in API as
( )
t
.
= 1,000,000
R
N 𝜃 = hoop force (lb/in.)
N 𝜙 = meridional force (lb/in.)
S = allowable tensile stress (psi)
(13.26)
This value is 1.8 times smaller than the value given by
Eq. (13.25). Accordingly, the limit of Eq. (13.26) is established as 15,000/1.8 = 8340 psi. Thus, ODEF shown in
Figure 13.11 is the criterion used for components having
compressive stress of equal magnitude in the meridional
and circumferential directions.
E = joint efficiency similar to that discussed in
Section 8.1
The API criteria for the components in compression
are as follows.
Figure 13.11 Allowable compressive stress. Source:
Courtesy of the American Petroleum Institute.
16 000
14 000
A
t
0( )
7,40 R
7
2
B
0+
0,15
σ=1
15 000
C
0( t
R )
12 000
00
,00
10 000
D
1,8
8000
6000
σ=
Compressive stress psi
266
4000
2000
σ=
0.002
0
1,
00
t )
0( R
0
,0
0.006
E
t )
,200 ( R
0 + 154
σ = 565
0.010
0.014
t
R
0.018
8340
0.022
F
Note:
If compressive stress is
latitudinal, use R = R 1
if compressive stress is
meridional, use R = R 2
18 000
0
N=0
A
12 000
S cs
Scs = 15,000 psi C
N=0
0.10
0.20
0.30
Do not extrapolate
beyond this line
0.40
10 000
0.50
0.60
=1
,80
0,0
00
( t
R )
8000
0.70
0.80
cs
6000
Values of tensile stress factor N
14 000
,15
= 10
t
( )B
400 R
7,
+ 27
4000
S
Coexistent compressive stress, Sc, for biaxial tension–compression (psi)
13.3 API 620 Tanks
0.90
2000
0
0
0.002
0.004
0.006
0.008
0.010
0.012
0.014
0.016
0.018
0.020
0.022
t/R ratio
Figure 13.12 Biaxial stress chart for combined tension and compression in steels of yield stress 30 000–38 000 psi. Source: Courtesy of the
American Petroleum Institute.
13.3.1.3 Compressive Stress with Unequal Magnitude
in the Meridional and Circumferential Directions
follows. Let
M=
The criteria for this case are based on the following
equations:
(larger stress) + 0.8(smaller stress)
stress determined from OABC in
Figure 13.11 using R for the larger force
≤ 1.0
1.8(smaller stress)
≤ 1.0.
stress determined from OABC in
Figure 13.11 using R for the smaller force
(13.27b)
The criteria are based on the assumption that the ability of a component to resist compressive force in a given
direction is reduced as the tensile force in the other direction increases. The governing relationship is derived as
actual tensile stress
.
allowable tensile stress
Then,
M2 + MN + N 2 = 1.0.
(13.28)
The interaction of this equation with Eq. (13.25) is
shown in Figure 13.12.
13.3.2
13.3.1.4 Compressive Stress in One Direction
and Tensile Stress in the Other Direction
allowable compressive stress
from OABC of Figure 13.11
and
N=
(13.27a)
actual compressive stress
Compression Rings
As shown in Example 13.3, there are unbalanced horizontal forces at the roof-to-shell and cone-to-shell junctions. These forces must be carried by a compression-ring
region at that location. The region can be in tension or
compression, depending on the direction of the discontinuity as well as the hoop forces. API 620 assumes that
267
13 Flat-Bottom Tanks
th
Roof of tank
)
Figure 13.13 Compression-ring region. Source: Courtesy of the
American Petroleum Institute.
–c
(t h
=0
Wh
.6
R2
R2
Wc = 0.6 Rc(tc–c)
268
Compression
ring region
Rc
tc
Cylindrical sidewall of tank
portions of the roof, shell, and cone shown in Figure 13.13
are part of the compression-ring region. The total force
given by the following equation is assumed to be supported by the ring region:
Q = N𝜙r Wh + N𝜃s Wc + HR,
(13.29)
where
Q = total force at ring region (lb)
N 𝜙r = meridional force in roof or cone (lb/in.)
Example 13.4
Determine the required thicknesses of the 20-ft shell and
the conical reducer in Example 13.3. Also, determine the
required stiffening ring area at point c. Let S = 20,000 psi,
E = 1.0, and CA = 0.0.
Solution:
20-ft Shell
From Example 13.3, the forces at point c are
N 𝜃s = circumferential force in shell (lb/in.)
N𝜙 = −2777 lb∕in.,
W h = effective length of roof or cone as determined
from Figure 13.13 (in.)
N𝜃 = 2940 lb∕in.,
W c = effective length of shell as determined from
Figure 13.13 (in.)
and the forces at point d are
N𝜙 = −2777 lb∕in.,
N𝜃 = 4236 lb∕in.
H = unbalanced horizontal force at junction (lb/in.)
R = radius of tank at junction (in.)
The total required area at the junction is determined
from
⎧ Q
when Q is compressive,
⎪ 15,000
(13.30)
A=⎨
⎪Q
when
Q
is
tensile,
⎩ SE
where
A = required area (in.2 )
S = allowable tensile stress (psi)
E = joint efficiency
Details of various ring attachments are shown in
Figure 13.14.
Thus, the forces at point d control. From Eq. (13.24),
4236
20,000 × 10
= 0.21 in.
t=
Let
t=
9
in.
16
Then,
t
= 0.0047,
R
4236
= 7530 psi,
0.5625
actual compressive stress
2777
=
= 4940 psi.
0.5625
actual tensile stress =
13.3 API 620 Tanks
Compression areas are shaded.
16t (max)
16t (max)
Wh
Wh
16t (max)
Wh
t
t
Wc
Wc
Wh
t
Wc
Wc
Bracket
Detail a
Detail b
Detail C
Alternate roof
connection
Permissible
Wh
Wh
L
Wh
t
C
L
L
L
t
t
Detail d
Wh
t
Wc
Wc
Wc
Wc
For details e through g, Wh ≥ L ≤ 16t and Wh ≤ 32t
Detail f
Detail f-1
Detail e
Detail g
Permissible
A
A
Neutral axis
of angle
Wh
B
A
Wh
B
B
t
Wc
16t (max)
Detail h
Wc
Wh
B
t
t
B≤A
B≤A
t
A
Wh
16t (max)
Wc
Neutral axis
of angle
Detail h-1
16t (max)
Neutral axis
of angle
Detail i
16t (max)
Wc
Neutral axis
of angle
Detail i-1
Permissible where roof (or bottom)
plate thickness is ≤ 1/4 in.
Neutral axis
of angle
Detail j
Neutral axis
of angle
Detail k
Detail 1
Not permissible
Figure 13.14 Some permissible and nonpermissible details of compression-ring-juncture construction. Source: Courtesy of the American
Petroleum Institute.
269
270
13 Flat-Bottom Tanks
The allowable compressive stress from Eq. (13.25) is
1.8 × 106 × (0.5625/120) = 8440 psi. From Eq. (13.28),
Compression Ring
From Example 13.3, the discontinuity force at point c is
7530
N=
= 0.38,
20,000
4940
M=
= 0.59,
8400
0.382 + 0.38 × 0.59 + 0.592 = 0.72 < 1.0.
9
-in.
16
Use t =
H = −3927 lb∕in.
(inwards),
Wc = 0.6(120)(0.5625)
= 4.93 in.,
)
(
120
(0.6875)
Wh = 0.6
0.707
= 6.48 in.,
shell.
Conical Transition Section
N𝜃s = 2940 lb∕in.,
N𝜙r = −3927 lb∕in.,
From Example 13.3, the forces at point b are
Q = (−3927)(6.48) + 2940(4.93)
N𝜙 = 849 lb∕in.,
+ (−3927)(120)
N𝜃 = 6847 lb∕in.,
= −482,190 lb.
and from Eq. (13.24),
From Eq. (13.30),
6847
20,000
= 0.34 in.
t=
482,190
15,000
= 32.15 in.2 = required area
A=
The forces at point c are given by
available area = (0.6875)(6.48)
N𝜙 = −3927 lb∕in.,
+ (0.5625)(4.93)
= 7.23 in.2
N𝜃 = 4158 lb∕in.
needed area = 31.25 − 7.23 = 24.02 in.2
Let
t=
11
in.
16
Use 2 in. × 12 in. ring.
Then,
t
0.6875
=
= 0.0057,
R
120
4158
actual tensile stress =
= 6050 psi,
0.6875
3927
actual compressive stress =
0.6875
= 5710 psi,
allowable compressive stress
from Eq. (13.25) =
6050
= 0.30,
20,000
5710
M=
= 0.78,
7290
0.302 + 0.30 × 0.78 + 0.782
= 0.9144
Use t =
11
16
13.4.1
(acceptable).
in. for conical transition section.
Design Rules
ANSI B96.1 Standard for Aluminum Tanks was withdrawn subsequent to the 1999 edition, and the rules
were incorporated in Annex AL of API 650. Differences
in various requirements between aluminum and steel
tanks are given in Table 13.1.
The design of dome roofs is obtained from Eq. (9.2b) for
axial compression and is based on an FS of 4.0. Hence,
1.8 × 106 × 0.6875
120
= 7290 psi,
N=
13.4 Aluminum Tanks
p=
0.0625E
,
(rh ∕th )2
or, in terms of thickness t h ,
th = 4rh (p∕E)0.5 ,
(13.31)
where
E = modulus of elasticity of aluminum, psi. It is
normally taken as 10 000 ksi at room temperature
P = roof dead and live loads (psi)
rh = roof radius (in.)
t h = roof thickness (in.)
13.5 AWWA Standard D100
The required area at the roof-to-shell junction is
obtained from Eq. (13.4):
Ej
=
joint efficiency of the longitudinal weld
𝛾
=
density of water, 0.0361 lb/in.3
(13.32)
G
=
specific gravity of liquid
H
=
height of liquid (in.)
S
=
allowable stress (psi)
t
=
thickness (in.)
A = (H)(0.5D)∕f
or
A = pD2 ∕(8f tan 𝜃),
(13.33)
where
The required section modulus Z for the top stiffening
ring is obtained from Eq. (13.20) using an FS of 2.0.
A = area at dome roof-to-shell junction (in.2 )
D = diameter of tank (in.)
f
= the allowable stress equal the lesser 0.6F y or
0.5F t (psi)
Ft
= tensile strength of the material (psi)
Fy
= yield stress of the material (psi)
p = roof dead and live loads (psi)
𝜃
= angle of the roof with respect to the
horizontal base
The required thickness of a self-supporting conical roof
is obtained from Eq. (6.43) as
p∕E =
2.6(t sin 𝜃∕D)2.5
.
(FS)(L∕2D)
(13.34)
This equation, when plotted similar to Figure 13.2
using an FS = 2.0, can be approximated by the following
equation used in API for establishing the thickness of an
aluminum conical roof
t = (2D∕ sin 𝜃)(p∕E)0.5 .
(13.35)
The required area at the cone roof-to-shell junction is
obtained from Eq. (13.9) as
A = D2
p
.
8f tan 𝜃
(13.36)
The design of aluminum shells is based on Eq. (13.15),
which is based on the “one-foot” method given by
t=
𝛾 GD(H − A)
+ CA,
2Ej S
where
A = 12 in.
CA = corrosion allowance (in.)
D = diameter of tank (in.)
PHD3
.
12Ec
(13.37)
(13.38)
The maximum height for the unstiffened shell is
obtained from Eq. (e).
(
)
(t∕D)2.5
2.42E
.
Pcr =
(1 − 𝜇2 )0.75 H∕D − 0.45(t∕D)0.5
(13.39)
Using an FS to obtain the allowable pressure, a value of
0.3 for Poisson’s ratio, and disregarding the small quantity
0.45(t/D)0.5 results in
p = 2.60E
Rearranging this equation and using L = (D/2)(tan 𝜃)
gives
t = (0.392D∕ sin 𝜃)(p∕E)0.4 (FS tan 𝜃)0.4 .
Z=
(t∕D)2.5
.
(FS)(H∕D)
(13.40)
Using an FS of 2.0, a modulus of elasticity of
10,000,000 psi, a wind pressure of 0.13 psi (19 psf ), and
solving for H results in the following equation listed in
API 650,
H = 2400t(1200t∕D)1.5 .
(13.41)
13.5 AWWA Standard D100
Most water storage tanks are built in accordance with the
American Water Works Association (AWWA) Standard
for Welded Steel Tanks for Water Storage [5]. The maximum allowable tensile stress in the plates as permitted
by the standard is generally 15,000 psi. The allowable
compressive stress in shell components is similar to
that in API 650. The allowable compressive stress in
structural components is similar to that in AISC. Other
design details given in AWWA D100-11 are very similar
to those in API 650 with the exception of those shown in
Table 13.1.
The minimum depth of a tank foundation is obtained
from Figure 13.15 for various locations in the United
States. Also, pipes connected to the tank must be buried
in the soil at a minimum depth of not less than that
shown in Figure 13.16.
271
272
13 Flat-Bottom Tanks
125°
5ʺ
120°
115°
110°
105°
70ʺ
95°
100°
80ʺ 90ʺ
100ʺ
45°
90°
85°
80°
75°
70°
65°
1009ʺ 0ʺ
80ʺ
100ʺ
90ʺ
100ʺ
90ʺ
45°
70ʺ
60ʺ
60ʺ
40°
50ʺ
40ʺ
60ʺ
50ʺ
35°
30ʺ
20ʺ
40ʺ
Consult lo
cal
records
for this ar
ea
40°
10ʺ
30ʺ
35°
5ʺ
0°
20ʺ
30°
30°
10ʺ
0ʺ
5ʺ
5ʺ
0ʺ
Oʹ
25°
0
118°
110°
105°
100°
95°
90°
25°
Kilometers
100 200 300 400 500
85°
80°
75°
Figure 13.15 Extreme frost penetration (in.). Source: Courtesy of AWWA.
7
5 5 1/2
6
4 4 1/2
6
3 3 1/2
2 2 1/2
7 1/2
71
/2
6 1/2
61
/
6 2
7
61
/2
6 1/2
6
7 1/2
7
6 1/2
6
5 1/2
5
4 1/2
4
3 1/2
3
2 1/2
2
5 1/2
5
4 1/2
4
3 1/2
3
2 1/2
2
0
200
Km
Figure 13.16 Depth of cover above top of pipe (ft). Source: Courtesy of AWWA.
400
Further Reading
References
1 (2013). Welded Tanks for Oil Storage, 12th ed., API
4 API ed. (2014). Design and Construction of Large,
Standard 650e. Washington, DC: American Petroleum
Institute.
2 Zick, L.P. and McGrath, R.V. (1968). Design of
large-diameter cylindrical shells. Presented at the
33rd Midyear Meeting of the American Petroleum
Institute, New York, 1968.
3 Karcher, G.G. (1981) Stresses at the shell-to-bottom
junction of elevated-temperature tanks.
Proceedings—Refining Department, 46th Midyear
Meeting, American Petroleum Institute (May 1981).
Welded, Low-Pressure Storage Tanks, 12th ed., API
Standard 620,e. Washington, DC: American Petroleum
Institute.
5 AWWA (2011). Standard for Welded Carbon Steel
Tanks for Water Storage, AWWA D100-11. New York:
American Water Works Association.
Further Reading
(1976). Steel Tanks for Liquid Storage—Steel Plate
Engineering Data, vol. Vol. 1. Washington, DC:
American Iron and Steel Institute.
273
Rod baffle heat exchangers. Source: Courtesy of the Nooter Corporation, St. Louis.
276
14
Heat-Transfer Equipment
Heat-transfer equipment is used in many applications
such as heat exchangers in the petrochemical industry
and condensers and evaporators in heating and refrigerating systems. Heat-transfer equipment varies from
miniature heat exchangers a few inches in diameter
to over 20 ft in diameter. This chapter presents the
theoretical background and design equations of heat
exchangers.
14.1 Types of Heat Exchangers
Heat exchangers in the United States are normally
designed according to the Standards of Tubular
Exchanger Manufacturers Association (TEMA) [1] and
the ASME Code, VIII. In general, TEMA requirements
are a supplement to the ASME requirements, for they
tend to include areas not discussed in the ASME. Most
of the TEMA design equations relate to tubesheet design
when affected by differential pressure and temperature,
expansion joints, bustles, and so on.
TEMA uses alphabetical designation to differentiate
between various types of frequently used components.
This is illustrated in Figure 14.1. The components can be
interchanged to form a wide variety of heat-exchanger
configurations, as shown in Figure 14.2.
Their rules, which apply to three different classes of
construction depending on the severity of service, are
referred to as R, C, or B. A summary of the differences
between these classes is given in Table 14.1.
Letting G = 2a and using a shape factor in bending of
1.5, the aforementioned equations can be written in
terms of thickness as
1.36G
(P∕S)0.5 for simply supported plate
T=
3
1.06G
T=
(P∕S)0.5 for fixed plate.
3
The TEMA equation for the design of a U-tube
tubesheet is
FG
(14.1)
(P∕𝜂S)0.5 ,
T=
3
where
for simply supported condition,
F = 1.2 0 < T∕D < 0.02
= (1∕15)[17 − 100(t∕D) 0.02 < T∕D < 0.05
= 1.0 0.05 < T∕D < 0.1,
for fixed condition,
F = 1.0 0 < T∕D < 0.02
= (1∕12)[17 − 100(t∕D)] 0.02 < T∕D < 0.05
= 0.8 0.05 < T∕D < 0.1,
D = diameter of shell
G = effective diameter of tubesheet
P = effective pressure
S = allowable stress
T = thickness of tubesheet
t = thickness of shell
The shearing stress in the tubesheet at the outer tube
perimeter must also be checked and kept below an
allowable stress. The total force W due to pressure in the
tubesheet shown in Figure 14.3 is
14.2 TEMA Design of Tubesheets
in U-tube Exchangers
The basic equation for the design of heat exchangers is
obtained from Examples 7.1 and 7.2 as
⎧ 1.24Pa2
⎪
2
𝜎=⎨ T 2
0.75Pa
⎪
⎩ T2
for simply supported plate
for fixed plate
.
W = PA.
The shear area As through the outer perimeter is
obtained from Figure 14.3 and is
(
)
do
As = CT 1 −
.
p
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
14.2 TEMA Design of Tubesheets in U-tube Exchangers
Front end
stationary head types
Rear end
head types
Shell types
E
A
L
Fixed tubesheet
like “A” stationary head
One-pass shell
Channel
and removable cover
F
M
Fixed tubesheet
like “B” stationary head
Two-pass shell
with longitudinal baffle
B
N
G
Fixed tubesheet
like “N” stationary head
Split flow
Bonnet (Integral cover)
P
H
C
Outside packed Floating head
Removable
tube
bundle
only
Channel integral with tubesheet and removable cover
Double-split flow
S
J
Floating head
with backing device
Divided flow
T
N
Pull through floating head
K
Channel integral with tubesheet and removable cover
Kettle-type reboiler
U
U-Tube bundle
D
X
W
Special High-pressure closure
Cross-flow
Externally sealed
floating tubesheet
Figure 14.1 Various TEMA components. Source: Courtesy of the Tubular Exchanger Manufacturers Association, Inc.
277
278
14 Heat-Transfer Equipment
36
3
4
31
5
34
6
34
12
29
8
7
18
28
27
36
32
40
36
9
15
16
1
5
3
34
35
10
35
12
11
34
13
33
17
AES
32
2
6
3
32
7
8
37
27
28
14
12
34
2
5
5
34
33
37
12
6
3
BEM
4
36
3
5
1
5
34
12
31
34
3
6
34
10 33
8
28
27
29
7
32
35
35
15
34
23
24
12
25
19
22
36
20
21
AEP
36
4
1
3
5
34
34
5
31
6
10
34
12
12 34
8
35
30
CFU
(a)
28
27
35
7
32
9
33
Figure 14.2 Typical heat-exchanger configurations. Source: Courtesy of the Tubular Exchanger Manufacturers Association, Inc.
14.2 TEMA Design of Tubesheets in U-tube Exchangers
8
34
39
12
9
36 4
3 34 5
31
15
17
36
38
16
1
34 3
5
6 10
34 12
35
27
35
7
28
12
34
39
AKT
36
34
4
3
5
3
1
10
6
8
28
7
34 12
12 34
35
35
23 24 26 24 23 15 1
27
34
12
34
5
3
36
4
AJW
(b)
Figure 14.2 (Continued)
Table 14.1 Some TEMA requirements of classes R, C, and B exchangers.
Paragraph
Class
R
C
B
Service
RCB-1.1
Severe
Moderate
General
Corrosion allowance
(carbon steel)
RCB-1.5
1/8 in.
1/16 in.
1/16 in.
Minimum thickness of
longitudinal baffle
RCB-4.4
1/4 in.
1/4 in. carbon steel
1/4 in. carbon steel
Minimum tie-rod
diameter
RCB-4.7
3/8 in.
1/4 in.
1/4 in.
Preferred gasket contact
surface tolerance
RCB-6.3
±1/32 in.
None
None
Minimum bolt size
RCB-11
3/4 in.
1/2 in.
5/8 in.
279
280
14 Heat-Transfer Equipment
Symmetric
1 (C) = 4 × 1” + 0.5” + 3 × 1” × 2
4
3”ΦHOLE
4
C = 34.97”
R = 5.59”
1”
1”
C
1”
2
R
a = 6”
Figure 14.4 Tubesheet with tube holes.
p
do
From Eq. (14.3) with
A = πR2 = 98.17 in.2
Figure 14.3 Tubesheet showing tube layout.
and
Hence, the shearing stress 𝜎 is expressed as
W
PA
𝜎=
=
.
As
CT(1 − do ∕p)
(14.2)
The allowable shearing stress in the ASME Code,
VIII-1, is given by
Thus, T min = 0.58 in.
𝜎 = 0.8S.
Thus, Eq. (14.2) becomes
0.31DL ( P )
,
T=
1 − do ∕p S
(14.3)
where
DL = 4A/C
A = area of tubesheet within outer tube perimeter
C = perimeter of outer tubes, as defined in Figure 14.3
do = outside diameter of tube
p = distance between tubes
Example 14.1
A tubesheet for a U-tube exchanger has a 12 in. diameter
and is subjected to a design pressure of 100 psi. If the tube
layout is as shown in Figure 14.4 and S = 17,000 psi, what
is the required thickness? Assume the edge to be simply
supported.
Solution:
From Eq. (14.1),
(1.25)(12)
T=
2
= 0.58 in.
C = 34.97 in.
(
)
0.31(4 × 98.17∕34.97)
100
T=
1 − 0.75∕1
17,000
= 0.08 in.
14.3 Theoretical Analysis of Tubesheets
in U-tube Exchangers
Gardner in 1960 published a paper [2] that explained the
interaction between the tubes and tubesheet in U-tube
heat exchangers. Gardner assumed the interaction to be
represented by Figure 14.5. Hence, the bending in the
tubes is
E I
(14.4)
MT = −Fn T T 𝜃,
l
where
MT = bending moment of tube
ET = modulus of elasticity of tubes
I T = moment of inertia of tube
l = baffle spacing as shown in Figure 14.6
𝜃 = rotation of tube at tubesheet junction
F n = factor relating effect of baffles on tube-end bending
moment given by Figure 14.6
√
100
17,000
Similarly, the radial bending moment in the tubesheet
is given by
NET IT Fn Δr
Mr =
𝜃,
(14.5)
πa2 l
14.3 Theoretical Analysis of Tubesheets in U-tube Exchangers
a
becomes
[
(
)]
Q
dw
dw
d 1 d
r
= ∗ + 𝜁2
.
(14.7)
dr r dr
dr
D
dr
This equation can be solved in terms of Bessel functions. For a uniformly loaded plate, the solution can be
expressed as
}
{ 2
Ua − U 2
Pa4 1
w= ∗
(U
)
−
I
(U)]
−
A[I
0
a
0
D 2Ua4
2
(14.8)
(
)
3
−1
dw Pa
[U − AI 1 (U)],
(14.9)
= ∗
dr
D
2Ua3
where
r
1 Δr
2
e
MT
MT
w = deflection
P = applied pressure
Figure 14.5 Tubesheet-to-tube junction.
Ua = 𝜁 a
U = 𝜁r
A = constant of integration
I 0 , I 1 = modified Bessel function of zero and first order,
respectively
D∗ =
ℓ1
ℓ1/ℓ
ℓ1
ℓ
E* = effective modulus of elasticity of perforated
tubesheet
Fn
N
E∗ T 3
12(1 − 𝜇∗2 )
T = thickness of tubesheet
2
3 or more
0
4.00
4.00
0.2
3.83
3.83
a = radius of tubesheet
0.4
3.69
3.70
r = radius of a given point on tubesheet
0.6
3.59
3.60
0.8
3.52
3.53
3.43
3.46
1
1.0
3.00
𝜇* = effective Poisson’s ratio of perforated tubesheet
d = tube diameter
p = tube pitch.
For fixed tubesheets, the rotation at the edge is zero and
Eq. (14.9) can be solved for A:
Ua
for fixed edge.
(14.10)
A=
I1 (Ua )
For simply supported tubesheets, the moment at the
edge is not zero because the outer tubes have a bending moment that is transferred to the tubesheet. For this
boundary condition, the value of A is given by
Ua
A=
I1 (Ua )
Figure 14.6 Baffle layout. Source: From Gardner 1960 [2].
where
Mr = radial bending moment in tubesheet
N = number of tube holes
Δr = radius increment as shown in Figure 14.5
a = radius of tubesheet
Define
𝜁=
(
NET IT Fn
πa2 lD∗
)1∕2
,
(14.6)
where D* is the modified flexural rigidity of the tubesheet
and N is the total number of tubes. The differential
equation of the bending of a plate as given by Eq. (7.7)
⎤
⎡
⎢ (π∕N)1∕2 U 2 + (1 + 𝜇∗ ) ⎥
a
⎥
×⎢
[
]
1∕2 2
⎥
⎢
U
(π∕N)
+
a
]
⎢[
∗ ⎥
⎣ Ua I0 (Ua )∕I1 (Ua ) − (1 − 𝜇 ) ⎦
for simply supported edge.
(14.11)
281
282
14 Heat-Transfer Equipment
With the value of A established for the two boundary
conditions, the values of Mr , Mt , and Q can be obtained
from Eqs. (7.3a), (7.4), and (7.10) as follows:
⎫
⎧
(1 + 𝜇∗ ) − A
]⎪
[
2 −1 ⎪
∗
1−𝜇
Mr = Pa
⎨
I (U) ⎬
2Ua2 ⎪ I0 (U) −
⎪
U 1
⎭
⎩
0.08
0.07
Simply supported
0.06
Fe
0.04
⎫
⎧
(1 + 𝜇∗ ) − A
]⎪
[
2 −1 ⎪
∗
1−𝜇
Mt = Pa
⎨ ∗
I (U) ⎬
2Ua2 ⎪ 𝜇 I0 (U) +
⎪
U 1
⎭
⎩
A
Q = Pa
I (U).
(14.12)
2Ua 1
The maximum value of Mr can be obtained from
Eq. (14.12) for various boundary conditions and U
values. Hence,
Mmax = Pa2 Fm ,
(14.13)
where F m is a coefficient obtained from Figure 14.7.
The maximum bending stress is given by
6F P ( a )2
,
(14.14)
𝜎= m
𝜂
T
where
𝜂 = ligament efficiency of perforated tubesheet in
bending
p−d
.
=
p
0.22
0.20
0.18
0.16
0.2
0.3
0.03
Cla
mp
0.02
ed
0.01
0
1
2
3
4
5
Ua
6
7
8
9
10
Figure 14.8 Moment factor F q . Source: From Gardner 1960 [2].
Also, the maximum value of MT in the tubes can be
expressed as
)
(
t 2 P πda3
(14.15)
Ua2 F𝜃 ,
max MT =
6
2NI T
where F 𝜃 is a coefficient obtained from Figure 14.8.
The interaction between the tubesheet thickness and
the pressure is obtained by combining Eq. (14.6) with
U a = 𝜁 a, giving
T
𝜆
= 2∕3
a
Ua
or
T
−2∕3
=
Ua
,
(14.16)
𝜆a
where
[
]1∕3
12(1 − 𝜇∗2 )Fn ET NI T
𝜆=
.
π𝜂
E∗ la3
Also, Eq. (14.14) can be expressed as
P
𝜆2
=
4∕3
𝜂𝜎
6Fm Ua
0.14
Fm
π/N = 0
0.05
0.12
or
𝜆2 𝜂𝜎
4∕3
(14.17)
= 6Fm Ua .
P
Eqs. (14.16) and (14.17) are combined in a plot shown
in Figure 14.9.
0.10
0.08
Cla
mp
0.06
ed
0.04
Simp
le– π
/N = 0
0.02
0
1
2
3
4
5
Ua
6
7
8
9
0.3
0.2
10
Figure 14.7 Moment factor F m . Source: From Gardner 1960 [2].
Example 14.2
Find the thickness of the tubesheet in Example 14.1 if
E∗ = 9.0 × 106 psi
𝜇∗ = 0.3
N = 88
IT = 0.0166
6
ET = 30 × 10 psi
l = 12 in.
Fn = 3.46
14.4 ASME Equations for Tubesheets in U-tube Exchangers
Solution:
Let T = 0.40 in. Then,
diameter, an adjustment is made to Gardner’s Eq. (14.13).
The tubesheet is assumed to consist of a perforated plate
of radius a and an outside ring of outer radius b. Accordingly, Figure 14.9 must be modified to reflect the ratio
of b/a.
The ASME, VIII-1, design procedure is outlined for
U-tube heat exchangers as follows.
(9.0 × 106 )(0.40)3
= 52,750
12(1 − 0.32 )
(
)0.5
88 × 30 × 106 × 0.0166 × 3.46
𝜁=
π × 62 × 12 × 52,750
= 1.46
D∗ =
14.4.1
Ua = 𝜁 a = 8.73
√
√
π∕N = π∕88 ≈ 0.2.
Nomenclature
The nomenclature is based on Figure 14.10 for tube-totubesheet geometry, Figure 14.11 for tube layout, and
Figure 14.12 for tubesheet edge attachment.
From Figure 14.7,
Fm = 0.03
1.0 − 0.75
𝜂=
= 0.25,
1.0
and from Eq. (14.14),
(6)(0.03)(100) ( 6 )2
𝜎=
0.25
0.40
𝜎 = 16,200 psi (acceptable).
A = outside diameter of tubesheet
AL = total area of untubed lanes
= U L1 LL1 + U L2 LL2 + · · · (limited to 4Do p)
C = diameter of bolt circle
ct = corrosion allowance on the tube side of the
tubesheet
Dc = inside diameter of channel
Do = equivalent diameter of outer tube limit circle,
Figure 14.10
14.4 ASME Equations for Tubesheets
in U-tube Exchangers
Ds = inside diameter of shell
d = diameter of tube hole
d* = effective tube hole diameter
The ASME code, VIII-1, uses the method discussed in
Section 14.3 for designing tubesheets. The effective modulus of elasticity and Poisson’s ratio for a perforated plate
are obtained from O’Donnell’s work [3] and are summarized in Figures 14.13 and 14.14. Because the diameter
of the outer tube row is normally less than the tubesheet
Figure 14.9 Plot of Eqs. (14.16) and (14.17).
Source: From Gardner 1960 [2].
E = modulus of elasticity of tubesheet material
Ec = modulus of elasticity of channel material
Es = modulus of elasticity of shell material
Et = modulus of elasticity of tube material
3.0
r)
nte
ce
(
ed
d
mp
rte
Cla
po
2.0
ly
mp
1.0
p
su
Si
0.8
1
λ
T
a
0.6
0.4
π/N
=
d
rte
0.2
o
pp
ply
su
0.3
e)
(
g
ed
1
0.
3
0.0
1
0.0
m
0.1
Si
0.2
0.3
0.6 0.8 1.0
1 p
λ2 ησ
2.0
3.0 4.0
6.0 8.010.0
283
284
14 Heat-Transfer Equipment
Do
ℓtx
h
ro
p
tt
dt
(a)
(b)
d
h
hg
h
tt
dt
(c)
(d)
Figure 14.10 Tube-to-tubesheet geometry. (a) Tubesheet layout; (b) expanded tube joint; (c) tube side pass partition groove depth; (d)
tubes welded to back side of tube sheet (dt − 2tt ≤ d < dt ).
E* = effective modulus of elasticity of
tubesheet in the perforated region
G1 = midpoint of contact between flange and
tubesheet
Gc = diameter of channel gasket load reaction
Gs = diameter of shell gasket load reaction
h = tubesheet thickness
hg = tube-side pass partition groove depth,
Figure 14.10c
hg ′ = effective tube-side pass partition groove
depth
LL1 , LL2 , … = length(s) of untubed lane(s),
Figure 14.11
𝓁 tx = expanded length of tube in tubesheet
(0 < 𝓁 tx < h), Figure 14.10b
𝜇
=
basic ligament efficiency for shear
𝜇
*
=
effective ligament efficiency for bending
𝜈c
=
Poisson’s ratio of channel material
𝜈s
=
Poisson’s ratio of shell material
𝜈
*
=
effective Poisson’s ratio in perforated region of
tubesheet
Ps
=
shell-side internal design pressure
Pt
=
tube-side internal design pressure
p
=
tube pitch
*
p
=
effective tube pitch
ro
=
radius of outermost tube hole center,
Figure 14.10a
𝜌
=
tube expansion depth ratio = 𝓁 tx /h (0 < 𝜌 < 1)
S
=
allowable stress for tubesheet material
14.4 ASME Equations for Tubesheets in U-tube Exchangers
UL2
Do
Do
LL1
UL1
UL1
AL = UL1 LL1
LL1
LL2 = LL1 – UL1
AL = UL1LL1 + UL2LL2
(a)
(b)
Do
LL2
UL2
UL1
LL1
UL3
LL3
AL = UL1LL1 + UL2LL2 + UL3LL3
(c)
Figure 14.11 Tube layout. (a) one lane; (b) Two lane; (c) Three lanes.
Sc = allowable stress for channel material
Ss = allowable stress for shell material
St = allowable stress for tube material
Sy,c = yield stress for channel material
Sy,s = yield stress for shell material
SPS,c = allowable primary plus secondary stress
for channel material
SPS,s = allowable primary plus secondary stress
for shell material
t c = channel thickness
t s = shell thickness
t t = tube wall thickness
U L1 , U L2 , … = center-to-center distance(s) between
adjacent tube rows of untubed lane(s)
but not to exceed 4p, Figure 14.11.
W c = channel flange design bolt load for the gasket
seating condition
W s = shell flange design bolt load for the gasket
seating condition
W max = maximum flange design bolt
load = max[(W c ), (W s )]
14.4.2
Preliminary Calculations
The dimensional quantities, based on tube-to-tubesheet
attachment method, are calculated as follows:
For tube attachment in accordance with Figure 14.10b,
Do = 2ro + dt
𝜇 = (p − dt )∕p
∗
(14.18a)
(14.18b)
d = max{[dt − 2tt (Et ∕E)(St ∕S)𝜌], [dt − 2tt ]}
(14.18c)
285
286
14 Heat-Transfer Equipment
tc
ts
ts
A
Ps
Pt
Dc
h
A
C
Ds
Gc
Pt
Ps
Ds
h
(a)
Parameters
ρs = Ds/Do, ρc = Dc/Do
βs = [12(1 – νs2)]1/4 /[(Ds + ts)ts]1/2, ks = (βs Es ts3)/[6(1 – νs2)]
λs = [6 Dsks/h3][1 + hβs + (h2 βs2/2)]
δs = [Ds2/4Es ts][1 – νs/2], ωs = ρs ks βs δs (1 + hβs)
βc = [12(1 – νc2)]1/4/[(Dc + tc)tc]1/2, kc = (βc Ec tc3)/[6(1 – νc2)]
λc = [6 Dc kc/h3][1 + hβc + (h2 βc2/2)]
δc = [Dc2/4Ec tc][1 – νc/2], ωc = ρc kc βc δc (1 + hβc)
K = A/Do
F = [(1 – ν∗)/E∗][λs + λc + E(ln K)]
M∗ = MTS – ωs Ps + ωc Pt
(b)
Parameters
ρs = Ds/Do, ρc = Gc/Do
βs = [12(1 – νs2)]1/4 /[(Ds + ts)ts]1/2, ks = (βs Es ts3)/[6(1 – νs2)]
λs = [6 Dsks/h3][1 + hβs + (h2 βs2/2)]
δs = [Ds2/4Es ts][1 – νs/2], ωs = ρs ks βs δs (1 + hβs)
K = A/Do
F = [(1 – ν∗)/E∗][λs + E(ln K)]
M∗ = MTS – ωs Ps – [Wc(C – Gc)/2πDo]
ts
A
G1
C
Gc
Pt
Ps
Ds
h
(c)
Parameters
ρs = Ds/Do, ρc = Gc/Do
βs = [12(1 – vs2)]1/4 /[(Ds + ts)ts]1/2, ks = (βs Es ts3)/[6(1 – vs2)]
λs = [6 Dsks/h3][1 + h βs + (h2 βs2/2)]
δs = [Ds2/4Es ts][1 – vs/2], ωs = ρs ks βs δs (1 + h βs)
K = A/Do
F = [(1 – ν∗)/E∗][λs + E(ln K)]
M∗ = MTS – ωs Ps – [Wc(G1 – Gc)/2πDo]
Figure 14.12 (a) Tubesheet integral with shell and channel; (b) tubesheet integral with shell and gasketed with channel, extended as a
flange; (c) tubesheet integral with shell and gasketed with channel, not extended as a flange; (d) tubesheet gasketed with shell and
channel; (e) tubesheet gasketed with shell and integral with channel, extended as a flange; (f ) tubesheet gasketed with shell and integral
with channel, not extended as a flange.
14.4 ASME Equations for Tubesheets in U-tube Exchangers
A (extended)
C
Gc
Pt
tc
A
(not
extended)
Ps
A
C
Gs
h
Ps
Pt
Dc
Ds
h
Gs
(d)
Parameters
ρs = Gs/Do, ρc = Gc/Do
K = A/Do
F = [(1 – v∗)/E∗][E(ln K)]
M∗ = MTS + [Wmax (Gc – Gs)/2πDo]
(e)
Parameters
ρs = Gs/Do, ρc = Dc/Do
βc = [12(1 – vc2)]1/4/[(Dc + tc)tc]1/2, kc = (βc Ec tc3)/[6(1 – vc2)]
λc = [6 Dc kc/h3][1 + h βc + (h2 βc2/2)]
δc = [Dc2/4 Ec tc][1 – vc/2], ωc = ρc kc βc δc (1 + h βc)
K = A/Do
F = [(1 – v∗)/E∗][λc + E(ln K)]
M∗ = MTS + ωc Pt + [Ws(C – Gs)/2πDo]
tc
A
G1
Dc
Pt
Ps
h
C
Gs
(f)
Parameters
ρs = Gs/Do, ρc = Dc/Do
βc = [12(1 – vc2)]1/4/[(Dc + tc)tc]1/2, kc = (βc Ec tc3)/[6(1 – vc2)]
λc = [6 Dc kc/h3][1 + h βc + (h2 βc2/2)]
δc = [Dc2/4 Ec tc][1 – vc/2], ωc = ρc kc βc δc (1 + h βc)
K = A/Do
F = [(1 – v∗)/E∗][λc + E(ln K)]
M∗ = MTS + ωc Pt + [Ws(G1 – Gs)/2πDo]
Figure 14.12 (Continued)
287
14 Heat-Transfer Equipment
0.7
0.8
0.7
h/p
≥2.00
1.00
0.6
0.50
0.6
0.5
0.5
0.4
0.25
0.3
0.15
v*
E*/E
288
0.4
0.3
0.2
h/p
0.2
≤0.10
0.25
0.50
≥2.00
0.1
0
0
0.1
0.1
0.2
0.3
μ*
0.4
0.5
0.6
0
≤0.10
0
0.1
0.2
0.3
μ*
0.4
0.5
0.6
(b)
(a)
Figure 14.13 Values of E * /E (a) and 𝜈 * (b) for equilateral triangular pattern.
Ck = 4 min[(AL ), (4Do p)]∕πDo
∗
p = p∕(1 − Ck )
2
0.5
(14.18d)
(14.18e)
𝜇 = (p − d )∕p
(14.18f)
Hg′
(14.18g)
∗
∗
∗
∗
(14.19c)
Step 1: Calculate the dimensional parameters in accordance with Eqs. (14.18) or (14.19) depending on
tube attachment.
Step 2: Estimate a tubesheet thickness. The thickness
should not be less than that required for shear
stress around the periphery of the tubesheet
given by the equation
0.5
(14.19d)
𝜇 = (p − d)∕p
h = [1∕(4𝜇)][Do ∕(0.8S)]|Ps − Pt |. (14.20)
∗
(14.19e)
Hg′
(14.19f)
Step 3: From Figures 14.13 and 14.14, calculate E* and
v* .
Step 4: Calculate the “parameter” values for any particular tubesheet edge attachment shown in
Figure 14.12.
Step 5: Calculate the quantity
= max[(hg − ct ), (0)].
For tube attachment in accordance with Figure 14.10d,
Do = 2ro + d
(14.19a)
𝜇(p − d)∕p
(14.19b)
Ck = 4 min[(AL ), (4Do p)]∕πDo
∗
p = p∕(1 − Ck )
∗
14.4.3
then a new tubesheet thickness and/or shell and channel
thicknesses are assumed, and the stresses are calculated.
The process continues until the calculated stresses are
within the allowable stress values.
∗
= max[(hg − ct ), (0)].
2
Design Equations
The design methodology consists of calculating first the
thicknesses of the shell, channel, and flange in accordance
with the procedures established in VIII-1 and described
earlier in this book. The thickness of the tubesheet is then
assumed. The stresses in the tubesheet, shell, and channel are then calculated in accordance with the following procedure. If any of the components is overstressed,
MTS = (Do 2 ∕16)
]
[
(𝜌s − 1)(𝜌s 2 + 1)Ps − (𝜌c − 1)
.
(𝜌c 2 + 1)Pt
(14.21)
14.4 ASME Equations for Tubesheets in U-tube Exchangers
0.4
0.8
h/p
0.7
≥2.00
1.00
0.3
0.6
0.50
0.25
v*
E*/E
0.5
0.4
0.3
0.2
0.15
h/p
≤0.10
0.25
0.50
≥2.00
0.2
0.1
0.1
0
0
0.1
0.2
0.3
μ*
0.4
0.5
0.6
(a)
0
≤0.10
0
0.1
0.2
0.3
μ*
0.4
0.5
0.6
(b)
Figure 14.14 Values of E * /E (a) and 𝜈 * (b) for square pattern.
Step 6: Calculate the maximum bending moment in the
tubesheet for each loading condition
Mp = [M∗ − (Do 2 ∕32)(F)(Ps − Pt )]∕
(14.22)
(1 + F)
2
∗
Mo = Mp + (Do ∕64)(3 + v )(Ps − Pt )
(14.23)
M = max[|Mp |, |Mo |].
(14.24)
Step 7: Calculate the bending stress in the tubesheet for
each loading condition
𝜎 = 6M∕[(𝜇∗ )(h − h′g )2 ].
(14.25)
If s < 2S, then the assumed thickness h is adequate. Otherwise, the thickness needs to be
increased and the analysis repeated.
Step 8: The longitudinal stress in the shell is calculated
from
𝜎s,m = (Ds )(Ps )∕[4ts (Ds + ts )]
2
(14.26a)
𝜎s,b = (6ks ∕ts 2 )
⎧𝛽 𝛿 P + 6[(1 − v∗ )∕E∗ ] ⎫
⎪ s s s
⎪
× ⎨(Do ∕h3 )(1 + h𝛽s ∕2)×
⎬
2
⎪[Mp + (Do ∕32)(Ps − Pt )]⎪
⎩
⎭
(14.26b)
𝜎s = |𝜎s,m | + |𝜎s,b | ≤ 1.5Ss .
(14.26c)
Step 9: The longitudinal stress in the channel is calculated from
𝜎c,m = (Dc 2 )(Pt )∕[4tc (Dc + tc )]
(14.27a)
𝜎c,b = (6kc ∕tc )
2
⎧𝛽 𝛿 P + 6[(1 − v∗ )∕E∗ ] ⎫
⎪ c c t
⎪
× ⎨(Do ∕h3 )(1 + h𝛽c ∕2)×
⎬
⎪[Mp + (Do 2 ∕32)(Ps − Pt )]⎪
⎩
⎭
(14.27b)
𝜎c = |𝜎c,m | + |𝜎c,b | ≤ 1.5Sc .
(14.27c)
Example 14.3
A U-tube heat exchanger with an integral tubesheet has
details as shown in Figure 14.12a. Check the stresses in
289
290
14 Heat-Transfer Equipment
the shell, channel, tubes, and tubesheet based on the following data.
From Eq. (14.18a),
Do = 2ro + dt
= 2(23.0) + 1.0 = 47.0.
Design
Shell side
Tube side
Temperature
60 psi
400 ∘ F
675 psi
400 ∘ F
Joint efficiency
0.70
1.0
S for shells and tubesheet material
18.0 ksi
18.0 ksi
Pressure
S for tube material
16.5 ksi
Additional data:
Ds = 59.25 in.
Dc = 52.0 in.
dt = 1.0 in.
t s = 0.50 in.
t c = 1.5 in.
t t = 0.035 in.
h = 5.25 in.
hg = 0.1875 in.
ro = 23.0 in.
p = 1.25 in.
U L1 = 6.0 in.
LL1 = 47.0 in.
E = Ec = Es = 29,000 ksi
Et = 27,000 ksi
𝜌 = 0.9
𝜈 s = 𝜈 c = 0.3
Tubes are on an equilateral triangular pitch.
Solution:
Channel Shell
From Eq. (8.1),
S = P(R + 0.6tc )∕(joint Efficiency)tc
= 675[(52.0∕2) + 0.6(1.5)]∕(1.0)(1.5)
= 12,100 psi < 18,000 psi acceptable.
Shell-side Shell
From Eq. (8.1),
S = P(R + 0.6ts )∕(joint Efficiency)ts
= 60[(59.25∕2) + 0.6(1.5)]∕(0.7)(0.5)
= 5130 psi < 18,000 psi acceptable.
Tubes
From Example 8.2,
S = P(R − 0.4tt )∕(joint Efficiency)tt
From Eq. (14.18b),
𝜇 = (p − dt )∕p
= (1.25 − 1.0)∕1.25 = 0.2.
From Eq. (14.18c),
d∗ = max{[dt − 2tt (Et ∕E)(St ∕S)𝜌], [dt − 2tt ]}
⎧[1.0 − 2(.035)(27∕29) ] ⎫
⎪
,⎪
= max ⎨ (16,500∕20,000) (0.9) ⎬
⎪[1.0 − 2(0.035)]
⎪
⎩
⎭
= max[0.952, 0.930]
= 0.952
AL = UL1 LL1
= (6.0)(47.0) = 282.0.
From Eq. (14.18d),
Ck = 4 min[(AL ), (4Do p)]∕πDo 2
= 4 min[282.0, (4)(47)(1.25)]∕π(47)2
= 4 min[282.0, 235]∕π(47)2
= 0.1355.
From Eq. (14.18e),
p∗ = p∕(1 − Ck )0.5
= 1.25∕(1 − 0.1355)0.5
= 1.344.
From Eq. (14.18f),
𝜇∗ = (p∗ − d∗ )∕p∗
= (1.344 − 0.952)∕1.344
= 0.292.
From Eq. (14.18g),
Hg ′ = max[(hg − ct ), (0)]
= 675[(1.0∕2) − 0.4(0.035)]∕(1.0)(0.035)
= max[(0.1875 − 0), 0]
= 9370 psi < 16,500 psi acceptable.
= 0.1875 in.
Tubesheet
Since the tube-side pressure is much higher than the
shell-side pressure, the design of the tubesheet will be
based on a tube-side pressure of 675 psi and a shell-side
pressure of 0 psi.
The preliminary calculations given by Eq. (14.18) will
be performed first.
For tube attachment in accordance with Figure 14.10b,
Next the parameters shown in Figure 14.12a are calculated as follows:
𝜌s = 1.2606
𝜌c = 1.1064
𝛽 s = 0.3326
𝛿 s = 5.1448 × 10
k s = 220,809
−5
𝜔s = 13.0794
𝜆s = 2,316,603
14.5 Theoretical Analysis of Fixed Tubesheets
𝛽 c = 0.2029
𝛿 c = 1.3209×10
k c = 3,637,578
−5
K = 1.2739
𝜆c = 20,649,653
𝜔c = 22.2803
h/p = 4.2
*
E* = 8,700,000 psi
E /E = 0.3 from
Figure 14.13
𝜈 * = 0.38
MTS = −21,233.0 M* = −6750.9
F = 2.1370
The moments are obtained from Eqs. (14.22) through
(14.24).
TEMA design equations for determining fixed tubesheet
thickness is based partly on the theoretical work done
by Gardner [4, 5] and Miller [6]. From Eq. (7.9), the
differential equation for the bending of circular plate is
given by
{
[
(
)]}
q
d 1 d
dw
1 d
(14.28a)
r
r
= .
r dr
dr r dr
dr
D
The next section shows that in a fixed tubesheet, the
quantity q, which is the local pressure at radius r, is not a
constant. Rather, it is a function of w given by [7]
q = C2 + K1 (m − 2w),
where
q = local pressure
From Eq. (14.22),
Mp = 28,415 in-lbs/in.
From Eq. (14.23),
Mo = 47,416 in-lbs/in.
C 2 = constant
From Eq. (14.24),
M = 47,416 in-lbs/in.
K 1 = tube bundle stiffness =
Stress Calculations
Tubesheet
The bending stress in the tubesheet is obtained from
Eq. (14.25),
S = 6M∕[(𝜇∗ )(h − hg ′ )2 ]
= 6(47,416)∕[0.292)(5.25 − 0.1875)2 ]
= 38,050 psi < 2(20,000) = 40,000 psi
acceptable.
Shell
The membrane stress is obtained from Eq. (14.26a) =
0 psi.
The bending stress is obtained from Eq. (14.26b) =
22,680 psi.
Total stress = 0 + 22,680 = 22,680 psi < 1.5(18 000) =
27,000 psi acceptable.
Channel
The membrane stress is obtained from Eq. (14.27a) =
5475 psi.
The bending stress is obtained from Eq. (14.27b) =
17,075 psi.
Total stress = 5475 + 17,075 = 22,550 psi < 1.5(18,000)
= 27,000 psi acceptable.
14.5 Theoretical Analysis of Fixed
Tubesheets
The stress analysis of fixed tubesheets in heat exchangers
is very complex due to the large number of variables that
affect the analysis, such as differences in tube and shell
strain, the ratio of shell and tubesheet stiffnesses, effective applied pressure, and the relative thermal expansion
of shell and tubes. The development of the simplified
Nt(do − t)ET
R2 L
r = bolt circle
L = length of tubes
do = outside diameter of tube
t = thickness of tube wall
m = distance tubesheet edges move with respect to each
other
w = deflection of tubesheet
Eq. 14.28a can be written for perforated tubesheets
subjected to q pressure as
x4
3
2
d4 q
dq
3d q
2d q
+
2x
−
x
+x
+ x4 q = 0,
4
3
2
dx
dx
dx
dx
where
(
x=
2K1
D∗
)1∕4
(r) = 𝛽r
and
E∗ T 3
.
12(1 − 𝜇∗2 )
This equation can be solved in terms of Bessel functions. For symmetric loads, the solution can be taken as
D∗ =
(14.28b)
q = C2 [Z1 (x) + HZ2 (x)]
C
w = 4 2 ∗ {Z1 (xa ) − Z1 (x) + H[Z2 (xa ) − Z2 (x)]}
𝛽 D
(14.28c)
−C2 1
𝜃 = 4 ∗ [Z1 (x) + HZ12 (x)]
(14.28d)
𝛽 D
[
(
]
)
⎧ Z (x) + 1 − 𝜇 Z 1 (x) − ⎫
1
C ⎪ 2
⎪
(x
]⎬ ,
)
Mr = 22 ⎨ [
1−𝜇
𝛽 ⎪
H Z1 (x) −
Z21 (x) ⎪
⎩
⎭
x
(14.28e)
291
292
14 Heat-Transfer Equipment
where the Z functions are as defined in Chapter 7, and C
and H are constants to be determined from the boundary
conditions.
Defining P as the average pressure acting on the
tubesheet, its total value is
Similarly, for simply supported tubesheet Ma = 0 and
Eq. (14.33) becomes
{
}
Z2 (xa ) + [(1 − 𝜇)∕xa ]Z11 (xa )
H=−
. (14.36)
Z1 (xa ) − [(1 − 𝜇)∕xa ]Z21 (xa )
a
1
2rqπ dr
πa2 ∫0
]
[
Z11 (xa )
Z21 (xa )
= 2C2
.
1−H 1
xa
Z2 (xa )
P=
(14.29)
The value of C 2 can then be determined from this
equation as
⎫
⎧
⎪
⎪
xa
⎪
⎪
C2 = P ⎨
[
]⎬ .
1
⎪ 2Z1 (x ) 1 − H Z1 (xa ) ⎪
⎪ 2 a
Z21 (xa ) ⎪
⎭
⎩
(14.31)
where qa is the local pressure at radius a. Then from Eqs.
(14.28b) and (14.29),
Z1 (xa )
+H
xa Z2 (xa ) Z2 (xa )
,
Fa =
2 Z21 (xa )
Z11 (xa )
1−H 1
Z2 (xa )
Mmax = Pa2 Fm ,
(14.30)
Now define
qa = Fq P,
For a given value of xa , the constants H can be calculated from Eqs. (14.35) and (14.36). The constant
C 2 can then be determined from Eq. (14.30). Once
H and C 2 are known, the magnitude of the bending
moment at any location in the tubesheet is obtained
from Eq. (14.28e). The maximum value of this Mr at
any given xa is obtained from Eq. (14.28e) and normally
expressed as
(14.32)
and from Eqs. (14.28d) and (14.28e),
]
[
⎧ Z (x ) + 1 − 𝜇 Z1 (x ) − ⎫
1 a
⎪ 2 a
⎪
xa
]⎪
⎪ [
1−𝜇 1
⎪ H Z1 (xa ) −
Z2 (xa ) ⎪
xa
Ma
⎪
∗⎪
= −𝛽D ⎨
⎬
1
1
𝜃a
Z1 (xa ) + HZ2 (xa )
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
⎭
where F m is obtained from Figure 14.15.
The maximum bending stress is
( )
6P a 2
Fm .
𝜎r =
𝜂 T
0.22
0.20
0.18
0.16
0.14
]
[
𝛽D∗ 𝜃a
1−𝜇 1
Z1 (xa )
Z2 (xa ) +
Ma
xa
H=−
].
[
∗
𝛽D 𝜃a
1−𝜇 1
1
Z2 (xa ) −
Z2 (xa )
Z1 (xa ) −
Ma
xa
(14.34)
Z11 (xa ) +
The value of H is based on the edge condition of the
tubesheet. For fixed tubesheet, 𝜃 a = 0 and Eq. (14.34)
gives
H=−
Z11 (xa )
Z21 (xa )
for fixed tubesheets.
(14.38)
For large values of xa , the Z values can be approximated
by those given in Table 7.1, and the quantities F q and F m
can be expressed as [5]
√
1
Fq = (1 + 2xa ) for fixed tubesheets
4√
2
(14.39)
Fm =
4xa
(14.33)
or
(14.37)
(14.35)
Fm
0.12
0.10
0.08
Cla
Sim
0.06
ply
0.04
mp
ed
sup
po
rte
d
0.02
0
0
1
2
3
4
5
xa
6
Figure 14.15 Values of moment factor F m .
7
8
9
10
14.6 ASME Equations for Fixed Tubesheets
and
Dc = inside diameter of channel
√
1
Fq = (1 + 2 2xa )
4
for simply supported tubesheets
Fm =
−e−π∕4
.
2xa
Dj = inside diameter of the expansion joint at its
convolution height
(14.40)
Example 14.4
Determine the stress at the edge of a fixed tubesheet
of thickness T = 0.50 in. if the geometry is as shown
in Figure 14.4 and K 1 = 52,800 lb/in.3 , E* = 9 × 106 psi,
𝜇* = 0.3, and P = 100 psi.
Solution:
(9 × 106 )(0.5)3
= 103,020
12(1 − 0.32 )
(
(
)
)0.25
2K1 0.25
2 × 52,800
𝛽=
=
= 1.0062
D∗
103,020
xa = 𝛽(6) = 6.037.
D∗ =
From Figure 14.15, F m = 0.059 and 𝜂 = 0.25. From Eq.
(14.38),
(6)(100) ( 6 )2
(0.059)
𝜎=
0.25
0.5
𝜎 = 20,390 psi.
Ds = inside diameter of shell
dt
= outside diameter of tubes
E
= modulus of elasticity of tubesheet material
Ec = modulus of elasticity of channel material
Es = modulus of elasticity of shell material
Et = modulus of elasticity of tube material
G1 = midpoint of contact between flange and
tubesheet
Gc = diameter of channel gasket load reaction
Gs = diameter of shell gasket load reaction
𝛾
= axial differential thermal expansion between
tubes and shell
h
= tubesheet thickness
J
= ratio of expansion bellows to shell axial rigidity
(J = 1.0 if no bellows)
K j = axial rigidity of expansion bellows, total
force/elongation
k
14.6 ASME Equations for Fixed
Tubesheets
The design procedure for fixed tubesheets in ASME follows the same procedure as that for U-tube exchangers.
The equations for fixed tubesheets are more complicated
than those for U-tube exchangers due to the interaction
of the tubes in supporting the tubesheet and the effect of
any expansion joints on the stiffness of the shell.
14.6.1
A
Nomenclature
= outside diameter of tubesheet
𝛼 s,m = mean coefficient of thermal expansion of shell
material
𝛼 t,m = mean coefficient of thermal expansion of tube
material
= constant accounting for the method of support
for the unsupported tube span under
consideration
= 0.6 for unsupported spans between two
tubesheets
= 0.8 for unsupported spans between a tubesheet
and a tube support
= 1.0 for unsupported spans between two tube
supports
L
= tube length between inner tubesheet faces
𝓁
= unsupported tube span under consideration<>
N t = number of tubes
𝜈
= Poisson’s ratio of tubesheet material
𝜈c
= Poisson’s ratio of channel material
𝜈s
= Poisson’s ratio of shell material
vt
= Poisson’s ratio of tube material
Pe = effective pressure acting on tubesheet
Ps = shell-side internal design pressure
ac
= radial channel dimension
Pt
= tube-side internal design pressure
ao
= equivalent radius of outer tube limit circle
S
= allowable stress for tubesheet material
as
= radial shell dimension
Sc = allowable stress for channel material
C
= diameter of bolt circle
Ss
= allowable stress for shell material
293
294
14 Heat-Transfer Equipment
St
= allowable stress for tube material
Sy
= yield stress for tubesheet material
Sy,c
= yield stress for channel material
Sy,s
= yield stress for shell material
SPS
= allowable primary plus secondary stress for
tubesheet material
SPS,c = allowable primary plus secondary stress for
channel material
SPS,s = allowable primary plus secondary stress for
shell material
T
= tubesheet design temperatures
Ta
= ambient temperature
Tc
= channel design temperature
Ts
= shell design temperature
Tt
= tube design temperature
T s,m = mean shell metal temperature along the shell
T t,m = mean tube metal temperature along the tube
tc
= channel thickness
ts
= shell thickness
tt
= tube wall thickness
W
= channel flange design bolt load for the gasket
seating condition
14.6.2
Preliminary Calculations
1) Calculate the dimensional quantities from Eqs. (14.10)
and (14.11) as applicable.
2) Calculate the quantities
ao = Do ∕2
𝜌s = as ∕ao
𝜌c = ac ∕ao
𝜒s = 1 − Nt (dt ∕2ao )2
𝜒t = 1 − Nt [(dt − 2tt )∕2ao ]2 ,
where, as and at are obtained from Figure 14.16.
3) Calculate the shell axial stiffness K s , tube axial stiffness K t , and stiffness factors K s,t and J.
Ks = πts (Ds + ts )Es ∕L
Kt = πtt (dt + tt )Et ∕L
Ks,t = Ks ∕(Nt Kt )
J = 1∕[1 + (Ks ∕Kj )]
14.6.3
Design Equations
Step 1: Determine the loading conditions to be evaluated, as follows.
Loading Case 1. Pt is applied, Ps = 0, differential
thermal expansion = 0.
Loading Case 2. Pt = 0, Ps is applied, differential
thermal expansion = 0.
Loading Case 3. Pt is applied, Ps is applied, differential thermal expansion = 0.
Loading Case 4. Pt = 0, Ps = 0, differential thermal expansion is applied.
Loading Case 5. Pt is applied, Ps = 0, differential
thermal expansion is applied.
Loading Case 6. Pt = 0, Ps is applied, differential
thermal expansion is applied.
Loading Case 7. Pt is applied, Ps is applied, differential thermal expansion is applied.
Step 2: Estimate a tubesheet thickness.
Step 3: From Figures 14.13 and 14.14, calculate E*
and 𝜈*.
Step 4: Calculate the “parameter” values for any particular tubesheet edge attachment shown in
Figure 14.16.
Step 5: Calculate the quantity
]1∕4
[
24(1 − 𝜈 ∗2 )Nt Et tt (dt − tt )ao 2 ∕
.
Xa =
(E∗ Lh3 )
Step 6: Obtain the quantities Zd , Z v , Z w , and Z m from
Figure 14.17.
Step 7: Calculate the values
K = A∕Do
F = [(1 − 𝜈 ∗ )∕E∗ ][𝜆s + 𝜆c + E(ln K)]
Φ = (1 + 𝜈 ∗ )F
Q1 = (𝜌s − 1 − ΦZv )∕(1 + ΦZm )
QZ1 = (Zd + Q1 Zw )Xa 4 ∕2
QZ2 = (Zv + Q1 Zm )Xa 4 ∕2
U = [Zw + (𝜌s − 1)Zm ]Xa 4 ∕(1 + ΦZm ).
Step 8: Calculate 𝛾 for the following loading conditions:
For loading cases 1–3: 𝛾 = 0.
For loading cases 4–7: 𝛾 = [𝛼 t,m (T t,m − T a ) −
𝛼 s,m (T s,m − T a )]L.
Step 9: Calculate the quantities
𝜔s = 𝜌s ks 𝛽s 𝛿s (1 + h𝛽s )
𝜔s ∗ = [ao 2 (𝜌s 2 − 1)(𝜌s − 1)∕4] − 𝜔s
𝜔c = 𝜌c kc 𝛽c 𝛿c (1 + h𝛽c )
}
{
[(𝜌c 2 + 1)(𝜌c − 1)∕4]−
𝜔c ∗ = ao 2
(𝜌s − 1)∕2
− 𝜔c .
14.6 ASME Equations for Fixed Tubesheets
Step 10: Calculate the equivalent pressures Ps ′ , Pt ′ , P𝛾 ,
Pw , Prim , and effective pressure Pe.
Step 12: Find the value of F m from Figure 14.18 or
14.19.
Step 13: Calculate the bending stress in the tubesheet.
⎧𝜒s + 2(1 − 𝜒s )𝜈t +
⎫
⎪(2∕Ks,t )(Ds ∕Do )2 𝜈s −
⎪
Ps = ⎨ 2
P
(𝜌s − 1)∕(JK s,t ) − (1 − J) ⎬ s
⎪ 2
⎪
⎩[Dj − (2as )2 ]∕(2JK s,t Do 2 )⎭
𝜎 = (1.5Fm ∕𝜇∗ )[2ao ∕(h − hg ′ )]2 Pe
′
For loading cases 1 − 3 ∶ |𝜎| < 1.5S
For loading cases 4 − 7 ∶
Pt ′ = [𝜒t + 2(1 − 𝜒t )𝜈t + (1∕JK s,t )]Pt
|𝜎| < min[3S, 2Sy ].
P𝛾 = [(Nt Kt )∕(πao 2 )]𝛾
Step 14: Calculate the shearing stress in the tubesheet at
the outer edge of the perforated region.
Pw = −[(U𝛾b )∕(2πao 2 )]W
Prim = −(U∕ao )(𝜔s Ps − 𝜔c Pt )
2
∗
∗
𝜏 = (1∕(2𝜇))(ao ∕h)Pe
Pe = {JKs,t ∕{1+JK s,t [QZ1 (𝜌s −1)QZ2 ]}}
′
For all loading cases ∶ |𝜏| < 0.8S.
′
(Ps − Pt + P𝛾 + Pw + Prim )
Step 15: Calculate the stress in the shell.
Step 11: Calculate the factors
𝜎s,m = [ao 2 ∕(2as + ts )ts ]
Q2 = [(𝜔s Ps − 𝜔c Pt ) + (𝛾b W ∕2π)]∕
∗
∗
[Pe (𝜌s 2 − 1)(Ps − Pt )]+
(1 + FZm )
Q3 = Q1 + [(2Q2 )∕(Pe ao 2 )].
Figure 14.16 (a) Tubesheet integral
with shell and channel; (b) tubesheet
integral with shell and gasketed with
channel, extended as a flange;
(c) tubesheet integral with shell and
gasketed with channel, not extended
as a flange; (d) tubesheet gasketed
with shell and channel.
[as 2 ∕(2as + ts )ts ]Pt
Parameters
as = Ds/2
βs = [12(1 – vs2)]1/4/[(Ds + ts) ts]1/2
ts
tc
ks = (βs Es ts3)/[6(1 – vs2)]
λs = [6 Ds ks / h3][1 + h βs + (h2 βs2 / 2)]
A
Dc
Pt
Ps
δs = [Ds2/(4 Es ts)][(1 – vs / 2)]
Ds
Dj
ac = Dc/2
βc = [12(1 – vc2)]1/4/[(Dc + tc) tc]1/2
h
kc = (βc Ec tc3)/[6(1 – vc2)]
λc = [6 Dc kc / h3][1 + h βc + (h2 βc2 / 2)]
δc = [Dc2/(4 Ec tc)][(1 – vc / 2)]
γb = 0
(a)
Parameters
ts
as = Ds/2
βs = [12(1 – vs2)]1/4/[(Ds + ts) ts]1/2
A
C
Gc
Pt
Ps
h
Ds
ks = (βs Es ts3)/[6(1 – vs2)]
λs = [6 Ds ks / h3][1 + h βs + (h2 βs2 / 2)]
δs = [Ds2/(4 Es ts)][(1 – vs / 2)]
ac = Gc/2
βc = 0
kc = 0
λc = 0
δc = 0
(b)
295
296
14 Heat-Transfer Equipment
Figure 14.16 (Continued)
Parameters
ts
as = Ds/2
βs = [12(1 – vs2)]1/4/[(Ds + ts) ts]1/2
ks = (βs Es ts3)/[6(1 – vs2)]
C
Gc
Pt
Ps
Gl
λs = [6 Ds ks / h3][1 + h βs + (h2 βs2 / 2)]
A
δs = [Ds2/(4 Es ts)][(1 – vs / 2)]
Ds
ac = Gc/2
h
βc = 0
kc = 0
λc = 0
δc = 0
γb = (Gc – G1) / Do
(c)
Parameters
as = Gs/2
A
(extended)
A
(not extended)
C
Gc
Pt
Ps
βs = 0
ks = 0
λs = 0
δs = 0
Ds Gs
ac = Gc/2
h
βc = 0
kc = 0
λc = 0
δc = 0
γb = (Gc – Gs) / Do
(d)
𝜎s,b = (6ks ∕ts 2 ){𝛽s [𝛿s Ps − (𝜈s as 𝜎s,m )∕Es ]
+ [6(1 − 𝜈 ∗2 )∕E∗ ][ao 3 ∕h3 ]
(1 + h𝛽s ∕2) × [Pe (Zv + Zm Q1 )
+ 2Zm Q2 ∕ao 2 ]}
𝜎s = |𝜎s,m | + |𝜎s,b |
For loading cases 1 − 3 ∶ |𝜎s | < 1.5Ss
For loading cases 4 − 7 ∶
|𝜎s | < min[3S, 2Sy,s ].
Step 16: Calculate the stress in the channel.
𝜎c,m = [ac 2 ∕(2ac + tc )tc ]Pt
𝜎c,b = (6kc ∕tc 2 )
⎧𝛽c 𝛿c Pt − [6(1 − 𝜈 ∗2 )∕E∗ ]⎫
⎪+ [ao 3 ∕h3 ](1 + h 𝛽c ∕2) ⎪
⎨× [P (Z + Z Q )
⎬
e
v
m 1
⎪
⎪
2
⎩+ 2Zm Q2 ∕ao ]
⎭
𝜎c = |𝜎c,m | + |𝜎c,b |
For loading cases 1 − 3 ∶ |𝜎c | < 1.5 Sc
For loading cases 4 − 7 ∶
|𝜎c | < min[3Sc , 2Sy,c ]
Step 17: Calculate the stress at the outmost tube row.
Fq = (Zd + Q3 Zw )Xa 4 ∕2
𝜎t,o = [(Ps 𝜒s − Pt 𝜒t ) − Pe Fq ]∕(𝜒t − 𝜒s ).
When the tube stress is in tension,
For loading cases 1 − 3 ∶ |𝜎t,o | < 1.5St
For loading cases 4 − 7 ∶ |𝜎t,o | < 2St .
When the tube stress is in compression, determine the allowable compressive stress:
𝓁t = K𝓁
rt = {[dt 2 + (dt − 2tt )2 ]1∕2 }∕4
Ft = 𝓁t ∕rt
Ct = [(2π2 Et )∕Sy,t ]1∕2
14.6 ASME Equations for Fixed Tubesheets
Figure 14.17 Values of Z d , Z m , Z v , and Z w .
0.80
0.70
Zd, Zv, Zw or Zm
0.60
0.50
0.40
Zd
0.30
0.20
Zm
0.10
0
Zv = Zw
0
2
4
6
8
10
12
Xa
Figure 14.18 Values for bending
factor F m for negative values of Q.
0.4
Q3 = –0.8
Q3 = –0.7
Q3 = –0.6
0.3
Fm
Q3 = –0.5
Q3 = –0.4
0.2
Q3 = –0.3
Q3 = –0.2
0.1
Q3 = –0.1
0
1.0
Q3 = 0.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0
Xa
Fs = max[(3.25 − 0.5Fq ), 1.25] ≤ 2.0
When Ct ≤ Ft , then
Stb = min{[(π2 Et )∕(Fs Ft 2 )], St }
When Ct > Ft , then
Stb = min{[(Sy,t ∕Fs )
(1 − (Ft ∕2Ct ))], St }.
The aforementioned stress equations are for
the outer tube row in a tubesheet where the
stress in normally, but not necessarily, the
highest. The stress in tubes at other locations
within the tubesheet can be determined in
a similar way. The ASME VIII-1 Code gives
the equations for calculating stress in tubes at
various locations within the tubesheet.
297
14 Heat-Transfer Equipment
0.7
0.6
0.5
Q3 = 0.8
0.4
Q3 = 0.7
Fm
298
Q3 = 0.6
0.3
Q3 = 0.5
Q3 = 0.4
0.2
Q3 = 0.3
Q3 = 0.2
0.1
0
1.0
Q3 = 0.1
Q3 = 0.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0 10.0 11.0 12.0 13.0 14.0 15.0 16.0
Xa
Figure 14.19 Values for bending factor F m for positive values of Q.
Example 14.5
A fixed heat exchanger with an integral tubesheet has
details as shown in Figure 14.12a. Check the stresses in
the shell, channel, tubes, and tubesheet based on the
following data.
𝛼 tm = 9.0 × 10−6 in./
in./∘ F
T a = 70 ∘ F
A = 47.0 in.
d = 1.012 in. N t = 1835
L = 264.0 in.
𝓁 = 36.0 in.
𝛼 s = 6.26 ×
10−6 in./in./∘ F
k = 1.0
Tubes are on an equilateral triangular pitch.
Design
Shell side
Tube side
Pressure
Temperature
370 psi
300 ∘ F
110 psi
300 ∘ F
Joint efficiency
1.0
1.0
S for shells and tubesheet material
20.0 ksi
20.0 ksi
S for tube material
14.2 ksi
Solution:
Channel Shell
From Eq. (8.1),
S = P(R + 0.6tc )∕(joint Efficiency)tc
= 110[(46.5∕2) + 0.6(0.25)]∕(1.0)(0.25)
= 10,300 psi < 20,000 psi acceptable.
Shell-side Shell
From Eq. (8.1),
Additional data:
S = P(R + 0.6ts )∕(joint Efficiency)ts
Ds = 46.0 in.
Dc = 46.5 in.
dt = 1.5 in.
t s = 0.50 in.
t c = 0.25 in.
t t = 0.083 in.
h = 0.375 in.
hg ′ = 0 in.
J = 1.0
ro = 21.5 in.
p = 1.875 in.
𝜌 = 0.9
U L1 = 0 in.
LL1 = 10.487 in.
E = Ec = Es
= 29,000 ksi
Et = 27,000 ksi 𝜈 s = 𝜈 c = 𝜈 t = 0.3
= 370[(46.0∕2) + 0.6(0.5)]∕(1.0)(0.5)
= 17,240 psi < 20,000 psi acceptable.
Tubes
From Example 8.2,
S = P(R − 0.4tt )∕(joint Efficiency)tt
= 110[(1.5∕2) − 0.4(0.085)]∕(1.0)(0.085)
= 926 psi < 14,200 psi acceptable.
14.6 ASME Equations for Fixed Tubesheets
Tubesheet
There are seven loading conditions to be evaluated as
detailed in Step 1 of Subsection 14.16.3:
Case 1: Pt = 110 psi
Case 2: Ps = 370 psi
Case 3: Pt = 110 psi and Ps = 370 psi
Case 4: Tube temperature (300 − 70 = 230 ∘ F)
and shell temperature 230 ∘ F.
Case 5: Pt = 110 psi, tube temperature = 230 ∘ F
and shell temperature = 230 ∘ F
Case 6: Ps = 370 psi, tube temperature = 230 ∘ F
and shell temperature = 230 ∘ F
Case 7: Pt = 110 psi, Ps = 370 psi, tube temperature = 230 ∘ F and shell temperature = 230 ∘ F
Calculations for Case 1 are shown as follows. The
results of Cases 2–7 are tabulated at the end of this
example.
The following parameters are given in Sections 14.4.2
and 14.16.2, and Figure 14.16a.
Shell
𝜌s = 1.0337
as = 23.0
𝛽s = 0.3770
ks = 250,299 𝜆s = 1,508,306,309
From Step 8,
γ = 0.
From Step 9,
𝜔s = 3.4525 𝜔s∗ = −3.1665
𝜔c = 1.8638 𝜔c∗ = −9.6931.
From Step 10, calculate the equivalent pressures:
Ps′ = 0
Pt′ = 988.09
Pw = 0
Prim = −692.35
P𝛾 = 0
Pe = −1.1390.
From Step 11, calculate
Q2 = 6.8505 Q3 = −0.1066.
From Step 12, calculate F m
Fm = 0.05328.
From Step 13, calculate the bending stress in the
tubesheet:
𝜎 = 5070 psi < 1.5(20,000) = 30,000 psi.
From Step 14, calculate the shearing stress at the outer
perforated region of the tubesheet:
𝜏 = 170 psi < 0.8(20,000) = 16,000 psi.
𝛿s = 3.1010 × 10−5
From Step 15, calculate the stress in the shell:
Channel
𝜌c = 1.0449
Ac = 23.25
𝛽c = 0.5317
kc = 44,128.6
𝜆c = 284,665,169
𝛿c = 6.3376 × 10−5
𝛾b = 0.0
< 20,000 psi acceptable
𝜎m = 2320 psi
𝜎b = 18,640 psi
Total stress = 20,960 psi < 3(20,000) =
60,000 psi acceptable.
Tubes
𝜇 = 0.2
∗
AL = 0
∗
p = 1.875
d = max[1.4012, 1.334]
use 1.4012
Ck = min[0, 0.2146]
use 0.0
𝜇∗ = 0.2527
𝜒s = −1.0850
𝜒t = −0.6490
The following preliminary calculations are obtained
from the aforementioned parameters:
K s = 8,023,556
K t = 37,788.3
K = 1.0562
h/P = 0.2
From Figure 14.13,
E* /E = 0.34
𝜈 * = 0.15
E* = 9 860 000
F = 154.7032
X a = 26.4981
From Figure 14.17,
Zd = 0.005995
Zv = 0.07218
Zw = 0.07218
Zm = 0.8695
From Step 7 of Section 14.16.3,
F = 177.9086 Q1 = −0.08226
Qz1 = 14.3374 Qz2 = 160.6790
U = 321.3580.
K s,t = 0.1157
From Step 16, calculate the stress in the channel:
𝜎m = 5090 psi
< 20,000 psi acceptable
𝜎b = 29,200 psi
Total stress = 34,290 psi < 3(20,000) =
60,000 psi acceptable.
From Step 17, calculate the stress in the outermost
tube:
S = −930 psi < −3020 acceptable.
The aforementioned calculations are made for loading case 1. Similar calculations must also be made for
loading cases 2 through 7. The results are summarized in
Table 14.2
The table indicates that the maximum stress in the shell
occurs in loading case 7 for the geometry, pressures, and
temperatures given. The maximum stress in the channel occurs in loading case 5, while the maximum stress
in the tubesheet occurs in loading case 6. The maximum
tensile stress in the outer tubes occurs in loading case 2,
299
300
14 Heat-Transfer Equipment
q/2
Table 14.2 Maximum stress, psi.
Shell
Channel
Tubesheet
Outer tubes
Loading case 1
20,960
34,290
5 070
Loading case 2
6 430
14,360
7 460
260
Loading case 3
25,660
19,110
2 630
−570
−930
Loading case 4
2 730
1 870
17,330
−170
Loading case 5
21,960
35,330
12,500
−990
Loading case 6
9 150
12 490
24,790
90
Loading case 7
28,390
20,980
19,960
−740
while the maximum compressive stress occurs in loading
case 5.
w
q/4
d
Figure 14.21 Dimensions of a flanged and flued expansion joint.
q
14.7 Expansion Joints
The two most common types of expansion joints are the
bellows [8] and the flanged and flued [9], as shown in
Figure 14.20. Bellows’ expansion joints are used when
the expansion is large and the pressure is low. The membrane stress in bellows’ expansion joints is obtained
from Figure 14.21. The total force due to pressure is
given by
[( ) ( )
( q ) ( )]
q
1
d
P.
+ (w)
F=
2
2
2
2
The total force resisted in the metal is
]
[( q ) ( )
π
(2) + [w − 2(4q)] .
F = (S)(t)
4
2
Equating the two force terms results in an expression
for the membrane hoop stress in a bellow expansion joint:
(d + w)P
S=
,
t[(π − 2) + 4w∕q]
w
M
V
V
H
M
H
Figure 14.22 Forces in a flanged and flued expansion joints.
where
d = diameter of cylindrical shell (in.)
P = applied pressure (psi)
q = convolution pitch (in.)
S = allowable stress (psi)
w = convolution depth (in.)
Similarly, the longitudinal membrane stress is given by
S = Pw∕2t.
Flanged and flued expansion joint
The bending stress in bellows’ expansion joints is
obtained from treating a single convolution as a beam
of unit width as shown in Figure 14.22. The longitudinal
bending stress due to pressure is given by
Pw2
,
2t 2
where K 1 = factor defined in Figure 14.23.
The longitudinal membrane stress due to deflection is
given by
S = K1
Bellows expansion joint
Figure 14.20 Various expansion joints.
S=
Et 2 𝛿
,
2w3 K2
14.7 Expansion Joints
Figure 14.23 Values of K 1 . Source:
Courtesy of ASME.
1.0
0.9
0.8
0.7
0.2
0.4
0.6
0.8
1.0
0.5
1.2
q
2.2 (d + w)t
K1
0.6
1.4
0.4
1.6
0.3
2.0
2.5
0.2
3.0
3.5
4.0
0.1
0
0
0.1
0.2
0.3
0.4
0.5
q
0.6
0.7
0.8
0.9
1.0
2w
where
K 2 = factor defined in Figure 14.24
E = modulus of elasticity (psi)
𝛿 = longitudinal movement (in.)
The longitudinal bending stress due to deflection is
given by
S=
5Et𝛿
,
3w2 K3
where K 3 = factor defined in Figure 14.25.
When the stress in a bellow expansion joint is too
high, reinforcing rings are normally added as shown in
Figure 14.26. The stress equations in the bellows and
rings are modified by the ratio of the stiffnesses of the
rings and bellows. These equations are given in the
ASME code, VIII-1.
Flanged and flued expansion joints are used when the
pressure is high and the expansion is low. Approximate
analysis of these expansion joints using plate and shell
equations can be performed by matching the deflections
and rotations of the various components due to pressure
or deflection. Referring to Figure 14.27, it is seen there
301
14 Heat-Transfer Equipment
Figure 14.24 Values of K 2 . Source:
Courtesy of ASME.
3
0.2
0.4
0.6
0.8
2
1.0
1.5
1.2
1.0
0.9
0.8
0.7
1.6
q
0.5
0.4
0.3
2.2 (d + w)t
1.4
0.6
K2
302
2.0
0.2
2.5
0.15
3.0
0.10
0.09
0.08
3.5
0.07
0.06
0.05
4.0
0
0.1
0.2
0.3
0.4
0.5
q
2w
0.6
0.7
0.8
are four forces and two bending moments that need to
be determined. These are F 1 , F 2 , H 1 , H 2 , M1 , and M2 .
The known quantities from the tubesheet analysis are
the vertical deflection at point A and/or axial force F 1 .
Assuming that the deflections at points A and B are
due to bending rather than stretching, the following six
compatibility equations can be written:
Vertical deflection at point
A = known quantity
(14.42)
Rotation of the expansion joint shell at point
B = rotation of the plate at point B
1.0
Horizontal deflection of the vessel shell
at point A = horizontal deflection of the
expansion joint shell at point B
(14.44)
H2 (a∕b)H1
(14.45)
F2 = (a∕b)F1 for the deflection case
(14.46)
F2 = F1 + pressure forces for the pressure case.
(14.47)
(14.41)
Rotation of the vessel shell at point
A = rotation of the plate at point A
0.9
(14.43)
The aforementioned Eqs. (14.41)–(14.47) are solved to
obtain the bending moments and forces at the flanges and
flued expansion joint.
A more exact analysis can be achieved by using numerical methods such as a finite-element analysis.
14.8 Tube-to-Tubesheet Junctions
3.0
1.2
1.0
1.4
2.8
2.6
0.8
0.6
0.4
0.2
2.4
2.2
1.6
2.0
1.9
1.8
1.7
1.6
1.5
1.4
2.0
q
K3
1.2
1.1
1.0
0.9
2.2 (d + w)t
1.3
2.5
0.8
0.7
3.0
0.6
0.5
3.5
4.0
0.4
0
0.1
0.2
0.3
0.4
0.5
q
0.6
0.7
0.8
0.9
1.0
2w
Figure 14.25 Values of K 3 . Source: Courtesy of ASME.
14.8 Tube-to-Tubesheet Junctions
The calculations for the strength of tubes, tubesheets,
and expansion joints are based on the assumption that
the tube-to-tubesheet junction is adequate to transfer
the loads. The junction is made by welding, expansion,
or a combination of the two. If the tubes are welded to
the tubesheet, the welds must be as strong as the tubes;
otherwise, a detailed evaluation of the weld strength
versus applied loads must be performed.
Expansion of the tubes into the tubesheet holes is
usually achieved by hydraulic expanders, pill rollers,
or a combination of the two. Research into the interaction between tubes and tubesheets due to hydraulic
expansion [10, 11] has shown that the tubes shorten
without significant change in thickness. An approximate
equation for tube shortening due to hydraulic expansion
is given by
L′ =
2ri + t
L,
2ri + t + 2c
303
304
14 Heat-Transfer Equipment
ℓf
Af
A
A-A
Equalizing ring
Reinforcing rings
q
A
+
End equalizing ring
+r
w
tm
Ar
Ar
General note:
Nominal r ≥ 3tm
d diam.
Reinforced bellows
Figure 14.26 Bellow expansion joint. Source: Courtesy of ASME.
F1
Pill rollers cause thinning of the tube wall with a
subsequent increase in the length. The elongation is
given by
(2ro − t)t
L,
L′ =
(2ro − t ′ )t ′
M1
where
a
H1
L = original length of pill-rolled section (in.)
M2
F1
A
H2
H1
F1
M1
F2
B
L′ = new length due to pill rolling (in.)
ro = outside radius of tube (in.)
F2
H2
t = original thickness (in.)
M2
b
Figure 14.27 Forces in a flanged and flued expansion joint.
where
2c = difference between hole diameter and outer
diameter of tube (in.)
L = original length of hydraulically expanded section
(in.)
L′ = new length due to hydraulic expansion (in.)
ri = inside radius of tube (in.)
t = thickness of tube (in.)
t′ = reduced thickness due to pill rolling (in.)
In both hydraulic expansion and pill rolling, pressure
is applied to the tubes, which forces them against the
tubesheet and causes a partial yielding of the tubesheet.
This autofrettaging process results in a compressive
stress in the tubes within the tubesheet area when the
expansion pressure is released. This compressive stress
acts as a holding force, which prevents the tubes from
slipping out of the tubesheet. Research has shown that in
some cases, this force can reach 90% of the tensile force
of the tubes. In these cases, the expansion method is as
effective as welding. The compressive force between the
tubes and tubesheet due to expansion is greatly affected
by such factors as the modulus of elasticity, length of
expansion, yield stress, and temperature. The following is
a summary of the effect of these parameters on strength.
1) Modulus of Elasticity. If the modulus of elasticity
of the tubes is significantly lower than that of the
Further Reading
tubesheet, then the expansion method is not very
effective, since upon release of pressure, the tubes
tend to spring back more than the tubesheet.
2) Length of Expansion. The actual strength of the joint
is proportional to the length of expansion up to a certain ratio of tube length to diameter within the joint.
Tubes in thin tubesheets have limited length of expansion and thus lower joint strength.
3) Yield Strength. When the yield strength of the tubes
exceeds that of the tubesheet by a significant amount,
the compressive stress in the tubes at the joint is
reduced in magnitude, since the tubesheet begins to
yield while the tubes are still in the elastic range.
4) Temperature. Large temperature gradients between
the tubes and tubesheet influence the compressive
stress of the tubes within the tubesheet area. If the
temperature in the tubesheet is higher than that in the
tubes at the joint, then weaker joints result, since the
tubesheet tends to grow more than the tubes, which
causes reduction of the contact pressure between
the two.
References
1 (2007). Tubular Exchanger Manufacturers Association,
2
3
4
5
6
Standards of Tubular Exchanger Manufacturers Association, 9the. New York: Tubular Exchanger Manufacturers Association.
Gardner, K.A. (1960). Heat-exchanger tube-sheet
design – 3. U-tube and Bayonet-Tube sheets. Journal
of Applied Mechanics 27: 25.
O’Donnell, W.J. and Slot, T. (1971). Effective elastic
constants for thick perforated plates with square and
triangular penetration patterns. Journal of Engineering
for Industry 93: 935–942.
Gardner, K. A., “Heat exchanger tube-sheet design,”
Journal of Applied Mechanics, 1948, vol. 70, p. 377.
Gardner, K. A., “Heat exchanger tube-sheet
design—2. Fixed tube sheets,” Journal of Applied
Mechanics, 1952, vol. 74, p. 159.
Miller, K.A.G. (1960). The design of tube plates
in heat exchangers. In: Pressure Vessel and Piping
Design—Collected Papers 1927–1959, 672. American
Society of Mechanical Engineers.
Further Reading
Rubin, F.L. (1980). What’s the difference between TEMA
exchanger classes? Hydrocarbon Processing, vol. 59(5),
p. 96.
Rubin, F.L. and Gainsboro, N.R. (1979). Latest TEMA
standards for shell-and-tube exchangers. Chemical
Engineering, vol. 86, p.11.
7 Memo from Byrne, Jr., G.P. (1964). Secretary of the
8
9
10
11
Tubular Exchanger Manufacturers Association, to
members of the Technical Committee, dated January 3, 1964, regarding standards background data
prepared by Karl Gardner.
Anderson, W.F. (1964). Analysis of Stresses in Bellows,
Design Criteria and Test Results, Part 1, Atomics
International Report NAA-SR-4527.
Expansion Joint Manufacturers Association (2015).
Standards of the Expansion Joint Manufacturers
Association, 10the. New York: Expansion Joint Manufacturers Association.
Jawad, M., E. Clarkin, and R. Schuessler, “Evaluation
of tube-to-tubesheet junctions,” ASME Journal of
Pressure Vessel Technology, 1987, vol. 109, p. 19.
Yokell, S. (1982). Heat-exchanger tube-to-tubesheet
connections. Chemical Engineering, p. 78–94.
305
A thick-wall layered vessel. Source: Courtesy of the Nooter Corporation, St. Louis, MO.
308
15
Vessels for High Pressures
15.1 Basic Equations
This chapter presents some design aspects of solid and
layered vessels with pressures in the range of 10,000 to
100,000 psi and higher. At these high pressures, prestressing, or autofrettaging, becomes an important consideration in the design. These considerations are detailed in
the ASME Section VIII-3. And while the rules in Sections
VIII-1 and VIII-2 have no limits on upper pressure application, the rules in VIII-3 are written specifically for pressures over 10,000 psi and include rules and limitations for
prestressing and autofrettage methods as well as required
fracture mechanics and fatigue considerations.
It was shown in Eq. (8.1) that
PR
t=
SE − 0.6P
is the design equation for vessel shells. As the quantity
SE − 0.6P approaches zero, the thickness approaches
infinity. In other words, as the pressure increases, the
allowable stress of the shell material must be increased
higher than 60% of the design pressure for the equation
to be valid. This increase in allowable stress requires
materials of high tensile and yield properties. The limitations of Eq. (8.1) for high pressures are usually overcome
by using a different equation that is based on the theory
of plasticity as discussed later in this chapter.
Eq. (8.1) is shown in Figure 5.6 as being very similar to
Lame’s Eq. (5.9) for thick vessels. Disregarding external
pressures, Eqs. (5.9) and (5.10) become
(
)
ro2
′
𝜎𝜃 = P 1 + 2
r
(
2)
r
𝜎r = P′ 1 − o2
r
(15.1)
𝜎l = P′ ,
where
(
P′ = P
ri2
ro2 − ri2
)
.
The stress distribution given by Eq. (15.1) is shown in
Figure 15.1 for a vessel with ro /ri = 2.2. The maximum
stress is in the hoop direction and is at the inner surface,
where r = ri . As the pressure is increased, the stresses
increase until they reach a maximum limiting stress
where failure is assumed to occur. For thin vessels, the
ASME Code assumes that failure occurs when the yield
point is reached. This failure criterion is convenient and
is called the maximum-principal-stress theory. In thick
vessels, the criterion usually applied for ductile materials
is the energy of distortion theory. This theory states
that the inelastic action at any point in a body under
any combination of stresses begins only when the strain
energy of distortion per unit volume absorbed at the
point is equal to the strain energy of distortion absorbed
per unit volume at any point in a bar stressed to the
elastic limit under a state of uniaxial stress as occurs in
a simple tension test. The equation that expresses this
theory is given by
W=
]
1+𝜇 [
(𝜎1 − 𝜎2 )2 + (𝜎2 − 𝜎3 )2 + (𝜎3 − 𝜎1 )2 ,
6E
(15.2)
where
W = strain energy
𝜇 = Poisson’s ratio
E = modulus of elasticity
𝜎 1 , 𝜎 2 , 𝜎 3 = principal stresses
For a bar stressed to the elastic limit in simple tension,
𝜎 2 = 𝜎 3 = 0, and the energy of distortion expression
becomes
(1 + 𝜇)𝜎y2
.
Wst =
3E
For a pressure vessel with the three principal stresses
given by Eq. (15.1), the energy of distortion expression is
)
(
(1 + 𝜇)(P′ )2 6ro4
Wp =
6E
r4
(
)2
( r )4
ri2
1+𝜇 2
o
=
.
P
2
2
E
r
ro − ri
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
15.2 Prestressing (Autofrettaging) of Solid-Wall Vessels
1.6
ic region
Elast
1.4
σθ
p
1.2
stic region
Pla
p
1.0
σ
p 0.8
σ
– pr
ri
0.6
ro
σl
p
0.4
0.2
Figure 15.2 Cylindrical shell with elastic and plastic regions.
0
0.45 0.5
0.6
0.7
r
ro
0.8
0.9
1.0
Elastic region:
( 2)
𝜎y 𝜌2
r
𝜎𝜃 = √ 2 1 o2
r
3 ro
(
)
𝜎y 𝜌2
ro2
𝜎r = √ 2 1 − 2
r
3 ro
𝜎y 𝜌2
𝜎l = √ 2 .
3 ro
Figure 15.1 Stress distribution in a thick cylinder.
Because W st = W p , the maximum pressure at which
yield is assumed to occur at the inner surface is given
by
( )
𝜎y ro2 − ri2 ri 2
P∗ = √
.
(15.3)
2
ro
3 ri
Plastic region:
)
(
𝜎y
𝜌2
r
𝜎𝜃 = √
1 + 2 + 2 ln
𝜌
ro
3
(
)
𝜎y
𝜌2
r
𝜎r = √
−1 + 2 + 2 ln
𝜌
ro
3
( 2
)
𝜎y 𝜌
r
+ 2 ln
𝜎l = √
.
2
𝜌
r
o
3
It is interesting to note that when the axial strain is
assumed to be zero, the axial stress becomes
𝜎l = p′ (2𝜇),
and the maximum pressure given by Eq. (15.3) becomes
]−1∕2
[
2
2
4
2
𝜎
r
−
r
r
(1
−
2𝜇)
y
o
o
i
+
.
(15.4)
P∗ = √
2
3
ri4
3 ri
In most applications, the difference between Eqs. (15.3)
and (15.4) is negligible.
(15.5)
(15.6)
The relationship between the applied pressure and the
elastic–plastic interface radius 𝜌 is given by
)
(
𝜎y
ri
𝜌2
P= √
,
(15.7)
1 − 2 − 2 ln
𝜌
ro
3
where
15.2 Prestressing (Autofrettaging)
of Solid-Wall Vessels
𝜎 𝜃 , 𝜎 r , 𝜎 l = hoop, radial, and longitudinal stress,
respectively (psi)
𝜎 y = yield stress of material (psi)
As the pressure in Eq. (15.3) is exceeded, the inner part of
the shell becomes plastic, whereas the outer part remains
elastic, as illustrated in Figure 15.2. The derivation of
the relationship between the elastic and plastic regions
is beyond the scope of this book. However, it suffices to
say that the derivation [1] is based on plastic analysis of
an incompressible material with zero axial strain. The
resultant equations are as follows:
ro = outside radius of shell (in.)
ri = inside radius of shell (in.)
r = radius at any point in the shell (in.)
𝜌 = elastic–plastic interface radius (in.).
Eq. (15.7) may be used to determine the lower bound
pressure P* at which yielding occurs by letting 𝜌 = ri .
309
15 Vessels for High Pressures
Hence,
𝜎y
P∗ = √
3
(
1−
ri2
ro2
from Eq. (15.5),
)
.
(15.8)
This equation gives results very close to those given by
Eq. (15.4). Eq. (15.7) can also be used to determine the
upper-bound pressure P* at which total yielding occurs,
by letting 𝜌 = ro :
−2𝜎y
r
P∗ = √ ln i .
r
o
3
(15.9)
Example 15.1
A pressure vessel with a solid wall has an ri of 8 in., ro of
17.6 in., and 𝜎 y = 80,000 psi. Plot 𝜎 𝜃 and 𝜎 r when P = P*
and when P = 60,000 psi.
Solution:
From Eq. (15.3),
)(
(
)
80,000 17.62 − 82
8 2
√
17.6
82
3
= 36,650 psi,
P∗ =
and from Eq. (15.1),
(
)
17.62
𝜎𝜃 = 9540 1 + 2
r
(
)
17.62
𝜎r = 9540 1 − 2
.
r
These two values are shown in Figure 15.3.
From Eq. (15.7) with P = 60,000 psi and 𝜎 y = 80,000 psi,
a trial-and-error calculation gives 𝜌 = 11.50 in. Hence
70
σ
θ
60
50
40
wh
en
P
σθ w
hen P
30
Stress, KSI
310
= 60
,000
psi
)
17.62
𝜎𝜃 = 19,720 1 + 2
r
(
)
17.62
𝜎r = 19,720 1 − 2
,
r
and from Eq. (15.6),
(
)
r
𝜎𝜃 = 46,190 1.427 + 2 ln
11.5 )
(
r
.
𝜎r = 46,190 −0.573 + 2 ln
11.5
A plot of these equations is shown in Figure 15.3.
The two plots in Figure 15.3 are significant because
they show the redistribution of the stress pattern as the
inner region of the cylinder becomes plastic. Also note
the reduction of the stress at the inner surface and the
increase of the stress at the elastic–plastic boundary as
the pressure is increased beyond P* .
Example 15.2
Plot the circumferential residual stress 𝜎 𝜃 when the
autofrettaging pressure mentioned in Example 15.1 is
reduced to zero. Also plot 𝜎 𝜃 when a design pressure of
45,000 psi is applied.
Solution:
The maximum autofrettaging pressure of 60 ksi is less
than twice the lower-bound pressure P* . Accordingly,
the stress distribution resulting from a pressure drop
of 60 ksi is in the elastic range, as shown in Figure 15.4.
From Eq. (15.1),
(
)
82
P′ = −60,000
17.62 − 82
P′ = −15,625 psi
(
)
17.62
𝜎𝜃 = −15,625 1 + 2
.
r
P
= P*
(
60 000 psi
2P* = 73 300 psi
20
36 650
10
0
P
hen
σ rw
–10
–20
–30
en
wh
–40
= P*
P
si
0p
,00
0
=6
ε
σr
–50
–36 650
–60
8
10
12
14
16
17.6
Radius (in.)
Figure 15.3 Stress in a cylinder with elastic and plastic regions.
Figure 15.4 Pressure–strain diagram due to uploading and then
downloading of pressure.
15.3 Layered Vessels
60
50
σθ at P = 45 KSI
40
Hoop stress, K.S.I.
30
20
(a) Concentric
Residual st
ress
10
(b) Spiral
0
–10
–20
–30
–40
–50
–60
8
10
12
14
16
17.6
(c) Shrink fit
(d) Coil wrap
Figure 15.6 Various types of layered cylindrical shells.
Figure 15.5 Circumferential stress in a cylinder with elastic and
plastic regions.
The stress distribution given by this equation is superimposed with that in Figure 15.3 for 𝜎 𝜃 at P = 60 ksi. The
resultant residual stress is shown in Figure 15.5.
The internal pressure of 45,000 psi gives a stress of
(
)
17.62
,
𝜎𝜃 = 11,720 1 + 2
r
and the total stress distribution due to this and residual
stress is given in Figure 15.5.
Prestressing (autofrettaging) of the shell by increasing
the internal pressure requires special attention to the
actual yield and tensile properties of the material. This is
especially true when heads are attached to the shell. The
stress in the heads during autofrettaging the shell must
carefully be monitored and kept below certain established limits. Secondary stresses that may be present in
the head, shell, or junctions have to be taken into consideration as well. ASME Section VIII-3 contains rules for
autofrettaging the shell and resulting residual stresses.
15.3 Layered Vessels
Layered vessels were developed in the United States and
Germany at about the same time during World War II.
In Germany, they were used in ammonia plants as well
as for producing gasoline from coal. In the United States,
they were used for ammonia-synthesis processes for
the ultimate production of nitrates. Since World War II,
the technology of building layered vessels has improved
substantially. Today, layered vessels are used in a wide
range of high-pressure applications in the petrochemical
industry such as heat exchangers, urea reactors, ammonia converters, autoclaves, and coal gasification reactors.
Layered vessels consist of a multitude of layers wrapped
tightly around an inner shell to form a pressure-retaining
envelope, as shown in Figure 15.6. The vent-hole system
is a safety feature. It consists of a multitude of small holes
drilled radially into the layers and extending from the outermost layer to and including the layer adjacent to the
inner shell. The holes are sized and spaced so that they do
not affect the structural integrity of the vessel. The venting system acts as a monitor of potential problems such
as erosion and corrosion that may occur in the inner shell
during the operation of the vessel.
Layered vessels are constructed by various methods.
The difference between these methods is in the thickness
of individual layers, wrapping procedure, and welding
technique. In general, layered-vessel construction can be
divided into three categories. The first is the concentricor spiral-wrapped method where the layers consist
of segments welded together in a spiral or concentric
fashion to form the required thickness, as shown in
Figure 15.6a, b. The second is the shrink-fit method
whereby layers are individually formed into cylinders
and shrunk on each other to form the required total
thickness (Figure 15.6c). The third is the coil-wrapped
method whereby a continuous sheet or strip is wound
in a spiral or helical fashion to form a cylinder as in
Figure 15.6d.
The earliest reference to layered vessels was made
in the 1951 API–ASME Code. In later years when the
API Standard and the ASME Code were separated, the
layered-vessel criterion was deleted from both. It was not
until January 1979 that layered vessels were included in
the ASME Code. In establishing the new layered-vessel
rules, consideration was given to the state of the art as
well as the experience and research accumulated by the
industry in the past 50 years. An effort was also made
311
312
15 Vessels for High Pressures
to provide rules to accommodate all types of known
layered-vessel construction.
Today, most layered vessels are constructed in accordance with the ASME Code, VIII. The majority of
the design equations given in the code for solid-wall
vessels are applicable to layered vessels. For fabrication, the ASME Code, VIII, gives additional rules for
layered-vessel construction. One criterion for controlling wrapping tightness of layered shells is that the area
of any gap between two adjacent layers, as measured
from the end of a shell section, must not exceed the
thickness of a layer expressed in square inches. This is
illustrated in Figure 15.7.
Another criterion used occasionally to measure the
tightness of layered shells is limiting the circumferential
expansion of the outer layer during hydrostatic testing to
a value not less than one-half that of an equivalent solidwall thickness. Hence, the stress at the outer layer due to
internal pressure P as given by Eqs. (5.9) and (5.10) is
𝜎𝜃 =
and the circumferential growth is given by
e=
h=
Pri2
ro2 − ri2
M0
.
2𝛽 2 D
(15.11)
Substituting
6M
𝜎b = 2 0
√t
3(1 − 𝜇2 )
4
𝛽=
r2 t2
and
𝜎r = 0
𝜎l =
D=
,
Et 3
12(1 − 𝜇2 )
into Eq. (15.11) gives
0.55r𝜎b
.
(15.12)
E
Eq. (15.12) cannot be used directly, because the quantity 𝜎 b is not readily known. This quantity, however,
can be related to an allowable stress by calculating the
and from Eq. (3.1),
1
𝜖𝜃 = [𝜎𝜃 − 𝜇(𝜎r + 𝜎l )]
E
or
(2 − 𝜇)Pri2
𝜖𝜃 =
.
E(ro2 − ri2 )
h=
The circumferential growth can be expressed as
Gap
e = 2π𝜖𝜃 ro
=
2πro (2 − 𝜇) Pr2i
E(ro2 − ri2 )
(15.10)
The actual measured growth must not be less than
one-half of the value given by Eq. (15.10).
A third criterion for determining the maximum permissible gap in layered shells is by relating the gap height
to a given stress level. Referring to Figure 15.8 and Eq.
(5.24) and assuming the end to be fixed against rotation,
it can be shown [2] that the gap h can be related to the
bending moment by the expression
2Pri2
ro2 − ri2
1.7πP(2Rm − t)2 (2Rm + t)
.
8ERm t
.
Let Rm be defined as the mean radius. Then,
t
ri = Rm −
2
t
ro = Rm + ,
2
h
(a)
Qo
Mo
t
t
t
h
r
R
Figure 15.7 Typical gap between two layers.
(b)
Figure 15.8 Forces in a layer due to a gap.
15.3 Layered Vessels
principal stresses at a layer as
𝜎𝜃 = S + 𝜇𝜎b
𝜎r = −P
S
𝜎l = 𝜎b + ,
2
where:
𝜎 𝜃 , 𝜎 r , 𝜎 l = principal stresses
S = hoop stress
𝜇 = Poisson’s ratio
𝜎 b = bending stress due to gap h
P = internal pressure.
The maximum stress intensity is given by
S
+ P.
(15.13)
2
The maximum stress intensity 𝜎 lr is limited by the
ASME Code, VIII-2, to 3Sm : in general terms,
𝜎lr = 𝜎b +
S
(15.14)
+ P ≤ KSm ,
2
where K = 3 for an indefinite number of cycles.
In cases where 𝜎 lr is greater than 3Sm but less than
3mSm , the rules of the ASME Code, VIII, paragraph
4-136.4, apply for a simplified elastic–plastic analysis. In
this case,
𝜎
Sa = K lr
2
or
2S
(15.15)
𝜎lr = a ,
K
where Sa is the alternating stress as defined in the ASME
Code, VIII, and K is given by
)
(
𝜎lr
1−n
−1 ,
(15.16)
K =1+
n(m − 1) 3Sm
𝜎b +
where
m = 3.0 for carbon steel
= 2.0 for low-alloy steel
= 1.7 for austenitic stainless steel
n = 0.2 for carbon and low-alloy steels
= 0.3 for austenitic stainless steels
Sm = allowable stress.
Eq. (15.16) can be expressed in terms of Sa by substituting Eq. (15.15) for 𝜎 lr and letting m = 3.0 and n = 0.2,
which yields
√
1 4 Sa
1
.
(15.17)
+
K =− +
2
4 3 Sm
Eq. (15.13) and (15.15) can be expressed as
2Sa
S
S .
𝜎b + + P =
2
Sm K m
(15.18)
Assuming S = Sm and
2Sa
.
N=
KSm
Eq. (15.18) becomes
S
𝜎b = NSm − m − P,
2
and Eq. (15.12) can be written as
]
[
0.55rSm
P
h=
,
N − 0.5 −
E
Sm
(15.19)
where
N = 2Sa /KSm
= 3 for an indefinite number of cycles
√
1
K =− +
2
1 4 Sa
+
4 3 Sm
Sa = maximum allowable alternating stress
Sm = allowable stress
E = modulus of elasticity
r = radius where gap is measured
P = design pressure.
Example 15.3
A layered vessel with a 42 in. inner diameter is constructed of carbon steel with E = 29,000,000 psi.
Determine the maximum allowable gap if P = 4000 psi,
Sm = 20,000 psi, and the maximum cycle life of the vessel
is 10,000 cycles.
Solution:
From Division 2 of the ASME Code, VIII, Figure 5-110.1,
Sa = 40,000 psi. From Eq. (15.19),
√
(
)
1
1 4 40,000
K =− +
+
2
4 3 20,000
= 1.21
2 × 40,000
N=
1.21 × 20,000
= 3.31
0.55 × 21 × 20,000
29,000,000
= 0.021 in.
h=
(
)
4000
3.31 − 0.5 −
20,000
Eq. (15.19) determines the maximum value of any one
gap in a layered vessel. Where there is more than one
313
15 Vessels for High Pressures
gap in a given cross section, a criterion is needed to take
accumulated strains of all gaps into account. This can be
accomplished by determining first the strain developed
in closing each gap. The total strain is then summed up
and compared with an allowable strain.
In getting the strain required to close one gap, refer to
Figure 15.9, where
into Eq. (15.24) yields
[
(
)]
R
a sin 𝛼
𝛼
𝜖 = 2 𝛼 + sin−1
− .
πR1
R2
π
B
h𝛼
A + sin−1 C,
πR1
π
𝜖=
X02 + Y02 = R21
and
(15.25)
The total strain required to close one gap is obtained
by substituting Eqs. (15.21) and (15.23) into 15.25, which
gives
(15.26)
where
X02 + (Y0 − a)2 = R22 .
2 + h∕R1
2(1 − cos 𝛼 + h∕R1 )
(2 + h∕R1 )(h∕R1 )
1
−
B=1+
R1 2(1 − cos 𝛼 + h∕R1 )
(2 + h∕R1 )(h∕R1 ) sin 𝛼
C=
.
2(1 + h∕R1 )(1 − cos 𝛼 + h∕R1 )
−(2 + h∕R1 )(h∕R1 )
A=1−
Combining these two equations gives
R22 − R21 = a2 − 2aY 0 .
(15.20)
Also from Figure 15.9,
R1 + h = R2 + a
(15.21)
Y0 = R1 cos 𝛼.
(15.22)
and
Substituting Eqs. (15.21) and (15.22) into Eq. (15.20)
gives
a=
h(h + 2R1 )
.
2(R1 + h − R1 cos 𝛼)
(15.23)
The circumferential strain determined from Figure 15.9
is expressed as
𝜖=
2R2 𝛽 − 2R1 𝛼
.
2𝜋R1
(15.24)
Substituting
𝛾 =𝛽−𝛼
and
c
R2
a sin 𝛼
=
R2
sin 𝛾 =
A plot of Eq. (15.26) shows a linear dependence of
strain on h/R1 . Hence, the ASME Code, VIII, Division 2,
approximated Eq. (15.26) by the quantity
𝜖 = 0.109(2𝛼)
(15.27)
Eq. (15.27) calculates the strain needed to close any
given gap. The total strain required to close all the gaps
is determined by summing all the individual strains
given by Eq. (15.27). The hoop stress in a layer due to all
accumulated gap strains is approximated by
E ∑
𝜖.
𝜎𝜃 =
1 − 𝜇2
The total hoop stress due to gap strains and internal
pressure can be expressed as
𝜎𝜃 =
Y
h
R1
or in terms of Figure 15.7
( )
hl
𝜖 ≤ 0.109 2 .
r
ro2 + ri2
ro2
−
ri2
P+
E ∑
𝜖.
1 + 𝜇2
The maximum radial stress is given by
h
𝜎r = −P.
Hence, the maximum stress intensity is
𝜎𝜃r = 𝜎θ − 𝜎r
∝
β
a
∝
c
XO,YO
R2
R1
or
ro2 + ri2
ro2
∝
314
X
−
P+
E ∑
𝜖 + P ≤ NSm
1 − 𝜇2
and
∑
Figure 15.9 Circumferential gap.
ri2
1 − 𝜇2
𝜖≤
E
(
NSm −
2ro2 P
ro2 − ri2
)
.
(15.28)
15.4 Prestressing of Layered Vessels
Example 15.4
The following gaps were measured after forming a layered vessel of ri = 40 in. and ro = 55 in.:
Pi + 1
Ri + 1
Ri + 2
Gap
Length (in.)
Height (in.)
At radius (in.)
1
5
0.008
43.25
2
9
0.005
43.25
3
7
0.010
48.50
4
6
0.009
53.00
R2 X
R1
Weld
Layer i
6
Determine if these gaps are acceptable if E = 30 × 10 psi,
P = 6000 psi, 𝜇 = 0.3, Sm = 20,000 psi, and N = 3.
Solution:
From Eq. (15.27),
5(0.008)
43.252
9(0.005)
= 0.109
43.252
7(0.010)
= 0.109
48.502
6(0.009)
= 0.109
53.002
= 10.30 × 10−6 .
Figure 15.10 Designation of various layers.
𝜖1 = 0.109
= 2.33 × 10−6
𝜖2
= 2.62 × 10−6
𝜖3
𝜖4
Total 𝜖
Referring to Figure 15.10, the total radial deflection due
to transverse shrinkage of a number of seams in one layer
is
ns
d=
,
(15.30)
2π
where
= 3.25 × 10−6
= 2.10 × 10−6
d = radial deflection
n = number of welded seams in a layer.
From Eq. (15.28),
)
(
2ro2 P
1 − 𝜇2
NSm − 2
E
ro − ri2
(
)
2(55)2 (6000)
0.91
3
×
20,000
−
=
30 × 106
552 − 402
= 10.47 × 10−6 .
Because total 𝜖 is less than 10.47 × 10−6 , the gaps are
satisfactory.
In the spiral or concentric method of fabrication, the
transverse weld shrinkage in the longitudinal seams
causes prestressing of the vessel layers. Such weld
shrinkage is influenced by many factors, such as speed
of welding, amount of heat input, and welding method.
The shrinkage [3] can generally be expressed as
s = kw,
where
s = transverse shrinkage in welds
k = coefficient of transverse shrinkage
Weld shrinkage decreases the diameter of a welded
layer. This causes a pressure between the welded layer
and the layers underneath it. The deflection compatibility
equation at the pressure interface is
d − di = do ,
(15.31)
where
di = radial deflection of layer i, shown in Figure 15.10,
due to interface pressure Pi + 1
do = radial deflection of all layers beneath layer i due
to interface pressure Pi + 1
15.4 Prestressing of Layered Vessels
w = width of seam weld.
Layer 1
Inner shell
(15.29)
Substituting Eq. (15.30) into Eq. (15.31) gives
ns
(15.32)
do + di = .
2
The deflection of layer i an all layers beneath i due
to pressure Pi + 1 can be obtained from Eq. (5.8) by
substituting 𝜎 i = 0 and disregarding the term in 𝜇2 . The
expressions for the deflection, using the terminology of
Figure 15.10, becomes
(
)
Pi+1 Ri+1 R2i+1 + R21
do =
− 0.3
(15.33)
E
R2i+1 − R21
)
(
Pi+1 Ri+1 R2i+2 + R2i+1
di =
+ 0.3 .
(15.34)
E
R2i+2 − R2i+1
315
316
15 Vessels for High Pressures
Substituting Eqs. (15.33) and (15.34) into Eq. (15.32)
and rearranging the terms yields
2
Pi+1 =
2
2
3
2
1
I.S.
2
nsE (Ri+1 − R1 )(Ri+2 − Ri+1 )
.
(R2i+2 − R21 )
4πR3i+1
(15.35)
The stress in layer i can be expressed as
20
"L
Pi+1 Ri+1
𝜎i =
.
(15.36)
t
Substituting Eqs. (15.29) and (15.35) into Eq. (15.36)
gives
𝜎i =
2
2
2
2
nkwE (Ri+1 − R1 )(Ri+2 − Ri+1 )
.
(R2i+2 − R21 )
4πtR2i+1
The stress in any layer due to welding other layers
around it is
(
)
R21 ∑ Pi+1 R2i+1
𝜎x = − 1 + 2
.
(15.38)
2
2
X
i Ri+1 − R1
Substituting Eqs. (15.29) and (15.35) into Eq. (15.38)
results in
)
(
R2 ∑ 1 R2i+2 − R2i+1
−nkwE
.
𝜎x =
1 + 12
4π
X
Ri+1 R2i+2 − R21
i
(15.39)
Eqs. (15.37) and (15.39) are necessary to determine the
precompressive stress in a layered vessel due to wrapping
by the concentric or spiral method of fabrication.
In the shrink-fit method, the precompression equations
are the same as Eqs. (15.37) and (15.39) except that the
quantity nkw/2π is replaced by 𝛼ΔT R. Hence,
2
2
2
2
E𝛼ΔT (Ri+1 − R1 )(Ri+2 − Ri+1 )
(15.40)
𝜎i =
2tRi+1
R2i+2 − R21
)
(
R21 ∑ R2i+2 − R2i+1
−E𝛼ΔT
𝜎x =
. (15.41)
1+ 2
2
X
R2i+2 − R21
i
In the coil-wrapped method, the initial stress in the
outer layer i is known. Hence, the applied pressure for
Eq. (15.18) is
𝜎t
Pi+1 = c ,
Ri+1
𝜎i = 𝜎c ,
and the stress in the inner layers is given by
(
)
R2 ∑ Pi+1 R2i+1
𝜎x = − 1 + 12
.
2
2
X
i Ri+1 − R1
.R.
"O
22
(15.37)
The stress in the layers below i due to welding i is
(
)
−Pi+1 R2i+1
R21
𝜎x = 2
1+ 2 .
X
Ri+1 − R21
the total stress in the outer layer is
.R.
Figure 15.11 Longitudinal welds in layers.
Example 15.5
Determine the wrapping stress in the vessel shown in
Figure 15.11 if n = 2, k = 0.1, E = 30 × 106 psi, w = 0.375 in.
Solution:
From Eq. (15.37), the stress in layer 1 due to wrapping of
layer 1 is
2(0.1)(0.375)(30 × 106 )
4π(0.5)(20.5)2
(20.52 − 202 )(212 − 20.52 )
×
212 − 202
= 8733 psi.
𝜎1 =
The stress in layer 2 due to wrapping of layer 2 is
2.25 × 106 (212 − 202 )(21.52 − 212 )
4π(0.5)(21)2
21.52 − 202
= 11,365 psi.
𝜎2 =
The stress in layer 3 due to wrapping of layer 3 is
2.25 × 106
4π(0.5)(21.5)2
(21.52 − 202 )(222 − 21.52 )
×
222 − 202
= 12,487 psi.
𝜎3 =
From Eq. (15.39), the stress in the inner shell due to
wrapping of all three layers is
)
(
−2.25 × 106
202
𝜎IS =
1+
4
20.252
(
2
2
21 − 20.5
21.52 − 212
×
+
20.5(212 − 202 ) 21(21.52 − 202 )
)
222 − 21.52
+
21.5(222 − 202 )
Nomenclature
Nomenclature
= −(179,049)(1.975)
× (0.0247 + 0.0163 + 0.0120)
The stress in layer 1 due to wrapping of layers 2 and 3
is
Et 3
12(1 − 𝜇2 )
E = modulus of elasticity
e = circumferential growth
h = gap
k = weld shrinkage
M0 = longitudinal bending moment
n = number of seams in a layer
P = pressure
R = inside radius as defined by ASME
Rm = mean radius
r = radius
ri = inside radius
ro = outside radius
S = stress
Sm = allowable stress given by ASME
t = thickness
w = width of weld seam
𝛼 = coefficient of thermal expansion
3(1 − 𝜇2 )
𝛽4 =
r2 t2
ΔT = change in temperature
𝜖 = strain
𝜇 = Poisson’s ratio
𝜌 = radius at the interface between the
elastic and plastic zones
𝜎 b = bending stress
𝜎 c = stress in coil
𝜎 i = stress at layer i
𝜎 l = longitudinal stress
𝜎 r = radial stress
𝜎 x = stress at layer x
𝜎 y = yield stress
𝜎 𝜃 = hoop stress
D=
= −18,737 psi.
(
)
202
(0.0163 + 0.0120)
𝜎1 = −179,049 1 +
20.752
= −9774 psi.
The stress in layer 2 due to wrapping of layer 3 is
(
)
202
𝜎2 = −179,049 1 +
(0.0120)
21.252
= −4052 psi.
Total stress in inner shell = −18,737 psi.
Total stress in layer 1 = 8733–9774 = −1041 psi.
Total stress in layer 2 = 11,365–4052 = 7313 psi.
Total stress in layer three = 12,487 psi.
ASME VIII-3 includes equations similar to those
shown in this section for prestressing of layered vessels.
15.5 Wire-Wound Vessels
Another technique for applying compression on the
inner surface of the shell is to use high-strength prestressed wires. These wires are normally 1/8 of an inch in
diameter or smaller with tensile stress over 180 000 psi.
The wires can also be supplied as flat strips rather than
round. The wires are wrapped around the outside surface
of the shell in a helical fashion. The effective thickness of
one layer is calculated from a “smeared” material for a
given length. The prestressed wires are applied to either
a solid shell or layered shell as shown in Figure 15.12.
The stress analysis for wire-wound vessels is similar as
for layered vessels with the thickness of each wire layer
determined from an equivalent “smeared” area over a
given length. ASME VIII-3 includes rules and equations
for flat wire-wound shells with open ends.
Figure 15.12 Layered shell with prestressed
wire on the outside.
Prestressed wire
Solid wall
R
Layered wall
(a)
R
(b)
317
318
15 Vessels for High Pressures
References
1 Prager, W. and Hodge, P.G. (1965). Theory of Perfectly
3 Jawad, M.H. (September 1972). Wrapping Stress and its
Plastic Solids. New York: Wiley.
2 Jawad, M. H., Layered Vessels in the Petrochemical Industry. Proceedings of the API 44th Midyear
Meeting, May 14, 1979, San Francisco, Vol. 58, p. 273.
Effect on Strength of Concentrically Formed Plywalls.
ASME Publication 72-PVP-7.
Further Reading
1 Armstrong, W.P. and Jawad, M.H. (November 1981).
Evaluation of thermal conductivity in layered vessels.
ASME Journal of Pressure Vessel Technology.
2 Brownell, L.W. and Young, E.H. (1959). Process Equip-
ment Design. New York: Wiley.
Tall vessel. Source: Courtesy of the Nooter Corporation, St. Louis, MO.
320
16
Tall Vessels
16.1 Design Considerations
Special design considerations are required for tall vessels that are installed in the vertical position. These
vessels may utilize support skirts, support rings, ring
girders, lugs, and other forms of support attachments
as described in Chapter 12. However, the vessel itself
requires special design considerations in selecting the
proper thicknesses and stiffening rings, if needed, to
adequately support the vessel and to resist the applied
loadings.
In addition to loadings from internal and external
pressures, tall vessels must be capable of withstanding
additional loadings from the dead load of the vessel, the
contents, internal parts, insulation, piping, and external
equipment, and from earthquake loading and wind
loading. The tall vessel, similarly to most other types
of vessels, may also be subjected to applied forces and
moments from thermal expansion of the piping. The
most critical combination of loadings that causes the
highest stresses may not occur when all of the loads
are applied at the same time. Certain loads may cause
critical stresses during the time of erection of the vessel,
whereas other combinations of loadings may cause critical stresses when the vessel is filled. The proper design
of the vessel may require examining several different
loading conditions to establish the proper thickness and
other requirements for a safe design.
Some of the combinations of loadings requiring careful
consideration are as follows:
1) Vessel installed in place but not operating (no contents, internals, or insulation) and not under an
applied earthquake or wind loading
2) Vessel under internal pressure with contents and
other dead loads with or without earthquake or wind
loading
3) Vessel under external pressure with contents and
other dead loads with or without earthquake or wind
loading
Other combinations, for a specific vessel, may be worse
than any of the listed conditions. The designer must be
certain that all conditions are examined to determine the
controlling condition.
The required thicknesses and other design requirements vary somewhat depending upon the design theory
chosen. The maximum-stress theory is used for the
design of most tall vessels. This theory is used in the
ASME Code, VIII-1, and the API 620 [1] and 650 [2]
design rules. The effects of using other theories are
discussed later.
The two external loadings that are important in the
design are those due to earthquake loadings and wind
loadings. Although the ASME Code, VIII-1, does not
specify design methods or design codes that are considered, application of the two commonly used design
rules is discussed. Remember, in the specific location
where the tall vessel is to be installed, as given in the
design specification or purchase order, the design rules
may be somewhat different from either of the two rules
described here. Local requirements are always to be
considered.
Once the external loadings and overturning moments
are determined, they are combined with the internal
and external pressures and any other loadings that
are applicable to the tall vessel. The following sections
describe different methods for establishing the forces,
moments, and overturning moments from external loadings. In addition, methods are given for combining those
loadings with other loadings in order to determine the
highest stresses and to satisfy UG-22 of the ASME Code,
VIII-1. Also included are methods of considering the
dynamic effects of the wind loading and wind velocity
on vortex shedding and ovalling vibrations.
16.2 Earthquake Loading
16.2.1
Lateral Loads
In tall vessels, one cause of stresses in the vessel wall
is the overturning moment from the lateral force of an
earthquake loading. Although most design standards
require vessels to withstand earthquakes, usually no
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
16.2 Earthquake Loading
specific applicable rules are given. The purchase order
or design specification should list the applicable code to
be used for earthquake design such as the ASCE 7-10
[3] standard. Additionally, the location of the installation
is required to determine the appropriate earthquake
factors. The procedure for determining the lateral earthquake loading is similar in most building standards,
although some coefficients in the formulas and in the
earthquake zone map may differ in various standards.
The procedure described in this section generally follows
the ASCE 7-10 Standard.
The lateral earthquake force equation is expressed as
Fp = 0.4ap SDS Wp (1 + 2z∕h)∕(Rp ∕Ip ).
where Ss is the mapped MCER (maximum considered
earthquake ground) spectral response acceleration
parameter at short periods determined from Figure 16.1.
F a is the site foundation coefficient. Values of F a may be
obtained from Table 16.2. Whereas the soil properties
are not known in sufficient detail to determine the site
class, Site Class D shall be used unless the authority
having jurisdiction or geotechnical data determines the
presence of Site Class E or F soils at the site.
A more precise equation for F p is obtained from the
following equation that considers the natural period of
vibration, T, of the vessel in consideration,
(16.1)
The limits of F p are
Fp = 0.4Ip Sa Wp (1 + 2z∕h),
(16.4)
where
(0.3)SDS Ip Wp < Fp < (1.6)SDS Ip Wp ,
Sa = SDS (0.4 + 0.6T∕To ) T < To
(16.5a)
To < T ≤ Ts
(16.5b)
Sa = SDS
where
ap = amplification factor. It is equal to 2.5 for pressure
vessels.
Sa = SD1 ∕T
Ts < T ≤ TL
Sa = SD1 TL ∕T
2
T > TL
(16.5c)
(16.5d)
SD1 = 0.667Fv S1 ,
F p = horizontal seismic load
h = height
and
I p = importance factor.
S1 = design, 5% damped, spectral acceleration
parameter at a period of 1 s obtained from
Figure 16.2.
Rp = response modification factor. It is equal to 2.5 for
pressure vessels.
T = natural period of the fundamental frequency of a
pressure vessel.
W p = weight of vessel
z = elevation of support from the ground level.
The value of the Importance Factor I p ranges from 1.0
to 1.5 depending on the service condition of the vessels.
Various standards assign deferent values to it. Table 16.1
lists some service conditions and the corresponding I p
factors used by some of the standards.
Based on the aforementioned definitions, Eq. (16.1)
becomes
Fp = 0.4Ip SDS Wp (1 + 2z∕h).
(16.2)
The value of SDS is defined by the equation
SDS = 0.667Fa Ss ,
(16.3)
T o = 0.2 SD1 /SDS
T s = SD1 /SDS
T L = long-period transition periods obtained from
Figure 16.3.
Whereas the soil properties are not known in sufficient
detail to determine the site class, Site Class D shall be
used unless the authority having jurisdiction or geotechnical data determines the presence of Site Class E or F
soils at the site.
The natural period of the fundamental frequency of a
pressure vessel with a uniform cross section is [4]
T = 0.091(wL4 ∕EI)0.5
Table 16.1 Importance factor Ip .
for vertical cantilever vessel
4
Category
Ip
Category C: Vessels with contents that may cause a
threat to public safety if released
1.50
Category B: Vessels with contents that may cause
discomfort to public safety if released
1.25
Category A: Vessels that don’t fall under category B
or C
1.00
T = 0.032(wL ∕EI)
(16.6a)
0.5
for vessel supported on two end saddles.
(16.6b)
The natural frequency and period for vessels having
different thicknesses along the length are determined
by numerical analyses such as the finite element and
Stodola methods. Section 16.2.2 introduces the Stodola
method.
321
322
16 Tall Vessels
120
110
100
45
40
35
Discussion
References
Maps prepared by United States Geological Survey (USGS)
in collaboration with the Federal Emergency Management
Agency (FEMA)-funded Building Seismic Safety Council
(BSSC) and the American Society of Civil Engineers (ASCE).
The basis is explained in commentaries prepared by BSSC
and ASCE and in the references.
Ground motion values contoured on these maps incorporate:
• a target risk of structural collapse equal to 1% in 50 years
based upon a generic structural fragility
• a factor of 1.1 to adjust from a geometric mean to the
maximum response regardless of direction
• deterministic upper limits imposed near large, active
faults, which are taken as 1.8 times the estimated median
response to the characteristic earthquake for the fault (1.8
is used to represent the 84th percentile response), but not
less than 150% g.
As such, the values are different from those on the uniformhazard 2008 USGS National Seismic Hazard Maps posted at:
http://earthquake.usgs.gov/hazmaps.
Larger, more detailed versions of these maps are not
provided because it is recommended that the corresponding
USGS web tool (http://earthquake.usgs.gov/designmaps or
http://content.seinstitute.org) be used to determine the
mapped value for a specified location.
Building Seismic Safety Council, 2009, NEHRP Recommended
Seismic Provisions for New Buildings and Other Structures:
FEMA P-750/2009 Edition, Federal Emergency Management
Agency, Washington, DC.
Huang, Yin-Nan, Whittaker, A.S., and Luco, Nicolas, 2008,
Maximum spectral demands in the near-fault region,
Earthquake Spectra, Volume 24, Issue 1, pp. 319-341.
Luco, Nicolas, Ellingwood, B.R., Hamburger, R.O., Hooper,
J.D., Kimball, J.K., and Kircher, C.A., 2007, Risk-Targeted
versus Current Seismic Design Maps for the Conterminous
United States, Structural Engineers Association of California
2007 Convention Proceedings, pp. 163-175.
Petersen, M.D., Frankel, A.D., Harmsen, S.C., Mueller, C.S.,
Haller, K.M., Wheeler, R.L., Wesson, R.L., Zeng, Yuehua,
Boyd, O.S., Perkins, D.M., Luco, Nicolas, Field, E.H., Wills,
C.J., and Rukstales, K.S., 2008, Documentation for the 2008
Update of the United States National Seismic Hazard Maps:
U.S. Geological Survey Open-File Report 2008-1128, 61p.
30
Figure 16.1 Ss , maximum considered earthquake (MCER ) ground motion parameter for 0.2 s spectral response acceleration, 5% of critical
damping, Site Class B in the contiguous United States. Source: Courtesy of ASCE.
16.2 Earthquake Loading
90
80
100
0
100
100
0
200
100
200
70
300
300
400
400
50
500 mi
500 km
45
40
35
Explanation
Contour intervals (%g)
300
200
150
125
100
90
80
70
60
50
40
35
30
25
20
15
10
5
30
Areas with a constant spectral
response acceleration of 150% g
10
10
10
10
Contours of spectral response
25
acceleration expressed as a percent
of gravity. Hachures point in
direction of decreasing values
+
Point value of spectral response
acceleration expressed as a percent
of gravity
16.9
Figure 16.1 (Continued)
323
324
16 Tall Vessels
Table 16.2 Values of F a and F v .
Values of F a
Site class
Ss ≤ 0.25
Ss = 0.5
Ss = 0.75
Ss = 1.0
Ss ≥ 1.25
A. Hard rock
0.8
0.8
0.8
0.8
0.8
B. Rock
1.0
1.0
1.0
1.0
1.0
C. Very dense soil and soft rock
1.2
1.2
1.1
1.0
1.0
D. Stiff soil
1.6
1.4
1.2
1.1
1.0
E. Soft clay soil
2.5
1.7
1.2
0.9
0.9
Ss = 0.4
Ss ≥ 0.5
Values of F v
Site class
Ss ≤ 0.1
Ss = 0.2
Ss = 0.3
A. Hard rock
0.8
0.8
0.8
0.8
0.8
B. Rock
1.0
1.0
1.0
1.0
1.0
C. Very dense soil and soft rock
1.7
1.6
1.5
1.4
1.3
D. Stiff soil
2.4
2.0
1.8
1.6
1.5
E. Soft clay soil
3.5
3.2
2.8
2.4
2.4
Note: Straight line interpolation is permitted between values.
If the designer is using the allowable stress design
(ASD) method, the value of Fp may be multiplied by 2/3.
Example 16.1
The vertical pressure vessel shown in Figure 16.4 has the
following design values:
Total weight = 252,000 lbs
T = 0.088 s (obtained from Example 16.3 in Section
16.2.2)
I p = 1.0
Location: vicinity of Portland, Oregon.
Base of the vessel is at ground level (z = 0.0).
Site class is B (rock foundation): F a = 1.0
Determine the earthquake lateral load using the ASD
method and
a) Equation (16.2)
b) Equation (16.4).
b)
S1 = 0.5 from Figure 16.2
TL = 16 from Figure 16.3
Ts = 0.5∕0.9 = 0.556
To = 0.2(0.556) = 0.111.
Since T is less than T o , Eq. (16.5a) is used
Sa = (0.6)[0.4 + 0.6(0.088)∕0.111] = 0.525.
From Eq. (16.4),
Fp = (2∕3)[0.4(1)(0.525)(252,000)(1)]
= 0.14(252,000)
= 35,280 lbs = 1176 lbs∕ft.
A software program and maps may be accessed for
determining the values of Ss and S1 more accurately at
the USGS Website “http://earthquake.usgs.gov/hazards/
designmaps/” based on the address of the vessel location.
Solution:
a)
Ss = 0.9 from Figure 16.1.
From Eq. (16.3),
SDS = 0.667(1)(0.9) = 0.6.
From Eq. (16.2),
Fp = (2∕3)[0.4(1)(0.6)(252,000)(1)]
= 0.16(252,000)
= 40,320 lbs = 1344 lbs∕ft of length.
16.2.2 Numerical Method for Calculating Natural
Frequency
The fundamental natural frequency, and hence the
natural period, of a vertical vessel comprising different
thicknesses, weights, and diameters can be obtained,
with good accuracy, from a numerical structural analysis
method developed by Stodola [5–7] for beams with
variable cross sections. The analysis, using this method,
converges rapidly, making it a very practical tool for
approximating the fundamental natural frequency of a
16.2 Earthquake Loading
120
110
100
45
40
35
Discussion
References
Maps prepared by United States Geological Survey (USGS)
in collaboration with the Federal Emergency Management
Agency (FEMA)-funded Building Seismic Safety Council
(BSSC) and the American Society of Civil Engineers (ASCE).
The basis is explained in commentaries prepared by BSSC
and ASCE and in the references.
Ground motion values contoured on these maps incorporate:
• a target risk of structural collapse equal to 1% in 50 years
based upon a generic structural fragility
• a factor of 1.3 to adjust from a geometric mean to the
maximum response regardless of direction
• deterministic upper limits imposed near large, active
faults, which are taken as 1.8 times the estimated median
response to the characteristic earthquake for the fault (1.8
is used to represent the 84th percentile response), but not
less than 60% g.
As such, the values are different from those on the uniformhazard 2008 USGS National Seismic Hazard Maps posted at:
http://earthquake.usgs.gov/hazmaps.
Larger, more detailed versions of these maps are not
provided because it is recommended that the corresponding
USGS web tool (http://earthquake.usgs.gov/designmaps or
http://content.seinstitute.org) be used to determine the
mapped value for a specified location.
Building Seismic Safety Council, 2009, NEHRP Recommended
Seismic Provisions for New Buildings and Other Structures:
FEMA P-750/2009 Edition, Federal Emergency Management
Agency, Washington, DC.
Huang, Yin-Nan, Whittaker, A.S., and Luco, Nicolas, 2008,
Maximum spectral demands in the near-fault region,
Earthquake Spectra, Volume 24, Issue 1, pp. 319-341.
Luco, Nicolas, Ellingwood, B.R., Hamburger, R.O., Hooper,
J.D., Kimball, J.K., and Kircher, C.A., 2007, Risk-Targeted
versus Current Seismic Design Maps for the Conterminous
United States, Structural Engineers Association of California
2007 Convention Proceedings, pp. 163-175.
Petersen, M.D., Frankel, A.D., Harmsen, S.C., Mueller, C.S.,
Haller, K.M., Wheeler, R.L., Wesson, R.L., Zeng, Yuehua,
Boyd, O.S., Perkins, D.M., Luco, Nicolas, Field, E.H., Wills,
C.J., and Rukstales, K.S., 2008, Documentation for the 2008
Update of the United States National Seismic Hazard Maps:
U.S. Geological Survey Open-File Report 2008-1128, 61p.
30
Figure 16.2 S1 , maximum considered earthquake (MCER ) ground motion parameter for 1 s spectral response acceleration, 5% of critical
damping, Site Class B in the contiguous United States. Source: Courtesy of ASCE.
325
326
16 Tall Vessels
90
80
100
0
100
100
0
200
100
200
70
300
300
400
400
50
500 mi
500 km
45
40
35
Explanation
Contour intervals (%g)
125
100
75
60
50
40
30
25
20
15
10
8
6
4
2
30
Areas with a constant spectral
response acceleration of 60% g
10
10
10
10
Contours of spectral response
acceleration expressed as a percent
of gravity. Hachures point in
direction of decreasing values
+
Point value of spectral response
acceleration expressed as a percent
of gravity
10.7
Figure 16.2 (Continued)
25
16.2 Earthquake Loading
120
110
100
45
40
35
30
Figure 16.3 T L long-period transition period in the contiguous United States. Source: Courtesy of ASCE.
327
328
16 Tall Vessels
90
80
100
0
100
100
0
200
100
200
70
300
300
400
400
50
500 mi
500 km
45
40
35
30
Contours of peak ground acceleration
expressed as a percent of gravity.
Hachures point in direction of
decreasing values
Explanation
Figure 16.3 (Continued)
25
16.2 Earthquake Loading
L1
L2
1
L3
2
3
4
#
240 000
L1
L2
L3
20′
C1
C2
t = 3″
C3
10′
C4
F2
F3
C5
F4
10′
t = 3/4″
#
12 000
Figure 16.5 Simply supported vessel with equivalent forces at the
nodal points.
Figure 16.4 Vertical pressure vessel.
pressure vessel. The method consists of assuming first a
deflected shape of the beam. Then the distributed mass
of the beam multiplied by the square of the frequency
𝜌 and by the assumed deflection is applied as an inertia
force. This distributed inertia force, at various segments
of the beam, is replaced by equivalent concentrated loads
at nodal points for ease of calculations as illustrated in
Figure 16.5. These concentrated forces are then used to
determine the shear and moment diagrams along the
beam. The moment diagram, divided by the quantity
EI along the length of the beam, is then assumed as
a distributed load applied on a conjugate beam. This
distributed load is replaced by equivalent concentrated
loads at the nodal points. The moment diagram obtained
from the concentrated loads on the conjugate beam
is taken as the new deflection of the beam. This new
deflected shape is compared with the original assumed
curve. If they are not the same, then a second iteration
is performed using the new calculated deflection as
a starting point. If the calculated deflection curve is
within reasonable accuracy of the assumed curve, then
the analysis is complete. The parameter of the original
assumed deflection is equated to the parameter of the
calculated deflection to obtain the natural frequency 𝜌.
The aforementioned procedure is applicable to vessels
with multiple thicknesses and weight distribution. The
procedure is straightforward and is easily adaptable to a
spreadsheet program. This procedure is described for a
vertical vessel divided into three sections for illustration
purposes, as follows.
Assume a cantilever beam divided into three lengths
L1 , L2 , and L3 as shown in Figure 16.5a. Assume the force
distribution in the beam is as shown in Figure 16.5b. This
force can be transferred to nodal points 2–4 using the
equations
F2 = C1 L1 ∕3 + C2 L2 ∕3 + C3 L2 ∕6
(16.7a)
F3 = C2 L2 ∕6 + C3 L2 ∕3
+ C4 L3 ∕3 + C5 L3 ∕6
F4 = C4 L3 ∕6 + C5 L3 ∕3.
(16.7b)
(16.7c)
These equations will be used in the following two
examples to illustrate the Stodola procedure for obtaining the natural frequency of a vertical vessel with
different thicknesses and weight distribution.
Example 16.2
Find the fundamental natural frequency and period
of vibration of a cantilever beam of uniform cross
section, Figure 16.6. Let L = length, in., EI = constant,
lbs-in.2 , w = uniform distributed weight, lbs/in. of length,
m = mass = w/386.4, lbs-s2 /in.2
Solution:
After a number of trials, the deflection of the beam is
assumed as shown in line (1) of Figure 16.6. The deflections at points 2–4 are assumed as 1.77, 5.60, and 10 in.,
respectively, multiplied by a constant K. The mass of the
beam multiplied by this deflection and multiplied by the
quantity 𝜌2 is assumed as an applied distributed inertia
load as shown in line (2), where 𝜌 is the natural frequency.
The equivalent concentrated forces due to the distributed
329
330
16 Tall Vessels
L/3
L/3
2
1
(1)
L/3
δ
3
4
K
1.77
5.60
10.00
Kmϱ2
(2)
1.77
5.60
10.00
F2 = 0.704
F3 = 1.896 F4 = 1.421
Kmϱ2L
(3)
V
(4)
1.421
Kmϱ2L
3.317
4.021
From structural analysis, the moment diagram divided
by EI is applied on a conjugate beam as an equivalent
load. The shear due to this load is the rotation of
the actual beam, and the moment due to this load is
the deflection of the actual beam. Also, the fixed and
free ends of the actual beam become free and fixed,
respectively, in a conjugate beam.
Line (6) shows the forces at points 1–3 of the equivalent M/EI load. Lines (7) and (8) show the rotation and
deflection diagrams. Point 4 of the deflection diagram
is adjusted to match the deflection assumed in line (1)
as shown in line (9), and all other points in line (8) are
adjusted accordingly as shown in line (9). The average
variation of the deflection at points 2 and 3 is about 5%.
Thus, the analysis is deemed complete. A slightly more
accurate result can be obtained by performing a second
cycle of analysis using the shape of line (9) as a starting
point.
The natural frequency for the beam is obtained by
equating the maximum deflections in lines (1) and (8).
10 K = 0.837 Km𝜌2 L4 ∕EI.
(5)
M
Kmϱ2L2
EI
0.473
1.578
2.917
F3 = 0.193
Kmϱ2L3
EI
(6)
1
2
3
0.953
Kmϱ2L3
EI
1.146
δ
0.138
0.455
0.837
(9)
δ′
1.65
and T = 1.8(mL4 /EI)0.5 s.
It is of interest to note that the frequency obtained from
theoretical analysis Eq. (16.6a) is
f = 0.56(EI∕mL4 )0.5 cps,
4
θ
0.414
(8)
𝜌 = 3.46 (EI∕mL4 )0.5 rad∕s
f = 0.55(EI∕mL4 )0.5 cycles∕s(cps)
F1 = 0.414 F2 = 0.539
(7)
or,
Kmϱ2L4
EI
K
5.44
10
Figure 16.6 Natural frequency calculation procedure in a simply
supported vessel.
load at points 2–4 are calculated from Eqs. (1)–(3) and
are shown in line (3). Lines (4) and (5) show the shear
and moment diagrams. In line (5), the moment is divided
by the quantity EI.
(16.8)
which is essentially the same as that obtained from the
numerical analysis.
Example 16.3
A vertical vessel has the dimensions shown in Figure 16.4.
The skirt weighs 12,000 lbs or 100 lbs/in., and the weight
of the shell, heads, and fluid is 240,000 lbs or 1000 lbs/in.
Moment of inertial of the shell is 2.074 × 106 in.4 ,
and moment of inertial of the skirt is 0.518 × 106 in.4 .
Modulus of elasticity of all components is 30 × 106 psi.
Determine the fundamental natural frequency and
fundamental period of the vessel.
Solution:
Mass of the shell = 1000/386.4 = 2.588 lbs-s2 /in.2
Mass of the skirt = 100/386.4 = 0.259 lbs-s2 /in.2
EI of shell = 6.222 × 1013 lbs-in.2
EI of skirt = 1.554 × 1013 lbs-in.2
After a number of trials, the deflection of the beam is
assumed as shown in line (1) of Figure 16.7. The mass
16.3 Wind Loading
120″
Mass
120″
1 0.259
(1)
δ
120″
2 2.588
2.588 4
3
K
2.00
5.88
10.00
(2)
0.518
K ϱ2
5.176
15.217
25.88
F2 = 532.1 F3 = 1838.5 F4 = 1339.5
K ϱ2
(3)
V
(4)
3178.0
3710.1
K ϱ2
1339.5
of the beam multiplied by this deflection and multiplied
by the quantity 𝜌2 is assumed as an applied distributed
inertia load as shown in line (2), where 𝜌 is the natural
frequency. The equivalent concentrated forces due to the
distributed load at points 2–4 are calculated from Eqs.
(1)–(3) and are shown in line (3). Lines (4) and (5a) show
the shear and moment diagrams. In line (5b), the moment
is divided by the quantity EI.
Line (6) shows the forces at points 1–3 of the equivalent M/EI load. Lines (7) and (8) show the rotation and
deflection diagrams. Point 4 of the deflection diagram
is adjusted to match the deflection assumed in line (1)
as shown in line (9), and all other points in line (8) are
adjusted accordingly as shown in line (9). There is no
deviation between the calculated and original deflection
curves. Thus, the analysis is complete.
The natural frequency for the beam is obtained by
equating the maximum deflections in lines (1) and (8).
10 K = 1947.6 K𝜌2 ∕106
or
(5a)
M
𝜌 = 71.7 rad∕s
K ϱ2
160,740
f = 11.4 cps.
542,100
987,310
(5b)
M
EI
and T = 0.088 s.
2.583
8.713
63.534
K ϱ2
109
Problems
34.884
16.1
What is the total lateral earthquake force using
Eq. (16.2) and the maximum bending moment
for the following vessel? A vertical vessel with flat
ends is supported on the lower head. The shell is
a seamless steel pipe with 36 in. inside diameter,
1.0 in. thick, and 40 ft. long with 3 in. thick flat
heads. The vessel contains gas at ambient temperature. The vessel is installed in an earthquake zone
with the following parameters: I p = 1.0, F a = 1.0,
Ss = 1.8, S1 = 0.9, T L = 12.0.
16.2
What is the total lateral earthquake force using Eq.
(16.4) for the vessel mentioned in Problem 16.1.
F1 = 3239.0 F2 = 3066.2 F3 = 380.9
(6)
1
(7)
2
4
θ
3239
(8)
3
δ
6305.2
388.7
K ϱ2
109
K ϱ2
109
6686.1
K ϱ2
106
1145.3
1947.6
(9)
δ′
2.00
K
5.88
10
Figure 16.7 Natural frequency calculation procedure in a simply
supported vessel with variable thickness.
16.3 Wind Loading
Two distinctly different kinds of design considerations
result from wind loading. First, the static force from the
wind-loading pressure against the vessel causes an overturning moment that must be considered in designing tall
vessels. The second consideration is the dynamic effect
from vortex shedding of wind passing around the vessel.
331
332
16 Tall Vessels
16.3.1
External Forces from Wind Loading
16.3.2
As with earthquake loading, many different design procedures determine the wind loading and its application to a
vessel. A commonly used standard is ASCE 7-10. When
ASCE 7-10 is used, the basic equation for determining
the velocity pressure qz at various heights z is calculated
from
qz = 0.00256 Kd Kz Kzt V 2
(psf),
(16.9)
where
K d = wind directionality factor and is taken as 0.95 for
pressure vessels
K zt = topographic factor obtained from ASCE 7-10
Section 26.8.2. It takes into consideration the
effect of escarpments and hills on wind velocity
and gusts.
K z = velocity pressure exposure coefficient
V
= wind velocity, mph.
If the analysis is by the ASD method, then the value of
qz may be multiplied by the factor 0.6.
The value of K z is obtained from the equations
Kz = 2.01(15∕zg )2∕𝛼
Kz = 2.01(z∕zg )2∕α
for z < 15 ft
for 15 ft ≤ z ≤ zg ,
where 𝛼 and zg are obtained from Table 16.3.
Charts for wind velocities for structure risk categories
I to IV are given in ASCE 7-10. Building categories are
defined as follows:
Category I: structures that represent low risk to human
life in the event of failure
Category II: structures not listed in categories I, III, or IV.
Category III: structures that represent a substantial risk
to human life in the event of failure
Category IV : structures designated as essential facilities
such as hospitals and police stations
Figure 16.8 shows an ASCE wind velocity chart for
Categories III and IV structures. For ASCE wind velocity charts for Category I and II structures, see Figures
26.5-1C and 26.5-1A of ASCE 7-10.
Table 16.3 Values of 𝛼 and zg .
𝜶
Z g (ft)
B
7.0
1200
C
9.5
900
D
11.5
700
Exposurea)
a) For the definition of the exposure
categories, see Section 26.7.3 of ASCE 7-10.
Dynamic Analysis of Wind Loads
When a laminar wind flows by a circular pressure
vessel, the air stream or wake behind the vessel is no
longer smooth. There is a region of pressure instability
in which vortices are shed in a regular pattern. These
vortices cause an alternating force perpendicular to
the wind direction that can make the vessel vibrate.
When the frequency of the vortex shedding coincides
with the natural frequency of the vessel, a resonance
is caused with increasing amplitude. To prevent this
condition, the natural frequency of the vessel is set
higher than the vortex-shedding frequency determined
by the maximum velocity of laminar wind at the vessel
location [8]. The resonant wind velocity is related to the
height-to-diameter ratio of a cylindrical vessel, as shown
in Figure 16.9.
The natural frequency of the vessel, f n , and the frequency of vortex shedding, f v , are given by the following
equations:
1
fn =
= natural frequency of vessel(cps)
T
(16.10)
0.2Vr
fv =
= frequency of vortex shedding (cps),
D
(16.11)
where
T = period of vibration from Eq. (16.6a)
V r = maximum velocity of laminar wind or resonant
velocity (ft/s) (see Figure 16.9)
D = outside diameter of vessel (ft).
Then we require
fn > vv .
(16.12)
In addition to the vortex shedding, the vessel is
examined for ovalling vibration as a ring. The ovalling
frequency f o will be at least twice the vortex-shedding
frequency f v as given by the following equations:
683t
= ovalling frequency (cps)
D2
t = nominal thickness of shell (in.).
fo =
(16.13)
If f o is less than 2f v , add a stiffening ring within 24 in.
of both shell-to-head junctions, and additional stiffening
rings evenly spaced to keep the span a maximum of 80 ft
and to make
fo > 2fv .
(16.14)
Example 16.4
A vertical vessel made with a cylindrical shell and hemispherical heads is to be installed out of doors near Corpus
Christi, Texas. The shell has 5 ft. 0 in. inside diameter,
16.3 Wind Loading
115(52)
1
15(52)
120(54)
20(54)
160(72)
150(67)
140(63)
130(58)
120(54)
115(51)
115(51)
120(54)
130(58)
140(63)
150(67)
165(74)
165(74)
Figure 16.8 Wind velocity for Categories III and IV structures. Source: Courtesy of ASCE.
333
334
16 Tall Vessels
120(54)
130(58)
140(63)
150(67)
160(72)
150(67)
120(54)
160(72)
170(76)
150(67)
160(72)
170(76)
180(80)
180(80)
190(85)
120(54)
160(72)
130(58) 150(67)
140(63)
200(89)
200(89)
Special wind region
Vmph
(m/s)
Guam
210
(94)
Virgin Islands
175
(78)
American Samoa
Special wind region statewide
Hawai –
170
145
(76)
(65)
Location
160(72)
170(76)
180(80)
Puerto Rico
Figure 16.8 (Continued)
1.0 in. nominal wall thickness, and 100 ft 0 in. length,
tangent to tangent. The contract specification requires
the vessel to be designed according to the ASCE 7-10
Standard. What are the lateral wind forces to be used for
design. Use K zt = 1.0 and Exposure D and Category III.
Solution:
Figure 16.8 shows that Corpus Christi, Texas, is located
in a 160 mph wind zone. Therefore, the wind forces at
various locations using the ASD method are calculated
as follows:
The wind pressure is obtained from Eq. (16.9) and
Figure 16.8 as
qz = (0.6)[0.00256(0.95)Kz (1.0)(160)2 ] = 37.4 Kz .
The hemispherical top head projection can be approxi′
mated as an equivalent cylinder with length L as follows:
Outside diameter Do = inside diameter + 2t = 5 +
2(1/12) = 5.167 ft.
Projected area of the hemispherical head = projected
area of the equivalent cylinder
πD2o ∕(4)(2) = Do L′
16.3 Wind Loading
53.7 psf
150
140
Maximum velocity of laminar wind, Vr (ft/s)
130
Vr =
120
t = 1″
15,106.7
(H/D)2
110
F2
100
90
100′
80
70
ID = 5″
60
h2
50
40
38.5 psf
F1
30
20
h1
10
0
10 12 14 16 18 20 22 24 26 28 30
Height to diameter ratio (H/D)
Figure 16.9 Resonant wind velocity, V r , versus H/D.
′
or, L = πDo /8 = 2.03 ft.
Overall length of the cylinder = 100 + 2.03 = 102.03 ft.
The values of K z are obtained from Table 16.3 as follows.
For z < 15 ft,
Kz = 2.01(15∕700)2∕11.5 = 1.030
qz = 37.4(1.03) = 38.5 psf.
F1 = 38.5Do L = 38.5[(5.0 + 2(1)∕12)](15)
= 2980 lbs.
Kz = 2.01(z∕700)2∕11.5 = 0.643 z0.174
) = 24.0 z
0.174
psf.
The distribution of this pressure is shown in Figure
16.10. The distribution is nonlinear. The force due to this
pressure can be obtained in one of two ways. The first
is to approximate the distribution by a series of straight
line segments and then calculate the area of each of these
segments. The second way is more direct and consists of
integrating the aforementioned equation over the length.
Hence,
102.03
F2 = 5.167
∫15
F = 2980 + 21 560 ≈ 24 540 lbs.
Example 16.5
What is the overturning moment due to the lateral
wind forces assuming that the vessel is supported at the
lower shell-to-head junction for the vessel mentioned in
Example 16.4.
h1 = 7.5 ft.
The moment arm for force F 2 is obtained by integrating
the equation
For 15 ft ≤ z ≤ 102.03 ft,
qz = 37.4(0.643 z
The total wind force is
Solution:
From Figure 16.10, the moment arm for force F 1 is
The force is
0.174
Figure 16.10 Vertical vessel subjected to wind load.
24.0 z0.174 dz = 21,560 lbs.
102.03
h2 = (5.167∕21,560)
∫15
24.0(z0.174 )(z)dz
= 60.6 ft.
The total bending moment is
M = h1 F1 + h2 F2
= (7.5)(2980) + (60.5)(21,560)
= 1,328,900 ft-lbs.
Example 16.6
Determine if the vessel described in Example 16.4 is adequate to resist vortex shedding and ovalling vibration.
335
336
16 Tall Vessels
Solution:
16.5
1) Determine the total dead load weight of the vessel:
Shell = 𝜋(312 − 302 )(1200)(490∕1728) = 65,210 lb
4
Heads = 𝜋(313 − 303 )(490∕1728) = 3 320 lb
3
Total weight = W = 68,530 lb
A pressure vessel is 10 ft in inside diameter, 2.0 in.
thick, and 150 ft tall. Determine if the vessel design
is adequate to resist ovalling vibration.
Answer:
The design is not adequate, because f v = 0.11
and f o = 0.09; consequently, f o does not equal or
exceed 2f v .
2) Determine these properties:
w=
68,530
W
=
= 57.1 lb∕in.
h
1200
6
E = 30 × 10 psi
16.4 Vessel Under Internal Pressure
Only
I = 0.049(624 − 604 ) = 89,000 in.4
√
(57.1)(1200)4
= 0.6047
T = 0.0908
(30 × 106 )(89,000)
1
1
fn =
=
= 1.65 cps
T
0.6047
(100)(12)
H
=
= 19.35
D
(62)
Vr = 40.33 ft∕s
0.2 Vr
(0.2)(40.33)
fv =
=
= 1.56 cps
D
5.17
fn > fv ∶ 1.65 > 1.56 (acceptable)
683(1)
fo =
= 25.55 cps
(5.17)2
fo > 2fv ∶ 25.55 > 3.12 (acceptable)
For a tall vessel under internal pressure only, the primary
additional consideration to the internal pressure is the
effect of fluid pressure head and the dead load. This is
especially important at the bottom of a vessel, where the
effects may combine. The fluid pressure head may occur
only during hydrostatic testing of the vessel, or it may be
a continuing load occurring during operation of the vessel. The additional pressure caused by the fluid head is
calculated as follows:
H𝛾
,
(16.15)
Pf =
144
where
Pf = additional internal pressure effect from fluid
pressure head (psi)
H = height of fluid column above point (ft)
𝛾 = density of fluid (lb/ft3 ).
Problems
16.3
A tall vessel constructed with a cylindrical shell
and flat closure ends is to be installed near Denver,
Colorado. The inside diameter of the cylindrical
shell is 8 ft, the nominal wall thickness is 1.0 in.,
and the straight length from head weld seam to
head weld seam is 125 ft. The flat heads have 6.0 in.
nominal thickness. What is the total lateral wind
force used for the design of the vessel following the
rules of the ASCE 7-10 Standard Code?
Answer:
Lateral wind force = 14 060 lb.
16.4
Based on a support line at the lower head, what
is the overturning moment from the lateral wind
force for the vessel mentioned in Problem 16.3?
Answer:
M0 = 11,682,400 in.-lb.
If the fluid head exists in the vessel during operation,
the value of Pf is added to the internal pressure when the
minimum required thicknesses are set. At the bottom of
the vessel, the stresses and minimum required thickness
are set by the total pressure. It may be possible to decrease
the thickness when the fluid-head effect is decreased in a
vessel where a variation in plate thicknesses is acceptable.
If the fluid head exists in the vessel only during hydrostatic testing, the primary membrane stress from the
combination of the hydrostatic test pressure and the
fluid-head pressure may go as high as the yield strength
of the vessel material at the test temperature. However, if
the resulting minimum required thickness from the combination is indicated as more than the thickness required
for the normal design conditions, substitution of a pneumatic test or a combination hydrostatic-pneumatic test
should be considered. In general, the minimum required
thickness of a vessel should never be set by the requirements of the hydrostatic head unless it is impossible to
test it any other way. Also, remember that a hydrostatic
test may use fluids other than water if water causes a
problem such as corrosion.
16.4 Vessel Under Internal Pressure Only
For a vessel under internal pressure only, there are,
of course, always loadings from not only the contents
of the vessel but also its own weight to be considered.
Whether the stresses caused by these additional loadings
are tensile or compressive depends upon the location of
the external supports or support skirt location.
In the actual design, the minimum required thickness
is initially set by the circumferential stress given in Eq.
(5.1) as follows:
PR
.
(16.16)
t
From the equation in UG-27(c)(1) of the ASME Code,
VIII-1, expressed in terms of SE, the new equation is
(
)
R
SE = P
+ 0.6 ,
(16.17)
t
where
𝜎𝜃 =
S = allowable tensile stress (psi)
P = internal design pressure (psi)
R = inside radius (in.)
t = minimum required thickness (in.).
Using this equation, a tentative minimum required
thickness is set, based on the circumferential stress.
When the final required thickness is determined, it may
be necessary to include the effect of the fluid head as
well as the internal design pressure. Based on membrane stress equations, the total longitudinal stress is
determined from the following equation:
PR
W
±
,
2t
πDm t
Example 16.7
A vertical vessel with a cylindrical shell and hemispherical heads is installed inside a building. The shell has 5 ft
inside diameter and 0.5 in. nominal thickness, and is 30 ft
from tangent to tangent. The vessel contains a fluid at
35 lbs/ft3 . Determine the total longitudinal stress in the
cylindrical shell above and below the support lines, which
is at the lower shell-to-head junction. SE = 15,000 psi.
Solution:
Assume that the internal pressure is set by Eq. (16.17) for
a value of SE = 15,000 psi. Rearranging the terms gives
15,000(0.5)
SEt
=
= 245 psi.
P=
R + 0.6t
30 + 0.6(0.5)
The dead load of vessel above the support line is
E = weld joint efficiency (E = 1.0 for seamless)
𝜎L = +
the terms are both tensile. For some arrangements, the
condition without internal pressure may be more critical
than when internal pressure is considered.
If 𝜎 L is positive, the actual stress is positive, and the
allowable stress is determined from the allowable-tensilestress tables. If the value of 𝜎 L is negative, the allowable
stress is determined by the method that establishes
the maximum allowable axial compressive stress in a
cylindrical shell.
(16.18)
where
𝜎 L = total longitudinal stress (psi)
W = total dead load of vessel and contents acting on
the plane being examined (lb). This includes static
head of contents.
Dm = mean diameter of shell (in.) = 2R + t
From the equation in UG-27(c)(2) of the ASME Code,
VIII-1, with the terms rearranged and the dead-load term
added, the following equation is obtained:
(
)
W
R
𝜎L = +P
− 0.2 ±
.
(16.19)
2t
πDm t
In both Eqs. (16.18) and (16.19), the dead-load term
may be either tension or compression, depending upon
the plane being examined. In general, above the support
line, this term is compressive, and the total longitudinal
stress is the difference between the internal-pressure
effect and the dead-load effect. Below the support line,
Shell = π(30.52 − 302 )(360)(490∕1728)
= 9700 lb.
Upper head = (4∕3)π(30.53 − 303 )
× (490∕1728)(1∕2) = 815 lb.
Total weight = 9700 + 815 = 10,515 lb.
The longitudinal stress using Eq. (16.19) is
(
)
10,515
30
− 0.2 −
𝜎L = +245
2 × 0.5
π(60.5)(0.5)
𝜎L = +7300 − 110 psi
⎧7190 psi tension
with internal
⎪
⎪
pressure
𝜎L = ⎨
.
110
psi
compression
without internal
⎪
⎪
pressure
⎩
The dead load of vessel and contents below the support
line is
Fluid in shell = π(30)2 (360)(35∕1728)
= 20,620 lb.
Fluid in heads = (4∕3)π(30)3 (35∕1728)
= 2290 lb.
Weight of lower head = 815 lb.
Total weight = 20,620 + 2290 + 815
= 23,725 lb.
337
338
16 Tall Vessels
The longitudinal stress, using Eq. (16.19), is
(
)
23,725
30
𝜎L = + 245
− 0.2 +
2 × 0.5
π(60.5)(0.5)
𝜎L = + 7300 + 250 = 7550 psi tension
with internal pressure.
Example 16.8
For the vessel described in Example 16.7, determine the
fluid pressure head for containing fluid at 35 lb/ft3 and
for hydrostatic testing using water at 62.4 lb/ft3 .
Solution:
Fluid pressure head is needed at both the lower
shell-to-head tangent line and the lowest point of
the lower head.
At the lower shell-to-head tangent line, the fluid height
is 30 ft + 2.5 ft = 32.5 ft.
(32.5)(35)
= 7.9 psi
For fluid Pf =
144
(32.5)(62.4)
For water Pf =
= 14.1 psi
144
At the bottom of the lower head, the fluid height is
30 ft + 5 ft = 35 ft.
(35)(35)
For fluid Pf =
= 8.5 psi
144
(35)(62.4)
For water Pf =
= 15.2 psi
144
Problems
16.6
What is the circumferential stress based on the
ASME Code, VIII-1, at the bottom of a tall vessel
that contains fluid at 50 lb/ft3 and an internal
design pressure of 400 psi? The vessel has 36 in.
inside diameter by 0.5 in. minimum wall by 45 ft
0 in. overall length with 3 in. thick flat heads on
each end. E = 1.0, and the vessel is supported at
the bottom.
Answer:
𝜎 𝜃 = 15,200 psi
16.7
What is the maximum stress in the shell of the vessel mentioned in Problem 16.6 during a standard
hydrostatic test of 1.5P?
Answer:
𝜎 𝜃 = 22,700 psi
16.8
What is the maximum longitudinal stress in the
vessel mentioned in Problem 16.6 with and without considering the internal design pressure?
Answer:
𝜎 L = 7110 psi tension with internal pressure considered
𝜎 L = 170 psi compression without considering
internal pressure
16.5 Vessel Under Internal Pressure
and External Loading
When a tall vessel supported in the vertical position is
subjected to internal pressure and external loading from
such sources as earthquake or wind, both the tension
and compression sides of the cylinder must be examined.
These items are similar to those for a tall vessel under
internal pressure only, except for the latter, only one of
the sides needs examination. The earthquake loading or
the wind loading is resolved into an overturning moment
that is further resolved into tensile and compressive
loads.
Using the membrane stress equations given in Eq.
(16.18), the total longitudinal stress is determined
from
4Me
PR
W
.
(16.20)
𝜎L = +
±
±
2t
πDm t πD2m t
And in terms of the present ASME Code, VIII-1, the
total longitudinal stress is obtained from
(
)
4Me
W
R
𝜎L = +P
.
(16.21)
− 0.2 ±
±
2t
πDm t πD2m t
As can be seen from this equation, many different
combinations of stresses must be examined to obtain
the maximum stress: with the first term equal to zero
when there is no internal pressure (vessel not operating)
or with internal pressure; with the second term being
either tension or compression; with the third term being
either tension or compression; and so on. There may
be more combinations because the value of W , the
dead load, depends upon the location of the plane that
is being considered. Usually, the maximum stress is
located at the support line; however, when the vessel
has various diameters, it may be necessary to examine
stresses at different cross sections. This, of course, is in
addition to any local stresses in the shell caused by the
support.
Example 16.9
The vessel described in Example 16.7 is to be supported at
the lower shell-to-head junction. What are the longitudinal stresses in the shell above and below the support line?
The internal pressure is 225 psi. Let the bending moment
Me = 864,000 in.-lb.
16.5 Vessel Under Internal Pressure and External Loading
Solution:
From Example 16.7, the dead load of the various components is
shell = 9700 lb
The overturning moment due to the wind loading at the
support line is determined from Example 16.5 as
M0 = (2,216,950)(12) = 26,603,340 in.-lb.
Determine the total longitudinal stress 𝜎 L using Eq.
(16.21) as follows:
(
)
66,870
30
𝜎L = + 475
− 0.2 ±
2×1
π(61)(1)
4(26,603,340)
±
π(61)2 (1)
𝜎L = + 7030 ± 350 ± 9100.
heads = 1630 lb
shell fluid = 20,620 lb
fluid, heads = 2290 lb.
The overturning moment is
Me = 864,000 in.-lb.
Using Eq. (16.21), the total longitudinal stress is
(
)
W
30
− 0.2 ±
𝜎L = +225
2 × 0.5
π(60.5)(0.5)
(4)(864,000)
±
.
π(60.5)2 (0.5)
The side of applied force above the support line has
dead load = shell + upper head = 9700 + 815
= 10,515 lb
𝜎L = + 6705 − 110 + 600
= 7200 psi tension.
The side of applied force below the support line has
dead load = lower head + contents
Windward-Side Stresses
1) Pressure + dead load + wind load = + 7030 − 350 +
9100 = 15 780 psi tension.
2) No pressure + dead load + wind load = 0 − 350 +
9100 = 8750 psi tension.
3) Dead load only = 350 psi compression.
Leeward-Side Stresses
1) Internal pressure + dead load + wind load = +
7030 − 350 − 9100 = 2420 psi compression.
2) No internal pressure + dead load + wind load = 0 −
350 − 9100 = 9450 psi compression.
Allowable Stress – Tension
S = 15,800 psi for SA-516 Grade 60 at 650∘ F
t
Allowable Stress – Compression
= 815 + 22,910 = 23,725 lb
A=
𝜎L = + 6705 + 250 + 0
= 6955 psi tension.
The opposite side of applied force above the support
line has
𝜎L = + 6705 − 110 − 600
{
5600 psi tension
=
710 psi compression
with pressure
.
without pressure
and from Figure 8.11 and also Figure CS-2 of the ASME
Code, II-D, the value of B is 11,500. This gives an allowable stress of Sc = 11,500 psi. All calculated stresses are
less than the allowable stresses.
Problems
16.9
A tall vessel is constructed of a cylindrical shell
with a flat head on each end. The shell has 4 ft
inside diameter by 2 in. thickness by 75 ft length
from end to end. The flat heads are 7.5 in. thick.
The vessel is supported on the bottom, which
rests on structural supports that are 75 ft from
the ground level to the support line of the vessel. The wind zone is 120 mph, and the ASCE
7-10 standard prevails. What are the longitudinal
stresses from wind loading on both the windward
and leeward sides above the support line?
16.10
Assume that the vessel described in Problem 16.9
is supported on the ground instead of 75 ft in the
air. What are the longitudinal stresses from the
wind loading in the shell above the support line?
Example 16.10
The vessel given in Example 16.3 is to be constructed
from SA-516 Grade 60 material and designed for 475 psi
at 650 ∘ F. The weld joint efficiency is E = 1.0. What are
the total longitudinal stresses on both the windward and
leeward sides at the support line? What are the allowable
tensile and compressive stresses?
Solution:
From 1A of the ASME Code, II-D, for SA-516 Grade 60 at
650 ∘ F, the allowable tensile stress is St = 15 800 psi. The
weight of the shell and upper head above the support line
is determined from Example 16.6 as
W = 65,210 + 0.5(3320) = 66,870 lb.
0.125 0.125
=
= 0.0040,
Ro ∕t
31∕1
339
16 Tall Vessels
16.6 Vessel Under External Pressure
Only
600
500
400
𝛼=
Pv
Po Do
Pv =
W
πDm
1.23D2o
.
(16.24)
L2
By applying Sturm’s equation [10] for the ratio between
the external pressure alone and the axial compressive
loadings, an equation can be developed that gives an
equivalent external pressure Po′ for the combined loading as a multiple of the base external pressure Po . This
equation is
m=
Po′ =
𝜂 2 − 1 + m + m𝛼
Po .
𝜂2 − 1 + m
Pve = 0.25Po Dm .
6
3
4
2
η=
30
20
0.2 0.3
0.5
(16.26)
Figure 16.11 Number of lobes, 𝜂, into
which a shell will collapse when
subject to uniform external pressure
on sides and ends.
60
50
40
10
0.1
(16.25)
For 𝛼 > 1.0, the vessel may fail by yielding and should
also be checked as a cantilever beam, including the
axial-stress effect due to the external pressure. The axial
load from the external pressure is
5
100
80
(16.23)
and the compression term m is
7
200
(16.22)
where the axial-compression unit load is
8
300
A comparison of results between pressure on the sides
only and pressure on the sides and ends indicates that the
value of 𝜂 changes very little for values of 𝛼 ≤ 1.0, where
the ratio 𝛼 is the axial-compression unit load divided by
the allowable external pressure that is permitted when it
is acting alone. Expressed in an equation,
9
1000
800
14 15
12 13
10 11
For a tall vessel under external pressure only, in addition to the basic considerations for external-pressure
design given in Chapter 8, the effect of fluid pressure
and dead load shall be considered. This is very similar to
the conditions considered in Section 16.3 for the vessel
under internal pressure only. When direct loads caused
by the fluid pressure and dead load create a compressive
loading, it has to be combined with the loading from the
external pressure.
The ASME Code, VIII-1, has methods to consider
each of these two types of loadings separately. External
pressure, of course, is described in UG-28, and axial
compressive stress on a cylindrical shell is described in
UG-23(b). However, there is no method given in the
ASME Code that describes how to consider both at the
same time. An arbitrary method to consider these two
loadings simultaneously and use the ASME Code, VIII-1
procedure was developed by Dr. E. O. Bergman in 1954
[9]. This method combines the effects of axial loadings
and external pressure by establishing an adjusted pressure used in the external-pressure calculation procedure
given in UG-28.
Based on the von Mises instability formulas as
discussed in Chapter 8 for a cylinder loaded with
both radial and axial pressure loading, a value of 𝜂 is
determined for values of L/Do and t/Do . This is plotted
in Figure 16.11, from which the value of 𝜂 can be easily
determined for various values of L/Do versus t/Do .
Diameter / thickness (Do /t)
340
0.8 1.0
2
3 4 5 6 8 10
Length to diameter ratio (L /Do)
20
30 40
60 80 100
16.7 Vessel Under External Pressure and External Loading
The axial load from external pressure in Eq. (16.26)
is combined with the axial load from dead loads in Eq.
(16.23) to give the total axial loading on the cylindrical
shell. When this loading is divided by the shell thickness,
the result is the total axial compressive stress on the
cylindrical shell. This actual stress is compared with the
allowable axial compressive stress determined according
to UG-23(b) of the ASME Code, VIII-1.
Example 16.11
A tall vessel is constructed with a cylindrical shell and
two hemispherical heads. The vessel is designed for
full vacuum (15 psi external pressure) at a design temperature of 550 ∘ F. The material is SA-516 Grade 60.
There is no corrosion allowance required. The vessel has
10 ft 0 in. inside diameter with 116 ft 8 in. from tangent
to tangent. The vessel is to be installed in the vertical
position and supported at the bottom tangent line. It
contains a fluid weighing 50 lb/ft3 . Three stiffening rings
are evenly spaced at 30 ft 0 in. with 28 ft 4 in. from each
tangent line. What are the longitudinal stresses above
the support line?
Solution:
Determine a preliminary thickness based on the
external pressure alone, using the procedure of the
ASME Code, VIII-1, UG-28(c)(1). Assume a thickness of t = 0.75 in. Then Do = 120 + 2(0.75) = 121.5 in.
Do /t = 121.5/0.75 = 162, and L/Do = 360/121.5 = 2.9630.
From Figure G of II-D, A = 0.00022.
2(0.000 22)(26.38 × 106 )
= 23.9 psi;
3(162)
3
t = in. (O.K.)
4
Determine the weight of the vessel and contents:
Pa =
shell ∶ 𝜋(60.75 − 60 )(1400)(490∕1728) = 112,950
4
heads ∶ 𝜋(60.753 − 603 )(490∕1728) = 9740
3
shell contents ∶ 𝜋(60)2 (1400)(50∕1728) = 458,150
4
head contents ∶ 𝜋(60)3 (50∕1728) = 26,180.
3
Determine the maximum compressive load (lb/in.)
using Eq. (16.23) as follows:
2
2
112,950 + 0.5(9740)
W
=
= 310.6 lb∕in.
πDm
𝜋(120.75)
Po = 15.0 psi
Pv
310.6
=
= 0.1704
𝛼=
Po Do
(15)(121.5)
1.23
1.23
=
= 0.1401
m=
(L∕Do )2
(2.963)2
Pv =
𝜂 = 3.0 from Figure 16.11 for L∕Do
= 2.963 and Do ∕t = 162
9 − 1 + 0.14 + 0.14(0.1704)
Po′ =
(15) = 15.04 psi.
9 − 1 + 0.14
As determined in step 1, the maximum allowable external pressure based on t = 0.75 in. is Po = 23.9 psi at 550 ∘ F.
Because the required pressure of 15.04 psi is less than the
permissible pressure of 23.9 psi, t = 0.75 in. is satisfactory.
Because 𝛼 is less than 1.0, no cantilever-beam check is
needed.
Problem
16.11
A tall vessel consists of a cylindrical shell with the
lower head hemispherical and the upper head
flat. Because the upper head must support a dead
weight of 20,000 lb from connecting equipment,
it is made 2.5 in. thick. The vessel is designed for
full vacuum at room temperature. The material
is SA-516 Grade 60, and there is no corrosion.
The vessel is to be hydrostatically tested in the
horizontal position and installed in the vertical
position. The vessel is supported at the lower
head-to-shell tangent line. No stiffening rings
are permitted. The vessel has 5 ft 0 in. inside
diameter by 60 ft 0 in. from tangent to tangent.
The vessel contains only a gas during operation.
What is the required thickness of the vessel?
Answer:
Required thickness = 5/8 in.
16.7 Vessel Under External Pressure
and External Loading
A tall, vertically supported vessel, which is subjected to
external pressure and external loading, contains stresses
similar to those in a vessel with external pressure only.
The method of combining loadings as developed by
Bergman and described in Section 16.6 is used to obtain
Po′ used in the analysis.
The only difference between Sections 16.6 and 16.7 is
that an expanded value and additional terms are used in
obtaining Po , which is given in Eq. (16.23):
Pv =
W
4We
4M
+
+
.
πDm πD2m πD2m
(16.27)
This new value of Pv is then used to obtain new values
of 𝛼 and Po′ from Eqs. (16.22) and (16.25).
341
342
16 Tall Vessels
References
1 ANSI/API Standard 620 Recommended Rules
2
3
4
5
for Design and Construction of Large, Welded,
Low-Pressure Storage Tanks. Washington, D.C:
American Petroleum Institute.
ANSI/API Standard 650 Welded Steel Tanks for
Oil Storage. Washington, D.C: American Petroleum
Institute.
ASCE Standard 7-10 (2010). Minimum Design
Loads for Buildings and Other Structures. Virginia:
American Society of Civil Engineers. Reston.
Young, W., Budynas, R., and Sadegh, A. (2012). Roark’
Formulas for Stress and Strain. New York: McGraw
Hill.
Godden, W.G. (1965). Numerical Analysis of Beam
and Column Structures. Englewood Cliffs, NJ:
Prentice Hall.
Bibliography
Windenburg, D.F. and Trilling, C. (1934). Collapse by
instability of thin cylindrical shells under external
pressure. Trans. ASME 56: 819–825.
6 Den Hartog, J.P. (1985). Mechanical Vibrations. Dover
Publications.
7 Biggs, J. (1964). Introduction to Structural Dynamics.
New York, NY: McGraw Hill.
8 DeGhetto, K. and Long, W. (1966). Check towers
for dynamic stability. Hydrocarbon Process. 45 (2):
143–147.
9 Bergman, E.O. (1960). The design of vertical pressure
vessels subjected to applied forces. In: Pressure Vessels
and Piping Design, 576–580. New York: ASME.
10 Sturm, R.G. (1941). A Study of the Collapsing
Pressure of Thin-Walled Cylinders. Bulletin No.
329, University of Illinois, Engineering Experiment
Station.
Rectangular header. Source: Courtesy of Ecodyne MRM Division.
344
17
Vessels of Noncircular Cross Section
17.1 Types of Vessels
Although many kinds of noncircular cross sections may
be used for process vessels, only a few configurations are
used widely. The ASME Code, VIII-1, limits the design
rules for vessels of rectangular cross section as shown
in Figure 17.1 and Figure 17.2 and for obround cross
sections. Some additional rules are given for vessels with
a circular cross section that utilize stay plates to give
added strength. The rules mentioned in this chapter are
limited to vessels with a straight longitudinal axis and
noncircular cross sections. Rules are given elsewhere
for vessels, with or without a circular cross section, that
have no straight longitudinal axis, such as a torus.
Some vessels contain very few openings, whereas
many others contain many openings. If there are only
a few openings, they usually are individually reinforced
by replacing the area removed as described in Chapter
11. In many vessels, multiple openings are calculated
according to the rules for ligaments. If the diameters of
the openings are uniform through the wall thickness, the
ligament efficiency is calculated very similarly to that
for a circular vessel. If the opening consists of several
Figure 17.2 C-shape headers with large-radius corners for
minimum stress concentration and flat weld joint for easy
radiography. Source: Courtesy of Ecodyne MRM Division.
Figure 17.1 Four-plate rectangular header utilizes weld joints at
each corner. Source: Courtesy of Ecodyne MRM Division.
different diameters through the vessel wall thickness,
then depending upon the type of calculated stress (either
direct membrane stress of bending stress), the effective
opening size and the ligament efficiency calculations
are determined in different ways. Vessels with rectangular or square cross sections may be built with many
different combinations of wall thicknesses. There are
combinations where two opposite sides have the same
wall thickness and where the other two opposite sides
have a different wall thickness from the adjacent sides.
This type is often used in air-cooled heat exchangers.
Other vessels may have all sides of different thicknesses,
whereas still others may utilize stay plates to stiffen the
flat sides of the vessel.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
17.4 Ligament Efficiency for Constant-Diameter Openings
Several different combinations are shown in Figure 17.3
for rectangular cross sections without stay plates, in
Figure 17.4 for rectangular cross sections with stay
plates, and in Figure 17.5 for obround and circular cross
sections with and without stay plates. Although the
analysis is similar for vessels with internal pressure and
with external pressure, the effect of stay plates on each
type is different. The dimension dj in the figure is the
distance from mid length of the plate to weld joint or
centerline of row of holes.
17.2 Rules in Codes
Specific rules for the design of vessels of noncircular
cross section have been mentioned in the ASME Code,
VIII-1, since the 1977 Addenda. Prior to that time, the
ASME Code applied several different methods for establishing the maximum allowable working pressure for
vessels of noncircular cross section. In 1963, the ASME
Code contained Code Case 1318 for noncylindrical
pressure vessels. This code case permitted the heads and
covers to be calculated according to UG-34, but required
that the side plates parallel to the longitudinal axis be
calculated according to U-2(g). This, of course, permits
the thickness or maximum allowable working pressure
to be determined analytically or by one of the proof tests
permitted by the ASME Code.
In addition to the rules for vessels of noncircular cross
section given in the ASME Code, VIII-1, design rules for
rectangular vessels are contained in the British Code, the
European Codes [1, 2], and several other codes.
The ligament efficiencies em and eb shall be applied only
to the calculated stresses for the plate containing the ligaments.
When em and eb are less than the butt-weld joint efficiency E, which would be used if there were no ligaments
in the plate, the membrane and bending stresses, which
are calculated based on the gross area of the section,
shall be divided by em and eb , respectively, to obtain
the stresses based on the net area for the section. The
allowable stresses for membrane and membrane plus
bending shall be calculated using E = 1.0.
When em and eb are greater than the butt-weld joint
efficiency E, which would be used if there were no ligaments in the plate, the stresses shall be calculated as
if there were no ligaments in the plate. The allowable
stresses for membrane and membrane plus bending
shall be calculated as described in Section 17.7 using the
appropriate E factor.
17.4 Ligament Efficiency
for Constant-Diameter Openings
For flat plates with constant-diameter openings in a regular pattern as shown in Figure 17.6, the ligament efficiency for both membrane and bending stresses is the
same. When the two openings being considered for setting the ligament efficiency have different diameters, it
is necessary to determine an equivalent diameter of the
openings by averaging their diameters as follows:
DE = 0.5(d1 + d2 ).
The ligament efficiency is then determined as
em = e b =
17.3 Openings in Vessels
with Noncircular Cross Section
For those vessels and headers that contain a limited
number of openings that are not in a regular pattern,
the method of reinforced openings may be applied. For
the flat sides of noncircular vessels, the reinforcing rules
for flat plates apply. If the openings are arranged in a
regular pattern, the method of ligament efficiency may
be applied.
Because the ligament efficiency appears in the
equations both for membrane stress and for bending
stress, for some opening configurations, it is necessary
to determine both the ligament efficiency for membrane stress and the ligament efficiency for bending
stress. In addition, it is necessary to determine the
weld joint efficiency relative to the degree of weld joint
examination.
(17.1)
p − DE
.
p
(17.2)
Example 17.1
What is the membrane and bending ligament efficiency
in a rectangular-cross-section header in which H = 6 in.,
h = 12 in., and t 1 = t 2 = 0.75 in., with a single row of 1.5 in.
diameter holes on 4 in. center-to-center spacing?
Solution:
Using Eq. (17.2), calculate the efficiency as
em = e b =
4 − 1.5
= 0.625.
4
Example 17.2
A single row of openings is alternately spaced on 4 and
3.5 in. center-to-center spacings. The opening diameters
also alternate, with a 1.5 in. diameter opening followed by
a 1.25 in. diameter opening. What is the minimum ligament efficiency for setting the thickness?
345
346
17 Vessels of Noncircular Cross Section
t1
t1
N
h/2
Q1
Q
h/2
P
h/2
t1
h/2
t1
l1
B
dj
t22
(b)
H1
t1
a
N
L
P
H
2
Pitch distance to next
reinforcing member
t1
C
Q
P
h/2
A
dj
h1
M
a > 3 t3
L
h/2
t2
t1
(c)
H
2
t2
(d)
dj
t
E
F
a
C
lL
L
D
θ
t1 = t2
L11
ls
B
P
dj
A
t2
t2
L
M
t2
t2
(a)
D
θ
dj
M1
H
2
l1
P
dj
M
t2
Q
t = t1 = t2
L11
t1 = t2
l11
l1
l1
l11
(e)
Figure 17.3 (a–e) Plain rectangular cross sections. Source: Courtesy of American Society of Mechanical Engineers, from Figure 13-2(a) of
the ASME Code, VIII-1.
17.4 Ligament Efficiency for Constant-Diameter Openings
t1
t1
N
h
N
O
t4
P
h
M
Stay
H
2
t2
h
P
t4
M
Stay
P
t2
P
M
Stay
h
t3
Q
P
h
t2
H
2
t2
t1
t1
(a)
(b)
t1
P
t4
t1
N
t4
η
N
h/2
O
Stay
O
Stay
P
h/2
P
M
M
H
2
h/2
t2
t2
t2
h/2
H
2
t4
P
t2
Stay
t4
P
Stay
P
t4
Stay
t1
t1
(c)
(d)
Figure 17.4 (a–d) Rectangular cross sections with stay plates. Source: Courtesy of American Society of Mechanical Engineers,
from Figure 13-2(a) of the ASME Code, VIII-1.
Solution:
Calculate the equivalent diameter DE to be used in the
ligament-efficiency equation as
Problems
17.1
DE = 0.5(1.5 + 1.25) = 1.375 in.
The ligament efficiency is based on the minimum
spacing of p = 3.5 in. using the equivalent diameter
DE = 1.375 in.:
3.5 − 1.375
= 0.607.
em = e b =
3.5
A header is designed for a ligament efficiency of
em = eb = 0.667. What is the minimum center-tocenter spacing for 1.75 in. diameter openings?
Answer:
d = 5.25 in.
17.2
The flat side plate of a rectangular header contains two rows of 7/8 in. diameter openings. The
rows are 3 in. apart, and the openings are on 8 in.
alternate longitudinal spacing along each row.
347
348
17 Vessels of Noncircular Cross Section
t1
t
C
C
O
B
B
O
t2
L
dj
dj
A
A
P
L
Pitch distance to next
rainforcing member
L
P
(a)
L
(b)
t1
t1
C
O
B
P
P1
t2
t3
L
O
A
Plate member
A
Stay
P
L
t3
(c)
P2
(d)
Figure 17.5 (a–d) Obround and circular cross sections with and without stay plates. Source: Courtesy of American Society of Mechanical
Engineers, from Figures 13-2(b) and (c) of the ASME Code, VIII-1.
The header also contains a longitudinal weld joint
along the center of the same flat side plate. The
weld joint, only visually examined, has a weld joint
efficiency of E = 0.70. What efficiency is used to
calculate the allowable stress?
Answer:
E = 0.70, since it is less than the ligament efficiency.
17.3
If the header mentioned in Problem 17.2 has the
weld joint examined by full radiography, what
efficiencies are used to set the minimum required
thickness?
Answer:
em = eb = 0.825, which is the ligament efficiency, is
used to calculate the applied stress. E = 1.0 is used
to calculate the allowable stress (see Section 17.3).
17.5 Ligament Efficiency for Multidiameter Openings Subject to Membrane Stress
Example 17.3
Determine the membrane ligament efficiency in a
header where t 1 = t 2 = 1.50 in. The header contains a
series of openings on 4 in. centers. The openings are
multidiameter, as shown in Figure 17.8.
Solution:
Calculate the equivalent diameter DE using Eq. (17.4) as
follows:
do
p
p = 4 in.
d0 = 1.625 in. T0 = 0.125 in.
d1 = 1.5 in.
T1 = 1.125 in.
d2 = 1.375 in. T2 = 0.25 in.
1.00
DE =
1.50
× (1.625 × 0.125 + 1.5 × 1.125 + 1.375 × 0.25)
= 1.490 in.
p
d1
Example 17.4
Tubes are expanded into a rectangular header that is
1.25 in. thick. The holes are 0.875 in. diameter with two
0.03125 in. deep grooves in the hole for holding power.
The grooves are 0.125 in. high with 0.25 in. spacing
between them. The top groove is 0.25 in. from the top
edge. What is the membrane ligament efficiency if the
openings are on 3 in. centers?
d2
Figure 17.6 Openings with constant diameter.
17.5 Ligament Efficiency
for Multidiameter Openings Subject
to Membrane Stress
There are many different arrangements of plates with
openings of more than one diameter as shown in
Figure 17.7. For use in air-cooled heat exchangers, the
diameters increase through the plate thickness, whereas
for rolled-in tubes, the larger diameters rolled inside
the tube for holding power. Any arrangement of various
diameters may be considered.
For membrane stress, the ligament efficiency is
em =
p − DE
,
p
DE =
1
(d T + d1 T1 + d2 T2 + · · · + dn Tn ). (17.4)
t 0 0
Solution:
Calculate the equivalent diameter DE using Eq. (17.4):
p = 3 in.
d0 = 0.875 in. T0 = 0.25 in.
d1 = 0.9375 in. T1 = 0.125 in.
d2 = 0.875 in. T2 = 0.25 in.
d3 = 0.9375 in. T3 = 0.125 in.
d4 = 0.875 in. T4 = 0.50 in.
1
DE =
(0.875 × 0.25 + 0.9375 × 0.125 + 0.875
1.25
× 0.25 + 0.9375 × 0.125 + 0.875 × 0.5) = 0.888.
(17.3)
where
Figure 17.7 Openings with more than one
diameter.
p
d0
d1
d2
b0 = P-d0
b1 = P-d1
b2 = P-d2
T0
T1
T2
t
349
17 Vessels of Noncircular Cross Section
p = 4ʺ
t = 1.5ʺ
T1 = 1.125ʺ
T0 = 0.125ʺ
Figure 17.8 Hole details for Example
17.3.
d2 = 1.375ʺ
d1 = 1.5ʺ
d0 = 1.625ʺ
T2 = 0.25ʺ
350
The membrane ligament efficiency is
em =
3 − 0.888
= 0.704.
3
Problems
17.4
A rectangular header contains a series of tube
holes on 3.5 in. centers. The holes are 1 in. diameter with the ends counterbored 0.25 in. deep to
a diameter of 1.125 in. If the plate is 1.5 in. thick,
what is the ligament efficiency for membrane
stresses?
Answer:
em = 0.702
17.5
A seamless square header that has 7 1/4 in. inside
measurement by 1.125 in. thick is to be formed so
that it will have a constant thickness. The header
contains a row of 2 in. diameter holes on 3 in.
centers. The holes have one groove for expansion
at the midthickness of the plate that is 0.125 in.
high and has a 2.125 in. diameter. The opposite
wall contains a series of handhole openings of
4.25 in. diameter on 7 in. centers with a seat on
the outside 0.125 in. deep by 4.75 in. diameter.
What are the ligament efficiencies for membrane
stresses?
Answer:
em = 0.328 on tube side
em = 0.384 on handhole side
17.6
A header is to be made from 1 in. thick plate.
The plate contains a row of tube holes on 3.5 in.
centers. The holes are alternately 2 and 2.5 in. All
holes are counterbored 0.25 in. and to a 0.25 in.
larger diameter. What is the ligament efficiency
for membrane stress?
Answer:
em = 0.339
17.6 Ligament Efficiency
for Multidiameter Openings Subject
to Bending Stress
For a flat plate that contains multidiameter openings, it
is necessary to determine an effective ligament efficiency
in bending by locating a neutral axis of the various diameters and thicknesses of the openings and the effective
moment of inertia. From structural mechanics, the basic
equations are
ΣAX
X=
(17.5)
ΣA
(17.6)
I = Σ A X2.
From Figure 17.7 and Eq. (17.5),
(
)
T0
+ T1 + T2 + · · · + Tn
Σ A X = b0 T0
2
(
)
T1
+ b1 T1
+ T2 + · · · + Tn
2
(
)
T2
+ b2 T2
+ · · · + Tn
2
( )
Tn
+ bn Tn
2
Σ A = b0 T0 + b1 T1 + b2 T2 + · · · + bn Tn
Eq.(17.7)
ΣAX
X=
=
.
ΣA
Eq.(17.8)
(17.7)
(17.8)
(17.9)
17.6 Ligament Efficiency for Multidiameter Openings Subject to Bending Stress
From Eq. (17.6),
I=
I = Σ I0 + Σ A X 2
(17.10)
3
3
3
3
b T
bT
bT
bT
I= 0 0 + 1 1 + 2 2 +···+ n n
12
12
12
12
)2
(
T0
+ b0 T0
+ T1 + T2 + · · · + Tn − X
2
)2
(
T1
+ b1 T1
+ T2 + · · · + Tn − X
2
)2
(
T2
+ b2 T2
+ · · · + Tn − X
2
(
)
T 2
+ bn Tn X − n
2
c = larger of X or t − X.
(17.11)
(17.12)
3
Because c = t/2 and I = bE t /12,
)
(
c
6
t
12
=
=
I
2 bE t 3
bE t 2
(17.13)
and
6I
.
t2 c
For bending stress, the ligament efficiency is
DE = p −
eb =
P − DE
.
p
+ (2.375)(0.125)
× (0.625 + 0.375 + 0.25 − 0.3683)2
+ (2.5)(0.375)(0.1875 + 0.25 − 0.3683)2
+ (2.625)(0.25)(0.3683 − 0.125)2
I = 0.0884
c = (the larger of 0.3683 or 0.75 − 0.3683)
= 0.3817
From Eq. (17.14), the equivalent diameter is
DE = 4 −
(6)(0.0884)
= 4 − 2.47 = 1.53 in.
(0.75)2 (0.3817)
The ligament efficiency for bending stress is calculated
from Eq. (17.15) as
The width of the ligament is
bE = p − DE .
1
[(2.375)(0.125)3 + (2.5)(0.375)3
12
+ (2.625)(0.25)3 ]
(17.14)
(17.15)
Example 17.5
Determine the bending ligament efficiency in the header
in Example 17.3.
eb =
4 − 1.53
= 0.617.
4
Example 17.6
The header mentioned in Example 17.4 is subjected to
bending stresses in the flat sides. What is the ligament
efficiency for the bending stress?
Solution:
p = 3 in.
T0 = 0.25 in.
d0 = 0.875 in.
b0 = 3 − 0.875 = 2.125 in.
T1 = 0.125 in. d1 = 0.9375 in.
b1 = 3 − 0.9375 = 2.0625 in.
T2 = 0.25 in. d2 = 0.875 in.
b2 = 3 − 0.875 = 2.125 in.
T3 = 0.125 in. d3 = 0.9375 in.
Solution:
p = 4 in.
T0 = 0.125 in. d0 = 1.625 in.
b0 = 4 − 1.625 = 2.375 in.
T1 = 0.375 in. d1 = 1.5 in.
b1 = 4 − 1.5 = 2.5 in.
T2 = 0.25 in. d2 = 1.375 in.
b2 = 4 − 1.375 = 2.625 in.
Σ AX = 2.375 × 0.125(0.0625 + 0.375 + 0.25)
+ 2.5 × 0.375(0.1875 + 0.25)
+ 2.625 × 0.25(0.125)
Σ AX = 0.6963
Σ A = 2.375 × 0.125 + 2.5 × 0.375 + 2.625
× 0.25 = 1.8906
0.6963
X=
= 0.3683 in.
1.8906
b3 = 3 − 0.9375 = 2.0625 in.
T4 = 0.5 in.
d4 = 0.875 in.
b4 = 3 − 0.875 = 2.125 in.
Σ AX = (2.125)(0.25)
× (0.125 + 0.125 + 0.25 + 0.125 + 0.5)
+ (2.0625)(0.125)
× (0.0625 + 0.25 + 0.125 + 0.5)
+ (2.125)(0.25)(0.125 + 0.125 + 0.5)
+ (2.0625)(0.125)(0.0625 + 0.5)
+ (2.125)(0.5)(0.25)
Σ AX = 1.6484
Σ A = (2.125)(0.25) + (2.0625)(0.125)
+ (2.125)(0.25) + (2.0625)(0.125)
+ (2.125)(0.5) = 2.6406
351
352
17 Vessels of Noncircular Cross Section
1.6484
= 0.6243
2.6406
1
I = [(2.125)(0.25)3 + (2.0625)(0.125)3
12
+ (2.125)(0.25)3
X=
+ (2.0625)(0.125)3 + (2.125)(0.5)3 ]
+ (2.125)(0.25)(0.125 + 0.125
+ 0.25 + 0.125 + 0.5 − 0.6243)2
+ (2.0625)(0.125)
× (0.0625 + 0.25 + 0.125 + 0.5 − 0.6243)2
+ (2.125)(0.25)
× (0.125 + 0.125 + 0.5 − 0.6243)2
+ (2.0625)(0.125)(0.0625 + 0.5 − 0.6243)2
+ (2.125)(0.5)(0.6243 − 0.25)2
I = 0.3451 in.4
c = (the larger of 0.6243 or 1.25 − 0.6243)
= 0.6257 in.
The equivalent diameter is equal to
DE = 3 −
(6)(0.3451)
= 3 − 2.118 = 0.882 in.
(1.25)2 (0.6257)
Thus,
3 − 0.882
= 0.706.
3
Comparing this with the efficiency from Example 17.4,
we see that this is a case where the grooves for expanding the tube have little to do with the efficiency, because
em = 0.704 and eb = 0.706.
eb =
Problems
17.7
The header mentioned in Problem 17.4 is subjected to bending stresses as well as membrane
stresses. What is the ligament efficiency for
bending stresses?
Answer:
eb = 0.702
17.8
What are the ligament efficiencies for bending
stresses mentioned in Problem 17.5?
Answer:
eb = 0.333 on tube side
eb = 0.367 on handhole side
17.7 Design Methods and Allowable
Stresses
Design rules given in this chapter are for vessels and
headers that have a straight longitudinal axis with a
noncircular cross section and closure plates on each
end. The formulas are based on assuming a unit-length
vessel section with no strengthening effect from the
longitudinal direction of the plate. However, for certain
uniform thickness vessels and headers, provisions are
given for plates with a length-to-width ratio of 2 or less to
compensate for the added strength from the longitudinal
direction.
The design rules in the ASME Code, VIII-1, provide
for vessels of rectangular and obround cross section
where different walls may have different thicknesses. The
method used in the ASME Code, VIII-1, combines plates
and shell theory and structural-design theory where it
is necessary to assume wall thicknesses and to calculate
stresses for comparison with allowable stress values.
These methods were described in Chapter 7.
For vessels and headers of uniform thickness,
equations can be developed that can be solved for
the minimum required thickness in terms of geometry
and applied loadings. Even in these cases, some assumptions are necessary because the thickness is directly
related to the primary membrane stress, but it is related
by the square of the thickness to the primary bending
stress. Both primary membrane and primary membrane
plus primary bending stresses are determined for the
various configurations at each point examined.
According to the ASME Code, VIII-1, a joint efficiency
is required for the longitudinal butt joints (and other butt
joints). Although the weld joint efficiency is applied to
the allowable design membrane stress at all points being
evaluated, it is applied to the allowable bending stress
only at the joint. If the point being evaluated is located
away from the weld joint, the allowable design bending
stress is not modified by the weld joint efficiency. Therefore, the E value used to calculate the allowable stress for
the combined membrane plus bending stress at points
away from the weld joint is E = 1.0.
Provision is made for taking account of holes with different diameters through the plate thickness. For those
cases, an equivalent diameter is determined for calculating the ligament efficiencies, which are used directly
in the equations to calculate the applied stresses. When
the ligament efficiencies em and eb are less than the weld
joint efficiency factor, the allowable stresses are calculated using E = 1.0 (see Section 17.3).
From the foregoing discussion, one realizes that care
is required in determining the actual calculated stresses
and the allowable stresses at any location. In general, it
is necessary to calculate the membrane stress and the
17.8 Basic Equations
bending stress separately at each location so that various
combinations of calculated stresses may be compared
with various combinations of allowable stresses.
The calculated primary membrane stress is limited
to the basic allowable tensile stress given in the applicable code multiplied by the weld joint efficiency, SE.
However, the combination of primary membrane plus
primary bending stress multiplied by the weld joint
efficiency for a plate assumed to be rectangular in cross
section is limited to the following:
1) At design temperatures where tensile strength and
yield strength govern, the lesser of the following
multiplied by the weld joint efficiency:
a) 1.5 times the basic allowable tensile strength at
design temperature.
b) The yield strength at design temperature.
2) At design temperatures where creep and rupture
strength govern, the lesser of the following multiplied
by the weld joint efficiency:
a) 1.25 times the basic allowable stress at design temperature.
b) The yield strength at design temperature.
For sections that are not rectangular in cross section
(such as composite reinforced bars or shapes and for plate
sections, etc.), the allowable stress is the lesser of the following:
1) 1.5 times the design stress SE at design temperature;
2) Two-thirds of the yield strength at design temperature, except that for materials where higher deformation is permitted, the yield-strength limits may be the
lesser of the following:
a) 90% of yield strength at design temperature.
b) Two-thirds of the specified minimum yield
strength at room temperature.
At design temperatures where creep and rupture strength govern, the same limits apply, but the
shape-factor multiplier is limited to 1.25 regardless of
the actual shape factor. For external pressure, where the
total stress may be compressive, a limitation is also set
based on buckling of the side plate.
The basic theory is the maximum stress theory that is
generally used in structural analysis.
As mentioned previously, it is necessary to calculate
stresses at various points in order to determine which
combination controls. Certain analysis methods, such
as that in the Swedish Pressure Vessel Code, combine
the membrane stress and the bending stress in the same
equation. It may be necessary to separate them for evaluation when different efficiencies apply to the membrane
stress and to the bending stress.
17.8 Basic Equations
Pressure vessels having a rectangular cross section,
Figure 17.1, subjected to internal pressure are analyzed by one of the methods used for indeterminate
structures [2, 3]. The structure shown in Figure 17.1 is
indeterminate to the third degree, and the method of
virtual work (Castigliano’s theory) is best suited for solving this kind of a problem. The formulation and solution
are presented in Section 17.11 for a general case where
the thicknesses and moduli of elasticity are different
for each side. ASME uses a symmetric condition for
the basic case where the opposite sides are identical in
thickness and modulus of elasticity. This simplifies the
design equations considerably. The remaining discussion
in this chapter is based on this symmetric assumption in
accordance with ASME.
For analysis purposes, the noncircular cross section of
the vessel is considered as a structural frame. Each component of the rectangular or obround frame contains a
load that causes a membrane stress and a moment that
causes a bending stress. The total stress at any point is
the summation of these two stresses.
As shown in Figure 17.9, the direct or membrane stress
at any point is caused by internal pressure loading against
the adjacent walls. This loading is resisted by the thickness, strength of material, and weld or ligament efficiency
of the walls that are carrying the load. The applied loading is (P)(h), and the resisting loading is (Sm )(2t 1 ). When
t1
t1
h
t2
I1
Internal pressure, P
H
t2
I2
MQ
MM
MN
MN
MM
Figure 17.9 Diagram of internal pressure loading and bending
moment for rectangular-cross-section header.
353
354
17 Vessels of Noncircular Cross Section
these are equated to each other and solved for Sm , the
membrane stress for the short side is found to be
Sm =
Ph
,
2t1
(17.16)
and for the long side,
Sm =
PH
.
2t2
(17.17)
From the theory of structural frames [2, 3], the basic
moment equations for a rectangular frame under internal pressure loading P when the two pairs of opposite
sides have equal thickness and equal length, as shown in
Figure 17.9, are as follows:
Bending moment at corner Q is
( 3
)
3
P h ∕I2 + H ∕I1
MQ =
.
(17.18)
12
h∕I2 + H∕I1
Bending moment at midpoint of long side M is
Ph2
.
(17.19)
8
Bending moment at midpoint of short side N is
MM = MQ −
PH 2
.
8
Moment of inertia I 1 for short side is
MN = MQ −
t13
.
12
Moment of inertia I 2 for long side is
I1 =
t23
.
12
Basic equation for bending stress is
I2 =
M Q t1
2I1
.
M Q t2
2I2
.
M N t1
.
2I1
Bending stress at midpoint of long side is
Clearing P/12 in Eq. (17.29) gives
)
( 3
3
P h I1 + H I2
− 1.5H 2 .
MN =
12
hI 1 + HI 2
Multiplying Eq. (17.30) through by −1,
)
(
h3 I1 + H 3 I2
P
2
.
1.5H −
MN =
12
hI 1 + HI 2
(17.29)
(17.30)
(17.31)
Substituting Eq. (17.31) into Eq. (17.25) and c for t 1 /2,
(
)
h3 I1 + H 3 I2
Pc
2
(Sb )N =
1.5H −
.
(17.32)
12I1
hI 1 + HI 2
1) Calculate the moment of inertia:
(1)3
t3
=
= 0.0833.
12
12
2) Calculate bending moment1 at the corner, MQ , using
Eq. (17.18):
I1 = I2 =
(17.23)
(
)
150 (14)3 ∕0.0833 + (7.25)3 ∕0.0833
12
14∕0.0833 + 7.25∕0.0833
= 1838.28.
MQ =
(17.24)
3) Calculate bending moment1 at midpoint of long side,
MM , using Eq. (17.19):
Bending stress at the midpoint of short side is
(Sb )N =
Clearing Eq. (17.28), we obtain
( 3
3 )
P h I1 + H I2
P
12H 2
−
×
.
MN =
12
hI 1 + HI 2
12
8
(17.28)
(17.22) Solution:
Bending stress in the corner of long side is
(Sb )Q =
Substituting Eq. (17.27) in Eq. (17.20) gives
( 3
3 )
P h I1 + H I2
PH 2
MN =
.
−
12
hI 1 + HI 2
8
(17.27)
(17.20) Example 17.7
Determine the adequacy of a rectangular-cross-section
header with a design pressure of 150 psi and made from
a seamless forging with an allowable stress S = 12 500 psi.
(17.21) The header inner dimensions are 14 in. by 7.25 in. with
a constant thickness of 1 in. All openings are reinforced.
Maximum temperature is 100 ∘ F.
Mc
Sb =
.
I
Bending stress in the corner of short side is
(Sb )Q =
follows: Cross-multiplying Eq. (17.18),
( 3
3 )
P h I1 + H I2
.
MQ =
12
hI 1 + HI 2
(17.25)
Ph2
= 1838.28
8
(150)(14)2
= −1836.72.
−
8
M M = MQ −
M M t2
.
(17.26)
2I2
Simplifying equations, the equation for the bending
1 Positive moment results in compression in outer fiber; negative
moment results in tension in outer fiber.
stress at the midpoint of the short side is determined as
(Sb )M =
17.8 Basic Equations
4) Calculate bending moment at midpoint of short side,
MN , using Eq. (17.20):
PH 2
= 1838.28
8
(150)(7.25)2
= 852.73.
−
8
5) Calculate bending stress at the corner of long side:
M N = MQ −
(1838.28)(0.5)
= 11,030 psi (T in).
0.0833
6) Calculate bending stress at midpoint of long side:
(Sb )QM =
(−1836.72)(−0.5)
= 11,020 psi (T out).
0.0833
7) Calculate bending stress at the corner of short side:
(sb )M =
8) Calculate bending moment at midpoint of short side:
(852.73)(0.5)
= 5120 psi (T in).
0.0833
9) Calculate membrane stress on long side:
(Sb )N =
(150)(7.25)
PH
=
= 540 psi.
2t
2(1)
3.75 − 1.5
= 0.60.
3.75
2) Calculate the moments of inertia of both the long
side and the short side:
em = e b =
I1 =
Ph (150)(14)
=
= 1050 psi.
2t
2(1)
11) Total stress at the corner of long side:
(St )QM = (Sm )M + (Sb )QM
= 540 + 11,030 = 11,570 psi
1) Calculate the ligament efficiency of the long side
with the tube holes:
t13
(0.625)3
= 0.0203
12
12
t3
(1)3
I2 = 2 =
= 0.0833.
12
12
3) Calculate the bending moment at the corner, MQ ,
using Eq. (17.18):
10) Calculate membrane stress on short side:
(Sm )N =
Example 17.8
A rectangular-cross-section header is made from two
C-sections and butt welded along the center of the short
sides. The weld joints are spot examined with the backing
strip left in place (from Table UW-12 of the ASME Code,
VIII-1, E = 0.8). The design pressure is 115 psi, and the
material is SA-515 Grade 70 with an allowable stress
S = 17,500 psi. The long side is 13.5 by 1 in. thick, and the
short side is 6 by 0.625 in. thick. One long side contains
a row of 1.5 in. diameter holes on 3.75 in. centers. Is the
design acceptable?
Solution:
(Sb )QN = 11,030 psi (T in).
(Sm )M =
Since there are no butt welds, E = 1.0, and SE =
12,500(1) = 12,500 psi. For membrane plus bending, the allowable stress is 1.5SE = 1.5(12,500)(1) =
18,750 psi. All stresses are less than allowable stress
values, so 1 in. thickness is satisfactory.
(T in).
12) Total stress at midpoint of long side:
(St )M = (Sm )M + (Sb )M = 540 + 11,020
= 11,560 psi (T out).
13) Total stress at the corner of short side:
=
[
]
115 (13.5)3 ∕0.0833 + (6)3 ∕0.0203
MQ =
12
13.5∕0.0833 + 6∕0.0203
= 841.35.
4) Calculate the bending moment at the midpoint of the
long side, MM , using Eq. (17.19):
Ph2
= 841.35
8
(115)(13.5)2
= −1778.50.
−
8
5) Calculate the bending moment at the midpoint of the
short side, MN , using Eq. (17.20):
M M = MQ −
(St )QN = (Sm )N + (Sb )QN = 1050 + 11,030
= 12,080 psi (T in).
14) Total stress at midpoint of short side:
(St )N = (Sm )N + (Sb )N = 1050 + 5120
= 6170 psi (T in).
MN = MQ −
(115)(6)2
PH 2
= 841.35 −
= 323.85.
8
8
355
356
17 Vessels of Noncircular Cross Section
6) Calculate the bending stress at the corner of the long
side with the holes:
(Sb )QM =
MQ c 2
I2
14) Total stress at the corner of short side:
(St )QN = (Sm )N + (Sb )QN
= 1240 + 12,950 = 14,190 psi
(841.35)(0.5)
=
= 5050 psi.
(0.0833)
Allowable stress = 1.5SE = (1.5)(17,500)(0.80)
= 21,000 psi.
7) Calculate the bending stress at the midpoint of the
long side with the holes (eb = 0.60; E = 1.0):
15) Total stress at midpoint of short side:
M c
(1778.50)(0.5)
(Sb )M = M 2 =
= 17,790 psi.
I2 eb
(0.0833)(0.6)
(St )N = (Sm )N + (Sb )N
8) Calculate the bending stress at the corner of the
short side (E = 1.0):
(Sb )QN =
MQ c 1
I1
=
(841.35)(0.3125)
= 12,950 psi.
(0.0203)
9) Calculate the bending stress at the midpoint of the
short side (E = 0.80):
(Sb )N =
= 1240 + 4990 = 6230 psi
Allowable stress = 1.5SE = (1.5)(17,500)(0.8)
= 21,000 psi.
MN c 1
(323.85)(0.3125)
=
= 4990 psi.
I1
(0.0203)
All calculated stresses are less than the allowable
stresses and are acceptable.
Problems
17.9
10) Calculate the membrane stress on the long side
(em = 0.60; E = 1.0):
(115)(6)
PH
=
= 580 psi
2tem
2(1)(0.6)
Allowable stress = SE = (17,500)(1.0)
= 17,500 psi.
(Sm )M =
11) Calculate the membrane stress on the short side
(E = 0.80):
Ph (115)(13.5)
=
= 1240 psi
2t1
2(0.625)
Allowable stress = SE = (17,500)(0.8)
= 14,000 psi.
(Sm )N =
A header is made from two C-sections that are
butt-welded along the centerline of the short side.
The weld is not examined, and the backing strip
remains in place; therefore, E = 0.65. The header
is 12 by 6 by 1 in. thick with a design pressure of
200 psi. What is the stress at the weld joint?
Answer:
Maximum stress = 6 600 psi
17.10
A square header is made from two C-sections
that are butt-welded along the centerline of two
opposite sides. The weld joint efficiency is 0.65.
The adjacent side contains a row of openings
that are 2 in. in diameter on 5.75 in. centers. The
design pressure is 110 psi, and the header is 7.25
by 0.75 in. thick. What is the maximum stress and
where is it located?
12) Total stress at the corner of long side:
(St )QM = (Sm )M + (Sb )QM = 580 + 5050
= 5630 psi
Allowable stress = 1.5SE = 1.5(17,500)(1.0)
= 26,250 psi.
13) Total stress at midpoint of long side:
Answer:
Maximum stress = 5670 psi at the inside corner of
the welded side
17.9 Equations in the ASME Code, VIII-1
Appendix 13 of the ASME Code, VIII-1, contains
many equations for determining stresses in various
(St )M = (Sm )M + (Sb )M = 580 + 17,790 noncircular-cross-section vessels. The basis for the
ASME equations is the same as for those in Section 17.8,
= 18,370 psi
but the equations in the ASME Code are expressed in
Allowable stress = 1.5SE = 1.5(17,500)(1.0)
different terms and are adjusted to be applicable to the
= 26,250 psi.
specific geometries shown.
17.9 Equations in the ASME Code, VIII-1
When the opposite sides have the same thickness, the
equations in the ASME Code, VIII-1, are the same as
those given in Eqs. (17.16) and (17.17). However, when
different thicknesses occur, or when there are rounded
corners, or when some other variation occurs in the
geometry, the ASME Code has a specific equation for
that configuration that was developed from the theory
of structural frames.
The equations for the bending stresses were similarly
developed. The principles of structural frames are applied
to the various combinations of geometry to give different
equations for each shape. For example, the equation for
the bending stress at the midpoint of the short side of a
rectangular-cross-section header is Eq. (3) of Article 13-7
of the ASME Code, VIII-1. The derivation of this same
equation as given in Eq. (17.32) is
[
]
Pc
1 + 𝛼2K
1.5H 2 − h2
(17.33)
(Sb )N =
12I1
1+K
H
𝛼=
(17.34)
h
I
K = 2𝛼
(17.35)
I1
[ 2
]
I
H
H
× 2×
1 + 𝛼2K = 1 +
h2
I1
h
3
3
h I1 + H I2
=
(17.36)
h3 I
[ 1
]
hI + HI 2
H I2
1+K =1+
(17.37)
= 1
×
h I1
hI 1
[
]
3
3
2
hI 1
21+ 𝛼 K
2 h I1 + H I2
−h
×
= −h
1+K
h3 I1
hI 1 + HI 2
h3 I1 + H 3 I2
=
.
(17.38)
hI 1 + HI 2
Substituting Eq. (17.38) into Eq. (17.33) gives
[
]
h3 I1 + H 3 I2
Pc
2
(Sb )N =
1.5H −
.
12I1
hI 1 + HI 2
(17.39)
Equation (17.39) is identical to Eq. (17.32) and shows
the relationship between the equations in Appendix 13
of the ASME Code, VIII-1, and those derived from basic
theory. The ASME Code contains extensive nomenclature in Article 13-5 for various configurations. Equations
for the bending stresses in Table 13-18.1 of the code are
shown in Table 17.1. In addition, equations for membrane stresses for various configurations in the ASME
Code, VIII-1, are summarized in Table 17.2.
Although all the formulas have been developed on
the basis of length-to-width ratios L1 /H and L1 /h of 4
or more – values for which there is no long-dimension
effect – provisions are given for the simple rectangular
header shown in Figure 17.3a to reduce stresses when the
aspect ratio is less than 2. The membrane stresses remain
the same, but the bending stresses are reduced by multiplying by the factors shown in Table 17.3. The stresses
are then obtained by using the following equations:
For the short side at the corner:
(Sb )Q = Eq. (17.23) × C2 .
(17.40)
For the short side at the midpoint:
(Sb )N = Eq. (17.25) × C1 .
(17.41)
For the long side at the corner:
(Sb )Q = Eq. (17.24) × C2 .
(17.42)
For the long side at the midpoint:
(Sb )M = Eq. (17.26) × C1 .
(17.43)
For those cases where either L1 /H or L1 /h is less than
1.0, it is necessary to reorient the axes of the header and
to recalculate all properties such as the moment of inertia of the wall. Dimensions are chosen so that the longest
dimension is L1 , the middle dimension is h, and the shortest dimension is H. This may result in a part that was originally considered to be an end closure becoming a wall
of the header. All calculations are based on this revised
configuration.
Vessels of noncircular cross section may be subjected
to external pressure. The membrane and bending stresses
are considered the same as for internal pressure unless
the resulting stresses are compressive, when buckling is a
possible mode of failure. Interaction equations are used
to examine the various plates for stability. Calculated
stresses are compared with critical buckling stresses with
a factor of safety applied. This is described in Article
13-14 of the ASME Code, VIII-1.
Example 17.9
The rectangular-cross-section header mentioned in
Example 17.7 is made according to ASME Code, VIII-1,
rules. What are the bending stresses at the midpoints of
both the short side and the long side?
Solution:
Knowns: H = 7.25 in., h = 14 in., t = 1 in., c = 0.5 in., and
E = 1.0. Calculate 𝛼, I, and K as follows:
H
7.25
=
= 0.5179
h
14
(1)3
t3
=
= 0.0833
I=
12
( ) 12
I
0.0833(0.5179)
= 0.5179.
K= 2 𝛼=
I1
0.0833
𝛼=
Calculate the bending stress at the midpoint of the
short side using Eq. (17.33) and Table 17.1, second
357
358
17 Vessels of Noncircular Cross Section
Table 17.1 Bending stress values in rectangular headers.
Figure
Location of
weld between
17.3a
M and Q
17.3a
N and Q
17.3b
M and Q
17.3b
M1 and Q1
17.3c
A and B
17.3c
D and C
17.3d
M and Q
17.3d
N and Q
17.3e
A and B
17.3e
B and C
17.3e
F and E
17.3e
E and D
17.5a
A and B
17.5b
A and B
Bending stress at joint, ±(Sb )j (psi)
{ [
(
}
)]
Pc
1 + 𝛼2 K
h2 1.5 −
− 6dj2
12I2
1+K
[
]
(1 + 𝛼 2 K)
Pc
1.5H 2 − h2
− 6dj2
12I1
1+K
{ 2
}
Pc
h
h2
[(K2 − k1 k2 ) + 𝛼 2 k2 (K2 − k2 )] −
+ dj2
2I22 2N
4
{ 2
}
h2
Pc
h
[(K1 k1 − k2 ) + 𝛼 2 k2 (K1 − k2 )] −
+ dj2
2I2 2N
4
(
)
Pd2j
c
MA +
I1
2
]
[
c
P
MA + (L2 + 2aL − 2al1 − l12 + dj2 )
I1
2
[
(
]
)
12dj2
1 + 𝛼12 k
Pph2
3−2
− 2
24Z21
1+k
h
[
(
]
)
2
1 + 𝛼1 k
Pp
2
2
2
3H − 2h
− 12dj
24Z1
1+k
(
)
pd2j
1
MA + P
Z21
2
(
)
pd2j
c
MA + P
I2
2
{
p
1
MA + P [(L + L11 )2 + 2a(L + L11 − l1 − l11 )
Z11
2
−(l1 + l11 )2 + dj2 ]}
{
p
c
2
− l12
MA + P [L2 + 2LL11 + L211 − 2l1 l11 − l11
I1
2
+2a(L + L11 − l1 − l11 ) + dj2 ]}
(
)
2
Pc −LC 1 dj
+
I2
6A
2
(
)
2
Pp −LC 2 dj
+
Z11
6A3
2
Source: Courtesy of American Society of Mechanical Engineers, from Table 13-18.1 of ASME Code VIII-1.
equation:
Example 17.10
[
] The header mentioned in Example 17.9 is to be built with
3
(150)(0.5)
1 + (0.5179)
a shortened length L1 = 18 in. What are the stresses at the
(Sb )N =
1.5(7.25)2 − (14)2
12(0.0833)
1 + 0.5179
midpoints now?
(Sb )N = 5120 psi.
Solution:
Calculate the bending stress at the midpoint of the long
L1 = 18 in., H = 7.25 in., and h = 14 in. Calculate the folside using Table 17.1, first equation:
lowing:
[
]
L1
18
(150)(0.5)(14)2
1 + (0.5179)3
=
= 2.48 C1 = C2 = 1.00
1.5 −
(Sb )M =
H
7.25
12(0.0833)
1 + 0.5179
L1
18
= 11,020 psi.
=
= 1.29 C1 = 0.79 and C2 = 0.83.
h
14
17.9 Equations in the ASME Code, VIII-1
Table 17.2 Membrane stress in rectangular headers.
Figure
17.3a
17.3b
Location
Short side
Ph/(2t 1 )
Long side
Ph/(2t 2 )
Short side
Ph/(2t 1 )
P
{4NH 2 − 2h2 [(K2 + k2 )
8NHt 2
−k1 (K1 + k2 ) + 𝛼 2 k2 (K2 − K1 )]}
Long side
17.3c
Short side
P(a + L)/t 1
Long side
P(l1 + a)/t 1
√
P( L2 + l12 + a)∕t1
Corner
17.3d
17.3e
17.5a
17.5b
Membrane stress (psi)
Short side
Php/2(A1 + pt 1 )
Long side
PHp/2(A2 + pt 2 )
Short side
P(L + L11 + a)/t 1
Long side
P(l1 + l11 + a)/t 2
Midpoint, curve
P(a + L)/t 1
End, curve
Pa/t 1
Side
Pa/t 2
Midpoint, curve
P(a + L)p/(A1 + pt 1 )
End, curve
Pap/(A1 + pt 1 )
Side
Pap/(A1 + pt 1 )
Table 17.3 Values of C 1 and C 2 as a
function of geometry.
L1 /H or L1 /h
Solution:
Calculate the header-geometry properties as follows:
L1
12
=
= 1.66
H
7.25
L1
12
=
= 0.86.
h
14
Since L1 /h < 1.0, the axes of the header must be reoriented for analysis. In the new terms, L1 = 14 in., h = 12 in.,
and H = 7.25 in.
The first assumption is that the original flat end closures, which have now become sides, are also 1 in. thick.
On that basis, I = 0.0833 remains. We have
7.25
H
=
= 0.6042
𝛼=
h
12
K = 0.6042.
Check if the strengthening factors apply:
L1
14
=
= 1.93 C1 = 0.99 and C2 = 0.99
H
7.25
L1
14
=
= 1.17 C1 = 0.62 and C2 = 0.68.
h
12
Strengthening factors apply to both short side and
long side.
Calculate the bending stress at the midpoint of the
short side using Eq. (17.41):
(150)(0.5)
12(0.0833)
[
]
1 + (0.6042)3
× 1.5(7.25)2 − (12)2
(0.99)
1 + 0.6042
(Sb )N = 2280 psi.
(Sb )N =
C1
C2
1.0
0.56
0.62
1.1
0.64
0.70
1.2
0.73
0.77
1.3
0.79
0.82
1.4
0.85
0.87
1.5
0.89
0.91
1.6
0.92
0.94
1.7
0.95
0.96
1.8
0.97
0.97
1.9
0.99
0.99
2.0
1.00
1.00
Calculate the bending stress at the midpoint of the long
side using Table 17.1, first equation:
(150)(0.5)(12)2
12(0.0833)
[
]
1 + (0.6042)3
× 1.5 −
(0.62)
1 + 0.6042
= 4950 psi.
(Sb )M =
Problems
The strengthening effect applies to the long side only
because the length-to-width ratio is less than 2.0. This
gives
(Sb )′N = (Sb )N × C1 = (11,020)(0.79) = 8710 psi.
Example 17.11
The header mentioned in Example 17.9 is to be built with
a shortened length L1 = 12 in. What is done to the header
axes for analysis, and what are the stresses at the midpoints of the short side and the long side?
17.11
A rectangular vessel with the cross section shown
in Figure 17.3c is constructed from SA-516 Grade
60 material. The design pressure is 75 psi at a
design temperature of 300 ∘ F. There is no corrosion, and full radiography is applied at all weld
joints. What are the total stresses in the long
side plates and in the corner section, using the
method in the ASME Code, VIII-1, Appendix 13?
Dimensions are as follows: t 1 = 0.75 in.; a = 3 in.;
L = 6 in.; l1 = 3 in.
359
360
17 Vessels of Noncircular Cross Section
d
d
c
2n
p
2m
p
Figure 17.11 Ligaments in a box header.
Figure 17.10 Box header.
where m and n are one-half of the lengths of the box
header as shown in Figure 17.10.
At the center of the long length 2m,
Answer:
At center, S = 14,100 psi
At end, S = 1480 psi
In bend at 26.6∘ , S = 10,720 psi
Y = |(1∕3)[(m3 + n3 )∕(m + n)] − 0.5 m2 |.
The value of Y when the holes are off center line as
shown in Figure 17.11 is
Y = |(1∕3)[(m3 + n3 )∕(m + n)] − 0.5(m2 − c2 )|,
17.10 Design of Noncircular Vessels
in Other Codes
In addition to the ASME Section VIII Code, design rules
for noncircular pressure vessels are contained in other
design codes as well as in various textbooks. In addition,
empirical design rules have been developed and used for
a number of years based on deflection and burst proof
tests. Among those design codes for noncircular vessels
is the British Code BS 1113 [4], the European codes EN
12952 [1] Part 3, and EN 13445 [2]. Although each set
of rules is based on the method of structural frames,
the specific applications and consequent answers are
different.
17.10.1
The required thickness of the box header shown in
Figure 17.10 is
(17.44)
where
f = allowable stress
𝜂 = membrane ligament efficiency
𝜂 ′ = bending ligament efficiency
P = internal pressure
t = thickness.
The value of Y in Eq. (17.44) is obtained as follows.
Near the corners,
Y = (1∕3)[(m3 + n3 )∕(m + n)],
𝜂 = (p − d)∕p
𝜂 ′ = (p − d)∕p
when d < 0.6m
𝜂 = (p − 0.6m)∕p when d ≥ 0.6m,
′
where p is the pitch of the holes and d is the diameter of
the holes.
British Standard BS 1113 also gives equations for staggered holes along the length and an equation for the minimum corner radii.
17.10.2 Method of the European Standards EN
12952 and EN 13445
Method of the British Code BS 1113
t = [Pn∕(2f 𝜂)] + [4YP∕(f 𝜂 ′ )]0.5 ,
where c is as defined in Figure 17.11.
The ligament efficiencies are given by
The methodology and equations for box headers in EN
12952 Part 3 are essentially the same as those in British
Standard BS 1113 with a slight change in terminology.
The equations for the bending stress in EN 13445 using
the terminology in Figure 17.12 are given as follows:
at point A,
(𝜎b )A = MA e∕2I1 ;
(17.45)
at point B,
(𝜎b )B = (e∕4I1 )(2MA + PL2 );
(17.46)
at point C,
(𝜎b )C = (e∕4I1 )[2MA + P(2aL − 2a𝓁1 + L2 )];
(17.47)
17.11 Forces in Box Headers due to Internal Pressure
1
The membrane stresses are given as follows:
θ
at point A,
D
C
(𝜎m )A = (𝜎m )B ;
(17.50)
at point B,
a
B
(𝜎m )B = P(a + 𝓁1 )∕e;
L
(17.51)
at point C,
A
(𝜎m )C = P(a + L)∕e;
e
(17.52)
at point D,
(σm )D = (σm )C ;
(17.53)
at the corner between points B and C,
(σm )B-C = (P∕e)[a + (L2 + 𝓁1 2 )0.5 ].
Figure 17.12 Box header terminology.
at point D,
(17.54)
Problems
(𝜎b )D = (e∕4I1 )[2MA + P(2aL − 2a𝓁1 + L − 𝓁1 )];
2
2
(17.48)
17.12
where
a = radius of inner corner
e = thickness
A box header with a length 2m = 18 in. and a
width 2n = 9 in. is subjected to an internal pressure of 20 psi. Determine the required thickness
in accordance with BS 1113. Let the allowable
stress = 20,000 psi and ligament efficiency = 1.0.
Answer:
t = 0.287 in.
I 1 = e3 /12
𝓁 1 = one-half of the length as defined in Figure 17.12
L = one-half of the length as defined in Figure 17.12
P = internal pressure
𝜎 b = bending stress.
The value of MA is given by
17.13
A box header with a length 2𝓁 1 = 18 in., width
2L = 9 in., and corner radius a = 0.5 in. is subjected to an internal pressure of 20 psi. Determine
the stress at point C in accordance with EN
13445. Let the thickness e = 0.287 in. How does
this stress compare to the allowable stress of
1.5S = 30,000 psi?
MA = P(−K3 ),
Answer:
𝜎 = 27,000 psi
where
K3 = 𝓁1 2 [(C1 + C2 )∕C3 ]
C1 = 6𝜙2 𝛼3 − 3𝜋𝜙2 + 6𝜙2 + 𝛼3 3
17.11 Forces in Box Headers due
to Internal Pressure
C2 = 3𝛼3 2 − 6𝜙 − 2 + 1.5π𝛼3 2 𝜙 + 6𝜙𝛼3
C3 = 3(2𝛼3 + 𝜋𝜙 + 2)
and
𝛼3 = L∕𝓁1
𝜙 = a∕𝓁1 .
The stress at the corner of the box header is given by
(𝜎b )B−C = (e∕4I1 ){2MA + P[2a(L cos 𝜃
−𝓁1 (1 − sin 𝜃)) + L2 ]}.
(17.49)
A rectangular header, Figure 17.13a, is an indeterminate
structure to the third degree. The forces and moments in
the structure can best be determined from Castigliano’s
theorem [[5, 6]]. The forces and moments are determined by cutting the structure at an arbitrary location,
Figure 17.13b, and unknown redundant forces H o , V o ,
and Mo are then placed at this location.
Deflection and rotation compatibility equations can be
written as follows since the free edge o is connected to
point A in the actual structure
361
362
17 Vessels of Noncircular Cross Section
E3
C
𝛿 MH = rotation at point o due to unknown force H o
(𝛿 MH = 𝛿 HM )
D
t3
E2 t2
𝛿 MV = rotation at point o due to unknown force V o
(𝛿 MV = 𝛿 VM )
t4 E4
P
𝛿 MM = rotation at point o due to unknown moment Mo .
t1
E1
B
The deflections and rotation due to P, H o , V o , and Mo
at point o are calculated from Castigliano’s equation,
A
(a)
𝛿=
L1
Vo
o
Mo
Ho
Mm
ds,
EI
(17.58)
where M is the moment expression along the structure
due to applied pressure P and m is the moment expression along the structure due to the redundant forces.
The values of H o , V o , and Mo are obtained by calculating each of the terms in Eqs. (17.55)–(17.57) using
Eq. (17.58) and then solving the three simultaneous
equations.
L2
(b)
∫
A
Figure 17.13 (a,b) General terms of a box header.
17.11.1
Deflection in the horizontal direction at point o = 0
Deflection in the vertical direction at point o = 0
Rotation at point o = 0.
Or,
𝛿H p + 𝛿HH Ho + 𝛿HV Vo + 𝛿HM Mo = 0
(17.55)
𝛿V + 𝛿VH Ho + 𝛿VV Vo + 𝛿VM Mo = 0
(17.56)
𝛿M p + 𝛿MH Ho + 𝛿MV Vo + 𝛿MM Mo = 0,
(17.57)
p
where
𝛿 H P = horizontal deflection at point o due to applied
loads
𝛿 HH = horizontal deflection at point o due to unknown
force H o
𝛿 HV = horizontal deflection at point o due to unknown
force V o
𝛿 HM = horizontal deflection at point o due to unknown
moment Mo
𝛿 V P = vertical deflection at point o due to applied loads
Square Headers
For simplicity, it will be assumed that all lengths (L),
thicknesses (t), and moduli of elasticity (E) are the same
for all members of the structure. Thus, the analysis can
be simplified by letting
L1 = L2 = L
t1 = t2 = t3 = t4 = t
E1 = E2 = E3 = E4 = E.
Solving Eqs. (17.55)–(17.58) results in the following
expressions:
Ho = PL∕2 Vo = PL∕2 Mo = PL2 ∕12.
With these expressions known, the remaining forces
and moments in the structure are obtained.
17.11.2
Rectangular Headers
The solution of rectangular headers follows the same procedure as that for square headers. However, in this case,
the values of t and E may be different for each part of the
structure. Accordingly, the equations for H o , V o , and Mo
become more complicated and are given by the following
expressions:
𝛿 VH = vertical deflection at point o due to unknown
force H o (𝛿 VH = 𝛿 HV )
Mo = K20 P
Vo = K21 P
(17.60)
𝛿 W = vertical deflection at point o due to unknown
force V o
Ho = K22 P,
(17.61)
𝛿 VM = vertical deflection at point o due to unknown
Moment Mo (𝛿 VM = 𝛿 MV )
𝛿 M P = rotation at point o due to applied loads
where
Ei = modulus of elasticity for member i
K 1 = E1 t 1 3 /12
K 2 = E2 t 2 3 /12
(17.59)
17.11 Forces in Box Headers due to Internal Pressure
K 3 = E3 t 3 3 /12
K 4 = E4 t 4 3 /12
With these expressions known, the remaining forces
and moments in the structure are obtained.
L3
LL2
L3
K5 = 2 + 1 2 + 2
3K2
K3
3K4
2
2
LL
L L
K6 = 1 2 + 1 2
2K2
2K3
2
−L2
LL
L2
K7 =
− 1 2− 2
2K2
K3
2K4
3
2
L
L L
L3
K8 = 1 + 1 2 + 1
3K1
K2
3K3
2
−L1
LL
L2
K9 =
− 1 2− 1
2K1
K2
2K3
L
L
L
L
K10 = 1 + 2 + 1 + 2
K1 K2 K3 K4
−L1 2 L2 2
L4
L L 3 L 3L
L4
K11 =
− 2 − 1 2 − 1 2− 2
4K2
8K2
2K3
6K3
8K4
4
3
3
4
2 2
−L1
L L
LL
L
L L
K12 =
− 1 2− 1 2 − 1 − 1 2
8K1
2K2
6K2
8K3
4K3
3
2
3
3
L
L L
L
L
LL2
L3
K13 = 1 + 1 2 + 2 + 1 + 1 2 + 2
6K1
2K2
6K2 6K3
2K3
6K4
K14 = K6 2 − K5 K8
K15 = K6 K7 − K5 K9
K16 = K5 K12 − K6 K11
K17 = K7 K8 − K6 K9
K18 = K7 K9 − K6 K10
K19 = K6 K13 − K7 K12
K K − K14 K19
K20 = 16 17
K15 K17 − K14 K18
K16 − K15 K20
K21 =
K14
K11 + K6 K21 + K7 K20
K22 = −
K5
ti = thickness of member i.
Example 17.12
The box header shown in Figure 17.14 is subjected to an
internal pressure of 20 psi. Sides 1 and 4 are constructed
of carbon steel plates with E = 30,000 ksi, and sides
2 and 3 are constructed of stainless steel plates with
E = 25,000 ksi. The four sides of the header have different
thicknesses due to some other load requirements. Calculate the forces and bending moments in the various
members due to internal pressure.
Solution:
20.00
P
L1
18.00
L2
9.00
t1
0.38
t2
0.50
t3
0.63
t4
0.75
E1
30.00
E2
25.00
E3
25.00
E4
30.00
K1
0.1318
K2
0.2604
K3
0.5086
K4
1.0547
K5
4 030.0646
K6
5 665.9046
K7
−512.4250
K8
29,765.0995
K9
−2 169.3850
K 10
215.0161
K 11
−59,219.8387
K 12
−247,406.1926
K 13
16,897.5821
K 14
−87,852,799.6921
K 15
5,839,410.6593
K 16
−661,528,989.6918
K 17
−2,960,851.6201
K 18
−106,613.7551
K 19
−31,037,019.6555
K 20
28.8116
Mo
576.23
K 21
9.4450
Vo
188.90
K 22
5.0791
Ho
101.58
Figure 17.14 Rectangular box header.
3
t3 = 0.625ʺ
L2 = 9ʺ
E3 = 25 × 106 psi
t2 = 0.50ʺ
2
t4 = 0.75ʺ
E2 = 25 × 106 psi
E4 = 30 × 106 psi
t1 = 0.375ʺ
E1 = 30 × 106 psi
1
L1 = 18ʺ
4
363
364
17 Vessels of Noncircular Cross Section
171.10
78.42
3
188.90
78.42
171.10
188.90
311.81
78.42
472.01
311.81
416.03
101.58
78.42
472.01
P = 20 psi
2
Knowing the values of Mo , V o , and H o , the remaining
forces and moments in the structure can easily be calculated and are shown in Figure 17.15.
4
576.23
101.58 416.03
101.58
576.23
171.10
101.58
188.90
1
171.10
188.90
Figure 17.15 Moments and forces in members.
References
1 European Committee for Standardization. Publication
EN 12952 Part 3.
2 European Committee for Standardization. Publication
EN 13445.
3 Parcel, J. and Moorman, R. (1955). Analysis of Stat-
ically Indeterminate Structures. New York, NY: John
Wiley Publishing 20th Printing 1976.
Further Reading
(1950). Allowable Working Pressures – Square and
Rectangular Headers. Barberton, OH: Babcock & Wilcox
(private communication).
Young, W., Budynas, R., and Sadegh, A. (2012). Roark’s
Formulas for Stress and Strain. New York: McGraw Hill.
4 British Standards Institution. Publication BS 1113.
5 Wang, C.K. (1953). Statically Indeterminate Structures.
New York, NY: McGraw Hill.
6 Blodgett, O. (1993). Design of Weldments. Cleveland,
OH: James F. Lincoln Arc Welding Foundation.
Large Central-Station Boiler. Source: Courtesy of Babcock & Wilcox.
366
18
Power Boilers
18.1 General
The American Society of Mechanical Engineers (ASME)
Boiler and Pressure Vessel Code contains rules for
the construction of both power boilers [1] and heating
boilers [2]. The ASME Code, I, contains rules for the construction of power boilers in which steam or other vapor
is generated at a pressure above 15 psi, for use external
to itself. In addition, ASME Code, I, contains rules for
watertube boilers, firetube boilers, feedwater heaters,
miniature boilers, electric boilers, and organic-fluid
vaporizer generators. The rules are applicable below a
15 psi limit for power boilers and to operation at pressures above 160 psi and/or temperatures above 250 ∘ F
for high-temperature water boilers. The scope includes
the boiler and its external piping, including such parts as
the superheaters, economizer, and other pressure parts
with no intervening valves between the parts and the
boiler.
18.3.1
Allowable Stress Values
The basic factor of safety of the ASME Code, I, is 3.5,
based on the ultimate tensile strength. However, the
allowable stress values are set by combinations of test
data, experience, and successful application of the values.
The basis for setting the allowable stress values essentially applies factors of safety to the strength properties
described in Section 4.1.2.
18.3.2
Cylinders under Internal Pressure
The ASME Code, I, contains two basic formulas for
determining the minimum required thickness of a
cylinder under internal pressure. One is for tubes up
to and including 5 in. outer diameter; the other is for
piping, drums, and headers (without a limitation on the
diameter).
For tubes up to and including 5 in. OD, the equation is
t=
18.2 Materials
Generally, the only materials listed in paragraphs PG-6
through PG-9 of the ASME Code, I, are permitted
for construction of pressure parts of power boilers.
Additionally, materials listed in Code Cases and materials qualified in accordance with paragraph PG-10 are
allowed. Gage glass bodies and connectors that have had
a pressure–temperature rating marked on them may also
be used without further proof.
18.3 General Design Requirements
General design requirements are given for most of the
major components; however, depending upon the specific type of boiler or boiler component, there may be
alternative formulas or charts, which may be used to set
the minimum required thickness of some components.
PD
+ 0.005D + e.
2Sw + P
(18.1)
For piping, drums, and headers, the equation is
t=
PD
PR
+C =
+ C,
2SE + 2yP
SE − (1 − y)P
where
t = required thickness
P = internal pressure
D = outside diameter
R = inside radius
E = lower of joint efficiency or ligament efficiency
factors (E = 1.0 for a seamless cylinder)
S = allowable stress at design temperature
C = factor for threading and stability
e = factor for expanded tubes
w = weld joint strength reduction factor
y = temperature coefficient as follows:
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
(18.2)
18.3 General Design Requirements
y
Temperature (∘ F)
Material
T ≤ 900
950
1000
1050
1100
1150
1200
≥1250
Ferritic
0.4
0.5
0.7
0.7
0.7
Austenitic
0.4
0.4
0.4
0.4
0.5
0.7
0.7
0.7
0.7
0.7
Alloy 800
0.4
0.4
0.4
0.4
0.4
0.7
0.4
0.5
0.7
Alloy 800H
0.4
0.4
0.4
0.4
0.4
0.4
0.5
0.7
Alloy 825
0.4
0.4
0.4
—
—
—
—
—
y is a correction factor to account for the redistribution of stress in a cylinder due to creep at temperatures
in the time-dependent regime [3] at an extended period of time. Values of y between temperatures may be
interpolated. The y table in the ASME Code, I, lists additional nonferrous materials.
Example 18.1
A boiler steam drum with an inside diameter of 60.0 in. is
subjected to an internal pressure of 1500 psi. The allowable stress of the drum plate is 17,500 psi; the design temperature is 650 ∘ F; the attached tubes are 4 in. nominal
diameter, so that C = 0. The weld joint efficiency is 0.90,
and the tube ligament efficiency is 0.62. Find the required
thickness according to ASME Code, I.
Solution:
From the chart in ASME Code, I, y = 0.4 and E is the lesser
of the joint efficiency and the tube ligament efficiency.
Using Eq. (18.2),
PR
+C
SE − (1 − y)P
1500 × 30
=
17,500 × 0.62 − (1 − 0.4) × 1500
t = 4.52 in.
t=
Example 18.2
A 3.5 in. OD seamless tube is to be used in the boiler
described in Example 18.1. The tube is strength-welded
into the drum. The allowable tube stress is 15,000 psi.
What is the minimum required thickness of the tube?
Let w = 1.0.
Solution:
When the outside diameter of the tube is 5 in. or less,
the required thickness is determined by using both Eqs.
(18.1) and (18.2). The lesser required thickness becomes
the minimum required thickness for that tube.
From Eq. (18.1),
PD
+ 0.005D + e
2S + P
1500 × 3.5
+ 0.005(3.5) + 0
=
2 × 15,000 × 1.0 + 1500
t = 0.184 in.
From Eq. (18.2),
PD
+C
t=
2SE + 2yP
1500 × 3.5
+ 0.065
=
2 × 15,000 × 1 + 2 × 0.4 × 1500
t = 0.233 in.
Since t from Eq. (18.1) is the lesser, it is used:
t = 0.184 in.
Problems
18.1
A superheater header is designed for 2000 psi at a
temperature of 1000 ∘ F. The material is ferritic, the
inside diameter of the header is 11.5 in., and the
allowable stress is 7800 psi. The header is seamless,
and all tubes are welded. Determine the required
thickness of the header.
Answer:
t = 1.60 in.
18.2
A steam drum has 48.0 in. inside diameter. The
design pressure is 1250 psi, and the design temperature is 650 ∘ F. The allowable stress value
is 15,000 psi. The drum is fully radiographed
(E = 1.0), but the ligament efficiency is 0.46. The
drum contains tubes, which are expanded into
tube seats and have 3 in. outside diameter. Determine the required thickness of the drum and the
tube. Design data for tube: C = 0.065; e = 0.04 if
thickness does not exceed 0.120 in.
t=
Answer:
t drum = 4.88 in.
t pipe = 0.135 in.
367
18 Power Boilers
18.4 Formed Heads under Internal
Pressure
t = 5PL∕4.8Sw,
5PL
5 × 1500 × 60
=
4.8Sw 4.8 × 17,500 × 1.0
t = 5.36 in.
t=
The method to calculate the required thickness of a
hemispherical head for ASME Code, I, is the same as
that described for ASME Code, VIII-1, in Section 9.2
and given in Eq. (9.1).
For unstayed dished heads, which contain a spherical
segment, ASME Code, I, uses a different equation than
that given in Eq. (9.9) for the ASME Code, VIII-1. It is
Problem
18.3
Example 18.3
Determine the required thickness of a seamless dished
head for the boiler steam drum given in Example 18.1.
The dish radius is equal to the inside diameter of
the drum.
What is the required thickness of a dished head
if D = 240 in., L = 240 in., S = 17 500 psi, W = 1.0,
and P = 50 psi?
Answer:
t = 0.72 in.
(18.3)
where the definitions of symbols are the same as those
that are given for Eq. (9.9). For other formed heads, such
as ellipsoidal, the thickness shall be not less than that
required for a seamless cylinder of the same diameter.
18.5 Loadings on Structural
Attachments
Limitations are given in ASME Code, I, for the maximum
load per inch of an attachment to a tube. The equation is
W
6We
(18.4)
P= r ± 2
l
l
and
Pa ≥ P,
(18.5)
where
Solution:
P = 1,500 psi; L = 60 in.; S = 17,500 psi; w = 1.0. From
Eq. (18.3),
5000
4500
4000
3500
3000
P = allowable load (lb/in.) from Figure 18.1
P = actual load
Figure 18.1 Chart for determination of
allowable loading on structural
attachments to tubes. Source: Courtesy of
the American Society of Mechanical
Engineers.
2 in. O.D.
3.1/4 in. O.D.
4 in. & 5 in. O.D.
2500
2000
Tension loading
1500
Allowable loading, (lb / in.)
368
Compression loading
all diameters
1000
900
800
700
600
550
500
450
400
350
300
1/2 in. O.D.
1 in. O.D.
250
200
150
100
90
80
74
6
7 8 9 10
15
20
25 30 35 40
50 60 70 8090 100
Outside diameter
(Wall thickness)2
150
200 250 300
400 450
18.6 Watertube Boilers
Problem
W
Wr
Lug
e
Tube
18.4
l/2
l
A 1200 lb. horizontal section of an economizer
tube is supported by a movable lug. When the
tube expands, the loading becomes eccentric
to the lug center line and the tube as shown in
Figure 18.3b. The tube is 3.0 in. OD × 0.300 in. Is
the design satisfactory, and what are the allowable
loads and the actual loads?
Answer:
Allowable unit loads are 950 lb. in tension; 720 lb.
in compression. Actual unit loads are 750 lb. in
tension; 150 lb. in compression.
Figure 18.2 Method of computation of attachments to tubes.
Source: Courtesy of the American Society of Mechanical Engineers.
W r = load component normal to tube axis (lb)
W = load applied to lug (lb)
18.6 Watertube Boilers
l = length of attachment to tube (in.)
e = eccentricity of W , (in.) (see Figure 18.2)
Power boilers are usually watertube boilers. Watertube
boilers are constructed so that water and steam are
circulated inside of the tubes and combustion gases are
circulated around and against the outer tube wall; thus,
heat is transferred through the wall to the fluid within
the tube. There are basically two types of watertube
boilers: the once-through, forced-flow boiler, as shown
in Figure 18.4, and the drum-type boiler, as shown in
Figure 18.5. The former is used exclusively for power
generation at large central stations, while the latter
is used not only for power generation at both central
stations and industrial companies, but also for the generation of steam used for processing industries. Some of
the advantages of watertube boilers are their ability to
generate high rates of steam, their ability for quick load
change, and their relatively short start-up time.
The ASME Code, I, lists equations representing the
curves shown in Figure 18.1 to facilitate the calculations.
Example 18.4
A 700 lb. load is supported from a lug on a vertical tube, as
shown in Figure 18.3a. The tube is 3.5 in. OD × 0.380 in.
Are the lug dimensions satisfactory?
Solution:
3.5
D
=
= 24.2.
t2
(0.380)2
Allowable unit load, Pa = 960 lb. in compression;
1300 lb. in tension. Actual width W = 700 lbs.
6We 6 × 700 × 2
P= 2 =
l
(3)2
= 930 lb in compression and tension.
18.6.1
The actual loading is less than the allowable loading;
therefore, the lug dimensions are satisfactory.
Special Design Requirements and Rules
In addition to the general design requirements given in
Part PG of ASME Code, I, special requirements for
Figure 18.3 Structural attachments to tubes.
2 in.
700 lb
1 in.
1200 lb
2 in.
3 in.
3.1/2 in. O.D.
x 0.380 in.
4 in.
3 in. O.D.
x 0.300 in.
(b)
(a)
369
370
18 Power Boilers
Sealingair fan
Attemperator
Reheat
superheater
Secondary
superheater
Primary
superheater
Reheat
superheater
Economizer
Gas
tempering
ports
Coal
silo
230′–0″
Coal
silo
Gasrecirculating
ports
feeder
Feeder
Secondaryair duct
Air heater
Dust
collector
Cyclone
furnaces
Cyclone
furnaces
Air
inlet
Gas
outlet
Gasrecirculating
fans
35–0″
27′–0″
16′–0″
27′–9″
Figure 18.4 Once-through forced-flow boiler. Source: Courtesy of Babcock & Wilcox.
23′–3″
40′–0″
18.6 Watertube Boilers
Gas
outlet
Forced-draft
fan
Induced-draft
fan
Air heater
Economizer
Superheater
41′–3″
Dust
collector
Attemperator
Coal
hopper
Stoker
16′–0″
19′–9″
Figure 18.5 Drum-type boiler. Source: Courtesy of Babcock & Wilcox.
23′–0″
371
372
18 Power Boilers
Steam nozzle
Safety-valve
connection
Feed pipe
Dry pipe
Braces
Through stay
Blowdown
connection
Grate
Ashpit
Figure 18.6 Firetube boiler.
Table 18.1 Maximum allowable working pressure for steel flues for firetube locomotive boilers.
Nominal wall thickness (in.)
0.095
0.109
0.120
Outer diameter
of flue (in.)
0.134
0.148
0.165
0.180
0.203
0.220
0.238
…
Maximum allowable working pressure (psi)
1
470
690
860
…
…
…
…
…
…
1.5
320
460
570
720
860
…
…
…
…
…
1.75
270
400
490
620
740
890
…
…
…
…
2
240
350
430
540
650
780
900
…
…
…
2.25
210
310
380
480
580
690
800
960
…
…
2.5
190
280
350
430
520
630
720
860
970
1080
3
160
230
290
360
430
520
600
720
810
900
3.25
…
210
270
330
400
480
550
660
750
830
3.5
…
200
250
310
370
450
520
620
690
770
680
4
…
180
220
270
330
390
450
540
610
4.5
…
160
190
240
290
350
400
480
540
600
5
…
…
180
220
260
320
360
430
490
540
5.375
…
…
160
200
240
290
340
400
450
500
5.5
…
…
…
200
240
290
330
390
440
490
6
…
…
…
180
220
260
300
360
410
450
General notes: 1. The steel specifications are listed in PG-9 of ASME Code, I.
2. The pressure values are obtained from the approximate Eq. P = 15,650 (t − 0.065)/D, where D is the outer diameter of tube, P is the pressure, and
t is the tube thickness.
Source: Courtesy of the ASME.
References
watertube boilers are given in Part PWT. The minimum
wall thickness is determined by Eqs. (18.1) and (18.2).
18.7 Firetube Boilers
Firetube boilers are constructed in a reverse manner
to watertube boilers. That is, the combustion gases
pass inside the tubes while the water and steam circulate
around the outside. The advantages are that construction
is simple and less water treatment is required. One type
is shown in Figure 18.6.
18.7.1
Special Design Requirements and Rules
The maximum allowable working pressure for steel flues
for firetube locomotive boilers is given in Table 18.1.
There are a number of other equations given in the
ASME Code, I, for determining maximum allowable
working pressure P depending upon size and application. These equations were developed in the early 1920s
and are based upon experimental test data. See Part PFT
of the ASME Code, I, for the equations and restrictions
on their use.
References
1 ASME Boiler and Pressure Vessel Code, Section I,
Power Boilers, ANSI/ASME BPV-I. New York: American Society of Mechanical Engineers.
2 ASME Boiler and Pressure Vessel Code, Section IV,
Heating Boilers, ANSI/ASME BPV-IV, American
Society of Mechanical Engineers, New York.
3 Jawad, M. and Jetter, R. (2009). Design and Analysis of
ASME Boiler and Pressure Vessel Components in the
Creep Range. New York: ASME Press.
373
375
A
Guide to ASME Code
Figure A.1 Guide to ASME Section VIII, Division 1. Source: Courtesy of The Hartford Steam Boiler Inspection and Insurance Company.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
376
A Guide to ASME Code
Figure A.1 (Continued)
Guide to ASME Code
Figure A.1 (Continued)
377
378
A Guide to ASME Code
Figure A.2 Guide to ASME Section I. Source: Courtesy of The Hartford Steam Boiler Inspection and Insurance Company.
Guide to ASME Code
Figure A.2 (Continued)
379
380
A Guide to ASME Code
Truncated Cone Shaped Combustion Chamber—PW-9,29,39;
PFT-14,23.
Radial Stay—PG-13, 46; PW-19;
PFT-23,25-27; Fig. A-8
Upper Tube Sheet and Lower Tube
Sheet—PG-8, 18, 22,46,76,77,
79; PW-9;, 11, 31,; PMB-5
PEB-5
Fire Tube—PG-9, 16; PFT-12.
Mud Ring, Water Wall, Water Leg—
PW-9, 31; PWT-12; PFT-5, 11, 21,
27; Fig A-8
Furnace Sheet & Throat Sheet—
PG-6; PFT-11, 20, 27
Furnace Staying—PG-13, 46-49;
PFT-26-28; Fig A-8.
Combustion Chamber or Fire
Box—PFT-13, 21, 23.
Ogee Attachment—PFT-5, 20, 27.
VERTICAL TUBE BOILER
Guide to ASME Code
Sting Stay—PG-13, 46; PFT-13, 30.
Wrappar Sheet—PG-6, 27; PFT-9,
11, 23, 41.
Radial Stay or Crown Stay—PG13,46; PW-19; PFT-23, 25-27;
Fig. A-8
Crown Stay—PG-6; PFT-13
23, 30.
Firebox Welding—PW-9, 26-38;
PFT-21, 27.
Furnace Sheet & Throat Sheet—
PG-6; PFT-11, 20, 27.
Furnace Staying—PG-13, 48-49;
PFT-26-28; Fig. A-8.
Mud Ring, Water Wall, Water Leg—
PW-9, 31; PWT-12; PFT-5, 11, 21,
27; Fig A-8
Base Plate—PG-6.
Combustion Chamber or Fire
Box—PFT-13, 21, 23.
Crown Bar (Arch Bar)—PG-6;
PFT-13, 30.
FIREBOX BOILER
381
383
B
Sample of Heat-Exchanger Specification Sheet
HEAT-EXCHANGER SPECIFICATION SHEET
1
JOB NO.
2
CUSTOMER
REFERENCE NO.
3
ADDRESS
4
PLANT LOCATION
5
SERVICE OF UNIT
6
SIZE
TYPE
(HORIZ.)
(VERT.)
CONNECTED IN
7
SQ. FT. SURF./UNIT
(GROSS)
(EFF.)
SHELLS/UNIT
SQ. FT.
SURF./SHELL
PROPOSAL NO.
DATE
ITEM NO.
8
PERFORMANCE OF ONE UNIT
9
SHELL SIDE
10
FLUID CIRCULATED
11
TOTAL FLUID
ENTERING
12
VAPOR
13
LIQUID
14
STEAM
15
(GROSS) (EFF.)
TUBE SIDE
NONCONDENSABLES
16
FLUID VAPORIZED OR
CONDENSED
17
STEAM CONDENSED
18
GRAVITY
19
VISCOSITY
20
MOLECULAR WEIGHT
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
384
B Sample of Heat-Exchanger Specification Sheet
BTU/LB.° F
BTU/LB.° F
BTU/HR-FT.° F
BTU/HR-FT.° F
BTU/LB
BTU/LB
TEMPERATURE IN
°F
°F
25
TEMPERATURE OUT
°F
°F
26
O P E R A T I N G P R E S S U RE
PS I G
PS I G
27
N O . P A S S E S P E R S HE L L
28
VELOCITY
FT/SEC
FT / S E C
29
PRESSURE DROP
PSI
PSI
30
FOULING RESISTANCE
(MIN.)
31
HEAT-EXCHANGEDBTU/HR
32
TRANSFER RATE—
SERVICE
21
SPECIFIC HEAT
22
THERMAL
CONDUCTIVITY
23
LATENT HEAT
24
MTD CORRECTED.° F
CLEAN
33
CONSTRUCTION OF ONE SHELL
34
DESIGN PRESSURE
PSI
PSI
35
TEST PRESSURE
PSI
PSI
36
DESIGN
TEMPERATURE
°F
°F
37
TUBES
NO.
O.D.
38
SHELL
I.D.
O.D.
39
CHANNEL OR BONNET
CHANNEL COVER
40
TUBESHEET—
STATIONARY
TUBESHEET–FLOATING
41
BAFFLES—CROSS
TYPE
FLOATING HEAD
COVER
42
BAFFLES—LONG
TYPE
IMPINGEMENT
PROTECTION
43
TUBE SUPPORTS
44
TUBE TO TUBESHEET JOINT
45
GASKETS
46
CONNECTIONS–SHELL
SIDE
IN
OUT
BWG
RATING
LENGTH
PITCH
SHELL
COVER
(INTEG)
(REMOV)
Sample of Heat-Exchanger Specification Sheet
47
CHANNEL SIDE
48
CORROSION
ALLOWANCE—SHELL
SIDE
49
CODE REQUIREMENTS
50
REMARKS
IN
OUT
RATING
TUBE SIDE
TEMA CLASS
51
52
53
54
Source: Courtesy of the Tubular Exchanger Manufacturers Association Tarrytown, NY.
385
387
C
Sample of API Specification Sheets
API Standard 650
Storage Tank Specification Data Sheet
______________________________________
______________________________________
General; Information (By Purchaser)
1. Purchaser/Agent _______________________________________________________
Address ________________________________________________________________
City ______________________ State ________________ Phone __________________
2. User _________________________________________________________________
3. Erection Site: Name of Plant ______________________________________________
Location _______________________________________________________________
4. Tank No. ____________ Tank Capacity (bbl): Nominal _____ Not Working _______
5. Pumping Rates: In ______________ bbl/hr.
Out _____________ bbl/hr.
6. Max. Operating Temperature _________________ F
7. Product Stored _____ Design Specific Gravity __________ @ _________________ F
Design Metal Temp. __________ F Vapor Pressure _____________________ in. water
8. Corrosion allowance (in.): Shell _________________ Roof _____________________
Bottom ________________________ Structurals _______________________________
9. Shell Design: Basic API 650_____App. A_____App. F_____Design Pressure______
10. Roof Design: Basic API 650 ____________ Floating Roof App. C ______________
11. Internal Floating Roof App. H ___________________________________________
Frangible Roof Joint: Yes __________________ No ______________
Roof Loads: Uniform Live (consider snow) ____________________________ lb/sq. ft.
Special Loading (provide sketch)____________________________________
12. Earthquake Design per App. E: Yes ______________ No _____________________
Seismic Zone (Figure E-1) _________________ Essential Facilities Factor__________
Zone Coefficient (Table E-1) _______________________________________________
Site Amplification Factor (Table E-2) ________________________________________
Roof Tie Rods (3.10.4.5): Yes _______________ No __________________
13. Wind Load: Velocity (mph) _____________________________________________
Provide Intermediate WindGirder (as per 3.9.9): Yes _____________ No __________
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
Sheet 1 of 3
File No. _____
388
C Sample of API Specification Sheets
14. Environmental Effects: Rainfall, max. ____________________________ in. per hr.
Snowfall, Total Accumulation ___________________________________________ in.
15. Diameter and/or Height Restrictions ________ Diameter, max. _______ Height, max.
16. Foundation Type: Earth _______________ Concrete Ringwall _________________
Other __________________________________________________________________
Remarks _______________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
Source: Courtesy of the American Petroleum Institute, Washington, D.C.
Sample of API Specification Sheets
Sheet 2 of 3
File No. _____
Construction Details (By Manufacturer and/or Purchaser, as Applicable)
1. Manufacturer __________________________________________________________
Address _____________________________________________________________
City ___________________ State ______________ Phone ____________________
Serial No. ___________________________________________________________
2. Fabricator_____________________________________________________________
Address _____________________________________________________________
City _______________ State _____________ Phone _________________________
Serial No. ___________________________________________________________
3. Material Specifications: Shell _____________________________________________
Roof _______________________________________________________________
Bottom _____________________________________________________________
Structurals ___________________________________________________________
4. Shell Courses (No. of) ___________________________________________________
5. Plate Width and Thickness (Including Corrosion Allowance)
1 ____________________ 2 ____________________ 3 ______________________
4 ____________________ 5 ____________________ 6 ______________________
7 ____________________ 8 ____________________ 9 ______________________
6. Tank Bottom: Plate Thickness ____________________________________________
Seams (Check One) ______________________ lap _______________ butt
Slope ________ in. per ft.
Check One: To _________ From ________ Center
7. Bottom Annular Plates Min. Width and Thickness (see 3.5)____________________
8. Roof to Shell Detail: Figure F-1 ____________________________________________
9. Intermediate WindGirder: Yes _______________ No _________________________
Top WindGirder (Use as Walkway): Yes _______________ No _______________
10. Roof Type: Supported _______________ Self-Supported _____________________
Slope or Radius ____________________ Floating __________________________
11. Roof Plate: Thickness ____________________ Lap Joint _____________________
Butt Joint ___________________________________________________________
12. Paint-Shell: Exterior–Yes _______ No ______ Interior–Yes _______ No _________
Bottom Interior–Yes _______ No _______ Underside–Yes _______ No ________
Surface Preparation ___________________________________________________
13. Tank Bottom Coating: Interior–Yes __________ No ________ Material _________
Application Specification ______________________________________________
14. Paint: Structural Steel Interior–Yes _____ No ____ Exterior–Yes _____ No ______
Specification _________________________________________________________
15. Inspection By: Shop _________________________ Field _____________________
16. Weld Examination: Radiograph __________________________________________
Supplementary Liquid Penetrant or Ultrasonic ___________________________
17. Films _____________________ Property of ________________________________
389
390
C Sample of API Specification Sheets
18. Leak Testing: Bottom ___________________ Roof __________________________
Shell _______________________________________________________________
19. Mill Test Reports Required: Yes _________________ No _____________________
Plate _________________ Structural Shapes _______________________________
20. Purchaser’s Reference Drawing __________________________________________
21. Tank Size: Diameter and Height in ft. _____________________________________
22. Date of Edition or Revision of API Standard 650 ____________________________
Remarks ________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
Sample of API Specification Sheets
Sheet 3 of 3
File No. _____
Appurtenances (By Manufacturer and/or Purchaser, as Applicable)
1. Stairway Style (Check One): Circular _______________ Straight _________________
Angle, Degree to Horizontal ____________________ Ladder ______________________
2. Walkway: Width _____________________ Length ___________________________
3. Drawoff Sump: Standard _________________ Special ________________________
4. Bolted Door Sheet (App. A only): ________ Raised-Type _____ Flush-Type ______
5. Scaffold Hitch _________________________________________________________
6. Internal Pipe: Swing Line _________________ Suction Nozzle __________________
Heating Coil Surface Area _________________________________________ sq. ft.
7. Roof Drain: Hose ___________ Jointed ___________ Siphon ___________________
8. Shell Manways: No. and Size _____________________________________________
9. Roof Manways: No. and Size _____________________________________________
10. Shell Nozzles (See Figure 3-4B, 3-5, 3-6, and Tables 3-8, 3-9, and 3-10)
Mark
Size
Flanged
SGL
DBL
Screwed
SPL
A
B
C
D
E
Orientation
N = 0*
Height From
Bottom
Service
11. Roof Nozzles (Including Venting Connection) (See Figure 3-4 and 3-15 and Tables 3-16 and 3-17)
Mark
Size
Flanged
Screwed
Reinf.
Orientation
N = 0*
Distance
from Center
Note: Sketch and/or separate sheet may be attached to cover special requirements.
Service
391
393
D
Sample of Pressure Vessel Design Data Sheets
Pressure Vessel Design Data Sheet
1) Requirements:
Working Pressure: 525 psig
Design Code: ASME Section VIII – Div. 1
Working Temperature: 500 ∘ F
Construction: All Welded
corrosion allowance: 0.063 in.
2) Material Specifications:
Material
Material specifications
Maximum allowable stress
Reference
P no.
Grp. no.
Plate
SA-204, Grade B
20,000 psi
Tbl. 1A, II-D
3
2
Castings
SA-216, Grade WCA
16,300 × 0.8 = 13,040
Tbl. 1A, II-D UG-24(a)
1
1
Forgings
SA-181
16,300
Tbl. 1A, II-D
1
1
Bolting
SA-193, Grade B16
20,000
Tbl. 1A, II-D
Piping
SA-106, Grade B
17,100
Tbl. 1A, II-D
1
1
Gaskets
1/16 in. mineral
y = 3700; m = 2.75
Tbl. 2–5.1
3) Sketch and General Dimensions of Vessel
4-in. outlet with welding neck flange
Seamless dished head, pressure on
concave side, unstayed
8-in. outlet with lap joint flange
8-in. manway
36 in.
nominal inside
diameter
Special cast steel welding
neck flange
Special forged steel blind
flange
1-1/2 in. blowdown outlet with welding
neck flange
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
394
D Sample of Pressure Vessel Design Data Sheets
Design of Head and Shell
Shell Thickness: UG-27 (c)(1)
Minimum required thickness, t =
1)
PR
(SE) − (0.6P)
(estimate E = 0.70, Tbl UW − 12)
(double welded butt joint)
(525)(18)
= 0.691
(20,000 × 0.70) − (0.6 × 525)
add corrosion allowance ∶ 0.691 + 0.063 = 0.753 in.
2) Check postweld heat treatment and radiograph requirements.
UCS-56, Tbl UCS-56 Note (3)(a): Postweld heat treatment required if thickness exceeds 5/8 in; because
0.753 > 0.625, postweld heat treatment required UW-11 (a), UCS-57 (note: cf. UW-52): radiograph required
P-3, Gr 1, 2, 3 if t > 0.750 in.; because 0.753 > 0.750, radiography is required;
3)
recalculate thickness required using E = 0.85(Tbl UW − 12, spot examined)
525 × 18
t+c=
+ 0.063 = 0.566 + 0.063 = 0.629
(20,000 × 0.85) − (315)
use plate thickness = 0.750
4) Check applicability of formula: UG-27 (c)(1)
0.385 SE = (0.385) (20,000) (0.85) = 7700; 525 < 7700 formula OK
0.5R = (0.5) (18) = 9.0;0.750 < 9.0 formula OK
Head Thickness: (UG-32)
1) A dished head will be used, UG-32 (j) requires that the inside knuckle radius shall be not less than 6% of the outside
diameter of the skirt, nor less than three times the head thickness. Also, the inside crown radius is to not exceed
the outside diameter of the skirt, thus try:
inside crown radius, L = 27 in. < 37.375 in.
inside knuckle radius, r = 4 in. > 0.06 × 37.375 = 2.24 in.
2) Thickness required: UA-4 (d)
PLM
t=
(2SE) − (0.2P)
t+c =
E = 1.00 for seamless head
for L∕r = 27∕4 = 6.75,
M = 1.40(Tbl UA − 4.2)
(525)(27)(1.40)
+ 0.063
(2 × 20,000 × 1) − (0.2 × 525)
= 0.500 + 0.063
t + c = 0.563 therefore use head plate thickness of 0.688 in.(11∕16)
3) Check minimum thickness: UG-32 (b).
For a joint efficiency of 0.85 between the head and shell, postweld heat-treated double-butt weld for the circumferential joint: UG-32 (f ).
[
]
1
PL
tmin =
(0.85) (2SE) − (0.2P)
[
]
(525)(18)
1
tmin =
; tmin = 0.237
0.85 (2 × 20,000 × 1) − (0.2 × 525)
0.237 < 0.688, thickness OK.
from UCS-16 (b): 3/32 < 0.688, thickness OK.
4) Check minimum inside knuckle radius: UG-32 (j)
(3)(0.688) = 2.063 4.0 > 2.063 knuckle radius OK
Sample of Pressure Vessel Design Data Sheets
5) Cylindrical skirt: UW - 13 (a) 0.688 < (1.25)(0.750) = 0.937
A cylindrical skirt is optional for the head butt-welded to the shell, a tangent of 1 1/2 in. length will be used to
provide a bevel for the butt weld; this will avoid cutting the bevel into the knuckle.
Sketch of the head and shell showing principal dimensions and tolerance.
11
16
3
4
7± 1
35 8 8
[UG–80]
1
1
a) Vessel to be postweld heat-treated at 1100 ∘ F for 3/4 h.
b) Maximum permissible offset of circumferential joints = 3/16′′ (UW-33).
c) Spot examination of welded joints per UW-52.
1 3/32′′ reinforcing bead: UW-35 (a), UW-51 (a)(1).
No offset preparation necessary between the head and shell because the thicknesses do not differ by more than 1/8 in.
UW-9(c).
Inside surface of head shell not deviate
more than 7/16 in. from 27 in. radius UG–81 (a)
4 in. R
27 i
1–1/2 in., bevel 30″ for welding
n. R
1
18 in.
4
14 in.
Data Sheet for Reinforcement Calculations (UG-37, −40)
1) Shell
Description: (longitudinal butt joint, double-welded, postweld heat-treated, spot radiography).
Design pressure
P
525 psig
Joint efficiency
E
0.85
Maximum allowable stress
S
20,000 psi
Corrosion allowance
c
0.063 in.
Inside radius of shell (specify) before corrosion allowance is added R, L, K 1 , D
Nominal thickness, exclusive of corrosion allowance
t
0.687 in.
Minimum required thickness (reference: UG-27)
tr
0.480 in.
tr =
(525)(18)
PR
=
(SE) − (0.6P) (20,000 × 1) − (0.6 × 525)
Excess thickness
t – tr
0.207
2) Nozzle
Material used: seamless steel pipe, 5A-106, Grade B (Schedule 80).
18 in.
395
D Sample of Pressure Vessel Design Data Sheets
Maximum allowable stress
S
17,100 psi
Inner diameter of finished opening in corroded condition
d
7.750 in.
Nominal thickness exclusive of corrosion allowance
tn
0.438 in.
Thickness required for hoop stress (UG-27)
t rn
0.117 in.
trn =
(525)(7.50∕2)
PR
=
(SE) − (0.6P) (17,100 × 1) − (0.6 × 525)
3) Sketch of reinforcement with dimensions and welding detail (UW-15, −16, −18)
8.625
tr n = 0.117
xs = 0.321
1 in.ϕ
Tell tale
4
Hole (UW-15c)
0.063
50
14.625 DIA
0·207
0.687
tn = 0.438
1.592
396
d = 7.750
15.50
Reinforcement Calculations for 8 in. Nozzle
1) Reinforcement required: UG-37
A = 3.720 sq. in.
A = (d)(t r )(F) = (7.750) (0.480)(1)
2) Metal considered to have reinforcing value: UG-40(d).
A1 = metal in the shell and in the nozzle wall within the shell wall thickness available for reinforcement:
Shell: (15.50–8.625)
(0.207)
= 1.423 sq. in.
Nozzle: (8.625–7.750)
(0.207)(17,100/20,000)
= 0.155 sq. in.
(cf UG-41a)
A1 = 1.578 sq. in.
A2 = metal in the nozzle wall outside the shell thickness available for reinforcement:
(0.438–0.117)(2)(h)
A2 = 0.642 h sq. in.
(0.438–0.117)(2)(1.595) see below for h
= 1.024 sq. in.
A3 = metal added as reinforcement and metal in attachment welds:
Added reinforcement (1/2 in. plate, 8–5/8 ID by 14–5/8 OD).
(0.500)(14.625–8.625)
= 3.000 sq. in.
One-half in. full fillet welds around outside of reinforcing plate and around nozzle wall:
(0.50)(0.50)(0.50)(4)
= 0.500 sq. in.
A3 = 3.500 sq. in.
Sample of Pressure Vessel Design Data Sheets
3) Width of area of reinforcement normal to vessel wall:UG-40(c)
(2.5)(0.687)
= 1.720 in.
or
(2.5)(0.438)
= 1.095.
Added reinforcement
= 0.500.
h = 1.595 (this governs)
4) Summary:
Reinforcement used:
A1 : shell and nozzle
1.578
A2 : nozzle
1.024
A3 : added reinforcement and welds
3.500
6.102 sq. in. used >3.72 required.
Attachment Welding for 8 in. Outlet (UW-15, -16)
1) Allowable stresses in welds: (Tbl. UW-15)
a) Combined end and side loading stress in butt welds
(20,000) (0.74) =
14,800 psi
b) Combined end and side loading stress in fillet welds
(20,000) (0.49) =
9800 psi
2) Strength of Welds:
a) Outer fillet weld between vessel wall and reinforcing plate
(π) (14.625/2) (0.500) (9800) =
112,600 lb.
b) Butt weld between vessel wall and nozzle wall
(π) (8.625/2) (0.687) (14,800) =
137,800 lb.
c) Fillet weld between nozzle and reinforcing plate
(π) (8.625/2) (0.500) (9800) =
66,400 lb.
d) Butt weld between nozzle wall and reinforcing plate
(π) (8.625/2) (0.500) (14,800) =
100,300 lb.
3) Loads:
a) Strength in tension of the plate removed (UG-41(b)(2)
(20,000) (0.480) (8.625) =
82,800 lb.
b) Strength of metal in vessel wall available for reinforcement
(20,000) (0.207) (15.50–8.625) =
28,460 lb.
c) Reinforcement load carried by the nozzle wall
(17,100) (2) (0.438) (0.687) =
10,290 lb.
+ (17,100) (2) (0.321) (1.595) =
17,510 lb.
27,800 lb.
397
398
D Sample of Pressure Vessel Design Data Sheets
d) Reinforcement load carried by reinforcing plate
82,800 − (28,460 + 27,800) =
26,540 lb.
4) Summary:
a) Load to be carried by attached reinforcement
82,800–28,460 =
54,340 lb.
Strength of attachment = 112,600 + 137,800 =
250,400 lb.
b) Load to be carried by nozzle wall =
17,510 lb.
Strength of attachment =
137,800 lb.
c) Load to be carried by reinforcing plate =
26,540 lb.
Strength of attachment =
112,600 lb.
Attachment welding is satisfactory.
8 in. Lap-Joint Flange Attachment and Selection
1) Check welding between the nozzle and shell and reinforcing for strength to withstand hydrostatic end force.
Permissible shear stress in butt welds (Tbl UW-15)
(20,000) (0.60) =
12,000 psi.
Shear strength of butt weld (cf sheet 4 for dimensions)
(π) (8.625) (0.687) (12,000) =
223,400 lb.
Hydrostatic end force to inner diameter of nozzle
(π/4) (7.750)2 (525) =
24,700 lb.
Thus, the butt weld alone is safe for hydrostatic end force
223,400 > 24,700 lb.
2) From ANSI B16.5-2017 Steel Pipe Flanges and Fittings, use 400 lb.
Rated at 665 psi at 500 ∘ F.
From vendor’s catalog:
Number of bolts
12
Size of bolts
1 in.
Bolt circle diameter
13 in.
3) Use lap-joint stub end, 8.625 OD by 7.625 ID by 8 in. long
4) From vendor’s catalog, use 1/16 in. gasket – 12 in. OD by 8 in. ID.
Preliminary Design of 18 in. Integral Flange
A = (27.125)
C = (24.625)
W
critical section
R
R
g1
B = 16.875
t
W
g1/2
3
h
1
Stress
g0 = 0.563
18 in. O.D.
Design pressure =
525 psi
Flange design stress = 12,000 psi
Bolt design stress =
20,000 psi
Try 1.250 in. Diameter bolts
Root area = 0.943 sq. in. (8 thrd)
Estimate g0 = 0.563 in.
Section for
Section for
Radial
M longitudinal
hub stress
Flange
M
Sample of Pressure Vessel Design Data Sheets
Trial Dimensions
1) Hub length, h ≅
√
√
Bg. = (16.875)(0.563) = 3.08 in.
try h = 3.00 in.
2) Center line bolt circle to hub ≅ (1.5)Db = (1.5)(1.25) = 1.875
try R = 2.00 in.
3) Minimum bolt circle diameter for nut clearance = 18 + (2)(1 + 2)
try C = 24.0 in.
4) Bolt spacing = 3 in. or (2.25 × Db ) = (2.25)(1.25) = 2.81 in.
5) Number of bolts = π(24.0)/2.81 = 26.8
try 24 bolts
6) Minimum flange OD = 24.0 + 2(1.25) = 26.5
try A = 26.5 in.
Flange Thickness Calculations
Assume that the hydrostatic end force, H, is effective to 22 in. diameter, the assumed location of the gasket reaction,
and use a factor of safety of 2 to assure a tight joint.
1) Design hydrostatic end force, H = (π/4)(222 )(525)(2) H = 400,000 lb
2) Required number of bolts to carry this load:
= (400,000)/(20,000)(0.943) = 21.2 (this compares favorably with 24)
3) g 1 /2 = (1/2)(0.563 + 1.0) = 0.782
4) For the critical section at 16.875 + (2)(0.782) = 18.438 diameter, the arc length of a sector containing one bolt will
be: arc length = (π)(18 438)/24 = 2.42 in.
5) Allowable bolt load on sector = (20,000)(0.443) W = 18,860 lb/bolt
6) Gasket reaction = 18,860 – (1/24)(π/4)(222 )(525) = 18,860 – 8330 = 10,530 lb/bolt
7) Bending moment at critical section
= (18,860)(1/2)(24.0 – 18.438) – (10,530)(1/2)(22.00 – 18.438)
= 52,500 – 18,750 = 33,850 lb in.
8) Calculated thickness for radial flange stress: Sf = Mc/I; (UA-52)
12,000 = (33,850)/(1/6)(2.42)(t 2 );t 2 = 7.00
t = 2.65 in.
try t = 2.75 in.
9) Calculated thickness for longitudinal hub stress (UA-52)
(12,000)(1.5) = (33,850)/(1/6)(2.42)(g 1 2 ) ∴ g 1 2 = 4.66 and g 1 = 2.16 in.
10) Revise estimated flange dimensions using:
g 1 = 2.00 in
C = 16.875 + 2(2.00 + 1.875)= 24.625 in.
R = 1.875 in
A = 24.625 + 2(1.25)
= 27.125 in.
h = (3)(2 – 0.563) = 4.31, say 4.50 in.
Preliminary Design of 18 in. Integral Flange
1) A 1/16 in. flat asbestos gasket will be used, 19 in. ID by 22 in. OD
2) From Table 2-5.2:
Basic gasket seating width, bo = (1/2)(22√
– 19)/2 = 0.750 in.
Effective gasket seating width, b = (1/2) 0.750 = 0.433 in.
3) Location of gasket load reaction, Table 2-5.2
G = 22.0 – 2(0.433) = 22.0 – 0.866
G = 21.133 in.
399
400
D Sample of Pressure Vessel Design Data Sheets
Calculation for 18 in. Flange, 2–4 to 2–8
Blind Flange for 18 in. Manway√
Flange
1) From UG-34 (c)(2): t = d CP∕S + 1.9WhG ∕Sd3 , Figure UG-34(j).
2) From Calculation sheet above: W = 453,000 lb.
hG = (1∕2)(24.625 − 21.133) = 1.746 in.
d = G = 21.133 in.
H = 184,000 lb
Sample of Pressure Vessel Design Data Sheets
3) From sheet 1: S = 16,300 psi (SA-181 forging).
4) C = 0.30 UG-34(d)
√
(0.3)(525) 1.9(184 000)(1.746)
= 2.467
t = (21.133)
+
16,300
(16,300)(21.133)3
adding corrosion allowance: 2.467 + 0.063 = 2.530
use 2–5/8 thickness.
5) Sketch of blind flange
2–1/4 in. dia. spot face for nuts
Bolt holes,
evenly spaced,
24 at
1–5/8 in. dia.
2–5/8 in.
63
22 in. outside diameter of raised face
1/4 in.
raised face
24–5/8 bolt circle diameter
27–1/8 in. dia. flange outside diameter
Data Sheet for Reinforcement Calculations (UG-37, -40)
1) Shell
Description: longitudinal butt joint, double-welded, postweld heat-treated, spot radiography.
Design pressure
P
525 psig
Joint efficiency
E
0.85
Maximum allowable stress
S
20,000 psi
Corrosion allowance
c
0.063 in.
Inside radius of shell before corrosion allowance is added
R, L, K 1 , D
18 in.
Nominal thickness, exclusive of corrosion allowance
t
0.687 in.
Minimum required thickness (reference UG-27) see sheet 3
tr
0.480 in.
Excess thickness
t – tr
0.207 in.
2) Nozzle
Material used: seamless steel pipe, SA-106, Grade B (Schedule 40).
Maximum allowable stress
S
17,100
Inside diameter of finished opening in corroded condition
d
17.0 in.
Nominal thickness exclusive of corrosion allowance
tn
0.500 in.
Thickness required for hoop stress (UG-27)
t rn
0.266 in.
trn =
(525)(17.0 × .5)
PR
=
(SE) − (0.6P) (17,100 × 1) − (0.6 × 525)
401
D Sample of Pressure Vessel Design Data Sheets
3) Sketch of reinforcement with dimensions and welding detail (UW-15, -16, -18).
18.0
tn = 0.500
trn = 0.266
1.72
402
27.5
0·207
Tell-tale hole,
1/4 IN-D/A
0.75
0.480
0.687
0.063
d = 17.00
34.00
Reinforcement Calculations for 18 in. Manway Alternate Method: Load Calculations Based on Nozzle
and Vessel O. D
1) Total load to be carried within area of reinforcement
(525) (37.375/2) (34)
= 333 000 lb.
2) Strength of vessel walls within area of reinforcement
(0.687) (34.0–18.0) (20,000)
= 219,800 lb.
3) Strength of nozzle wall within vessel wall thickness
(0.687) (18.0–17.0) (17,100)
= 11,700 lb.
4) Load carried by nozzle wall due to pressure in nozzle
(525) (18) (h)
(est. h = 1)
(525) (18) (1.72)
= 9450 h lb.
= 16 260 lb. corr.
5) Strength of nozzle wall outside of vessel wall thickness
(0.500) (2) (h) (17,100)
= 17,100 h lb.
(0.500) (2) (1.72) (17,100)
29,400 lb. corr.
6) Strength of nozzle wall available for reinforcement
(11,700) + (17,100 – 9450) (h)
= 19,350 lb. est.
(11,700) + (7650 × 1.72) = 11,700 + 13,200
24,900 lb. corr.
7) Load to be carried by added reinforcement
333 000 – 219,800 – 19,350
= 93,850 lb. est.
333 000 – 219,800 – 24,900 (h = 1.72 in.)
70,400 lb. corr.
8) Area required of added reinforcement
93,850/20,000
= 4.693 in.2 est.
Data Sheet for Reinforcement Calculations (UG-37, -40)
9) Area of reinforcement used
3/4 in. plate, 18 in. ID by 27.5 in. OD
= 7.12 in.2
Two 3/4 by 3/4 fillet welds, (0.75) (0.75) (2)(1/2)
= 0.56 in.2
= 7.68
10) Strength of reinforcement added
(7.68) (20,000)
= 153,600 lb.
11) Width of area of reinforcement, h
(2.5)(0.687)
=
1.72 in. (this governs)
(2.5)(0.500)
=
1.25
Added reinf.
=
0.75
h = 1.72 in.
OR
2.00
Summary
12) Load calculated to have existed in the metal removed:
(525) (37.375/2)(18)
= 176,200
13) Strength of metal in vessel wall available for reinforcement:
(0.207)(34 – 18)(20,000)
= 66,200 lb.
14) Strength of nozzle wall available for reinforcement
= 24,900 lb.
15) Strength of added reinforcement
= 153,600 lb.
= 244,700 lb.
Total strength
Because the total strength exceeds the load calculated to have existed in the metalremoved, the design is
satisfactory.
Data Sheet for Reinforcement Calculations (UG-37, -40)
1) Head
Description: dished, 27 in. crown radius, 4 in. knuckle radius.
Design pressure
P
525 psig
Joint efficiency
E
1.0
Maximum allowable stress
S
20,000 psi
Corrosion allowance
c
0.063 in.
Inside radius of shell, or inner crown radius, or equivalent spherical radius (specify)
before corrosion allowance is added
R, L, K 1 , D
27.0 in.
Nominal thickness, exclusive of corrosion allowance
t
0.625 in.
Minimum required thickness (reference: UG-37(b), UA-4(d)
tr
0.355
E = 1, M = 1
t=
Excess thickness
(525)(27)
PLM
=
(2SE) − (0.2P) (2 × 20,000) − (0.2 × 525)
t − tr
0.270
403
D Sample of Pressure Vessel Design Data Sheets
2) Nozzle
Material used: seamless steel pipe, SA-106, Grade B (Schedule 80).
Maximum allowable stress
s
17,100 psi
Inner diameter of finished opening in corroded condition
d
3.951 in.
Nominal thickness exclusive of corrosion allowance
tn
0.274 in.
Thickness required for hoop stress (UG-27)
t rn
0.0618 in.
trn =
(525)(3.951 × .5)
PR
=
(SE) − (0.6P) (17,100 × 1) − (0.6 × 525)
3) Sketch of reinforcement with dimensions and welding detail (UW-15, -16, -18).
4.500
tn = 0.274
trn = 0.0618
0.686
404
5
8
0.355
0.688
11
16
0.063
d = 3.951
7.902
Reinforcement Calculation for 4 in. Nozzle in Dished Head
1) Total load to be carried within area of reinforcement:
(525)(1/2)(27 + 1.25)(7.902)
= 58 600 lb.
2) Strength of head within area of reinforcement
(0.625)(7.902 – 4.50)(20,000)
= 42,500 lb.
3) Strength of nozzle wall within head wall thickness
(0.625)(4.50 – 3.951)(17,100)
= 5870 lb.
4) Load carried by nozzle wall due to pressure in nozzle
(525)(4.50)(h)
= 2360 h lb.
5) Strength of nozzle wall outside of head thickness
(0.274)(h)(2)(17,100)
= 9370 h lb.
6) Strength of nozzle wall available for reinforcement
5870 + (9370 – 2360)(h)
(est. h = 1)
5870 + 4810
= 12,880 lb. est.
= 10,680 lb. corr.
7) Load to be carried by added reinforcement
58 600 – (42,500 + 12,880)
= 3220 lb. est.
58 600 – (42,500 + 8840)
= 7260 lb. corr.
Summary
8) Area required of added reinforcement
3220/17,100
= 0.188 sq. in. est.
9) Area of reinforcement used: two 5/8 in. fillet welds inside and out plus extension on nozzle inside of head
Welds: (0.625)(0.625)(0.5)(4)
= 0.781 sq. in.
Nozzle: (0.625)(0.274)(2)
= 0.342 sq. in.
10) Strength of reinforcement added
(0.781)(20,000) + (0.342)(17,100)
= 21,500 lb.
11) Width of area of reinforcement, h
(2.5)(0.274)
h = 0.685 in.
Summary
12) Load calculated to have existed in the metal removed
(525)(1/2)(27 + 1.25)(4.50)
33 400 lb.
13) Strength of metal in head available for reinforcement
(0.270)(7.902 – 4.50)(20,000)
18,370 lb.
14) Strength of nozzle wall available for reinforcement
10,680 lb.
15) Strength of added reinforcement
21,500 lb.
16) Total strength
50,550 lb.
Because the total strength of 50,550 lb. exceeds the load of 33 400 lb. the design is satisfactory.
405
407
E
Sample Materials for Process Equipment
Material Specifications
Tubes
Seamless
Carbon Steel.
ASME SA-179 cold drawn.
ASME SA-210, specify grade.
Low-Alloy Steel.
ASME SA-209.
High-Alloy Steel.
ASME SA-213, specify grade.
ASME SA-268, specify grade.
Nickel and Nickel Alloy.
ASME SB-163, specify alloy and temper.
Aluminum and Aluminum Alloy.
ASME SB-234, specify alloy and temper.
Copper and Copper Alloy.
ASME SB-111, specify alloy and temper.
ASME SB-395, specify alloy and temper.
Welded
Carbon Steel.
ASME SA-214, electric resistance welded.
High-Alloy Steel.
ASME SA-249, specify grade.
Shells, Channels, Covers, Floating Heads, Tubesheets, and Flanges
Pipe
Carbon Steel.
ASME SA-106 seamless, Grade B or Grade A.
ASME SA-53 Grade B or Grade A.
Low-Alloy Steel.
High-Alloy Steel.
ASME SA-335, specify grade.
ASME SA-376, specify grade.
ASME SA-312, specify grade.
Aluminum and Aluminum Alloy.
Copper and Copper Alloy.
ASME SB-241, specify alloy and temper.
ASME SB-42.
ASME SB-43, specify temper.
Plate
Carbon Steel.
ASME SA-285 Grade C for plates up to 2 in. thick.
ASME SA-515, specify grade.
ASME SA-516, specify grade.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
408
E Sample Materials for Process Equipment
Low-Alloy Steel.
ASME SA-204 firebox quality, specify grade.
ASME SA-203 Grade B firebox quality for plates up to 6 ft thick.
ASME SA-387, specify grade.
High-Alloy Steel.
ASME SA-240, specify type.
Nickel and Nickel Alloy.
ASME SB-162, specify temper.
ASME SB-127, specify temper.
ASME SB-168, specify temper.
Aluminum and Aluminum Alloy.
ASME SB-209, specify alloy and temper.
Copper and Copper Alloy.
ASME SB-11, specify type.
ASME SB-96, specify alloy.
ASME SB-169, specify alloy and temper.
ASME SB-171, specify alloy.
Castings
Carbon Steel.
ASME SA-216, specify grade.
ASME SA-352.
Low-Alloy Steel.
ASME SA-217, specify grade.
ASME SA-352, specify grade.
High-Alloy Steel.
ASME SA-351, specify grade.
Copper Alloy.
ASME SB-61 valve bronze.
Gray Iron.
ASME SA-278 Class 30.
Aluminum and Aluminum Alloy.
ASME SB-26.
ASME SB-62 cast brass.
Forgings
Carbon Steel.
ASME SA-105 Grade I or II.
ASME SA-181.
ASME SA-266 Class 1 or 2.
Low- and High-Alloy Steel.
ASME SA-182, specify grade.
ASME SA-336, specify class.
Nickel and Nickel Alloy.
ASME SB-160, specify temper.
ASME SB-164, specify temper and class.
ASME SB-166, specify temper.
Aluminum and Aluminum Alloy.
ASME SB-247, specify alloy and temper.
Bolting
Studs and Stud Bolts
Alloy Steel.
ASME SA-193, specify grade.
Nickel and Nickel Alloy.
ASME SB-160, specify temper.
ASME SB-164, specify temper and class.
ASME SB-166, specify temper.
Aluminum and Aluminum Alloy.
ASME SB-211, specify alloy and temper.
Material Specifications
Nuts
Carbon Steel.
ASME SA-194 Grade 2H, minimum requirement.
Alloy Steel.
ASME SA-194, specify grade.
Nickel and Nickel Alloy.
ASME SB-160, specify temper.
ASME SB-164, specify temper and class.
ASME SB-166, specify temper.
Aluminum and Aluminum Alloy.
ASME SB-211, specify alloy and temper.
409
411
F
Required Data for Material Approval in the ASME Code
Approval of New Materials Under the American Society of Mechanical Engineers (ASME) Boiler and Pressure
Vessel Code
The following is a summary of the requirements of Mandatory Appendix 5 of ASME II-D regarding data needed in
order to add new materials into the code.
A) Code Policy
1) It is the policy of the ASME Boiler and Pressure Vessel Committee to adopt for inclusion in Section II only such
Specifications that have been adopted by National or International Standard – developing organizations.
2) It is expected that requests for Code approval will normally be for materials for which there is an existing
Specification. For other materials, request should be made to National or International Standard – developing
organizations to develop a Specification, which can be presented to the Code Committee.
B) Application
The inquirer needs to identify
1) The section of the code in which the new material is to be used.
2) The product form desired
3) The intended temperature range
4) Whether cyclic service is to be considered
5) Whether external pressure is to be considered
C) Chemical composition
The inquirer shall specify the chemical composition of the material and any elements that significantly influence
the strength, ductility, toughness, weldability, and post weld heat treatment.
D) Heat treatment
A description of the heat treatment temperature ranges shall be provided including heating and cooling rates and
heat treating medium.
E) Time-independent properties
The time-independent properties at and above room temperature shall be furnished at 100F temperature increments up to 100F above the intended service. The data shall include
1) Tensile strength
2) Yield strength
3) Reduction of area
4) elongation
F) Time-dependent properties
The following data is needed when the expected material is used in the time-dependent regime
1) Time-dependent data shall be furnished starting at 50F below the temperature where the time dependent temperature may govern and extending at least 100F above the maximum intended temperature at increments of
100F.
2) Rupture time at each temperature increment shall be provided. The longest rupture time at each temperature
increment shall be in excess of 10,000 hours for each required material heat.
3) Minimum creep rate for at least two heats shall be provided for at least two test stresses at each test temperature
resulting in a minimum creep rate below 3x10-4 %/hr.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
412
F Required Data for Material Approval in the ASME Code
G) Low-temperature properties
Toughness data shall be supplied if the material is expected to be used in components operating at low temperatures. The type of data shall be as required by the requested construction code.
H) Stress-strain curves.
Stress-strain curves shall be provided for materials in equipment operating under compressive loads. The curves
shall be provided for three heats of materials at 100F increments starting at room temperature and ending 100F
greater than the required design temperature.
I) Fatigue date
Fatigue data shall be furnished when the material is under cyclic service. The range of data is from 103 to 106 cycles
over the range of desired temperatures.
J) Physical properties
Physical properties shall be provided for at least three heats over the desired temperature range. The properties
include the coefficient of thermal expansion, thermal conductivity and diffusivity, modulus of elasticity, density,
and Poisson’s ratio.
K) Weldability
1) Data must be supplied on the weldability, weld metal, and weldment properties of the material. Also needed are
the procedure qualification tests in accordance with the requirements of ASME Section IX.
413
G
Procedure for Providing Data for Code Charts for External-Pressure Design
On occasion, the ASME Boiler and Pressure Vessel Committee is requested to provide for a new material chart for
external-pressure design such as those in Subpart 3 of Section II-D. The Subgroup on External Pressure (SGEP) requires
reliable data upon which to base the construction of charts. The SG is not in a position to develop or evaluate the
required data. Consequently, the SGEP recommends the following procedures to be followed in providing the SG with
adequate and reliable data.
1) The compilation and evaluation of material data are rightfully the responsibility of the Committee on Materials.
The subgroups involved would be the Subgroup on Nonferrous Alloys, the Subgroup on Strength-Ferrous Alloys,
and the Subgroup on Physical Properties.
2) Upon receipt of an inquiry for a new chart, the secretary should refer the inquiry to the appropriate SG. The SG
shall determine whether or not adequate data are available or whether the inquirer shall be requested to supply the
required data. The SG should screen and evaluate the data and forward them to the SGEP with their comments or
recommendations. The Materials SG should clearly identify the material and define its use as to product form and,
where applicable, any restrictions on the method of fabrication of the completed pressure vessel (i.e., heat treatment
or welding limitations).
3) It is suggested that, to expedite processing of inquiries, a specific individual might in some cases be designated as
the member responsible for liaison with SGEP. This member would be responsible for the transmission of approved
data to SGEP.
A description of the data required for proper preparation of the design charts follows. It is felt that these are minimum requirements for the preparation of reliable charts. The use of so-called typical stress–strain curves based on
a statistically significant volume of data may be satisfactory if the region between the proportional limit and the yield
strength is accurately represented. The development of the tangent modulus in this region is a critical step. It is suggested that this description be prepared in a form suitable for attachment to any requests for material data from an
inquirer.
A copy follows of a description of the method used to derive the material curves on the charts directly from the
laboratory stress–strain curves. It was felt that this procedure might enable the Materials Subgroups to better evaluate
our data requirements. The balance of the lines on the chart is a function of the geometry of the vessel, and so it does
not change with the material of construction.
Data Needed by the SG External Pressure for the Preparation of Code Charts
for External-Pressure Design
1) The minimum specified yield strength or yield point (state which) as given in the specifications for the material.
2) Stress–strain curves representative of the material at the following temperatures:
a) Ambient (room) temperature.
b) The highest temperature for which coverage is desired.
c) One or more intermediate temperatures as may be desirable to facilitate interpolation on the chart.
Temperatures at some multiple of 100 ∘ F are preferred.
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
414
G Procedure for Providing Data for Code Charts for External-Pressure Design
3)
4)
5)
6)
The stress–strain curves should extend to at least the 0.3% offset point (to ensure being able to obtain reliable
values of the tangent modulus to 0.2% offset). Consideration should be given to extending tests to higher values
of strain for possible future use with stress intensity values in the elastoplastic range. (This is much less expensive
than to run additional tests at a later date.)
Stress–strain curves in compression are preferred. It is recommended that compression tests be made in accordance with ASTM Specification E-9, Standard Methods of Compression Testing of Metallic Materials.
Stress–strain curves from tension tests will be acceptable if there is sufficient background of information to
show that there is no substantial difference between the stress–strain characteristics of the material in tension
and compression. Data should indicate whether tension or compression tests were made.
The expected properties of the material at the temperatures described earlier, for material having the minimum
specified properties, are as follows:
a) Yield strength or yield point (state which).
b) Proportional limit.
c) Elastic modulus (state whether by the dynamic method or from stress–strain curves).
The condition of the material as stated in the specifications, for example, annealed, hot finished, cold drawn, temper,
and so on.
Stated whether intended for welded construction. The aforementioned data should properly include the effect of
the heat of welding on the properties of the material. It is acceptable in such cases to use data for the material in the
annealed condition.
The inquirer should supply data from at least three specimens at each temperature and that these specimens should
preferably be taken from more than one “production lot” or “heat.”
415
H
Corrosion Charts
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
Corrosion data–salts.
MEDIA
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Acetyl
chloride
Cold and
boiling
B
C
C
B
A
A
A
A
A
Aluminum
acetate
Saturated
Aluminum
chloride
5%
Room
AB
A
B
B
A
A
Aluminum
fluoride
5%
Room
AB
A
A
A
A
A
Aluminum
hydroxide
Saturated
A
A
A
A
AB
A
A
A
Aluminum
oxalate
A
B
B
A
A
AB
AB
C
B
B
AB
A
A
A
A
A
AB
B
B
C
C
C
A
B
A
AB
B
A
C
B
C
C
A
A
AB
AB
A
A
A
A
A
A
A
AB
A
Aa)
A
AB
A
Aa)
A
Aa)
A
A
Aa)
A
AB
A
Aa)
A
Ba)
A
A
A
A
A
Aluminum
potassium
sulfate
2%
Room
AB
A
A
A
Aluminum
potassium
sulfate
10%
Room
AB
A
A
A
Aluminum
potassium
sulfate
10%
Boiling
C
A
AB
AB
B
A
AB
AB
Ba)
A
Ca)
A
A
Aluminum
potassium
sulfate
Saturated
Boiling
C
A
AB
AB
B
A
AB
AB
Ba)
B
Ca)
A
A
Aluminum
sulfate
10%
Room
BC
A
B
B
A
A
A
A
AB
A
B
Aa)
A
Ba)
A
A
A
Aluminum
sulfate
10%
Boiling
A
AB
AB
A
A
AB
A
B
C
AB
Ba)
A
Ca)
A
A
A
A
Ca)
A
A
A
A
Aluminum
sulfate
Saturated
Room
Aluminum
sulfate
Saturated
Boiling
B
A
A
A
A
AB
AB
A
A
A
B
A
BC
AB
BC
A
Aa)
B
A
C
B
C
AB
C
B
Ca)
AB
A
A
A
A
C
A
A
AB
A
C
A
A
B
A
Ammonia
(anhydrous – dry)
A
A
A
A
A
A
A
A
A
AB
Ammonium
alum
B
A
C
B
A
A
A
A
AB
B
Ammonium
alum (slightly
ammoniacal)
AB
A
C
C
A
A
A
A
AB
BC
BC
B
Corrosion data–salts.
MEDIA
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
ammonium
bicarbonate
Hot
A
A
C
C
A
A
B
B
B
C
A
AB
AB
BC
B
A
A
B
A
Ammonium
bromide
5%
Room
AB
A
BC
BC
A
A
A
A
Ammonium
carbonate
All conc.
Hot & cold
A
A
B
B
A
A
A
A
A
A
A
A
C
Aa)
A
A
A
B
A
A
A
A
A
Aa)
A
A
A
Aa)
A
A
A
A
A
A
A
A
A
A
A
A
Ammonium
chloride
1%
Room
AB
A
B
BC
A
A
Aa)
Ammonium
chloride
10%
Boiling
C
A
C
C
A
A
Aa)
A
Aa)
AB
AB
C
B
B
Ammonium
chloride
28%
Boiling
C
A
C
C
A
A
Ba)
A
Aa)
B
B
C
C
B
Ammonium
chloride
50%
Boiling
C
A
C
C
A
Ba)
A
AB
AB
C
C
Ammonium
hydroxide
AB
A
C
C
A
A
A
A
C
BC
C
A
A
Ammonium
monosulfate
C
A
B
B
A
A
A
A
B
A
A
BC
A
A
A
A
C
A
A
B
A
A
AB
A
Ammonium
nitrate
5%
Room
A
A
C
BC
C
A
A
A
AB
C
Ammonium
oxalate
5%
Room
A
A
C
B
A
A
A
A
AB
AB
Ammonium
persulfate
5%
Room
B
A
B
B
C
A
A
A
AB
C
C
B
A
A
A
A
Ammonium
phosphate
5%
Room
A
A
B
B
A
A
A
A
AB
AB
AB
B
A
A
A
A
Ammonium
sulfate
1–5%
Agitated,
aerated
Room
BC
A
B
B
A
A
A
A
AB
AB
AB
B
A
A
A
A
A
Ammonium
sulfate
10%
Boiling
C
A
B
B
A
B
B
A
B
AB
B
B
C
Aa)
A
B
Ammonium
sulfate
Saturated
Boiling
C
A
B
B
A
B
B
A
B
AB
B
B
C
Ammonium
sulfate
Saturated
A
C
C
A
A
C
C
C
C
B
A
A
Amyl acetate
10–100%
Cold &
boiling
200∘
A
A
A
A
A
A
A
A
A
A
A
A
A
A
BC
A
A
A
A
A
A
AB
AB
AB
A
A
A
AB
B
C
C
C
A
A
AB
A
C
C
C
A
A
A
A
A
A
A
A
Amyl chloride
Aniline
5%
hydrochloride
Room
C
B
B
Antimony
trichloride
Room
B
A
A
Barium
carbonate
Room
AB
A
A
A
A
A
A
B
AB
AB
A
A
AB
C
A
A
A
A
(continued)
Corrosion data–salts.
MEDIA
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zircotration (∘ F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Barium
chloride
5% to sat.
Room
C
A
B
B
Barium
chloride
Aqueous
sol.
Hot
C
A
BC
BC
C
A
A
A
Barium
hydrate
A
A
A
A
A
A
B
A
A
Aa)
AB
AB
B
B
AB
AB
AB
BC
Ba)
Aa)
A
AB
A
AB
A
A
A
A
A
AB
B
A
A
A
A
AB
A
A
A
A
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Barium nitrate Aqueous
sol.
Hot
A
A
B
B
Barium sulfate
Room
A
A
A
A
Room
A
A
A
A
A
A
Room
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
AB
B
A
A
A
AB
AB
AB
AB
A
A
Aa)
A
AB
AB
A
Bb)
Aa)
C
A
A
A
Butyl acetate
100%
Calcium
carbonate
Calcium
chlorate
Dilute
Hot or cold
A
A
A
A
A
A
A
Calcium
chloride
Dil. or conc. Room
AB
Calcium
hydroxide
10–20%
Boiling
B
A
A
A
A
A
A
A
A
A
A
A
A
A
Calcium
hydroxide
50%
Boiling
C
A
A
A
A
A
C
A
AB
AB
AB
A
C
B
Calcium
hypochlorite
2%
Room
C
C
C
B
C
A
Ba)
Aa)
C
C
B
C
Ba)
Aa)
Ba)
Calcium
sulfate
Saturated
Room
AB
A
A
A
A
A
A
A
A
A
A
Room
A
A
A
A
A
A
A
AB
A
A
A
A
A
A
A
AB
Pure
Room
AB
A
AB
A
A
A
A
A
A
A
A
AB
Aa)
A
Aa)
Carbon
tetrachloride
5–10%
Aqueous
sol.
Room
B
A
AB
A
A
A
A
A
A
A
A
AB
C
B
C
Chlorobenzene (pure)
Concentrated
Room
AB
B
AB
A
AB
AB
AB
AB
AB
AB
A
A
A
B
A
A
A
A
C
A
B
B
A
A
A
A
AB
B
B
A
A
A
A
Sat. Sol.
C
A
A
A
A
A
A
A
AB
A
A
A
A
A
A
A
Copper
chloride
1% Agitated Room
& aerated
C
C
B
A
A
Ba)
Aa)
B
B
B
C
Ba)
Aa)
Ba)
A
A
Copper
chloride
5% Agitated Room
C
C
C
A
A
BC
B
C
C
C
C
Ca)
Ba)
Ba)
A
A
C
Ca)
Ca)
Ca)
A
A
5% Aerated Room
C
C
C
A
A
C
C
C
A
A
Copper
carbonate
Copper
chloride
Room
A
A
Carbon
tetrachloride
Copper acetate Saturated
A
A
Aa)
Carbon
bisulfide
A
Bb)
AB
Corrosion data–salts.
MEDIA
Copper
cyanide
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Saturated
Boiling
C
B
C
C
A
A
A
A
C
C
C
C
A
A
A
A
Copper nitrate 1–5%
Room
C
A
C
C
C
A
A
A
BC
C
C
C
A
A
Copper nitrate 50%
Aqueous
Room
C
A
C
C
C
A
A
A
C
C
C
C
A
A
Copper Sulfate 5%
Room
C
A
BC
B
A
A
A
A
B
B
B
BC
A
A
Copper sulfate Saturated
Boiling
105∘
C
A
BC
BC
A
A
C
C
C
BC
A
A
C
C
C
BC
C
A
C
C
C
C
C
C
C
C
Cupric
chloride
Cupric nitrate
A
C
C
A
C
C
C
A
A
A
C
C
C
C
A
A
Room
A
A
A
A
A
A
AB
A
A
A
A
A
A
A
Ethyl chloride 5%
Room
AB
A
A
AB
A
A
A
A
AB
AB
A
A
A
Ethylene
chloride
Room
AB
A
B
B
A
A
A
A
B
A
A
Aa)
Aa)
Ba)
Ethyl acetate
100%
B
A
A
A
A
A
A
A
A
A
A
A
A
C
A
A
AB
C
A
A
A
A
Ferric chloride 1% Still
Room
C
C
C
C
C
A
Ba), d)
C
C
C
C
Ba),d)
A
A
Ferric chloride 1% Still
Boiling
C
C
C
C
C
A
C
C
C
C
C
C
C
C
C
A
A
C
Ferric chloride 5% Still
Room
C
C
C
C
C
A
C
C
C
C
C
C
C
C
C
A
A
C
Ferric chloride 5% Agitated Room
C
C
C
C
C
A
C
C
C
C
C
C
C
C
C
A
A
C
Ferric chloride 5% Aerated Room
C
C
C
C
C
A
C
C
C
C
C
C
C
C
C
A
A
C
A
B
B
C
A
A
A
A
AB
AB
B
A
A
A
A
Ferric
hydroxide
Room
Ferric nitrate
1–5%
Room
C
A
C
C
C
A
A
A
C
C
C
C
A
A
A
A
A
Ferric sulfate
1–5%
Room
C
A
C
C
C
A
AB
A
C
B
C
C
Aa)
A
A
A
A
Ferrous
chloride
10%
Room
C
C
B
B
A
A
Ba)
C
C
C
B
C
C
A
A
AB
A
C
C
C
B
A
A
A
A
Ferrous sulfate Dilute
Room
Ferrous
ammonium
citrate
Hydrogen
peroxide
B
to 5%
Room
A
A
B
B
A
A
B
B
A
A
BC
BC
B
A
A
A
AB
AB
A
Hydrogen
peroxide
to 5%
Boiling
A
A
C
C
B
A
B
A
Hydrogen
sulfide
Dry
Room
A
A
A
A
B
A
A
A
B
B
A
A
A
A
A
A
Hyposulfite
soda (hypo)
A
Lactic acid
salts
AB
A
A
A
AB
A
B
AB
AB
A
BC
Ac)
Ac)
Ac)
A
A
C
Bc)
Ac)
Bc)
A
A
A
AB
AB
B
A
A
A
Ac)
A
AB
AB
AB
B
A
A
B
A
AB
B
A
A
A
A
A
Lead acetate
BC
A
B
B
A
A
B
AB
AB
B
A
A
A
A
A
Manganese
carbonate
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
B
BC
BC
A
A
AB
AB
AB
BC
A
A
A
A
Manganese
chloride
10–50%
Aqueous
sol.
Boiling
A
(continued)
Corrosion data–salts.
MEDIA
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zircotration (∘ F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Magnesium
carbonate
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Magnesium
chloride
1–5% Still
Room
B
A
B
B
A
A
A
A
A
A
A
B
Aa)
A
C
A
A
A
Magnesium
chloride
1–5% Still
Hot
C
A
B
B
A
A
A
A
A
A
A
B
B
Ba)
C
A
A
A
Magnesium
hydroxide
Thick
Room
suspension
C
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
B
B
C
A
A
A
AB
B
B
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
AB
AB
A
A
A
A
C
Aa)
Aa)
Magnesium
nitrate
Magnesium
sulfate
5%
Hot
A
A
A
A
A
A
A
A
Methylene
chloride
40%
Room to
boiling
BC
A
A
A
A
A
A
A
Mercuric
bichloride
0.07%
Room
C
C
C
C
C
Aa)
Aa)
Mercuric
chloride
Dilute
Room
C
C
C
C
C
A
C
C
C
C
C
C
C
C
Mercuric
cyanide
C
A
C
C
C
A
A
A
C
C
C
C
A
A
Mercurous
nitrate
C
A
C
C
C
A
A
A
B
C
C
C
A
A
A
A
C
A
BC
B
C
A
AB
A
B
B
Aa)
C
A
C
A
A
A
A
Nickel
chloride
Room
Nickel nitrate 10%
Room
BC
A
B
B
AB
A
A
A
C
C
C
B
A
A
A
A
Nickel sulfate 10%
Room
BC
A
A
A
A
A
A
A
B
B
B
A
A
A
B
A
A
A
A
A
C
A
A
C
A
C
Room
A
A
A
A
A
A
A
A
A
A
BC
A
BC
BC
C
A
A
A
AB
BC
A
A
A
Nitrous oxide Dry
Phosphoric
anhydride
Dry
Phosphorous
trichloride
AB
A
A
A
A
Potassium
bichromate
Neutral
Room
A
A
A
A
C
A
A
A
AB
A
AB
A
A
A
Potassium
bromide
5%
Room
B
A
A
A
A
A
A
A
AB
A
AB
A
Ba)
Aa)
Potassium
carbonate
1%
Room
BC
A
A
A
A
A
A
A
AB
A
AB
A
A
A
AB
A
AB
B
C
A
AB
C
C
C
AB
A
A
Ba)
A
Aa)
Potassium
chlorate
Potassium
chloride
1–5%
Room
AB
A
A
A
A
A
Aa)
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Aa)
Aa)
A
A
A
AB
Corrosion data–salts.
MEDIA
Potassium
chloride
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zircotration (∘ F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
1–5%
Boiling
Potassium
cyanide
B
A
BC
BC
C
A
Aa)
A
Ba)
AB
AB
BC
Aa)
Aa)
Aa)
A
C
A
C
C
A
A
A
A
B
AB
B
C
A
A
A
A
A
A
A
A
C
A
A
A
AB
B
B
A
A
A
A
A
Potassium
dichromate
Neutral
Potassium
ferricyanide
5%
Room
A
A
A
A
A
A
A
A
B
B
B
A
A
A
A
A
Potassium
ferrocyanide
5%
Room
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
C
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Potassium
hydrate
Potassium
hydroxide
5%
Room
C
A
B
A
A
A
A
A
A
A
A
B
A
A
Potassium
hydroxide
27%
Boiling
C
A
BC
BC
A
A
A
A
AB
A
A
BC
A
Potassium
hydroxide
50%
Boiling
C
A
B
B
A
A
A
A
AB
A
A
B
Potassium
hypochlorite
C
A
C
B
C
A
C
BC
C
C
C
Potassium
iodide
B
A
A
A
A
A
AB
A
AB
B
A
A
A
A
C
A
A
A
AB
AB
Potassium
oxalate
B
A
A
A
A
A
A
A
Potassium
Neutral
permanganate
A
A
A
A
C
A
A
A
AB
A
A
A
A
Potassium
nitrate
5%
Room
Potassium
sulfate
1–5%
Room
A
A
A
A
Potassium
sulfate
1–5%
Hot
B
A
B
B
A
A
Potassium
sulfide (salt)
A
C
C
A
A
Quinine
bisulfate (dry)
A
Ae)
Ae)
B
A
A
Ae)
Ae)
Quinine
sulfate (dry)
Silver bromide 10%
A
A
AC
A
A
C
AC
A
B
A
C
A
C
C
C
B
A
A
A
AB
A
A
A
AB
A
A
AB
AB
A
B
AB
B
AB
AB
AB
C
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
B
A
A
A
A
C
A
A
A
A
AB
Ae)
B
A
B
A
A
A
B
A
A
A
A
AB
Ae)
Ba)
A
AB
C
Ca)
Ca)
C
A
C
C
C
C
A
A
C
A
A
A
A
A
A
C
A
C
C
A
A
C
A
C
C
C
A
Silver cyanide
C
A
C
C
A
A
A
A
Silver nitrate
C
A
C
C
AB
AB
A
A
AB
C
C
C
A
A
A
A
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
Room
A
A
Silver chloride
Sodium
5%
acetate (moist)
Room
A
A
A
AB
(continued)
Corrosion data–salts.
MEDIA
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zircotration (∘ F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Sodium
benzoate
A
Sodium
bicarbonate
All conc.
Sodium
bichromate
Neutral
150∘
Sodium
bisulfate
Sodium borate
A
A
A
A
AB
AB
AB
AB
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
AB
B
A
A
A
A
AB
A
A
A
A
A
A
A
A
A
A
A
A
C
A
A
A
AB
AB
A
B
B
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
A
A
A
A
Sodium
bromide
5%
Room
B
A
A
A
A
A
A
A
B
AB
B
A
A
A
Sodium
carbonate
All conc.
Room
AC
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
Sodium
chlorate
25%
AB
A
AB
AB
C
A
A
A
AB
AB
A
A
Aa)
A
Aa)
A
A
A
A
A
Aa)
Aa)
Sodium
chloride
5% Still
Room to
150∘
AB
A
A
A
A
A
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Aa)
Ba)
A
A
A
A
Sodium
chloride
20%
Aerated
Room
B
A
A
A
A
A
Aa)
Sodium
chloride
Saturated
Room
B
A
A
A
A
A
Aa)
A
A
A
A
A
Aa)
Aa)
A
A
Sodium
chloride
Saturated
Boiling
B
A
A
A
A
A
Aa)
A
A
A
A
A
Ba)
Aa)
A
A
A
Sodium citrate
C
A
A
A
A
A
A
A
AB
Sodium
ferricyanide
A
A
A
A
A
A
Aa)
A
AB
Sodium
ferrocyanide
A
A
A
B
BC
A
A
A
Sodium
fluoride
5%
Room
Sodium
hydrosulfite
AB
AB
A
A
A
A
A
A
Aa)
Aa)
Aa)
A
A
A
A
B
B
B
A
A
AB
AB
A
A
A
A
A
B
A
Ba)
B
C
10%
Room
C
A
B
A
A
A
A
A
A
A
A
B
A
A
A
Sodium
hypochlorite
5%
Room
BC
B
C
C
C
A
Ba)
Aa)
C
C
C
C
Ba)
Aa)
C
Sodium
hyposulfite
Dilute 5%
Room
A
B
B
A
A
AB
A
AB
AB
AB
B
Cc)
C
Bc)
C
A
A
A
A
A
A
A
A
A
A
A
A
A
A
C
A
A
A
A
A
A
Sodium nitrate All conc.
Room
AB
A
AB
AB
C
A
A
Sodium
hydroxide
Sodium lactate
A
C
A
A
A
B
A
A
A
A
A
Corrosion data–salts.
MEDIA
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Sodium nitrite
A
Sodium
peroxide
Sodium
phosphate
5%
A
A
A
A
B
B
B
A
A
A
A
212∘
AB
A
B
B
C
A
A
A
AB
AB
AB
B
A
A
C
Room
AB
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
AB
A
A
A
A
A
A
A
A
A
A
A
A
Sodium
silicate
AB
A
A
Sodium sulfate 5% Still
Room
A
A
A
A
A
A
A
A
AB
AB
B
A
A
A
A
A
A
Sodium sulfate Concentrated
Room
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
AB
A
C
Ca)
Ba)
A
B
A
A
B
A
A
AB
C
A
A
C
A
A
AB
AB
B
C
B
C
A
A
C
A
Sodium sulfide Saturated
Room
BC
A
C
B
A
A
AB
A
Sodium sulfite 5%
Room
A
A
B
B
C
A
A
A
Stannic
chloride
5%
Room
C
A
C
C
A
A
C
B
B
AB
Stannous
chloride
5%
Room
C
A
B
B
A
A
AB
A
AB
Sulfur chloride Dry
AB
B
C
A
A
A
A
A
AB
AB
A
C
C
Sulfur dioxide Dry
Room
A
A
A
A
C
A
A
A
A
A
A
A
A
A
Sulfur dioxide Moist
Room
B
A
B
B
C
A
B
A
A
A
A
B
B
A
C
A
B
B
C
A
B
A
Titanium
tetrachloride
A
A
AB
A
C
A
A
A
A
A
A
Zinc chloride 5% Still
Room
C
A
B
B
A
A
B
A
AB
AB
AB
B
A
A
C
A
A
A
Zinc chloride 5% Still
Boiling
C
A
B
B
AB
AB
C
A
B
B
AB
B
C
C
C
A
A
A
Zinc sulfate
5%
Room
AB
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
Zinc sulfate
Saturated
Room
AB
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
Zinc sulfate
25%
Boiling
AB
A
B
B
A
A
A
A
AB
AB
AB
B
A
A
A
A
A
A
Corrosion data–acids.
Media
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tan- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 bronze steel 304 steel 316 steel 430 talum Grade nium
Acetic acid
5%
Room
Unaerated
A
A
A
AB
A
A
A
A
AB
A
AB
A
A
A
A
A
A
A
Acetic acid
20%
Room
Unaerated
A
A
A
A
A
A
A
A
AB
AB
B
A
A
A
A
A
A
A
Acetic acid
50%
Room
Unaerated
A
A
AB
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
Acetic acid
50%
Boiling
Unaerated
B
A
C
BC
A
A
B
A
AB
AB
BC
C
B
A
A
A
Acetic acid
100%
Room
Unaerated
A
A
A
A
A
A
B
A
AB
AB
A
A
A
A
A
A
Acetic acid
100%
Boiling
Unaerated
C
A
C
C
A
A
C
A
C
AB
B
C
C
B
A
A
A
Acetic
Anhydride
Unaerated Room
A
A
A
AB
A
A
A
A
AB
AB
A
A
A
A
A
A
A
A
Acetic
anhydride
Unaerated Boiling
B
A
C
C
A
A
A
A
B
B
A
C
A
A
C
A
A
AB
A
B
B
A
A
A
AB
AB
B
B
C
C
C
A
A
A
Acetic vapors 100%
Hot
Unaerated
Arsenic acid 90%
225∘
B
B
B
B
B
B
B
Benzoic acid
5%
Room
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
Boric acid
5%
Boiling
A
A
A
A
A
A
A
A
B
AB
B
A
A
A
A
A
A
Butyric acid
5%
Room
Carbonic acid
B
B
A
A
A
A
A
A
A
A
AB
B
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
Carbolic acid,
C. P.
Room
A
A
Ae)
Ae)
A
A
A
A
AB
AB
AB
Ae)
Aa)
Aa)
C
A
Chloroacetic
acid
Room
C
C
BC
BC
A
A
C
C
AB
AB
AB
BC
C
C
C
A
Chloric acid
Room
C
C
C
A
A
C
C
C
C
C
C
C
Chlorosulfonic 10%
acid
C
A
BC
C
A
A
A
A
A
B
A
A
Chromic acid 5%
Room
AB
A
C
C
C
A
A
A
AB
C
C
C
A
A
B
A
A
Chromic acid, 10%
C. P.
Boiling
AB
C
C
C
C
A
C
BC
C
C
C
C
C
B
C
A
A
Chromic acid 50%
C
C
C
C
B
C
C
C
C
C
C
C
C
C
A
B
5% Still
Boiling
150∘
AB
Citric acid
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
A
Citric acid
15%
Room
A
A
A
A
A
A
AB
A
AB
AB
AB
A
A
A
A
A
A
A
Citric acid
15%
Boiling
BC
A
B
B
A
A
AB
A
AB
AB
AB
B
B
A
AB
A
A
A
Citric acid
Concentrated
Boiling
BC
A
B
B
A
A
BC
AB
B
B
B
B
C
B
A
A
AB
A
A
A
A
A
A
A
A
A
AB
AB
A
Formic acid
5% Still
Room to
150∘
C
A
A
A
A
A
B
A
B
B
AB
A
Ba)
Aa)
C
A
B
A
Gallic acid
5%
Room To
boiling
A
A
A
A
A
A
B
A
B
B
B
A
A
A
A
A
Fatty acids
A
Corrosion data–acids.
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tan- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 bronze steel 304 steel 316 steel 430 talum Grade nium
Media
Hydrobromic
acid
Boiling
C
C
C
C
A
B
C
C
C
C
C
C
C
C
C
A
Hydrochloric 5%
Room
acid
Unaerated
C
B
C
BC
B
A
C
B
B
AB
B
BC
C
C
C
A
B
A
Hydrochloric 10%
Room
acid
Unaerated
C
B
C
BC
A
A
C
B
C
B
B
C
C
C
C
A
B
A
Hydrochloric 20%
Room
acid
Unaerated
Hydrochloric All
100∘
C
B
C
BC
A
A
C
B
C
C
C
C
C
C
C
A
C
A
C
C
C
C
A
A
C
BC
C
C
C
C
C
C
C
A
C
A
acid
Hydrochloric All
acid
122∘
C
C
C
C
B
B
C
BC
C
C
C
C
C
C
C
A
C
A
Hydrochloric All
acid
160∘
C
C
C
C
A
B
C
C
C
C
C
C
C
C
C
A
C
A
Hydrochloric Concenacid fumes
trated
100∘
C
C
C
BC
C
BC
C
C
C
C
C
C
C
A
C
C
A
A
C
C
A
A
A
A
C
A
A
C
A
A
All
C
C
C
BC
A
A
C
C
A
A
A
C
C
C
C
C
C
Hydrofluoric
acid, vapors
212∘
C
C
C
BC
A
A
C
C
A
A
A
C
C
C
C
C
Hydrofluosilicic5%
acid
70∘
A
C
BC
A
A
B
A
B
B
C
C
C
C
Hydrofluosilicic
acid vapors
212∘
C
C
C
C
Lactic acid
5%
Lactic acid
5%
Room
150∘
Lactic acid
10%
Hydrocyanic
acid
Hydrofluoric
acid
All
Cold & Hot
Molybdic acid 5%
Room
Muriatic acid
C
C
C
AB
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
A
BC
A
A
A
A
A
B
A
B
C
C
A
B
A
B
A
A
A
A
AB
AB
A
A
C
A
B
C
C
AB
C
B
A
A
AB
A
C
BC
AB
AB
AB
C
A
A
150∘ to
Boiling
Malic acid
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
A
B
A
Room
C
C
C
BC
A
A
B
C
C
C
C
C
C
C
A
C
Nitric acid
5%
Room
BC
A
C
C
C
A
A
A
C
C
C
C
A
A
A
A
A
A
Nitric acid
20%
Room
C
A
C
C
C
A
AB
A
B
C
C
C
A
A
A
A
A
A
Nitric acid
50%
Room
C
A
C
C
C
A
AB
A
AB
C
C
C
A
A
A
A
A
A
Nitric acid
50%
Boiling
C
A
C
C
C
C
AB
A
C
C
C
C
A
A
A
A
A
A
Nitric acid
65%
Boiling
C
A
C
C
C
C
B
A
C
C
C
C
B
B
C
Nitric acid
95%
Room
A
A
C
C
C
A
C
A
C
C
C
C
Nitric acid
Concentrated
Room
A
A
C
C
C
A
A
A
C
C
C
C
A
A
Nitric acid
Concentrated
Boiling
C
A
C
C
C
C
B
A
C
C
C
C
C
C
Nitric acid
Fuming
Room
A
A
C
C
C
A
C
C
Nitrous acid
5%
Room
AB
A
B
B
A
A
C
A
A
AB
A
B
A
A
A
A
A
A
A
A
C
A
B
A
A
C
A
A
A
A
C
B
A
A
A
A
A
(continued)
Corrosion data–acids.
Media
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tan- Comm’l Zircotration (∘ F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 bronze steel 304 steel 316 steel 430 talum Grade nium
Oleic acid
A
Aa)
A
Ba)
A
A
A
A
A
A
C
B
A
A
A
A
B
AB
B
AB
C
C
A
C
AB
A
AB
A
B
Ac)
Ac)
Ac)
A
A
A
A
AB
AB
A
B
A
A
A
A
A
A
A
A
AB
AB
AB
B
A
A
B
A
A
A
A
A
A
B
B
AB
B
A
A
B
A
A
A
A
A
A
A
B
AB
B
B
A
A
B
A
A
A
A
A
A
A
C
C
C
C
A
A
A
A
A
A
A
Room
A
A
A
AB
A
A
A
A
Oxalic acid
5%
Cold & Hot
B
A
A
A
A
A
A
A
A
AB
Oxalic acid
10%
Room
B
A
A
A
A
A
A
A
AB
AB
Oxalic acid
10%
Boiling
C
A
AB
AB
B
A
A
AB
Phosphoric
acid
1%
Room
C
A
B
BC
A
A
A
A
Phosphoric
acid
5%
Room
C
A
B
BC
A
A
A
Phosphoric
acid
10% Still
Room
C
A
B
BC
A
A
Phosphoric
acid
10%
Agitated
Room
C
A
B
BC
A
Phosphoric
acid
10%
Aerated
Room
C
A
B
BC
Picric acid
Concentrated
Room
A
A
C
C
AB
AB
A
Pyrogallic acid
A
A
A
A
A
A
A
A
AB
AB
AB
A
Salicylic acid
A
A
A
A
A
A
B
A
AB
AB
AB
A
200∘
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
AB
A
C
A
AB
A
AB
BC
C
B
C
A
B
A
AB
C
AB
BC
BC
C
C
C
A
C
A
Stearic acid
Concentrated
Succinic acid
Molten
A
A
A
B
B
Sulfuric acid
5%
Room
BC
A
BC
BC
A
A
Sulfuric acid
5%
Boiling
C
A
BC
BC
A
B
Sulfuric acid
10%
Room
B
A
BC
BC
A
A
Sulfuric acid
10%
Boiling
C
B
C
C
A
B
Sulfuric acid
50%
B
C
A
A
A
A
A
A
A
A
AB
AB
C
AB
C
BC
C
B
C
A
B
A
AB
C
B
C
C
C
C
C
A
C
A
Room
C
A
BC
BC
A
A
C
AB
C
B
C
BC
C
C
A
B
A
Sulfuric Acid 50%
Boiling
C
C
C
C
A
C
C
BC
C
C
C
C
C
C
C
A
C
A
Sulfuric acid
Concentrated
Room
A
A
C
C
A
A
A
A
C
A
A
A
A
C
C
Sulfuric acid
Concentrated
Boiling
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
Sulfuric acid
Concentrated
300∘
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
A
C
C
A
A
A
C
C
AB
Sulfuric acid
Fuming
Room
A
Sulfuric
anhydride
Dry
Room
A
Sulfurous acid Saturated
375∘
B
Sulfurous
spray
Room
B
Room
150∘
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
A
A
A
AB
A
A
A
A
A
B
A
B
B
B
A
A
A
B
A
A
A
Room
150∘
BC
A
A
A
A
A
A
A
AB
A
AB
A
A
A
A
A
A
B
BC
A
A
A
A
A
B
A
B
B
B
A
A
A
C
A
A
Room
C
BC
B
C
C
B
BC
C
C
C
C
A
A
A
A
A
A
A
A
A
A
Tannic acid
10%
Tannic acid
All
Tartaric acid
10%
Tartaric acid
10%
Trichloroacetic
acid
Uric acid
Concentrated
C
C
C
A
B
C
C
C
C
C
C
B
B
A
A
C
B
C
C
C
B
A
C
C
C
A
C
A
C
C
C
C
C
B
A
B
B
C
A
C
A
C
C
C
B
C
C
A
A
A
AB
C
C
A
A
A
A
B
Corrosion data–miscellaneous.
Media
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Acetone
Boiling
A
A
A
A
A
A
A
A
A
A
A
A
A
A
B
A
A
Alcohol–
methyl, propyl,
butyl, ethyl
Room
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
C
Molten
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
AC
Room
AB
A
C
A
A
A
A
B
B
B
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Alkaform
Aluminum
Aniline
Conc.
Baking oven
gases
A
A
A
A
Ae)
Ae)
A
A
A
A
A
A
A
Ae)
A
A
Af )
A
A
Benzene
Room
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Benzol
Hot
A
A
A
A
A
A
A
A
AB
A
AB
A
A
A
A
A
Hot
C
AB
B
B
C
A
BC
B
C
C
B
BC
AB
Cold
A
A
Ae)
Ae)
A
A
Aa)
A
A
A
A
Ae)
Aa)
A
Aa)
A
Fused
A
C
Beer
Bleaching
powder
Solution
Blood (meat
juices)
Borax
Bromine
Dry
Bromine
Water
Room
C
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
C
A
C
A
A
A
A
A
C
C
C
A
C
AB
C
BC
C
A
C
B
C
BC
C
C
C
C
C
A
Ae)
Ae)
A
A
A
A
A
A
A
AB
Ae)
A
A
A
Camphor
A
A
A
A
A
A
A
A
B
A
AB
A
A
A
A
A
Carbonated
beverages
AC
A
Ae)
Ae)
A
A
A
A
A
A
A
Ae)
A
A
Af )
A
AC
A
A
A
A
A
A
A
A
A
A
A
A
C
Buttermilk
Room
Carbon
monoxide gas
900∘
Cadmium
Molten
Caustic lime
Caustic soda
A
A
C
C
C
C
C
C
C
C
C
C
C
C
C
A
AC
A
A
A
A
A
A
A
A
A
A
A
A
A
A
AC
C
AC
C
A
AB
A
A
A
A
A
A
A
A
AB
A
A
A
C
Room
AC
A
C
C
C
A
C
A
C
C
C
C
C
Ba)
C
A
A
C
Chlorine
gas – dry
Room
A
A
A
A
A
A
C
A
A
A
A
A
C
C
C
A
C
B
Chlorine
gas – moist
Room
C
C
C
C
C
A
C
C
AB
AB
C
C
C
C
A
A
C
Chlorine
gas – moist
212∘
C
C
C
C
C
C
C
C
BC
BC
C
C
C
C
A
A
C
Chlorinated
water
Saturated
BC
(continued)
Corrosion data–miscellaneous.
Media
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zirco∘
tration ( F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Chloroform
Room
Chromium
plating bath
Room
Cider
Room
Coffee
Boiling
A
A
A
AB
C
C
B
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Ae)
Ae)
A
A
A
A
A
AB
AB
Ae)
A
A
A
A
Copal varnish
A
A
A
A
A
A
A
A
A
AB
AB
A
A
A
A
A
Cream of
tartar
A
A
A
A
A
A
A
A
A
A
A
A
A
AB
A
A
A
A
A
AB
AB
AB
A
Aa)
A
A
Creosote (coal
tar)
Hot
Crude Oil
Developing
solutions
A
Room
Distillery wort
Room
Ether
Room
Flue gases
Room
Food pastes
Formaldehyde
A
A
A
A
A
A
A
A
A
AB
AB
C
C
C
C
AB
AB
A
A
A
A
AB
AC
AB
A
A
A
A
AB
Ac)
Ac)
Ac)
A
A
Ae)
Ae)
A
A
A
A
A
A
A
Ae)
A
A
B
A
A
Ae)
Ae)
A
A
A
A
A
Ae)
A
A
A
Ac)
A
A
A
A
AB
A
A
A
A
A
A
A
A
AB
AB
B
AB
BC
BC
B
A
A
AB
AB
BC
A
BC
BC
C
C
C
A
A
A
C
C
Ae)
Ae)
A
A
A
A
A
A
A
Ae)
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Aa)
Aa)
Aa)
A
A
A
AB
A
AB
A
BC
A
A
A
A
C
C
A
AB
C
Fuel oil
Hot
A
A
A
A
A
A
A
A
B
AB
A
A
A
A
A
Fuel oil
(containing
H2 SO4 )
Hot
BC
A
AB
AB
A
A
B
A
C
B
C
AB
C
B
A
Fruit juices
Room
AB
A
Ae)
Ae)
A
A
A
A
AB
AB
AB
Ae)
A
A
Furfural
A
A
A
A
A
A
A
A
B
A
B
A
A
A
Gasoline
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Glauber’s Salt Solution
Hot
A
A
A
A
A
A
A
A
A
A
A
Glue – dry
Room
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Glue – solution
acid
Hot
B
A
AB
AB
A
A
A
A
AB
A
A
AB
Ba)
A
Glycerine
Room
A
A
A
A
A
Gypsum
A
A
A
A
A
A
A
A
A
A
A
Ae)
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Hydrocarbons
A
A
A
A
A
A
Ae)
AB
AB
AB
A
Ink
AC
A
A
A
A
A
A
A
AB
AB
AB
A
Bc)
A
Iodine
AB
C
C
AC
C
A
C
C
C
A
A
C
C
C
A
C
A
A
A
A
A
A
Room
A
A
A
Dyewood,
liquor
Fluorine
AB
A
Corrosion data–miscellaneous.
Media
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zircotration (∘ F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Iodoform
C
A
Kerosene
Room
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Ketchup
Room
B
A
Ae)
Ae)
A
A
AB
A
A
A
A
Ae)
Aa)
A
Aa)
A
A
Lard
Room
A
A
Lead
Molten
A
Linseed oil
A
A
Ae)
Ae)
A
A
A
A
Ae)
A
A
A
A
A
C
C
A
A
C
C
C
C
C
C
C
C
C
A
AC
A
A
A
AB
AB
A
A
A
A
A
A
A
AB
A
A
A
A
A
C
A
AB
A
A
A
A
A
A
A
A
AB
A
A
B
C
C
Lysol
230∘
212∘
AC
A
AB
A
A
A
AB
A
A
A
A
AB
Aa)
Aa)
C
A
Mayonnaise
Cold & Hot
BC
Ae)
Ae)
A
A
AB
A
AB
AB
AB
Ae)
Aa)
A
Meats
(unsalted)
Room
A
Ae)
Ae)
A
A
AB
A
A
AB
Ae)
Aa)
A
Aa)
A
Ae)
Ae)
A
A
A
A
Ae)
A
A
C
C
A
A
A
A
A
AB
AB
C
A
A
A
A
A
Ae)
Ae)
A
A
A
A
A
B
B
Ae)
A
A
A
A
A
B
B
A
A
Aa)
A
A
C
C
B
Aa)
Aa)
A
A
A
A
A
A
A
Lye (caustic)
34%
Mash
Hot
Mercury
A
C
Milk – fresh or
sour
Hot or cold
Mine
water – acid
Mustard
Room
Naphtha
Fused
A
A
A
A
A
A
AB
Ae)
Ae)
A
A
A
A
A
A
A
Nitre Cake
A
A
B
Molasses
A
A
B
A
AB
A
A
AB
AB
A
AB
A
A
A
A
A
A
A
Ae)
Aa)
Aa)
C
A
A
A
A
A
A
A
A
B
A
B
C
Oils – crude
Hot & cold
A
A
AC
AB
A
A
A
A
A
A
A
AB
Ac)
Ac)
Ac)
A
A
Oils –mineral
– vegetable
Hot and
cold
A
A
AB
AB
A
A
A
A
A
A
A
AB
Ac)
A
Ac)
A
A
Paraffin
Molten
A
A
Paregoric
compound
A
A
A
A
A
A
A
A
A
A
A
A
Ae)
Ae)
A
A
A
A
A
Ae)
A
A
B
A
A
A
A
A
A
A
A
A
A
Petroleum
ether
A
A
A
A
A
A
A
A
A
A
Phenol
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Phenolic resins
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Pine tar oil
Potash
Solution
Rosin
A
A
A
A
A
A
A
A
A
A
A
Hot
C
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Molten
A
A
A
A
A
A
B
B
AB
A
A
A
A
A
Sal ammoniac 20%
Boiling
C
A
C
C
A
A
B
A
C
B
B
Salt
Saturated
Room
AB
Aa)
A
A
A
A
Aa)
A
A
A
AB
A
A
Bb)
Aa)
Salt brine
Saturated
Hot
B
Aa)
AC
A
A
A
AB
A
AB
AB
AB
AB
Aa)
Aa)
C
B
A
A
B
C
A
A
A
A
A
(continued)
Corrosion data–miscellaneous.
Titanium
Concen- Temperature Alu- Carpenter
Cupro- Hastelloy Hastelloy Incoloy Incoloy Inconel Monel Nickel Silicon– Stainless Stainless Stainless Tanta- Comm’l Zircotration (∘ F)
minum No. 20 Cb-3 Copper nickel B-2
C-276
800
825
600
400 200 Bronze steel 304 steel 316 steel 430 lum Grade nium
Media
Sea water
AB
Sewage
AB
Soaps
Room
Soy bean oil
Soda pulp
Starch
A
AC
A
B
AB
AB
AB
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
B
B
Sugar juice
A
A
Ae)
Ae)
A
A
Tin
Molten
Tomato juice
Room
A
A
A
Molten
A
A
B
Sulfur – wet
Aa)
A
BC
Steam
Sulfur – dry
A
A
C
Solution
A
A
AB
A
AB
A
AB
A
A
A
AB
Aa)
Aa)
B
Ac)
Ac)
AB
A
A
A
A
A
C
A
A
A
A
A
A
A
A
A
A
A
A
A
A
C
A
A
A
A
A
A
A
A
A
A
Ae)
A
A
A
A
A
A
A
A
C
C
A
B
B
C
C
A
A
A
A
A
C
C
C
A
A
A
A
A
B
C
C
B
Ba)
Aa)
Ba)
A
C
C
C
C
C
C
C
C
C
A
B
A
Ae)
Ae)
A
A
Aa)
A
Aa) B
AB
AB
Ae)
Aa)
Aa)
Turpentine oil
A
A
A
A
A
A
A
A
AB
AB
AB
A
A
A
A
Tung oil
A
A
A
A
A
A
AB
A
A
A
A
AB
A
A
A
B
C
AB
A
AC
A
Varnish
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Vegetable
juices
AB
A
Ae)
Ae)
A
A
A
A
AB
AB
AB
Ae)
A
A
A
A
Vinegar – still
Room
AB
A
Ae)
Ae)
A
A
A
A
A
A
A
Ae)
A
A
A
A
Vinegar – agitated
Room
AB
A
Ae)
Ae)
A
A
A
A
A
A
A
Ae)
A
A
A
A
Vinegar – aerated
Room
A
A
A
A
AB
A
AB
AB
A
A
A
A
AB
A
A
A
A
Vinegar – fumes
AB
A
AB
AB
A
A
A
AB
AB
Ba)
Aa)
Ba)
A
Vinegar and
salt
C
A
B
B
A
A
A
AB
B
Water
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Water –hot
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
Water – salt
B
A
AC
A
A
A
Aa)
A
AB
AB
AB
AB
Aa)
Aa)
C
A
A
A
Water – sea
A
A
AC
A
A
A
Aa)
A
C
A
A
A
Whiskey
A
Ae)
Ae)
A
A
C
C
C
A
AC
Zinc
Molten
A – Fully resistant. B – Slightly attacked. C – Unsatisfactory.
a) Subject to pitting at air line or when allowed to dry.
b) Keep solution alkaline.
c) May attack when sulfuric acid is present.
d) May attack when hydrochloric acid is present.
e) Tin-coated.
f ) Not recommended for use with beverages.
Source: Courtesy of the Nooter Corporation, St. Louis, MO.
C
C
C
C
A
AB
AB
AB
AB
Aa)
Aa)
A
A
A
Ae)
A
A
C
C
C
C
C
C
C
A
431
I
Various ASME Design Equations
ID formulas
Part
Thickness t (in.)
Pressure P (psi)
Stress S (psi)
Cylindrical shell
PR
SE − 0.6P
SEt
R + 0.6 t
P(R + 0.6 t)
t
Hemispherical shell
PR
2SE − 0.2P
2SE t
R + 0.2 t
P(R + 0.2 t)
2t
2 : 1 Ellipsoidal head
PD
2SE − 0.2P
2SE t
D + 0.2 t
P(D + 0.2 t)
2t
SE t
0.885L + 0.1 t
P(0.885L + 0.1 t)
t
Flanged and dished head 6% knuckle
3 + (L∕r)1∕2
0.885PL
M=
4
SE − 0.1P
Other-size knuckles
PLM
2SE − 0.2P
2SE t
LM + 0.2 t
P(LM + 0.2 t)
2t
Cone
PD
2 cos 𝛼(SE − 0.6P)
2SE t cos 𝛼
D + 1.2 t cos 𝛼
P(D + 1.2t cos 𝛼)
2t cos 𝛼
R = inside radius
D = inside diameter
OD formulas
Part
Thickness
Pressure
Stress
Cylindrical shell
PRo
SE + 0.4P
SE t
Ro − 0.4 t
P(Ro − 0.4 t)
t
Hemispherical shell
PRo
2SE + 0.8P
2SE t
Ro − 0.8 t
P(Ro − 0.8 t)
2t
2 : 1 Ellipsoidal head
PDo
2SE + 1.8P
2SEt
Do − 1.8t
P(Do − 1.8t)
2t
2SE t
1.77L − 1.57 t
P(1.77L − 1.57t)
2t
Flanged and dished head 6% knuckle
3 + (L∕r)1∕2
1.77PL
M=
4
2SE + 1.57P
Other-size knuckles
PLM
2SE + P(M − 0.2)
2SE t
LM − t(M − 0.2)
PLM − P t(M − 0.2)
2t
Cone
PDo
2 cos 𝛼(SE + 0.4P)
2SE t cos 𝛼
Do − 0.8 t cos 𝛼
P(Do − 0.8 t cos 𝛼)
2 cos 𝛼
Ro = outside radius
Do = outside diameter
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
433
J
Joint Efficiency Factors
The criteria for joint efficiency requirements in the American Society of Mechanical Engineers (ASME) VIII-1 Code
were completely modified in the Winter 1986 Addenda. A summary of these requirements is given in this Appendix.
Figure J.1 shows the weld categories as defined in Section VIII-1. Table J.1 lists the required joint efficiencies for various
weld categories. The requirements of Paragraphs UW-11 and UW-12 of Section VIII-1 are summarized in Table J.2
for cylindrical shells and Table J.3 for heads. Figure J.2 shows 19 examples that illustrate the applicability of the rules
to various vessel configurations.
Figure J.1 Welded-joint
categories.
See UW-3(b)
C
B
C
α
A
D
C
A
A
A
A
A
B
A
A
A
B
D
B
D
B
B
A
D
B
A
C
C
D
Table J.1 Maximum allowable joint efficienciesa) for arc- and gas-welded joints.
Degree of radiographic
examination
Type no.
Joint description
Limitations
Joint
category
a
Full
b
Spot
c
None
(1)
BUTT JOINTS AS ATTAINED
BY DOUBLE WELDING OR BY
OTHER MEANS, WHICH WILL
OBTAIN THE SAME QUALITY
OF DEPOSITED WELD METAL
ON THE INSIDE AND
OUTSIDE WELD SURFACES
TO AGREE WITH THE
REQUIREMENTS OF UW-35,
WELDS USING METAL
BACKING STRIPS, WHICH
REMAIN IN PLACE, ARE
EXCLUDED.
NONE
A, B, C, & D
1.0
0.85
0.70
(Continued)
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
434
J Joint Efficiency Factors
Table J.1 (Continued)
Degree of radiographic
examination
Joint
category
a
Full
b
Spot
c
None
(a) NONE EXCEPT AS SHOWN
IN (b) BELOW
A, B, C, & D
0.90
0.80
0.85
(b) CIRCUMFERENTIAL BUTT
JOINTS WITH ONE PLATE
OFFSET, SEE UW-13(c) AND
FIGURE UW-13.1 (k).
A, B, & C
0.90
0.80
0.85
Type no.
Joint description
Limitations
(2)
SINGLE-WELDED BUTT
JOINT WITH BACKING STRIP
OTHER THAN THOSE
INCLUDED UNDER (1)
(3)
SINGLE-WELDED BUTT
JOINT WITHOUT USE OF
BACKING STRIP
CIRCUMFERENTIAL BUTT
JOINTS ONLY, NOT OVER
5/8 IN. THICK AND NOT OVER
24 IN. OUTSIDE DIAMETER
A, B, & C
NA
NA
0.60
(4)
DOUBLE FULL-FILLET LAP
JOINT
LONGITUDINAL JOINTS NOT
OVER 3/8 IN. THICK
A
NA
NA
0.55
CIRCUMFERENTIAL JOINTS
NOT OVER 5/8 IN. THICK
B&C
NA
NA
0.55
(a) CIRCUMFERENTIAL
JOINTSb) FOR ATTACHMENT
OF HEADS NOT OVER 24 IN.,
OUTSIDE DIAMETER TO
SHELLS NOT OVER 1/2 IN.
THICK
B
NA
NA
0.50
(b) CIRCUMFERENTIAL
JOINTS FOR THE
ATTACHMENT TO SHELLS OF
JACKETS NOT OVER 5/8 IN. IN
NOMINAL THICKNESS
WHERE THE DISTANCE FROM
THE CENTER OF THE PLUG
WELD TO THE EDGE OF THE
PLATE IS NOT LESS THAN
1-1/2 TIMES THE DIAMETER
OF THE HOLE FOR THE PLUG.
C
NA
NA
0.50
(a) FOR THE ATTACHMENT
OF HEADS CONVEX TO
PRESSURE TO SHELLS NOT
OVER 5/8 IN. REQUIRED
THICKNESS. ONLY WITH USE
OF FILLET WELD ON INSIDE
OF SHELLS, OR
A&B
NA
NA
0.50
(5)
(6)
SINGLE FULL-FILLET LAP
JOINTS WITH PLUG WELDS
CONFORMING TO UW-17
SINGLE FULL-FILLET LAP
JOINTS WITHOUT PLUG
WELDS
(b) FOR ATTACHMENT OF
HEADS HAVING PRESSURE
ON EITHER SIDE. TO SHELLS
NOT OVER 24 IN. INSIDE
DIAMETER AND NOT OVER
1/4 IN. REQUIRED THICKNESS
WITH FILLET WELD ON
OUTSIDE OF HEAD FLANGE
ONLY.
a) E = 1.0 FOR BUTT JOINTS IN COMPRESSION.
b) JOINTS ATTACHING HEMISPHERICAL HEADS TO SHELLS ARE EXCLUDED.
Joint Efficiency Factors
Table J.2 Joint efficiencies for cylindrical shells.
YES
FULL RT
PER UW - 11
NO
YES
SEAMLESS
SHELL
NO
J.E. FOR
LONG IT.
STRESS:
TYPE 1 = 1.0
TYPE 2 = 0.9
SPOT RT
J.E. FOR
FOR TYPE
CIRCUMF. YES
“B” & “C”
STRESS
BUTT WELDS
= 1.0
SELECT RT
FOR TYPE
“B” & “C”
BUTT WELDS
SELECT RT
FOR TYPE
“A” & “D”
BUTT WELDS
NO
J.E. FOR
CIRCUMF.
STRESS
= 0.85
SELECT J.E.
PER UW - 12
CALCULATE
REQUIRED
THICKNESS
Table J.3 Joint efficiencies for heads.
YES
FULL RT
PER UW - 11
NO
SELECT HEAD TYPE
HEMISPHERICAL
ELLIPSOIDAL
SELECT RT
FOR TYPES
“A” & “D”
BUTT WELD
SELECT J.E.
PER UW - 12
CALCULATE
REQUIRED
THICKNESS
TORISPHERICAL
FLAT
SELECT RT
FOR TYPE “A”
BUTT WELD
435
436
J Joint Efficiency Factors
Figure J.2 Joint efficiencies. Source:
Courtesy of Mr R. J. Cepluch.
C
(a)
B
B
A
D
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUTT WELD FULL RT
B TYPE 1 BUTT WELD FULL RT
SEE UW-11(a)(5)(a)/UW-12(a)/TABLE UW-12(a)
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR)
E = 1.00 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD FULL RT
B TYPE 1 BUTT WELD SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(a), (b) & (d)/TABLE UW-12(a) & (b)
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.85 LONGITUDINAL STRESS CALCULATIONS
C
B
A
A
D
A
A TYPE 1 BUTT WELD FULL RT
SEE UW-11(a)(5)(a)/UW-12(a)/TABLE UW-12(a)
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 HEAD CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD IN SHELL FULL RT
A TYPE 2 BUTT WELD HEAD TO SHELL FULL RT
SEE UW-11(a)(5)(a)/UW-12(a)/TABLE UW-12(a)
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.90 HEAD CALCULATIONS (JOINT EFFICIENCY)
E = 0.90 LONGITUDINAL STRESS CALCULATIONS
SEAMLESS
HEMISPHERICAL
Joint Efficiency Factors
Figure J.2 (Continued)
(b)
C
B
A
A
D
A
SEAMLESS
HEMISPHERICAL
A TYPE 1 BUTT WELD IN SHELL FULL RT
A TYPE 2 BUTT WELD HEAD TO SHELL SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(a), (b) & (d)/TABLE UW-12(a) & (b)
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.80 HEAD CALCULATIONS (JOINT EFFICIENCY)
E = 0.80 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD IN SHELL NO RT
A TYPE 2 BUTT WELD HEAD TO SHELL NO RT
SEE UW-12(c) & (d)/TABLE UW-12(c)
E = 0.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.65 HEAD CALCULATIONS (JOINT EFFICIENCY)
E = 0.65 LONGITUDINAL STRESS CALCULATIONS
C
B
A
A
D
A
SEAMLESS
HEMISPHERICAL
A TYPE 1 BUTT WELD IN SHELL SPOT RT
A TYPE 2 BUTT WELD HEAD TO SHELL SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b) & (d)/TABLE UW-12(b)
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.80 HEAD CALCULATIONS (JOINT EFFICIENCY)
E = 0.80 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD IN SHELL SPOT RT
A TYPE 2 BUTT WELD HEAD TO SHELL NO RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b), (c) & (d)/TABLE UW-12(b) & (c)
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.65 HEAD CALCULATIONS (JOINT EFFICIENCY)
E = 0.65 LONGITUDINAL STRESS CALCULATIONS
437
438
J Joint Efficiency Factors
(c)
Figure J.2 (Continued)
C
B
B
A
D
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUTT WELD SPOT RT
B TYPE 1 BUTT WELD SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b) & (d)/TABLE UW-12(b)
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.85 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD SPOT RT
B TYPE 1 BUTT WELD NO RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b), (c)/TABLE UW-12(b) & (c)
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.85 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.70 LONGITUDINAL STRESS CALCULATIONS
C
B
B
A
D
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUTT WELD SPOT RT
B TYPE 2 BUTT WELD SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b) & (d)/TABLE UW-12(b)
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.80 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD SPOT RT
B TYPE 2 BUTT WELD NO RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b), (c) & (d)/TABLE UW-12(b) & (c)
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.85 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.65 LONGITUDINAL STRESS CALCULATIONS
Joint Efficiency Factors
Figure J.2 (Continued)
(d)
C
B
B
A
D
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 3 BUTT WELD NO RT
B TYPE 3 BUTT WELD NO RT
SEE UW-11(a)(5)(a) & (b)/UW-12(c) & (d)/TABLE UW-12(c)
E = 0.60 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.85 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.60 THE LONGITUDINAL STRESS CALCULATIONS
A TYPE 3 BUTT WELD FULL OR SPOT RT
B TYPE 3 BUTT WELD FULL OR SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(a),(b) & (d)/TABLE UW-12(a) & (b)
E = 0.60 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.60 LONGITUDINAL STRESS CALCULATIONS
C
B
B
A
D
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUTT WELD NO RT
B TYPE 1 BUTT WELD NO RT
SEE UW-12(c) & (d)/TABLE UW-12 (c)
E = 0.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.85 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.70 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD NO RT
B TYPE 1 BUTT WELD SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b), (c) & (d)/TABLE UW-12(b) & (c)
E = 0.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.85 LONGITUDINAL STRESS CALCULATIONS
439
440
J Joint Efficiency Factors
(e)
Figure J.2 (Continued)
C
B
B
A
D
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUTT WELD NO RT
B TYPE 6 SINGLE FILLET WELD
SEE UW-11(a)(5)(a) & (b)/UW-12(d)/TABLE UW-12(c)
E = 0.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.85 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.45 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD SPOT RT
B TYPE 6 SINGLE FILLET WELD
SEE UW-11(a)(5)(a) & (b)/UW-12(b) & (d)/TABLE UW-12(b) & (c)
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.85 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.45 LONGITUDINAL STRESS CALCULATIONS
B TYPE 4 DOUBLE FILLET WELD SAME AS B TYPE 6 EXCEPT
E = 0.55 LONGITUDINAL STRESS CALCULATIONS
C
B
B
D
SEAMLESS SHELL
B
SEAMLESS SHELL
B TYPE 1 BUTT WELD FULL RT
SEE UW-11(a)(5)(a)/UW-12(a) & (d)/TABLE UW-12(a)
E = 1.00 SHELL CALCULATIONS (QUALITY FACTOR)
E = 1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E = 1.00 LONGITUDINAL STRESS CALCULATIONS
SEAMLESS SHELL
B TYPE 1 BUTT WELD SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b) & (d)/TABLE UW-12(c)
E = 1.00 SHELL CALCULATIONS (QUALITY FACTOR)
E = 1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.85 LONGITUDINAL STRESS CALCULATIONS
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
Joint Efficiency Factors
Figure J.2 (Continued)
(f)
C
B
B
D
SEAMLESS SHELL
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
SEAMLESS SHELL
B TYPE 6 SINGLE FILLET WELD
SEE UW-11(a)(5)(a)/UW-12(d)/TABLE UW-12(c)
E = 0.85 SHELL CALCULATIONS (QUALITY FACTOR)
E = 0.85 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.45 LONGITUDINAL STRESS CALCULATIONS
B TYPE 4 DOUBLE FILLET WELD SAME AS B TYPE 6 EXCEPT
E = 0.55 LONGITUDINAL STRESS CALCULATIONS
C
B
B
A
D
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A ERW PIPE
B TYPE 1 BUTT WELD SPOT RT
SEE UW-11(a)(5)(a) & (b)/UW-12(b) & (d)/TABLE UW-12(b)
E = 1.00 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E = 1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.85 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
A ERW PIPE
B TYPE 1 BUTT WELD NO RT
SEE UW-11(a)(5)/UW-12(c) & (d)/TABLE UW-12(c)
E = 0.85 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E = 0.85 HEAD CALCULATIONS (QUALITY FACTOR)
E = 0.70 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
NOTE 1: THE QUALITY FACTOR SHOWN IN THE SHELL CALCULATIONS
IS IN ADDITION TO THE CATEGORY A JOINT FACTOR (E=0.85)
INCLUDED IN THE ALLOWABLE STRESS VALUE OBTAINED FROM
THE APPLICABLE STRESS TABLE.
NOTE 2: DIVIDE THE ALLOWABLE STRESS VALUE OBTAINED
FROM THE APPLICABLE STRESS TABLE BY 0.85 BEFORE APPLYING
A JOINT EFFICIENCY IN THE LONGITUDINAL STRESS CALCULATION
(eg, SEE NOTE 26, TABLE UCS-23)
441
442
J Joint Efficiency Factors
(g)
Figure J.2 (Continued)
C
B
A
B
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
B
D
D TYPE 1 BUTT WELD
SEE UW-11(a)(5) FULL RT IF PART IS DESIGNED E = 1.00
RT NOT REQUIRED FOR tr or trn CALCULATIONS E = 1.00
E = APPLICABLE EFFICIENCY IF OPENING THROUGH CATEGORY A WELD
D FULL OR PARTIAL PENETRATION CORNER WELD
RT NOT REQUIRED E = 1.00
RT NOT REQUIRED FOR tr or trn CALCULATIONS E = 1.00
E = EFFICIENCY OF THE BUTT WELD PENETRATED
C
B
A
B
B
D
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
B & C TYPES 1 OR 2 BUTT WELD IN NOZZLES/COMMUNICATING CHAMBERS
RT NOT REQUIRED TO CALCULATE trn
RT MAY BE REQUIRED FOR
SERVICE RESTRICATION
MATERIAL THICKNESS
USER/DESIGNATED AGENT
C
B
C
A
D
C
FLAT HEAD
C IS PARTIAL OR FULL PENETRATION CORNER JOINT FOR ALL CASES
E FOR C IS NOT ESTABLISHED BY CODE RULES
A TYPE 1 BUTT WELD FULL RT - E = 1.00
A TYPE 1 BUTT WELD SPOT RT - E = 0.85
A TYPE 1 BUTT WELD NO RT - E = 0.70
A ERW BUTT WELD - E = 1.00 (JOINT EFFICIENCY IN STRESS VALUE)
A SEAMLESS - E = 1.00
NOTE : FLAT HEAD FORMULA HAS BUILT-IN STRESS MULTIPILER
REGARDLESS OF TYPE OF JOINT OR EXAMINATION. A FACTOR FOR
E WILL APPLY IN THE FLAT HEAD FORMULA IF A CATEGORY A
JOINT DOES EXIST IN THE HEAD
Joint Efficiency Factors
Figure J.2 (Continued)
C
(h)
B
C
A
D
C
TUBE SHEET
OR
FLANGE
A TYPE 1 BUTT WELD FULL RT
C TYPE 1 BUTT WELD FULL RT
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 1.00 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD FULL RT
C TYPE 1 BUTT WELD SPOT RT
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.85 LONGITUDINAL STRESS CALCULATIONS
C
B
C
A
D
C
TUBE SHEET
OR
FLANGE
A TYPE 1 BUTT WELD FULL RT
C TYPE 1 BUTT WELD NO RT
E = 0.85 SHELL CALCULATIONS (QUALITY FACTOR)
E = 0.70 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD SPOT RT
C TYPE 1 BUTT WELD NO RT
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E = 0.70 LONGITUDINAL STRESS CALCULATIONS
443
444
J Joint Efficiency Factors
(i)
C
B
C
A
D
C
TUBE SHEET
OR
FLANGE
A TYPE 1 BUTT WELD FULL RT
C WELD FULL OR PARTIAL PENETRATION CORNER JOINT
E = 1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E FOR C IS NOT ESTABLISHED BY CODE RULES
A TYPE 1 BUTT WELD SPOT RT
C WELD FULL OR PARTIAL PENETRATION CORNER JOINT
E = 0.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E FOR C IS NOT ESTABLISHED BY CODE RULES
A TYPE 1 BUTT WELD NO RT
C WELD FULL OR PARTIAL PENETRATION CORNER JOINT
E = 0.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E FOR C IS NOT ESTABLISHED BY CODE RULES
C
B
C
A
D
C
TUBE SHEET
OR
FLANGE
A ERW PIPE
C TYPE 1 BUTT WELD SPOT RT
E = 1.00 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E = 0.85 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
A ERW PIPE
C TYPE 1 BUTT WELD NO RT
E = 0.85 SHELL CALCULATION (QUALITY FACTOR), NOTE 1
E = 0.70 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
A ERW PIPE
C TYPE 2 BUTT WELD NO RT
E = 0.85 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E = 0.65 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
A ERW PIPE
C FULL OR PARTIAL PENETRATION CORNER JOINT
E = 1.00 SHELL CALCULATION (QUALITY FACTOR), NOTE 1
E FOR C IS NOT ESTABLISHED BY CODE RULES, NOTE 2
NOTE 1: THE QUALITY FACTOR SHOWN IN THE SHELL CALCULATIONS IS IN
ADDITION TO THE CATEGORY A JOINT FACTOR (E = 0.85) INCLUDED IN THE
ALLOWABLE STRESS VALUE OBTAINED FROM THE APPLICABLE STRESS TABLE
NOTE 2: DIVIDE THE ALLOWABLE STRESS VALUE OBTAINED FROM THE APPLICABLE
STRESS TABLE BY 0.85 BEFORE APPLYING A JOINT EFFICIENCY IN THE
LONGITUDINAL STRESS CALCULATIONS (eg. SEE NOTE 26, TABLE UCS-23)
Figure J.2 (Continued)
445
K
Simplified Curves for External Loading on Cylindrical Shells
Figure K.1 Moment M(x, 𝜙) (do /Mc ) due to an external
circumferential moment Mc on a circular cylinder.
1.0
Mϕ
Mx
M(x,ϕ) (d o /Mc)
0.10
Bending stress = Kb [M(x,ϕ) (d o /Mc)]
6Mc
d oT 2
0.01
0.001
0.10
λ=
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
1.0
do
DmT
10.0
K Simplified Curves for External Loading on Cylindrical Shells
Figure K.2 Membrane force N(x, 𝜙) (do T/Mc ) due to an external
moment Mc on a circular cylinder.
1.0
Nx
Nϕ
0.10
N(x,ϕ) (d oT/Mc)
446
Membrane stress = Kn [N(x,ϕ) (d oT/Mc)]
0.01
0.001
0.10
λ=
1.0
do
D mT
Mc
d oT 2
10.0
Simplified Curves for External Loading on Cylindrical Shells
Figure K.3 Moment M(x, 𝜙) (do /ML ) due to an external
longitudinal moment ML on a circular cylinder.
1.0
Mx
Mϕ
0.10
M(x,ϕ) (d o /ML)
Maximum value off axis
of symmetry
Value on axis of
symmetry
Bending stress = Kb [M(x,ϕ) (d o /ML)]
0.01
0.001
0.10
λ=
1.0
do
DmT
6ML
d oT 2
10.0
447
K Simplified Curves for External Loading on Cylindrical Shells
Figure K.4 Membrane force N(x, 𝜙) (do T/ML ) due to an external
moment ML on a circular cylinder.
1.0
Nϕ
Nx
0.10
N(x,ϕ) (d oT/ML)
448
Membrane stress = Kn [N(x,ϕ) (d oT/ML)]
0.01
0.001
0.10
λ=
1.0
do
DmT
ML
d oT 2
10.0
Simplified Curves for External Loading on Cylindrical Shells
Figure K.5 Bending moment M(x, 𝜙) due to an external radial
load P on a circular cylinder. Mx /P on longitudinal axis, M𝜙 /P on
transverse axis.
1.0
0.10
M(x,ϕ)/P
Dm /T = 10
30
Bending stress = Kb (M(x,ϕ) /P)
0.01
6P
T2
100
200
600
0.001
0.10
1.0
λ=
do
D mT
10.0
449
K Simplified Curves for External Loading on Cylindrical Shells
Figure K.6 Bending moment M(x, 𝜙) due to an external radial
load P on a circular cylinder. M𝜙 /P on longitudinal axis, Mx /P on
transverse axis.
1.0
0.10
M(x,ϕ) /P
450
Bending stress = Kb (M(x,ϕ) /P)
6P
T2
Dm /T = 10, 30
0.01
100
200
600
0.001
0.10
λ=
1.0
do
D mT
10.0
Simplified Curves for External Loading on Cylindrical Shells
Figure K.7 Membrane force N(x, 𝜙) T/P due to an external radial
load P on a circular cylinder (transverse axis).
1.0
Nx
0.10
N(x,ϕ) T/P
Nϕ
Membrane stress = Kn (N(x,ϕ) T/P)
0.01
0.001
0.10
λ=
1.0
do
RmT
P
T2
10.0
451
K Simplified Curves for External Loading on Cylindrical Shells
Figure K.8 Membrane force N(x, 𝜙) T/P due to an external radial
load P on a circular cylinder (longitudinal axis).
1.0
Nϕ
0.10
Nx
N(x,ϕ) T/P
452
Membrane stress = Kn (N(x,ϕ) T/P)
0.01
0.001
0.10
λ=
1.0
do
RmT
P
T2
10.0
453
L
Conversion Tables
Conversion to metric units.
Multiply customary units
By factor
To get metric units
Inches
0.0254
Meters
US gallons
0.003785
m3
ft3
0.02832
m3
Pounds mass
0.4536
Kilograms
Pounds force
4.448
Newtons
psi pressure
6 894.8
Pascals
Bars
100,000
Pascals
Btu
1 005.056
Joules
Horsepower (550 ft-lb/s)
√
ksi − in. fracture toughness
∘F
745.7
Watts
√
MPa- m
∘C
1.1
(∘ F − 32)/1.8
Other conversions.
Multiply
By factor
To get
ft3
7.48
US gallons
Bars
14.50
psi
Miles
5280
Feet
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
454
L Conversion Tables
Conversion to customary units.
Multiply metric units
By factor
To get customary units
Meters
39.370
Inches
3
m
264.201
US gallons
m3
35.311
ft3
Kilograms mass
2.205
Pounds mass
Newtons
0.225
Pounds force
Pascals
0.000145
psi pressure
Pascals
0.00001
Bars
Joules
0.0009478
Btu
Watts
√
MPa- m fracture toughness
∘C
0.001341
Horsepower (550 ft-lb/s)
√
ksi − in.
∘F
0.9091
1.8(∘ C) + 32
455
Index
a
Aluminum alloys 15, 31, 32, 407–409
Aluminum tanks 270–271
Anchor chair design 239–241
Annular plates in flat bottom tanks
261–263
ANSI B31.3 chemical plant & pet. ref.
piping, nozzle reinforcement
6, 14, 212
ANSI B31 piping code, organization
6
ANSI B31.1, power piping code 210
allowable stresses 212
nozzle reinforcement 211
ASCE 7. Minimum design loads 321,
332
ASME code
allowable external pressure
stresses 13–14
allowable tensile stresses 12–13
design criteria 18–23
external pressure design 121–126
fracture criteria 12, 18
guide to Sect. I 378–381
guide to Sect. VIII-1 375–377
history 4–7
nozzle reinforcement, Section I
195–198
nozzle reinforcement,
Section VIII-1 198–204
nozzle reinforcement,
Section VIII-2 204–210
organization 4–7
Australian code 16
Autofrettaging. See Prestressing
Axial compression 13, 123, 132, 146,
147, 270, 340
b
Base ring design 234–241
Bijlaard stresses (WRC 107) 218,
219, 222, 223, 225, 242
Blind flanges 150–184, 400, 401
Bolt dimensions 236
Bolted flanged connections
flange bolting design 161–163
flange contact facings 155
British code 345, 360
Brittle fracture
ASME fracture criteria 35–49
crack shape factor 39
factors influencing 42–43
theory 39, 45
Buckling
conical heads 98, 99
cylinders 69–72, 98
hemispherical heads 91–93
c
Charpy V-notch test 36–37, 43,
45
China code 15
Circular flat plates 151–153. See also
Flat heads and covers, types
Compression ring 96, 267–270
Concrete properties 237
Conical heads and transitions
cone-cylinder junction 96–97
discontinuity analysis 97–98
external pressure 98–99
Conversion tables 453–454
Copper alloys 15, 32, 407, 408
Corrosion 10, 11, 26, 30–32, 34, 155,
255, 258, 271, 279, 283, 311,
336, 341, 385, 387, 389,
393–396, 401, 403, 404,
415–430
Corrosion charts 416–430
Creep rate 12–15, 28, 411
Cylinders
buckling
end load 72
side and end loads 71–72
side load 70–71
discontinuity analysis
long cylinders 61–66
short cylinders 66–69
stress due to internal pressure
ASME equations for cylinders
56–60
thermal stress 72–80
d
Design criteria 12, 18–23, 143, 254
Design equations, various ASME 431
Design specifications 10–11, 320,
321
Discontinuity stresses
conical heads and transitions 136
cylinders 60–69
gross 115
hemispherical heads and transitions
60–69
local 115
Drop weight test (DWT) 37, 38
e
Earthquake loading, tall vessels
320–332, 338
Ellipsoidal heads
ASME design for external pressure
93–95
ASME equations for internal
pressure 93–95
Equivalent stress 19, 115, 118, 119
European code 14
Expansion joints 7, 276, 293,
300–304
External loading, simplified curves for
cylinders 445–452
External pressure design
conical heads and transitions
98–99
cylinders 13, 60, 129–132
ellipsodial heads 98–99
hemispherical heads 137–139
Structural Analysis and Design of Process Equipment, Third Edition. Maan H. Jawad and James R. Farr.
© 2019 American Institute of Chemical Engineers, Inc. Published 2019 by John Wiley & Sons, Inc.
456
Index
External pressure design (contd.)
procedure for providing data for
code charts 413–414
tall vessels 340–341
torispherical heads 143
f
Ferrous alloys 26, 31, 34, 35, 413
Firetube boilers 4, 5, 366, 372, 373
Fixed tubesheets 281, 291–300
Flanges
blind 150–184, 400, 401
flat-face with metal-to-metal
contact 177
full-face gasket 154, 171–176
reverse 154, 170–173
ring-type gaskets 154, 164–170,
173
Flange calculation sheets
reverse flange with ring-type gasket
176, 180
ring flange with ring-type gasket
176, 179
slip-on or lap-type with ring-type
gasket 176, 178
slip-on with full-face gasket 177,
181
welding-neck flange with full-face
gasket 177, 182
welding neck flange with ring-type
gasket 176, 180
Flanged and dished heads 136,
138–143, 147, 431. See also
Torispherical heads
Flat bottom tanks
aluminum tanks 270–271
API620 263–270
API650 254–263
AWWA 271–272
D100 271–272
Flat-face flange with metal-to-metal
contact 177
Flat heads and covers, types
ASME formulas 153–154
circular plates, uniform loading
151–153
comparison of theory and ASME
formulas 154
Flat plates
circular 151–153
elastic foundation 107–108
rectangular 106–107
Force-stress equations 22–23
Fracture analysis diagram 37–39
Fracture toughness factor 39
French code 14
Full-face gasket flange 154, 171–177
g
Gaskets
compressed fiber 158
delta 155, 159, 160
design rules, ASME 160–161
double-cone 155, 159–160
flat metal 155, 158
high-pressure type 155, 158–159
jacketed 155, 158
lens ring 159
metallic O-and C-rings 155–156
metal ring 155, 158
rubber O-rings 155
spiral wound 155, 158
Guide to ASME code 375–381
Gust factor 332
h
Heat exchangers 11, 26, 31, 107,
158, 275, 276, 280, 283, 289,
291, 298, 311, 344, 349,
383–385
Heat transfer equipment 276–305
Heat treating 31, 35, 45, 129, 411
Hemispherical heads and transition
sections
ASME equations for internal
pressure 137–139
buckling strengths 91–93
discontinuity analysis 60–69
thermal stress 72–80
High-pressure vessels 6
Hydrogen embrittlement 50
Hydrostatic test, for determining
maximum defect size 42
l
Layered vessels 307, 308, 311–317
prestressing of 315–317
Ligament efficiency of openings
215–217
in vessels of noncircular cross
section 345–352
Lug design 320
m
Material cost 30, 136
Material specifications 6, 28, 45, 389,
393, 407–409
Material strength properties
creep rate 28
rupture strength 28
specified minimum yield strength
12, 13
n
Natural frequency 321, 324–332
Nelson chart 49, 50
Nickel alloys 12, 13, 15, 32–34,
407–409
Noncircular cross section vessels
allowable stresses 352–353
ASME code, Section VIII-1
356–360
basic equations 353–356
design methods 352–353
Nonferrous materials and alloys 6,
10, 367
Nonmetallic vessels, ASME Section X
26
Nozzles
external loadings 218–230
fatigue evaluation 217–218
stresses 188–192, 218–230
o
i
Openings
ligament efficiency 188, 195,
215–217, 345–352
in vessels of noncircular cross
section 345–352
stresses and loads 188–192
Out-of-roundness of cylinders external
pressure 129–132
Indian code 7
Internal pressure
heads 368
shells 56–60
tall vessels 336–339
j
Japanese codes 15
Joint efficiency factors
433–444
k
K Factors
39, 140
114, 352,
p
Peak stress 19, 115, 132, 188, 217,
218, 223
Power boilers
design requirements 373
Index
firetube 373
loading on attachments 368–369
watertube 369–373
Pressure vessel usage 4
Prestressing
layered-wall vessels 311–315
solid-wall vessels 309–311
Primary stress 19, 116, 118, 188, 192,
193, 204, 208, 210, 223
Purchase orders 10–11, 320, 321
r
Rectangular flat plates 106–107
Rectangular vessels 345, 359
analysis of 345, 359
Reinforced openings. See also
Openings
reinforcement limits 193–214
reinforcement rules, ANSI/ASME
B31.1 210–212
reinforcement rules, ANSI/ASME
B31.3 212–214
reinforcement rules, Section I
195–198
reinforcement rules, Section VIII-1
198–204
reinforcement rules, Section VIII-2
204–210
theory 192–193
Required data for material approval in
ASME-VIII 411–412
Reverse flanges 154, 170–173
Ring girders 12, 243–246, 248, 320
Roofs 15, 16, 50, 136, 195, 254–258,
261–264, 267, 268, 270, 271,
345, 360, 366, 387, 389, 391
Rupture strength 12, 14, 18, 30,
353
s
Saddle design 245–248
Sample of, specification sheets
API 387–391
heat exchangers 383–385
pressure vessel design 393–405
various materials for process
equipment 407–409
Secondary stress 18, 19, 115, 116,
118–121, 223, 285, 294, 311
Selection of vessels 10–16
Shell design
API 258–261
ASME 114–132
Skirt design 235
Spherically dished covers 150, 151,
177–184
Stiffening rings
allowable gaps 129
design, ASME 126–129
Strain-deflection equations 20–22
Strength theories 18–23
Stress-strain relationships 19–20,
147
Supports, vessel
legs 241–242
ring girders 234, 243–245
saddles 245–248
skirts 234–241
t
Tall vessels
design considerations 320
earthquake loading 320–331
external pressure and external
loading 341
external pressure only 340–341
internal pressure and external
loading 338–339
internal pressure only 336–338
wind loading 331–336
Tanks. See Flat bottom tanks
Tensile strength, specified minimum
12, 13, 200
Thermal stress
cylinders 72–80
general 19
hemispherical heads 91
local 115
Titanium 15, 25, 26, 28, 30, 31,
33, 34, 42, 136, 158, 416–450
Torispherical heads
ASME design for external pressure
95
ASME equations for internal
pressure 95
Tubesheets
ASME design for fixed 291–293
ASME design for u-tube
283–291
theoretical design for fixed
291–293
theoretical design for u-tube
280–283
Tube-to-tubesheet joints 384
Tubular Exchanger Manufacturers
Association (TEMA) 6, 11,
276–280, 291, 385
u
Uniform building code 6
U-tube exchangers 276–291, 293
v
Vibration analysis 332
Vortex shedding 320, 331, 332, 335
w
Watertube boilers 366, 369–373
Wind loading, tall vessels 320,
331–336, 338
Worldwide presure vessel codes
Australia 16
China 7, 15–16
Europe 14
India 7
Japan 7, 15
United Kingdom 7
y
Yield strength, specified minimum
12, 13, 353
z
Zirconium 26, 30, 31, 33, 158,
416–430
457
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