GUIDEBOOK
FOR THE DESIGN OF
ASME SECTION VIII
PRESSURE VESSELS
Second Edition
by
James R. Farr
Wadsworth, Ohio
Maan H. jawad
Nooter Corporation
St. Louis, Missouri
ASME Press
New York
2001
Copyright © 200 I
The American Society of Mechanical Engineers
Three Park Ave., New York, NY 10016
Library of Congress Catetogtng-tn-Publtcattcn Data
Parr, James R.
Guidebook for the design of ASME Section
R. Farr, Maan H. Jawad.-2nd ed.
p. em.
vm pressure vessels/by
James
Includes bibliographical references and index,
ISBN 0-7918-0172-1
I. Pressure vessels-Design and construction.
2. Structural engineering. 1. Jawad, Maan H.
II. Title.
TA660. T34 F36 2001
681 '.76041-dc21
2001046096
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To our children,
Katherine, David, Susan, Nancy, and Thomas
Jennifer and Mark
Cover Photo Courtesy of Nooter Corp.
PREFACE TO
SECOND EDITION
The ASME Boiler and Pressure Vessel Code, Section VIII, is a live and progressive document. It strives
to containthe latest, safe and economical rules for the design and construction of pressure vessels, pressure
vessel components, and heat exchangers. A major improvement was made within the last year by changing
the design margin on tensile strength from 4.0 to 3.5. This reduction in the margin permits an increase in
the allowable stress for many materials with a resulting decrease in minimumrequired thickness. This was
the firstreduction in this design margin in 50 years andwas based upon the many improvements in material
properties, design methods, and inspection procedures during that time.
Chapters and parts of chapters bave been updated to incorporate the new allowable stresses and improvements which have been made in design methods since this book was originally issued. Some of these
changes arc extensive and some are minor. Some of the examples in this book have changed completely
and some remainunchanged. This book continuesto be an easy reference for the latest methods of problem
solving in Section VIII.
James R. Farr
Wadsworth, Ohio
Maan H. Jawad
St. Louis, Missouri
July 2001
v
ACKNOWLEDGMENTS
We are indebted to many people and organizations for their help in preparing this book, Special thanks are
given to the Noorer Corporation, fellow Committee Members; and to former coworkers for their generous
support during the preparation of the manuscript. We also give thanks to Messrs. Greg L. Hollinger and
George B. Komora for helping with tbe manuscript, and to our editor Ray Ramonas at ASME for having
great patience and providing valuable suggestions.
vii
CONTENTS
Preface
Acknowledgments
List of Figures
List of Tables
~
:
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Chapter 1
Background Information
1.1
Introduction
1.2
Allowable Stresses
.,
1.3
Joint Efficiency Factors
1.4
Brittle Fracture Considerations
1.5
Fatigue Requirements
1.6
Pressure Testing of Vessels and Components
1.6.1 ASME Code Requirements
1.6.2 What Does a Hydrostatic or Pneumatic Pressure Test Do?
1.6.3 Pressure Test Requirements for VIlJ-l
1.6.4 Pressure Test Requirements for VIII-2
Chapter 2
Cylindrical Shells
2.1
Introduction
2.2
Tensile Forces, VIII-I
2.2.1 Thin Cylindrical Shells
2.2.2 Thick Cylindrical Shells
2.3
Axial Compression
2.4
External Pressure
2.4.1 External Pressure for Cylinders with Dolt :2: 10
2.4.2 External Pressure for Cylinders with Dolt < 10
2.4.3 Empirical Equations
2.4.4 Stiffening Rings
2.4.5 Attachment of Stiffening Rings
2.5
Cylindrical Shell Equations, Vlll-2
2.6
Miscellaneous Shells
2.6.1 Mitered Cylinders
2.6.2 Elliptical Shells
'"
.
..
.
.
.
.
.
.
..
..
.
.
.
.
.
.
,
Chapter 3
Spherical Shells, Heads, and Transition Sections
3.1
Introduction
3.2
Spherical Shells and Hemispherical Heads, VIII-1
3.2.1 Internal Pressure in Spherical Shells and Pressure on Concave Side of
Hemispherical Heads
,
3.2.2 External Pressure in Spherical Shells and Pressure on Convex Side of
Hemispherical Heads .
3.3
Spherical Shells and Hemispherical Heads, VIII-2
ix
.
.
..
..
.
..
..
..
..
..
..
.
.
v
vii
xiii
xvii
1
1
2
3
9
19
22
22
22
23
24
27
27
27
27
33
36
42
43
46
47
48
50
53
54
54
55
..
.
.
57
57
57
.
57
.
61
64
x Contents
3.4
3.5
3.6
3.7
3.8
Ellipsoidal Heads, VIII-I
3.4.1 Pressure on the Concave Side
3.4.2 Pressure on the Convex Side
Torispherical Heads, VlIl-1
3.5.1 Pressure on the Concave Side
3.5.2 Pressure on the Convex Side
Ellipsoidal and Torispherical Heads, VnI~2
Conical Sections, VIII~ 1
3.7.1 Internal Pressure
3.7.2 External Pressure
Conical Sections, VIII-2
..
.
..
.
.
.
.
.
.
.
.
"
Chapter 4
Flat Plates, Covers, and Flanges
4.1
Introduction
4.2
Integral Flat Plates and Covers
4.2.1 Circular Flat Plates and Covers
4.2.2 Noncircular Flat Plates and Covers
4.3
Bolted Flat Plates, Covers, and Flanges
4.3.1 Gasket Requirements, Bolt Sizing, and Bolt Loadings
4.4
Flat Plates and Covers With Bolting
4.4.1 Blind Flanges & Circular Flat Plates and Covers
4.4.2 Noncircular Flat Plates and Covers
4.5
Openings in Flat Plates and Covers
4.5.1 Opening Diameter Does Not Exceed Half the Plate Diameter
4.5.2 Opening Diameter Exceeds Half the Plate Diameter
4.6
Bolted Flange Connections With Ring Type Gaskets
4.6.1 Standard Flanges
4.6.2 Special Flanges
4.7
Spherically Dished Covers
4.7.1 Definitions and Terminology
4.7.2 Types of Dished Covers
.
..
.
..
..
..
..
..
.
.
.
..
..
..
.
.
..
..
..
Chapter 5
Openings
5.1
Introduction
5.2
Code Bases for Acceptability of Opening
5.3
Terms and Definitions
5.4
Reinforced Openings-General Requirements
5.4.1 Replacement Area
5.4.2 Reinforcement Limits
5.5
Reinforced Opening Rules, VIII~l
5.5.1 Openings With Inherent Compensation
5.5.2 Shape and Size of Openings
5.5.3 Area of Reinforcement Required
5.5.4 Limits of Reinforcement
5.5.5 Area of Reinforcement Available
5.5.6 Openings Exceeding Size Limits of Section 5.5.2.2
5.6
Reinforced Opening Rules, VIII-2
5.6.1 Definitions
5.6.2 Openings Not Requiring Reinforcement Calculations
5.6.3 Shape and Size of Openings
5.6.4 Area of Reinforcement Required
5.6.5 Limits of Reinforcement
5.6.6 Available Reinforcement
.
.
..
.
..
.
..
.
..
.
.
..
.
..
.
.
..
..
..
.
..
,
"
65
65
67
68
68
71
72
74
74
85
95
101
101
101
101
104
105
105
106
106
107
107
107
108
108
109
118
124
125
125
133
133
133
134
134
134
134
136
136
137
137
140
140
lSI
153
153
153
ISS
ISS
155
157
Contents
5.7
5,6.7 Strength of Reinforcement Metal .." ,.,."."
5.6.8 Alternative Rules for Nozzle Design
Ligament Efficiency Rules. VIII-l
",,'
, ,
,,' ,.,., ",.""
".."."""""
"..".",
Chapter 6
Special Components, Vlff-1
6.1
6.2
6.3
6.4
6.5
Introduction
"
,.,.,
,
,
,'
",
Braced and Stayed Construction
6.2.1 Braced and Stayed Surfaces
,.,
6.2.2 Stays and Staybo1ts
Jacketed Vessels
6.3.1 Types of Jacketed Vessels
,
,
63.2 Design of Closure Member-for.Jacket to Vessel ";
,
6.3.3 Design of Openings in Jacketed Vessels
"
"."
Half-Pipe Jackets
6.4.1 Maximum Allowable Internal Pressure in Half-Pipe Jacket
6.4.2 Minimum Thickness of Half-Pipe Jacket
Vessels of Noncircular Cross Section
6.5.1 Types of Vessels
"
6.5.2 Basis for Allowable Stresses
6.5.3 Openings in Vessels of Noncircular Cross Section
6.5.4 Vessels of Rectangular Cross Section
,
,
.
.
.
..
.
,
,
,, ""'"
.
.
,..,
..
,
..
..
,
..
..
"
"
Chapter 7
Design of Heat Exchangers
7.1
7.2
7.3
7.4
Introduction
Tubesheet Design in Ll-Tube Exchangers
7.2.1 Nomenclature
"
7.2.2 Design Equations for Simply Supported Tubesheets
7.2.3 Design Equations for Integral Construction
7.2.4 Design Equations for Integral Construction With Tubesheet Extended as a Flange
Fixed Tubesheets
,
7.3.1 Nomenclature
,
7.3.2 Design Equations
Expansion Joints
.
Chapter 8
Analysis of Components in
8.1
8.2
8.3
8,4
8.5
VIII~2
Introduction
Stress Categories
Stress Concentration
Combinations of Stresses
Fatigue Evaluation
References
,
Appendices
Appendix A·-Guide to VIII-1 Requirements ,...
Appendix B-Material Designation
Appendix C-Joint Efficiency Factors
Appendix D-F1ange Calculation Sbeets
Appendix Be-Conversion Factors
Index
,
,
,
,
,
,
..
xi
157
157
164
169
169
169
169
172
173
174
175
179
181
181
182
186
187
187
187
197
.
201
.
..
..
..
.
..
.
.
.
..
201
.
.
.
.
.
..
233
233
233
239
240
245
201
201
207
209
212
213
213
217
230
249
.
,..................................................................
..
25]
253
255
277
283
285
LIST OF FIGURES
Figure
Number
i.i
1.2
Welded Joint Categories (ASME VIII-I)
Category C Weld
.
.
Some Governing Thickness Details Used for Toughness (ASME VIII-I)
.
ELl
1.3
E1.2
1.4
1.5
1.6
1.7
2.l
2.2
2.3
2.4
2.5
E2.8
2.6
2.7
E2.l3
2.8
2.9
2.10
3.1
E3.4
3.2
3.3
3.4
3.5
3.6
3.7
E3.11
E3.12
E3.l3
3.8
3.9
3.10
3.11
3.12
4.1
4.2
Impact-Test Exemption Curves (ASME VIII-I)
.
Charpy Impact-Test Requirements for Full Size Specimens for Carbon and Low Alloy Steels
With Tensile Strength of Less Than 95 ksi (ASME VllI-I)
..
Reduction of MDMT Without Impact Testing (ASME VIII-I)
.
Fatigue Curves for Carbon, Low Alloy, Series 4XX, High Alloy Steels, and High Tensile Steels
for Temperatures Not Exceeding 700'F (ASME VIII-2)
.
Comparison of Equations for Hoop Stress in Cylindrical Shells
.
5
7
8
11
14
15
16
17
20
28
29
30
Chart for Carbon and Low Alloy Steels With Yield Stress of 30 ksi and Over, and Types 405 &
410 Stainless Steels
.
C Factor as a Function of R!T (Jawad, 1994)
.
39
Geometric Chart for Cylindrical Vessels Under External Pressure (Jawad and Farr, 1989)
Some Lines of Support of Cylindrical Shells Under External Pressure (ASME VIII-I)
..
.
40
43
45
Some Details for Attaching Stiffener Rings (AS"ME VIII-I)
Mitered Bend
,
Elliptical Cylinder
..
.
.
38
49
,
"
,
51
55
56
59
62
66
69
70
73
75
78
79
83
92
95
Inherent Reinforcement for Large End of Cone-to-Cylinder Junction (ASME VIII-2)
Values of Q for Large End of Cone-to-Cylinder Junction (ASME VllJ-2)
Inherent Reinforcement for Small End of Cone-to-Cylinder Junction (ASME VIII~2)
Values of Q for Small End of Cone-to-Cylinder Junction (ASME Vill-2)
Some Acceptable Types of Unstayed Flat Heads and Covers
..
Multiple Openings in the Rim of a Flat Head or Cover With a Large Central Opening
xiii
..
.
.
.
96
97
98
99
.
103
109
xiv
List of Figures
E4.5
E4.6
E4.7
4.3
E4.8
5.1
5.2
5.3
E5.1
E5.2
E5.3.1
E5.3.2
E5.4
5.4.1
5.4.2
Ring Flange Sample Calculation Sheet
Welding Neck Flange Sample Calculation Sheet
Reverse Welding Neck Flange Sample Calculation Sheet
Spherically Dished Covers With Bolting Flanges (ASME VlII-l)
Example Problem of Spherically Dished Cover, Div. 1
Reinforcement Limits Parallel to Shell Surface
Chart for Determining Value of F for Angle
Determination of Special Limits for Setting t, for Use in Reinforcement Calculations
Example Problem of Nozzle Reinforcement in Ellipsoidal Head, Div. 1
Example Problem of Nozzle Reinforcement of 12 in. X 16 in. Manway Opening, Div. 1
Example Problem of Nozzle Reinforcement of Hillside Nozzle, Div. 1
Example Problem of Nozzle Reinforcement of Hillside Nozzle, Div. 1
Example Problem ofNozzle Reinforcement With Corrosion 'Allowance.Triv. ·1··
5.5
Nozzle Nomenclature and Dimensions (Depicts General Configurations Only)
Limits of Reinforcing Zone for Alternative Nozzle Design
Example Problem of Nozzle Reinforcement in Ellipsoidal Head, Div. 2
Example Problem of Nozzle Reinforcement of 12 in. x 16 in. Manway Opening, Div. 2
Example Problem of Nozzle Reinforcement of Series of Openings, Div. 1
Typical Forms of Welded Staybolts
Typical Welded Stay for Jacketed Vessel
Some Acceptable Types of Jacketed Vessels
Some Acceptable Types of Closure Details
Some Acceptable Types of Penetration Details
Spiral Jackets, Half-Pipe and Other Shapes
Factor K for NPS 2 Pipe Jacket
Factor K for NPS 3 Pipe Jacket
Factor K for NPS 4 Pipe Jacket
Vessels of Rectangular Cross Section
Vessels of Rectangular Cross Section With Stay Plates
Vessels of Obround Cross Section With and Without Stay Plates and Vessels of Circular
Section With a Stay Plate
,
Plate With Constant-Diameter Openings of Same or Different Diameters
Plate With Multidiameter Openings
Example Problem of Noncircular Vessel, Div. 1
Various Heat-Exchanger Configurations (TEMA, 1999)
Some Typical Tubesheet Details for If-Tubes (ASME, 2001)
.
Tubesheet Geometry
.
Effective Poisson's Ratio and Modulus of Elasticity (ASME, 2001)
..
Chart for Determining A (ASME. 2001)
Fixity Factor. F (ASME, 2001)
Some Typical Details for Fixed Tubesheet Heat Exchangers (ASME, 1995)
Z" Z,. and Z.. versus X, (ASME, 2001)
Values of Q3 Between 0.0 and 0.8
Values of Q3 Between -0.8 and 0.0
Bellows-Type Expansion Joints
Flanged and FIued Expansion Joints
5.6
E5.5
E5.6
E5.7
6.1
6.2
6.3
6.4
6.5
66
6.7
6.8
6.9
6.10
6.11
6.12
6.13
6.14
E6.8
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
7.10
7.11
7.12
E8.1
8.1
E8.4
8.2
8.3
A.1
.
.
.
.
.
.
e
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llO
ll5
119
125
128
135
138
139
141
143
146
147
149
152
152
154
158
159
162
165
171
171
174
176
180
182
183
184
185
188
190
Cross
.
.
.
..
.
191
192
193
198
202
203
205
206
.
208
209
214
221
.
.
.
222
..
.
, .
.
223
231
231
238
Linearizing Stress Distribution
Model of a Finite Element Layout in a Flat Head-to-Shell Junction
Fatigue Curves for Carbon, Low Alloy, 4XX High Alloy, and High Strength Steels for
Temperatures Not Exceeding 700°F (ASME VllI-2)
Cyclic Curves
240
243
.
..
.
..
246
247
252
List of Figures
Cl
C2
C.3
CA
C5
C6
C7
c.S
C.9
C.lO
C.ll
C.12
C.B
C.14
C.15
C.16
C.l7
C.18
C.19
C.20.E
D.1
D.2
D.3
DA
D.5
D.6
Fig.
Fig.
Fig.
Fig.
Fig.
Fig.
D.l-Ring Flange With Ring-Type Gasket
D.2-Slip-On or Lap-Joint Flange With Ring-Type Gasket
D.3-Welding Neck Flange With Ring-Type Gasket
DA-Reverse Welding Neck Flange With Ring-Type Gasket
D.5-Slip-On Flange With Full-Face Gasket
D.6-Welding Neck Flange With Full-Face Gasket
_
.
..
.
.
.
.
xv
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
277
278
279
280
281
282
LIST OF TABLES
Table
Number
i.i
1.2
1.3
104
1.5
ELl
1.6
1.7
2.1
3.1
3.2
3.3
304
E3.14
6.1
6.2
6.3
8.1
8.2
8.3
8.4
8.5
8.6
E8A
B.1
B.2
B.3
BA
B.5
Criteria for Establishing Allowable Stress Values for VIII-l (ASME II-D)
Criteria for Establishing Design Stress Intensity Values for VITI-2 (ASME II-D)
Stress Values for SA-5IS and SA-516 Materials
.
.
"
.
Allowable Stress Values for Welded Connections
,
".,
Maximum Allowable Efficiencies for Arc- and Gas-Welded Joints
,
,,"
,
.
Stress Categories ,
,
,..,..,
.
Assignment of Materials to Curves (ASME VIII-I)
Minimum Design Metal Temperatures in High Alloy Steels Without Impact Testing
Tabular Values for Fig. 2.4
.
.
Factor K o for an Ellipsoidal Head With Pressure on the Convex Side ....
Values of.6. for Junctions at the Large Cylinder Due to Internal Pressure
Values of n. for Junctions at the Small Cylinder Due to Internal Pressure
.
Values of 11 for Junctions at the Large Cylinder Due to External Pressure
Allowable Stress and Pressure Data
, .
Example of Pressure Used for Design of Components
..
Closure Detail Requirements for Various Types of Jacket Closures
Penetration Detail Requirements
..
,'..,
Primary Stress Category
,
.
.
Structural Discontinuity,
,..........
.. ,
.
Thennal Stress
" .....
Stress Categories and Their Limits (ASME VII1~2)
.
Classification of Stresses (ASME VIII-2)
Some Stress Concentration Factors Used in Fatigue
Summary of Finite Element Output , .
Carbon Steel Plate
"
.
Chrome-Moly Steel Plate Specifications, SA-387 ....
Chrome-Moly Steel Forging Specifications, SA-182
Chrome-Moly Steel Forging Specifications, SA-336 .." .." ........
Quench & Tempered Carbon and Alloy Steel Forgings, SA-50S
xvii
,
.
"
.
3
3
4
5
6
9
10
18
37
67
76
77
85
96
173
179
181
234
234
235
236
237
239
244
253
253
253
254
254
CHAPTER
1
BACKGROUND.. INFO.RMATION
1.1
INTRODUCTION
In this chapter some general concepts and criteria pertaining to Section VIII are discussed. These include
allowable stress, factors of safety, joint efficiency factors, brittle fracture, fatigue, and pressure testing.
Detailed design and analysis rules for individual components are discussed in subsequent chapters.
Since frequent reference will be made to ASME Section VIII Divisions land 2, tbe following designation
will be used from here on to facilitate such references. ASME Section VIII, Division 1 Code will be
designated by VIII-I. Similarly, VIlI-2 will designate the ASME Section VIII, Division 2 Code. Other
ASME code sections such as Section II Part D will be referred to as II-D. Eqnations and paragraphs
referenced in each of these divisions will be called out as they appear in their respective Code Divisions.
Many design rules in VIIl-l and VIII-2 are identical. These include flange design and external pressure
requirements. In such cases, the rules of VIII-I will be discussed with a statement indicating that the rules
of VIII-2 are the same. Appendix A at the end of this book lists the paragraph numbers in VlIl-1 that
pertain to various components of pressure vessels.
Section VIll requires the fabricator of the equipment to be responsible for its design. Paragraphs UG22 in VIII-l and AD-110 in VIIl-2 are given to assist the designer in considering the most commonly
encountered loads. They include pressure, wind forces, equipment loads, and thermal considerations. When
the designer takes exceptions to these loads either hecause they are not applicahle or they are unknown, then
such exceptions must be stated in the calculations. Similarly, any additional loading conditions considered by
the designer that are not mentioned in the Code must be documented in the design calculations. Paragraphs
U-2(a) and U-2(b) of VIII-l give guidance for some design requirements. VIII-2, paragraph AD-I 10 and
the User's Design Specifications mentioned in AG-30l provide the loading conditions to be used by the
manufacturer.
Many design rules in VIII-I and VIII-2 are included in the Appendices of these codes. These rnles are
for specific products or configurations. Rnles that have been substantiated by experience and used by
industry over a long period of time are in the Mandatory Appendices. New rules or rules that have
limited applications are placed in the Non-Mandatory Appendices. Non-Mandatory rules may eventually
he transferred to the Mandatory section of the Code after a period of nse and verification of their safety
and practicality. However, guidance-type appendices will remain in the Non-Mandatory section of the Code.
The rules in VIlI-I do not cover all applications and configurations. When rules are not available,
Paragraphs U-2(d), U-2(g), and UG-IOl must he used. Paragraph U-2(g) permits the engineer to design
components in the absence of rules in VIll-i. Paragraph UG-lOI is for allowing proof testing to estahlish
maximum allowable working pressure for components. In VIII-2 there are no rules similar to those in
UG-lOI, since VIII-2 permits design by analysis as part of its requirement'. This is detailed in Paragraphs
AD-JOO(b), AD-J40, AD-ISO, aud AD-l60 of VIlI-2.
1
2 Chapter 1
1.2 ALLOWABLE STRESSES
The criteria for establishing allnwahle stress in VIII-I are detailed in Appendix P of VllI-I and Appendix I
ofIJ-D and are summarized in Table 1.1. The allowahle stress at design temperature for most materials is
the lessor of 1/3.5 the minimum effective tensile strength or 2/3 the minimum yield stress of the material
for temperatures helow the creep and rupture values. The controlling allowahle stress for most bolts is
1/5 the tensile strength. The minimum effective tensile stress at elevated temperatures is ohtained from the
actual tensile stress curve with some adjustments. The tensile stress value obtained from the actual curve
at a given temperature is multiplied by the lessor of 1.0 or the ratio of the minimum tensile stress at room
temperature obtained from ASTM Specification for the given material to the actual tensile stress at room
temperature obtained from the tensile strength curve. This quantity is then multiplied hy the factor 1.1. The
effective tensilestress is then equal-to the lessor ·of this-quantity-or-the-minimum tensile stress· at . room
temperature given in ASTM. This procedure is illustrated in example 4.1 of Jawad and Farr (reference 14,
found at hack of hook).
The 1.1 factor discussed above is a constant estahlished hy the ASME Code Committee. It is hased on
engineering judgment that takes into consideration many factors. Some of these include increase in tensile
strength for most carbon and low alloy steels between room and elevated temperature; the desire to maintain
a constant allowable stress level hetween room temperature and 500"F or higher for carhon steels; and the
adjustment of minimum strength data to average data. Above approximately 5000F or higher the allowahle
stress for carbon steels is controlled hy creep-rupture rather than tensile-yield criteria. Some materials may
not exhibit such an increase in tensile stress, hnt the criterion for 1.1 is still applicahle to practically all
materials in VIlJ-1.
Table 1.I also gives additional criteria for creep and rupture at elevated temperatures. The criteria are
hased on creep at a specified strain and rupture at 100,000 hours. The 100,000 hours criterion for rupture
corresponds to ahout eleven years of continual use. However, VIII-I does not limit the operating life of
the equipment to any specific numher of hours.
The allowahle stress criteria in VJJl-2 are given in JI-D of the ASME Code. The allowahle stress at the
design temperature for most materials is the smaller of 113 the tensile strength or 2/3 the yield stress. The
design temperature for all materials in VJJJ-2 is kept helow the creep and rupture values. Tahle 1.2
summarizes the allowable stress criteria in VIII-2.
A sample of the allowahle stress Tables listed in Section ll-D of the ASME Code is shown in Tahle 1.3.
It lists the chemical composition of the material, its product form, specification numher, grade, Unified
Numhering System (UNS), size, and temper. This information, with very few exceptions, is identical to
that given in ASTM for the material. The Tahle also lists the P and Group numbers of the material. The
P numbers are used to cross reference the material to corresponding welding processes and procedures
listed in Section IX, "Welding and Brazing Qualifications," of the ASME Code. The Tahle also lists the
minimum yield and tensile strengths of the material at room temperature, maximum applicable temperature
limit, External Pressure Chart reference, any applicable notes, and the stress values at various temperatures.
The designer may interpolate hetween listed stress values, hut is not permitted to extrapolate heyond the
puhlished values.
Stress values for components in shear and bearing are given in various parts of VllJ-l, VIll-2, as well
as II-D. Paragraph UW-15 of VIII-I and AD-132 of VIJI-2 lists the majority of these values. A summary
of the allowable stress values for connections is shown in Table 1.4.
Some material designations in ASTM as well as the ASME Code have been changed in the last 20 years.
The change is necessitated by the introduction of subclasses of the same material or improved properties.
Appendix B shows a cross reference between older and newer designations of some common materials.
The maximum design temperatures allowed in VIII cannot exceed those puhlished in Section ll-D. VIII-]
defines design temperature as the mean temperature through the cross section of a component. VIII-2 defines
design temperature as the mean temperature in the cross section of a component, but the surface temperature
cannot exceed the highest temperature listed in JI-D for the material. This difference in the definition of
temperature in VIJI-I and VIIl-2 can he substantial in thick cross sections suhjected to elevated temperatures.
Background Information
3
TABLE 1.1
CRITERIA FOR ESTABLISHING ALLOWABLE STRESS VALUES FOR VIII-1 (ASME II-D)
Below Room
Temperature
ProduetlMaterial
Tenshe
Slrell9th
Wrought or cast
ferrO\l5 and
nonferrous
~
Welded p"pe or
tube, ferrous and
nonferrous
~Sr
a.s
Room Temperatureand Above
Yield
SlrenSth
'1, s-
Tensile
Strength
~
VI >< 0.85 s,
~S
3.5
% s,
!2 SrRr
'-'
a
"
Yield
Strength
r
'I, S"R y
F."SR",
'f,
(F.... "
Or O.qSyR y
[Note nll
~SrRr
';; ><
O,85Sy
'5
i em,
Rate
Stress
Rupture
>< O.85S yRy
cr 0.9 x O.85S yRy
O.85lSR",
I
().85 R",,"
1.05,
{O.8 " O.851S R",",
0.85S e
[Note u»
NOTE:
(i) . Twcisets of allowable stress values-may be provided.tn .Table·lA .for austenntc.rnatenals and in .Ta,bl e.1B.f{}rspec:ific nonferrous alloys~
The lower values are not specifically identified by a footnote, These lower values do not exceed two-thirds of the minimum yield strength
at temperature. The higher alternative allowable stresses are ldentifled by a footnote. These higher stresses may exceed two-thirds but do
not exceed 90% of the minimum yield strength at temperature. The higher values should be used only where slightlyhigher deformation is
not in itself objectionable. These higher stresses are not recommended for the design of flanges or for other strain sensitive applications.
Nomenclature
RT = ratio of the average temperature dependent trend curve value of tensile strength to the room temperature tensile strength
Ry = ratio of the average temperature dependent trend curve value of yield strength to the room temperature yield strength
SRaVQ = average stress to cause rupture at the end of 100,000 hr
SRmin = minimum stress to cause rupture at the end of 100,000 hr
Be = average stress to produce a creep rate of 0.01%/1000 hr
ST = specified minimum tensile strength at room temperature, ksi
Sy = specified minimum yield strength at room temperature
TABLE 1.2
CRITERIA FOR ESTABLISHING DESIGN STRESS INTENSITY VALUES FOR VIII-2
(ASME II-D)
Product/Material
Wrought or cast, ferrous and nonferrous
Tensile Strength
1/3
s,
Yield Strength
% SyRyor
2/~ Sy
.!.:.! s-e,
O.9S yRy[Note
3
Welded pipe or tube, ferrous and nonferrous
0.85
3
s-
{I,I x 0.85j S R
3
T T
0.85
1.5
s,
u»
0.85 SyRyor
1.5
(0.9 x 0.85)S yRy
[Note (Ill
NOTE:
(1)
Two sets of allowable stress values may be provided In Table IA for austenitic materials and in Table 18 for specific nonferrous alloys.
The lower values are not specifically identified by a footnote. These lower values do not exceed two-thirds of the minimum yield strength
at temperature. The higher alternative allowable stresses are identified by a footnote. These higher stresses may exceed two-thirds but do
not exceed 90% of the minimumyield strength at temperature. The higher values should be used only where slightly higher deformation is
not in itself objectionable. These higher stresses are not recommended for the design of flanges or for other strain sensitive applications.
1.3 JOINT E}'FICIENCY FACTORS
In the ASME Boiler Code, Section I, as well as in VIII-2, all major longitudinal and circumferential butt
joints must be examined by full radiograpby, with few exceptions. VIII-I, on the other hand, permits various
levels of examination of these major joints, The examination varies from full radiographic to visual,
depending on various factors specified in VIII-l and by the user. The degree of examination influences the
required thickness through the use of Joint. Efficiency Factors, E. The Joint Efficiency Factors, which are
sometimes referred to as Quality Factors or weld efficiencies, serve as stress multipliers applied to vessel
components when some of the joints are not fully radiographed. These multipliers result in an increase in
the factor of safety as well as the thickness of these components. In essence, VIII-I vessels have variable
4 Chapter 1
TABLE 1.3
STRESS VALUES FOR SA-515 AND SA-516 MATERIALS
Line
Nominal
Product
Spec
Alloy
Classl
DesigJ
Cond.!
Size/
Temper
Thick, in.
Group
P·No
No.
No.
Composition
Form
28
CS
Plate
SA·515
70
K03101
2
29
CS
Plate
SA·516
70
K02700
2
No.
Type/Grade
UNS No.
External
Applic. & max. Temp. Limits
Min.
Min.
Tensile
Yield
(NP
Line
Strength
Stress
(SPT ~ Supports only)
No.
ksi
ksl
III
VIII-1
No.
Notes
28
70
38
1000
700
1000
CS·2
G10,S1,T2
29
70
38
850
700
1000
CS·2
G10,S1,T2
~
Pressure
Not Permitted)
Chart
Maximum Allowable Stress, ksi, for Metal Temperature, "F, Not Exceeding
Line
No.
-20 to 100
150
200
300
400
500
600
650
700
750
800
850
900
28
20.0
20.0
20.0
20.0
20.0
20.0
19.4
18.8
18.1
29
20.0
20.0
20.0
20.0
20.0
20.0
19.4
18.8
18.1
14.8
12.0
9.3
6.7
14.8
12.0
9.3
6.7
Note:
G10, 81, T2 are described in ll-D and pertain to metallurgical information
factors of safety, depending on the degree of radiographic examination of the main vessel joints. As an
example, fully radiographed longitndinal butt-welded joints in cylindrical shells have a Joint Efficiency
Factor, E, of 1.0. This factor corresponds to a safety factor of 3.5 in tbe parent material. Nonradiographed
longitudinal butt-welded joints have an E value of 0.70. This reduction in Joint Efficiency Factor corresponds
to a factor of safety of 5.0 in the plates. This higher factor of safety dne to a nonradiographed joint results
in a 43% increase in the required thiekness over that of a fully radiographed joint.
ASME VIII-I identifies four joint categories that require E factors. They are Categories A, B, C, and D
as shown in Fig. 1.1. Category A joints consist mainly of longitudinal joints as well as circumferential
joints between hentispherical heads and shells. Category B joints are the circumferential joints between
various components as shown in Fig. 1.1, with the exception of circumferential joints between hemispherical
heads and shells. The attachment of flanges to shells or heads is a Category C joint. The attachment of
nozzle necks to heads, shells, and transition sections is categorized as a Category D joint.
The four joint categories in VIII-l do not apply to items such as jacket closure bars, tubesheet attachments,
and ring girders. The degree of examination of the welds attaching these components to the shell or head
is not covered in VIII-I. Most designers assign an E value of 1.0 when calculating the shell or head thickness
at such junctions. This is justified since in most cases the strain in the hoop direction, and hence hoop
stress, is close to zero at the junction due to the restraint of tubesheet or bars.
Background Information
5
TABLE 1.4
ALLOWABLE STRESS VALUES FOR WELDED CONNECTIONS
VIIH
Component
Fillet weld
Fillet weld
Groove weld
Groove weld
Nozzle neck
Dowel bolts
Any location
Type of Stress
Stress Value
tension
shear
tension
shear
shear
shear
bearing
0.55S·
0.49S
0.74S
0.60S
0.70S
0.80S
1.60S
Reference
UW-18(d)
UW-15(c)
UW-15(c)
UW-15(c)
UG-45(c)
II-D
II-D
*$ .~.. allowable stress tor vltt-t construction
VIII-2
Component
Fillet weld
Fillet weld
Groove weld
Groove weld
Nozzle neck
Any location
Type of Stress
tension
shear
tension
shear
shear
bearing
Stress Value
O.58 m*
o.ss,
0.75Sm
O.758 m
o.es,
S,
Reference
AD-920
AD-920
AD-920
AD-920
AD-132.2
AD-132.1
*5 m = stress intensity values for VIIl-2 construction
FIG_ 1.1
WELDED JOINT CATEGORIES (ASME VIII-1)
The type of construction and joint efficiency associated with each of joints A, B, C, and D is given in
Table 1.5. The categories refer to a location within a vessel rather than detail of construction. Thns, a
Category C weld, which identifies the attachment of a flange to a shell, can be either fillet, comer, or butt
welded, as illustrated in Fig. 1.2. The Joint Efficiency Factors apply only to tbe butt-welded joint in sketch
(c). The factors do not apply to sketches (a) and (h) since tbey are not butt welded.
The Joint Efficiency Factors used to design a given component are dependent on the type of examination
performed at the welds of tbe component. As an example, tbe Joint Efficiency Factor in a fully radiographed
longitudinal seam of a shell course is E = 1.0. However, this number may have to be reduced, depending
on the degree of examination of the circumferential welds at either end of the longitudinal seam. Appendix B
shows some typical components and their corresponding Joint Efficiency Factors.
6 Chapter 1
TABLE 1.5
MAXIMUM ALLOWABLE JOINT EFFICIENCIES'" FOR ARC- AND GAS-WELDED JOINTS
Degree of
Radiographic
Examination
b - -ca
Joint
Joint
Type
-Description
Limitations
Category
Full' Spot' None
No.
(1)
Butt joints as attained by
double-welding or by other
means which will obtain the
same quality of deposited
weld metal on the inside and
None
A,B,C&D
1.0
0.85
0.70
(a) None except as shown in
(b) below
A,B,C&D
0.90
0.80
0.65
(b) Circumferential butt joints
with one plate offset, see
UW-13(c) and Fig. UW13.1(k).
A, B &C
0.90
0.80
0.65
A, B& C
NA
NA
0.60
A
NA
NA
0.55
NA
NA
0.55
B
NA
NA
0.50
C
NA
NA
0.50
(a) For the attachment of
heads convex to pressure to
shells not over 5/8 in.
required thickness. Only with
use of fillet weld on inside of
sneus, or
A&B
NA
NA
0.45
(b) For attachment of heads
having pressure on either
side. To shells not over 24
in. inside diameter and not
over 1/4 in. required
thickness with fillet weld on
outside of head flange only.
A&B
NA
NA
0.45
outside weldaurtaces tc
agree with the requirements
of UW-35. Welds uSing metal
backing strips which remain
in place are excluded.
(2)
Single-welded butt joint with
backing strip other than
those included under (1)
(3)
Single-welded butt joint without
use of backing strip
Circumferential butt joints only,
not over 5/8 in. thick and not
over 24 in. outside diameter
(4)
Double full fillet lap joint
Longitudinal joints not over
3/8 in. thick
Circumferential joints not over
5/8 in. thick
(5)
Single full fillet lap joints with
plug welds conforming to
UW-17
(a) Circumferential joints- for
attachment of heads not over
24 in. outside diameter to
shells not over 1/2 in. thick
(b) Circumferential joints for the
attachment to shells of
jackets not over 5/8 in. in
nominal thickness where the
distance from the center of
the plug weld to the edge of
the plate is not less than
1-1/2 times the diameter of
the hole for the plug.
(6)
Single full fillet lap joints
without plug welds
Notes:
(1) The single factor shown for each combination of joint category and degree of radiographic examination replaces both the stress
reduction factor and the joint efficiency factor considerations previously used in this Division.
(2) See UW-12(a) and UW-51.
(3) See UW-12(b) and UW-52.
(4) Joints attaching hemispherical heads to shells are excluded.
(5) E = 1.0 for butt joints in compression.
(6) For Type NO.4 Category C joint, limitation not applicable for bolted flange connections.
Background Information 7
.AI
,
I
,I
I
,
I
I
I
I
I
""
I
I
,
(b)
,I
I
.
I
I
...
I
FIG. 1.2
CATEGORY C WELD
8 Chapter 1
Example 1.1
Problem
Determine the category and Joint Efficiency Factor of the joints in the heat exchanger shown in Fig. E1.l.
The channel side is spot radiographed. The longitudinal seam, b, is a single-welded butt joint with a backup
COVER
I
0
CHAN~EL
PASS
PARTITION
d
At
b
u
\
f
9
C
h
r
SHELL
J
n .....
=6
..
TUBE
~
~
1
"il
u
JAC KET
~ ~
F
m
I
I
.......-.:::f"HEAD
.....
0
SKIRT
FIG. E1.1
Background Information
9
bar. The shell side is not radiographed. The longitndinal and circumferential seams m and I are singlewelded butt joints witb backup bars. The jacket longitudinal seam, n, is a single-welded butt joint witbout
backup bar.
Solution
The joint categories of tbe various joiuts can be tabulated as given in Table ELI:
TABLE E1.1
STRESS CATEGORIES
Location
Joint
Category
Joint Efficiency
(a)
Ohannel-to-ftanqe connectlon
C
Does not apply
(b)
Longitudinal channel seam
A
0.80
(e)
Channel-to-tubesheet weld
C
Does notapply
(d)
Nozzle-to-channel weld
D
Does not apply
(e)
Flanqe-to-nczzle neck
C
Does not apply
(f)
Pass partitlon-to-tubesheet weld
None
Does not apply. See also
UW~15(c)
and
UW-18(d) of VIII-1
(9)
Tube-to-tubesheet weld
None
(h)
Shell-to-tubesheet weld
C
Does not apply
(i)
Jacket bar-to-inner-shell weld
None
Does not apply
OJ
Jacket to bar-to-outer-shell weld
None
Does not apply
(k)
Nozzle-to-jacket weld
D
Does not apply
(I)
Longitudinal shell seam
A
0.65
(m)
Head-to-shell seam
B
0.65
(n)
Longitudinal jacket seam
A
0.60
(0)
Sklrt-tc-head seam
None
Does not apply. See also UW-20 of vul-t
Does not apply
1.4 BRITTLE FRACTURE CONSIDERATIONS
Both VIII-I and V111-2 require tbe designer to consider brittle fracture rules as part of tbe material and
design selection. The rules for carbon steels are extensive and are discussed. first. VIII-l has two options
regarding toughness requirements for carbon steels. The first is given in Paragraph UG-20(f) and allows
tbe designer to exempt tbe material of construction from impact testing when all of the following criteria
are met:
1. The material is limited to P-No. I, Gr. No. I or 2.
2.
3.
4.
5.
6.
Maximum thickuess of 1/2 in. for materials listed in Curve A in Table 1.6.
Maximum thickness of I in. for materials listed in Curves B, C, and D of Table 1.6.
The completed vessel shall be hydrostatically tested per UG-99(b), (e), or 27-3.
Design temperature is between - 20°F and 650°F.
Thermal, mechanical shock, and cyclical loadings do not control the design.
The above requirements are intended for relatively thin carbon steel vessels operating in a service that
is neither severe in thermal and pressure cycling nor in extreme cold temperatures. Vessels of low alloy
10 Chapter 1
TABLE 1.6
ASSIGNMENT OF MATERIALS TO CURVES (ASME Vm·1)
GENERAL NOTES ON ASSIGNMENT OF MATERIALS TO CURVES:
(a) Curve A applies to:
(1)
(2:)
all carbon and all low alloy steel plates, structural shapes, and bars not listed In Curves S, C, and 0 below;
SA-lIb Grades WeB and wee If normalized and tempered or water-quenched and tempered, SA-217 Grade Web If
normalized and tempered or water-quenched ami tempered,
(b) Curve 8 applies to:
(ll SA-2lb Gradf! WCA if normalized and tempered or water-quenched and tempered
SA-21b Grades WeB and wee for thicknesses not elCceedlng 2: tn., if produced to fine grain practlce and water-quenched and
tempered
SA-217 Grade WC9 if normalized and tempered
SA·285 Grades A and B
SA-414 Grade A
SA-SIS Grade 60
SA-Sib Grades 65 and 70 If not normalized
SA.b iau not normalized
SA-662 Grade B If not normalized;
(2) except lor cast steels, all materials of Curve A if produced to fine grain practlce and normalized which are not listed in Curves
C and D below;
(3) all pipe, fittings, forgings and tubing not listed for Curves C and D beloW;
(4) parts permitted under us-n shall be included in Curve B even when fabricated from plate that otherwise would be assigned to
a dlllerent curve.
(c) Curve C
Ul SA-IB2 Grades 21 and 22 if normalized and tempered
SA-302 Grades C and D
SA-33u F21 and F22 If normallzed and tempered
SA-387 Grades 21 and 22 If normalized and tempered
SA-Slu Grades 55 and 60 II not ncrmauzeo
SA-533 Grades Band C
SA-bul Grade A;
(2) all material 01 Curve B If produced to fine grain practice and ncrmaltaed and not listed for Curve D below.
(dl Curve 0
SA-203
SA-SOB Grade 1
SA-51u if normalized
SA-524 Classes 1 and 2
SA-537 Classes 1, 2, and 3
SA·bI2 if ncrmallzed
SA...f:,b2 If normalized
SA-BB Grade A
SA-738 Grade A with Cb and V deliberately added in accordance with the provisions of the material specification, not cctoer
than _20°F (_29°C)
SA-BB Grade B not colder than _20°F (_29°C)
(e) For bolting and nuts, the 10llowlng Impact test exemption temperature shall apply:
Bolting
Impact Test
Exemption Temperature, OF
Spec. No.
Grade
SA-193
SA-193
B5
B7 (2Y~ in. dia. and under)
(Over 2Y2 in. to 7 In., Incl'>
SA-193
SA-193
SA-307
SA-320
SA-325
SA-354
SA-354
SA-449
SA-540
B7M
B16
-'0
·55
-40
·55
-'0
-'0
B
L7, L7A, 17M, L43
impact tested
"a
·'0
Be
o
BO
+20
·'0
623/24
+10
Nuts
Spec. No.
Grade
Impact Test
Exemption Temperature, of
SA-194
2, 2H, 2HM, 3, 4, 7, 7M,
and 16
-55
SA-540
823fB24
-55
(f) When no class or grade is shown, en classes or grades are included.
(g) The following shall apply to all material assignment notes.
(l) Cooling rates faster than those obtained by cooling ill air, followed by tempering, as permltted by the material specification, are
considered to be equivalent to normalizing or normalizing and tempering heat treatments.
(2} Fine grain practice is defined as the procedure necessary to obtain a fine austenitic grain size as described in SA·20.
NOTES:
Tabular values for this Figure are provided in Table UCS-66.
(2) Castings not listed in General Notes (a) and {b} above shall be impact tested.
(l)
Background Information
11
steel or those with carbon steel operating beyond the scope of Paragraph UG-20(f) require an evaluation
for brittle fracture in accordance with the rules of UCS-66. The procedure consists of
1. Determining the governing thickness in accordance with Fig. 1.3.
2. Using Fig. 1.4 to obtain the temperature that exempts the material from impact testing. If
the specified Minimum Design Metal Temperature, MDMT, is colder than that obtained
from the figure, then impact testing in accordance with Fig. 1.5 is required. The specified
MDMT is nsually given by the user, while the calculated MDMT is obtained from VIlI-1.
The calculated MDMT is kept equal to or colder than the specified MDMT.
2
Section x-x
tg1 "" tA
tff2 "" tA (seamless) or ta (welded)
--<1.(8)
Butt Welded Components
II
I
<t.
tc
!r1I
+1 I--tc
2
1
tB
110
1-1'" .
<1. I
I
tg 1 "" the thinner
of tA or tc
tg3 = the thinner
tff2 ""the thinner
of tBor tc
NOTE: Using tgl' t g 2. and tg 3' determine the warmest MDMT and use that
assembly.
oftAortB
8S
the permissible MDMT for the welded
(bl Welded Connection with Reinforcement Plato Added
FIG. 1.3
SOME GOVERNING THICKNESS DETAILS USED FOR TOUGHNESS (ASME VIII-1)
12 Chapter 1
tg 1 =
t:
Cl.
(For
'A
0
rt-.
tg1 "" 4" (For ~ welded
welded
or nonwelded)
or nonwelded)
tg2 =
tc
The governing thickness
of
0
is the greater
of f g 1 or tQ2
(c) Bolted ~Iat Head or Tubtnlheet and Flange
(d) Integral Flet Head or Tubesheet
Cl.
tg 1
'A f:\ .
= T {For 0./ welded
or nonweldedl
trn
'" thinner of
fA
or fa
The governing thickness
of
0
is the greater
of t0 1 or tg2
(8)
Flat Head or Tubesheet With 8 Corner Joint
FIG. 1.3
(CONT'D)
BackgroundInformation
13
Pressure part
tg1 = thinner of fA or fa
(ft Welded Attachments
8S
Defined in UCs.&6(a)
FIG. 1.3
(CONT'D)
3, The temperature obtained from Fig, 104 may be reduced in accordance with Fig, 1.6 if fhe
component operates at a reduced stress, This is detailed in Paragraph UCS-66(b) of VIlI- 1.
At a ratio of 0.35 in Fig, 1.6, the permitted temperature reduction drops abruptly, At this
ratio, the stress in a component is about 6000 psi. At this stress level, experience has shown
fhat brittle fracture does not occur regardless of temperature level.
4, The rules in VIlI-l also allow a 30'F reduction in temperatnrc below that obtained from
Fig, 104 when the component is post-weld heat tteated but is not otherwise required to be
post-weld heat treated by VIlI-l rules,
The toughness rules for ferritic steels with tensile properties enhanced by heat treatment are given in
Paragraph UHT-6 of VIlI-1. The rules require such steels to be impact tested regardless of temperature,
The measured lateral expansion as defined by ASTM E-23 shall be above 0,015 in,
The toughness rules for high alloy steels are given in Paragraph UHA-51 of VIlI-I. The permissible
Minimum Design Metal Temperature for base material is summarized in Table 1.7. Similar data are given
in VIII-I for fhe weld material and weld qualifications, Thermally heated stainless steels may require impact
testing per the requirements of UHA-51(c),
The rules for toughness in VIlI-2 are different fhan those in VIII-I. However, the concepts of exemption
curves and Charpy impact levels are similar in VIlI-2 and VIII-I. The toughness requirements for carbon
and low alloy steels are given in Paragraph AM-218 of VIII-2, High alloy steels are covered in Paragraph
AM-213.
14 Chapter 1
Example 1.2
Problem
Determine the Minimnm Design Metal Temperature, MDMT, for the reactor shown in Fig. E1.2. Let the
shell, head, pad, and ring material be SA-516 Or. 70 material. Flange and cover material is SA-105. Pipe
material is SA-106. The required shell thickness is 1.75 in., and the required head thickness is 0.86 in. The
required nozzle neck thickness is 0.08 in. Assume a joint efficiency of 1.0 and no corrosion allowance.
Solution
Shell
SA,5 I6 specifications require the material to be normalized when the thickness exceeds 1.5 in. Thus, from
Table 1.6, Curve D is to be used for normalized SA-516 Or. 70 material. Using Fig. 1.4 and a governing
thickness of 2.0 in., we get a minimum temperature of - 5°F. The ratio of required thickness to actual
thickness is 1.75/2.0 = 0.88. Using Fig. 1.6 for this ratio, we obtain 12°F. Hence, MDMT = - 5 - 12
= - J7 op .
Head
For a I in-thick head, SA-516 specifications permit a non-normalized material. Thus, from Table 1.6.
Curve B is used. Using Pig. 1.4 aud a governing thickness of 1.0 in., we get a minimum temperature of
300P. The ratio of reqnired thickness to actual thickness is 0.86/1.0 = 0.86. Using Pig. 1.6 for this ratio,
we obtain 14°F. Hence, MDMT = 30 - 14 ~ 16°P.
t=I.0'
3' X
'J,:;'
RING
4' RAD,
2~'
300" WN FLANGE
X 2' PAD
SA 105
-~"ci-'-
4' SCH. 40 PIPE
FIG. E1.2
--
~ ~ I t=2~'
Background Informarion
140
120
100
I
I
II
I
I
I
80
A~
I
40
20
o
- 20
- 40
- 55
- 60
-80
I
1/ / /
~
--"""'" J--
Y
/'
II
J
/
! II /
i~ /
V
V
---of
/'
/
8
.....-.>
/'
I
I
60
15
....
..........
.»
~
~
s,
.....V
~ """
....
--
~
V
1/
i
--, 1I
0.394
-- f-- -
- - ... .,.-- - - - - - - -
--
--
TP'" "Sj,n. requlT
2
3
4
--
6
Nominal Thickness, in.
(Limited to " in. for Welded Construction)
FIG. 1.4
IMPACT-TEST EXEMPTION CURVES (ASME VIII-1)
Stiffener
For a 0.75-in. stiffener, Curve B of Table 1.6 is to be used. Using Fig. 1.4 and a governing thickness of
0.75 in., we obtain a ntinimum temperature of 15°F. Since stresses cannot be established from VlIl-l rules,
the MDMT = IsoF.
Pad
The material will be normalized since it is 2.00 in. thick. Curve D of Table 1.6 is used. Prom Pig. 1.4 and
a governing thickness of 2.0 in., we obtain a minimum temperature of - 5°F. Since stresses cannot be
established from VIIl-l rules, the MDMT = - SOp.
16 Chapter 1
! I
0.3~4 in.
50
l
I
I
40
I
Minimum specified
ylelclm~ngth
I
I
"0
•E
•i);
I
'0
•~
-S
'0
30
•
~
E
~
I
./
£
20
./
!
./
I
15
I
10
!
i.>
55 ksi
' / ...V
I
I
I
~
"o:>
65 ksi
./
I
~
I
/
./
.
50 ksi
~~
~ ~.... ~
45 ksl
,38 ksi
I
I
I
I
I
I
o
1.0
2.0
;;., 3.0
Maximum Nominal Thickness of Material or Weld, in.
GENERAL NOTES:
(a) Interpolation between yield strengths shown is permitted.
(bl The minimum impact energy for one specimen shall not be less than 2}3 of the average energy required for
three specimens.
le) Materials produced and impact tested in accordance with SA-320, SA·333, SA-334, SA-350, SA-3S2, SA·42Q
and SA-765 do not have to satisfy these energy values. They are acceptable for use at minimum design metal
temperature not colder than the test temperature when the energy values required by the applicable
specification are satisfied.
Idl For materials haVing a specified minimum tensile strength of 95 ksi or more, see UG"84(cH4)(bl.
FIG. 1.5
CHARPY IMPACT-TEST REQUIREMENTS FOR FULL SIZE SPECIMENS FOR CARBON
AND LOW ALLOY STEELS WITH.TENSILE STRENGTH OF LESS THAN 95 ksi
(ASME VIII-1)
Background Information
1.00
.,ro
0
a:
~
>
.~
E
0.80
!
<i'
5
"\
-,
I'\.
'"
~
5
~
ro
"0
0.60
c
m
E
0
z
~
<, r-,
~
til
~ 0040
I
0.35
.s
•IJJ
(;
"i
0.20
a:
0.00
<,
r--
v/c //// t:-;//c 'l//: f"////: '///'///'/// '///'/// V// V/C V/C t:-;//'/// rr>>
~~
~~
~
~~ l% ~~~~~~~~
~ ~
~~ l%
~
.;.
17
'///////V///'///////X///:~
~/ See UCS-66Ibl(3) when ratios are 0.35 and smaller ~
////'////V/// ' / / / ' / / / ,///////X///,
%§ ~~~~
~
~~~~~
~~~
o
20
40
60
80
of [See UCS-66Ib)J
100
120
140
Nomenclature (Note references to General Notes of Fig. UCS~66"2.)
t r = required thickness of the component under consideration in the corroded
condition for all applicable loadings [General Note (2)J, based on the
applicable joint efficiency £ (General Note (3)], in.
t n = nominal thickness of the component under consideration before corrosion
allowance is deducted, in.
c = corrosion allowance, in.
E* = as defined in General Note (3).
Alternative Ratio = $* E* divided by the product of the maximum allowable stress value
from Table UCS~23 times E, where 5* is the applied general
primary membrane tensile stress and E and E* are as defined in General
Note (3).
FIG. 1.6
REDUCTION OF MDMT WITHOUT IMPACT TESTING (ASME VIII-1)
Nozzle Neck
From Table 1.6, Curve B is to be used for a nozzle neck of 0.258-in. thickness. From Fig. lA, minimum
temperature is - 20'F. The ratio of required thickness to actual fhickness is 0.08/0.258 X 0.875 = 0.36.
Using Fig. 1.6 and this ratio, we get 130'F. Hence, MDMT = - 20 - -130 = - 150°F.
18 Chapter 1
TABLE 1.7
MINIMUM DESIGN METAL TEMPERATURES IN HIGH ALLOY STEELS WITHOUT
IMPACT TESTING
Base Material
Austenitic chromiummanganese-nickel
stainless steel (200 series)
Is material
• Austenitic ferritic duplex steel
with t « 3/8 in.
• Ferrltic chromium stainless steel
with t < 1/8 in.
• Martensitic chromium stainless
steel with t « 1/4 in.
Flange
Since the flange is ANSI BI6.5, it is good to -20 oP.
Cover
Prom Pig. 1.3(c), the controlling cover thickness is 2.5/4 = 0.625 in. Curve B applies for this material,
and the MDMT = 5°P.
Therefore, the MDMT for this reactor is governed by the head with a value of 16°P. A colder value can
be obtained by impact testing the various components. Thus, assuming a specified MDMT of -15°P is
required, then the head, stiffener, pad, and cover need impact testing.
Background Information 19
1.5 FATIGUE REQUIREMENTS
Presently, VIII-l does not list any rules for fatigne evaluation of components. When fatigue evaluation of
a component is required in accordance with UG-22 or U-2(g) of VIII-I, the general practice is to use the
VIII-2 fatigue criteria as a guidance up to the temperature limits of VIII-2. At temperatures higher than
those given in VIII-2, the rules of II1-H are followed for general guidance. Other fatigue criteria, such as
those given in other international codes and ASME B31.3, may also he considered as long as the requirements
of U-2(g) of VIII-l are met.
VIII-2 contains detailed rules regarding fatigue. Paragraph AD-160 gives criteria regarding the need for
fatigue analysis. The first criterion is listed in Paragraph AD-160.l and is based on experience. Vessels
that have operated satisfactorily in a certain environment may be cited as the basis for constructing similar
vessels operating under similar conditions without the need for fatigue analysis.
The second criterion for vessel components is based on the rule that fatigue analysis is not requited if
all of Condition A or all of Condition B is satisfied, as noted below.
Condition A
Fatigue analysis is not required for materials witb a tensile strength of less than 80 ksi when the total
number of cycles in (a) through (d) below is less than 1000.
a. The design number of full range pressure cycles including startup and shutdown.
b. The number of pressure cycles in which the pressure fluctuation exceeds 20% of the
design pressure.
c. Number of changes in metal temperature between two adjacent points. These changes are
multiplied by a factor obtained from tbe following chart in order to transform them to
equivalent cycle number.
Metal Temperature Differential, OF
50
51
101
151
251
351
Factor
o
or less
to 100
to 150
to 250
to 350
to 450
1
2
4
8
12
Higher than 450
20
d. Number of temperature cycles in components that have two different materials where a
difference in the value (uJ - (2)b.T exceeds 0.00034. Where, U is the coefficient of thermal
expansion and AT is the difference in temperature.
Condition B
Fatigue analysis is not requited when the following items (a) through (f) are met:
20
Chapter 1
a. The number of full range pressure cycles, including startup and shutdowu, is less than the
uumber of cycles determined from the appropriate fatigue chart, Fig. 1.7, with an S, value
equal to 3 times the allowable design stress value, Sm.
b. The range of pressure fluctuation cycles during operation does not exceed P(1/3)(S,1Sm),
where P is the design pressure, S, is the stress obtained from the fatigue curve for the number
of significant pressure cycles, and Sm is the allowable stress. Significant pressure cycles are
defined as those that exceed the quantity P(l/3)(SI Sm)' S is defined as
S = S, taken at 10' cycles when the pressure cycles are 0; 10'.
S = S, taken at actual number of cycles when the pressure cycles are > 106
c. The temperature difference between adjacent points during startup and shutdowu does not
exceed S,I (2E<x), where S, is the value obtained from the applicable design fatigue curve
for the total specified uumber of startup and shutdown cycles.
d. The temperature difference between adjacent points during operation does not exceed
S,I (2E<x), where S, is the value obtaiued from the applicable design fatigue curve for the
total number of significant fluctuations. Siguificant fluctuations is defiued as those exceeding
the quantity S/(2E<x), where S is as defined in (b) above. Adjacent points are defiued in AD160.2, Condition A, Paragraph (c) of VIII-2.
e. Range of significant temperature fluctuation in components that have materials with different
coefficient of expansion or modulus of elasticity aud that do not exceed the quantity
S,I [2(E, <x, - E, <X,)], where <X is the coefficient of thermal expansion and E is the modulus
of elasticity. Significant temperature fluctuation is that which exceeds the value S I [2(E, <X,
- E 2 <X,)], where S is as defined in (b) above.
f. Range of mechanical loads does not result in stress intensities whose range exceeds the Sa
value obtained from the fatigue chart.
ror
.11
E",30 x 10'p$",
rL... (1)
NOTES!.
_
12l Interpolate for UTS 80-115 kll.
(3) Table 6--110.1 contains tabu1eled values and 8 formula for an accurate interpolation of
the" curves.
-
::
a
~
/
Fo' UTS .. 80 k$i
~--
...... 1'-.--
For UTS 1'5·130 k,i
-- '-:"" ---
r
I
I
""I
I
,
I "
"
,
I
1111111
Numb~, (If
,
I
I
--- -- --
I II "
I
I
I'";'ii"
eveh'l
. FIG. 1.7
FATIGUE CURVES FOR CARBON, LOW ALLOY, SERIES 4XX, HIGH ALLOY STEELS, AND
HIGH TENSILE STEELS FOR TEMPERATURES NOT EXCEEDING 700'F (ASME VIII-2)
Background Information
21
The third criterion for nozzles with nonintegral reinforcement is given in Paragraph AD-160.3 of VllI-2
and is very similar to Conditions A and B detailed above.
Example 1.3
Problem
A pressure vessel consisting of a shell and two hemispherical heads is constructed from SA 516-70 carbon
steel material. The self-reinforced nozzles in the vessel are made from type SA 240-304 stainless steel
material. The vessel is shut down six times a year for maintenance. At start-up, the full pressure of 300
psi and full temperature of 400"P are reached in two hours. The maximum /IT between any two points
during start-up is 250"P. At normal operation, the /IT is negligible. At shutdown, the maximum /IT is
100oP. Determine the maximum number of years that this vessel can be operated if a fatigue evaluation is
not performed. Let the coefficient of expansion for carbon steel be 6.5 X 10- 6 in.lin.l"P and that for
stainless steel be 9.5 X 10- 6 in.lin./"F.
Solution
Prom Condition A, determine the number of cycles in one year.
a. Number of full pressure cycles for one year is 6.
b. This condition does not apply for this case.
c. Prom the chart, the 250 0P difference in temperature during start-up corresponds to 4 cycles.
The 1000P difference in temperature during shutdown corresponds to I cycle. Thus total
equivalent cycles due to temperature in one year is (4 + I) 6 = 30 cycles.
d. At nozzle attachments, the quantity (9.5 X 10- 6 - 6.5 X 10- 6) 400 is equal to 0.0012.
Since this value is greater than 0.00034, the equivalent cycles per year = 6.
Total cycles per year due to (a), (e), and (tl) = 6 + 30 + 6 = 42.
Number of years to operate vessel if fatigue analysis is not performed = 1000/42
23.8 years.
Example 1.4
Problem
A pressure vessel has an inside diameter of 60 in., internal pressure of 300 psi, and design temperature of
500 oP. The shell thickness is 1/2 in. at an allowable stress level of 18,000 psi (material tensile stress =
70 ksi). The thickness of the hemispherical heads is 1/4 in. at an allowable stress level of 18,000 psi.
Integrally reinforced nozzles are welded to the sbell and are also constructed of carbon steel with an
allowable stress of 18,000 psi. At start-up, the full pressure of 300 psi and full temperature of 500 0P are
reached in eight hours. The maximum flT between any two points during start-up is 60 oP. At normal
operation, the /IT is negligible. At shutdown, the maximum /IT is 50 oP. Determine if the shell and heads
are adequate for 100,000 cycles without the need for fatigue analysis. Prom II-D, the coefficient of expansion
for carbon steel is 7.25 X 10- 6 in.lin./"P and the modulus of elasticity is 27.3 X 106 psi. Use Pig. 1.7
for a fatigue chart.
Solution
Condition B is to be used.
a. Three times allowable stress at the nozzle location is = 3(18,000)
1.7 for this value gives a fatigue life of 4200 cycles.
b. This condition does not apply.
=
54,800 psi. Using Pig.
22
Chapter 1
c. From Fig. 1.7, with 100,000 cycles, the value of S, = 20,000 psi. The value of S,/(2Ea)
= 20,000/(2 X 37.3 X 106 X 7.25 X 10- 6) = 51°F. Since this value is less than 60°F,
the specified cycles are inadequate. The designer has two options in this situation. The first
is to perform fatigue analysis, which is costly. The second option, if it is feasible, is to
reduce the AT at startup to 51°F.
d. This condition does not apply.
e. This condition does not apply.
f. This condition does not apply.
Example 1.5
Problem
In Example lA, determine the required thickness of the shell and heads for 1,000,000 cycles without the
need to perform fatigue analysis.
Solution
From Fig. 1.7, with a cycle life of 1,000,000, the value of S, = 12,000 psi. From Condition B, subparagraph
(a), the maximnm stress value for the shell is (12,000/3) = 4,000 psi. The needed shell thickness = 0.5
X 18,000/4000 = 2.25 in. The required head thickness = 0.25 X 18,000/4000 = 1.13 in.
The maximum AT at start-up or shutdown cannot exceed S,/(2Ea) = 12,000/(2 X 27.3 X 106 X 7.25
X 10'6) = 31°F, otherwise a fatigue analysis is necessary.
1.6
PRESSURE TESTING OF VESSELS AND COMPONENTS
1.6,1
ASME Code Requirements
Pressure vessels that are designed and constructed to VIII-l rules, except those tested in accordance with
the requirements of UG-IOI, are required to pass either a hydrostatic test (UG-99) or a pneumatic test (UG100) of the completed vessel before the vessel is U-stamped. Pressure vessels that are designed and
constructed to VIIl-2 rules also are required to pass either a hydrostatic test (Article T-3) or a pneumatic
test (Article T-4) before the U2-stamp is applied. Each component section of the ASME Boiler and Pressure
Vessel Code has a pressure test requirement thatcalls for a pressure test at or above the maximum allowable
working pressure indicated on the nameplate or stamping and in the Manufacturer's Data Report before
the appropriate Code stamp mark may be applied.
Under certain conditions, a pneumatic test may be combined with or substituted for a hydrostatic test.
When testing conditions require a combination of a pneumatic test with a hydrostatic test, the requirements
for the pneumatic test shall be followed. In all cases, the term hydrostatic refers not only to water being
an acceptable test medium, but also to oil and other fluids that are not dangerous or flammable; likewise,
pneumatic refers not only to air, but also to other nondangerous gases that may be desirable for "sniffer" detection.
1.6.2 What Does a Hydrostatic or Pneumatic Pressure Test Do?
There is always a difference of opinion as to what is desired and what is accomplished with a pressure
test. Some persons believe that the pressure test is meaut to detect major leaks, while others feel that there
sbould be no leaks, large or small. Some feel that the test is necessary to invoke loadings and stresses that
are equivalent to or exceed those loadings and stresses at operating conditions. Others feel that a pressure
test is needed to indicate whether a gross errorhas been made in calculations or fabrication. In some cases,
it appears that the pressure testing may help round out corners or other undesirable wrinkles or may offer
some sort of a stress relief to some components.
Background Information
1.6.3
23
Pressure Test Requirements for YIn·1
1.6.3.1 Hydrostatic Test Requirements. A hydrostatic pressure test is the preferred test method. A
pneumatic test or a combination of pueumatic/hydrostatic test is conducted only when a hydrostatic test caunot
be done. Except for certain types of vessels that are discussed later. the hydrostatic test pressure at every point
in the vessel shall be at least 1.3 times the maximum allowable working pressure multiplied by the ratio of the
allowable tensile stress value at test temperature divided by the maximum allowable tensile stress value at
design temperature. As an alternative, a hydrostatic test pressure may be detemtined by calculations agreed
upon by the user and the manufacturer. In this case, the MAWP (maximum allowable working pressure) of
each element is determined and multiplied by 1.3 and then adjusted for the hydrostatic head. The lowest value
is used for the test pressure, which is adjusted by the test temperature to design temperature ratio. In any case,
the testpressure is limited to that pressure which will not cause any visible permanent distortion (yielding) of
any element The metal temperature of the vessel or component to be tested is recommended to be at least
30°F above the MDMT to be marked on the vessel but need not exceed 120°F. Also, a liquid pressure relief
valve set at 1 1/3 times the test pressure is recommended in the pressure test system.
After the test pressure is reached, the pressure is reduced to the test pressure divided by 1.3, and welded
joints, connections, and other areas are visually examined for leaks and cracks. The visual examination
may be waived if a gas leak test is to be applied, if hidden welds have been exantined ahead of time, and
if the vessel will not contain a lethal substance. Venting shall be provided at all high locations where there
is a possibility of air pockets fornting during the filling of the vessel for testing. The general rules for
hydrostatic testing do not call for a specific time for holding the vessel at test pressure. The length of this
time may be set by the Authorized Inspector or by a contract specification.
1.6.3.2 Pneumatic Test Requirements. For some vessels, it is necessary to apply a pneumatic test
in lieu of a hydrostatic test. This may be dne to any number of reasons, including vessels designed and
supported in such a manner that they cannot be safely filled with liquid and vessels that cannot tolerate
any trace of water or other liquids. If a vessel is to be pneumatically tested, it shall first be examined
according to the requirements ofUW-50. This paragraph requires that welds around openings and attachments
be examined by MT or PT before testing. Except for certain vessels, the pneumatic test pressure at every
point in tbe vessel sball be l.l times the maximum allowable working pressure multiplied by the ratio of
the allowable tensile stress value at test temperature divided by the allowable tensile stress value at design
temperature. For pneumatic testing. the metal temperature of the vessel or component shall be at least 30°F
above the MDMT to be marked on the vessel.
The test pressure shall be gradually increased to no more than half of the full test pressure and then
increased in steps of one-tenth of the test pressure until the full test pressure is reached. After that, the
pressure shall be reduced to the test pressure divided by 1.1 and all areas are to be examined. All other
requirements for hydrostatic testing shall be observed, including the waiving of the visual examination,
provided the same limits are met.
1.6.3.3 Test Requirements for Enameled or Class-lined Vessels. The maximum test pressure for
enameled and glass-lined vessels does not have to be any greater than 1.0 MAWP unless required by the
Authorized Inspector or by a contract specification. Higher test pressure may damage the enameled or glass
coating. All other rules for hydrostatic testing apply.
1.6.3.4 Test Reqnirements for Vessels Built to the Rules of Parts UCI or UCD. For those vessels
designed and constructed to the rules of Part UCI for Cast Iron and Part UCD for Cast Ductile Iron, where
the factor of safety on tensile strength to set the allowable tensile stress values is 10 and 5. respectively,
the multiplier for the hydrostatic test pressure is set differently. For Part UCl, the test pressure shall be 2.5
MAWP. but is not to exceed 60 psi for a design pressure less than 30 psi and 2.0 MAWP for a design
24 Chapter I
pressure equal to or greater than 30 psi. Fnr Part UCD, the test pressure shall be 2.0 MAWP. With these
changes, the remaining rules of UO-99 are followed.
1.6.3.5 Test Requirements fnr Vessels Built to the Rnles of Part ULT. Alternative rules for the
design and construction of vessels to operate at cold temperatures as low as - 320 0 P are given in Part
ULT. These rules permit the use of increased allowable tensile stress values at temperatures colder than
ambient temperature to as low as 320'F for 5%, 8%, and 9% nickel steels, 5083 aluminum alloy, and
Type 304 stainless steels. Other materials listed in both Section II and Subsection C may be used for vessels
and parts for design at cold temperature with the allowable tensile stress values set by the value at WO°F.
When the vessel is desigued and constructed to Part ULT rules, special hydrostatic testing requirements
are . n~(;essfl1"Y . due .to .t~~.fa(;tt1l~t. . the . lllateri:ll.. is.str0ll~t3r . at.. desi~n . . t~lTIP~ratllre .thaIl.at . . aIIllJieIltt~st
temperature. The vessel shall be hydrostatically tested at ambient temperature with the test pressure held
for 15 minutes and either of the following criteria may be applied:
a. A standard hydrostatic test as described in 1.6.3.1 is used, but with the ratio of allowable
stresses not applied and the test pressure shall be 1.4 MAWP, if possible, instead of 1.3
MAWP.
b. In applying (a), the membrane stress in the vessel shall not exceed 0.95 of the specified
minimum yield strength nor 0.5 of the specified minimum tensile strength. In complying
with these stress limits, the ratio of bydrostatic test pressure divided by the MAWP may be
reduced below 1.4, but it shall be not less than 1.1 MAWP. If the value comes ant less than
1.1 MAWP, a pneumatic test shall be conducted using the rules of UO-lOO, but omitting
the adjustment for the allowable tensile stress ratio.
A vessel to be installed vertically may he tested in the horizontal position, provided the test pressure is
applied for 15 minutes at not less than 1.4 MAWP, including a pressure equivaleut to the liquid head in
operating position.
1.6.3.6 Proof Testing to Establish MAWP. In addition to the hydrostatic or pneumatic pressure test
of the completed vessel, a pressure proof test is permitted to establish the MA WP of vessels and vessel
parts for which the strength cannot be calculated with assured accuracy. The rules for such a pressure proof
test are given in UO-1OI of VlIl-l and may be based on yielding or on bursting of the vessel or vessel
part. Proof tests must be witnessed by the Authorized Inspector, who indicates acceptance by signing the
Manufacturer's Data Report Form. Duplicate or similar parts to that part which has had its MAWP establi.shed
by a proof test according to the reqnirements of UO-1OI(d) of VIII-l may be used without a proof test of
their own, but shall be given a hydrostatic or pneumatic pressure test as part of the completed vessel
pressure test.
1.6.4
Pressure Test Requirements for V1II·2
1.6.4.1 Hydrostatic Test Requirements. Except for glass-lined and enameled vessels, the hydrostatic
test pressure at every point in the vessel shall be 1.25 times the design pressure (or MAWP) to be marked
on the vessel multiplied by the ratio of the design stress intensity valne at test temperature divided by the
design stress intensity value at design temperature. For glass-lined or enameled vessels, the hydrostatic test
pressure shall be at least equal to, but need not exceed, the design pressure (or MA WP). Similar to the
alternative in VIII-I, the hydrostatic test pressure may be determined by calculations agreed upon between
the User and the Manufacturer and shall be described in the Design Report.
The hydrostatic test pressure shall not exceed a value that results in the following:
a. A calculated primary membraue stress intensity Pm of 90% of the tabulated yield strength
S; at test temperature;
Background Information
25
b. A calculated primary membraue plus primary bending stress intensity F; + P, not to exceed
tbe limits given below:
[Pm
[Pm
+
+ Pb] '" 1.35 S" when Pm'" 0.67 S"
Pb] '" 2.35 S, - 1.5 p.. when 0.67 S, < P; '" 0.90 S,
(1.1)
(1.2)
1.6.4.2 Pneumatic Test Requirements. A pneumatic test is permitted only when one of tbe following prevails:
a. Vessels cannot be safely filled with water due to their design and support system;
b. Vessels in which traces of testing liquid cannot be tolerated:
When a pneumatic test is permitted in lieu of a hydrostatic test, except for glass-lined and enameled
vessels, the pneumatic test pressure at every point iu the vessel shall be 1.15 times the design pressure (or
MAWP) to be marked on the vessel multiplied by the ratio of the design stress intensity value at test
temperature divided by the desigu stress intensity value at design temperature. For glass-lined or enameled
vessels, the pneumatic test pressure shall be at least equal to, but need not exceed, the design pressure
(or MAWP).
The pneumatic test pressure shall not exceed a value that results in the following:
a. A calculated primary membrane stress intensity P; of 80% of the tabulated yield strength
S, at test temperature;
b. A calculated primary membrane plus primary beuding stress intensity Pm + Ph not to exceed
the following limits:
+ F.] '" 1.20 s; when r; '" 0.67 S,
(1.3)
+ Pb] '" 2.20 S, - 1.5 p.. when 0.67 S, < P; '" 0.80 S,
(1.4)
[Pm
[Pm
CHAPTER
2
CYLINDRICAL SHELLS
2.1 INTRODUCTION
The rules for cylindrical shells in VIII-l and VIII-2 take into consideration internal pressure, external pressure,
and axial loads. The rules assume a circular cross section with uniformthickness in the circumferential and
longitudinal directions. Design reqnirements are not available for elliptic cylinders or cylinders with variable
thicknesses and material properties. However, such construction is not prohibited in VIII in accordance
with Paragraphs U-2Cg) of VIII-l and, AG-lOOCb) and (d) of VIII-2. The design and loading conditions
given in VIII-I are discussed first in this chapter, followed by the rules in VIII-2.
2.2 TENSILE FORCES, vm.i
The governing equations and criteria for the design of cylindrical shells under tensile forces are given in
several paragraphs of YIIl-I. The tensile forces arise from various loads such as those listed in Paragraph
UG-22 and include internal pressure, wind loads, and earthquake forces.
2.2.1 Thin Cylindrical Shells
The required thickness of a cylindrical shell due to internal pressure is determined from one of two equations
listed in Paragraph UG-27. The eqnation for the required thickness in the circumferential direction, Fig.
2.1Cal, due to internal pressure is given as
t = PR/ CSE - 0.6P),
when
t
< O.5R
or
P < 0.385SE
(2.1)
where
E=
P=
R=
S=
Joint Efficiency Factor
internal pressure
internal radius
allowable stress in the material
t = thickness of the cylinder
This equation can be rewritten to calculate the maximum pressure when the thickness' is known. It takes
the form
P
=
SEI/ (R
27
+ O.6t)
(2.2)
28 Chapter 2
,,
--1,
--,,,
,
--' .. "
,
(bl
FIG. 2.1
It is of interest to note the similarity between Eq. (2.1) and the classical equation for circumferential
membrane stress in a thin cylinder (Beer and Jobnston, 1992), given by
t
~
(2.3)
PRISE
The difference is in the additional term of O.6P in tbe denominator. This term was added by the ASME to
take into consideration the nonlinearity in stress that develops in thick cylinders, i.e., when the thickness
of a cylinder exceeds a.1R. This is demonstrated in Fig. 2.2 for circnmferential stress calculated by three
different methods. The first is from Eq. (2.3), the theoretical equation for thin cylinders; the second is from
Eq. (2.1); and the third is from Lame's theoretical equation for thick cylinders and is discussed later as
Eq. (2.12).
Similarly, the equation for the required thickness in the longitudinal direction, Fig. 2.1(b), due to internal
pressure is given as
t = PRI(2SE
+
OAP),
with
t
< O.5R
or
P
< 1.25SE
(2A)
Cylindrical Shells 29
20.0 -I-r
10.0H-
~
SIP
l\
i~~ ........,
1"-r-, -e-,
to
-~
--..........
.........
--
I--~
--
---;;
Eq.(2.12)
Eq.(2.1 )
1"--...
~
1--k
Eq.(2.3)
.10
to
1.5
2.5
2.0
3.0
3.5
4.0
(R+t) I R
FIG. 2.2
COMPARISON OF EQUATIONS FOR HOOP STRESS IN CYLINDRICAL SHELLS
or in terms of pressure,
P
~
2SEtl(R - OAt)
(2.5)
Notice again the similarity between Eq. (2.4) and the classical equation for longitudinal stress in a thin
cylinder given by
t
=
PRI2SE
(2.6)
Equations (2.1) and (2.4) are in tenus of the inside radii of cylinders. In some instances, the outside radius
of a shell is known instead. In this case, the governing equation for circumferential stress i~ expressed in
30
Chapter 2
terms of the outside radius Ro. This equation, which is obtained from Eq. (2.1) by substituting (Ro - t)
for R, is given iu VIIl-I, Appendix 1, Article 1-1, as
I
=
PRo/(SE
+
with
O.4P),
I
< O.5Ro
or
P < 0.385SE
P = SEI/(R a - O.4t)
(2.7)
(2.8)
VIIl-1 does not given an equation for the thickness in the longitudinal direction in terms of outside radius
Ro. Such an expression can be obtained from Eq. (2.4) as
PRol (2SE
+
l.4P)
(2.9)
P = 2SEI/(R o - 1.4/)
(2.10)
I =
Equations (2.1) through (2.10) are applicable to solid wall as well as layered wall construction. Layered
vessels consist of thin cylinders wrapped around each other to form a thick cylinder, Fig. 2.3. At any given
cross section, a-a, the total thickness consists of individual platematerial as well as weld seams. The Joint
Efficiency Factor for the overall thickness of a layered vessel is calculated from the ratio
E
a
= (:l:
E,I,)II
-,
-,
a
-+-+--1-+-+--+--+++++
FIG. 2.3
(2.11)
Cylindrical Shells 31
where
E = overall Joint Efficiency Factor for the layered cylinder
E, = Joint Efficiency Factor in a given layer
t = overall thickness of a layered cylinder
t, = thickness of one layer
The rules in VIII-l assnme that the longitudinal welds in varions layers are staggered in such a way that
E in Eq. (2.11) is essentially equal to 1.0.
Example 2.1
Problem
A pressure vessel is constructed of SA 516-70 material and has an inside diameter of 8 fl. The internal
design pressure is 100 psi at 450'F. The corrosion allowance is 0.125 in., and the joint efficiency is 0.85.
What is the required shell thickness if the allowable stress is 20,000 psi?
Solution
Refer to Paragraph UG-27 of VIII-I. The qnantity 0.3855E = 6545 psi is greater than the design pressure
of 100 psi. Thus, Eq. (2.1) applies. The inside radius in the corroded condition is equal to
R = 48
~
t ~ [PR/(SE - 0.6P)]
+ 0.125
48.125 in.
+ corrosioo
[100 X (48.125)/(20,000 X 0.85 - 0.6 X 100)] + 0.125
~
0.41 in.
The calculated thickness is less than O.5R. Thns, Eq. (2.1) is applicable.
A check of Eq. (2.4) for the required thickuess in the longitndinal direction will result in a t = 0.27 in.,
including corrosion allowance. This is about 60% of the thickness obtained in the circumferential direction.
Example 2.2
Problem
A pressure vessel with an internal diameter of 120 in. bas a shell thickness of 2.0 in. Determine the
maximum pressure if the allowable stress is 20 ksi. Assume E = 0.85.
Solution
For the circumferential direction, the maximum pressure is obtained from Eq. (2.2) as
P
= 20,000 X 0.85 X 2.0/ (60 +
~
556 psi
0.6 X 2.0)
32 Chapter 2
For the longitudinal direction, the maximnm pressure is obtained from Eq. (2.5) as
p
~
2 X 20,000 X 0.85 X 2.0/(60 - 0.4 X 2.0)
~1]49psi
Thus, the maximum pressure permissible in the vessel is 556 psi.
Example 2.3
Problem
A vertical boiler is constructed of SA 516-70 material and built in accordance with the requirements of
VIII-I. It has an outside diameter of 8 It and an internal design pressure of 450 psi at 709°F. The corrosion
allowance is 0.125 in., and the joint efficiency is 1.0. Calculate the required thickness of the shell if the
allowable stress is 17,500 psi. Also, calculate the maximum allowable additional tensile force in the axial
direction that the shell can withstand at the design pressure.
Solution
From Eq. (2.7), the required thickness is
t ~
450 X 48/(17,500 X 1.0 + 0.4 X 450) + 0.125
1.222 + 0.125
1.35 in.
From Eq. (2.10), the maximum allowahle axial pressure is
p
~
2 X ]7,500 X 1.0 X 1.222/(48 - 1.4 X 1.222)
= 924.0 psi
Subtracting from this value the internal pressure of 450 psi results in the additional equivalent pressure P',
that can be applied to the cylinder during operation.
P' = 924.0 - 450
~
474.0 psi
Total corroded metal area of cylinder = 'IT(R; - R2 )
~
1T(48' - 46.778')
= 363.9 in.'
Hence, total allowable force in cylinder during operation is
F
~
474.0 X 363.9
= 172,500Ib
Cylindrical Shells 33
Example 2.4
Problem
What is the required thickness of a layered cylinder subjected to an internal pressure of 1400 psi? Let
R = 72 in., S = 18 ksi, t, = 0.25 in. The longitudinal seams of the layers are staggered circumferentially
so that any cross section will have only one longitudinal joint with an efficiency of 0.65.
Solntion
This problem must be solved by trial and error. Let E = 1.0. Then from Eq. (2.1),
I
=
1400
~
5.87 in.
x
72/(18000
x
l.0 - 0.6
x
1400)
Try 24 1/4 in. layers with a total thickness of 6.0 in. The joint efficiency from Eq. (2.11) for the total
cross section is
E = (23
~
x
l.00
+ 1 x 0.65)/24
0.985
Using this Joint Efficiency Factor, recalculate the required thickness:
t
x
~
1400
=
5.97 in.
72/(18000
x
0.985 - 0.6
x
1400)
Since this thickness is less than the assnmed thickness of 6.0 in., the solution is complete. Hence, 24
1/4 in. layers are adeqnate.
2.2.2 Thick Cylindrical Shells
The VIlI-l code is routinely referenced in constructing vessels with internal pressures higher than 3000
psi. Special consideration must be given to details of construction, as specified in Paragraph U-I(d) of VIlI-1.
AB the ratio of II R increases beyond 0.5, the tbickness given by Eq. (2.1) becomes nonconservative, as
illustrated in Fig. 2.2. A more accurate equation that determines the thickness in a thick cylinder, called
Lame's equation, is given by
SE = P(R~
+
where Ro and R are outside and inside radii, respectively. By substituting the relationship R o = R
this expression, Eq. (2.12) becomes
t = R(Z,n - 1)
= (SE
+
P)/(SE -
+
t
into
(2.13)
where
z
(2.12)
R') I (R;, - R')
P)
34
Chapter 2
Equation (2.13) is used in Appendix 1-2 of VIII-l to determine the required thickness in thick cylinders
for the conditions t > O.5R or P > O.385SE. This equation can also be written in terms of pressure as
P
~
SE[(Z - 1)1(Z
+
(2.14)
I)J
where
+
t)IRJ'
t>
O.5R
Z ~ [(R
For .1ongitudinaJ .Stre:ss,
t = R(ZII2 -
1),
with
or
P> 1.25SE
(2.15)
where
~
Z
(PIS£)
+1
Equation (2.15) can be written in terms of pressure, P, as
P
~
SE(Z -
(2.16)
1)
where
+
Z ~ [(R
t)IRJ'
The thick cylinder expressions given by Eqs. (2.12) throngh (2.16) can be expressed in terms of ontside
radii as follows.
For circumferential stress,
with
t
>
O.5R
or
P > 0.3855E
(2.17)
where
Z
~
(SE
+
P)I(SE - P)
or in terms of pressure,
P
~
SE[(Z -
1)1(Z
+
(2.18)
1)]
where
Z ~ (Rol R)'
=
[Rol(R o -
t)J'
For longitudinal stress with t > O.5R or P > 1.25SE,
(2.19)
Cylindrical Shells 35
where
Z
= (PIS£) +
1
or in terms of pressure, P,
P = SE(Z -
(2.20)
1)
where
Z = (RoIR)'
= [Ro/(R a -
t)]'
All of the equations given so far are in terms of internal pressure only. VIII-1 does not give any equations
for calculating stresses in cylinders resnlting from wind and earthquake loads. One method of calculating
these stresses is given in Section 2.3.
Example 2.5
Problem
Calculate the required shell thickness of an accumulator with P
psi, and E = 1.0. Assume a corrosion allowance of 0.25 in.
=
10.000 psi, R
=
18 in., 8
=
20,000
Solution
The quantity 0.3858£ = 7700 psi is less than the design pressure of 10,000 psi. Thus, Eq. (2.13) is applicable.
Z
= (SE +
=
P)/(SE - P)
(20,000 x 1.0
+ 10,000) I (20,000 X 1.0 - 10,000)
= 3.0
t = R(Z'12 - 1)
= (l8.25)(3.0~'
=
Total t
13.36
- 1.0)
13.36 in.
+
0.25
=
13.61 in.
Example 2.6
Problem
What is the required thickness in Example 2.5 if the design pressure is 7650 psi and the corrosion allowance
is zero?
36 Chapter 2
Solution
The quantity 0.385S£ = 7700 psi is greater than the design pressure of7650 psi. Thus, Eq. (2.1) is applicable.
t = PRI(SE - 0.6P)
= 7650
x
18/(20,000
x 1.0 - 0.6 x 7650)
= 8.94 in.
It is of interest to determine the accuracy of Eq. (2.1) by comparing it with the theoretical Eq. (2.13),
which gives
Z = (SE + P)/(SE - P)
= (20.000
=
x
1.0
+ 7650) I (20,000 x 1.0 - 7650)
2.239
t = R(Z;/2 - 1)
18(2.239 0.5
-
1.0)
8.93 in.
This comparison demonstrates the accuracy of the "simple-to-use" Eq, (2.1) over a wide range of Rlt ratios.
Example 2.7
Problem
What is the maximum stress in a layered vessel subjected to an internal pressure of 15.000 psi? The outside
diameter is 24 in., and the inside diameter is II in.
Solution
The thickness of 6.50 in. is greater than O.SR. Thus, either Eq. (2.17) or Eq. (2.13) may be used, since
both the outside and inside diameters are given. Both of these equations are in terms of the quantity Z,
which is a function of stress S. Solving for S in these equations is not easy. However, since both of these
equations were derived from Eq. (2.12), we can use it directly to solve for S. Thus,
SE = 15,000 (12 2
+
5.52)/(12' - 5.52)
= 22,980 psi
2.3 AXIAL COMPRESSION
Vessel components are frequently subjected to, axial compressive stresses caused by such items as wind,
dead loads, earthquake, and nozzle loads. The maximum compressive stress is limited by either the allowable
tensile stress, using a Joint Efficiency Factor of 1.0, or the allowable compressive stress, whichever is less.
TIle allowable tensile stress controls thick cylinders, while the allowable compressive stress controls thin
Cylindrical Shells 37
cylinders. The procedure for calculating the allowable axial compressive stress in a cylinder is given in
Paragraph UG-23 of VIII-l and is based on a theoretical equation with a large LID ratio (Jawad, 1994).
It consists of calculating the quantity
A
~
O.12S/(R o / t)
(2.21)
where
A = strain
Ro
= outside radius of the cylinder
t = thickness
andthenusingastress",straindiagramfurnishedbytheASMEtodeterminethepermissibleaxial.compressive
stress, B. The ASME plots stress-strain diagrams, called External Pressure Charts, for various materials at
various temperatures on a log-log scale. One such chart for carbon steel is shown in Fig. 2.4. The strain,
A, is plotted along the hotizontal axis, and a stress, B, along the vertical axis. The majority of the materials
listed in the stress tables of II-D or VlII-l construction have a corresponding External Pressure Chart (EPC).
Tabular values of the curves in these charts are also given in II-D, for example those shown in Table 2.1
for Fig. 2.4.
If the calculated value of A falls to the left of the stress-strain line in a given External Pressure Chart,
then B must be calculated from the equation
B
~
(2.22)
AE/2
TABLE 2.1
TABULAR VALUES FOR FIG. 2.4
OF
300
500
A
0.100
0.765
0.800
0.900
0.100
0.200
0.300
0.400
0.500
0.25
0.100
-04
-03
0.100
0.663
0.900
0.100
0.250
0.300
0.800
0.100
0.150
0.200
0.272
0.100
-04
-03
-·02
-01
+00
-02
-01
+00
B, psi
OF
0.145 +03
0.111 +05
0.114
0.119
0.123
0.150
0.163
0.170
0.172
0.178
0.178
700
0.135 +03
0.895 +04
0.965
0.101 +05
0.121
0.124
0.143
0.147
0.155
0.163
0.170
0.170
800
900
A
0.100
0.559
0.100
0.300
0.100
0.250
0.100
-04
-03
-02
0.100
0.499
0.100
0.150
0.200
0.300
0.300
0.100
-04
-03
-02
0.100
0.247
0.100
0.150
0.200
0.300
0.800
0.300
0.100
-04
-03
-02
-01
+00
-01
+00
-01
+00
B, psi
0.124 +03
0.665 +04
0.808
0.101 +05
0.122
0.139
0.139
0.114 +03
0.569 +04
0.717
0.805
0.849
0.897
0.124 +05
0.124
0.104 +03
0.444 +04
0.605
0.689
0.742
0.795
0.927
0.112 +05
0.112
38 Chapter 2
GENERAL NOTE: See Table 2.1 for tabular values.
--
.- .-
25,000
U3~t
20.000
500 F
18,000
I I
./
/
,,
700 F
800 F
I
16.000
14,000
I
900 F
12.000
!-
10,000
9,000
~
8,000
0
7.000
~
~
~
6.000
E
I-+-+++-++E
m
D
29.0
x We <,
ar.o x 10~
I-+-+++-++E",
CL+,H-t+1f-++++-+-H-+t+t--/--i-H-I,-++t+1-l
I
3,500
f--,:::oj~~++I+.\+l-+-+++-+-t-++I++-+-+-H-H+++1-l
3.000
r---.
24-5 X 10
1
20,8 x 1(1
I 1111
5,000
-++t+H--I~H+-++++++t---+-+-++++-I+t-H 4.000
1
E 22.8 X 101
1-- 1--+-1--+-1-+Em
e
ml£
I
2345678923456789
.00001
0.0001
3456789
0.001
0.01
3
456789
2,500
0,1
FACTOR A
FIG. 2.4
CHART FOR CARBON AND LOW ALLOY STEELS WITH YIELD STRESS OF 30 ksl AND
OVER, AND TYPES 405 & 410 STAINLESS STEELS
where
E = modulus of elasticity of the material at design temperatme
The modulus of elasticity, E, in Eq. (2,22) is obtained from the actual stress-strain diagrams furnished
by the ASME, such as those shown in Fig, 2,4,
It is of interest to note tbat the stress B in the External Pressure Chart, Fig, 2,4, has a value of half the
stress obtained from the actual stress-strain curve of the given material, This was done by the ASME in
order to utilize these charts for other loading conditions, such as external pressure on cylindrical shells as
well as axial compression and vacuum on heads with various shapes. Thus, the stress, B, from Fig. 2.4 for
carbon steel at room temperature correspoudiug to a straiu, A, of 0,1 in.rm. is 17,8 ksi, This is half the
actual yield stress of 35,6 ksi for this material, as obtained from the actual given stress-strain curve, Also,
the value of the modulus of elasticity obtained from the elastic portion of the curve by finding the slope
between any two points along the curve is half the actual indicated value,
If we substitute Eq. (2,21) into Eq. (2,22), we find that in the elastic range, the buckling equation for
design becomes
B = EI16(R olt)
This can also be written as
B
= O,0625EI(Rolt)
(2.23)
Cylindrical Shells 39
The theoretical equation for the critical axial buckling stress used by the ASME is given by
<To ~
(2.24)
O.6E/(R o / t )
A comparison of the design Eq. (2.23) and the critical axial. buckling stress Eq. (2.24) indicates that a
factor of safety of about ten was used by the ASME. However, experiments performed subsequently to the
publication of Eq. (2.24) have shown that a more realistic critical axial buckling stress equation is of the form
<T. ~
(2.25)
O.6CE/(R o / t)
where C is obtained from Fig. 2.5.
A comparison of Eq. (2.23) with Eq. (2.25) indicates that the factor of safety varies from a conservative
value of 10.0 for small Ral t ratios to an unconservative value of 1.0 for large Roll ratios. This fact should
be considered when designing cylinders with large diameter to thickness ratios.
VIIl-l allows an increase of 20% in the value of B obtained from Fig. 2.4 or calculated from Eq. (2.22)
when live loads, such as wind and earthquake, are considered. Wind and earthquake loads are usually
obtained from various standards, such as ASCE-7 and the Uniform Building Code.
Example 2.8
Problem
The tower shown in Fig. E2.8 has an empty weight of 60 kips. The contents weigh 251 kips. Deterntine
the required thickness of the supporting skirt. Allowable tensile stress is 16 ksi. Use Fig. 2.4 for axial
compression calculations. The temperature of the skirt is 200°F at the hase and 800eF at the top.
1.0
0.8
0.6
C 0.4
0.2
lh
I\~I
,~,. 'i,
~~~.
. .1. R
....
~
- C= 1-0.90l1-e
I I I I
400
16
t
r
•
•
-. 1-- .s: .-- r--• f - oj ...
d~
.~
~<
u
. c
II
800 1200 1600 2000 2400 2800 3200 3600 4000
R/t
FIG. 2.5
C FACTOR AS A FUNCTION OF RfT (Jawad, 1994)
40
Chapter 2
I
i
I
32 psf
>-----
p,f
-1------
i
r--
Lin
20
N
psf
FIG, E2.8
Solution
Assume t = 3/8 in.
Axial force
=
Axial compressive stress
60
+ 251 = 311 kips
=
force/area of material in skirt
= 311,000/-rr
=
X 96 X 0.375
2750 psi
The wind moment at the bottom of skirt using a vessel projected area of 8 It is
M
=
32 X 8 X 36 X (36/2 + 34 + 26) + 24 X 8 X 34 X (34/2 + 26)
+ 20 X 8 X 26 X (26/2)
Cylindrical Shells 41
~
718,848 + 280,704 + 54,080
~
1,053,632 ft-lb
Notice that in many applications, the projected area must be increased beyond 8 ft to take into consideration
such items as insulation, ladders, and platforms. Also, the moment may have to be modified for shape and
drag factors.
The bending stress is obtained from the classical equation for the bending of beams:
Stress
~
Mell
where
c = maximum depth of the cross section from the neutral axis
I = moment of inertia
M = applied moment
and for thin circular cross sections, this equation reduces to
Stress ~ M / ('ITR~t)
= 1,053,632 X 12/('IT X 48' X 0.375)
~
4660 psi
Total compressive stress = 2750
+ 4660
= 7410 psi
The allowable compressive stress is calculated from Eq. (2.21).
A
~
0.125/(48/0.375)
~
0.00098
From Fig. 2.4 with A = 0.00098 and temperature of 200'F, we get B = 12,000 psi, which is the allowable
compressive stress. Thus, the selected thickness is adequate at the bottom of the skirt. Note that the thickness
would have been inadequate if the temperature at the bottom was 800°P'
Now let us check the thickness at the top of the skirt. The axial stress due to dead load stays the same.
The bending moment becomes
M
=
32 x 8 X 36 X (36/2 + 34 + 26 - 16) + 24 X 8 X 34 x (34/2 + 26 - 16)
+ 20 X 8 X 10 X (10/2)
= 571,392 +
~
176,256 + 8000
755,648 ft-lb
42 Chapter 2
Bending stress
~
755,648 X 12/('iT X 482 X 0,375)
= 3340 psi
Total compressive stress = 2750
=
+
3340
6090 psi
From Fig, 2,4 with A = 0,00098 and temperature of 800'F, we get B
thickness is adequate at the top of the skirt,
7,000 psi. Thus, the selected
Maximum tensile stress at bottom of skirt = 4660 - 2750
= 1910 psi
Maximum tensile stress at top of skirt
=
3340 - 2750
= 590 psi
Both of these values are less than 16,000 psi, which is the allowable tensile stress for the skirt,
These calculations show that t = 3/8 in, for the skirt is satisfactory, This thickness may need to be
increased in actual construction to take into account such items as opening reinforcements, corrosion, outof-roundness considerations, and handling factors.
Example 2,9
Problem
What is the allowable compressive stress in an internal cylinder with Do = 24 ft,
temperature = 900'F? Use Fig, 2,4 for the External Pressure Chart,
t
= 3/16 in" and design
Solution
From Eq, (2.21),
A = 0,125/(12010,1875)
=
0.0002
From Fig, 2.4, this A value falls to the left of the cnrve for 900'F, Therefore, Eq. (2,22) must be used, The
value of E is obtained from Fig, 2,4 as 20,8 X 106 psi for 900'F, Hence, allowable compressive stress B is
B
= 0,0002
X 20,800,000/2
= 2080 psi
2.4 EXTERNAL PRESSURE
External pressure on cylindrical shells causes compressive forces that could lead to buckling, The equations
for the bnckling of cylindrical shells under external pressure are extremely cumbersome to use directly in
Cylindrical Shells 43
design (Jawad, 1994). However, these equations can be simplified for desigu purposes by plotting them so
that the minimum buckling strain is expressed in terms of length, diameter, and thickness of the cylinder.
These plots are utilized by the ASME as discussed next. The rules and factors of safety in VTII-l and Vill-2
are identical for external pressure. Accordingly, references in this section are made to paragraphs in Vill-l
only.
2.4.1 External Pressure for Cylinders with Dolt
2:
10
The ASME uses plots to express the lowest critical strain, A, in terms of the ratios LIDo and Dolt of
the cylinder, as shown in Fig. 2.6. The designer calculates the known quantities LIDo and D,1t and then
uses the figure to determine buckling strain, A. To correlate buckling strain to allowable external pressure,
the designer uses the stress-strain diagram of Fig. 2.4 to obtain a B value. The allowable external pressure
can then be determined from this B value, as explained below. Accordingly, the procedure in ASME Vill-I
Paragraph UG-28 for determining the allowable external pressure for cylinders with Dolt ratios equal to
or greater than ten consists of the following steps:
1. Assnme a value of t for the cyliuder.
2. Calculate the quantities LIDo and Dolt.
3. Use Fig. 2.6 with the ca!cnlated values of LIDo and Dolt and establish an A value.
e
LIDo
-•
,
e e
e
,
e
s
...
'..
A
FIG. 2.6
GEOMETRIC CHART FOR CYLINDRICAL VESSELS UNDER EXTERNAL PRESSURE
(Jawad and Farr, 1989)
44 Chapter 2
4. Use an External Pressure Chart such as Fig. 2.4 with an A value and determine a B value
from the appropriate temperature chart.
5. Calculate the allowable external pressure from the equation
p
~
(4/3)(B)/(D olt)
(2.26)
6. When A falls to the left of the curves, the value of P is determined from
p
~
2AEI3(Dolt)
(2.27)
where
E = modulus of elasticity
Note that the curves in Fig. 2.6 are based on a thin cylinder simply supported at the ends with external
pressure acting laterally and on the ends. These curves can also be used, conservatively, for cases where
the pressure is on the sides only. as is the case with jacketed vessels. These curves can also be used,
conservatively, for cylinders with fixed rather than simply supported ends.
The effective length of a cylinder, L, needed to use Fig. 2.6, can sometimes be difficult to establish when
the cylinder is attached to other components, such as heads and transition sections. Figure 2.7 is provided
by VIII· 1 to define the effeetive length of some commonly encountered cylinders. The effective length of
cylinders with spiral stiffeners or variable thicknesses between supports is not addressed by VIII·\.
The factor of safety for the allowable external pressure obtained by using Eq. (2.26) or (2.27) is three
against buckling and also against yield.
Example 2.10
Problem
What is the required thickness of a cylindrical shell with length equal to 20 ft and outside diameter equal
to 5 ft? The cylinder is subjected to an external pressure of 15 psi at 500"F? Use Fig. 2.4 for the External
Pressure Chart
Solution
Try t = 0.50 in. Then LIDo = 4.0 and Dol t = 120.00. From Fig. 2.6, we obtain a value of A
Then from Fig. 2.4, B = 3000 psi. The allowable pressure is obtained from Eq. (2.26) as
=
0.00022.
P = (4/3) (3000)/120.00
~
33.3 psi
Since this value is higher than 15 psi, try a new thickness of 0.375 in. Then LIDo = 4.0 and Dolt =
160.00. From Fig. 2.6, we get A = 0.00017, and from Fig. 2.4, we see that A falls to the left of the curves.
Thus, from Eq. (2.27)
P
~
2 X 0.00017 X 27.0 X ]0'/(3 X 160.00)
=
19.1 psi
A new trial of t = 5/16 in. results in an unacceptable allowable pressure of 12.2 psi. Thus, the required
thickness to be used is 3 I 8 in.
Cylindrical Shells 45
r:
T
T
1
hl3
h/3
L
L
L
h/3
-l
h
(a - 1)
18- 21
Ibl
[Notes OJ and (21]
[Note (311
---i
T r rt
1 1 1
L
L
te -
1)
h
T
Ic - 2)
{Notes Il} and (21J
T
L
L
L
Idl
tel
[Note (311
r:
L
1
IfI
[Note 13H
NOTES:
(1) When the cone-to cylinder or the knuckle-to-cylinder junction is not 8 line of support, the nominel thiokness of the cone,
knuckle, or toriconical section shall not be less than the minimum required thickness of the adjacent cylindrical shell.
(2} Calculations shall be made using the diameter and corresponding thickness of each section with dimension L as shown.
(3) When the cone-to-cylinder or the kmrckle-to-cvlinder junction is a line of support, the moment of inertia shall be provided in
accordance with 1-8.
FIG. 2.7
SOME LINES OF SUPPORT OF CYLINDRICAL SHELLS UNDER EXTERNAL PRESSURE
(ASME Vm·1)
46 Chapter 2
The allowable compressive hoop stress is then
19.1
x
3010.375
1530 psi
2.4.2 External Pressure for Cylinders with Dolt < 10
When Dol I is less than 10, the allowable external pressure is taken as tbe smaller of tbe valnes determined
from the following two equations:
[2.167/(D olt) - 0.0833] B
2S
Po' ~ Dolt [1 - lI(Dolt))
(2.28)
(2.29)
wbere B is obtained as discussed above. For values of (Dol I) of less than or equal to 4, the A value is
calculated from
A ~ 1.1/(Do l t)'
(2.30)
For values of A greater than 0.10, use a value of 0.10. The value of S is taken as the smaller of two times
the allowable tensile stress, or 0.9 times the yield stress of the material at the design temperature. The yield
stress is obtained from the External Pressure Cbart of the material by using twice the B value obtained
from the extreme right-hand side of the termination point of the appropriate temperature curve.
The factor of safety in Eqs. (2.28) and (2.29) varies from 3.0 for Dol I = 10 to about 1.67 for Dol I =
2. This gradual reduction in the factor of safety as the cylinder gets thicker is justified since buckling ceases
to be a consideration and the factor of safety for external pressure is kept the same as that for internal
pressure, which is 2 I 3 s;
Example 2.11
Problem
The inside cylinder of a jacketed vessel has an outside diameter of 20 in., a length of 72 in., and a thickness
of 5 in. What is the maximum allowable jacket pressure? Use Fig. 2.4 for an External Pressure Chart. Let
the design temperature be 300°F. The allowable stress from tension is 17,500 psi.
Solution
Calculations give LIDo = 3.60 and Doll = 4.0. And since Dolt = 4.0, Eq. (2.30) must be used. Hence,
A ~ 1.1/(4.0)'
= 0.0688
From Fig. 2.4, B
17,800 psi.
Cylindrical Shells 47
From Eq. (2.28).
P"
~
[(2.167/4.0) - 0.0833] 17.800
~
8160 psi
The yield stress of the material is (0.9)(2B) or 32,040 psi. Twice the allowable stress is 35,000 psi. Hence
S = 32,040 psi is to be nsed. From Eq. (2.29),
P"
~
(2 X 32,040/4.0)(1 - 1/4.0)
= 12,020 psi
Therefore, the allowable jacket pressure in accordance with VUl-1 is 8160 psi. Notice, however, that this
pressure is greater than 0.385S, an indication that thick-shell equations may have to be used. Such equations
for external pressure are not in VUl-I yet.
2.4.3 Empirical Equations
It is of interest to note that Fig. 2.6 can only be used for (Dolt) of up to 1000. Larger values are not
permitted presently by the ASME. One approximate equation (Jawad, 1994) that is frequently used by
designers for large (Dol t) ratios was developed by the U.S. Navy and is given by
P
~
0.866E/(LlDo)(Dolt)25
(2.31)
where
E
P
~
~
modulus of elasticity, psi
allowable external pressure, psi
This equation incorporates a factor of safety of 3 and a Poisson's ratio of 0.30.
Many pressure vessels are subjected routinely to vacuum as well as axial loads from wind and dead load.
Section VU1 does not give any method for calculating the allowable compressive stress due to combined
effect of vacuum and axial loads. One such method is given by Bergman (Bergman, 1955). It uses an
equivalent external pressure to account for the axial compression effect on external pressure. Another
method that is used to combine axial and external pressure is that of Gilbert (Gilbert and Polani, 1979).
This method uses an interactive equation sintilar to the one used for calculating the buckling of beam columns.
Example 2.12
Problem
Solve Example 2.10 using Eq. (2.31).
Solution
From Example 2.10, t = 0.375 in., LIDo = 4.0, Dolt
P
~
160.00. andE = 27,000 ksi. Theu from Eq. (2.31),
~
0.866 X 27,000,000/(4.0)(160)25
~
18.1 psi
This approximate value differs from the answer in Example 2.10 by about 6%.
48
Chapter 2
2.4.4 Stiffening Rings
The required thickness of a shell with a given diameter under a specified external pressure can normally
be reduced by sbortening the shell's effective length. The length can be reduced by providing stiffening rings
at various intervals, as shown in Fig. 2.7. The required moment of inertia of suchrings is determined from
I,
=
[D~L,(r
+
(2.32)
A,IL,)A] 114
or
+ A,/4)A]/l0.9
(2.33)
where
Is = required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its neutral axis, in."
I' = required momeut of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell about
s
their combined ueutral axis, in.' The effective length of the shell is taken as l.lO(D ot,)!l 2.
L, = half the distance from the center line of the stiffening ring to the next line of support on one side,
plus half the distance from the center line of the ring to the next line of support on the other side.
A line of support is (1) a stiffening ring, (2) jacket bar, (3) circumferential line on a head at one,
third the depth of the head, (4) cone-to-cylinder junction.
As = area of the stiffening ring, in.?
t = minimum thickness of the shell, in.
t, = nominal thickness of the shell, in.
To design the stiffening ring
1. Assume first an area, A" of the stiffening ring and calculate the available moment of inertia.
I, or Is.
2. Calculate B from the equation
B = 0.75 [PDo/(l
+ A,IL,)]
(2.34)
3. Use the appropriate External Pressure Chart and determine an A value.
4. If B falls below the left end of tbe temperature line, calculate A from
A
=
2B/E
(2.35)
5. Solve Eq. (2.32) or (2.33) for the required moment of inertia.
6. The furnished moment of inertia must be greater than the required one.
Example 2.13
Problem
Calculate the required thickness of the shell and the required moment of inertia of the stiffening ring shown
in Fig. E2.13(a). The shell and ring material are SA 285,C. External pressnre is 12 psi, and the design
temperature is lOOoP.
I
I
Cylindrical Shells 49
Solution
Try t = 1/4 in. Then Dolt = 240.0 and LIDo
0.00016, and from Eq. (2.27),
= 12(10 +
2.5/3)/60
2.17. From Fig. 2.6, A
P = 2 X 0.00016 X 29,000,000/3(240)
= 12.9 psi
Thus, a shell thickness of 1 14 in. is adequate.
For the stiffeniug ring, try a 3 X 1 14 in. hard way bar, as shown in Fig. E2.13. For ease of calculations,
assume that the stiffening ring is not integral with the shell. Hence, Eq. (2.32) can be used. The moment
of inertia of the bar is bd31 12. Thus,
I = 0.25 X 3.0'/12
= 0.56 in."
:
I
I
l-r-r--r-
I
I
I
5' 0.0.
I
~
I
I
I
I
-
I
I
I
I
(
-..
.. '"
I
I
I
I
.671'
r---
I
~
"
•
:
;
FIG. E2.13
tJ
50
Chapter2
From Eq. (2.34),
B = 0.75[12 X 60.0/(0.25
=
+ 0.25 X 3.0/140.0)]
2110 psi
Since this value falls below the left end of the material line in Fig. 2,4, we use Eq, (2.35):
A
~
2 X 2110/29,000,000
= 0.000146
From Eq. (2.32),
t,
= [60.0' x 140.0(0.25
=
+ 0.25 x 3.0/140.0) 0.000146] /14
1.34 in.'
Since this number is larger than the actual moment of inertia of the ring (0.56 in.'), the assumed ring is
inadequate and a larger ring is required. However, before such a new ring is chosen, let us use the effective
moment of inertia of the existing ring and shell and compare that to Eq. (2.33).
From Eq. (2.33),
t; =
=
[60.0' x 140.0(0.25 + 0.25 x 3.01l40.0) 0.000146]/10.9
1.72in.4
The effective centroid of the shell-ring section, Fig. E2.13(h), is
h = [4.26 X 0.25 X 0.125
=
+ 0.25 X 3.0 (1.5 + 0.25)]/(4.26 X 0.25 + 0.25 X 3.0)
0.796 in.
The actual moment of inertia is
t
= 4.26 X 0.253/12
~
+ 4.26 X 0.25 X 0.671' + 0.25 X 3.03/12 + 0.25 X 3.0 X 0.954'
0.006 + 0.480 + 0.563 + 0.683
= 1.73 in.'
Thus, using the composite section results in a 1/4 in. X 3 in. stiffener that is adequate.
2.4.5 Attachment of Stiffening Rings
Details of the attachment of stiffening rings to the shell are given in Fig. DO-30 of VIII-I, which is
reproduced in Fig. 2.8. The welds must he able to support a radial pressure load from the shell of PL,. This
is based on the code assumption that the stiffening rings must support the total lateral load if the shell
segments between the rings collapse. Also, the code requires that the welds support a shear load of 0.01
PL»o. This shear load is arbitrary and is based on the assumption that if the rings buckle, bending moments
Cylindrical Shells
('..- .f'1
p--..,..
I
2in.min.
In-Line
St8f98red
Intermittent
Intermittent
Weld
ContinuousFillet Weld
One Side, Intermittent
Other Side
Wald
S '" St external stiffeners
S -c 12 t internal stiffeners
t
(al
(el
(bl
w
fdl
fal
FIG. 2.8
SOME DETAILS FOR ATTACHING STIFFENER RINGS (ASME VIII-1)
51
52 Chapter2
occur and generate shear forces. VITI-l also has other requirements pertaining to stitch welding and gaps
between the rings and the shelL These requirements are given in Paragraphs UG-29 and UG-30 of VIII-!.
Example 2.14
Problem
Calculate the required size of the double fillet welds attaching the stiffening riug shown in Fig. E2.13(b)
of Example 2.13 to the shelL Let the allowable tensile stress of SA 285-C at loo'F be 15,700 psi.
Solution
The radial load, FI, on the rings is equal to PL,.
PL, = 12.0 x 140
=
1680 lb I in. of circumference
Allowable tensile stress in the fillet weld from Table 1.4 is 0.55S
weld is
8635 psi. The total load carried by
Total load = number of welds attaching ring X size of weld X allowable stress
=2XWX8635
Hence, the required weld size, W, is
W = 1680/ (2 x 8635)
=
0.10 in.
Use 2 1/4 in. continuous fillet welds, in accordance with the minimum requirements of UG-30(t).
Shearing force, V, on the weld is
V
=
om PL,Do
= 0.01 x
12
x
140
x
60
= 1008lb
Allowable shearing stress in fillet weld from Table 1.4 is 0.55S = 8635 psi. From strength of materials,
the equation for shear stress is given by
-r = VQ/lt
wbere Q is at the location of the weld, as shown in Fig. E2.13(b), and is given by
Q
= 4.26
x 0.25(0.796 - 0.125)
= 0.71
in.'
Cylindrical Shells 53
Hence,
T
=
1008
=
827 psi < 8635 psi
x
0.71 I 1.73(2
x
0.25)
2.5 CYLINDRICAL SHELL EQUATIONS, VIII·2
The equation for the design ofthin cylindrical shells is given in Section AD-20l ofVIII-2 and is expressed as
t
where
P=
R=
5=
t=
= PRI(S -
O.5P),
for
P
<
OAS
(2.36)
internal pressure
inside radius
allowable stress at design temperature
required thickness
This equation is derived from the basic equation for thin cylindrical shells, t = PRIS, by substituting
for the inside radius, R, the quantity R + t /2. Equation (2.36) is applicable as long as the axial tensile
force, F, is not larger than O.5PR. When F exceeds 0.5PR, then the required thickness of the cylinder is
governed by the equation for longitudinal tensile stress as follows:
t
=
(O.5PR
+
(2.37)
F) I (S - 0.5P)
For values of P > OAS, the ratio of t IR increases beyond the scope of Eq, (2.36) and the design eqnation
in VIIl-2 is based on the plastic theory of thick cylindrical shells (Prager and Hodge, 1965). This equation
is given by
t = R(eP'S - I)
for
P
>
OAS
(2.38)
Example 2.15
Problem
Calculate the required thickness of a cylindrical shell with an inside diameter of 60 in. and an allowable
stress of 20 ksi. Let the pressure be (a) 1000 psi, (b) 8500 psi.
Solution
(a) Since P
< OAS, Eq. (2.36) applies. Then,
t
=
1000
=
1.54 in.
x
30/(20,000 - 0.5
x
1000)
54 Chapter 2
(b) Since P
> OAS, Eq. (2.38) applies. Then,
t = 30(e85OO120000
=
-
I)
15.89 in.
2.6 MISCELLANEOUS SHELLS
2.6.1
Mitered Cylinders
Mitered cylinders, Fig. L9,are used in nozzle connections, transition sections, and reducers. Neither VITI-l
nor VIII-2 give design rules. The piping code (ASME B31.3, 1993) gives design equations for various
miters. The basic equation for the allowable pressure in a shell with a single miter is
P = (SEt/R){l/[1
+
0.643(Rlt)"2 tan
P = (SEtIR){I/[1
+
1.25(Rlt)'12 tan
en
for
e<
22.5'
(2.39)
ell
for
e>
22.5'
(2.40)
where
E = Joint Efficiency Factor
P = allowable pressure, psi
R = radius of the shell in accordance with Fig. 2.9, in.
S = allowable stress
t = shell thickness, in.
a = angle of change in the direction of the miter joint, Fig. 2.9
e = «t ;
ASME B31.3 gives further information, such as equations for multiple miters, curved miters, and length
of tapers.
Example 2.16
Problem
A mitered cylinder has an inside radius of 24 in., a of 40', a design pressure of 500 psi, an allowable
stress of 15,000 psi, and a joint efficiency of 0.85. Determine the required thickness.
Solntion
From Eq. (2.1) for a straight cylinder,
t
= 500 x
=
Try t
=
1 in.
e = 40/2
24/(15,000
x
0.85 - 0.6
x
500)
0.96 in.
= 20'. Thus, Eq. (2.39) governs.
P = (15,000 x 0.85 x 1.0/24) (I I [I
= 248 psi
+ 0.643(24/1.0)112 tan
20])
Cylindrical Sliells 55
FIG. 2.9
MITERED BEND
Tliis is inadequate. Try
t ~ 1.8125
in.
P = (15.000 X 0.85 X 1.8125/24){lI[1 + 0.643(24/1.8125)'/2 tan 20])
=
Use t
520 psi> 500 psi
113/ 16 in.
2.6.2 Elliptical Shells
Elliptical shells. Fig. 2.10, are encountered occasionally by the pressure vessel designer. The stresses, away
from discontinuities, in the shell due to internal pressure can be approximated by using the membrane
theory of elliptical cylinders (Flugge, 1967). Tlie hasic equation for hoop stress is expressed as
(2.41)
where
a = major radius of the ellipse, in.
b = minor radius of the ellipse, in.
E = Joint Efficiency Factor
P = design pressure, psi
S = allowable stress, psi
t ~ thickness, in.
<!> = angle as defined in Fig. 2.10
56 Chapter 2
,.....--------.....,- - -
::::;;;;;;--r--~,
- - - - -...11--.....
,
FIG. 2.10
ELLIPTICAL CYLINDER
Example 2.17
Problem
Solve Example 2.1 if the cylinder is elliptical in cross section with the major diameter equal to 100 in. and
the minor diameter equal to 92 in.
Solution
a = 50.125 in.• b = 46.125 in., P
For
q,
~
100 psi, S = 20,000 psi, and E = 0.85
= 0", Eq. (2.41) gives
t
=
100 X 50.125' X 46.125'/20,000 X 0.85(50.125' sin' 0 + 46.125' cos' 0)3;2 + 0.125
~
0.320 + 0.125
= 0.45 in.
For
q,
= 90", Eq. (2.41) gives
t = 100 X 50.125' X 46.125'/20,000 X 0.85(50.125' sin' 90
= 0.250
=
+ 46.125' cos' 90)3;2 + 0.125
+ 0.125
0.375 in.
Use t = 0.45 in.
This thickness is about 10% higher than that for a cylinder with a circular cross section having an average
diameter of 96 in.
CHAPTER
3
SPHERICAL SHELLS, HEADS, AND
TRANSITION SECTIONS
3.1 INTRODUCTION
Sections VIII-l and VIII-2 contain rules for the design of spherical shells, heads and transition sections,
Head configurations include spherical, hemispherical, torispherical, and ellipsoidal shapes, Transition sections include cortical and toriconical shapes, The design rules for most of these shapes differ significantly
in VIII-l and VIII-2, This difference is due to the design approach used in developing the equations for
VIII-l and VIII-2, In this chapter a brief description of the various kinds of heads is given.
3.2 SPHERICAL SHELLS AND HEMISPHERICAL HEADS, VIII·1
3.2.1 Internal Pressnre in Spherical Shells and Pressure on Concave Side of
Hemispherical Heads
The required thickness of a thiu spherical shell due to internal pressure is listed in Paragraph UG-27 and
is given by
t
~
PRI(2SE - Oo2P),
when
t
< 0.356R
or
P < O.665SE
(3.1)
where
E = Joint Efficiency Factor
P = internal pressure
R = internal radius
S = stress in the material
t = thickness of the head
This equation can be rewritten to calculate the maximum pressure when the thickness is known. It then
takes the form
P
~
2SEt I (R
57
+ 0.2t)
(3.2)
58
Chapter 3
Notice the similarity between Eq. (3.1) and the classical equation for the membrane stress in a spherical
shell (Beer and Johnston, 1992), given by
(3.3)
t = PRI2SE
The difference is in the additional term of 0.2? in the denominator. This term was added by the ASME
to take into consideration the nonlinearity in stress that develops in thick spherical shells.
The designer should be aware that Eq. (3.1) determines the thickness based on pressure only. Large
spherical shells for liquid storage usually have low internal pressure. Thus, the governing thickness is
controlled by the liquid weight rather than Eq. (3.1). One method for determiniug the thickness in such
spheres is given in (API 620, 1990).
In some instances, the outside radius of a shell is known rather than the inside radius. In this case the
governing equatiou is obtained from Eq. (3.1) by substituting (Ro - t) for R. The resulting equation is
given in VlII-l, Appendix 1, Article 1-1, as
t = PRol (2SE
+
0.8P),
1< 0.356R o
with
or
P
<
0.665SE
(3.4)
or
P
=
2SEI/(Ro - 0.81)
(3.5)
As the ratio of tlR increases beyond 0.356, the thickness given by Eq. (3.1) becomes noneonservative.
This is similar to the case for cylindrical shells discussed in 2.2. The ASME VIIl-I eqnation for thick
spherical shells is given by
I = R(f't; -
1)
(3.6)
where
Y = 2(SE
+
P)/(2SE - P)
Equation (3.6) is used in Appendix 1-3 of VIII-I to determine the required thickness in thick spherical
shells for the conditions where I > 0.356R or P > 0.665SE. This equation can also be written in terms of
pressure as
P = 2SE[(Y -
1)/(Y
+
(3.7)
2)]
where
Y = [(R
+
I)IR)'
The thick shell expressions given by Eqs. (3.6) and (3.7) can be stated in terms of outside radii as
with
t
>
O.356R
where
Y = 2(SE
+
P)/(2SE - P)
or
P> 0.665SE
(3.8)
Spherical Shells, Heads, and Transition Sections
59
or in terms of pressure,
P
= 2SE[(Y -
Il/(Y
+ 2)J
(3.9)
where
Y = (RoIRl' = [Rol(R a - I)J3
Equations (3.1) through (3.9) are also applicable to hemispherical heads with pressure on the concave
side. This is illustrated in Fig. 3.1. For an applied internal pressure in compartment A, the hemispherical
heads abc and def are subjected to concave pressure and Eqs. (3.1) through (3.9) may be used. Paragraph
UGe32(f) of Vlfl- I gives the rules for the design of hemispherical heads due to pressure on the concave side.
Example 3.1
Problem
A pressure vessel is constructed of SA 516 e70 material and has an inside diameter of 8 ft. The internal
design pressure is 100 psi at 450°F. The corrosion allowance is 0.125 in. and the joint efficiency is 0.85.
What is the required thickness of tbe hemispherical beads if the allowable stress is 20,000 psi?
Solution
The quantity 0.6658E = 11,300 psi is greater than the design pressnre of 100 psi. Thus, Eq. (3.1) applies.
The inside radius in the corroded condition is equal to
b
1l------IIc
A
SHELL A
d II-----llf
INTERMEDIATE
HEAD
e
SHELL B
B
SKIRT
FIG, 3.1
60 Chapter 3
R = 48
+ 0.125
= 48.125 in.
The total head thickness is
t
= PR / (25£
=
100 X (48.125)/(2 X 20,000 X 0.85 - 0.2 X 100) + 0.125
= 0.142
=
- 0.2P) + corrosion
+ 0.125
0.27 in.
The calculated thickness is less than 0.356R. Thus, Eq. (3.1) is applicable.
Example 3.2
Problem
A pressure vessel with an internal diameter of 120 in. has a head thickness of 1.0 in. Determine the
maximum pressure if the allowable stress is 20 ksi. Assume E = 0.85.
Solution
The maximum pressure is obtained from Eq, (3.2) as
P
~
2 x 20,000 X 0.85 X 1.0/(60 + 0.2 X 1.0)
= 565 psi
Example 3.3
Problem
A vertical unfired boiler is constructed of SA 516-70 material and built in accordance with the requirements
of VIII-I. It has an outside diameter of 8 ft and an internal design pressure of 450 psi at 550 "F. The
corrosion allowance is 0.125 in. and the joint efficiency is 1.0. Calculate the required thickness of the
hemispberical head if the allowable stress is 19,700 psi.
Solution
From Eq. (3.4), the required head thickness is
t = 450 X 48/(2 X 19,700 x 1.0
= 0.543
+ 0.125
= 0.67 in.
+ 0.8
X 450)
+
0.125
Spherical Shells, Heads, and Transition Sections
61
Example 3.4
Problem
Calculate the required hemispherical head thickness of an accumulator with P
S = 15,000 psi, and E = 1,0, Assume a corrosiou allowance of 0.25 in.
=
10,000 psi, R
=
18 in"
Solution
The quantity 0.665SE = 9975 psi is less than the design pressure of 10,000 psi. Thus, Eq. (3.6) applies.
Y = 2(SE
=
+
2(15,000
P)/(2SE - P)
x
1.0
+ 10,000)/(2 x 15,000 x 1.0 - 10,000)
= 2.5
t
= R(y"3
- 1)
=
(18.25)(2,5'!3 - 1.0)
=
6.52 in.
Total head thickness = 6.52 + 0.25 = 6.77 in.
The required thickness of the shell for this vessel is calculated in Example 2.5. Attaching the head to
the shell requires a transition with a 3:1 taper, as shown in Fig. UW-13.1 of VIII-I, This taper, however,
is impractical to make in this case since the thickness of the head is abouttwo-thirds the radius. One method
of attaching the head to the shell is shown in Fig. E3.4.
3.2.2
External Pressure in Spherical Shells and Pressure on Convex Side of
Hemispherical Heads
The procedure for calculating the external pressure on spherical shells is given in Paragraph UG-28(d) of
VIII-I and consists of calculating the quantity
A = 0.125/(R o / t)
(3.10)
where
A = strain
R o = outside radius of the spherical shell
t = thickness
and then using a stress-strain diagram similar to Fig. 2.4 to determine a B value. The allowable external
pressure is calculated from
Po
=
B/(Ro/t)
(3.11)
62 Chapter 3
t - 6.77"
..--t =13.61"
FIG. E3.4
If the calculated value of A falls to the left of the stress-strain line in a given External Pressure Chart,
then P, must be calculated from tbe equation
P, ~ O.0625E/(R o / t)2
(3.12)
where
E = modulus of elasticity of the material at design temperature
The modulus of elasticity, E, in Eq. (3.12) is obtained from the actual stress-strain diagrams furnished
by the ASME, such as those shown in Fig. 2.4.
Equations (3.10) and (3.11) are also applicable to hemispherical heads with pressure on the convex side,
as mentioned in Paragraph UG-33(c) ofVllI-1. Tbis is illustrated in Fig. 3.1. For an applied internal pressure
in compartment B, the hentispherical head def is subjected to convex pressure and Eqs. (3.10) and (3.11)
may be used.
Spherical Shells, Heads, and Transition Sections
63
Example 3.5
Problem
Determine the required thickness of the head in Example 3.1 due to an external pressure of 10 psi.
Solution
From Example 3.1, the required thickness for internal pressure is 0.14 in. We will use this thickness as
our assumed t. Then from Eq. (3.10),
A
~
0.125/[(48 + 0.125 + 0.14)/0.14]
= 0.00036
From Fig. 2,4, B = 4,700 psi. And from Eq. (3.11),
Po = 4,700/(48.265/0.14)
13.6 psi
=
Since this pressure is larger than the design pressure of 10 psi, the minimum calculated thickness of 0.14
in. is adequate.
Example 3.6
Problem
What is the required thickness of a hemispherical head subjected to external pressure of 15 psi? Let Ro
150 in. and design temperature = 900°F. The material is SA 516-70.
Solution
Assume t = 0.25 in. Then from Eq. (3.10),
A
= 0.125/(150/0.25)
=
0.00021
Since the A value is to the left of the 900°F material liue in Fig. 2,4, we have to use Eq. (3.12).
Po = 0.0625
~
x
20,800,000/(150/0.25)'
3.6 psi
Since this value is less than 15 psi, a larger thickness is needed. Try t
A
~
0.125/(150/0.50)
= 0.00042
= 0.50 in.
=
64
Chapter 3
From Fig. 2.4, B = 4500 psi. And from Eq. (3.11),
Po = 4500/(15010.50)
=
15 psi
The selected thickness of 0.50 in. is adequate for the 15 psi external pressure. The thickness may have to
be increased due to handling and fabrication requirements.
3.3 SPHERICAL SHELLS AND HEMISPHERICAL HEADS, VIII-2
The required thickness of a spherical shell due to internal pressure is giveu in Paragraph AD-202 of
VIII-2 as
I = 0.5PRI(S -
P < 0.48
when
0.25P),
(3.13)
where
P = internal pressure
R = internal radius
S = stress in the material
t = thickness of the hemisphere
As the ratio of P /S increases beyond 0.4, the thickness given by Eq. (3.13) becomes nonconservative.
This is similar to the case for cylindrical shells discussed in 2.2. The VIII-2 equation for thick hemispherical
heads is given by
In [(R +
P > 0.48
when
1)1R] = 0.5P/S,
(3.14)
This can be written also as
t
= R(eO.5PIS
-
1),
when
P
> O.4S
(3.15)
When meridional forces, F (for instance, wind and earthquake loads) are present on the head, then Eq. (3.13)
is modified as follows:
I = (0.5PR
+ F)/(S
- 0.25P),
when
P < O.4S
(3.16)
where
F = Meridional force, lh/in. of circumference. F is taken as positive when it is in tension and negative
when it is in compression. When F is larger than 0.5PR, then buckling could occur and the rules
for external pressure must be considered.
The rules for calculating the required thickness of hemispherical heads subjected to pressure on the
concave side are given in Paragraph AD-204.l of VIII-2. The rules are identical to those for spherical
shells given by Eqs. (3.13) through (3.16).
The procedure and the factors of safety for calculating the a1lowahle external pressure on spherical shells
in VIll-2 are given in Paragraph AD-320. The rules and factors of safety are identical to those given in
VIII-!. Similarly, the rules in VIII-2 for calculating the allowable pressure on tbe convex side of hemispherical
heads are given in Paragraph AD-350.1. They are identical to those given in VIII-I.
Spherical Shells, Heads, and Transition Sections
65
Example 3.7
Problem
Determine the required thickness for a hemispherical head subjected to an internal pressure of 10,000 psi.
Let S = 20 ksi, R = 20 in.
Solution
PIS = 0.5. Since this ratio is larger than 0.4, Eq. (3.15) must be used.
t
=
20 (e°.5
=
5.68 in.
x 10,000/20,000 -
1)
3.4 ELLIPSOIDAL HEADS, VIU·l
3.4.1 Pressure on the Concave Side
A commonly used ellipsoidal head has a ratio of base radius to depth of 2:1 (Fig. 3.2a). The shape can be
approximated by a spberical radius of 0.9D and a knuckle radius of O.17D, as sbowu in Fig. 3.2(b). The
required thickness of 2:1 heads due to pressure on the concave side is given in Paragraph UG-32(d) of
Vlll-L Tbe thickness is obtained from the following eqnation:
t
= PD / (2SE -
0.2P)
(3.17)
or in terms of required pressure,
P
~
2SEt/(D
+ 0.2')
(3.18)
where
D = inside base diameter
E ~ Joint Efficiency Factor
P = pressure on the concave side of the head
S = aJlowable stress for the material
t = tbickness of the head
Ellipsoidal heads with a radius-to-depth ratio other than 2:I may also be designed to the requirements
of VIII-I. The governing equations are given in Appendix 1-4 of VIII-I as
t = PDK/ (2SE - 0.2P)
(3.19)
where
K = (116)[2
+
(D/2h)']
and D /2h varies between 1.0 and 3.0. The 1.0 factor corresponds to a hemispherical head. The K equation
is given in Article 1-4(c) of Appendix I of Vlll-I.
Equation (3.19) can be expressed in terms of the required pressure as
P
= 25Et/ (KD +
0.2t)
(3.20)
66
Chapter 3
= 2:1
D/2h
o
o
(b)
FIG. 3.2
These equations can also be written in terms of the outside diameter, Do> Thus,
=
+
2P(K - 0.1)]
(3.21)
P = 2SEtl[KDo - 21(K - 0.1)]
(3.22)
t
PDoKl[2SE
or in terms of required pressure
It is of interest to note that VIII-I does not give any PIS limitations for the above eqnations. Nor does
it have any rules for ellipsoidal heads when the ratio of PIS is large.
Spherical Shells, Heads, and Transition Sections
67
3.4.2 Pressure on the Convex Side
The thickness needed to resist pressure on the convex side of an ellipsoidal head is given in Paragraph
UG-33 of VIII -I. The required thickness is the greater of the two thicknesses determined from the steps below.
1. Multiply the design pressnre on the convex side by the factor 1.67. Then use this new
pressnre and a joint efficiency of E = 1.0 in the appropriate equations listed in Eqs. (3.17)
through (3.22) to determine the required thickness.
2. Determine first the crown radius of the ellipsoidal head. Then use this value as an equivalent
spherical radius to calculate a permissible external pressure in a manner similar to the
procednre given for spherical shells in Section 3.2.2. The procednre consists of calculating
the quantity
A
= 0.125/(KoDolt)
(3.23)
where
A = strain
Ko
Do
t
= function of the ratio D o/2ho and is obtained from Table 3.1
=
=
outside base diameter of the ellipsoidal head
thickness
Then, using a stress-strain diagram similar to Fig. 2A, determine the B value. The allowable
pressure is calculated from
P,
~
BI(KoDolt)
(3.24)
If the calculated value of A falls to the left of the stress-strain line in a given External Pressure Chart,
then P, must be calculated from the equation
P,
= 0.0625E/(KoDolt)'
(3.25)
where
E = modulus of elasticity of material at design temperature'
The modulus of elasticity, E, in Eq. (3.25) is obtained from the actual stress-strain diagrams, such as
those shown in Fig. 2.4, furnished by the ASME.
Example 3.8
Problem
Calculate the required thickness of a 2.2:1 head with an inside base diameter of 18 ft, design temperature
of lOO°F, concave pressure of 200 psi, convex pressnre of 15 psi, allowable stress is 17,500 psi, and joint
efficiency of 0.85. The head is made of low-carbon steel.
TABLE 3.1
FACTOR Ko FOR AN ELLIPSOIDAL HEAD WITH PRESSURE ON THE CONVEX SIDE
Dol2h o
Ko
Dol2ho
Ko
3.0
1.36
1.8
0.81
2.8
1.27
1.6
0.73
2.6
1.18
1.4
0.65
2.4
1.08
1.2
0.57
2.2
0.99
1.0
0.50
2.0
0.90
68 Chapter 3
Solution
For Concave Pressure
From Eq. (3.19), with K = (1/6)[2
+
(2.2)2] = 1.14,
t = (200 X 216.0 X 1.14)/(2 X 17,500 X 0.85 - 0.2 X 200)
=
1.66 in.
For Convex Pressure
1. First calculate the pressure and thickness.
P
=
1.67 X 15
=
25.1 psi
t = (25.1 X 216.0 X 1.14)/(2 X 17,500 X 1.0 - 0.2 X 25.1)
= 0.18 in.
2. For external pressure, we determine K; from Table 3.1 as 0.99. Let minimum t = 1.66 in.
Do = 216 + (2 X 1.66) = 219.32
From Eq. (3.23),
A = 0.125/(0.99 X 219.32/1.66)
= 0.00096
From Fig. 2.4, B = 12,000 psi.
P,
=
12,000/(0.99 X 219.32/1.66)
=
91.7 psi
Thus, minimum t = 1.66 in.
3.5 TORISPHERICAL HEADS, VIII-l
3.5.1
Pressure on the Concave Side
Shallow heads, which are commonly referred to as Flanged and Dished heads, or F&D, can also be built
to VIII-l rules, in accordance with Paragraph UG-32(e). The most commonly used F&D heads can be
approximated by a spberical radius, L, of 1.0D and a knuckle radius, r, of 0.06D, as shown in Fig. 3.3.
The required thickness of such heads due to pressure on the concave side is obtained from
t
=
0.885PLI (SE - O.IP)
(3.26)
Spherical Shells, Heads, and Transition Sections
I·
o
69
·1
L=O
FIG. 3.3
or in terms of required pressure,
P = SEt I(0.885L + O.lt)
(3.27)
where
E =
L =
P=
S=
t
Joint Efficiency Factor
inside spherical radius
pressure on the concave side of the head
allowable stress for the material
= thickness of the head
Torispherical heads with various spherical and knuckle radii may also be designed to the requirements
of VIII-I. The governing equations are given in Appendix 1-4 as
t
=
PLMI(2SE - 0.2P)
(3.28)
70 Chapter 3
where
M ~ (114)[3
+
(Llr)'I']
and Llr varies between 1.0 and 16.67. The 1.0 ratio corresponds to a hemispherical shell. The M equation
is given in Article l-4(d) of Appendix 1 of VIII-I.
Equation (3.28) can be expressed in terms of the required pressure as
P
~
2SEt / (LM
+ 0.2t)
(3.29)
These equations can also be written in terms of the outside radius, La, as
t
~
PLoM/[2SE
+ P(M
- 0.2)]
(3.30)
2SEt/[MLo - t(M - O.2)J
(3.31)
or in tenus of required pressure,
P
~
The theoretical membrane stress distribution in the circumferential, N e, and meridional, N<j>' directions in
shallow heads due to internal pressure are shown iu Fig. 3.4. Both the circumferential and meridional
stresses at the crowu of the head are tensile with a maguitude of S ~ Pa' / 2bt. However. at the base of
the head, the meridional stress is tensile with magnitude S = Pa / 2, while the circumferential stress is
compressive with a value of S = (Pa/2t)[2 - (alb)']. This compressive stress, which is not considered
by Eq. (3.28), could canse buckling of the shallow head as the ratio of Dlt increases. One way to avoid
such failure is to calculate the thickness based on an equation (Shield and Drucker, 1961) that takes buckling
into consideration and is expressed as
nP IS, = (0.33
+ 5.5rID)(t1L) + 28(1 - 2.2rID)(tlL)' -
Q
FIG. 3.4
0.0006
Spherical Shells, Heads, and Transition Sections
71
where
D
= base diameter of head, in,
L
n
P
r
= spherical cap radius, in.
= factor of safety
S,
t
= design pressure, psi
= knuckle radius, in.
= yield stress of the material, psi
= thickuess, in.
This equation normally results in a thickuess that is greater than that calculated from Eqs. (3.26), (3.28),
or (3.30) for shallow heads with large D/t ratios.
Paragraph UG-32(e) of VIII-l states that the maximum allowable stress used to calculate the required
thickuess of torispherical heads cannot exceed 20 ksi, regardless of the strength of the material. This
requirement was added in the code to prevent the possibility of buckling of the heads as the thickuess is
reduced due to the use of materials with higher strength.
3.5.2
Pressure on the Convex Side
For pressure on the convex design, the buckling rules for calculating F&D head thicknesses are the same
as those for ellipsoidal heads, with the exception that the outside crown radius of the F&D head is used
in lieu of the quantity Kof) oExample 3.9
Problem
Calculate the required thickness of an F&D head with an inside base diameter of 18 ft, design temperature
of lOooF, interual (concave) pressure of 200 psi, external (convex) pressure of 15 psi, allowable stress is
17,500 psi, and joint efficiency of 0.85. The head is made of low-carbon steel.
Solution
For Concave Pressure
Using L = 216.0 in., r = 0.06
from Eq. (3.28)
t
~
X
216
13.0 in., and M = (1/4)[3
(200 X 216.0 X 1.77)/(2 X 17,500 X 0.85
+ (216/13)112]
0.2 X 200)
= 2.58 in.
For Convex Pressure
1. Find the pressure and the thickuess.
P
=
1.67 X 15
=
25.1 psi
t = (25.1 X 216.0 X 1.77)/(2 X 17,500 X 1.0 - 0.2 X 25.1)
~
0.27 in.
1.77, we get
72 Chapter 3
2, Let t = 2.58 in, Then
Outside radius = 216 + (2
X
2,58) = 221.16 in,
From Eq, (3.23),
A = 0.125/(221.16/2.58)
= 0,0015
From Fig. 2.4, B = 14,000 psi.
P, = 14,000/(221.16/2.58)
163 psi> IS psi
Thus, t = 2.58 in.
3.6 ELLIPSOIDAL AND TORISPHERICAL HEADS,
vm-z
The required thickness for ellipsoidal as well as torispherical heads is obtained from Paragraph AD-204
and Article 4-4 of Vlll-Z. The procedure utilizes a chart, Fig. 3.5, which takes into consideration the
possibility of buckling of thin shallow heads, as discnssed in the previous section, The design consists of
calculating the quantities PIS and riD first and then using Fig. 3.5 to obtain the quantity t/L, and thus t.
The thickness for 2:1 ellipsoidal heads is obtained by using the riD = 0.17 curve, while the thickness
for a standard F&D head is obtained by using the riD = 0.06 curve. Figure 3.5 is plotted from the
following equation:
(3.32)
where
A =A, + A 2 + A3
A[ = -1.26176643 - 4.5524592 (riD) + 28,933179 (riD)'
A 2 = [0.66298796 - 2.2470836 (rID) + 15.682985 (rID)2] [In(P / SJ]
A] = [0,26878909 X 10- 4 - 0.42262179 (rID) + 1.8878333 (rID)2][ln(PlS)]'
D = base diameter, in.
L = crown radius, in.
P = design pressure, psi
r = crown radius, in.
S = allowable stress, psi
t = thickness
Example 3.10
Problem
An F&D head with a 6% knuckle is subjected to 40 psi of pressure. What is the required thickness if D =
168 in.? Use Vlll-2 and then VlII-1 rules. S = 20.000 psi for VIIl-2, and 15,000 psi for VIIl-1.
73
Spherical Shells, Heads, and Transition Sections
0.10
0.09
0.08
/-
0.07
0.065
0.06
0.055
0.05
1/1
7
~ !-
7TI V//
/ WI,
0
0.045
~
0.04
Z
0.035
z
~/.
0.025
£!j
0.02
,10
0.015
\
1\
0.20
0.01
PIS
\
0.009
0.008
\
»;:
\ 1\
~~~
/,
IX//, V/, 10: l/jW
"z)) Ih
W
«
\ \-:; 0~ «;; 0: ~
\
r~
~ @ ~W
@ ~ ~ :fj
:;f~ ~ W
0.005
0.002
0.06\
1\
\
0.006
0.003
\
0.15
~/,
~
0.10
\
0.007
0.004
k%
0.17 (2:1 Ellipsoidal head)
h
L'
IA ~ ~
~,
,
t
I
0/2
--:J
oat heads use L '" 0.90
to calculate tiL
,
I
N
ci
M
0
0
1
0
s0
0
'"00
ci
80
ci
~
0
0
ci
co
0
0
ci
0
'"
0
in
0
ci ci
tiL
FIG. 3.5
- -
--
-
- !-
0.001
0
0
---
-
NOTE: For 2: 1 Elhpsoi-
I'.
--
0
ci
N
'"
0
N
M
0
ci
ci
0
0
in
M
0
ci
sci ..'"ci '"ci
0
0
74
Chapter 3
Solution
Using VIIl-2 rules,
PIS
= 0.002
tiL = 0.003
Using VIIl-l rules and Eq. (3.26),
and
and
riD = 0.06
From Fig. 3.4,
t = 0.885
=
x
t
= 0.003
x 168
= 0.50 in.
40 X 168/(15,000 X 1.0 - 0.1 X 40)
0.40 in.
Note that normally the thickness obtained for a given component is 33% bigher in VIIl-l than that obtained
from VIIl-2, since tbe allowable stress in VIIl-l is 4.0 while that in VIIl-2 is 3.0. However, as is illnstrated
above, in the case of F&D beads, this may not be so. The reason is in the safety factors imbedded in the
equations of VIII-l and VIIl-2.
3.7 CONICAL SECTIONS, VIll-l
Conical shells and transition sections have a variety of configurations, as shown in Fig. 3.6. The required
thicknesses of the conical and knuckle regions are calculated in a different manner. In addition, conical
sections without a knuckle that are attached to shells result in an unbalanced force at the junction that must
be considered by the designer. Vlll-l provides rules for the design of the junctions. These rules differ for
internal and external pressure.
3.7.1 Internal Pressnre
For internal pressure, the design equation for a conical section is given by
t = PD/[2 cos CY. (SE - 0.6P)],
where
(3.33)
where
t = required thickness, in.
P = internal pressure, psi
D = inside diameter of conical section under consideration, in.
S = allowable tensile stress, psi
E = Joint Efficiency Factor
Equation (3.33) can be expressed in terms of internal pressure as
P = 2SEt cos CY./(D + 1.21 cos CY.)
(3.34)
Equations (3.33) and (3.34) can also be expressed in terms of outside diameter as
t = PDal [2SE cos CY.
+ P(2 - 1.2 cos CY.)]
(3.35)
Spherical Shells, Heads, and Transition Sections
75
,
Portion of a cone
rt
t
t
L
D,
(al
(hi
DL,
-.-
DU
DL
Le
t
DL
D,
D"
D"
Iel
(dl
(.1
FIG. 3.6
p
~
2SEt cos at/[D a - t(2 - 1.2 cos at)]
(3.36)
Equations (3.33) to (3.36), which are applicable at any angle at, are limited by VIII-I to ex :5 30". When
the angle at exceeds 30", then VIII-I requires a knuckle at the large end, as showu in Fig. 3.6(c) and (e).
This type of construction will be discussed later in this section.
After determining the thickness of the cone for internal pressure, the designer must evaluate the coneto-shell junction, The cone-to-shell junction at the large end of the cone is in compression due to internal
pressure, in most cases, The designer must check the junction for required reinforcement needed to contain
the unbalanced forces in accordance with Paragraph 1-5 of Appendix I of VIII-I. The required area is
obtained from
A" ~ (k Q, R,fS, E,) (l - /)./ at) tan at
where,
A r L = required area at the large end of the cone, in.
E, = Joint Efficiency Factor of the longitudinal joint in the cylinder
(3.37)
76 Chapter 3
E; =
E, =
E, =
k=
QL =
R, =
S, =
S, =
S, =
y=
LI. =
modulus of elasticity of the cone, psi
modulus of elasticity of the reinforcing ring, psi
modulus of elasticity of the cylinder, psi
1 when additional area of reinforcement is not required
y I S,E,. but not less than 1.0 when a stiffening ring is required
axiaJload at the large end, lb/ in., including pressure end-load
large radius of the cone, in.
allowable stress in the cone, psi
allowable stress in the reinforcing ring, psi
allowable stress in the cylinder, psi
S,E, for the reinforcing ring on the shell
S,E, for the reinforcing ring on the cone
angle obtained from Table 3.2
The area calculated from Eq. (3.37) must be furnished at the junction. Part of this area may be available
at the junction as excess area. This excess area can be calculated from the equation
(3.38)
where
AeL = available area at the junction, in.'
t = minimum required thickness of the shell, in.
t; = nominal cone thickness, in.
t, = minimum required thickness of the cone, in.
t., = nominal shell thickness, in.
If this excess area is less than that calculated from Eq. (3.37), then additional area in the form of stiffening
rings must be added.
The cone-to-shell junction at the small end of the cone is in tension due to internal pressure, in most
cases. The designer must check the junction for required reinforcement in accordance with Paragraph 1-5
of Appendix I of VIll-1. The required area at the small end of the cone is obtained from
A"
=
(3.39)
(k Q, R,IS, E,)(l - LI.! c) tan a.
where
A" = required area at the small end of the cone, in,"
Q, = axial load (including pressure end load) at small end, Ib I in.
R, = small radius of the cone, in.
LI. = angle obtained from Table 3.3
TABLE 3.2
VALUES OF LI. FOR JUNCTIONS AT THE LARGE CYLINDER DUE TO
INTERNAL PRESSURE
LI, deg.
0.001
11
0.002
15
0.003
18
0.004
21
PIS,E,
LI, deg,
0.006
25
0.007
27
0.008
28.5
0.009'
30
P/SsE1
NOTE:
(1) I:!.. = 30° for greater values of PISsE1 .
0.005
23
Spherical Shells, Heads, and Transition Sections
77
The area calculated from Eq. (3.39) must be furnished at the junction. Part of this area may be available
at the junction as excess area. This excess area can be calculated from the equation
A"
=
0.78(R,I,)'" [(t, - I)
+
(3.40)
(I, - 1,)/eos e]
If this excess area is less than that calculated from Eq. (3.39), then additional area in the form of stiffening
rings must be added.
When the angle a exceeds 30°, VIII-I requires a knuckle at the large end, as shown in Fig. 3.6(c) and
(e). The required thickness for the knuckle (called a flange) at the large end of the cone is obtained from
the equation
t = PLMI(2SE - O.2P)
where
M = (1/4)[3
+
(Llr)1/2]
L = D;!2 cos a
D,
=
inside diameter at the knuckle-to-cone junction
cos a)
inside knuckle radius, in.
= D - 2r (I -
r =
Eqnation (3,41) can be expressed in terms of the required pressure
P
= 2SEII(LM + 0.2t)
(3.42)
Equations (3,41) and (3,42) can also be written in terms of the outside diameter, Do, as
= PL,MI[2SE +
P(M -- 0.2)J
(3.43)
P = 2SEII[ML, - !(M - 0.2)]
(3.44)
t
or in terms of required pressure,
When a knuckle is used at the cone-to-shell junction, the diameter at the large end of the cone is slightly
less than the diameter of the cone without a knuckle, as shown in Fig. 3.6. Thus, the design of the cone
as given by Eq, 3.33 is based on diameter D, ratber than on the shell diameter.
ASME VIII-I does not give rules for the design of knnckles (flues) at the small end of cones. One design
method uses the pressure-area procedure (Zick and Germain, 1963) to obtain the required thickness. Referring
TABLE 3.3
VALUES OF b. FOR JUNCTIONS AT THE SMALL CYLINDER DUE TO
INTERNAL PRESSURE
0.002
4
PfSsEj
~,deg.
0.08
24
NOTE:
(1) A = 30° for greatervalues of PI SsE1
0.005
6
0.10
27
0.010
9
0.125'
30
0.02
12.5
0.04
17.5
78 Chapter 3
O2
tf
01
I
ex
t
•
s
.I
I
FIG. 3.7
to Fig. 3.7 for terminology, we can determine the required thickness based on membrane forces in the flue
and adjacent cone and shell areas from
If
~
(180/""r)[P(K,
where
E = Joint Efficiency Factor
K, = 0.125 (2r + DJl' tan" - ""lTr'/360
K, = 0.28D,(D,t,)'12
K, = 0.78Ks(Kst,)'12
K4 = 0.78tJK,t,)1I2
K, = 0.55t,(D,t,)1I2
K, = [D, + 2r(1 - cos a)]/2 cos a
P = internal pressure, psi
+ K, + K,)/1.5SE - K, - K,]
(3.45)
Spherical Shells, Heads, and Transition Sections
79
S = allowable stress, psi
t, = thickness of the cone, in.
If = thickness of the flue, in.
I, = thickness of the shell, in.
a = flue angle, deg. The flue angle is normally the same as the cone angle.
Example 3.11
Problem
Determine the required thickness of the cone, the two cylinders, and the area at the cone-to-cylinder junctions
shown in Fig, E3.l 1. Let axial compressive load at cone vicinity from mounted equipment
50 kips.
=
Allowable stress, psi
Joint Efficiency Factor
Modulus of elasticity, ksi
Pressure, psi
Small Cylinder
Cone
Large Cylinder
Reinforcing Ring
15,000
0.85
27,000
100
16,000
l.0
29.000
100
17,500
0.85
25,000
100
13,000
•
Rs
= 5'-0'
I
•
. FIG. E3.11
30,000
80
Chapter 3
Solution
Small Shell
The required thickness from Eq. (2.1) is
t
=
100 X 60/(15,000 X 0.85 - 0.6 X 100)
= 0.47 in.
Use t = 1/2 in.
Cone
From Eq. (3.33), the cone thickness is calculated as
t = 100 X 2 X 7 X 12/[2 cos 28(16,000 X 1.0 - 0.6 X 100)]
= 0.60 in.
Use t
=
5/8 in.
Large Shell
Again, using Eq. (2.1), we get
t = 100 X 7 X 12/(17,500 X 0.85 - 0.6 X 100)
= 0.57 in.
Use t = 5/8 in.
Large Cone-to-Shell lunction
Assume that a reinforcing ring, if needed, is to he added to the shell. Then from Eq. (3.37), we calculate
the stiffness ratio, k, as
k
=
17,500 X 25,000,000/(13,000 X 30,000,000)
= 1.12
The axial loads are given by
QL = PRLI2 - axial equipment load
=
100 X 84/2 - [50,000/(27l'84)]
=
4105 Ib/in.
Spherical Shells, Heads, and Transition Sections
81
Next, we need to calculate the need for reinforcement in accordance with Table 3.2.
PlS,El = 100/17,500 X 0.85
= 0.0067
From Table 3.2, Ii. = 26.4". Reinforcement is needed since a
The amount of reinforcement is calculated from Eq. (3.37):
A'L
=
28".
=
(1.12 X 4105 X 84/17,500 X 0.85)(1 - 26.4/28) tan 28
=
0.79 in.'
The available area in the shell and cone due to excess thickness is calculated from Eq. (3.38):
=
(0.625 - 0.57)(84
= 0.040
= 0.59
X
0.625)112
+ (0.625 - 0.60)(84
X
0.625/c08 28)112
+ 0.193
in.'
0.79 - 059
The additional area needed at the large junction
rolled the hard way.
0.20 in.' Use a 2 in. X 1/4 in. bar
Small Cone-to-Shell Junction
Assume that a reinforcing ring, if needed, is to be added to shell. Then, the stiffness ratio is obtained from
k
=
15,000
=
1.04
X
27,000,000/(13,000
X
30,000,000)
The axial loads are equal to
Q" = PR)2 - axial equipment load
=
100 X 60/2 - [50,Ooo/(27l'60)]
=
2867 Ib/in.
The need for reinforcement is obtained from Table 3.3.
PI $,£, = 100/15,000 X 0.85
=
0.0078
82
Chapter 3
From Table 3.3, Ll = 7,68°. Since this is less than 28°, reinforcement is required in accordance with Eq. (3.39),
A" = (1.04 X
=
2867 X 60115,000 X 0.85)(1 - 7.68/28) tan 28
5.41 in.2
In order to determine what excess area, if any, is available at the cone-to-shell junction, we mnst calculate
the required thickness of the cone at the small junction. This information is needed because the cone
thickness used so faris based on the large diameter rather than on the small one. From Eq. (3.33), the
minimum cone thickness at the small end is
t = 100 X 2 X 5 X 12/[2 cos 28(16,000 X 1.0 - 0.6 X l00)J
= 0.43 in.
From Eq. (3.40), the available area is
A" = 0.78(60 X 0.50)<12 [(0.50 - 0.47)
=
+
(0.625 - 0.43)/C08 28J
1.07 in.2
The additional area needed at the small junction = 5.41 - 1.07 = 4.34 in.' Therefore, use a 4.5 in. X 1 in.
bar rolled the hard way.
Example 3.12
Problem
Determine the required thickness of the cone, knuckle, fine, and the two cylinders shown in Fig. E3.12.
Let P = 100 psi, S = 16,000 psi, and joint efficiency = 0.85.
Solution
Small Shell
From Eq. (2.1).
t = 100 X 60/(16,000 X 0.85 - 0.6 X 100)
= 0.44
Use t
in.
= 1/2 in.
Large Shell
From Eq. (2.1),
t
=
100 X 7 X 12/(16,000 X 0.85 - 0.6 X 100)
= 0.62 in.
Use
t
= 5/8 in.
Spherical Shells, Heads, and Transition Sections
Rs == 5'-0"
•
'Ii'Thick
4" Radius
I
•
I
I
I
I
•
I
•
I
10" Radius
•
FIG. E3.12
Knuckle at Large End
The thickness of the knuckle is obtaiued from Eq. (3.41).
D; ~ 7
x 2 x 12 - 2 x 10(1 - cos 25)
= 166.13 in.
L ~ 166.13/(2 X cos 25)
= 91.65 in.
M
~
(1/4)[3 + (91.65110)11']
=
1.51
83
84
Chapter 3
t
=
100 X 91.65 X 1.51/(2 X 16,000 X 0,85 - 0,2 X 100)
=
0.51 in,
Use t = 9/16 in.
Cone
From Eq. (3.33), with D = D, = 166.13 in.,
t = 100 X 166.13/[2 cos 25(16,000 X 0.85 - 0.6 X 100)]
=
0.67 in.
Use t = 11/16 in.
Flue
From Eq. (3.45), the required thickness of the flue is
K, = 0.125(2 X 4
K,
K6
K3
K"
s;
+ 120)'
tan 25 - 25-rr4'/360
=
951.51
=
0.28 X 120(120 X 0.50)'"
~
260.26
=
[120
=
66.62
=
0.78 X 66.62(66.62 X 0.6875)>12
~
351.67
=
0.78 X 0.6875(66.62 X 0.6875)>12
~
3.62
+ 2 X 4(1 - cos 25)]/2 cos 25
= 0.55 X 0.50(120 X 0.50)'"
~
2.13
Spherical Shells, Heads, and Transition Sections
t
~
85
(l80/251T4)[100(95J.51 + 260.26 + 35J.67)/1.5 X 16.000 X 0.85 - 3.62 - 2.13J
1.10 in.
3.7.2 External Pressure
The design of conical shells for external pressure follows the same procedure as that for cylindrical shells
given in sections 2.4.1 and 2.4.2, with the following exceptions:
Item
Cone
Thickness
Diameter
Length
t of cylinder
Do of cylinder
t, = (t of cone)(cos a)
D L = outside large diameter of the cone
L< = (L!2)(1 + D,/D r), where L is obtained from Fig. 3.6.
L of cylinder
(3.46)
0.47)
(3.48)
After designing the cone for external pressure, the cone-to-shell junctions must be evaluated. Due to
external pressure, the cone-to-shell junction at the large end of the cone is tension, in most cases. The
designer must check the junction for required reinforcement in accordance with Paragraph 1-8 of Appendix 1
of VlII-1. The required area is obtained from
(3.49)
where all terms are the sarue as those in Eq, (3.37) and !l is obtained from Table 3.4.
The area calculated from Eq. (3.49) must be furnished at the junction. Some of this area may be available
as excess area at the junction. This excess area can be calculated from the equation
fl.L
= 0.55(DLO'" (t, + tJ cos a)
(3.50)
If this excess area is less than that calculated from Eq, (3.49), then additional area in the form of stiffening
rings must be added.
In addition to having a sufficient reinforcement area, the cone-to-shell junction must have an adequate
moment of inertia to resist external pressure forces when the junction is considered as a line of support.
The required moment of inertia is calculated as follows:
1. Determine the quantity An from the equation
(3.5t)
TABLE 3.4
VALUES OF!l FOR JUNCTIONS AT THE LARGE CYLINDER DUE TO
EXTERNAL PRESSURE
P/S,E,
A, deg.
a
0
0.002
5
0.005
7
0.010
10
0.02
15
0.15
40
P/S,E,
1:., deg.
0.04
21
0.08
29
0.10
33
0.125
37
P/ S,E,
1:., deg.
0.20
47
025
52
0.30
57
0.35 1
60
NOTE:
(1) 1l = 60° for greater values of PISsE1
86 Chapter 3
where
As = area of the stiffening ring
LL = effective length of the shell
L, = effective length of the cone
[L 2 + (RL - R,)2]1/2
L = axial length of the cone
2. Calculate the quantities
M = (Rr - R!)/(3R, tan e) + L,/2 - (R, tan ,,)12
(3.52)
and
F L = PM + (axial forces other than pressure)(tan a)
(3.53)
3. Calculate B from the equation
B = O.75(F,D,/An)
(3.54)
4. Enter the appropriate EPC and determine an A value.
5. If B falls helow the left end of the temperature line, calculate A from the equation
(3.55)
where Ex is the smaller of Eel En or £8"
6. Calculate the moment of inertia from one of the following equations:
Is = ADrAn/14
(3.56)
or
I;
= ADi,AnI1O.9
(3.57)
where
I, = required moment of inertia, Fig. 2.8(a), of the cross section of the ring ahout its
neutral axis, in."
I;
= required moment of inertia, Fig. 2.8(h), of the cross section of the ring and effective
shell about their combined neutral axis, in."
7. The required momeut of inertia must be greater than the furnished one.
The cone-to-shell junction at the small end of the cone due to external pressure is in compression, in
most cases. The designer must check the junction for required reinforcement in accordance with Paragraph
1-8 of Appendix I of Vlll-I. The required area is obtained from
A" = (kQR,IS.,E,) tan "
(3.58)
Spherical Shells, Heads, and Transition Sections
87
The area calculated from Eq. (3.58) must be furnished at the juuction. Some of this area may be available
at the junction as excess area. TIlls excess area can be calculated from the equation
A" = 0.55(D,t,)Jl2[(t, - t)
+
(I, - t,) 1cos 0:]
(3.59)
If this excess area is less than that calculated from Eq. (3.58), then additional area in the form of stiffening
rings must be added.
In addition to having a sufficient area. the cone-to-shell junction must have an adequate moment of
inertia to resist external pressure forces when the junction is considered as line of support. The required
moment of inertia is calculated as follows:
1. Determine the quantity A" from the equation
(3.60)
where
A, = area of the stiffening ring
L s = effective length of the shell
L, = effective length of the cone
= [L' + (RL - R,)'pt'
L = axial length of the cone
2. Calculate the quantities
N = (Rr - Rj)I(6R, tan 0:)
+
+
(R, tan 0:)12
(3.61)
+ (axial forces other than pressure)(tan 0,')
(3.62)
L,12
and
F, = PN
3. Calculate B from tbe equation
B = 0.75(F,D,IA,,)
(3.63)
4. Enter the appropriate EPC and determine an A value.
5. If B falls below the left end of the temperature line, calculate A from the equation
(3.64)
where Ex is the smaller of En En or E;
6. Calculate the moment of inertia from one of the following equations;
(3.65)
or
(3.66)
where
I, = required moment of inertia, Fig. 2.8(a), of the cross section of the ring about its
neutral axis, in."
88 Chapter 3
I;
= required moment of inertia, Fig. 2.8(b), of the cross section of the ring and effective shell
about their combined neutral axis, in."
7. The required moment of inertia must be greater than the furuished one.
When the cone is fianged and flued, then the required thickness of the cone is determined as before,
except that Eq, (3.48) is replaced by
L, = r, sin ts:
+
(4/2)[(D,
L, = r,(D"IDL) sin a:
L, = [r,
+
+
+
(L,/2)[(D,
r,(D,,1D u)] sin rx
+
+
+
(3.68)
for sketch (d) in Fig. 3.6
DL)IDJ
(L,/2)[(D,
(3.67)
for sketch (c) in Fig. 3.6
D,)1D1"J
DJlDuJ
for sketch (d) in Fig. 3.6
(3.69)
Example 3.13
Problem
Check the calculated thicknesses in Example 3.11 due to an external pressure of IS psi. The axial compressive
load at cone vicinity from mounted equipment = 50 kips. Figure 2.4 applies for external pressure. Notice
that the modulus of elasticity, shown in Fig. 2.4, for external pressure calculations must be used and is
different from the values listed below for junction reinforcement. The design temperatnre is lOO°F.
Allowable stress, psi
Joint Efficiency Factor
Modulus of elasticity, ksi
External P, psi
Effective L
SmaU Cylinder
Cone
Large Cylinder
Reinforcing Ring
15.000
0.85
27,000
15
10 ft
16,000
1.0
29,000
15
17,500
0.85
25,000
15
20 It
13,000
Solntion
Small Shell
From Example 3.11, use I
=
1/2 in.
Do
=
2(60
+
0.5)
=
121 in.
Then LIDo = 0.99 and Doll = 242.
From Fig. 2.6, A
0.00038.
From Fig. 2.4, B = 5500 psi.
From Eq. (2.26),
P = (4/3)(5500)1(242) = 30.3 psi> 15 psi
By trial and error, it can be shown that for I
=
0.40 in., P
=
15 psi.
30,000
Spherical Shells, Heads, and Transition Sections
Large Shell
From Example 3.11, try
I
89
= 518 in.
Do
=
2(84
+ 0.625) = 169.3 in.
Then LIDo = 1.42 and D,It = 271.
From Fig. 2.6, A = 0.00021.
From Fig. 2.4, B = 3050 psi.
From Eq. (2.26),
P = (413)(3050)1(271) = 15.0 psi
Cone
From Example 3.11, use t = 518 in.
Do = 2(84
From
From
From
From
+ 0.625)
=
169.3 in.
Eq. (3.46), I, = 0.625 cos 28 = 0.552 in.
Eq. (3.47), o; = 169.3 in.
Fig. E3.11, L = (7 - 5)ltan 28 = 3.76 ft.
Eq. (3.48),
L, = (3.76
x 1212)(1 + 1211169.3) = 38.68 in.
Then L,ID L = 0.23 and DLI t, = 307.
From Fig. 2.6, A = 0.00121.
From Fig. 2.4, B = 13,000 psi.
From Eq. (2.26),
P
=
(4/3)(13,000)1(307)
By trial and error, it can be shown that for
I =
=
56.5 psi> 15 psi
0.33 in., P = 15 psi.
Large Cone-to-Shell function
Assume that a reinforcing ring, if needed, is to be added to the shell. Then from Eq. (3.49), we calculate
the stiffness ratio, k, as
k
=
17,500 X 25,000,000/(13,000 x 30,000,000)
=
1.12
The axial loads are given by
Qr. = PR 1./2
~
axial equipment load
-15 X 84.625/2 - [50,000/(2'IT84.625)]
-728.7 Ib/in.
90 Chapter 3
Next, we must calculate the need for reinforcement in accordance with Table 3.4.
PlS,E, = 15/(17,500 X 0.85)
=
0.001
From Table 3.4, Li = 2.5°. Hence, reinforcement is needed.
The amount of reinforcement is calculated from Eq. (3.49);
A,L = (1.12 X 728.7 X 84.625/17,500 X 0.85) {l -
(0.25)[15 X 84.625
+
728.7)/728.7](2.5/28)} tan 28
= 2.32 in.'
The available area in the shell and cone due to excess thickness is calculated from Eq. (3.50).
A,L = 0.55(84.625 X 0.625)H2 (0.625
+
0.6251 cos 28)
= 5.33 in.'
Therefore, the junction is inherently reinforced and no additional rings are required.
Next, determine the required moment of inertia at the largejunction needed for external pressure. From
Eq. (3.51), and assuming a ring area of 1.0 in',
An = (240)(0.625)/2
+ (3.67
X 12)(0.625)/2
+ 1.0
= 89.8 in.
From Eq. (3.52),
M
= (84.625'
=
- 60.52)/(3 X 84.625 tan 28)
+ (240/2) - (84.625 tan 28)/2
123.44
From Eq. (3.53),
FL
=
15 X 123.44
=
1901.6
+
(50,000/27r84.625)(tan ex)
From Eq. (3.54),
B
= 0.75(1901.6
=
X 84.625/89.8)
1344
From Eq. (3.55),
A = 2 X 1344/25,000,000
= 1.08 X 10-'
Spherical Shells, Heads, and Transition Sections
91
From Eq. (3.57),
I;
~
1.08 x 10- 4 X 84.625' x 89.8/10.9
= 6.37 in.'
A trial run indicates that the availahle moment of inertia is inadequate without a stiffening ring. Assume
that a 2 in. X 1/2 in. ring rolled the hard way will be used. The available moment of inertia is obtained
from Fig. E3.13(a) and (b). The neutral axis is at
Xl
= [(5.64
x
+
0.625'/2)
(5.64
X
+ [2 X 0.5(-1.0)J/[(5.64 x
= (1.10
= 0.59
+ 4.67 -
1.00)/(3.53
0.625
x
0.625)
+ (5.64 X
+ 3.53 +
2.65/2)
0.625)
+ (2 X
0.5)]
1.0)
in.
X,
= 2.65/2
X,
= 1
- 0.59
+ 0.59
~
0.74
in.
= 1.59 in.
The moments of inertia of the cone about its two major axes are
h ~ 5.64' X 0.625/12
= 9.34 in.'
I; ~ 0.625' X 5.64/12 = 0.12 in.'
I"i"j = 0.0
The moment of inertia of the cone around an axis y through its centroid is
Iy
=
1-; sirf a + ly cos' a + 2l;y sin a cos a
= 9.34
sin' 28 + 0.12 cos' 28 + 0.0
= 2.15 in.'
The total moment of inertia of the composite section = (l of shell about its neutral axis) + (area of shell)
(distance to composite section neutral axis)' + (l of cone about the y axis throngh its centroid) + (area
of cone) (distance to composite section neutral axis)' + (l of the stiffener about its neutral axis) + (area
of stiffener) (distance to composite section neutral axis).'
I ~ 0.625.3 X 5.64112
+ (5.64 X
+ 2.15 + (5.64 X
0.625)(0.59 - 0.625/2)'
0.625)(0.74)'
+ 2.0' X 0.5112 + (2.0 X 0.5)(1.59)'
92 Chapter 3
y
I
I
·
·
·
O.55 ..m;:tS
5.64'
-J1-0.625·
5.64'
2' X
\.2' RING
I
5.64'
I
I
•
I
,•
O.55 JOL tc
5.64'
2.65'
,
(a)
y
(bl
1:1
'.55~
4.76'
2.23"
,I
.
I
I
O.55 JOsts
I
4.26'
.II
,
I
.
Os
4.,26"
•
"I
II
r
XI
(d)
(e)
FIG. E3.13
= 0.12
=
+ 0.27 + 2.15 + 1.93 + 0.33 + 2.53
7.33 in.'
Hence, use 2 in. X 1/2 in. ring at the junction.
Small Cone-to-Shell Junction
Assume that a reinforcing ring, if needed, is to be added to shell. Then, the stiffness ratio is obtained from
k
=
15,000 X 27,000,0001(13,000 X 30,000,000)
=
1.04
Spherical Shells, Heads, and Transition Sections
93
The axial loads are equal to
Q,
=
PR,/2 - axial equipment load
-15 X 60.5/2
[50,000/(21160.5)]
- 585.3 Ib/in.
Reinforcement is required in accordance with Eq. (3.58).
A,.,
585.3 X 60.5/(15,000 X 0.85)] tan 28
= [1.04 X
= 1.54 in.?
From Eq. (3.59), the available area is
A,.,
= 0.55(2
X
60.5
X
0.5)"2[(0.5 - 004)
+
(0.625 - 0.33)/ c0828]
= 1.86 in.'
Hence, the small junction is inherently reinforced and no additional rings are needed.
Next, determine the required moment of inertia at the small junction needed for external pressure. From
Eq. (3.60),
An = (120)(0.5)/2 + (3.76 X 12)(0.625)/2 + 0.0
=
44.1 in.
From Eq. (3.61),
N
=
(84.6252 - 60.52)/(6 X 60.5 tan 28) + 120/2 - (60.5 tan 28)/2
= 62.06 in.
From Eq. (3.62),
F, = 15 X 62.06
+ (50,000/21160.5)(tan 28)
= 1000.8 Ibs/in.
From Eq. (3.63),
B
= 0.75(1000.8
=
1030 psi
X 60.5/44.1)
94 Chapter 3
From Eq. (3.64),
A
=
2
=
7.63 X 10-'
1030/27,000,000
X
From Eq. (3.66),
I;
=
7.63 X 10- 5 X 60.5' X 44.1/10.9
=···1;13··in.4
The available moment of inertia is obtained from Fig. E3.l3(c) and (d). The neutral axis is at
x, = [(4.26 X 0.5'/2) + (4.76 X 0.625 X 2.23/2)]/[(4.26 X 0.5) + (4.76 X 0.625)]
X2
=
0.75 in.
=
0.365 in.
The moments of inertia of the cone about its two major axes are
I, = 4.763 X 0.625112 = 5.62 in:
I;
= 0.625'
X
4.76/12
=
0.10 in.'
Ii; = 0.0
The moment of inertia of the cone around an axis y through its centroid is
Iy = Ix sin' a + 1y cos' a
= 5.62 sin' 28
=
I
= 0.53
+ 2I-;.y sin a
+ 0.10 cos' 28 + 0.0
1.32 in.'
X 4.26112 + (4.26 X 0.5)(0.75 - 0.5/2)'
+ 1.32 + (4.76 X 0.625)(0.365)'
= 0.04 +
=
cos a
0.53 + 1.32 + 0.4
2.29 in.'
Hence, an adequate moment of inertia is available at the junction.
Spherical Shells, Heads, and Transition Sections
95
3.8 CONICAL SECTIONS, VIll-2
The required thickness of a conical shell in VIIl-2 suhjected to iuternal pressure is obtained from Eqs. (2.36)
throngh (2.38). These equations, which are for cylindrical shells, are used with the radius R perpendicnlar to
the surface of the cone, as shown in Fig. 3.8. VIIl-2 does not list any rules for toriconical heads.
The need for reinforcement at the large end of a cone-to-cylinder junction subjected to internal pressure
is obtained from Fig. 3.9. The figure is used to determine a maximum angle a when the PIS value is known.
If the actnal angle a is less than that obtained from the figure, then no additional reinforcement is needed
and the original thickness is adequate. On the other hand, if the actnal angle a is larger than that obtained
from the figure, then additional reinforcement is needed in accordance with Fig. 3.10. The Q factor obtained
from Fig. 3.10 is multiplied by the value of the original shell thickness at the large end to establish a new
thickness at the junction.
I
•
I
•
I
•
I
•
I
•
R
I
•
I
FIG. 3.8
96 Chapter 3
0.02
I
,
I
,
..I
I
NOTE: CurvelJOV8med bV maximumstress
intensitY at surface (primarily due to
axial bending 11"*1. limited to 3 S
=
0.012
0.010
0.008
./
0.006
Adoquote
0.005
PIS
~
/
.-
....
Not
I
,-
,
V
/V /\nc_Th~kn"
0.004
Permitted
~
-
i
-:
0.003
0.002
0.001
10
15
20
25
30
35
Maximum Angle, a'
FIG. 3.9
INHERENT REINFORCEMENT FOR LARGE END OF CONE-TO-CYLINDER JUNCTION
(ASME VIII-2)
The same procedure is utilized at the small end of the cone, but with Figs. 3.11 and 3.12. The Q factor
in this case is multiplied by the required thickness of the shell at the small end of the cone.
Figures 3.9 through 3.12 apply to internal pressure only. For external pressure, the mles and factors of
safety in VlII-2 are identical to those in VIII-I.
Example 3.14
Problem
Determine the required thickness of the cone, the two cylinders, and the area at the cone-to-cylinder junctions
shown iu Fig. E3.1J. Allowable stress and pressure data is given in Table E3.14.
TABLE E3.14
ALLOWABLE STRESS AND PRESSURE DATA
Allowable stress, psi
Pressure, psi
Small Cylinder
Cone
Large Cyli nder
Reinforcing Material
15,000
100
16,000
100
17,500
100
13,000
100
Spherical Shells, Heads, and Transition Sections
97
2. 2
~
"'\
'"
-, "
2.0
1.•
\.
NOTE: Curva gowrned by maximum ,tr." intenlity.t
surface (primarilv due to .'liel bending ltre,,)
limited to 3$
1\".30 .....
~
'\.
-. "\. -,
""-.
"-,
I"'"
r-,
-. <,.......
\.
'," at
__ T
~deg.
Q 1.6
'<;
1.2
i'-..
'" <;
~
1
'
317
2.~!
l- r
t,
I,
l~ f-----
<,
<,
..
,
<,
"
r-,
---*-A ,
,,
./
'"<, -,
",-
---1
"
2.~\
- - \ ')
cos
r-, """ <, ~"
~.
r-, J".., r-....
"
......
........
0:
>.-/,
./
1.0
0.001
0.0015
I, 1,
Q
....
10 .....
0.003
0.002
,
I
<,
<,
0.004
0.005
0.006
0.008
0.01
0.015
PIS
FIG. 3.10
VALUES OF Q FOR LARGE END OF CONE·TO·CYLINDER JUNCTION (ASME VIII-2)
Solution
Small Shell
The required thickness from Eq. (2.36) is
t
=
100 X 60.0/(15,000 - 0.5 X 100)
= 0040 in.
Use t = 7116 in.
Cone
From Eq. (2.36), with R = 7 X 12/cos 28 = 95.14 in., the cone thickness is calcn1ated as
t
Use t = 5/8 in.
"
=
100 X 95.14/(16,000 - 0.5 x 100)
=
0.60 in.
0.02
98 Chapter 3
0.10
0.08
0.06
0.04
./
Adequate\
0.02
V
./
PIS
1/
0.010
\
In"eat'e
Thi'kne~
0.008
0.006
r
0.004
:/
I
0.002
NOTE; Curve governed bv membrane stress intensitY (due to average circumferential tension
stress and average radial compression stress) limited by 1.1 Sm at 0.5 "radius X t either
side of junction; where radius 'sIRS+~tl2l cn evt lnder side and ( R
t 121f cos a 0., •cone
side
s+..
I
I
I
0.001
o
2
6
4
Maximum Angle,
10
8
UO
FIG. 3.11
INHERENT REINFORCEMENT FOR SMALL END OF CONE·TO·CYLINDER JUNCTION
(ASME VIII-2)
Large Shell
Again, nsing Eq. (2.36), we get
t = 100
~
Use t
=
x
7
x
12/(17,500 - 0.5
x
100)
0.48 in.
1/2 in.
Large End of Cone-to-Shell function
The PIS ratio is
PIS = 100/17,500 = 0.0057
From Fig. 3.9, maximum u = 19°. Hence, reinforcement is required. From Fig. 3.10, Q
required thickness at the large end of the cone-to-shell junction is
t, = Q
x
t ratio of shell stress to reinforcement stress (must be equal to or greater than 1.0)
1.30 X 0.48 X 17,500113,000
= 0.84 in.
Use t
7/8 in., distributed as detailed in Fig. 3.10.
1.30. The
Spherical Shells, Heads, and Transition Sections
3.4~
NOTE: Curves governed by membrane stress intensitY ldue to average circumferential teneton
stress end average radial compression stresst limited by 1.1 Sm at 0.5 J radius X tr
either side of junction, where radius is (R s + tr / 2) on cylinder side and IRs + tr / 21 cos
I
f\
I
a on ccne stde.
99
-+-+-+-1
I I II
t
3.0 ~-+~-t---+-+-l.......jH-++t----.l-_~_.L--'--'--'-+ti
I
',-0'
a· ....
30i\:
\
/'
t2.6 [ \
25~
'\
$'
"\r-. ","'-"
~
- , 20d..: \ " ,
2.2
"'
Q
_
15~
~
1.4
~
5d...
I"'"
)/'
.....
r-.....'
0.002
I
tt-
-,-rG\
'"
"."
I"--
~\:4Y.::'~ i
...
i'-............. t-... .... ~
14.r.;;-R 1
. v..,', :
~
L
e-\
IS:
'"-"\ ',-
_L_''pf4-~ \
t
,
_
-"I,I---~~--l
t--
0.003
0.008 0.01
0.005
0.02
0.03
0.05
0.07
0.1
PIS
FIG. 3.12
VALUES OF Q FOR SMALL END OF CONE-TO-CYLINDER JUNCTION (ASME VIII-2)
Small End of Cone-to-Shell Junction
Here the PIS ratio is
PIS
~
100115,000
~
0.0067
From Fig. 3.11, maximum (Y = 1.5'. Hence, reinforcement is required. From Fig. 3.12, Q
required thickness at the small end of the cone-to-shell junction is
t,
x
=
2.15
=
0.99 in.
0.40
Use t = 1 in., distributed as detailed in Fig. 3.12.
x
15,000/13,000
2.15. The
CHAPTER
4
FLAT PLATES, COVERS,
ANI> FLANGE.s
4.1
INTRODUCTION
Flat plates, covers, and flanges are used extensively in pressure vessels. Circular plates are used for most
applications; however, there are some applications where the flat plate is ohround, square, rectangular, or
some other shape. When a flat plate or cover is used as the end closure or head of a pressure vessel, it
may be an integral part of the vessel by virtue of having been formed with the cylindrical shell or welded
to it or it may be a separate component that is attached by bolts or some quick-opening mechanism utilizing
a gasketed joint attached to a companion flange on the end of the shell.
Flat plates and covers may contain no openings, a single opening, or multiple openings. To satisfy the
loadings and allowable stresses, the plate may need to be of an increased thickness or it may reqnire
reinforcement from attachments. The equations for the design of unstayed plates and covers based on a
uniform thickness with a uniform pressure loading over the entire surface are described in UG-34 of VIII-I
and Article D-7 of VIII-2. For flat plates and covers with either single or multiple openings, design
requirements are given in UG-39 of VIII-I and in AD-530 of VIIl-2. Also, design requirements are given
in Nonmandatory Appendix AA of VIIl-I for the design of tubesheets with multiple openings.
For the design of flat plates and covers which are attached hy bolting that causes an edge moment due
to the gasket and bolt loading actiou, both gasket seating loads and operating loads shall be considered in
a similar manner to that required for determining the acceptability of a bolted, flanged joint. Since the
loadings and dimensions required for analysis of bolted, flat plates and flanges in both VIIl-I and VIlI-2
are very similar, if not the same, they will be treated together. Spherically-dished covers are considered in
section 4.7 and in Appendix 1-6 of VITI-I.
4.2
INTEGRAL FLAT PLATES AND COVERS
Since the design rules of circular flat plates and covers is the same in VITI-I and VIII-2, only the references
in VITI-I will be nsed. Design rules for noncircular flat plates and covers are given only in VIII-I.
4.2.1 Circular Flat Plates and Covers
In UG-34, the minimum required thickness of a circular, flat plate which is integrally formed with or
attached to a cylindrical shell by welding or a special clamped connection is calculated by using the
following equation:
t
= d(CP/SE)'12
101
(4.1)
102 Chapter 4
where
d = effective diameter of the fiat plate, in.
C = coefficient between 0.10 and 0.33, depending on comer details
P = design pressnre, psi
S = allowable stress at design temperature, psi
E = butt-welded joint efficiency of the joint within the fiat plate
t = minimum required thickness of the fiat plate, in.
For further description of the terms given above, refer to Fig. 4.1 (a) through (i), or Fig. UG-34 of VIII-I.
Depending on the shape and welding details of the comer, a value of C is selected. A value for E, the buttweld joint efficiency within the fiat plate, is required if the diameter of the head is sufficiently large that
the head needs to be made of more than one piece. The value ofE depends on the degree of NDE' performed.
It is not a weld efficiency of the head-to-shell comer joint!
Example 4.1
Problem
Using the rules of UG-34 of VIII-I, determine the minimum required thickness of an integral fiat plate
with an internal pressure of P ~ 2,500 psi, an allowable stress of S = 17,500 psi, and a plate diameter
of d = 24 in. There are no butt welded joints within the head. There is a corrosion allowance, c.a.
3/16 in. The comer details conform to Fig. 4.1 sketch (b-2) assuming tbat m = I.
Solution
From Fig. 4.1, sketch (b-2),
C
= 0.33
d
= 24,
d,
=
24
X m
=
0.33(1)
+ (2)(c.a.)
= 0.33
= 24.375 in.
From Eq. (4.1),
t, = (24.375)[(0.33)(2500)/(17,500)(1)]'''
= 5.29 in.
t, = to + c.a. = 5.48 in.
Example 4.2
Problem
Determine the minimum corner radius to make Example 4.1 valid.
Solution
First, the minimum required thickness of the cylindrical shell, t; must be calculated using Eq. (2.1).
t, = PRJ(SE - 0.6P)
=
(2500)(12.1875)/(17,500 x 1 - 0.6 x 2500)
lNDE is the nondestructive examination of the butt joint.
=
1.904 in.
Flat Plates, Covers, and Flanges
tttt
;... Center of weld
,,..y~
Lttt\l
~
t
ts
time
r .. 3t
'!'
Taper
min.
I
d
I
..L-[J'.'"
t,mm -2t$
Tangent
I
.
C" 0.17
C-O.l10r
C" 0.10
"1ft:
1:'T
.....y!-
Tengen,
..., :,hne
',"3'
d ' m i n ." 0.25 t$ for
t
1 1/2·
mm.
l!
-
~
r .. 3t t
.
In,
$ /.
d
but need not be greater
than 3/4 in.
C"O.33m
C min ... 0.20
0"
d
~
t
,
.
~
t
d
0.71,
J
r,
leI
min. nor less than 1.25 t$
f,
.l
T
..~
Continuation
of shell optional
d
,hen'
Projection
,,"yoM weld
l'"
d
t
t
o o
o
C .. 0.13
'.
'r
'M oeee no, ........"
a7t
.';.<
LA
rmn.
C .. 0.30
~~I
t w <02
,e".,
mm,
_.
C"O.20orO.13
·'" "ITi fiff
fi m
1~11
leI
Cemer of lap
I
0 375 in
min·'
forts" 1·1/2 in,
T
t(
d
I
too-
~
J
103
t
~s optional.
Bevel optional
- 45 deg. max.
Sketches leI. ttl, and (91 circular covers. C'" O.33m, C min. '" 0.20
Id}
Ie)
IfI
See Fig. UW·13.2 sketches Ie} to {gl.
inclusive. for details of welded joint
t s not lets than 1.25t,
---j
.L
"rITl
d
•
See Fig. UW·13.2 sketches la) to (91.
inclusive, for detail of outside
welded joint
~ dJ
h
~
O'.l.rnt·
f"'
f:J
,
f
d
t
I
0
o
C = 0.33
0
C"0.33m
Cmin." 0.20
Ih}
_.rmt
1.1
.Reteiniog ,in.
Im
d
_
C"O.3D
c = 0.3
C- 0.3
:
(Use Eq, (21 or (511
I Use Eq. (21 Or (511
Ii)
Ik)
T h...... ,in.
I '
C" 0.30
1m}
C" 0.30
lo}
Inl
Seal weld
30 deg. mm.
rrs.
-+---1- t+I-
.S.
fZl~~~.~1:
C" 0.75
C" 0.25
Ipl
NOTE. When pipe threads are
used, set Table 00-43
1.1
C - 0.33
1<1
deg.max'~3/4tmin.
mu\.t1"rort$
wtllchever
is greater
:.
.
1
,
-oT
d
.
0.8ts min,
r.s
C" 0.33
lal
FiG. 4.1
SOME ACCEPTABLE TYPES OF UNSTAYED FLAT HEADS AND COVERS
G
104
Chapter 4
For t > 1 \12 in.,
rmi. ~ 0.25/,
= 0.25(1.904)
~
0.476 in.
4.2.2 Noncircular Flat Plates and Covers
When the fiat plate or cover is sqnare, rectangular, elliptical, obround, or any shape other than circular, the
minimum required thickness is calculated in the samemanner as for a circular plate,except for the addition
of a factor to compensate for the lack of uniform membrane support obtained in a circular plate, This factor,
Z,is related to the ratio of the length of the short dimension, d, to the length of the long dimension, D,
and is determined by
Z
~
(4.2)
3,4 - (2Ad/D) '" 2.5
Using this value of Z, the minimum required thickness of an integrally attached noncircular fiat plate is
calculated from the following equation:
(4.3)
/ = d(ZCPlSF/"
where all terms are the sarue as in Eq. 4.1.
Example 4.3
Problem
Using the rules in UG-34 of VIII-I, determine the minimum required thickness of the fiat end plate of a
rectangular box header which is 8 in. X 16 in. and has an internal pressure of P = 350 psi and S =
15,000 psi. The plate is integrally welded into place. There is no corrosion allowance and no butt-welded
joints in the plate.
Solution
(I) Using Eq. (4.2), determine the stress multiplier, Z, and use it to solve Eq. (4.3).
(2) From Eq. (4.2),
Z = 3.4 - [2.4(8116)J = 2.2
(3) Using Eq. (4.3),
t
~
8[(2.2)(0.33)(350)/(15,000)(1)]'''
~
1.04 in.
Example 4.4
Problem
Determine the maximum permissible length of a rectangular plate which is 8 in. wide by 1/2 in. thick and
has P = 100 psi, and S = 15,000 psi. There is no corrosion allowance, and no butt-welded joints in the plate.
Flat Plates, Covers, and Flanges
105
Solution
(I) Rearrange Eq. (4.3), above, to solve for Z as follows:
Z
= SEt'ICPd'
=
(15,000)(1)(0.5)' 1(0.33)(100)(8)'
= 1.78
(2) Rearrange Eq. (4.2), above, to solve for D as follows:
D = 2.4dl(3.4 - Z)
[(2.4)(8)]/[3.4 -
1.78]
11.9 in.
A plate that is 8 in. wide by 1/2 in. thick can have a maximum length of 11.9 in.
4.3 BOLTED FLAT PLATES, COVERS, AND FLANGES
Bolted connections are used on pressure vessels because theypermit easy disassembly of components. The
bolted connection may consist of a flat plate or so-cal1ed blind flange, a loose-type flange, or an integraltype flange. An early method to analyze bolted flange connections with gaskets entirely within the circle
enclosed by the bolt holes was developed by Taylor Forge in 1937 (Waters, 1937). These methods were
further developed by the Code Committee for use in various sections of the ASME Code. In their latest
form, flange rules are in Appendix 2 of VIII-I and Appendix 3 of VIII-2. Rules for pairs of flanges with
metal-to-metal contact outside of the bolt circle are given in Appendix Y of VIII-I.
4.3.1 Gasket Requirements, Bolt Sizing, and Bolt Loadings
Appendix 2 of VIII-l provides design rules for flanges under internal and external pressure. Determination
of gasket requirements, bolt sizing, and bolt loading is the same for a bolted f1ate plate or blind flange,
loose-type flange, and integral-type flange. Loadings arc developed for gasket seating or bolt-up condition
and for hydrostatic end load or operating condition. Guidance is given for the selection of the gasket and
design factors, m, the gasket factor, and y, the gasket unit seating load, psi. Once the gasket material and
sizing is determined, the bolt-up and operating loads are determined, bolting is selected, and the design
bolt loading is calculated.
4.3.1.1 Gasket Design Requirements. The selection of gasket type and material is set by the designer
after considering the design specifications. Once the gasket is chosen, the m and y factors may be selected
from Appendix 2 of VIII-I. The vaiues of m and y in the table are nonmandatory, and different values of
m and y, wbich are either higher or lower, may be used if data are available to indicate acceptability. After the
gasket material and type are selected, the effective gasket width, b, is determined by the fol1owing procedure:
1. The basic gasket seating width, b.; is selected from Table 2-5.2 of VIII-I.
2. When b., :5 1/4 in., b = bo, and when b., > 1/4 in., b = O.5(bo) 1I2
From b, the values of G and he can be determined.
106
Chapter 4
4.3.1.2 Bolt Sizing and Bolt Loadings. The required bolt load for operating condition, Wm1, is
determined as follows:
Wm1
= H + H, = O.785G'P +
(2b x 3.14GmP)
(4.4)
The reqnired bolt load for gasket seating condition, Wm" is determined as follows:
W", = 3.14bGy
(4.5)
Once the required bolt loads are determined, the required bolt area for each loading coudition can be
calculated as follows:
(4.6)
and
(4.7)
The total required cross-sectional area of the bolts, Am' is the greater of A m1 or A m2 • The bolts are selected
so that A b , the actual bolt area, is equal to or greater than Am. The bolt load used for the design of flanges,
W, is theu determined from the following.
For operating condition,
(4.8)
For gasket seating condition,
W = O.5(A m
+ Ab)S,
(4.9)
wbere S, shall be not less than the allowable tensile stress value in II-D, psi.
In addition to safety, Eq. (4.9) provides some protection from overbolting during gasket seating at
atmospheric temperature before the internal pressure is applied. Where additional protection is desired or
required by the design specifications, the following equation is used:
(4.10)
4.3.1.3 Check for Gasket Crushout. Although not considered in VIII-I, it is prndent to design
against crushout of the gasket by determining the minimum gasket width using the following:
(4.11)
with changes of m and! or y being permitted, as described in 4.3.1.1.
4.4 FLAT PLATES AND COVERS WITH BOLTING
4.4.1
Blind Flanges & Circular Flat Plates and Covers
Before calculating the minimum required thickness of a blind flange, flat plate, or cover, determine if there
is a suitable blind flange available from any of the flange standards listed in Table U-3 of VIII-I and Table
AG-150.1 of VIII-Z. Staudard flanges are acceptable without further calculations for diameters aud pressure!
Flat Plates, Covers, and Flanges
107
temperature ratings in the respective standards when of the types shown in Fig. 4.1, sketches (j) and (k).
When there is no standard flange available, the minimum required thickness of tbe circular flat plate is
calculated by using the following equation:
+
, = d[(CP/SE)
(1.9WhGISEd')]112
(4.12)
where the definitions of terms are as given in 4.2.1 and the determination of Wand h G is as given in 4.3.1.1.
4.4.2
Noncircular Flat Plates and Covers
When the flat head is square, rectangular, elliptical, obround, or some other noncircular shape and utilizes
bolting, the minimnm required thickness of the noncircular flat head is calculated by using the following equation:
t = d[(ZCP/SE)
+
(6Wh G/ SELd 2)Jll2
(4.13)
where the terms are as explained in earlier sections.
4.5 OPENINGS IN FLAT PLATES AND COVERS
Rules for compensation or reinforcement required for openings in flat plates and covers are given in
UG-39 of VIII-I. Single, small openings which do not exceed the size limits UG-36(c)(3)(a) & (b) and are not
greater than one-fourth the plate or cover diameter are integrally reinforced and do not require reinforcement
calculations. Note: Particular care should be taken when standard blind flanges are used so as to not exceed
the size permitted by the Standard. When the opening size exceeds these limits, the pressure/temperature
ratings are no longer valid and calculations are required.
4.5.1
Opening Diameter Does Not Exceed Half the Plate Diameter
For a single opening when the opening diameter does not exceed half the plate diameter, standard reinforcement calculations may be made, keeping in mind that the total reinforcement required is:
A,
~
O.5dt,
(4.14)
where
AT
d
= required area of reinforcement, in.'
= opening diameter in the flat plate, in.
t =
minimum required thickness of the flat plate, in.
The reason for the factor of 0.5 instead of 1.0 is that, unlike the situation in cylindrical shells and formed
heads, the stress distribution through the thickness of a flat plate is primary bending stress instead of primary
membrane stress.
For multiple openings in which no diameter is greater than half the plate diameter, no pair of openings
has an average diameter greater than one-fourth the plate diameter, and the spacing between pairs of
openings is no less than twice the averagediameter, Eq. 4.14 for the minimum required thickness of a
plate with a single opening may be used. If the spacing between pnirs of openings is less than twice the
average diameter, but not less than 1V4 the average diameter, the amount of reinforcement is calculated by
108 Chapter4
adding the required reinforcement of the pair to no less than 50% of the required reinforcement hetween
the pair. If the spacing is less than I V4 the average diameter, U-2(g) applies. In all cases, the width of the
ligament hetween the pair of openings shall be no less than one-fourth of the diameter of the smaller of
the pair, and the edge ligament between an opening and the edge of the plate shall be no less than
one-fourth of the diameter of that opening.
In VIII-I, as an alternative to the rules for reinforcement of a single opening, the following procedure
may be used to determine the minimum required thickness of a flat plate.
1. In Eqs. (4.1) and (4.3), a value of 2C or 0.75, whichever is less, may be used, except for
sketches (b-I), (b-2), (e), (f), (g), and (i) of Fig. 4.1, for which 2C or 0.50, whichever is
less, must be used.
2. In Eqs. (4.12) and (4.13), the quantity under the square root sign must be doubled before
solving for t.
As an alternative to the rules for multiple openings, when the spacing for all pairs of openings is equal
to or greater than twice the average diameter of that pair, the alternative rules for single openings may be
used. When the spacing is less than twice but equal to or more than 11/ 4 the average diameter of the pair,
the required plate thickness shall be determined by the alternative rule for single openings multiplied by a
factor, h, that is defined as
h = (0.5/ e)1n
(4.15)
(4.16)
where
e = smallest ligament efficiency of all pairs
p = center-to-center spacing of a pair to get e
davg = average diameter of the same pair to get e
Again, in all cases, the width of the ligament between the pair of openings shall be no less than one-fourth
of the diameter of the smaller of the pair, and the edge ligament between an opening and the edge of the
plate shall be no less than one-fourth of the diameter of that opening..
4.5.2 Opening Diameter Exceeds Half the Plate Diameter
When the opening is a single, circular, centrally-located opening in a circular flat plate, the plate shall be
designed according to Appendix 14 of VIII-I.
For small openings which are located in the rim of the flat plate surrounding a large opening, as shown
in Fig. 4.2, and the plate is to be analyzed according to Appendix 14, the rules in section 4.5.1 (above) for
single openings and for multiple openings shall be followed, using as the minimum required thickness a
plate thickness which satisfies the rules of Appendix 14. As an alternative, the thickness determined according
to Appendix 14 may be multiplied by v'2 = 1.414 for a single opening in the rim or for multiple openings
which satisfy the ligament spacing for a flat plate which had its thickness set by the v'2 rule.
4.6 BOLTED FLANGE CONNECTIONS WITH RING TYPE
GASKETS
As described earlier in section 4.3, Appendix 2 of VIIl-I and Appendix 3 of VIII-2 contain rules for the
design of bolted flange connections with gaskets entirely within a circle enclosed by the bolt holes. Selection
Flat Plates, Covers. and Flanges 109
u"
l~
p • spaclhll. ceMel{O-¢IIhter. be\Wfth Ope<Mhll'
Uj, . .. • IipMm widttl
• dan • -801 di.....
of pili, of opIt'lin;a
IIIf.,
FIG. 4.2
MULTIPLE OPENINGS IN THE RIM OF A FLAT HEAD OR COVER WITH A LARGE
CENTRAL OPENING
of the gaskets and determination of the bolt sizes, bolt loading, and loading moment arms are obtained in
the sarne manner for flat heads or blind flanges as they are for integral and loose flanges. Using these
loadings and moment arms, gasket seating moments and operating moments are determined and stresses
are calculated and compared with allowable stress values. The procedure for calculating stresses and
acceptability of stresses is essentially the sarne for welding neck flanges as for loose, slip-on, or ring-type
flanges. For Code consideration, there are three types of flanges: loose, integral, and optional (see Fig. 4.3).
Loose means that, for calculations, the flange ring provides the entire strength of the flange, even though
the ring is attached to the vessel or pipe by threads or welds. Integral means that, for calculations, the ringand-shell or ring-and-pipe combination provides the strength of the flange, and the assumption is made that
the connection between these parts has enough strength that they act together. Optional means that the
connection is basically integral; however, it may be calculated as loose, which requires only one stress to
be calculated. The difference in these various flanges is the line of load application and the magnitude of
the loads. However, the applied moments are determined in a similar manner.
4.6.1 Standard Flanges
Typical of the loose-type flanges are the ring flange, which usually is made from a flat plate formed into
a ring, and the slip-on or lap-joint flange. Either of these may have a hub, but the weld size and strength
are not great enough to have the flange and vessel or pipe act together.
110 Chapter 4
Loadings are determined in the same manner as described in sections 4.3.1 and 4.4.1. Example 4.5
presents an example problem and filled-in Sample Calculation Sheet for a ring flange with a ring-type
gasket. See Appendix D for blank fill-in Sheet D.I (Ring Flange with Ring-Type Gasket) and blank fillin Sheet D.2 (Slip-on or Lap-Joint Flange with Ring-Type Gasket).
Integral-type flanges usually contain a tapered hub and flange ring which may be integrally formed or
the hub and neck are welded together and act integrally so that each carries part of the loadings. Example
4.6 gives an example problem and filled-in sheet for a welding neck flange with a ring-type gasket. See
Appendix D for blank fill-in Sheet D.3 (Welding Neck Flange with Ring-Type Gasket).
Example 4.5
Problem
Using the rules in Appendix 2 of Vlll-l, determine the minimum required thickness of a ring flange, shown
in Fig, E4.5, with the following desigu data:
Design pressure = 2000 psi;
Design temperature = 650"F;
Flange material is SA-105;
Bolting material is SA-325 Gr. 1;
Gasket is spiral-wound, fiber-filled,stainless steel, 13.75 in. l.D. X 1.0 in. wide;
No corrosion allowance.
1
2.000 1'5L
'SA -105'
13.15"1.0." ,"wid,c,.
SA-32S' Gr-i Gc l>,i'+(ZX'')-(2J<.~~ l5',d
flong_ Mat.rlol
&oiling Material
No Cer re ~j"o .... 4
COi'tosion AUowanee
&011;"'0
-
1"1,100
:ZOO, >'00
A.
O,d,," Temp., Sit
2(),100rf<
N
.:.. (;%.,..'/4 _
W
20,106 rs-'
3",500
Atm. Temp., S.
w.., _
CONDITION
5
LOAD
-b'll"Gy
HI'
+H=
HO=W.I-H_
200,5'00
he _ .5fC- Gl
Hr=H-Hll=
100, '00
HG=W
=
liT - .5( hD
':;102,5'00
.)
+ fIG}
1,,_ = .51C- Ol
o
,"f-c
j2A.
cZ2.S'~
26,5"
ho
B'
6
S:~81 I H-~
1-",. _~Hr
20@ \ ~ ;8.'t!
I "
I
12J5"
G'.
1>'0'1)
21.5
'/:>,'
MOMENT
~
=
4.815
?>,12'l
Me
= Haho
'i,,0l-
Mr
=
Mo
=
~.12q
SHAPE CONSTANTS
H,nr
M,
K = AlB
+
1,1'.1.\5,000
Hohll ~
~
141,100
'l?>1.2.00
2, L{ 'l'i, 000
2,OQ'O,000
• 2.018
r= 2,812-
If bolt ,pacinG ucuch 20
I, multiply ~ 8011 IPOein;
M. end ~ in I equotionl by:
2a
I
"H
~:
no;
:;o:r w..,/S. or w..l/S~_ 8,3 or
= 2S,2
ZO~ \
q>
= .5("' .. + A.15. _
><'2,,"00
M.
S.OI'''';
15',OQ3
\0,000
3,0
A.. _
LEVER ARM
X
hD _ .51C
7:11' / 4
G
y
m
'5~;'G, 0 00
2S'5, '100
H.
Operating
l
- 2b...G mf'_
11,&DOr" w..,
J
1.0
LOAD AND BOLT CALCULATIONS
H,
A'm. Temp..
N
•
Ro.i sed. f ..<e.
S,. 20 oaD " .i.
OeNonTemp.. S,.,
Flange
3
fACE
GASKET
Sf'ira.\ Wo... Kd. "lIe"'..I,
.j:; ~.... f; \ \ed.,st.;,,,le...~(
65'0 OF
De$i;n Temp,rat"".
il
I-
2
DESIGN CONDmONS
0.';0" Prellur•• ,.
7
+
OPERATING
.
~ ~='fi,l{
.=
t=
OF
SEATING
GREATER
.,
,=~
M.r ~~ gj
SlaB
•
•
•
COftIputed_ _ _ Dote
Chechd
FIG, E4.5
RING FLANGE SAMPLE CALCULATION SHEET
Number
)
Flat Plates, Covers, and Flanges
111
Solution
(1) The allowable tensile stress of the bolts from II-D at gasket seating and operating conditions (design
temperature) is S, = S, = 20.2 ksi.
(2) The allowable tensile stress of the flange from II-D at gasket seating is Sf' = 20.0 ksi and operating
conditions is Sf' = 17.8 ksi.
(3) The diameter of the gasket's line-of-action, G, is determined as follows:
bo
= NI2 =
0.5 in.
G = 13.75
(4) With N = I, b
of bolts are
=
0.3535, m
=
+
and
= 0.5(bO) 1I2 = 0.3535
(2 X I) - (2 X 0.3535)
=
3.0, and y
H
b
=
in.
15.043 in.
10,000, the bolt loadings and the number and diameter
= (7f/4)G'p
= (7f14)(15.043)'(2000)
Hp
=
355,5001b
=
2b'ITGmp
=
2(0.3535)7f(15.043)(3.0)(2000)
= 200,500
lb
= 355,500
+ 200,500
=
556,000 Ib
Wm2 = 7fbGy
=
7f(0.3535)(15.043)(10,000)
=
167,1001b
or
Wm2/S, = 556,000/20.2 = 27.5 in.'
Using 20 bolts of I V2-in. diameter,
Ab = actual bolt area
= 20(1.41) = 28.2 in.2
= 0.5(27.5
+ 28.2)(20,200)
112 Chapter 4
Wo
=
562,6001b
=
Wm1
= 556,000
lb
(5) Using Eg. (4.10), the gasket crushout width is
=
(28.2)(20,200)/2(10,000)(7T)(l5.043)
= 0.6 in. < 1.0 in. actual
(6) The total flange moment for gasket seating condition is:
Flange Load
=
HG
Wo
=
562,600 lb
Lever Arm
hG = 0.5(C - G)
=
3.729 in.
Flange Moment
= (562,600)(3.729)
= 2,098,000 in-Ib
(7) The total flange moment for operating condition is:
Flange Loads
HD
(7T/4)B'p
=
= (7T/4)(l2.75)'(2000)
=
HG
255,4001b
= H; =
200,500 lb
Flat Plates, Covers, and Flanges
H; = H - H p
= 355,500 - 255,400 = 100,100 Ib
Lever Arms
hD
=
0,5(C - B)
= 0.5(22.5
hG = 0.5(C -
- 12.75)
h, = O.5(h D
=
4.875 in.
G)
0.5(22.5 - 15.043)
=
=
+
= 3.729 in.
hG)
0.5(4.875 + 3.729)
=
4.302 in.
Flange Moments
= (255,400)(4.875) = 1,245,000 in.-lb
= (200,500)(3.729) = 747,700 in.-lb
= (100,100)(4.308) = 431,200 in.-lb
=
2,424,000 in.-lb
(8) Shape factor from Appendix 2 of VIII- I for
K
=
AlB
=
26.5/12.75
=
2.078
From Fig. 2-7,1 of VIII-I, Y = 2,812,
(9) The minimum required thickness of the flange is the larger tmi" of:
For gasket seating condition:
'mi"
=
[(MGsY)/(Sj)l)]in
=
[(2,098,000)(2.812)/(20,000)(12,75)] in
=
4,81 in.
113
114 Chapter 4
For operating condition:
[(2,424,000)(2.812)/(17,800)(12.75)]1n
~
5.48 in.
Example 4.6
Problem
Using the rules of Appendix 2 of VIII-l, determine the minimum required thickness of'aweldingneck
flange, of the type shown in Fig. E4.6 with the following design data:
Design pressure = 2000 psi;
Design temperature
=
650oP;
Flange material is SA-lOS;
Bolting material is SA-325 Gr. I;
Gasket is spiral-wound, fiber-filled, stainless steel, 13.75 in. J.D. X 1.0 in. wide;
No corrosion allowance.
Note: This flaoge has facing details, gasket size, and bolting that are the same as those given in Example
4.S, except that this is a welding neck flange instead of a ring flange.
Solution
(I) The allowable tensile stress of the bolts from IT-D at gasket seating and operating conditions (design
temperature) is S, = S, = 20.2 ksi.
(2) The allowable tensile stress of the flange from IT-D at gasket seating is Sf' = 20.0 ksi and operating
conditions is Sf' = 17.8 ksi.
(3) The diameter of the gasket's line-of-action, bolt loadings, bolt number and diameter, and crushout
width are the same as in Steps 3-5 of Example 4.5.
(4) The total flange moment for the gasket seating condition is the same as in Step 6 of Example 4.5.
M GS
~
2,098,000 in.-lb
(5) The total flange moment for the operating condition is:
Flange Loads
H D ~ (-rr/4)B'p
= (-rr/4)(10.75)'(2000)
He
~
181,5001b
~
H, =
200500lb
H, = H - H o
= 355,500 -
181,500 = 174,000Ib
Flat Plates, Covers, and Flanges
115
FACE
1.0
b
G
'."i", M.,.,i.,
CorrOI;on AUO"'<>tlc~
SA-32'S Gr.1 6- e, 1:'.'5 "2-.101<\!>.O~'"
N Q Col"yo'$ ioV\. 4
LOAD AND 10LT CALCULATIONS
CONDITION
5
LOAD
H.
Operating
I4
'l!"Il I P
He -W.. l-ti
-
Hr =H - tlo =
LEVER ARM
X
.. I~l, 70
:!-OO
0
IILj,o<:>O
3.0
MOMENT
ho _ 11+ .5g\
4. II!>
hG -
'!>.l'2.
hr
.5lC -Gl
= ..5ll! +
I!'
+ h(;1
=
Mo
'"0,
\
1'1 ,100
Hohp
00
4. 02
2,343,600
M.
Hc=W
he;
= .51C -
Gl
3.12'1
K AND HUB fACTORS
STRESS CALCULATION-Operating
«
S,.
S,.
Tong. fIg., ST
'i '{QO
\4 4~O
_m.Y/r' -2S. -
:;·.~'·.51SH + S~)or.51SI!
+ Sr)
-
STRESS CALCULATION
1.5 51.
51.
S,.
S,.
long. !i<Jb, 59
fm(l/Ag,1l
Radial fIg., Sa -
{3ma/Afl
2·~"5
A/,
f
V
$;"
z
\."q
r~j.,
y
2."1..'1
sr/s,
h.
...}89.
7
Tang. fIg" ST -11\0 V!I'-ZSH
F/h~
"!>."l>1S
3-'11'\
d
Vhog.1
u
0.114
207.7l:>
11.0
+1
{3 = 4/3 te
"t
'>4'i'1I--"-1+--B' 10.1 5 "
1.0
e _
STRESS FORMULA FACTORS
a _Ie
9,· 3.,,15"
0.04
2.',I
u
$ealing
t. '\0("
0.91
hl"~
1.315
+I
\·25,,",
_a/T
rno _ Mo/S _ operollng_
mo ::::: MQ/E
= ,eating =
%IB 000
+
If bolt .padng flxceedt 20
f, multiply
mo end flIQ in above equaliont by:
..
Bolt ~pgc;ing
2"
f
+
~
GwW Taylor·Bonney DiViSion.
"4.0"
LI.--.!-t--'--~,..-'
"
Compvted'
001.,,
Ch.,ckeO
Number
FIG. E4.6
WELDING NECK FLANGE SAMPLE CALCULATION SHEET
_
116 Chapter 4
Lever Anns
hD
=R +
0.58,
= 2.5 +
= 4.188 in.
0.5(3.375)
ho = 0.5(C - 0)
= 0.5(22.5 -
+ 8, +
hr = O.5(R
15.043) = 3.729 in.
hG)
+ 3.375 + 3.7285)
= 0.5(2.5
~
4.802 in.
Flange Moments
Mo
=
(181,500)(4.188)
=
Ho
~
(200,500)(3.729) = 747,700 in-Ib
~
(174,000)(4.802) = 835,500 in-lb
X
~
760,100 in-lb
ho
Mo = MD + Mo
+ MT
= 2,343,000 in-Ib
(6) Shape factors from Appendix 2 of VIII-I for K are
K
~
AlB
~
26.5/10.75
~
2.465
From Fig. 2-7.1 of VIII-I,
T
=
1.35 Z
8,/80
~
~
1.39
y
~
3.375/1.0
2.29
=
U
3.375
= [(1O.75)(1.0)J'" = 3.279
hlhD = 6.25/3.279 = 1.906
~
2.51
Flat Plates, Covers, and Flanges
117
From Appendix 2 of VIII-I,
F
= 0.57
e
= 0.04 f =
V
1.0
= F/h o
= 0.57/3.279 = 0.174
d = (UlV)hog~
= (2.51/0.04)(3.279)(1)2 = 205.76
= 2,098,000 in.-Ib and Sf' = 20.0 ksi and M, = 2,343,000 in.-Ib and Sf' = 17.8 ksi. Since at
gasket seating condition, the moment is smaller and tbe allowable stress is larger, only the operating
(7) M GS
condition is calculated.
Assume a flange thickness of t = 4.0 in.
L = [(Ie
=
+
1.256
1)/(7)
+
+ (I)'/a]
0.311
=
1.567
Longitudinal hub stress:
= (1)(2,343,200)/ (1.567)(3.375)'(10.75)
=
12,210 psi
Radial flange stress:
SR = [(4/3)le
+
1]M,ILt 2B
=
(1.928)(2,343,200)/ (1.567)(4)'(10. 75)
=
16,760 psi
Tangential flange stress:
Sr
=
[(YM,/ ,'B) - ZSR]
= {[(2.29)(2,343,200)/(4)'(l0.75)] - (1.39)(l6,760)}
=
7900 psi
I
I
118
Chapter 4
Combined stresses:
O.5(SH
0.5(SH
+
+
SR) =
S,) =
0.5(12,210 + 16,760) = 14,490 psi
0.5(12,210 + 7900) = 10,060 psi
(8) Allowable stresses:
SH ,; 1.5Sf:
12,210 psi < 26,700 psi
S, ,; Sf:
16,760 psi < 17,800 psi
S,'; Sf:
7900 psi < 17,800 psi
0.5(SH
+
SR) '" Sf:
14,490 psi < 17,800 psi
O.5(SH
+
S,) ,; Sf:
10,060 psi < 17,800 psi
Since all actual stresses are less than the allowable stresses, the selection of t ~ 4.0 in. is adequate. If an
optimum minimum thickness of the flange is desired, calculations must be repeated with a smaller value
of t until one of the calculated stresses or stress combinations is approximately equal to the allowable stress,
even though other calculated stresses are less than the allowable stress for that calculated stress.
4.6.2 Special Flanges
Rules for special flanges with different geometry and/or loading are given in Appendix 2 of VITI-I. Included
are: 2-9 for split loose flanges, 2-10 for noncircular shaped flanges with a circular hore, 2-11 for flanges
subject to external pressure, 2-12 for flanges with nut-stops, and 2-13 for reverse flanges. Flanges with
other geometry and loading shall follow U-2(g).
4,6.2.1 Reverse Flanges. Reverse flanges are described in Appendix 2-13 of VITI-I. They are similar
to standard flanges, except some of the loads on the flange ring cross section may he applied at different
locations and in a reverse direction, possibly causing a reverse moment. Vlll-I has chosen to use the term
aD to convert a standard flange to a reverse flange. Example 4.7 gives an example problem and a filled-in
sheet for a reverse welding neck flange with a ring-type gasket. See Appendix D for a blank fill-in Sheet
DA (Reverse Welding Neck Flange with Ring-Type Gasket).
The method of analysis for a reverse flange is similar to that used for an integral flat head with a large,
single, circular. centrally-located opening, as given in Appendix 14 of VIII-I. For both analyses, a special
limitation of the geometry is given. When K :5 2, calculated stresses are acceptable: however, when K
> 2, calculated stresses become increasingly conservative. For this reason, use of the analysis procedure
should be limited to K :5 2.
Example 4.7
Problem
Using the rules of Appendix 2 of VIIl-l, determine the minimum required thickness of a reverse welding
neck flange, shown in Fig. E4.7, with the following design data:
Flat Plates, Covers, and Flanges
1
G50" f
<JA ~ \05
""'''' T......."'"
..",.,"'••••,.,
Conoliof\AllowaN:.
flong.
GASKET
fACE
SriV'Q.\ w."w.\. \.oeo1e;1:...\,
2 1 0 0 0 YIA
"_g. Mat.rial
•
2
DESIGN CONDmONs
o.MllI n Pr••..,r •• ,.
tiM.f;lI...!., st.'.W!SHt"" I,
\;'15" I .P ?t \ " w'.o\e.
G, \?1S"+Z-.1d1, 1<;.O~3U
SI\·~25 ur.1
No Cor'(0i'U:I1'''L 4
o.lign Temp., 5,. \1.e-OD,~
Rll.iSE.d..f~
b
G
LOAD AND BOLT CALCULATIONS
W..1
20000 "'.
119
\00
brGy
'~1
2b.c.' -
1.00, SOD
:;.~. W.2/S.
.A.
~.'3 or Z1.5
20@ 1Y-z" 4>
or W.l/S. _
.. - 28.:L
i11---+-:------:=;;:,-'~~;=.-r--="=-----;;'~~~-I-"'--'="-:------;;~;:-T*-'----1
OelignTemp.,S. '1o,'ZOO"IA H _GlTPI4 ~S5'. . '500
w -.51A.+~ls.- '502>bOO
A.... T...... S..
:I
Bolting
H.
-
r.::.::c::-:::::c::=~"-;;:?'5+C-:--=-::..:.:;.c.:.-=--_;:;;'zT~_SE_+"-'::..::=-'..;""''-=-=~''-''='----AIm. Temp••S.
CONDITION
'2t> 200PfA
W.I - H,
LOAD
+ H_
X
5'"5'fvJooO
=
LEVER ARM
MOMENT
5
H,_W••-H=
H,_H
H.=
1.00,500
-llbb,f.jOO
'._.S(C-G)
'5.12.'1
h,-.'(C-'t G)_ \.1'1\
1'!1100
""_H,h, M,_H,,",
-SU"OOO
-
Seating
STRESS
CALCULATION-Dperat~·ng
6
K AND HUB FACTORS
1.5 5,.
S,.
9
1.5
00
h/h.
F
=
v
s,
-
I
-
S,.
•
Tong.F1Q.. ST- moYaN -ZSIl.IO,67te+lj/.B=
Z
\.C.1
-
,)
.....
S,.
S,.
RadiaIFIIlI·.S.
Pmcnt·
Tal'lQ. Fig., Sot
may.N
Tang. fIg .. St {AT 8'l
=7
?',1 eo ffJ..
L\,.'-\O
Z5R(O.67Ie+l)/,B
[V
f'f;'"
~~:(lt~·te)J =- \1,0~Of#.
- \·1,41
1.3Mo
91/go
1.0
•• - v;;:g;; ".9~O
STRESS FORMULA
7
,
a. .=-te + 1
.8=.4/3t_+ 1
r ;;:; «ITt!
6 _,'/d
l_Y+5
m.
h:
0
L[>
hr+Hr
1.~I'2.S"--.gof4--': 2.2..~"S'
20- I ~'~8""
::
A"
2(,.5"
\.0
O.I.\\q
0.851
All....l>l•
S,.
S,.
S,.
0.<;<;0
O.I~I
Fill•
]
MO'l
_M./8'
d
u~ h
I
V 090
f6.~~3
FACTORS
- 4.0
= I. !>2'!
= I. ~"'1
= 1.11$
-
1.1:11
2.'10"-
-
~
~
Comput.d
Check.d
_
Oal.
Number
FIG. E4.7
REVERSE WELDING NECK fLANGE SAMPLE CALCULATION SHEET
_
120 Chapter 4
Design pressnre = 2000 psi;
Design temperature = 650 o P;
Flange material is SA-I05;
Bolting material is SA-325 Gr. I;
Gasket is spiral-wound, fiber-filled, stainless steel, 13.75 in. LD. X 1.0 in. wide;
No corrosion allowance.
Note: This flange bas facing details, gasket size, and bolting that are the same as those given in
Example 4.5; however, this is a reverse welding neck flange with different flange dimensions.
Solution
(I) The allowable tensile stress of the bolts from lI.D at the gasket seating and operating conditions (design
temperature) is S, = S» = 20.2 ksi.
(2) The allowable tensile stress of the flange from lI-D at the gasket seating is Sf' = 20.0 ksi and operating
conditions is Sft> = 17.8 ksi.
(3) The diameter of the gasket line-of-action, bolt loadings, bolt number and diameter, and the crushont
width are the same as in Steps 3-5 of Example 4.5.
(4) The total flange moment for the gasket seating coudition is the same as in Step 6 of Example 4.5.
MGS
=
2,098,000 in.-Ib
(5) The total flange moment for operating condition is:
Flange Loads
HD
=
('IT/4)B'p
=
('IT/4)(22.875)'(2,000)
=
H,
=
~
~
556,000 - 355,500
821,900 Ib
200,500 Ib
H - Hn
= 355,500 - 821,900
~
-466,400Ib
Lever Arms
ho =
~
=
0.5(C + g, 0.5(22.5
+
2go - B)
1.8125 - 2 X 1.8125 - 22.875)
-1.094 in.
he = 0.5(C - G)
~ 0.5(22.5 -
= 3.729 in.
15.()43)
Flat Plates, Covers, and Flanges
+
hr = 0.5[C - 0.5(B
G)]
= 0.5[22.5 - 0.5(22.875
=
+ 15.043)]
1.771 in.
Flange Moments
=
(821,900)( -1.094) = - 899,200 in-Ib
=
=
(200,500)(3.729) = 747,700 in.-lb
(-466,400)(1.771)
=
-826,000 in-lb
- 977,500 in.-lb
Use the absolute value in the calculations.
(6) Shape factors from Appendix 2 of VIII-l for K are
K
=
AlB' = 26.5/13.25 = 2.0
From Fig. 2-7.1 of VIII-I; assuming fL = 0.3:
T
=
1.51
Z
=
a, = (11K'){l
Y
1.67
+
[3(K
= 2.96
+
+
=
0.857
=
1.241
fL)(Z -
=
3.26
1)(1 - fL)]/1TY)
= 0.419
T, = [(Z
U
fL)] a,T
121
122
Chapter4
~
1.366
= [(26,5)(1,8125)]'" ~ 6,930
From Appendix 2 of VIII-I:
F
~
0,909
f
V = 0550
= 1,0
e = Fl h,
= 0,909/6,930
d
~
~
0,131
(U,IV)h,g;
= 56543
= 2,098,000 in-Ib and Sf' = 20,0 ksi and M, = 977,500 in-Ib and Sf" = ]7,8 ksi. Since the
moment at operating condition is less than 0.5 times the moment at gasket seating condition with a
slightly less allowable stress, only the gasket seating condition is calculated,
Assume a flange thickness of t = 4,0 in,
(7) M GS
L ~ «(Ie
= 1,778
+
+
1)/T,]
+
(l"ld))
Ll31 = 2,909
Longitudinal hub stress:
= [(1)(2,098,000)] I [2,909)(1,8125)'(13,25)J
~
16,570 psi
Flat Plates, Covers, and Flanges 123
Radial flange stress:
SR ~ ([(4/3)te
+
I]MGsl/Lt'B'
~
[(1.699)(2,098,000)] I [(2.909)(4)'(13.25)]
~
5780 psi
Tangential flange stress:
5,
~ [(YRMGs/t'B') -
2S,(0.67te
+
I)]/~
~
{[(1.24I)(2,098,000)1 (4)'(13.25)] - [(1.67)(5,780)(1.351)] 11(1.699)
~
4610 psi
Combined stresses:
0.5(SH
+
S,)
O.5(SH
+
S,) ~ 0.5(16,570
~
0.5(16,570
+
5,780)
+ 4610)
~
1l,180 psi
= 10,590 psi
Tangential flange stress at B':
S; ~
(M Gslt 'B')[Y - [2K '(0.67te
+
1)/(K' -
I)L])
[(2,098,000)1 (4)'(13.25)]{2.96 - [2(2)'(1.351)1 (3)(2.909)])
17,040 psi
(8) Allowable stresses:
SH'" 1.5Sf
:
16,570 psi
<
26,700 psi
S, '" Sf:
5780 psi
< 17,800 psi
S, '" Sf:
46JO psi
<
17,800 psi
0.5(SH
+
S,) '" SJ'
ll,180 psi
< J7,800 psi
0.5(SH
+
S,) '" Sf:
10,590 psi
< 17,800 psi
S; '" Sf:
17,040 psi
<
17,800 psi
4.6.2.2 Full-Face Gasket Flanges. Although Fig. 4.1 (p), shows a flange with a full-face gasket
which perntits part of the gasket to lie outside the bolt circle, no design procedure exists for such a flange.
This type of gasket may be used with either a loose or an integral flange. See Appendix D for blank fillin Sheets D.5 (Slip-on Flange with Full-Face Gasket) and D.6 (Welding Neck Flange with Full-Face Gasket).
124 Chapter4
One of the basic differences with a full-face gasket is that a reverse moment is generated from tbat part
of the gasket loading outside the bolt circle.
Most often, a full-face gasket is used where the m aud y factors are relatively low, so that the bolt loading
is kept within acceptable limits. A full-face gasket design generally results in the total moments from gasket
seating aud from operation to be fairly low, and consequently, only a nominal flauge thickness is required.
However, bolt loads are usually higher.
4.6.2.3 Flat-Face Flange with Metal-to-Metal Contact Across the Face or at the Outer Edge. Appendix Y of VIII-I contains rules for the design of a flat-face flange with metal-to-metal contact across
the whole face or with a metal spacer added to the outer edge between pairs of flauges. Gasket loadings
usually are small, as most gaskets are of the self-sealing type. Inorder to make an analysis easier, assemblies
are classified and individual flanges are categorized. Once this is established, the rules for analysis are
given in VIII-I.
Classification of Assemblies
Class I:
Class 2:
Class 3:
A pair of flanges which are identical except for the gasket groove
A pair of nonidentical flauges in which the inside diameter of the reducing flange exceeds half
the bolt circle diameter
A flange combined with a flat head or a reducing flauge with au inside diameter that is small
and does not exceed half the bolt circle diameter
Categories of Flanges
Category I:
Category 2:
Category 3:
An integral flange or an optional flange calculated as an integral flange
A loose-type flange with a hub that is considered to add strength
A loose-type flauge with or without a hub-or au optional type calculated as a loose
where no credit is taken for the hub
type~
The analysis of au Appendix Y flange is similar to that made for an Appendix 2 flange, except for the
additional load and moment caused by the contact or prying effect. The contact force, He, aud its moment
arm, he, involve an interaction between the bolt elongation and the flange deflection and the moments from
the bolt loading and pressure loading.
The bolt loading for the operating condition is
WmI
=H +
He + He
(4.17)
4.7 SPHERICALLY DISHED COVERS
Rules are given in Appendix 1-6 of VIII-I for designing spherically dished covers with a bolting ring,
acting in the same manner as a flange ring, attached integrally to a segment of a sphere. The equations
given in VIII-l are approximate and may be conservative because they do not take into account the
discontinuity condition which exists at the intersectiou of the ring and head aud that would distribute forces
aud moments between the two parts relative to theirresistauce (stiffness). The Code procedures and equations
assume the entire loadings at the intersection are taken by the bolting ring alone. The rules of VIII-I permit
the use of a discontinuity analysis, should the designer so choose.
Flat Plates, Covers, and Flanges
125
4.7.1 Definitions and Terminology
Symbols and terms used for spherically dished covers are
t = minimum required thickness of the spherical head segment, in.
L = inside spherical or crown radius, in.
P = internal design pressure or MAWP, psi
S = maximum allowable tensile stress value, psi
T = flange thickness, in.
M, = total moment applied to the ring, in.-Ib
13, = angle formed by the tangent to the center line of the dished cover thickness at its point of intersection
with the flange ring
13,=carc sin [B/(2L + tl]
A = outside diameter of the flange, iu.
B = inside diameter of the flange, in.
C = bolt circle diameter, in.
4.7.2 Types of Dished Covers
Spherically dished covers may be either one piece, where the ring and head are one continuous thickness
of plate, or two pieces, where the ring and head are separate pieces which are welded together (no joint
efficiency is required) as shown in Fig. 4.3.
Not less Thin
2r and in No
CaI8 Less Th.n
1/2 in.
~,;:::;:=r=;
lL
L
Knuckle \
Radiu.
[U+","".."Jf'\
Gasket
Loooo Fie... Typo
-r
Ring'
Integr.' F'• • Type
Ge.ket
Shown
te)
r-".
I'
----1/2 A
f-----l/2A
!•PreferablY
T> t
1/2 C
1 2tmin.
I
1/2 c--+-l
L
\
1/28--+-.i
(e)
------+t
1'~---1/41A + 8/--01
Point of H 0
Full Penetration Weld
• Action
t
T
~~i
HD
L
Hr
T
1
\
Shown as Welded.
Smooth Weld Both
Centroid
~-\._-1
Ring
G_ket
Shown
O.7tmin.
Ibl
Skle.
~1/28
Use Any Suitable
f----l/2 C----\--';
Type of Gnket
tdl
FIG. 4.3
SPHERICALLY DISHED COVERS WITH BOLTING FLANGES (ASME VIII-1)
126 Chapter4
4.7.2.1
Ring and Head of Uniform Thickness.
This is the type of cover shown in Fig. 4.3(h).
(al Head thickness is
= (5PL)/(6S)
t
(4.18)
(b) Flange ring thickness using a ring gasket is
T = ([M,/SB]/(A
+ B)/(A - B)]}'"
(4.19)
(e) Fiange ring thickness using a full face gasket is
+
T = 0.6([PlS]/[B(A
4.7.2.2
B)(C - B)/(A - B)]}'/2
One Piece With Uniform Thickness.
(4.20)
This is the type of cover shown in Fig. 4.3(c).
(a) Head thickness is
t
=
(5PL)/6S
(4.21)
(b) Fiange ring thickness using a ring gasket and round bolt holes is
T
=
+
Q
([1.875M,(C
+
B)]/[SB(7C - 5B)]}'/2
(4.22)
+
B)/(7C - 5B)]
(4.23)
where
Q
=
(PL/4S)[(C
(e) Flange ring thickness using a ring gasket and bolt holes slotted through the edge is
([1.875M,(C
+
B)]/[SB(3C - B)]}'/2
(4.24)
Q = (PL/4S)[(C
+
B)/(3C - B)]
(4.25)
T = Q
+
where
(d) Flange ring thickness using a full-face gasket and round bolt holes:
T = Q
+
(Q'
+
[3BQ(C - B)/L]}'/2
(4.26)
+
(4.27)
where
Q = (PL/4S)[(C
B)/(7C - 5B)J
(e) Flange ring thickness using a full-face gasket and bolt holes slotted through the edge
T = Q
+ (Q' +
[3BQ(C - B)/L]}'"
(4.28)
where
Q = (PL/4S)[(C
+
B)/(3C - B)J
(4.29)
Flat Plates, Covers, and Flanges
4.7.2.3 Ring and Head.
(a) Head thickness:
127
This is the type of cover shown in Fig. 4.3(d).
(4.30)
t = (5PL)/(6S)
(b) Flange ring thickness:
The flange ring thickness is determined by combining the circumferential ring stress and the tangential
bending stress, as follows:
(i) Circumferential ring stress
(2) Tangential ring stress
=
S,
=
(4.31)
PDI2T
= S, = YM,/T 2B
(4.32)
(3) Total ring stress = S2 = S, + S, = (2S/ T)(F) + (S/ T')(J)
(4.33)
(4) Rearranging terms and solving for the flange-ring thickness, T,
T
=F +
(F T
+
J)'/2
(4.34)
-
B') V2I/[8S(A - B)l
(4.35)
+
B)]/[SB(A - B)l
(4.36)
where
F = [PB(4L 2
and
J = [M,(A
These rules may be used for either internal pressure or external pressure. The term P is an absolute value
for either internal or external pressure. The value of M; is determined by combining the moments from
bolt loading and gasket loading with the moment caused by the intemal pressure loading at the head-taring intersection. When M; is used in the equations, the absolute value is used. Example 4.8 shows the
analysis of a spherically-dished cover of the ring-and-head type that matches the flange in Example 4.6.
Example 4.8
Problem
A spherically-dished cover of the type shown in Fig. E4.8 is to be attached to the flange described in
Example 4.6. Determine the minimum required thickness of the head and flange ring. There is no corrosion
allowance, and no joint efficiency is required. The flange material is SA-lOS, and the head material is
SA-SI6 Or. 70. The dish radius is 0.9 LD.
Solution
(i) The allowable tensile stress for SA-SI6 Gr. 70 at 6S0'F is S, = 18.8 ksi and for SA-lOS at 100'F is
Sf' = 20.0 ksi and at 6S0'F is Sf' = 17.8 ksi.
128 Chapter4
A = 26.5"
0.5(A + B) = 18.625"
Full-penetration weld
T
1
L
T
Centroid
B = 10.75"
C = 22.5"
FIG. E4.8
EXAMPLE PROBLEM OF SPHERICALLY DISHED COVER, DIV. 1
(2) The dish radius, L, of the spherical head segment is
L = 0.9S = 0.9(10.75) = 9.675 in.
(3) The minimum required thickness of the head segment, using Eq. (4.17), is
Ih
=
5PLl6S,
= (5)(2,000)(9.675)/(6)(18,800)
= 0.858 in.
Assume that the thickness of the head is T, = 1.0 in.
(4) The head-to-ring angle, i3;, using the equation given in VUl-l, 1-6(h) is
13,
= arc sin[SI(2L +
=
I)J
arc sin{IO.75/[2(9.675) + I])
(5) The total flange moment for the gasket seating condition is the same as in Step 4 of Example 4.6.
MGS
=
2,098,000 in-Ib
(6) The total flange moment for the operating condition is:
Flange Loads
Same flange loads as in Example 4.6, plus an additional load occurs from the horizontal component,
H" due to the internal pressure load on the spherical head.
Flat Plates, Covers, and Flanges
HD
~
181,500 in.-Ib
HG
~
200,500 in.-lb
129
H, = 174,000 in.-Ib
=
(181,500)(1.607)
=
291,700 in.-lb
Lever Arms
hD = 0.5(C - B)
= (22.5 - 10.75) = 5.875 in.
ho = 3.729 in.
h, = 4.802 in.
h, is obtained by trial, using an assumed flange thickness and the perpendicular head thickness at the
head-to-ring intersection. The thickness of the head parallel to the flange ring face is:
= (1)(1.1778) = 1.178 in.
h, = 0.5(T - tp )
Assume that the thickness of the flange ring is T = 5.375 in.
h,
~
0.5(5.375 - 1.178)
~
2.099 in.
Flange Moments
M D = H D X hD
= (181,500)(5.875) = 1,066,000 in.-lb
~
(200,500)(3.729) = 747,700 in.-lb
4
~
(174,000)(4.802) = 835,500 in-Ib
=
(291,700)(2.099)
=
2,037,000 in-Ib
= 612,300 in-Ib
(7) The minimum required thickness is the larger of the thicknesses determined for the gasket seating
condition and for the operating condition by using the equations in Appendix 1·6(g)(2) as follows:
For gasket seating condition:
F = [PB(4L' - B')U'J/[8SiA - B)J
=
{(2,000)(10.75)[4(9.675)' - (l0.75)'J'n}/[8(20,000)(26.5 - 10.75)]
= 0.137
J
+
~
[MGs/(S")(B)][(A
=
[2,098,000/(20,000)(10.75)][(26.5
B)/(A - B)J
+
10.75)/(26.5 - 10.75)J
= 23.079
T = F
+
(F'
= (.137)
=
+
J)'"
+ [(.137)' + 23.079J'"
4.943 In.
For operating condition:
F ~ [PB(4L' - B')U'J/[8Sf"(A - B)J
=
{(2,000)(1O.75)[4(9.675)' - (10.75)'J"'j/[8(17,800)(26.5 - 1O.75)J
= 0.154
Flat Plates, Covers, and Flanges
.I
=
[M"/(S~)(B)][(A
=
[2,037,000/(17,800)(10.75)][(26.5 + 10.75)/(26.5 - 1O.75)J
+
131
B)/(A - B)J
= 25.177
T = F
+
(F'
= (.154) +
+
.1)1"
[(.154)' + 25.177Jw
= 5.174 in.
Since this is less than the 5.375 in. assumed, the thickness is acceptahle.
As with all calculated stress where a value of thickness, T, is assumed, a lesser value of required thickness
may be determined by repeated assumptions of thickness and further calculations until the assumed thickness
and the calculated thickness are the same.
CHAPTER
5
OPENINGS
5.1 INTRODUCTION
Openings through the pressure boundary of a vessel require extra care to keep loadings and stresses at an
acceptable level. Loads may be generated from both internal and external pressure and from applied external
loadings. An examination of the pressure boundary may indicate that extra material is needed near the
opening to keep stresses from loadings at an acceptable level. This may be provided by increasing the wall
thickness of the shell or nozzle or by adding a reinforcement plate around the opening. At some openings,
there may be a nozzle to which is attached external piping generating external forces and moments from
dead loads or thermal expansion. At other openings only a blind flange or flat cover with little or no
available reinforcement may exist.
In designing openings, two types of stresses are important: primary stresses, including both primary
membrane stress and primary bending stress; and peak stresses for fatigue evaluation. Altbough UG-22 of
VIII-l and AD-110 of VIII-2 require that both types be considered when evaluating loadings, in VlII-l
rules are given only for calculating the primary membrane stresses.
5.2
CODE BASES FOR ACCEPTABILITY OF OPENING
VIII-l gives two methods for examining the acceptability of openings in the pressure boundary for pressure
loading only. Other loadings shall be considered separately. The first method, the reinforced opening or
area replacement method, is used when that area which was to carry the primary membrane stress is missing
due to tbe opening. To replace this area, close-in substitute areas are called upon to carry the stress. The
second method is called the ligament efficiency method. This method examines the area of metal remaining
between adjacent openings compared with the area of metal that was there before the openings existed.
The primary membrane stress and shear stress are then examined for acceptability, Curves have been
developed to simplify this examination. For single openings, only the reinforced opening method is used,
while for multiple openiugs, either the reinforced opening method or the ligament efficiency method may
be used. Although the reinforced opening method and the ligament efficiency method are not developed
on the same basis, VIII-I permits either one to be used. It is appropriate to use whichever method is more
liberal, that is, the method giving the lower value for the increase in thickness. Consequently, both methods
may require examination.
Article D-S of VIII-2 contains reinforced opening rnles for a satisfactory design for pressure loading
only when a fatigue analysis is not required: It does not contain provisions for added loadings from piping
or other loadings which may be imposed. In lieu of meeting the reinforced opening rules of Article D-S,
an opening may be considered subject to the requirements of Appendices 4 and S of VIII-2.
133
134 Chapter 5
5.3 TERMS AND DEFINITIONS
Many terms and definitions used for openings, reinforcements, and ligaments are the same for VIII-I and
VIII-2, Some of the most common terms and definitions are given below, while others are given where
they are used,
d = diameter or chord of the opening in the plane being examined
db d-, d; = diameters or chords of various or adjacent openings
D = inside diameter of the cylindrical shell
D" = outside diameter of the cylindrical shell
t" = nominal wall thickness of the nozzle
t = nontinal wall thickness of the shell
t, = height of the reinforcement base (see Fig. UG-40 of VITI-l)
5.4 REINFORCED OPENINGS-GENERAL REQUIREMENTS
5.4.1 Replacement Area
5.4.1.1 Design for Internal Pressure. When there is an opening through the shell, except for fiat
heads, primary membrane stresses which develop from the pressure loading over the area formed by the
opening diameter and the minimum required thickness are interrupted. A substitute pathway is required.
For flat heads, the situation is similar, except primary bending stresses are interrupted. The assumption is
made that since primary bending stresses are maximum at the surfaces and zero at the centerline of the
thickness while primary membrane stresses are uniform across the wall thickness, the replacement area for
fiat heads needs to be only half the area required for cylindrical shells and formed heads.
Themethod presented for determining any needed reinforcement examines theregionaround theopening
for available areas to carry the primary stress around the opening. Since stress is related to the load and crosssectional area, areas: aresubstituted when makingcalculations. Placementandlocation of the replacementarea
is important. The replacement area should be close to the opening; but care should be taken, if temperature
is a consideration, not to generate an area of high thermal stresses. If it is not too difficult to place some
of the replacement area inside as well as outside of the vessel wall, try to place about two-thirds of the
replacement area on the outside and one-third on the inside.
5.4.1.2 Design for External Pressure. Although the procedure for evaluating stresses for external
pressure is based on a buckling and stability analysis, the method for determining the reinforcement
requirements for openings in shells under external pressure is very similar to that for shells under internal
pressure, but with the following changes:
(a) The minimum required thickness of the shell is based on the externalpressure requirements
and may be called t" instead of t..
(b) The replacement area required is 50% of that required for internal pressure.
5.4.2 Reinforcement Limits
The stress analysis basis used in the ASME Code to analyze the nozzle reinforcement is called Beams on
Elastic Foundation (Hetenyi, 1946). This method deterntines the effectiveness of the material close to the
opening for carrying loads. Reinforcement limits are developed parallel and perpendicular to the shell
surface near the opening, Although the method is a simplified application of the elastic foundation theory,
experience has shown that it does a good job,
Openings 135
Values from two equations are used to set the reinforcement limits measured along the vessel wall
surface, The greater value sets the horizontal limit for that opening, The first value is equal to d, and the
second value is equal to O.5d + t + t, as shown in Fig, 5,1. The relationship of the nozzle wall thickness
compared to the opening diameter or chord dimension, as appropriate, usually decides which of the two
values controls.
I, - t,
,
I: I,,
t
.4~~
I
d
; ;
-
d
O,5d
+ t + t"
d
t
d
O,5d
+ t + to
FIG. 5.1
REINFORCEMENT liMITS PARALLEL TO SHELL SURFACE
136 Chapter 5
For Section VIll-1, the reinforcement limits measured perpendicular to the shell surface are also set by
two limits; however, in this case, the smaller value is used. Using the beam on elastic foundation theory
for a cylindrical shell, the damping wavelength is a function of 1/ fl, where fl = 1.285/ (rt)112 for f1 =
0.3. When this vertical limit was set by the ASME Code committee years ago, the assumption was made
that rlt of 10 was appropriate. This gave
L
~
1/ 13
~
(rt)'" / 1.285
~
(0.lr 2)''' / 1.285
= 0.246r
~
2.46t
(5.1)
where rand t are basic terms of nominal dimensions related to either the shell or nozzle.
For application in various Code sections, the value was rounded to 2.5t. For VIII-!, it remains the smaller
of 2.5T or 2.5T, where T is the nominal shell or head thickness and T, is the nominal uozzle wall thickness.
For VIII-2, instead of assuming a fixed value of r/t, the actual values of rand t are used to set the vertical
limits which depend on the nozzle details, as shown in Fig. AD-540.1 of VIII-2.
5.5 REINFORCED OPENING RULES,
vm-i
In calculating the nozzle reinforcement requirements for VIII-I, it should be recognized that many of tbe
requirements weredevelopedyearsago, basedon the information availableatthattime.However,engineering
experience and additional data have shown that these rules are satisfactory for most designs and so are
still used.
As mentioned previously, reinforced opening rules are for pressure loading only. Other loadings that
need to be evaluated shall be considered separately using such methods as those given iu Welding Research
Council Bulletin No. 107, among others.
5.5.1 Openings with Inherent Compensation
Openings in vessels which are not subjected to rapid fluctuations in pressure do not require reinforcement
calculations [UG-36(c)(3)] if the following dimensional requirements are met:
(1) When using welded and brazed nozzles with a diameter not larger than:
(a) 3-1/2 in. diameter in a plate with a thickness :s 3/8 in.
(b) 2-3/8 in. diameter in a plate with a thickness> 3/8 in.
(2) When using threaded, studded, or expanded nozzles with a diameter not larger than
(a) 2-3/8 in. diameter in all plate thicknesses.
(3) When two openings are used, their centers shall be no closer than (d, + d,).
(4) When two openings in a cluster of three or more are used, their centers shall be no closer
than:
(5.2)
(a) For cylinders and cones: (l + 1.5 cos <!»(d1 + d2)
(b) For double-curved shells and beads: 2.5(d 1 + d,)
(5.3)
where
<!> = angle between the line connecting the centerlines of the two openings being considered and the
longitudinal axis.
Openings 137
5.5.2 Shape and Size of Openings
5.5.2.1 Shape of Opening. Openings in cylindrical shells and formed heads are usually circular,
elliptical. or obround. The latter shape is often developed for a nonradial nozzle opening. However, any
other shape is also permitted, but there may be no method of analysis given in the Code.
5.5.2.2 Size of Opening. For openings in a cylindrical shell, the rules given in UG-36 through
UG-42 of VIII-l are limited to the following sizes:
1. In shells 60 in. and less in diameter, the opening shall not exceed 0.5D or 20 in.
2. In shells over 60 in. in diameter, the opening shall not exceed 0.33D or 40 in.
When the size of the opening meets these limits, the rules given in UG-36 through UG-42 and in sections
5.5.3 through 5.5.5 (below) shall be used. When the size of the opening exceeds these limits, the rules
given in Appendix 1-7 of VIII-l and in section 5.5.6 shall be met as well.
5.5.3 Area of Reinforcement Required
5.5.3.1 Opening in a Cylindrical Shell (Except Nonradial Hillside). The total cross-sectional area
of reinforcement required for any plane through the center of the opening is determined by:
A
= dt.F
(5.4)
where
t, = minimum required thickness of the seamless shell, based on the circumferential stress calculated
by Eq. (2.1)
F = correction factor to obtain minimum required thickness of the shell on the plane being examined
f= 1.0, except for integrally reinforced openings listed in Fig. UW-I6.1, where F = 0.5(cos 28 + 1)
is permitted
8 = angle of the plane being examined from the longitudinal plane
The value of the F-factor vs. 8 is plotted in Fig. 5.2. The F-factor corrects the minimum required thickness
for all planes between 0° (the longitudinal plane) and 900 (the circumferential plane). This correction is
necessary to adjust for a minimum required thickness when the plane being examined is somewhere between
the longitudinal plane and the circumferential plane.
5.5.3.2 Opening in a Cylindrical Shell (Nonradial Hillside), The total cross-sectional area of reinforcement required for a plane through the center of the opening at a nonradial hillside nozzle is determined by:
A = ds,
(5.5)
where
d = chord length at the midsurface of the thickness required, excluding the excess thickness available
for reinforcement
t, = minimum required thickness of a seamless shell on the plane being examined
If the longitudinal plane is being examined, the value of t, is determined by Eq. (2.1). If the circumferential
plane is being examined, t, is determined by 0.51, or by Eq. (2.4).
138 Chapter 5
1.00
....
J
0.95
• _
~Ongitudinal
,/
8
shell axis
~
0.90
0.85
0.80
..."-
.
0
'"
iO
::J
0.75
>
0.70
,
~,
0.65
T
0.60
"
0.55
I""ho..
0.50
o
10
20
30
40
50
60
70
80
Angle 8. deg.• of plane with longitudinal axis
FIG. 5.2
CHART FOR DETERMINING VALUE OF FFOR ANGLE 6
90
Openings 139
5.5.3.3 Opening in a Spherical Shell or Formed Head. The total cross-sectional area of reinforcement
required for a plane through the center of an opening in a formed head is determined by
(5.6)
A = dt.
where
d = diameter or chord dimension of the opening
t, = minimum reqnired thickness of the spherical shell or formed head
(l) When the opening and its reinforcement are entirely within the spherical part of a torispherical
head [see Fig. 5.3(a)], r.is the minimum required thickness for a torispherical head using M = L
(1) When the opening is in a cone or conical shell, t, is the minimum required thickness of a
seamless cone of diameter D, measured where the nozzle centerline pierces the inside wall of
the cone.
(3) When the opening and its reinforcement are in an ellipsoidal head and are located within a
circle at the center of the head and the circle has a diameter equal to 0.8 shell diameter [see
Fig. 5.3(b)], t, is the minimum required thickness of a seamless spherical shell of radius KID,
where D is the shell diameter and KI is given in Table UG-37 of VllI-l.
spherical part· special limit for t r
(al Limits for Torispherical Head
0.8D = special limit for t r
h
~--------D
(b) Limits for Ellipsoidal Head
FIG. 5.3
DETERMINATION OF SPECIAL LIMITS FOR SETTING
CALCULATIONS
t, FOR USE IN REINFORCEMENT
140 Chapter 5
5.5.4 Limits of Reinforcement
As described in 5.4.2, limits of reinforcement are determined in both the vertical and the horizontal direction.
Excess cross-sectional area of material within these limits is available for reinforcement.
5.5.4.1 Parallel to Shell Surface. When the opening dimensions are within the limits given in section
5.4.2, the horizontal limits are the greater of:
(1) d
or
(2) 0.5d
+ t + t,.
5.5.4.2 Perpendicular to Shell Surface.
vertical limits are the smaller of:
When the opening is within the limits in section 5.4.2, the
(1) 2.5t
or
(2) 2.5t,
+
t..
5.5.5 Area of Reinforcement Available
When the reinforcing limits do not extend outside of an area where the required thickness and limits are
available equally on each side of the opening centerline, the following equations may be used to determine
the area of reinforcement available:
(1) Area available in vessel wall, A" is the larger of:
A, = (2d - d)(Et - Ft,)
(5.7)
or
A, = [2(O.5d + t
+
t,) - dJ(Et - Ft,)
(5.8)
(2) Area available in nozzle wall, A 2, is the smaller of:
A 2 = (5t)(t" - tm )
(5.9)
or
(5.10)
Example 5.1
Problem
Using the rules of VIII-I, determine the reinforcement requirements for an 8 in. l.D. nozzle which is
centrally located in a 2:1 ellipsoidal head, as shown in Fig. E5.1. The nozzle is inserted through the head
and attached by a fnll penetration weld. The inside diameter of the head skirt is 41.75 in. The head material
is SA-5l6 Gr. 70, and the nozzle material is SA-106 Gr. C. The design pressure is 700 psi, and the design
temperature is 400°F. There is no corrosion allowance, and the weld joint! quality factor efficiency is 1.0.
Openings 141
d = 41.76"
ASCU-LIMIT OF REINFORCEMENT
FIG. E5.1
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT IN ELLIPSOIDAL HEAD, DIV, 1
Solution
(1) The allowable tensile stress for both SA-5l6 Gr. 70 and SA-I06 Gr. C at 400°F is 20.0 ksi. Therefore,
I =
1.0.
(2) Using UG-32(d), the minimnm required thickness of a 2:1 ellipsoidal head without an opening is:
t, = (PD)/(2SE - 0.2P)
=
(700 x 41.75)/(2 x 20000 x 1.0 - 0.2 x 700)
=
0.733 in.
Nominal thickness used is 0.75 in.
(3) According to Rule (3) of t, in UG-37(a), when an opening and its reinforcement are in an ellipsoidal
head and are located entirely within a circle the center of which coincides with the head and the
diameter is equal to 80% of the shell diameter, t, is the thickness required for a seamless sphere of
radius K]D, where D is the shelll.D. and K is 0.9 from Table UG-37 of VlIl-1. For this head, the
opening and its reinforcement shall be withiu a circle with a diameter ofO.8D = (0.8)(41.75) = 33.4 in.
(4) The radius is
R = K,D = 0.9(41.75) = 37.575 in.
This radius is used in UG-32(f) to determine the t, for reinforcement calculations as:
t,
=
(PR)/(2SE - 0.2P)
=
[(700)(37.575)]/[2(20,000)(1.0) - 0.2(700)] = 0.625 in.
(5) Using UG-27(c)(I), the minimum required nozzle thickness is:
=
[(700)(4)]/[(20,000)(1.0) - 0.6(700)]
=
0.143 in.
Nominal thickness used is 1.125 in.
142
Chapter 5
(6) Limit parallel to head surface:
x
or (O.5d + t + t,), whichever is larger.
= d
~
+
8 in. or (4
+
.75
1.125 = 5.875 in.)
Use X = 8 in.
(7) Limit perpendicular to head surface:
Y
=
2.St
~
2.5(.75)
or
2.5tn, whichever is smaller.
~
1.875 in. or 2.5(1.125) = 2.81 in.
Use Y = 1.875 in.
(8) Size limit of the opening is
2X
2(8)
~
~
16 in.
This is less than the limit of 33.4 in. determined in (3). Therefore, the provision to use the spherical
head rule is valid.
(9) Reinforcement area required by UG-37(c) of VIll-1 is
+
A, = dt,F
2t,t,F(1 - f,)
= (8)(.625)(1)
+a
= 5.00 in.'
(10) Reinforcement area available in the head is
A, = d(Et - Ft,) - 2t,(Et - Ft,)(1 - f,,)
1.0, the second term becomes zero. Therefore,
A,
~
d(t - t,)
= (8)(.75 -
.625) = 1.00 in.'
(II) Reinforcement area available in nozzle is
A,
~
2Y(t, - tm)
=
(2)(1.875)(1.125 - .143)
=
3.68 in.'
(12) Reinforcement area available in fillet welds is:
=
2(.5)(.75)' = 0.56 in.'
Openings 143
(13) Total reinforcement area available in head, nozzle, and welds is:
=
1.00 + 3.68 + .56
=
5.24 in.'
Area available of 5.24 in.' is larger than the area required of 5.00 in.'
(14) Determination of weld strength and load paths:
According to UW-15(b). weld strengtb and load path calculations for pressure loading are not required
for nozzles which are like the one shown in Fig. UW-16.I(c) of VIII-I. Since this nozzle is similar
to that detail, no calculations are required.
Example 5.2
Problem
Using the rules of VIII-I. determine the reinforcement requirements for a 12 in. X 16 in. opening for a
manway as shown in Fig. E5.2. The man way forging is inserted through the vessel wall and attached by
a full penetration weld. The 12 in. dimension lies along the longitudinal axis of the vessel. The manway
cover seals against the outside surface of tbe manway forging. The LD. of the shell is 41.875 in. The shell
material is SA-516 Gr. 70 and the manway forging is SA-105. The design pressure is 700 psi. and the
design temperature is 400°F. There is no corrosion allowance. and all joint efficiencies I quality factors are
E = 1.0.
14t" 00
0
AI
U
\'
M
--
/
II
_373-
I
V/AJ
l
~
.~ W/#/-Y///
~~
•
I
I
__ c- - -
D
-
0
-
=.B
I
I
•
"'.
""!
'"
-
!
..,1",
m
X=12"
2X
=
Y
=
1.87 5"
-rl
Y =
t = 0.75-
1_8 75-
l,
X=12"
24"
ABCD-LIMIT OF REINFORCEMENT
FIG. E5.2
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF 12 in. x 16 in. MANWAY
OPENING, DIV. 1
144
Chapter 5
Solution
(I) The allowable tensile stress for both SA-516 Gr. 70 and SA-lOS at4000p is 20.0 ksi. Therefore,
(2) Using UG-27(c)(I), the minimum required thickness of the shell is:
f,
= 1.0.
t, = (PR)/(SE - 0.6P)
= [(700)(20.9375)]/[(20,000)(1.0) - 0.6(700)]
= 0.749 in.
Nominal thickness used is 0.75 in.
(3) The manway forging is elliptical. Since there are rio equations for determining the minimum required
thickness of an elliptical shell in VIII-I, the rules of U-2(g) are followed. For an elliptical shell, equation
(2.41) for minimum required thickness is given in section 2.6.2 of this book. The maximum value of
minimum required thickness is used for all planes as follows:
=
[(700)(8)2(6)2] 1{(20,000)(1.0)[(8)2(1)2 + (6)'(0)']3I2}
= 0.373 in.
NoJl1inal thickness used is 1.375 in.
(4) Examination of the longitudinal plane.
(a) Limit parallel to shell surface whichever is larger.
x
= d
=
or
(O.5d
+
I
+
I,)
12 in. or (6 + .75 + 1.375
=
8.125 in.)
Use X = 12 in.
(b) LiJl1it perpendicular to shell surface
Y
=
2.5t or 2.5tm whichever is smaller.
= 2.5(.75) = 1.875 in. or 2.5(1.375) = 3.437 in.
Use Y = 1.875 in.
(e) Reinforcement area required by UG-37(c) of VIIJ-I is
A, = dt.F
+
2t,I,F(1 - 1,)
=
(12)(.749)(1.0) + 0
=
8.988 sq. in. whcn j;
1.0.
(d) Reinforcement area available in the shell is:
A, = d(E,1 - FI,) - 21,(E,1 - Ft,)(1 - 1,,)
Openings
When!,t
145
1.0, the second term becomes zero; therefore,
At = (12)(.75 - .749) = 0.012 in.'
(e) Reinforcement area available in nozzle is:
Outward:
=
2(1.875)(1.375 - .373) = 3.758 in.'
Inward:
= 2(1.875)(1.375) = 5.156
in.'
(f) Reinforcement area available in fillet welds is:
A, = 2(.5)/,/
=
2(.5)(.75)' = 0.562 in.'
(g) Total reinforcement area available from shell, nozzle, and welds is:
=
0.012
+
3.758
+
5.156
+
0.562 = 9.488 in.'
Area available of 9.488 in.' is larger than required area of 8.988 in."
(5) Examination of the circumferential plane.
The opening has a 16 in. dimension on this [liane, but F = 0.5 and j, = 1.0.
(a) Reinforcement area required by UG-37(c) of VIII-I is:
= (16)(.749)(0.5) = 5.992 in.'
(b) Total reinforcement area available from shell and nozzle is:
AT =
> 5,992 in.'
Area available of >9.488 in.' is larger than area required of 5.992 in.' Note that the increase in
diameter from 12 in. to 16 in. would increase the limit parallel to the shell surface, X. However,
since AT is larger than Ar, the design is satisfactory and there is no need to consider the increased
limit in our evaluation of this design.
(6) Determination of weld strength and load paths.
According to UW-15(b), weld strength arid load path calculations for pressure loading are not required
for nozzles like the one shown in Fig. UW-16.1(c) of VUI-l. Since this nozzle is similar to that one,
no calculations are required.
146 Chapter 5
Example 5.3
Problem
Using the rules of VIII-I, determine the reinforcement requirements for a 5,625 in, LD, nozzle which is
located on a hillside or non-radial position on the circumferential plane as shown in Figs, E5,3,1 and E5,3,2,
The nozzle wall abuts the vessel wall and is attached by a full penetration weld, The LD, of the shell is
41.875 in, The shell material is SA-516 Or, 70, and the nozzle material is SA-106 Or. B. The design
pressure is 700 psi, and the design temperamre is 400°F. There is no corrosion allowance, and all joint
efficiency1quality factors are E = 1.0,
Solution
(I) At 400°F, the allowable tensile stress for SA-516 Or. 70 is 20,0 ksi and for SA-106 Or.
ksi, Therefore.j, = 17,1/20,0 = 0,855
(2) Using UG-27(c)(1), the minimum required thickness of the shell is:
I, ~ (PR)/(SE ~
it is 17,1
0,6P)
[(700)(20,9375)J/[(20,000)(1.0) - 0,6(700)J ~ 0,749 in,
Nominal thickness used is 0,75 in,
(3) Using UO-27(c)(1), the minimum required thickness of the nozzle is:
t.,
~
(PR,)/(SE - 0,6P)
~
[(700)(2,8125)J/[(17,100)(1.0) - 0,6(7oo)J
~
0,118 in,
Nominal thickness used is 1.5 in,
8.622" 00
:I
I
A
•
~e
II.!
B
d- ,.625"
I
). f;-.
.;.
~
~////0"//';:;
•
x·
'"M
r0OJ
---lIT.
~.625"
tit"
I
~---
x·
'r. W"
0.118"
y
=
1.875"
c
0.75"
k~
.... "'''..
W/#.i.% r0c.
t
,.625"
2X- 11.2$"
ci
'"
II
II
~
DA
BCD• 1I t.I IT 0 F R £INFOR(EME NT
c
FIG. E5.3.1
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF HILLSIDE NOZZLE, DIV. 1
Openings 147
t.= 0.749"
t" = 1.5"
t.n = O.11S"
FIG. E5.3.2
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF HILLSIDE NOZZLE, DIV. 1
(4) Examination of the longitudinal plane:
(a) Althongh allowable stresses of the shell and nozzle are different, since the nozzle weld is a full
penetration weld abutting the shell wall.j, = 1.0 for both A, and A, audj; = 0.855 for A2 .
(b) Liutit parallel to shell surface:
x=
=
d
or (O.5d
5.625 in. or
+
t
+
(2.81
t n) , whichever is larger.
+ .75 + 1.5 = 5.062 in.)
Use X = 5.625 in.
(c) Liutit perpendicular to shell surface:
Y = 2.5t
=
or 2.5t n, whichever is smaller.
2.5(.75)
=
1.875 in. or 2.5(1.5)
=
3.75 in.
Use Y = 1.875 in.
(d) Reiuforcement area required according to VIII-I, UG-37(c) is:
=
(5.625)(.749)(1.0)
= 4.213 in.'
(e) Reinforcement area available in the shell is:
A, = (2X - d)(t - t,)
=
(11.25 - 5.625)(.75 - .749)
=
0.005 in.'
148
Chapter 5
(f) Reinforcement area available in the nozzle is:
~
2(1.875)(1.5 - .118)(.855)
~
4.431 in.2
(g) Reinforcement area available in fillet welds is:
= 2(.5)(.75)' = 0.562 in.2
(h) Total reinforcement area available from shell, nozzle, and welds is:
~
.005 + 4.431 + .562
~
4.998 in.'
Area available of 4.998 in.' is larger than area required of 4.213 in.'
(5) Examination of the circumferential plane:
(a) Since this is a nonradial plane, it is necessary to determine the chord length measured diagonally
across the opening (chord length 1-2 in Fig. E5.3.2) hased on the midpoint of the minimum required
thickness of the shell, t; = 0.749 in. Based on the geometry, chord length 1-2 = 12.217 in.
Therefore, d' = 12.217 in. and F = 0.5 on the circumferential plane.
(b) Reinforcement area required using the chord length 1-2 is:
= (12.217)(.749)(0.5) = 4.575 in.2
(c) Based on the longitudinal plane, total reinforcement area available from shell and head is:
AT = 4.998 in.2
Area available of 4.998 in.' is larger than the required area of 4.575 in.' Since AT based on the
longitudinal plane is larger than A; even without any consideration of an increase in the limit
parallel to the shell surface due to the iucreased d', the design is satisfactory.
(6) Determination of weld strength and load paths.
According to UW-15(b), weld strength and load path calculations for pressure loading are not required
for nozzles of the type shown in Fig. UW-16.I(a) of VIII-I. Since this nozzle is similar to that one in
detail, no calculations are required.
Example 5.4
Problem
Using the rules of VIII-I, determine the reinforcement requirements for a 6.0 in. I.D. nozzle wltich is
located in a cylindrical shell, as shown in Fig: E5.4. The nozzle abuts the vessel wall and is attached by a
fun penetration weld. The J.D. of the shell is 30 in. The shell material is SA-516 Gr. 60, and the nozzle
is SA-106 Gr. B. The design pressure is 1000 psi, and the design temperature is 100°F. The corrosion
Openings 149
d
T. = 1.37S"
=6.0"
A
0= 6.2S"
~"= 0.189"
I
T = 1.12S"
FIG. ES.4
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT WITH CORROSION ALLOWANCE,
DIV.1
allowance is 0.125 in.• and all joint efficiency I quality factors are E = 1.0. When a corrosion allowance
is considered, all calculations are based on the corrosion allowance being fully corroded away.
Solution
(1) The allowable tensile stress for SA-516 Gr. 60 and SA-106 Gr. B is 17.1 ksi, Therefore.j, = 17.11
17.1 = 1.0.
(2) Using UG-27(c)(1), the minimum required thickness, In of the shell is:
R = r +
~
1,_ = (PR)/(SE -
15.0
c.a.
+
0.125 = 15.125 in.
0.6P)
= [(1000)(15.125)]/[(17,100)(1.0) - .6(1000)] = 0.917 in.
T, = 0.917
+
0.125 = 1.042 in.
Nominal thickness used is 1.125 in.
(3) Using UG-27(c)(1), the minimum thickness. t.; of the nozzle is:
R,
=
r. + c.a.
=
3.0
+
0.125
=
3.125 in.
150 Chapter 5
~
[(I000)(3.125)J/[(17,100)(1.0) - 0.6(1.000)]
~
0.189 in.
T;
= 0.189
+
0.125 = 0.314 in.
Nominal thickness used is 1.375 in.
(4) Limit parallel to the shell surface:
x=
~
d
or
(O.5d
6.25 in. or
+
t
+
[3.125
tn) , whichever is larger.
+ (1.125 - 0.125) + (1.375 - 0.125)]
~
5.375 in.
Use X = 6.25 in.
(5) Limit perpendicular to the shell surface:
Y
=
2.5t or 2.5tm whichever is smaller.
=
2.5(1.125 - 0.125) = 2.50 in. or
2.5(1.375 - 0.125) = 3.125 in.
Use Y = 2.50 in.
(6) Reinforcement area required according to VIII-1. UG-37(c) is:
~
(6.25)(0.917)(1.0)
~
5.731 in.'
(7) Reinforcement area availahle in the shell is:
A,
= (2X
~
- d)(t - ',)
(6.25)[(1.125 - 0.125) - 0.917J
~
0.519 in.'
(8) Reinforcement area available in the nozzle is:
A,
~
2Y(," - tm)
~
(2)(2.5)[(1.375 - 0.125) - 0.189]
(9) Total reinforcement area available from shell and nozzle is:
~ 0.519
+ 5.305
~ 5.824 in.'
This is larger than area required of 5.731 in.'
~
5.305 in.'
Openings 151
(10) Determination of weld strength and load paths.
According to UW-15(b), weld strength and load path calculations for pressure loading are not reqnired
for nozzles of the type shown in Fig. UW-16.I(a) of VIII-I. Since this nozzle is similar to that one
in detail, no calculations are required.
5.5.6 Openings Exceeding Size Limits of Section 5.5.2.2
When the opening diameter is large compared to the diameter of the shell or head in which this opening
is located, experience shows that some of the reinforcing area should be placed close to the edge of the
opening. Therefore, for such openings, the following special reqnirements shall be met in addition to those
listed in sections 5.5.3, 5.5.4, and 55.5,
5,5.6.1 Area of Reinforcement Required. The total cross-sectional area of reinforcement required
for any plane through the center of the opening is determined by:
A
5.5,6.2
~
(5. t 1)
0.67 dt.F
Limits of Reinforcement Parallel to Shell.
The horizontal limits are the larger of:
(I) 0.75d or
(2) 05d + I + I,.
5.5.6,3 Limits of Reinforcemeut Perpendicular to Shell.
as those given in section 5.5.4.2.
The vertical limits are exactly the same
5.5.6.4 Stresses When Nozzle Radius/Shell Radius 0; 0.7. In VIII-I, a method is provided for
examining membrane and bending stresses in radial nozzle openings when the nozzle radius I shell radius
0; 0.7. Other alternatives given in Code Case 2236 may also be used. This method is as
follows:
Membrane stress, Sm, and bending stress, S; for either Case A for a nozzle with a reinforcing pad or
Case B for a nozzle with integral reinforcement are calculated and compared to the allowable stress value.
(I) The membrane stress using the limits given in Fig. 5.4.1 is calculated for Eq. (5.12) for
Case A or by Eq. (5.13) for Case B as follows:
(5.12)
Case A:
Case B:
s; ~
P(R(R, + t, + (R.t)"'] + R,[I + (R,mt,)U2]}IA,
(5.13)
(2) The bending stress using the greater of the limits given in Fig. 5.4.1 or Fig. 5.4.2 is calculated
by using Eqs. (5.14), (5.15), and (5.16) for both Cases A and B as follows:
M ~ P{[(R,)'/6]
+ RR,e)
a = e + t/2
s,
~
Mall
(5.14)
(5.15)
(5.16)
152 Chapter 5
Rn
e
ea,aB Nonie With Integral
Case A Nozzle With
Reinforcing Pad
Type Reinforcament
GENERAL NOTE:
When any part of a flange is located within the greater oftha JR';;';r;;+ teor 1etn + te limit as indicated in Fig. 1·7-1 or
Fig. 1-7-2 Case A, or the greater of ./Rnm tn or 16t n for Fig, '-7-' or Fig. 1-7-2 Case B. the flange may be included as part
of the section which resists bending moment.
FIG. 5.4.1
R
n
/~
"
L
T- --- ,- ---1;,-1
...........
f4-
t
Neunat exts
L
of shaded area
:I:
e
L
r
J
te + 16t n
e
;;;d--+
~~'6r--1--r-rr
r
_+1--"
--* t
I
Shell center line
~ Nozzle center line
Case A Nozzle With
Reinforcing Pad
case B Nozzle With Integral
Type Reinforcement
GENERAL NOTE:
When erw part of a flange is located within the grealer of the )Rnm ttl + te or 16tn + fa limit as indicated in Fig. 1-7-1 or
Fig. 1-7-2 Case A, or the greater of~or 16t n for Fig. 1-7-1 or Fig. 1-7-2 Case B, the flange may be included as part
of the section which resists bending moment.
FIG. 5.4.2
Openings 153
where in addition to the definitions in section 5.3,
A, = shaded area in Fig. 5.4.1 or Fig. 5.4.2, Case A or Case B, in.'
I = moment of inertia of the shaded area abont the neutral axis, in.'
a = distance between the neutral axis of the shaded area and the inside surface of the
vessel wall, in.
Rm = mean radius of the shell, in.
Rnm = mean radius of the nozzle neck, in.
e = distance between the neutral axis of the shaded area and the midwall of the shell,
in.
Sm = membrane stress calculated by Eq. (5.12) or Eq, (5.13), psi
S» = bending stress calculated by Eq. (5.16), psi.
(3) Calculated stresses are compared with allowable stress, S, as follows:
S; + Sb
S
1.5S
5.6 REINFORCED OPENING RULES, VIII-2
Although there is a close similarity between the reinforcement rules in VllI-1 and VIIl-2, there are some
differences. Reinforcement limits and spacing are based upon the damping length of a beam on an elastic
foundation using the actual dimensions. In lieu of using these rules, the rules of Appendix 4 and Appendix
5 of vm-2 may be used.
5.6.1 Definitions
Symbols and terms used in opening calculations in VllI-2 are
d = diameter or chord dimension of the opening in the given plane, in.
t,
=
minimum required thickness of the shell or head without the opening, in.
F = 1.00 when the plane under consideration is in the spherical portion of a head or when the plane
contains the longitudinal axis of a shell. For other planes through the shell, the value of F from
Fig. 5.2 is used, except when a reinforcing pad is used.
Rm = mean radius of the shell or head at the opening, in.
t = nominal thickness of the shell or head at the opening, in.
t.; = minimum required thickness of the nozzle, in.
t;
nominal thickness of the nozzle, in.
= mean radius of the nozzle, in,
ri = transition radius between the nozzle and the vessel, in.
fp = nominal thickness of the connecting pipe, in.
=
r = r + O.5t"
In
K = 0.73r, when a transition radius r, is used and the smaller of the two legs when a fillet weld
transition is used, in.
For h, L, x, and other dimensions, see Fig. 5.5 in this text and Fig. AD-540.1 of VTII-2.
5.6.2 Openings Not Reqniring Reinforcement Calcnlations
No reinforcement calculations are required when the following criteria are met for circular openings:
(1) Single openings with d=, = 0.2(R m t)112 and two or more openings within a circle with a
diameter
S;
2.5 (Rmt) l l2. The sum of the diameters is
S;
0.25 (Rm t ) l l2.
154 Chapter 5
.'90deg.T -
Alternate nouHt
to pipo transrtion
1.1
'3
'-t---looi
/m -11--0111
,-+-......'"
r
t
r
L'
rm.r+o.5t'J
'2
-+.. . . . . ++-""'1
t
Shell thickness
Ibl
lei
Alternate with
fillet, transition
r
/m
tn
Alternative with
pad plata and fillet
'.
'dl
FIG. 5.5
NOZZLE NOMENCLATURE AND DIMENSIONS (DEPICTS GENERAL CONFIGURATIONS
ONLY)
Openings 155
(2) Center-to-center spacing <': 1.5(d1 + d,).
(3) Center-to-edge distance of another local stressed area, where PL is greater than US" is
equal to 2.5 (R mf ) l J2
5.6.3 Shape and Size of Openings
Openings are usually circular or elliptical or of a shape formed by the interaction of a circular or elliptical
cross section with another surface. The limits given below also apply:
(1) The ratio of the large to small dimension of the opening is limited to 1.5.
(2) 'Iheratio dID :5 0.50.
(3) The arc length between centerlines of openings is limited to no less than
(a) Three times the sum of the radii for formed heads and longitudinal axis of a cylindrical
Shell;
(b) Two times the sum of the radii for the circumfereutial direction.
(4) Rules shall be satisfied for all planes.
For an opening with a shape and size not within these limits, design-by-analysis shall be used.
5.6.4 Area of Reinforcement Required
In determining the area of reinforcement required for an opening in an Vill-2 vessel, each opening shall
be examined by two criteria: the entire area provided within limits and two-thirds of the area provided
within more restrictive horizontal limits.
The total cross sectional area of reinforcement required for any plane through the center of the opening
is determined by:
5.6.5 Limits of Reinforcement
5.6.5.1
Parallel to the Shell Surface
For 100% of the Required Reinforcement
The borizontal limits are the greater of:
(1) d or
(2) O.5d + t
+
fo'
For 2/3 of the Required Reinforcement Area
The horizontal limits are the greater of:
+ O.5(Rm f)lI'
O.5d + t + fo'
(I) O.5d
(2)
or
156 Chapter 5
5.6.5.2
Perpendicular to the Shell Surface
For the Nozzles Shown iu Fig. 5.5(a) and (b):
When h < 2.5t, + K, the perpendicular limits are the greater of
(1) 0.5(rmt,)112
+
K
or
(2) 1.73x
+ 2.5tp +
K
In either case, the limit shall not exceed either 2.5t or L + 2.5tp •
When h "" 2.5t, + K, the perpendicular limits are the greater of
or
(2) 2.5t,
In either case, the limit shall not exceed 2.5t.
For the Nozzle Shown in Fig. 5.5(c):
e "" 30 deg., the perpendicular limits are the greater of
When 45 deg, >
or
(2) L' + 2.5tp
When
In either case, the limit shall not exceed 2.5t.
e < 30 deg., the perpendicular limits are the greater of
or
(2) 1.73x + 2.5t,
In either case, the limit shall not exceed 2.5t.
For the Nozzle Shown in Fig. 5.5(d):
The perpendicular limits are the greater of
(1) 0.5(rm t,)[/2
+
t,
+
K
+ K
or
(2) 2.5/,
+
t,
In either case, the limit shall not exceed 2.5t.
Openings 157
In all cases, t, :'5 1.5t and :'5 1.73W, where
W = width of the reinforcing element, in.
t, = thickness of the reinforcing element, in.
5.6.6 Available Reinforcement
Metal contributing to the required area of reinforcement shall lie within the reinforcement limits given in
section 5.6.5 and shall he limited to a material which meets the following criteria:
(a)
(b)
(c)
(d)
(c)
(f)
Area in excess of that required to carry the primary membrane stress,
Area of nozzle wall in excess of that required to carry the primary membrane stress,
Weld metal within the shell and nozzle wall which required PQR,
Full-penetration weld joining weld pad to nozzle neck,
Other weld areas meeting requiremeuts of AD-570,
Metal meeting the following requirement:
[(a, - a,.)
(5.17)
L>Tj '" 0.0008
where,
e, = mean coefficient of expansion of the reinforcing metal, in.! in. OF
cc, = mean coefficient of expansion of the vessel metal, in. / in. OF
6..T = operating temperature range from 70 0 P to operating temperature, OF
5.6.7 Strengtb of Reinforcement Metal
(a) S"/ S, 2': 0.8,
(b) For S"/S, > 1.0, use S"/S, = 1.0 maximum
where
S; = design stress intensity value of the nozzle material, ksi
S, = design stress intensity value of the vessel material, ksi
5.6,8 Alternative Rules for Nozzle Design
An acceptable alternative to the regular reinforcement requirements may be used subject to special limitations
and other reinforcement requirements.
5.6.8.1
Limitations
(a) The reinforcement will have a circular cross section and he perpeudicular to shell.
(b)The reinforcement will have all integral construction using corner fillet radii.
5.6.8.2
Required Reinforcement Area, A r •
The required minimum reinforcement area related to
d/(Rt,)l/2 is:
<0.20
2::0.20 & <OAO
~OAO
In Cylinders
In Spheres & Heads
None except rz required
{4.05[d/(Rt r ) 112] 112 - 1.81sdi,
O.75dt,
None except r2 required
{5AO[d/(Rt,)I12]I12 - 2A1 }dt,
dt, cos <V
where <V = sin-1(dID)
158 Chapter 5
5.6.8.3 Limits of Reinforcing Zone. Metai included in meeting the reinforcing area requirements
shail be located in the zone boundary shown in Fig. 5.6.
Example 5.5
Problem
Using tbe rules of Article D-5 of VIII-2, determine the reinforcement requirements of an 8 in. J.D. nozzle
which is centrally located in a 2:1 ellipsoidai head as shown in Fig. E5.5. The nozzle is iuserted through
the head and attached by a full penetration weld. The inside diameter of the head skirt is 41.75 in. The
head materiai is SA-516 Gr. 70. and the nozzle materiai is SA-106 Gr. C. The design pressure is 700 psi,
and the design temperature is 500'F. There is no corrosion allowance.
Plane of NouN and
V_A... """"'"
in Cylindricol Shells
fa) Reinforcing Zone Limit
(1)
Lc • 0.945 ffr IR)2J3 R for nozzles in cylindrical shells.
(2)
Ln· 1.26 ItrlRl2/3 [R frlR + O,5)} for nozzles in heads.
(31 The center of Lc or L n is.t the junctureof the outside
surfaces of the shell and nozzles of thickneues t f and rrlt'
Zone
bound....
R
+
Tfllftll'VOf'lO
Ib) Rein_ Atu
I t I Hatched are. represent ..... ilable reinforcementarea A.,
121 Metal are. within the zone boundary. in IXcessof the ....
formedby the intersection of the basic shells, shallbe considered
as contributing to the fllQuired area At· The basic shells are
defined as having inside radius R. thickness tr,inside radius r,
thickness tN r
P''''I
Cyltndrtcal Sheela
I
{4} In constructions where the lone boundary passes thrau;h
• uniform thickneu wall segment. the zoneboundary may be
considered .. Lc or Ln through the thickness.
All PI. . . . " ' "
(31 The available reinforcement area A, shall be It lean
equal to A,/2 on each side of the noule center line and in every
plane containingthe nozzle axis.
R
\
FIG. 5.6
LIMITS OF REINFORCING ZONE FOR ALTERNATIVE NOZZLE DESIGN
Openings
2l4.t
159
SA-106 C
= 2.5"
.1
FIG. E5.5
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT IN ELLIPSOIDAL HEAD, DIV. 2
Solution
(I) The allowable stress intensity for SA-5l6 Gr. 70 is 20.5 ksi, and for SA-106 Gr. C it is 21.6 ksi.
Since the nozzle material is stronger than the head material, no adjustment is required.
(2) Using Fig. AD-204.l, the minimum required thickness of a 2:1 ellipsoidal head is:
P /S
= 700/20,500
~
0.034
which gives
II L
L
~
I
~
0.021
0.9D = 0.9(41.75) = 37.575 in.
=
0.021(37.575)
~
0.789 in.
Nominal thickness used is 1.0 in.
(3) Using AD-20l(a), the minimum required nozzle thickness is:
1m ~ PRI(S -
0.5P)
= (700)(4)/[21.600 - 0.5(700)]
~
0.132 in.
Nominal thickness used is 1.125 in.
(4) Limits parallel to head surface are
(a) For 100% of the required reinforcement:
x =
~
Use X = 8 in.
d or (O.5d + t
8 in. or (4
+
til)' whichever is larger
+ 1 + 1.25
=
6.125 in.)
160 Chapter 5
(h) For 2/3 of required reinforcement:
K'
+ O.5(Rmt) 1/2 or (O.5d + t +
= r
= [4
+
0.5(38.075 X 1)1/2
=
tn) , whichever is larger
7.085 in.] or 6.125 in.
Use X' = 7.085 in.
(5) Limits perpendicular to the head surface are calculated as shown below.
(a) Determine which limits given in section 5.6.5.2 for Fig. 5.4(a) apply by:
h < 2.5t, + K
(i)
or
(ii)
h '" 2.5t,
+
K
(h) h = 2.50 in.
2.5t, = (2.5)(1.125)
2.81 in.
=
K = 0.25 in.
2.5t,
+K=
+
0.25
=
3.06 in.
+
2.5tp
+
K, whichever is larger,
2.81
Limit (5)(a)(i) applies where h < 2.5t, + K
(c) Determine the perpendicnlar limit as follows:
Y
= 0.5(r",£")112
+
k
or 1.73x
but not more than 2.5t nor (L + 2.5tp)
= 0.5[(4.5625)(1.125)]'"
= 0
=
+ 2.5(1.125) + 0.25
2.5(1)
= (4)
=
+ 0.25
=
= 1.383 in.
= 3.063 in.
2.5 in.
+ 2.5(1.125)
= 6.813 in.
2.5 in.
(6) Reinforcement area required according to AD-520 of VTII-2 is:
=
dt,F
=
(8)(0.789)(LO)
=
6.312 in.'
2/3A, = 2/3(6.312)
=
4.208 in.'
A,
(7) Using the 100% limit, reinforcement area available in the head is
Openings
A,
~
(r -
~
(1.0 - 0.789)(2 X 8 - 8)
161
te)(2X - d)
~
1.688 in.'
(8) Reinforcement area available in the nozzle is
A,
~
2Y(t, -
~
2(2.5)(1.125 - 0.132)
1m )
~
4.965 in.'
(9) With the 100% limit, total reinforcement area available in the head and nozzle is:
~
1.688 + 4.965
~
6,653 in.'
Area available of 6.653 in." is larger than area required of 6.312 in.'
(10) With the 2/3 limit, reinforcement area available in the head is:
A,
~ (I -
~
t,)(2X' - rI)
(1.0 - 0.789)(2
X
7.085 - 8)
~
1.302in.'
(11) With the 2/3 limit, total reinforcement area available in the head and nozzle is
AT ~ A,
+
AT = 1.302
+ 4.965
~
6.267 in.'
Area available of 6.267 in.' is larger than the required area of 4.208 in.'
Example 5.6
Problem
Using the rules of VIII-2, determine the reinforcement requirements for a 12 in. X 16 in. opening for the
manway shown in Fig. E5.6. The manway forging is inserted through the vessel wall and attached by a
full penetration weld. The 12 in. dimension lies along the longitudinal axis of the vessel. The manway
cover seals against the outside surface of tbe manway forging. The I.D. of the shell is 41.875 in. The shell
material is SA-516 Gr. 70. and the manway forging is SA-lOS. The design pressure is 700 psi, and the
design temperature is SOoop. There is no corrosion allowance.
Solution
(I) The allowable stress intensity for SA-516 Or. 70 is 20.5 ksi, and for SA-lOS it is 19.4 ksi. An adjustment
of t: ~ 19.4/20.5 = 0.946 is required for A,.
(2) Using AD-201(a), the minimum required thickness of the shell is:
t, = (PR) / (SE - 0.5P)
~
[(700)(20.9375)]/[(20500)(1.0) - 0.5(700)]
~
0.728 in.
Nominal thickness used is \.0 in.
1.25"
0.385"
AI
:!
!
I
/
1~/MI0/,
y =
2lz'
t " 1"
~~////,Y///
'\
I
D
Y =
2lz'
-
X=12"
X=12"
I
2X
24"
ABCD=L~mit of Re~nforcement
FIG. E5.6
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF 12 in. x 16 in. MANWAY
OPENING, DIV. 2
(3) The manway forging is elliptical. Since there are no eqnations for determining the minimum required
thickness of an elliptical shell in VIII-2, oue ueeds to be located. For an elliptical shell, Eq. (2.41) for
minimum required thickness is given in section 2.6.2. The maximum value of the minimum required
thickness is used for all planes being examined as follows:
=
[(700)(8)'(6)']/ {(l9,400)(1.0)[(8)'(l)2
+
(6)'(0)2]312}
= 0.385 in.
Nominal thickness used is 1.0 in.
(4) Examination of the longitudinal plane:
(a) Determine the limit parallel to the sbell surface.
(i) For 100% of reinforcement area required:
x
=
d or
= 12 in.
(O.5d + t + t,J, whichever is larger
or (6 + I + 1 = 8.0 in.)
Use X = 12 in.
(ii) For 2/3 of reinforcement area required:
X' = r
=
Use 8.315 in.
+
O.5(R",t)l!2
or
(O.5d
[6 + 0.5(21.4375 XI)''']
+t+
=
tn) , whichever is larger
8.315 in. or 8.0 in.
Openings 163
(b) Determine the limit perpendicular to the shell surface.
(i) Determine which limits given in section 5.6.5.2 for Fig. 5.4(a) apply by
(a)
h
< 2.51. + K
or
(b)
h '" 2.51.
+
K
(ii) h = 2.50 in.
2.51. = (2.5)(1.0)
2.50 in.
~
K = 0.25 in.
2.5t.
+
K
~
Limit (5)(a)(i)(a) applies where h < 2.51.
(iii) Determine the perpendicular limits
2.75 in.
+
K
Y = O.5(rmt,y n + K or 1.73x + 2.5tp + K, whichever is larger,
but not more than 2.5t nor (L + 2.5t p )
=
0.5(6.625 X 1.0)'12
= 0
+ 0.25
+ 2.5 X 1.0 + 0.25
=
2.5(1) = 2.50 in.
~
3
=
2.50 in.
+ 2.5(1.0)
=
=
~
1.537 in.
2.75 in.
5.50 in.
(c) Reinforcement area required is
For 100% reinforcement:
A,
~
dtP
+
2',1,(1 - j;)
= (12)(0.728)(1)
+ 2(1.0)(0.728)(1 - 0.946)
~
8.814 in.'
For 2/3 A: 2/3A, = 5.876 in.
(d) Using the 100% limit, reinforcement area available in the shell is:
A,
~
(I - 1,)(2X - d)
= (1 - 0.728)(2 X 12 - 12)
(e) Reinforcement area available in tbe nozzle wall is:
~
3.264 in.'
164
Chapter 5
Outward:
A'l ~ 2(2.5)(1.0 - 0.385)(0.946) ~ 3.075 in.'
A" ~ 2(2.5)(1.0)(0.946) ~ 4.730 in.'
Inward:
(f) Total reinforcement area available from the shell and nozzle is:
= 3.264
+ 3.075 + 4.730
= 11.069 in.'
Area available of 11.069 in.' is larger than area required of 8.814 in.'
(g) Using the 2/3 limit, reinforcement area available in the shell is
A1
~ (I -
0.728)(2 x 8.315 - 12) ~ 1.259 in.'
AT ~ 1.259
+ 3.075 + 4.730
=
9.064 in.'
Area available of 9.064 in.' is larger than area required of 5.876 in.'
(5) Examination of circumferential plane:
The opening has a 16 in. dimension on this plane, but F = 0.5.
(a) Reinforcement area required according to AD-520 of VIII-2 is
A, ~
drF
=
(16)(0.728)(0.5)
~
5.824 in.'
(b) Total reinforcement area available from the shell and nozzle is
AT ::::: >9.064 in.'
Area available of 9.064 in.' is larger than area required of 5.824 in.' Since area is acceptable without
any adjustment of limits, the design is satisfactory.
5.7 LIGAMENT EFFICIENCY RULES, VIII·!
Section VIII-l permits two methods for calculating the replacement metal removed at openings. They are
the reinforced opening method and the ligament efficiency method. The ligament efficiency method considers
the metal area removed from the pressure boundary and the metal remaining between the two or more
openings in the pressure boundary. No metal is cousidered from any uozzle attached at the opening.
The ligament efficiency curves apply only to the cylindrical shell of a pressure vessel where the circumferential stress has twice the intensity of the longitudinal stress. Once this was established, Rankine's Ellipse
of Stress was used to determine the tension stress and the shear stress on any diagonal ligament plane.
Using these data for tension and shear, curves were developed with respect to the longitudinal plane
(circumferential stress) for various values of e, p' d, and t. Equations were developed to make the determination of the ligament efficiency an easy task. These equatious and a plot of the curves are given in Figs.
UG-53.5 and UG-53.6 of VIII-I.
Calculations should be made using both the reinforced opening method and the ligament efficiency
method: the thinner required plate thickness should be selected.
Openings 165
Example 5.7
Problem
Using the reinforcement requirements give in UG-37 through UG-42 of VIII-I, determine the minimum
required thickness of a 36 in. LD. cylindrical shell that has a series of openings in the pattern shown in
Fig. E5.7. The openings are 2.50 in. diameter on a staggered pattern of three longitudinal rows on 3.0 in.
circumferential spacing and 4.50 in. longitudinal spacing. The design pressure is 600 psi at the design
temperature of 500°F. Shell material is SA-516 Gr. 70, and nozzle material is SA-21O Gr. C. There is no
corrosion allowance. The openings are not located in or near any butt weldedjoint.
Solution
(1) The allowable tensile stress for both SA-516 Gr. 70 and SA-21O Gr. C at 500'F is 20.0 ksi. Therefore,
f, = 1.0.
(2) Using UG-27(c)(l) and assuming a searoless shell with E = 1.0, the minimum required thickness of
the shell for reinforcement calculations is
I, = (PR / (SE -
0.6P)
= [(600)(18)]/[(20,000)(1.0) - 0.6(600)J = 0.550
in.
(3) For comparison with the ligament efficiency method, determine the reinforcement requirements based
only on the shell thickness (without consideration of the nozzle thickness). Since the reinforcement
area available comes only from the shell, the shell thickness will have to be increased. A trial thickness
will be assumed and verified. Try t = 21, = 2(0.550) = 1.100 in. Assume a nominal thickness of
t = 1.25 in.
(4) Limit parallel to the shell surface is X = d or (0.5d + I), whichever is larger. X = 2.50 in. or (1.25
+ 1.25 = 2.50 in.). Use X = 2.50 in.
2'\4"
2'\4"
All holes
2Y2" dia.
LONG.
AXIS
FIG. E5.7
EXAMPLE PROBLEM OF NOZZLE REINFORCEMENT OF SERIES OF OPENINGS, DIV. 1
166 Chapter 5
(5) Examine the longitudinal plane, 1-2.
(a) With an actual center-to-center spacing of 4.5 in., the reinforcing limits of 2X = 2(2.5) = 5.0
in. exceeds the actual spacing of 4.5 in. Therefore, the reinforcement limits overlap, and the rules
of UG-42 apply. Those limits state that no reinforcement area shall be used more than once.
(b) Reinforcement area required is:
A, = dt,F
= (2.5)(0.550)(1.0) = 1.375 in.'
(c) Reinforcement area available in the shell is:
A,
=
(spacing - d)(t - t,)
~
(4.5 - 2.5)(1.25
0.550)
~
1.400 in.'
(6) Examine the diagonal plane, 2-3.
(a) With a circumferential spacing of 3 in. and a longitudinal spacing of 4.5 in., the diagonal centerto-center spacing is p' = [(3)' + (2.25)2Jlf2 = 3.75 in. And,
= tan:" (3/2.25); = 53.13'.
With a diagonal spacing of 3.75 in., the reinforcement limits of 5.0 in. exceed the spacing.
Therefore, the limits overlap, and the rules of UG-42 apply. From Fig. UG-37 of VIII-I, F =
0.68 for
= 53.13'.
(b) Reinforcement area required is
e
e
e
A,
~
dt,F
= (2.5)(0.550)(0.68)
~
0.935 sq. in.
(e) Reinforcement area available in the shell is
A,
=
(spacing - d)(t - Ft,)
~
(3.75 - 2.5)(1.25 - 0.68
x
0.550)
~
1.095 in.'
(7) Both the longitudinal and diagonal planes are satisfactory with t = 1.25 in., and the nozzle thickness
is not considered.
(8) As an alternative, the reinforcement requirements may be based on a combination of shell area and
nozzle area.
(9) As determined from VIII-I, UG-37(c)(l), the minimum required thickness of the nozzle with E =
1.0 is:
tm
~
(PR)/(SE - 0.6P)
~
[(600)(1.25)]/[(20,000)(1.0) - 0.6(600)]
~
0.038 in.
(10) With the spacing that close, it is doubtful that very much thickness over the minimum required
thickness could be added and be able to weld the tubes. Therefore, most, if not all, of the reinforcement
area will come from the shell.
Openings 167
Example 5.8
Problem
Using the rules of VIII-I, determine the minimum required thickness of the same shell given in Example
5.7 using the ligament efficiency rules of section 5.7 instead of the reinforcement rules of section 5.5.
Solution
(1) Determine the minimum longitudinal ligament efficiency or equivalent longitudinal efficiency and
compare it with the longitudinal butt joint efficiency. The lesser efficiency is used to calculate the
minimum required thickness of the shell.
(2) Determine the longitudinal ligament efficiency based on the longitudinal spacing of 4.5 in. as follows:
E ~ (1' - d)/1'
~
(4.5 - 2.5)/(4.5) = 0.44
(3) Determine the equivalent longitudinal ligament efficiency from the diagonal efficiency using Fig.
UG-53.6 of VIII-l as follows:
p' = p
~
3.75 in.; e
~
53.13"; d = 2.5 in.; 1'ld = 1.5, which gives E = 0.38.
(4) Using the minimum efficiency of E ~ 0.38 (which assumes that ligament efficiency is less than butt
joint efficiency), calculate t, from UG-27(c)(l) as follows:
t, = (PR)/(SE - 0.6P) = [(600)(18)]/[(20,000)(0.38) - 0.6(600)]
~
1.492 in.
CHAPTER
6
SPECIAL COMPONENTS,
VIII.. 1
6.1 INTRODUCTION
To meet the design and loading requirements of UG-22 of VIII-I, many design equations, charts, tables,
and curves for most standard components such as shells and heads are provided. However, there are
some design rules for components with special geometries and configurations which require additional
consideration. ForVIII-2,procedures aregiven for design by analysis of componentswith special geometries.
Consequently, very few rules for design of special geometries are given. This chapter contains guidance
for the design of some of those special geometries in VITI-I as follows:
(I) Braced and Stayed Construction (UG-47 through UG-50, and UW-19)
(2) Jacketed Vessels (Appendix 9)
(3) Half-pipe Jackets (Appendix EE)
(4) Vessels of Noncircular Cross Section (Appendix 13)
6.2 BRACED AND STAYED CONSTRUCTION
Rules for hraced and stayed construction are in UG-47 through UG-50 and in UW-19 of VIII-I. Stays are
used in pressure vessels to carry part or all of the pressure loading when it is desirable and possible to
reduce the span and! or thickness of a tubesheet, sideplate, or other pressure component. Opposite stayed
surfaces are "tied" together by staybolts, tubes, or baffles, which carry pressure loading as tension memhers.
Depending on the numher of ties, the thickness of braced and stayed surfaces may he less than wheu the
surfaces are not stayed, hecause the loading is now resisted by both bending moments and bending strength
and hy tensile strength of the stays.
6.2.1 Braced and Stayed Surfaces
For hraced and stayed surfaces tied together with threaded-end or welded-in stayholts of uniform diameter
and symmetrical spacing, the following formulas apply for determining the minimum required thickness
or internal design pressure:
t = p(P/SCl'"
P
=
t'SC/p'
where
t = minimum required thickness of the stayed plate, in.
P = internal design pressure (or MAWP), psi
169
(6.1)
(6.2)
170 Chapter 6
S = maximum allowable stress value in tension at design temperature, psi
p = maximum pitch hetween staybolts, in.
C = constant, the value of which depends on details of the staybolt end design as follows:
C = 2.1 for welded-in stays or threaded-end stays screwed through plates ,,; 7/16 in. thickness
with the threaded ends riveted over;
C = 2.2 for welded-in stays or threaded-end stays screwed into or through plates > 7/16 in.
thickness with threaded-ends riveted over;
C = 2.5 for threaded-end stays screwed through plates and fitted with single nuts outside the plate,
or with inside and outside nuts without washers, and for stays screwed into plates not less
than 1.5 times the diameter of the stayholt measured on the outside of the stayholt diameter.
If washers are used, they shall be at least half as thick as the plate being stayed.
C "".2.8 for threaded-end stays with heads not less,than ,1.3 times the diameter of the stays screwed
through plates or made with a tapered fit and having the heads formed on the stays hefore
installing them and with threaded-ends not riveted over;
C = 3.2 for threaded-end stays fitted with inside and outside nuts and outside washer, where the
diameter of washer is not less than OAp and the thickness of washer is not less than the
thickness, t, of the surface being stayed.
6.2.1.1
Special Limitations for Threaded-End Stay Construction
(a) Minimum thickness of plate to which stays cao be attached, other than outer cylindrical or
spherical plates, is 5116 in.
(b) When two plates are stayed together and oniy one requires staying, the C value is set by
the plate requiring staying.
(c) Maximum pitch for threaded-end staybolts is 8-lI2 in.
(d) Wben the spacing is unsymmetrical due to interference, half of the spacing on each side
of the stay being considered measured to the adjacent stay shall be used for loading.
6.2.1.2
Special Limitations for Welded-In Stay Construction
(a) Required thickness of the plate shall not exceed I 1/2 in. When plate thickness is greater
than 3/4 in., the pitch shall be 20 in. or less.
(b) The maximum pitch for welded-in stays is I5dn where d, is the diameter of the staybolt.
(c) Welded-in stay details shall conform to one of those shown in Fig. 6.1.
(d) Welds do not require radiography.
(e) Welds may require postweld heat treattoent; see US-40(f) of VITI-I.
6.2.1.3 Welded Stays for Jacketed Vessels.
the vessel meets the following criteria:
Welded stays, shown in Fig. 6.2, are permitted when
(a) Vessel design pressure is ,,; 300 psi.
(b) Required thickness of the plate does not exceed lI2 in.
(c) Minimum fillet weld size is not less than the plate tbickncss.
(d) Allowable fillet weld load is calculated according to UW-18(d), and inside welds are visually
examined before assembly.
(e) Maximum diameter or width of tbe bole in the plate is I lI4 in.
6,2.1.4 Welded Stays for Dimpled and Embossed Assemblies. Welded stays may be used in construction of a dimpled or embossed assembly where a dimpled or embossed plate is welded to another
dimpled or embossed plate or to a plain plate and the following rules from Appendix 17 of VIll-1 are met:
Special Components, VIII-1
171
Roundanchor
block
0.7 r min.
1'1
Ihl
101
Idl
tmin.
momp,,,,
penetration
lei
Diameter used 10
satisfy UG-50
requirements
ttl
Diameter used to
satisfy UG-50
r ; nomina! thickness of the thinner stayed plate
requirements
Ihl
191
FIG. 6.1
TYPICAL FORMS OF WELDED STAYBOLTS
... ,..-- ...
II:~lII
/D', --'"
I
, .,
,
I
'\
I t
I I
\.'.....__. _....<,/
I \
!
~ = 1% in.
rnax.
I
Min. width
2dmin.
Stay ber
e
d
I
FIG. 6.2
TYPICAL WELDED STAY FOR JACKETED VESSEL
(a) A welded attachment is made by fillet weld aronnd the edge of the opening. or when the
plate thickness with the opening is OS; 3/16 in. and the hole diameter is OS; I in., the hole
may be filled with weld metal. The allowable load for the weld shall be equal to the product
of the thickness of the plate containing the opening, the perimeter of the opening, the
allowable stress of the weaker of the plates being joined, and a fillet weld joint efficiency
of 0.55.
(b) When MAWP is determined by a DO-lOl proof test of the dimpled or embossed assembly,
a representative panel may be used that is rectangular with at least 5 pitches in each direction
and not less than 24 in. in either direction.
172 Chapter 6
(c) For a plain plate welded with one of the methods listed below, the minimnm required
thickness or the maximum allowable workiug pressure shall be determined by Eq. (6.1) or
(6.2) using a value of C = 3.0. The welding procedures are:
(I) Resistance seam welding
(2) gas tungsten arc seam welding without filler metal
(3) plasma arc seam welding without filler metal
(4) submerged-arc seam welding with filler metal
(d) For a plain plate with other methods of welding than those listed above, the minimum
required thickness or maximum allowable working pressure is calculated by using Eqs.
(6.1) and (6.2) with the appropriate C value.
6.2.2 Stays and Staybolts
6.2.2.1
Load Carried by a Stay
(a) The area supported by a stay is based on the full pitch dimensions with the area of the stay
subtracted. The load carried by that stay is the product of tbe area supported by the stay
times the internal design pressure (MAWP).
6.2.2.2
Minimum Reqnired Area of a Stay
(a) The minimum required area of a stay at its ieast cross section, the smaller of the area at
the root of the threads, or at any lesser cross section, is obtained by dividing the load carried
by a stay (from the calculation in section 6.2.2.1) by the allowable stress of the stay material
at design temperature and multiplying this result by 1.10.
(b) Stays made from two or more parts joined by welding shall have the minimum required
area of cross section of a stay determined the same way as in (a), above, but using a buttweld joint efficiency of 0.60.
6.2.2.3
Special Requirements for Threaded-End Stays
(a) Stays screwed through a plate shall extend two threads minimum and shall be riveted over
or upset, or they shall extend through with enough threads to be fitted with a threaded nut.
(b) If the stay end is upset for threading, it shall be fully annealed.
Example 6.1
Problem
Using the rules in UG-47 through UG-50 and Appendix I7 of VIII-I, determine the maximum allowable
working pressure (MAWP) of a dimpled plate/plain plate assembly which is resistance seam welded on
5 in. centers. Both plates are I /4 in. thick SA-285 Gr.A with a design temperature of 150°F and no
corrosion allowance.
Solution
The allowable tensile stress for II-D for SA-285 Gr.A at 150°F is 12.9 ksi. Since the joint is resistance
seam welded. according to section 6.2.1.4(c), C = 3.0.
Special Components, VIII-I
173
Using Eq, (6.2),
P = [(0.25)'(12,900)(3)/(5)'J = 96.8 psi
Example 6,2
Problem
A flat plate is stayed by welded-in stays. The design pressure is liS psi at a design temperature of 100"F
using SA-516 Gr. 60. Stays are located on 8 in. centers. There is no corrosion allowance. What is the
minimum required thickness of the stayed plate?
Solution
(I) The allowable tensile stress from IT-D for SA-516 Gr. 60 at 100°F is 17.1 ksi.
(2) If the plate thickness is > 7116 in., C = 2.2. If the plate thickness is s: 7/16 in., C = 2.1.
Using C = 2.2 in Eq. (6.1), the minimum required thickness of the plate is
t
=
(8)[(115)/(l7,100)(2.2)]112
= 0.442 in.
The minimum required thickness of 0.442 in. is greater than 0.438 in. (7/16 in.) and C = 2.2 is the
correct factor to lise.
6.3 JACKETED VESSELS
Jacketed vessels, as considered in Appendix 9 of VIII-I, applies also to the jacketed portion of the vessel,
including the wall of the inner vessel and the wall of the jacket, the closure between the inner vessel and
jacket, and other components, such as stiffeners, which carry pressure stresses. Jacketed vessels usually
are chosen to provide a chamber or annulus region in which a liquid or gas under pressure or vacuum is
used to heat and/ or cool the inner vessel contents and to provide an insulation chamber. Half-pipe jackets
attached around the ontside of the vessel are considered separately, in section 6.5.
Although pressure within the inner vessel or in the annulus may be equal to or less than 15 psi, where
that pressure or vacuum combines with a pressure or vacuum within the inner vessel or annulus to produce
a combined loading on the inner vessel waJl or the jacket wall which is greater than the individual loading,
the combined loading is considered within the scope of VIII-I. Table 6.1 shows various combinations of
design pressure for the inner vessel and for the jacket with the actual pressure to be used for design of
components.
TABLE 6,1
EXAMPLE OF PRESSURE USED FOR DESIGN OF COMPONENTS
Design Pressure
in Inner Vessel
-15
+15
+200
+100
Design Pressure
in Annulus
+200
+100
-15
+15
Pressure Used for
Design of Inner Vessel
-215
+15&-100
+215
+100&-15
Pressure Used for
Design of Jacket
+200
+100
-15
+15
174 Chapter 6
6.3.1 Types of Jacketed Vessels
Jacketed vessels are categorized into Types, which provides a way to assign closures and other design
requirements. These Types are shown in Fig. 6.3 and are defined as follows:
Type I-Jacket of any length confined entirely to the cylindrical shell
Type 2-Jacket covering a portion of the cylindrical shell and oue head
Type 3-Jacket covering a portion of the head
Type 4-Jacket with added stay or equalizer rings to the cylindrical shell portion to reduce effective length
Type 5-Jacket covering the cylindrical shell and any portion of either head
--+-.. ,-
-,,
I
I
:
-T
,
T
!
I
L
,,
I
I
'-
L
I
I
.....L
./
I
-'-
I
I
I
I.
,
L
I
I
I
Type 4
I
I
I
,
I
I,
I,
I
./
I
-
TVpe3
I
I
.....L
Type 3 - Jacket covering a
portion of head.
Type 4 - Jacket with addition
of stay or equalizer
rings to the cylindrical
I
'"
cylindrical shell.
shell and one head.
I
,
Type 1 - Jacket of sny length
confined entirely to
Type 2 - Jacket covering 8
portion of cylindrical
I
I
./
"
I
,
I
L
,
,I
I
I
~
,-
/1.
--,,=-
I
I
I
I
L
-L.
,r-
I
-t--
I,
Type 2
,
,-
I,
I
TVpe'
-, ,-
I
I
I
I,
_i..
--t--
... r--~,
,
/.
"
-, /"
./
TypeS
shell portion to reduce
effective length.
Type 5 - Jacket covering
cylindrical shell and
any portion of
eitherhead.
FIG. 6.3
SOME ACCEPTABLE TYPES OF JACKETED VESSELS
Special Components, VIII-I
175
6.3.2 Design of Closure Member for Jacket to Vessel
6.3.2.1
Nomenclature.
The symbols and terms used to design jacket closures are
t, = nominal thickness of the inner vessel wall, in.
t., = minimum required thickness of the jacket wall, in.
= minimum required thickness of the closure member, in.
nominal thickness of the closure member, in.
Ij = nominal thickness of the jacket wall. in.
t; = nominal thickness of the nozzle wall, in,
r = corner radius of the torus closure, in.
Rs·=··outside·.radiusofthe.innervessel,·.in.
R, = inside radius of the jacket, in.
R; = radius of the opening in the jacket at the penetration, in.
P = design pressure in the jacket annulus, psi
S = maximum allowable tensile stress value, psi
j = jacket space, in. This is equal to the inside radius of the jacket minus the outside
radius of the inner vesseL
L = design length of the jacket section as follows:
1. Distance hetween the inner vessel tangent line plus one-third of the head
depth if no stiffeners exist
2. Center-to-center distance between adjacent stiffening rings or jacket closure
3. Distance from first stiffening ring or closure to tangent line plus one-third of
the head depth
a, b, c, Y, and Z = minimum weld dimensions for the attachment of closure members, in.
f rc
te =
6.3.2.2
Closure Design Details.
Closure members between the jacket and inner vessel are designed
as various combinations of simple cantilevers or guided cantilevers, depending on the rigidity of the
attachment details. Specific design requirements shall be met depending upon the type of jacket and type
of closure used. Some acceptable types of closure details are shown in Fig. 6.4. Table 6.2 gives the required
closure member size and weld size details for the various types of jacket closures permitted.
Closure Detail Dimension Requirements for Various Types of Jacket Closures:
;:;::: t rj ; r 2: 3fc; t-. :=; 5/8 in.; Y 2: O.7fc
I,,;=: Ifj; r;=: 31,; 1,,:5 5/8 in.; Y;=: 0.831,
I" ;=: Ifj
I" ;=: 0.707j(PlS)lf2; permits fillet weld with throat;=: 0.71,
Iff;=: 0.707j(P /S)ll2; requires groove weld with throat of If
Iff [from VIII-I, UG-32(g), Eq. (4)] ;=: I fj;
:5 30°
Iff;=: larger of 21f} or 0.707j(P/S)1I2; I f} :5 5/8 in.; Y;=: smaller of 0.751f or 0.751,; Z;=: I}
I,,;=: larger of 21fj or 0.707j(P/S)lf2; Y;=: smaller of 1.51, or LSI,
I,,;=: 1.414[(PR,j)/S]lf2;j = {[(2SID/(PR)] - 0.5(1, + Ij )) ; Y;=: smaller of 1.51, or 1.51,
Meet the details and dimensions shown in the sketch
I f } :5 5/8 in.
I f } :5 5/8 in.; welds shall meet the details shown in Fig. 6.4, sketches (i-I) or (i-2),
which is the same as Fig. 9-5 in VIII-l
[13] For conical and toriconical jackets shown in Fig. 6.4, sketches (f-I) through (f-3) and (g-l)
through (g-6).
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
[11]
[12]
t rc
e
(6.3)
(6.4)
(6.5)
(6.6)
(6.7)
(6.8)
(6.9)
(6.10)
(6.11)
(6.12)
(6.13)
176 Chapter 6
TVpo 1 Jackets
dlmimslon
temln.
,,<Y"O,7tc
:;-Tb
MIn. throBt
TVpes 2 and 4 Jackets
• t
c
Mln.2fc but need
net exceed 1/210.
fc
-1,6f
C
,..
a.83f min.
c
r min. '" 3f c
(Elongated to
maintain min.
threat dimension)
lal
1.25
f
c
min.
1.25fC min.
rmln." 3fc
o
max.
'" BOdeg.
see Nete (1) to
See Note (1) to
sketch e-u
sketch (b·l)
t;
r
(b.31
(b·l)
NOTE (sketch b-fl:
(11 Cloture Bod ,heH one piece cceerruencn
or full penetration butt weld.
Backing strip may be used.
b
() max. = 30 deg.
t,
~
-Jv--l.f--'v-~=-'i
rcmin.
t
R,:..-J
(f·l)
(f-21
FIG. 6.4
SOME ACCEPTABLE TYPES OF CLOSURE DETAILS
Special Components, VIII-l
O.7Y min.
O.7Y min.
(d·l)
(d·2)
y
Backing strip
may be used
(e-2)
FIG, 6,4
(CONT'D)
177
178 Chapter 6
Weld Detail Pet
Fig. UW-13.2 (d)"
Weld Detail Per
Fig, UW·13.2 (el.
,,
,
,
min.
,,
.
tJ nun.
1-7-_,),146 de•.
t-~+-fi
~
, Backing Strip
_..J
2tl min.
Weld Detail Per
Fig. UW-13.2 (e)
May be Used
'I
Y""s+b
1.·21
11·31
_ 0.7 tlmln.
1
J-- 't min.
,...,P-tr ~_ __
'--'---1
Not Less
'e
. :::r. • . ~ ­
-:1- -~'
Than
II
"Plug Weld Per
UW·17
-.'_t
t
/
min.
't
0.83
min, Throat Dimension" fJ
_'t
tl min. ----,
./
,
;>
1.5 tl
(Elongated to
30 deg, max.
Maintain min.
Throat Dimension)
_ tj
R· --
Rj ---
J
1.-61
1.-6'
Sea Welding Details
(1·1) and 0-2)
Torispherlcal ellipsoidal and hemispherical
heads (0. D. of jacket head not greater than 0, D.
of vessel head, or I. O. of jacket head
nominally equal to O. D. of vessel heedl
See Details
Ihl
If·'1 to 1f·31
and
A
A
'3
'3
5/8 in.
'3
\.
\
'r: 5/8 in. max,
'r: 5/8 in. max,
max.
A>B
(/')
B
'i:
[i-1 tal)
I"'~_-/
,
/-~
A
"see Note
V ·See Note
B
_"
li-1 (b)J
A=8
0·2)
A<B
NOTE: *Full Penetration Welds
Conical and Toriconical
Ikl
FIG. 6.4
(CONT'D)
III
/
\I
Special Components, VIII-1
179
TABLE 6.2
CLOSURE DETAIL REQUIREMENTS FOR VARIOUS TYPES OF JACKET CLOSURES
Closure Detail
Fig.6.4(a)
Fig.6.4(b-1)
Fig. 6.4(b-2)
Fig. 6.4(b-3)
Fig. 6.4(c)
Fig.6.4(d-1)
Fig. 6.4(d-2)
Fig. 6.4(6-1)
Fig. 6.4(6-2)
Fig.6.4(f-1)
Fig. 6.4(f-2)
Fig. 6.4(f-3)
Fig.6.4(g-1)
Fig. 6.4(g-2)
Fig. 6.4(g-3)
Fig. 6.4(g-4)
Fig. 6.4(g-S)
Fig. 6.4(g-6)
Fig.6.4(h)
Fig. 6.4(k)
Fig. 6.4(1)
Type 1
Type 2
[1]
[3]
[2]
[3J
[4J
[6J
[7J
[7]
[7]
[7J
[8J
[8J
[8J
[10]
[10J
[10J
[11J
[11J
[11]
Type 3
Type 4
Type 5
[2]
[3J
[3J
[3]
[3J
[3J
[5J
[SJ
[S]
[3J
[3J
[SJ
[9J
[9J
[9J
[9J
[9J
[9J
[9J
[9J
[9J
[9J
[9J
[9J
[10]
[10J
[10J
[11J
[11J
[11J
[10J
[10J
[10J
[11J
[11]
[11J
[12]
[10J
[10J
[10]
[11J
[11]
[11J
[10]
[10J
[10J
[3J
[11J
[11]
[11J
[13J
[13J
NOTES:
[ ] Indicates the dimensional requirements listed below for that combination.
- Indicates that the combination is not permitted.
6.3.3 Design of Openings in Jacketed Vessels
The design of openings in jacketed vessels includes reinforcement for the opening and nozzle detail for
the innervessel and construction details for openings in the jacket. For openings (penetrations) in the jacket
of the type shown in Fig. 6.5, the jacket is considered as stayed and needs no reinforcement calculations.
Only pressure membrane loading is considered for the openings and penetration in these rules. Other
loadings given in UG-22 of VlIl-1 shall be considered.
6.3.3.1 Openings in Inner Vessel. The design of openings in the inner vessel shall be according to
therules given in Chapter 5 for combinationsofloadings due to internal pressure, external pressure (vacuum),
or both. No consideration shall be made for cross-sectional area from the jacket and closure.
6.3.3.2 Jacket Openings and Penetrations. No reinforcement is required for openings in the jacket
when the penetrations are as shown in Fig. 6.5. Jacket penetrations shall conform to that shown in Fig. 6.5
and Table 6.3.
Special Penetration Detail Requirements
[1] Jacket welded to nozzle wall, with details as shown in Fig. 6.5(a)
[2] t~ determined for the shell under external pressure
[3] i; ~ tu
(6.14)
180 Chapter 6
t
Vessel
Wall
t,
I
Attach per
Fig. UW-13.2 leI. If) or Ig}
,
t
I,
I
t,
I
I
,
I'
I--b \, t1'
Jacket
WaH
I
Attach per
Fig. UW-13.2 tel, ttl or (g)
I
I,
• .L iF.'IiFt=
:j1~~'
I
'j
B~'king Strip
Backing Strip
May be Used
Mav be Used
1i'"'2fjmhi;
a»
tj min.
(.1
t
I
Iel
Ibl
Fig.
\
\ tj
,,
Full Penetration
'e2
a e 2tj
Butt Weld Backing
Strip May be Used
Backing Strip
May Be Used
b '" tj
C =- 1.25tc 1
1.·1)
[d}
Co
,I
. Attach per
Fig, UW-13.2 (e), If} or Ig)
~~b
a'" 2t c2
b
»
t e2
C'" 1.25 tel
(8-21
______ R
p
----~
(fl
FIG. 6.5
SOME ACCEPTABLE TYPES OF PENETRATION DETAILS
Special Components, VIII-I
181
TABLE 6.3
PENETRATION DETAIL REQUIREMENTS
Special Requirements
Penetration Detail
[1]
[2J
Fig. 6.5(a)
Fig. 6.5(b)
Fig.6.5(c)
Fig.6.5(d)
Fig.6.5(e-1)
Fig.6.5(e-2)
Fig. 6.5(1)
[3]
[2]
[4J
[4]
[5]
NOTE:
Indicates special penetration detail requirements.
[4] t"l determined for shell under external pressure
t,,2 determined from the following formulas:
When no tubular section exists between jacket and torus,
t-a
= Pr/(SE
-
O.6P)
(6.15)
When tubular section exists between jacket and torus,
t-a
= PRp/(SE - O.6P)
(6.16)
where E = weld efficiency from Table UW-12 of VIII-l for either the circumferential weld in the
torus for the equation using r or for any weld in the opening closure for the equation using Rp ,
radius of penetration.
6.4 HALF·PIPE JACKETS
The rnles in Appendix EE of VIII-l for half-pipe jackets are given only for the design condition where
there is positive pressure inside the head or shell and positive pressure inside the half-pipe jacket and for
NPS 2, NPS 3, and NPS 4 pipe size jackets with vessel diameters between 30 in. and 170 in. Obviously,
there are other combinations of pressure loading that need to be considered separately from these rules.
Some combinations are
(I)
(2)
(3)
(4)
(5)
Positive pressure inside the shell and negative pressure inside the jacket
Negative pressure inside the shell and positive pressure inside the jacket
Negative pressure inside the shell and inside the jacket
Wind or earthquake loading with various combinations
Addition of cyclic loading to any combination
Some of these combinations may be more severe than the conditions considered here.
In addition to the half-pipe jacket configuration, there may be jackets of other geometries such as angles,
channels, and circular segments, as shown in Fig. 6.6.
6.4.1 Maximum Allowable Internal Pressure in Half-Pipe Jacket
The maximum allowable internal pressure, P', in a half-pipe jacket attached to a cylindrical shell is
determined as follows:
P'
= F/K
(6.17)
182
Chapter6
]
R
R
r
R
R
L
FIG. 6.6
SPIRAL JACKETS, HALF-PIPE AND OTHER SHAPES
where
pi
=
maximum allowable internal pressure in the jacket, psi
F = J.5S - S' (but F shall not exceed J.5S), psi
(6.18)
S = maximum allowable tensile stress at design temperature of the vessel, psi
S' = actual longitudinal stress in the shell or membrane stress in the head due to internal pressure and
other axial forces, psi. When axial forces are negligible, S' shall be taken as PR2t. When the
combination of axial force and pressure stress (PR2t) is such that S' would be a negative number,
then S' shall be taken as zero.
K = factor from Fig. 6.7 for NPS 2, from Fig. 6.8 for NPS 3, and from Fig. 6.9 for NPS 4
P = internal design pressure of the vessel, psi
R = inside radius of the shell or head, in.
D= 2R
6.4.2 Minimum Thickness of Half-Pipe Jacket
The minimum thickness of a half-pipe jacket is determined from
T = (P,r)/(O.85S, - O.6P,)
where
T = minimnm thickness of the half-pipe jacket, in.
r = inside radius of the jacket, in. (see Fig. 6.6)
(6.19)
Special Components, VUH
183
'000
9
8
7
6
5
4
3
2
Shellthickness
100
9
8
3/16 in.
.-
7
6
5
4
lILn.
V
........
31B in.
2
........
1/J in.
-"
'0
9
7
3/4 in.
6
1
5
I.
1
4
In.
3
2
z In.
1
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
D
FIG. 6.7
FACTOR K FOR NPS 2 PIPE JACKET
s, = allowable stress of tbe jacket material at design temperatnre,
psi
P, = internal design pressure in the jacket, psi
The weld thickness attaching the half-pipe jacket to the vessel shall have a throat thickness not less than
the smaller of the jacket or shell thickness. Special consideration of fillet welds may be required for vessels
in cyclic service.
184
Chapter 6
1000
9
8
7
6
6
4
3
Shell thickness
I
2
100
9
8
7
6
5
V
4
.,.
3
2
3f1~ in.
-- ---
vlin.
318 in.
1fd in.
V
3/4 in.
I
10
9
B
7
6
6
1 in.
4
3
2 in.
2
1
30
40
50
60
70
80
90
100
110
120
130
D
FIG. 6.8
FACTOR K FOR NPS 3 PIPE JACKET
140
150
160
170
Special Components, VIII-1 185
1000
9
8
7
8
6
4
Shell th1ickness
3
1
3116 in,
2
l-- ~
1/}in.
i"""
100
9
8
7
8
3/8 in.
4
....
3
.......
2
.......
10
}
I
.--
6
1/2 in.
-
3/4 in.
J
1 m.
8
7
8
6
4
2 in.
3
2
1
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
D
FIG,6.9
FACTOR K FOR NPS 4 PIPE JACKET
Example 6.3
Problem
Using the rules of Appendix EE of VIII- I, determine the minimnm required thickness of a cylindrical shell
with an interual design pressure of 350 psi and an externally attached NPS 4 schedule lOS half-pipe jacket
at an internal design pressure of 500 psi in noncyclic service. The inside diameter of the shell is 36 in. and
186 Chapter 6
E = 1.0. The design temperature is 100°F. The shell material is SA-5l6 0r.70. The jacket material is SA53 Or.SIA. There is no radiography of any jacket joints and no corrosion allowance.
Solution
(1) From II-D, aliowable stresses are: SA-516-70 = 20.0 ksi and SA-53-S/A = 13.7 ksi
(2) Minimum required thickness of the cylindrical shell using UO-27(c)(1) of VIII-l is
I,
= (PR)/(SE - 0.6P)= (350)(18)]/[20,000 X 1.0 - 0.6 X 350]= 0.318 in.
Nominal thickness used is
I
= 3/8 in.
(3) From Fig. 6.9 with D = 36 in. and t = 3/8 in.,K = 40.
S'
= PRI2t=
(350 X 18)/(2 X 0.375)
=
8400 psi
P' = F/K= (1.5 X 20,000 - 8400)/40 = 540 psi> 500 psi
(4) Assuming NPS 4 Schedule lOS,
Ij
= 0.120 X 0.875
rj = [(4.5/2) -
=
0.105 in.
0.105] = 2.145 in.
(5) Minimum thickness of the half-pipe jacket using Eq. (6.19) is
t,j = [(500)(2.145)J/[0.85 X 13,700 - 0.6 X 500] = 0.095 in.
< 0.105 in. actual
(6) Minimum fillet weld size is 0.120 X 1.414 = 0.170 in. Use 3/16 in.
(7) Summary: Use shell thickness of 3/8 in.; half-pipe jacket of NPS 4 Schedule lOS; and 3/16 in. fillet
weld size to attach half-pipe jacket to shell.
6.5
VESSELS OF NONCIRCULAR CROSS SECTION
The rules in Appendix 13 of Vlll-l for vessels of noncircular cross section are limited to vessels having
rectangular or obround cross sections. As such, the walls are subject to both tension and bending due to
internal pressure. Stresses are determined in the walls and end plates from pressure loadings, including
effects of stiffening, reinforcing, and staying members. The rules in this section are limited to vessels of
noncircular cross section with a straight longitudinal axis. Cross sections which do not have a straight
longitudinal axis, such as a torus, are not contained in this section.
Often, vessels of noncircular cross section contain openings of various diameters. In addition, the opening
may be of uniform diameter through the wall thickness of the vessel or may be of several different diameters
through the wall thickness. Consideration of the openings is made by ligament efficiency procedures. In
addition to considering many different opening sizes and the effects of stiffeners, stayplates, and reinforcement, this analysis considers vessels with different thicknesses of plate on various sides of the vessel cross
section. Furthermore, using this procedure can enable an engineer to consider not only the effects of opening
efficiency, but also the joint efficiency of butt welded joints. There are no rules given for calculating
openings by the reinforced opening method. If that method is chosen for noncircular cross section vessels,
U-2(g) shall be followed.
A structural frame analysis is used where moments of inertia and stiffness of various members are
determined and equations are developed by equating the end rotations and deflections. From this analysis,
Special Components, VIII-!
187
shears and moments are obtained, which then are used to calculate the membrane stress and bending stress,
comparing them with an allowable stress. For vessels of noncircular cross section that do not have equations
given in Appendix 13 of VIII-I, U-2(g) shall be followed.
6.5.1 Types of Vessels
Although there are many vessels with noncircular cross section, the rules of Appendix 13 are limited to
single wall vessels with essentially rectangular and obround cross sections and one circular vessel with a
single diametrical stay plate. Vessels of rectangular cross section are shown in Fig. 6.10; vessels of
rectangular cross section with stay plates are shown in Fig. 6.11; and vessels of obround cross section witb
and without stay plates and a circular cross section with a single diametrical stay plate are shown in Fig.
6.12. Detailed equations for each of these cross sections plus example problems are given in Appendix 13
of VIII-I.
6.5.2 Basis for Allowable Stresses
The calculated primary membrane stress shall not exceed the allowable tensile stress at design temperature,
(S) given in Il-D, for the material multiplied by the butt-welded joint efficiency (E) when applicable.
When the calculated primary membrane stress and the calculated primary bending stress are combined,
the following limits shall be met:
(1) For a rectangular cross section, 1.5 times the allowable tensile stress at design temperature
multiplied by the butt weld joint efficiency when applicable, 1.5 SE;
(2) For other cross sections (such as structural shapes), the lesser of
(a) 1.5 times tl,e allowable tensile stress at design temperarure multiplied by the butt weld
joint efficiency when applicable, 1.5 SE, or
(b) 0.67 times the yield stress at design temperature, 0.67 Sy, except wben greater deformation is acceptable where 0.90 times tbe yield strength at design temperature-but not
to exceed 0.67 specified minimum yield strength-may be used.
6.5.3
Openings in Vessels of Noncircular Cross Section
As stated previously, the only method for considering openings in vessels of noncircular cross section is
the ligament efficiency method. The reinforced operting method is not described nor considered, except by
U-2(g).
When the ligament efficiency method is used, it appears in the equations for both the membrane stress
and the bending stress. If the vessel or header is welded, the butt weld joint efficiency, E, also must be
determined. The ligament efficiencies, em and es; are applied only to the plates in which the openings
are located.
When both em and eb are less than E, the membrane sttess and bending stress are calculated on the gross
area of the section using E = 1.0. Those membrane and bending stresses are tben divided by em and ei,
respectively, to get the stresses based on the net area.
When both em and e; are equal to or greater than E, the membrane stress and bending stress are
calculated on the gross area of the section using the appropriate E, which depends on the butt weld joint
examination chosen.
188 Chapter 6
r
"
!
a,
a
T
hi'
a
p==
-..
h/2
r
-
h/2
1
d·
d·
+
M,
M
i
hi.
H
I+-
•
:.-
T
"
"
111
'21
{See Notes IH and (4)]
[see Notes (11 and (4)1
Pitch distance to next
L,
L,
-J
D
dl
reinforcing member
'I
-+
C
a
P
dj
+
hi'
h,
A
L,
+
-I-
'31
........
'.
hi'
'41
FIG. 6.10
VESSELS OF RECTANGULAR CROSS SECTION
Special Components, VIII-I
'01
pi,
R INot.11l1
I
t:
A
LL,,'
t
t
.ore
It I
n...~
_
'2· '1 "
(6b)
FIG. 6.10
(CONT'D)
\
\
'.
.1
btl the _ I n eight placoa..
(6al
,,
189
190 Chapter 6
"
t
0
N
•
P
h
N
0
P:;t
h
M
St!JV
'3
P
I
h
Stay
'3
-..
,
-~-
f-.
t
h
'.
M
'2
I-
h
P~
M
Stay
".,
"
"
t
te'
"
•,
'I
P:;;
+',
StAv
t
N
P~
P
"
Stay
N
h/2
0
«n
-l-
"
'2
'2
"
Stl'l¥
"
--
.
+
'2
...-
h/2
0
..
i+-!!..,
-
'2
SIRy
Stay
t
~
,
P=1
"
'I
t91
(101
FIG. 6.11
VESSELS OF RECTANGULAR CROSS SECTION WITH STAY PLATES
!-
Special Components, VllI-!
191
-- r' R---B .
'2
Plteh dl,tlW'left to M_t
d,
-t-1---4-
,..Inforc"",·membtr
l~.
I
I
L,
P"
I- ---
'21
"I
"
'2
StB¥
A
Plate member
L,
131
FIG. 6.12
VESSELS OF OBROUND CROSS SECTION WITH AND WITHOUT STAY PLATES AND
VESSELS OF CIRCULAR CROSS SECTION WITH A STAY PLATE
192 Chapter 6
6.5.3.1 Ligament Efficiency for Constant-Diameter Openings. For plates with constant-diameter
openings, shown in Fig. 6.13. the ligament efficiency for both membrane stress and bending stress is the
same. When the diameters of the openings are different, it is necessary to determine the equivalent diameter
by averaging the two diameter as follows:
D, = O.5(d, + d,)
(6.20)
The ligament efficiencies then are
(6.21)
6.5.3.2 Ligament Efficiency for Multi-Diameter Openings for Membrane Stresses. For many
applications of openings in plates. the opening may have more than one diameter through the plate thickness,
as shown in Fig. 6.14. For example, in air-cooled heat exchangers, the diameter increases through the
thickness, and in rolled-in tube arrangements, lands of slightly larger diameter are used for holding the
tube in place. The equations for determining the ligament efficiency for multidiameter openings for membrane
stress are similar to those for a constant diameter opening, with the major difference being the need to
determine the equivalent diameter of each multi-diameter opening.
14------p
1 4 - - - - - - P -----.....;
FIG. 6.13
PLATE WITH CONSTANT-DIAMETER OPENINGS OF SAME OR DIFFERENT DIAMETERS
Special Components, VIII-l
193
!+-------p--------..{
FIG. 6.14
PLATE WITH MULTIDIAMETER OPENINGS
For each multi-diameter opening, an equivalent diameter, DEl, DEl., etc., is determined as follows:
(6.22)
This calculation is repeated for an adjacent tube, and tbe results are then combined to obtain an equivalent
diameter for the combination of the two multidiameter openings as follows:
(6.23)
If d, and d-a are equal, DE = d..
The ligament efficiency for the membrane stress is:
em
=
(6.24)
(p - DE)/p
6.5.3.3 Ligament Efficiency for Multi-Diameter Openings for Bending Stress. Determining the
ligament efficiency for multi-diameter openings in order to calculate the bending stress requires locating
the neutral axis of the various sets of diameters and thicknesses and determining the effective moment of
inertia. From this, the ligament efficiency for bending stress is then determined. The basic structural
mechanics equations for doing this are
x=
:l: AX/:l:A
(6.25)
I = :l: AX'
(6.26)
where
:l: AX = b oTo(O.5To + n + T, + .. + T,)
+
b,T,(O.5T,
+
b,T,(O.5T,)
+ .. +
7~)
(6.27)
194
Chapter 6
(6.28)
x
~ SAX/SA
(6.29)
Also,
(6.30)
1
~
(boTb)1l2 + (b,Tl)1l2 + (b2Tll1l2 + .. + (b,T;)1l2
+ boTo(O.5To + T, + T2 + .. + T, - X)'
+ b,T,(0.5T, + T, + .. + T. - Xl
+ b2T,(0.5T, + .. + T. - Xl2
+
(6.31)
b.T.(X - 0.5T.)'
c
=
largerofXort - X
(6.32)
The width of a ligament is:
(6.33)
Other properties are:
~
D.5f
(6.34)
(b,JJ)/1I2
(6.35)
e
1
~
And solving:
ell
~
(b,f)
(6.36)
P - [(61)/(t2e)]
(6.37)
With
DE
~
Special Components, VIII-1
195
The ligament efficiency for bending stress is:
e, = (p - DE)/p
(6.38)
Example 6.4
Problem
Using the rules of Appendix 13 of VIII-I, determine the membrane and bending ligament efficiencies in
a pressure vessel of square cross section, in which H = h = 6 in. and t 1 = t-i = 0.75 in. and there is a
single row of 1.5 in. diameter holes on 4 in. center-to-center spacing.
Solution
Using Eq. (6.21), calcnlate the ligament efficiencies as foliows:
em
= eb = (4 - 1.5)/(4) = 0.625
Example 6.5
Problem
A pressure vessel contains a single row of openings that are alternately spaced on 4 in. and 3 in. centerto-center spacings. The opening diameters also alternate, with the first one being 1.5 in. diameter and the
next one being 1.25 in. diameter. Assuming the butt-joint efficiency is higher than the ligament efficiency,
determine the minimum ligament efficiency for setting the thickness of the vessel.
Solution
Using Eq. (6.20), determine the equivalent diameter, DE. as foliows:
DE
=
0.5(1.5 + 1.25)
=
1.375 in.
The ligament efficiency is based on the minimum spacing of p = 3 in. by using the equivalent diameter
of DE = 1.375 in. as foliows:
em = ev
=
(3 - 1.375)/(3)
=
0.542
Example 6.6
Problem
Using the rules in Appendix 13 of VIII-I, determine the membrane ligament efficiency of a pressure vessel
which is 1.50 in. thick and contains a row of multi-diameter openings on 3.50 in. center-to-center spacing,
as shown in Fig. 6.14. The dimensions of ali of the multidiameter openings are
do = 1.625 in.
d, = 1.5 in.
d, = 1.375 in.
To = 0.125 in.
T, = 1.125 in.
T,
=
0.25 in.
1%
Chapter 6
Solution
Using Eq. (6.22), determine the equivalent diameter, d.; of the opening as follows:
d,
=
[(1.625)(0.125) + (1.5)(1.125) + (1.375)(0.250)J/(1.5)
=
1.490 in.
Since the equivalent diameter of each opening is the same, d, = DE, and ligament efficiency is determined
by using Eq. (6.21) as follows:
em = (3.5 - 1.490)/(3.5) = 0.574
Exampie 6.7
Problem
Determine the bending ligament efficiency of the pressure vessel in Example 6.6, above.
Solution
Using the dimensions given in Example 6.6, determine the values for b as follows:
p = 3.5 in.
To = 0.125 in.
T, = 1.125 in.
T, = 0.25 in.
do = 1.625 in.
d, = 1.5 in.
d, = 1.375 in.
b"
=
3.5 - 1.625 = 1.875 in.
1.5 = 2.0 in.
3.5 - 1.375 = 2.125 in.
b, = 3.5 -
b,
=
Then
~ AX =
1.875 X 0.125(0.0625 + 1.l25 + 0.25)
+ 2.0 X 1.125(0.5625 + 0.25)
+ 2.t25 X
0.25(0.125)
= 2.2314
:s A
=
(1.875 X 0.125) + (2.0 X 1.125) + (2.125 X 0.25)
=
3.0156
X = (2.2314)/(3.0156) = 0.7400
I
=
(1/12)[(1.875)(0.125)' + (2.0)(1.125)' + (2.125)(0.25)3J
+ 1.875 X 0.125(0.0625 + 1.125 + 0.25 - 0.74)'
+ 2.0 X 1.125(0.5625 + 0.25 - 0.74)'
+ 2.125 X 0.25(0.74 - 0.125)'
= 0.5672
Special Components, VIII-1 197
c
~
DE ~
larger of 0.74 or (1.50 - 0.74)
3.5 - [(6)(0.5672)]/(1.5)2 (0.76)
~
~
0.76 in.
~
1.526in.
(3.5 - 1.526)/3.5 = 0.564
6.5.4 Vessels of Rectangular Cross Section
One of the least complex vessels of noncircular cross sectionis one with a rectangular cross section. Basic
equations are given here for that cross section, which is shown in Fig. 6.10, sketch (1). Eqnations for other
cross sections are given in Appendix 13-7 of VIlI-l.
(l) Membrane stress:
(a) Short-side plate:
(6.39)
(h) Long-side plate:
(6.40)
S; = PH/2tz
(2) Bending stress
(a) Short-side plate:
(S')N ~ (PeIl2I,)! -
1.5H 2
+
(Sb)Q ~ (Ph 2eIl2I,)[(l
h2[(1
+
+
a.'K)/(1
a.'K)/(1
+
+
K)])
(6.41)
(6.42)
K)]
(b) Long-side plate:
(S')M ~ (Ph2e/12I,)( -1.5
+
(Sb)Q ~ (Ph'cIl21 2)[(1
[(1
+
+
a.2K)/(1
a.'K)/(1
+
+
K)]
K)Jj
(6.43)
(6.44)
(3) Total stress
(a) Short-side plates:
(ST)N
= Eq.
(ST)Q
~
(6.39) + Eq. (6.41)
(6.45)
Eq. (6.39) + Eq. (6.42)
(6.46)
Eq. (6.40) + Eq. (6.43)
(6.47)
+ Eq. (6.44)
(6.48)
(h) Long-side plates:
(ST)M =
(ST)Q = Eq. (6.40)
198 Chapter 6
Example 6.8
Problem
A noncircular cross section vessel has outside dimensions of 7.25 in. X 7.25 in. and is 0.625 in. thick, as
shown in Fig. E6.8. Material is SA-516 Or. 70. Design temperature is 650°F, and design pressure is 150
psi. There is no corrosion allowance. All bull weld joints are radiographed, so that E = 1.0. One side of
the vessel contains a single row of openings which are 2.53 in. diameter on a center-to-center spacing of
3.5 in. Is the assumed thickness of t = 0.625 in. adequate to satisfy the rules of Appendix 13 of Vill-I?
If not, what thickness is required?
Solution
Following the rules in section 6.5.4,
t = 1'1 = 12 = 0.625 in.
h = H = 7.25 - 2(0.625) = 6.00 in.
J = bd 3112 = (l )(0.625)3112 = 0.0203 in.'
(I) The membrane and bending ligament efficiencies according to Eq. (6.21) are as follows:
= (3.5 - 2.53)/(3.5) = 0.277
(2) Determine the stresses according to section 6.5.4 as follows:
(a) membrane stress, using Eq. (6.39) or Eq. (6.40), is
Sm = (150)(6)/(2)(0.625) = 720 psi
7.25"
~I
3.5"
f•
r--
J
L
0.625 ••
FIG. E6.B
EXAMPLE PROBLEM OF NONCIRCULAR VESSEL, DIV. 1
Special Components, VIII-l
(b) Bending stress at midpoint of the side, using Eq, (6.41) or Eq. (6.43) aud
(Sb)N =
(X
=
I and K
=
199
I, is
[(150)(6)'(0.3125)112(0.0203)][ -1.5 + (I + 1)/(1 + I)] = 3460 psi
(c) Bending stress at the comer, using Eq. (6.42) or Eq. (6.44) and
(Sb)Q =
(X
=
1 and K
=
1, is
[(150)(6)'(0.3125)/12(0.0203)][(1 + 1)/(1 + 1)] = 6930 psi
(d) Total stress at the midpoint is
Sm
+
(Sb)M =
720 + 3460
=
4180 psi
(e) Total stress at the comer is:
Sm + (Sb)Q = 720 + 6930 = 7650 psi
(f) From lI-D, allowable stress for SA-516 0r.70 at 650'F is S = 18.8 ksi. Allowable design stresses are:
SE = (18,800)(0,277)
USE
=
=
(1.5)(18,800)(0.277)
5200 psi
=
7810 psi
(g) Calculated stresses vs. allowable stresses:
S; sSE: 720 psi < 5200 psi
S; + S, ,; USE: 7650 psi < 7810 psi
All calculated stresses are less than allowable design stresses; therefore, t = 0.625 in. is OK.
CHAPTER
7
DESIGN OF HEAT EXCHANGERS
7.1 INTRODUCTION
Heat exchangers are considered the workhorse in chemical plants and refineries. They come in all shapes,
sizes, and configurations and are essential in extracting or adding heat to various process fluids. Figure 7.1
(TEMA, 1999), shows various shapes of commonly used heat exchangers.
Design rules for heat-exchanger components are covered in various parts of VIII-I. Details of tube-totubesheet welds are given in UW-20 and Appendix A of VIII-I. The rules for tubesheet design are given
in Appendix AA of VIII-I for U-tube and fixed tubesheets. Three types of tubesheets are covered by the
design. They are simply supported, integral, and extended as a flange. Design rules for flanged and flued
expansion joints are given in Appendix CC of VIII-1 and those for bellows-type expansion joints are given
in Appendix 26 of VIII-I.
Other rules that govern the construction ofheat exchangers are given in the Tubular Exchanger Manufacturers Association standards (TEMA, 1999). These rules govern design, tolerances, baffle construction, and
other details of heat exchangers not listed in VIII-I. Design and construction of expansion joints are given
in the Expansion Joint Manufacturers Association Standard (EJMA, 1998) and Appendix 26 of VIII-I. The
rules govern bellows-type expansion joints and include design, fabrication tolerances, and fatigue.
Design of the tubesheet in a U-tube heat exchanger is based On the classical theory of the bending of a
circular plate subjected to pressure. The perforation of the plate is taken into consideration in VIII-I and
so is the stiffening effect of the attached tubes. Design of the tubesheet in a fixed-tube heat exchanger is
based on the classical theory of the bending of a circular plate on an elastic foundation. The theoretical
design equations in this case are extremely complicated. VIII-l rules include numerous assumptions and
simplifications in order to provide practical design equations. Design equations for tubesheets of U-tube
and fixed heat exchangers are given in the remainder of this chapter.
7.2
TUBESHEET DESIGN IN U-TUBE EXCHANGERS
VIII-I gives the design equations for fonr types of tubesheets. The four types are simply supported, fixed,
extended as a flange and welded to the channel side, and extended as a flange and welded to the shell side.
These four types are shown in Fig. 7.2. The design of tubesheets in U-tube exchangers takes into consideration
the ligament between the tubeholes as well as the stiffness of the attached tubes. The design of the tubesheet
is lengthy but straightforeword.
7.2.1 Nomenclature
The following nomenclature is used in the design equations of tubeshccts with If-tubes. The nomenclature
is based on Fig. 7.2 for tubesheet configuration and Fig. 7.3 for tubesheet geometry. The tube pattern layout
201
202 Chapter 7
FRONT END
pJL
I
E
It.
-- - .....c---
L,
ID
I
ONE PASS SHELL
L.
CHANNEL
F
AND REMOVABLE COVER
~
m-}·------·u---... w
G
" "
t'..I_CJ
~I hm--l---n ~
.~
t::
.."
~r
H
SHEET AND REMOVABLE COVER
~
-e
:..~~
S?
~
._~--
J
10U
TUBE SHEET
~~~
liKE "8" STATIONARY HEAD
N
'~IT
FIXED TUBE SHEET
LIKE "N" STATIONARY HEAD
TUBE~
K
I'
'.1_1-'
SPECIAL HIGH PRESSURE CLOSURE
I
T
~5IT
OUTSIDE PACKED FLOATING HEAD
W
4\'\\
.''-..::>.....:>...
_-_-~_t.:'~.d==::
FLOATING HEAD
WITH BACKING DEVICE
T
'~\'\,,,
:,,~f~{'.L.~-
PULL THROUGH FLOATING HEAD
,,
,,
1
1.
KETIlE TYPE REBOllER
U
U
~
U-TUBE BUNDLE
P
'iJ-: r~
p
5
T
~L~
"
,I
M
DIVIDED FLOW
SHEET AND REMOVABLE (OVER
"
1:--
nxso
LIKE "Ali STATIONARY HEAD
spur HOW
I
N
CHANNel INTEGRAL WITH
--i- I]
T
:::'J
dl.lC
T
1.
DOUBLE
CHANNel INTEGRAL WITH TUBE-
D
~~IT
SPliT FLOW
BONNET (INTEGRAL COVER)
...
L
flXEO TUBE$HEET
,J
",,..""'
""'"
••.o
..",
ON" __
~
TWO PASS SHEll
WITH LONGITUDINAL BAfflE
"","----~-
B
C
REAR END
HEAD TYPES
SHEll TYPES
51ATIONARY HEAD TYPES
X
~I
~ J]
CROSS fLOW
W
~
EXTERNALLY SEALED
FLOATING TUBESHEET
FIG. 7.1
VARIOUS HEAT-EXCHANGER CONFIGURATIONS (TEMA, 1999)
Design of Heat Exchangers
Channel \ .
.--
....-
r
,,r--"l,
1
Shell
",.
I
,(
II
P, :::;
:=P
s
hfJ ...............
"
-
i"'-=
h
G
A
A
,(
t
~
L:J
'--
1......1
(a) Typical Simply Supported If-Tube Tubesheet Arrangement
Channel
's
Shell
Ds
(b) Typical Integral U-Tube Type Tubesheet Arrangement
FIG. 7.2
SOME TYPICAL TUBESHEET DETAILS FOR U-TUBES (ASME, 2001)
203
204
Chapter 7
Chann.I ' \
r:-
\~-
--
rr:
r--t
r-
I
~
A
PI
C
G
A
----
~P,
==::
-..
n
Dc
l
-
D,
-
I.
tc
t
t-
-L..:..-
........I
(c) Typlcallnlegral Channel-Tubesheel If-Tube
Tubesheet Arrangement
Channel
t,
(d) Typlcallnlegral Shell-Tubesheel U-Tube
Tubesheet Arrangement
FIG. 7.2 (CONT'D)
Shell
Design of Heat Exchangers
205
00(j)
ooobo
JP 0,0 . :
ocFCDoo
00000
000
FIG. 7.3
TUBESHEET GEOMETRY
is normally on a triangular or square pattern. Subsequent equations will refer to the following symbols
and definitions:
c, = tubesheet corrosion allowance on the tube side, in.
C = bolt circle diameter of flange, in.
D I = diameter of the cylinder which is integral with the tuhesheet (either D, or D,), in.
Do = equivalent diameter of outer tube limit circle, in.
= 2r o + de
D, = inside shell diameter, in.
D, = inside channel diameter, in.
d, :::= nominal outside diameter of tube, in.
d* = effective tube hole diameter, in.
d* =
MAX
{[d' - 2t, (~)(%)
p} [d, - 2t,1}
E = modulus of elasticity for tubesheet material at design temperature, psi
E, = modulus of elasticity for channel material at design temperature, psi
E, = modulus of elasticity of cylinder which is integral with the tubesheet (either E, or E,), psi
E.• = modulus of elasticity for shell material at design temperature, psi
E, = modulus of elasticity for tube material at design temperature, psi
E* = effective modulus of elasticity of tubesheet in perforated region, psi, obtained from Fig. 7.4
1 - v*
Fe =
E*/E In K
G = diameter at location of gasket load reaction, in.
h = tubesheet thickness, in.
h g = tube-side pass partition groove depth, in.
h, = required tubesheet thickness, in.
K= MAX [ (DI ;, 2/'). (~) ]
206 Chapter 7
0.7
0.'
0.5
i;'"
0.4
0.3
0.2
Np
so.ic
0.25
0.1
0.1
0.50
~.OO
:!OO.10
o
0
0.1
0
0.2
0.3
0.4
0.5
o
0.'
0.1
0.2
0.3
0.4
0.6
0.5
p'
p'
v* (Equilateral Triangular Pattern)
E*JE {Equllat&fal Triangular Pattern}
0.4
0.8
, III
W_"'P
~.Oo
, !1.00
~:TI
o.3
II
III
0.25
0.5
'"'"
~
0.4
0.2
I
0.15
0.3
I
",p
~
SO'.lO
0.2
0.25
0.50
'~2.0C
0.1
0
0
0.1
0.2
0.3
0.4
p'
0.5
~
0.1
,
I
II
o
0.'
o
g).10
0.1
0.2
03
[@
0.4
0.5
p'
E* IE {Square Pattern}
FIG. 7.4
EFFECTIVE POISSON'S RATIO AND MODULUS OF ELASTICITY (ASME, 2001)
0.8
Design of Heat Exchangers
207
itt = expanded length of tube in tubesheet (0 :5 itt :5 h), in.
MAX [(a), (b), (c), ...] = greatest of a, b, c, ...
P = MAX [(P'd)' (P,d)J
PG = pressure acting on the side of the tubesheet which is gasketed (either P'd or P'd)' psi
PI = pressure acting on the side of the tubesheet which is integral (either P'd or P,,), psi
P s = shell-side internal design pressure, psi. For shell-side vacuum use a negative value for P;
PI = tube-side internal design pressure, psi. For tube-side vacuum use a negative value for PI'
Psd = most severe shell-side coincident design pressure
= MAX rep,), 0]. When either tube-side vacuum exists or differential pressure design is specified by the user, use P'd = MAX [(P, - P,), OJ.
Pld
most severe tube-side coincident design pressure
= MAX [(P,), OJ. When eithershell-sidevacuum exists or differential pressure design is specified by the user, use PM = MAX [(P, - P,), 0].
=
P = tube pitch, in.
p* = effective tube pitch, in.
p
(
I _ 8r, UL) 1l2
TfD;
r; = radius to outermost tube hole center, in.
S = allowable stress for tubeshect at tubesheet design temperature, psi
S, = allowable stress for tube material at tubesheet design temperature, psi. For a welded tube, use
the allowable stress for an equivalent seamless tube.
S, = allowable stress for shell material at design temperature, psi
S, = allowable stress for channel material at design temperature, psi
Sf = allowable stress for cylinder which is integral with the tubesheet (either S, or S,), psi
II = thickness of the cylinder which is integral with the tubesheet (either t, or I,), in.
t, = nominal tube wall thickness, in,
t, = shell thickness, in.
tc =
channel thickness, in.
U; = largest center-to-center distance between adjacent tube rows, but not to exceed 4p, in.
W = flange design bolt load, Ib
ex = tube pattern factor
= 0.39 for equilateral triangular pattern
= 0.32 for sqnare pattern
)" = factor from Fig. 7.5
fL = basic ligament efficiency
= p - d,
p
fL* = effective ligament efficiency
p* - d*
p*
v* = effective Poisson's ratio in perforated region of tubesheet, obtained from Fig. 7.4
p = tube expansion depth ratio = itt/h, (0 :5 P :5 I)
7.2.2 Design Equations for Simply Supported Tubesheets
The general geometry of the heat exchanger is usually establisbed by the process engineer. The inside
diameter of the shell and nnmber of required tubes to handle the heat-transfer requirements are obtained
208 Chapter 7
10 0
.,/'
500
./
-"""
v/ V V
v/V-""" ~
/'
v/
~~V V ~
V
/
10
/
7
/
./
/
/
//
/
/
/
.,-/
//
'/
I I V// '//
I
ViJ V~ /'
//; ~
1.0
~
1 1/
I
'IJ
'/ 1/
iIJ
If!
0.1
""2.335.!L(~)1~
hb Dz
o
0.1
0.2
0.3
0.4
0.6
2.825
r:
hb
0.8
0.9
(lLf(EL 3.631(!Lf-
0.7
hb
2t[
FIG. 7.5
CHART FOR DETERMINING A (ASME, 2001)
1.0
200
100
60
20
10
Design of Heat Exchangers
209
by the process engineer. Design of the tubes is based on cylindrical equations as outlined in Chapter 2.
Tubesheet design is governed by one of two equations. The first is based on bending of the tubesheet, and
the second is based on shear of the tnbesheet around its periphery. The bending eqnation is given by
h, = 0.556
(
g)
) In
",(IfilL*)
C.5:*S
G
(7.1)
While the shear eqnation is given by
(7.2)
7.2.3 Design Equations for Integral Construction
Equations (7.1) and (7.2), with slight modification, are also applicable to integral tubesheets. The thickness
hi calculated from Eq. (7.1) is reduced by the stiffness of the shell. This is accomplished by using a reduction
factor, FJ, obtained from Fig. 7.6. Also, the gasket diameter, G, in the various expressions is replaced by
the inside diameter as follows.
h
bs =
0.556 (
~:)
(
0""" )
D"
)
1.::*S F
in
(7.3)
I
(7.4)
where,
h b., = tubesheet thickness for bending due to shell-side pressure.
= tubesheet thickness for bending due to tube-side pressure.
hill
1.0
<, .......
0.9
<,
0.8
<,
0.7
0.6
0,01
0.02
0.03
0.04
0.05
tm/D m
FIG. 7.6
FIXITY FACTOR, F(ASME, 2001)
0.06
0.07
210
Chapter 7
and F 1 is obtained from Fig. 7.6 by using
o;
~
(D,
+
D,)/2
+ (t, + tJ/2
Example 7.1
Problem
A U-tube heat excbanger with a fixed tubesheet has details as shown in Fig. 7.2(b). Check the thicknesses
of the shells, tubes, and tubesheetif the design data are as shown in the table below.
Design
Pressure
Temperature
Shell material
Joint efficiency for shells
Tube material
Tubesheet material
S of shells and tubesheet material
S of tube material
Additional Design Data:
D, = 59.0 in.
t, = 0.035 in.
E = 28,000 ksi
e = 0.39
D, = 52.0 in.
r, = 23.0 in.
E, = 27,000 ksi
UL = 3.5
Shell Side
Tube Side
50 psi
300'F
SA 516-70
0.70
SA 688-304
SA 266-Cl 2
700 psi
3QO°F
SA 516-70
1.0
psi
16,100 psi
20,000 psi
20,000
I.,
=
0.4375 in.
t, = 1.125 in.
d, = 1.0 in.
I' = 1.25 in.
S = 20,000 psi
assume a trial h
S, = 16,100 psi
7.625 inch
Solution
Channel Shell
From Eq. (2.2),
P = 20,000 x 1.0 x 1.125/(26
~
+ 0.6 x 1.125)
843 psi> 700 psi
Shell-Side Shell
From Eq. (2.2),
P
~
20,000 x 0.7 x 0.4375/(29.5 + 0.6 x 0.4375)
= 206 psi > 50 psi
Tubes
Check the I in. 0.0. X 0.035 in. thick tubes for internal and external pressure.
For internal pressure: from Eq. (2.8),
Design of Heat Exchangers 211
p
x
~
16,100
~
1160 psi
1.0
x
0.035/(0.5 - 0.4 X 0.035)
This pressure is greater than the applied internal pressure of 700 psi.
For external pressure: The 50 psi external pressure does not control in this case,
Tube Sheet
Tube Side: The following factors have to be determined in order to calculate the required thickness in
accordance with Eqs. (7.3) and (7.4)
fJ, ~
p
~
-
1.0)/1.25 = 0.2
317.625 = 0.393
Pv ~ 1.25/(1 - (8 X 23 X 3.5)/(1T(47)2)05 = 1.312
d*
= d,
2t, = 1.0 - 2(0.35)
-
=
0.930
or
d* = d, .- 2t, (E,IE)(S,IS)p
~
1.0 - 2(0.035)(27,000/28,000)(16,100/20,000)
= 0.979
use d* = 0.979
fJ,*
=
(p* - d*)/p*
o;
=
(59
=
(1.312 - 0.979)/1.312
+ 52)/2 + (0.4375 + 1.125)/2
t; = (0.4375'"
=
=
0.254
56.281
+ 1.125''')"5 = 1.166
tmlDm = 0.021
From Fig. 7.6, F[ = 0.99
The maximum pressure on the tubeshcet is from the tube side. Hence, the required thickness due to
bending is given by Eq. (7.4).
hb, = 0.556(52/47)039(}90''') X 52 X (700/(1.5 X 0.254 X 20,000»°5 (0.99)
=
0.556 X 0.947 X 52 X 0.303 X 0.99
=
8.21 inch.
The required thickness due to shear is given by Eq. (7.2)
212
Chapter 7
h,
= 700
=
X 47 I (3.2 X 0.22 X 20,000)
2.57 inch.
Tube Side: By inspection, the shell-side pressure condition does not control due to low design pressure.
Hence, tbe required thickness of the tubesheet is 8.21 inch.
7.2.4 Design Eqnations for Integral Constrnction With Tubesheet Extended as
a Flange
The design equations for tubcsheets extended as flanges must include the effect of the bolt load in the
flange. They also include the effect of tube-side and shell-side pressures. The Vlfl-Lprocedure consists of
calculating equivalent moments given by
MG =
MI
2:~o (1 - ~)
we
= -
2'ITDo
(1
G)C
- -
+ PG ~l
(go - 1) (~; + 1)
D;
(D2
16 Do
-
PI -
-
-
) (D7
- + 1)
D;
1
(7.5)
(7.6)
From these two moments, we can determine the effective moments
MG
M,
-
1
D;
32
FaPa
(7.7)
+ r,
and
(7.8)
as well as
M, = M,
+
~1 (3 +
v*)PG
(7.9)
M(3 + v*)P,
(7.10)
and
M, = M2
-
D2
If we define M, to be the larger of M3 and M 4 , then the required thickness of the tubesbeet, h, is the larger of
(7.11)
or
Design of Heat Exchangers
h
PD,
213
(7.12)
a = 3.2fLS
Equation (7.11) does not take into consideration the stiffness of the attached shell. When the thickness
of the attached shell is large, then its stiffness could be utilized to reduce the thickness of the tubesheet.
The design procedure is based on a trial and error basis and begins by using the thickness from Eq. (7.11)
to calculate a factor F given by
1 - v*
E*/E
F~---
(E-A+lnK
E
1
)
(7.13)
Applying F obtained from this equation in lieu of F" to calculate M, through M4 in Eqs. (7.7) through
(7.10). Then calculate the bending stress in the tubesheet from the equation
(7.14)
The designer iterates through Eqs. (7.13) and (7.14) with various tubesheet thicknesses until the value
in Eq. (7.14) is as close as possible to the quantity 1.5S. Using the new value of tb calculate the tubesheet
to shell junction stress as follows
ITb
(7.15)
DO)
6 (
--.J..M--"'P
t} '+' 2
32 [
(7.16)
(7.17)
Tbe stresses from Eqs. (7.16) and (7.17) are limited to the quantity 1.5S. If they are larger, then the
tubesheet and/or the shell thickness must be increased and the iteration started from the beginning.
7.3 FIXED TUBESHEETS
The terms "fixed tubesheets" apply to those beat exchangers in which two tubesheets are used with the
tubes acting as stays and the shell acting as a support at the outside circumference. The rules in VIII· 1
give design equations for various types of edge support for the tubesbeets. They are simply supported,
fixed, and flanged and are shown in Fig. 7.7. The design of such tubcsheets is based on the theory of plates
on elastic foundation and is more complicated than that for tubesheets in U-tube construction.
7.3.1 Nomenclature
Symbols and terms used for fixed tubesheets include the following:
a; = radius of the perforated region, in.
a; = radial channel dimension, in.
as = radial shell dimension, in.
A = outside diameter of tubesheet, in.
214
Chapter 7
t
8e
t
....-Sheli
...----.
Channel """"""
r
as
P,
-
h _
g
e=p,
h ;.....
--... 1-
-
t
+
-
J
A
2
Os
t
--! - ro
+-
e
.
hi
t
tt
s
(a) Tvplcal Two-Side Integral Type Tubesheet Construction
A
as 2
Channel'
Shell
Ib) Typical Shell-Side Integral and Channel-Side Gasketed Type Tubesheet Construction
With Tubesheet Extended as a Flange
FIG. 7.7
SOME TYPICAL DETAILS FOR FIXED TUBESHEET HEAT EXCHANGERS (ASME, 1995)
Design of Heat Exchangers
0,
(e) Typical Shell-Side Integral and Channel·Side Gasketed Type 'rubesbeet Construction
With Tubesheet Not Extended as a Flange
r-
-
-
-
0,
as
P,
.;;.
2
[Lps
-
hg :::::
!
Channel
- -
Os
,-- I--
-
~L
h
r
I,
Jt 1..... ........
- =I
t
+'s
'-Shell
ld) Typical Two-Side Gasketed Type Tubesheet Construction
FIG, 7.7 (CONT'D.)
215
216
Chapter 7
c, = tubesheet corrosion allowance on tubeside
C = bolt-circle diameter, in.
d, = tube outside diameter, in.
d* = effective tube hole diameter, in.
D; = inside channel diameter, in.
D, = inside shell diameter, in.
E = modnlus of elasticity for the tubesheet materialat T, psi
E, = modulus of elasticity for the channel material at T" psi
E, = modulus of elasticity for the shell material at T" psi
E, = modulus of elasticity for tube material at T" psi
E* = effective modulus of elasticity of tnbesheet in perforated region, psi
h. = tubesheet thickness, in.
k, = tube-side pass partition groove depth, in.
k, = height of expansion joint outside of shell envelope, in.
.I = expansion joint factor (.I = 1.0 if no joint).
K, = axial rigidity of expansion joint, total force/elongation, lb/in.
e,., = expanded length of tube in tubesheet, in. (0 :5 f", :5 k)
L, = tube length between onter tubesheet faces, in.
L = L, - 2k, tube length between inner tubesheet faces
N, = number of tubes
p = nominal tube pitch, in.
P, = shell-side design pressure, psi (use a negative value for vacuum)
P, = tube-side desigu pressure, psi (use a negative value for vacuum)
r, = radius to outermost tube hole center, in.
S = allowable stress for tubesheet material at T, psi
S, = allowable stress for channel material at T" psi
S, = allowable stress for shell material at T" psi
S, = allowable stress for tube material at T" psi
t, = channel thickness, in.
t, = shell thickness, in.
t, = nominal tube wall thickness, in.
T ~ tubesheet design temperature, OF
T' = tubesheet metal temperature at the rim, OF
T, = channel design temperature, OF
T; = channel metal temperature at the tubesheet, OF
T, = shell design temperature, OF
T"m = mean shell metal temperature along shell length, OF
T; = shell metal temperatnre at the tubesheet, OF
T, = tubes design temperature, OF
Tt,m = mean tube metal temperature along tube length, OF
UL = largest center-to-center distance between adjacent tube rows, in., but not exceed 4p
(U, = 0.0 if no pass partition)
W = channel flange design bolt load, lb
Cls.m = mean coefficient of thermal expansion of shell material at Ts,m, in. / in. / OF
(X/,m = mean coefficient of thermal expansion of tube material at Tt,ml in.lin.l°F
c ' = mean coefficient of thermal expansion of tubesheet material at T', in./in./°P
ex; = mean coefficient of thermal expansion of channel material at T;, in.lin./op
CY.,~ = mean coefficient of thermal expansion of shell material at T:, in. / in. / OF
'Y = axial differential thermal expansion between tubes and shell
11 = flexural efficiency
Design of Heat Exchangers 21.7
'" = basic ligament efficiency
'"* =
effective ligament efficiency
v
v*
Poisson's ratio of tubesheet material
effective Poisson's ratio in perforated region of tubesheet
Poisson's ratio of channel material
Poisson's ratio of shell material
Poisson's ratio of tube material
=
=
Vc =
Vs =
VI =
p = e",lh, tube expansion depth ratio (0
7.3.2
:5
P :5 I)
Design Equations
The design of fixed tnbesheets is complicated by the fact that the thickness is needed in order to calculate
the stress. Accordingly, the thickness is assumed, and the stress is calculated. If the stress is high, a new
thickness is assumed, and the process is repeated. The procedure consists of the following steps:
In this step, geometric parameters are obtained. It is assumed that the material and the diameter and thickness
of the shell, bonnet, and tubes are known. Also, the tube layout and tube-to-tubesheet junction are assumed
to be known. Then the following parameters are obtained:
L.. *_' _-::-=d_*
!J.* = P
p*
a,
a,
p, = -
a,
a,
Pc = -
218 Chapter 7
where
1(, = shell axial stiffness
t,. (D.. + tJ E.r
L
'IT
K,
=
tube axial stiffness
'IT t r
(dt
t r) E1
-
L
V
2
~ = 3 (I - v,.)
yD
s
+
s
2
t't
'
If the shell is gasketed, set
13,
= O. If the channel is gasketed, set
1
J=--
1 + Ks
KJ
J = 1.0 if there is no expansion joint
'1
x
a
E*
= -
1 - v2
1-
X
E
r-
= [24 (1 _ V*2) N E, t, (d
ll)
I
E* L h"
v = ...2.-3 [13. Ii E.• a, (1
Eh
1 - v;
Q _ p, 1-
where
1
1 - <P 4
+
(1)Zm
a;]'"
13,
=
o.
Design of Heat Exchangers
219
I-V'[ (A) ]
<l>~-'lj-In
2a,
+V
and Z is obtained hom Fig. 7.8.
In this step the equivalent pressures are determined. Assume a tubesheet thickness, h, and calculate the
following parameters:
Q
_ (Zd + Q, Z,)
x:
Q
_ (~ + Q, Zm)
2
X:
zt -
2
Z2 -
0 for configuration a in Fig. 7.7
"Yb =
= Pc - C/(2a o ) for configuration b
=
Pc for configuration c
=
Pc - p, for configuration d
~* ~
IS
"YO'
=
*_
"Yc
-
0'
I-' s
p' (1 + ~, h)
s 6 (l __ v;)
t2
,
(p; - 1) (p, - I)
*
4
- "Is
Q2
I-'c
t'
3
c pc
= (p; -
=
l~
+
IHp,
4
"Yc
U
(I
+
~, h)
6 (1 _ v;)
+ I) _ pi - p, + ,,*
2
Ie
(p, - I) z"JX:
1 + q,Zm
the average temperature of the unperforated rim T, is
T
,
=
T'+T'+T'
s
c
3
T*=T;+Tr
,
T*c
2
~
T~+Tr
-2-
220 Chapter 7
For conservative values of Pt and Pt , use T, = T',
Pt = E, t,
a,
= u (p'! 'Y'f -
P"I*
(x
P' =
s
'
PI
ra; (n -
+
=
P, =
(
1
Xr
P'!
Ti,
"In
v + 2p; v _ p; K',l
r
JK,,(
+ JKAQZl +
+
V,
2
2ao
Po - 1)
I _ (I - J) ~ (D, + h;l) P
JK"t
S
N(d, - t,) (d, - 2t,)
+
and T;
70) - a' (T, - 70)J
N(d, - t,) d,
2a;
s
r7 = T',
lK,f
D,
a;
S
I ) P
JK
I
&,1
Q J X (P; - P;
+
P,
Z2
+
P"
+
e; + Prim)
Step 3
In this step, the effective pressnre and stress are obtained. First calculate Q3' Then with known values of
X, and Q3, use Figs. 7.9 and 7.10 to determine F m• Then calculate the stress
2(Ft "Ie
+ P! "I; + e, 'Ys + P'f "'In + ~
'lTa~
P,(I
Tubesheet Stress
The bending Stress is given by
=
IT
(1.5 *F
m
!L
)
(~)2
P
h _ h;
e
+
q,Zm)
Design of Heat Exchangers
0.80
gLng
0.70
~
0.60
a
>i
I
~
\
0.50
*
",e
0'.40
~
J.
z;
Curves
Zd'
Zm'ar. valid for
0.4.
They are sufficiently accurate to be used for other
values of V--,
!'-J..
\
'\
Zd
0.30
\.
~
0.20
~c
I
~
I
I
~
Zm
~
0.10
A
,,~
--l
,
o
o
~
-~
4
2
8
FIG. 7.8
X, (ASME, 2001)
where
h; = MAX [(h g
-
c,), (0)] for pressure load only
h; = 0 for pressure +
thermal load combined, or thermal load only
For pressure loads, [o ] :5 L5S. For pressure and/or thermal loads, [o ] :5 3S.
The sbearing stress is given by
"
rv:
6
z; z; and z; VERSUS
-
10
12
221
222 Chapter 7
0.7
0.6
\
_\ \
\\ ~
\\ ~
\' t\..
0.5
0.4
"\
0.3
...
\,
O:J ~ o.7
!\"-
O:J - o.
I
O:J = o.6
"' l\'-
......
0.2
O:J = o.8
!
O:J = o. 4
!
\.\"'- .....
O:J = o.3
~ <;
0.1
<;
o
1.0
2.0
3.0
O:J
-
4.0
J.2
I
O:J - O.
I
O:J = o. o
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
13.0
14.0
15.0
16.0
X.
FIG. 7.9
VALUES OF Q 3 BETWEEN 0.0 AND 0.8
where
f'
p -
d,
~--
'r :s;
p
O.8S
Tube Stress
The axial stress is given by
where
Fq = (Zd
+
Q3 Z,) X:12
ICJ"t,o I s: fit s,
If ITt,o is negative, the tubes must be checked against buckling. The maximum permissible buckling stress
limit S"bl for tubes is given by the followiug equatiou.
Design of Heat Exchangers
o. 4
223
0,= ..0.8
I
0,= ..0.7
I
0,= ..0.6
0.3
I
0, = -0 .5
I
......
0,=..0 .4
\
\
0,=..0 .3
\\
r-......
0.1
I
i\
<,
\
I
0,=..0 .2
~ ........
I
0,=..0 .1
I
0, = o. o
o
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
11.0
12.0
X,
FIG. 7.10
VALUES OF Q, BETWEEN -0.8 and 0.0
When Cc
e
i
::::;-,
r
St,bk
When Cc
=
e
i
>-,
r
NOTE: S,.b' shall not he greater than S,.
where
8 y, 1
=
yield stress for tube material at Tn psi.
F1 x
,
1T
2
E
--',
(f~",)
13.0
14.0
15.0
16.0
224
Chapter 7
r=
Vd; +
(d, - 2 t,)2
4
equivalent unsupported buckling length of the tube, in. The largest
value considering unsupported tube spans shall be used.
e=
unsupported tube span, in.
k = 0.6 for unsupported spans between two tubesheets,
0.8 for unsupported spans between a tubesheet and a tube support,
1.0·for unsupported spans between two tube supports.
F, = factor of safety given by
F, = MAX[(3.25 - 0.5 Fq ) , (1.25)]
NOTE: F, need not be taken greater than 2.0.
Shell Stress
VD, t, adjacent to the tubesheets. Using
The shell shall have a thickness t, for a minimum length of 1.8
the calculated values of X, and Q3, calculate QZ2*'
(2, + Q3 2 m ) X;
2
Then calculate memhrane, bending, and total stresses in the shell due to the joint interaction, using the
following eqnations:
12 (1 + 13, h). (§.) (13, t,) (~) PQ *
.
,
1
(.:. PS
I-v;
~ ---.
The value of
0',
E
2··
-
hJ X~·
. 11·
V s fT"m -t,
as
+.
p*)
,
R2
1-'.,
,
Z1
a.21
must be kept below 3.0 S,
Channel Stress
The channels shall have a thickness t, for a minimum length of 1.8 vr;:t; adjacent to the tubesheets.
Calculate membrane, bending, and total stresses in the channel due to the joint interaction, using the
following equations:
Design of Heat Exchangers
225
The value of (T, must be kept below 3.0 S,.
In complying with the above three steps, the designer must consider all loading combinations, such as
shell- and tube-side pressures, metal temperatures, and various mechanical loads. TEMA (TEMA, 1999)
gives a table of various load combinations, which may be evaluated by the designer.
Example 7.2
Problem
A fixed-tubesheet heat exchanger, with the tubesheet extended as a flange, has details as shown in Fig.
7.7(c). The shell has an expansion joint, as shown in Fig. 7.7(b). Check the thicknesses of the shells, tubes,
and tubesheet of the fixed-tubesheet heat exchanger. The design data are as shown in the table below.
Design
Tube Side
Shell Side
Pressure
Temperature
Shell material
Joint efficiency for shells
Tube material
Tubesheet material
S of shells and tubesbeet material
370 psi
3000 P
110 psi
3000P
SA 516-70
1.0
SA 249-316L
SA 516-70
SA 240-316L
0.85
20,000 psi for shell
16,700 psi for channel
20,000 psi for tubesheet
14,200 psi
S of tube material
Additional Design Data
D, = 46.0 in.
tl = 0.083 in.
E = 28,000 ksi
UL = 0
a,
= 9.0
f =
X 10- 6 in.lin.lF
36 in.
46.0 in.
21.5 in.
E, = 27,000 ksi
D,
r,
=
=
€I.X = 3.0 in.
L = 264 in.
Sy.1 = 19,000 psi
t, = 1.0 in.
1.875 in.
E, = 27,000 ksi
6
(XI = 9.0 X 10in./in.lF
N, = 1835
p =
Solution
Channel Shell
From Eq. (2.2),
P = 16,700 X 0.85 X 0.4375/(23 + 0.6 X 0.4375)
=
267 psi> 110 psi
t, = 0.4375 in.
d, = 1.5 in.
E., = 28,000 ksi
a, = 6.26 X 10- 6 in.lin.lF
J = 1.0
226
Chapter 7
Shell-Side Shell
From Eq. (2.2),
p
~
20,000 X 1.0 X 1.01(23.0 + 0.6 X 1.0)
= 847 psi> 370 psi
Tubes
Check the 1.5 in. a.D. X 0.083 in. thick tnbes for internal and external pressure.
For internal pressure: from Eq. (2.8),
P
=
16,700 X 1.0 X 0.0831(0.75 - 0.4 X 0.083)
=
1934 psi
This pressure is greater than the applied internal pressure of 110 psi.
For external pressure: External pressure calculations (made using appropriate External Pressure Chart from
the VIll-1 code) resulted in P = 525 psi, which is greater than the applied external pressure of 370 psi.
Tubesheet
A trial tubesheet thickness of 4.5 inch was initially investigated. It met all of the stress criteria except an
overstress in the tubesheet-to-shell junction. After a number of trials, a thickness of h = 6.75 inches will
be checked.
Step 1: The following parameters are calculated
=
p
d*
~
~
lih
3.0016.75
= 0.444
[1.5 - 2(0.083)(27128)(14.22120)0.444]
1.450 in.
or,
d* = 1.5 - 2(0.083)
~
1.334 in.
use d* = 1.450 in.
Qo
~
21.5 + 1.512 = 22.25 in.
p*
=
1.8751(1 - (2 X 21.5 X 0)1('IT(22.25)2)))v2
fL* ~ (1.875 -
1.450)11.875
+ 1.012
=
=
1.875
0.227
a,
~
46.012
p,
=
23.5122.25
a,
~
46.012 + 0.437512 = 23.219
p,
=
23.219122.25
x,
~
1 - 1835((1.5 - 2(0.083»12(22.25)2 = -0.649
~
=
23.50
1.0562
=
1.0436
Design of Heat Exchangers
x,
~
1 - 1835(1.512(22.25»)2
K, = 'JT(1.0)(46
+
~
-1.085
1)(28 X 10')/264 = 15.6604 X lO'
=
K, ~ 'JT(0.083)(1.5 - 0.083)(27 X 10')/264
K,., = 15.6604
~, ~
B,
lO'/(1835)(37,788)
X
(3(1 - 0.3'))"'/«(46
- 0.3'»)'14/«(46
=
227
~
37,788
0.2258
+ 1.0)/2)(1.0»'" = 0.2652
+ 0.4375)12)(0.4375))'"
~
0.4033
J = 1.0
hlp
=
From Fig. 7.4, v* = 0.42
6.75/1.875
=
3.6
=
£*1£
and
0.19.
E* = 0.19 X 28.0 X 106 = 5.23 X 10'
N
x,
~
~
0.19(1 - 0.3')/(1 - 0.42 2 )
= 0.21
[24(1 - 0.42 2)(1835)(27 X 0.083 X (1.5 - 0.083) X 22.25')/(5.23 X 264 X 6.75')]'"
(
= 3.404
v=
(2/(28 X 6.75'))[«0.2652 X 1.0' X 28 X 23.5)/(1 - 0.3'»)(1
+
~
«0.4033 X 0.4375' X 27 X 23.219)/(1 - 0.3 2)(1
(2/(28 X 6.75'))[191.7600(4.3923)
+
+
+
0.2652 X 6.75
+
0.4033 X 6.75
+
0.4033' X 6.75'/2)]
23.2667(7.4277)]
= 0.2358
<P
=
+
«1 - 0.3 2)/0.21)[ln(48.0/(2 X 22.25)
From Fig. 7.8, Zd
=
0.04, Z,
Q,
~
=
0.08, Z;
=
0.2358]
~
1.350
0.43.
(1.0562 - 1 - 1.35(0.08))/(1
+ 1.35(0.43»)
~
-0.033
Step 2: The effective pressure is calculated as follows
'I
~
Q"
~
(0.04
+
(-0.033)(0.08))(3.404)'/2
~
2.50
Q,2
= (0.08
+
(-0.033)(0.43)(3.404)'/2
~
4.42
[9 X 10-'(300 - 70) - 6.26 X 10-'(300 - 70)](264)
'Yb = 0
0.2652' X 6.75'/2)
=
0.166
228
Chapter 7
'It = 0.2652'(1.0)'(1.0562)3(1 + 0.2652(6.75»/6(1 - 0.3')
'I,
=
(1.0562' - 1)(1.0562 - 1)/4 - 0.0423
'If
~
0.4033'(0.4375)'(1.0436)3(1
'I,
=
(1.0436' - 1)(1.0436
U
=
+
(0.08
=
0.0423
-0.0407
+ 0.4033(6.75»/6(1 - 0.3') = 0.0241
+ 1)/4 - (1.0436 3
-
(1.0562 - 1)(0.43))(3.404'/(1
T, = (300
~
+ 300 + 300)/3
n
= (300
+ 300)/2
= 300
n
= (300
+ 300)/2
= 300
1.0562)/2
+
+ 0.0241 = 0.0294
1.350(0.43»
~
8.85
= 300
Pf = «27)(0.4375)/23.219)/[9(300 - 70) - 6.26(300 - 70)] = 320.61
Pf = «28)(1.0)123.5)/[6.26(300 - 70) - 6.26(300 - 70)] = 0.0
Pi
=
8.85[0(0.0423) - 320.61(0.0241)]
P;
=
[-1.085
=
-68.38
+ (1835(1.5 - 0.083) 1.5/2(22.25)')(0.3) + 2(1.0562)'(0.3)/0.2258
- (1.0562' - 1)/(1.0)(0.2258) - [(1 - 1)/(1.0(0.2258»][(23.5)(0)/22.25'][(46
+
0)/46])
P;
=
(-1.085
+ 1.1818 + 2.9643 - 0.5118 - 0)370
P;
=
[-0.649
+ ((1835)(1.5 - 0.083)(1.5 - 2 X 0.083)/2(22.25)')(0.3) + 1/(1)(0.2258)] 110
P;
~
531.38 psi
p,
=
«1835)(37,788)17r(22.25)')0.166
p"
=
-
Prim
=
-C
[(0)/(27r(22.25)')](8.85)(0)
8.85[370( -0.0407)
P, = «1.0)(0.2258)/[1.0
+
=
=
=
943.24 psi
7400.96
0
+ 110(0.0294)]
= 104.65
+ (1.0)(0.2258)(2.51 + (1.0562 - 1.0)(4.42)]} (943.24 - 531.38
7400.96 - 68.38
+
0
+
104.65)
= (0.2258/1.6228)(7849.09) = 1092 psi
Design of Heat Exchangers
Step 3: Determine the actual stress in various components as follows
Q, = -0.033
+ {2[(Jl0)(0.0294) + (320.61)(0.0241) + (370)( -0.0407) + (0)(0.0423)]
+ (0)(0)l7r(22.25)2)/(1098)(1 + (1.350)(0.43»
Q3
=
-0.033 + [2(3.234 + 7.7267 - 15.059 + 0) + 0]/1735.39
Q3
=
-0.033 - 0.0047
From Fig. 7.10 with X,
=
-0.0377
= 3.404 and Q, =
~0,0377,
we get F",
= 0.065
Tubesheet Stress
o,
=
(1.5 X 0.065/0.227)(2 X 22.25/6.75)2(1092)
=
20,385 psi < 3S
I" = (1.875 - 1.5)/1.875 = 0.2
7 =
<Y,
=
0.8S = 16,000 psi
(11(2 X 0.2»(22.25/6.75)(1092) = 9000 psi < 16,000 psi
Tube Stress
F,
=
(0.04)
+
(-0.0377)(0.08»(3.404'/2)
=
2.4828
S", = {[370(-1.085) - 110(-0.649)] - 1092(2.4828)}/«-0.649) - (-1.085»
=
(-401.450 + 71.390 - 2711.22)/0.436
=
-6975 psi
Since this stress is in compression, buckling allowable stress must be determined.
r
=
[1.52 + (1.5 - 2(0.083»2]'"/4
kIlr
=
(1.0)(36)/0.5018 = 72
C,
=
[(2)(7T)2(27,000,000)/19,000]'" = 167.5
F,
=
3.25 - 0.5(2.4828)
=
= 0.5018
2.024
or
F, = 1.25
use F,.
=
2.0 by definition
S,,", = (19,00012.0)[1 - (36/0.5018)/(2)(167.7)]
= 7465 psi> 6975 psi
229
Chapter 7
230
Shell Stress
Q,,*
~
[0.08
+ (- 0.0377)(0.43)](3.404)'/2
= 4.2823
«.; = (22.25' X 1092)/(2)(23.5)(1.0) + (110)(23.5)/(2)(1.0)
+
«370 - 110)/2)(22.25/1.0562 X 1.0)(1.0562' - 1.0)
11,566
'Y,b
~
+ 1293 + 316
= 13,175 psi
12(1 + (0.2652)(6.75)/2)(28/28)(0.2652 X 1.010.21)(22.253/6.75 3 X 3.404')(1092)(4.2823)
+ (1 -
0.3')~'[370
- 0.3(13175)(1.0/23.5)
12(1.8952)(1)(1.2629)(0.2668)(1092)(4.2823)
",.m
=
35,830
=
(T,.,
=
+ OJ(0.2652)'(23.5)'
+ (1.0989)(201.8) (0.0703) (552.25)
+ 8609 = 44,440 psi
13,175
+
=
44,440
57,615 psi < 60,000 psi
Channel Stress
",.m
= (110)(23.219)/(2(0.4375) = 2918 psi
(T,.b
= -12(1
+
+ (1 -
(0.4033)(6.75)/2)(27/28)(0.4033
0.3')~'[(l
- 0.3/2)110
x
0.4375/0.21)(22.25 3/6.75 3 X 3.404')(1092)(4.2823)
+ 320.61](0.4033)'(23.219)'
-12(2.3611)(0.9643)(0.8402)(0.2668)(1092)(4.2823)
- 28,640
",.m +
+ 39,916
+ 11,276
'Y,.b
=
2918
38
=
3 X 16,700
=
=
~
+ (1.0989)(414.11)(0.1627)(539.122)
11,276 psi
14,194 psi
50,100 psi
7.4 EXPANSION JOINTS
Two types of expansion joints are covered in VIIJ~ 1. The first is bellows-type and is given in Mandatory
Appendix 26 of VIII~1. Some typical bellows-type expansion joints are shown in Fig. 7.11. They inclnde
unreinforced as well as reinforced construction. The required thickness of an unreinforced bellows due to
pressure is obtained from the equation
t = P(d
+
w)18(1.l4
+
4w/q)
(7.18)
where t, d, w, and q are defined in Fig. 7.11. The stress in the point due to expansion is obtained from
any structural analysis method. VITI-I also gives rules for the design of reinforcing rings as well as
fatigue analysis.
The second type of expansion joints are flanged and flued expansion joints, shown in Fig. 7.12. They
are discussed in Non-Mandatory Appendix CC of VITI~ 1. The method of analysis for pressure and expansion
is not explicitly given in VIII-I, but is left to the discretion of the designer. Some simplified equations
are given in VIII-I for fatigue analysis. However, more experimental data are needed to substantiate
these equations.
Design of Heat Exchangers 231
f
ddlam.
(al Unfeinforced Bellows
I
A
/~_'
A,
I
\
f
Reinforcing rings
L,..A
I
V
A-A
Equaliling ring
I
End equaliling ring
A,
GENERAL NOTE:
Nominal r" 3tm
(b)
~f=r
t
Reinforced Bellows
FIG. 7.11
BELLOWS·TYPE EXPANSION JOINTS
(a) Flanged Only
(b) Flanged and Flued
FIG. 7.12
FLANGED AND FLUED EXPANSION JOINTS
dd"m.
CHAPTER
8
ANALYSIS OF COMPONENTS
IN
VIII"2
8.1 INTRODUCTION
Section VIII-2 requires stress analysis of vessel components when explicit design formulas are not given.
This includes flued-in heads, head-to-shell junctions, expansion joints, thermal stresses, and stresses in
componentsdueto loads otherthanpressure. In performing the stressevaluation,the designermust determine
the maximum stress at a given point or location. When compnter programs such as ANSYS and NASTRAN
are used to determine the stress, the output usually consists of the total combined stress at a given point.
This stress must then be separated into its components of membrane, bending, and peak stresses, TIlls is
necessary in order to compare each of these components to a corresponding allowable stress given in Vill-2
or to properly establish an allowable fatigue life. In this chapter only stress categories, stress concentrations,
combinations of stresses, and fatigue evaluation are discussed in accordance with the definitions and
requirements of VIlI-2.
8.2 STRESS CATEGORIES
Stress in any component and location is classified by VIIl-2 as one of three categories-primary, secondary,
and peak stresses. Primary stress, such as hoop stress in a cylinder due to internal pressure, is developed
by tbe imposed loading and is necessary to satisfy the laws of equilibrium. It is not self-limiting in that
gross distortion or failure of the structure will occur if its value substantially exceeds the yield stress. This
primary stress is divided into two subcategories in VIlI-2. They are primary membrane and primary bending
stresses. The longitudinal and circumferential stresses in a cylinder due to internal pressure are classified
as primary membrane stress. The primary membrane stress is again subdivided into two categories in Vlll-2.
They are referred to as general primary membrane and local primary membrane stresses. Examples of these
primary stresses are given in Table 8.1. Primary bending stress in VlII-2 refers to such items as the bending
of a flat cover or a dished head due to internal pressure.
Secondary stress is developed when the deformation of a component due to applied loads is restrained
by other components. Secondary stress is self-limiting in that local yielding can redistribute the stress to a
tolerable magnitude without causing failnre. An example of secondary stress is the bending stress that
develops at the attachment of a body fiange to the shell. This attacbment is referred to as a gross structural
discontinuity. Other examples of gross structural discontinuity are given in Table 8.2. Another example of
secondary stress is certain thermal stresses. These are referred to as general thermal stress. A typical example
of this stress is the longitudinal bending stress that occurs along a vessel skirt due to temperature gradients
along the length of the skirt. Other examples of general thermal stress are given in Table 8.3.
233
234 Chapter 8
TABLE 8.1
PRIMARY STRESS CATEGORY
I
Primary Stress
I
I
I
Membrane
I
I
I
I
I
General
I
Bending
I
I
I
Local
.I
I
I
Examples of a local primary
A genera! primary membrane
stress is one that is so
distributed in the structure
that no redistribution of load
occurs as a result of yielding.
membrane stress are the membrane
stress in a shell produced by
external load and the moment at a
permanent support or at a nozzle
connection.
An example is the stress in a
circular cylindrical shell due
to internal pressure.
An example is the bending stress
in the central portion of a flat head
due to pressure.
-
TABLE 8.2
STRUCTURAL DISCONTINUITY
I
Structural
Discontinuity
I
I
I
Gross Structural
Discontinuity
I
This is a source of stress or strain
intensification that affects a relatively
large protion of a structure and has a
significant effect on the overall stress or
strain pattern. Examples of gross
structural discontinuities are
head-to-shell and tlanqe-to-sheil
junctions, nozzles, and junctions
between shells of different diameters or
thicknesses.
I
Local Structural
Discontinuity
I
This is a source of stress or strain
intensification that affects a relatively
small volume of material and does not
have a significant effect on the overall
stress or strain pattern or on the
structure as a whole. Examples of
local structural discontinuities are
small-nozzle-to-shell junction,
local lugs, and platform supports.
I
Analysis of Components in VIII-2 235
TABLE 8.3
THERMAL STRESS
Thermal Stress
I
I
This is a self-balancinq stress produced by a
non-unltorrn distribution of temperature or by
different thermal coefficients of expansion.
Thermal stress is developed in a solid body
whenever a volume of material is prevented
from assuming the size and shape that it
normally should under a change in temperature.
I
I
General
I
General thermal stress is classified as a
secondary stress. Examples of general
thermal stress are
1, Stress produced by an axial termperature
distribution in a cylindrical shell.
2. Stress produced by the temperature
difference between a nozzle and the shell to
which it is attached.
3. The equivalent linear stress produced by
the radial temperature distribution in a
cylindrical shell.
I
I
Local
I
I
Local thermal stress is associated with
almost complete suppression of the
differential expansion and thus produces
no significant distortion. Examples of local
thermal stresses are
1. Stress in a small hot spot in a vessel wall.
2. The difference between the actual stress
and the equivalent linear stress.
3. Thermal stress in a cladding material.
The third category of stress defined in VIIl-2 is peak stress. Peak stress is so local that it does not cause
any noticeable distortion in a component, but it may cause fatigue cracks or brittle fracture. Examples of
peak stress are notch concentrations; local hot spots; local structural discontinuity, as defined in Tahle 8.2;
and local thermal stress. as defined in Table 8.3
VIIl-2 establishes limits for the tbree stress categories discussed so far. These limits are given in Table
8.4. The rationale for these limits are given in various publications (see such references as ASME, 1968;
ASME, 1969; and Jawad and Parr, 1989). VIIl-2 also lists the stress categories for some commonly
encountered loading conditions and vessel components. These are given in Table 8.5.
Example 8.1
Problem
A cylindrical shell with a fiat cover, see Fig. E8.1, is subjected to an internal design pressure of 800 psi
and an internal operating pressure of 700 psi. The allowable stress intensity value for the material from
II-D is 20 ksi. What stress intensity values should be calculated at sectious a-a and b-b, and what are the
allowable stress intensities at these locations?
Solution
Section a-a
From Table 8.5, fiat heads develop general primary membrane stress, Pm, and primary bending stress, Pb,
at the central region due to the internal design pressure of 800 psi. From Table 8.4 the allowable general
236
Chapter 8
TABLE 8.4
STRESS CATEGORIES AND THEIR LIMITS (ASME VIII-2)
Stress
Category
Description
Primary
General Membrane
Average primary
(For ex-
stress across
amples,
solid section,
Excludesdiscon-
see Table
4-120.11
tinuities and
concentrations.
Produced only by
mechanical loads.
Local Membrane
Average stress
across any
solid section,
Considersdiscontinuities
but not concentrations.
Produced only
by mechanical
loads~
Bending
Component of
primary stress
proportional
to distance
from centroid
of solid
Seeon ary
Membrane
plus Bending
Self-equilibrating
III Increment added
to primary or secondto satisfy conary stress by a continuityof structure.
centration (notch).
stress necessary
Occurs at structural discontinuities. Can be
caused by mechan-
section. Excludes discontinuities and
ical load or by
differerifiallher'"
mal expansion.
concentrations;
Produced only
by mechanical
Symbol
(2) Certain thermal
stresses which may
cause fatigue but
not distortion of
ve!i!iel·snape~
Excludes local
loads.
stress concentrations.
P,
[NOI. (31]
Peak
Combination
of stress
components
Q
F
I
,
,I
I
I
I
and allowI
able limits
of stress
intensities.
I
I
I
t..,..- __ J.
I
•
I
I
I
I
I
I
I
•
I
L-----...,.I+----
J
I
I
Use design loads
- -
-
_
Use operating loads
GENERAL NOTES,
(a) The stresses in Category Q are those parts of the total stress which are produced by thermal gradients, structural
discontinuities, etc., and do not include primary stresses which may also exist at the same point. It should be noted,
however, that a detailed stress analysis typically gives the combination of primary and secondary stresses directly and,
when appropriate, this calculated value represents the total of Pm (or PL) + Pb + Q and not Q alone. Similarly, if the
stress in Category F is produced by a stress concentration, the quantity F is the additional stress produced by the
notch, over and above the nominal stress. For example, if a plate has a nominal stress intensity, S, and has a notch
with a stress concentration factor, K, then Pm = 8, Pb = 0, Q = 0, F = Pm (K - 1) and the peak stress intensity
equals Pm + Pm (K ~ 1) = KP m•
(b) the kfactors are given in Table AD·150.1.
NOTES,
(1) This limit applies to the range of stress intensity (see 5·110.3). The quantity 38m is defined as three times the average
of the tabulated 8 m values for the highest and lowest temperatures during the operation cycle. In the determination of
the maximum primary-plus-secondary stress intensity range, it may be necessary to consider the superposition of
cycles of various origins that produce a total range greater than the range of any of the individual cycles. The value
of 38m may vary with the specific cycle, or combination of cycles, being considered since the temperature extremes
may be different in each case. Therefore, care must be exercised to assure that the applicable value of 38m for each
cycle, and combination of cycles, is not exceeded except as permitted by 4-136,7.
(2) Sa is obtained from the fatigue curves, Figs. 5-110.1, 5-110.2, and 5-110.3. The allowable stress intensity for the full
range of fluctuation is 2Sa (see 5-110.3).
(3) The symbols Pm, PLo PI» Q, and F do not represent single quantities, but rather sets of six quantities representing the
six stress components 0'1' Ui, 0'" 'Tff, "if> and 1'11-
Analysis of Components in Vlll-2
237
TABI.E 8.5
CI.ASSIFICATION OF STRESSES (ASME VIII·2)
Vessel Compoftfllt
Cylindrical or
spherical shell
Any shell or
head
Dished head or
conical head
Flat head
Type of Stress
Internal pressure
Gt!Oeral membrane
Gradiern··thtough
plate thickness
Axial thermal
gradient
Membrane
Bending
Q
Q
Junction with head
or flange
Internal pressure
Membrane
Bending
Q
Any section across
entire vessel
External load or
moment, or intemal pressure
General membrane
averaged across full
section. Stress compcnent perpendicular
to cross section
External load or
moment
Bending across full
section. Stress compcnent perpendicular
to cross section
S,hj!lI.plilter.emote
from discontinuities
P,
r;
e;
Near rczrle or
other opening
External load
moment, or in~
temal pressure
Local membrane
Bending
Peak (fillet Or comer>
Q
Any location
Temp. diff. between shell and
head
Membrane
Q
Bending
Q
Crown
Internal pressure
Membrane
Bending
P,
Membrane
Bending
Q
Membrane
Bending
P,
Membrane
Bending
Q [Note (2)]
Knuckle or junction
to shelf
Internal pressure
Center region
Internal pressure
Junction to shell
Perforated head
or shell
Classification
Origin of Stress
Location
Internal pressure
Typical ligament
in a uniform
pattern
Pressure
Isolated or atypical
ligament
Pressure
P,
F
e;
P t [Note u)]
r:
P,
Membrane (avo thru
cross section)
Bending (avo thru
width of !ig., but
gradient thru plate)
Peak
p.
Membrane
Bending
Peak
Q
P,
F
F
F
238 Chapter 8
TABLE 8.5 (CONT'D)
Vessel Component
Nozzle
location
Origin of Stress
Type of Stress
Cross section
perpendicular to
Internal pressure
General membrane
nozzle axis
Nozzle waH
Cladding
Any
or external load
(av. across full section).
p.
or moment
Stress component perpendicular to sectlon
See 4-138
External load or
moment
Bending across
nozzle section
See 4-138
Internal pressure
General membrane
Local membrane
Bending
P", See 4-138
P,
Q
Peak
F
Differential
expansion
Membrane
Q
Q
Bending
Peak
A",
Any
F
F
F
Equivalent linear
stress [Nole {4Jl
Q
temperature
distribution
Nonlinear portion
F
Radial
[Note (.)}]
Any
p.
Membrane
Bending
Differential
expansion
Any
Classification
of stress distribution
Any
Stress concentration
F
(notch effect)
NOTES:
(1) Consideration must also be given to the possibility of wrinkling and excessive deformation in vessels with large diameterto-thickness ratio.
(2) If the bending moment at edge is required to maintain the bending stress in the center region within acceptable limits,
the edge bending IS classified as p/}; otherwise, it is classified as Q.
(3) Consider possibility of thermal stress ratchet.
(4) Equivalent linear stress is defined as the linear stress distribution which has the same net bending moment as the
actual stress distribution.
67!l' THK,
e
b_
I' THK.
48' 1.0,
FIG. E8.1
_b
Analysis of Components in
VIII~2
239
primary membrane stress intensity, Pm, is equal to S; (20 ksi). The allowable primary bending stress intensity,
Ph, is equal to 1.5Sm (30 ksi).
Section b-b
From Table 8.5, flat heads develop local primary membrane stress, PL , and secondary stress, Q, at the
junction with the sbell due to internal pressure. From Table 8.4 the allowable local primary membrane
stress, PL , due to the design pressure of 800 psi is equal to 1.5Sm (30 ksi). The total allowable stress due
to local primary membrane plus secondary stresses (h + Q) is equal to 3Sm (60 ksi), It should be noted
that the two stress values, PL + Q, must be calculated at the operating pressure of 700 psi rather than at
the design pressure when comparing them to 3Sm , as sbown in Table 8.4.
8.3 STRESS CONCENTRATION
The stress concentration at a given location must be included in the stress analysis in order to establish the
fatigue life of that component. VIII-2 lists a few stress concentration factors for fillet welds and nozzle
penetrations due to internal pressure, as shown in Table 8.6. All other stress concentration factors are
usually obtained from handbooks such as Peterson (Peterson, 1974), experimental data, or a detailed stress
evaluation using finite element analysis. It must be remembered that substantial inaccuracies could occur
in using the finite element analysis if a large mesh is used near a stress concentration. The designer must
exercise great judgment in establishing the correct mesh size near such concentrations in order to obtain
accurate results.
Example 8.2
Problem
Categorize the stresses at section b-b in Example 8.1 if a stress concentration factor (SCF) of 4.0 is used
at that location due to weld details.
Solution
Section b-b
From Example 8.1, flat heads develop PL and Q at the juuction with the shell due to internal pressure.
Similarly, PL due to design pressure is equal to 1.5Sm (30 ksi). The quantity PI. + Q due to operating
pressure is equal to 3Sm (60 ksi). Also, from Table 8.4, the quantity (SCF)(PL + Q) at the operating pressure
of 700 psi must be used to find the quantity S, in determining the fatigne life of this section.
TABLE 8.6
SOME STRESS CONCENTRATION FACTORS USED IN FATIGUE
Location
Stress
Co ncentratlon
Fillet welds
NOZZle in spherical segment
NOZZle in cylindrical segment
Tangential nozzle in cylinder
Backing strips
4.0
2.2
3.3
Bolts
4.0
5.0
Crackllke defect
5.5
2.0 membrane
2,5 bending
VIII-2 Paragraph
5-112
4-612
4-612
4-614
AD-412.1
5-122
5-111
240 Chapter 8
8.4 COMBINATIONS OF STRESSES
In order to compare the actual stress at a given location to the allowable stress in VIII-2, the designer must
categorize the calculated stress as primary, secondary, or peale The designer must then combine the
categories in the appropriate fashion in order to compare them to the allowable stresses given in Table 8.4.
Identifying the stress category may be difficult sometimes, since the output of many finite element calculations
are prograrmned to display only the principal, or effective, stress at a location. Thus, the designer has to
either instruct the program to itemize the stresses or manually separate them into various categories.
Separating the stress output from a finite element (FE) program into its components is called "linearization."
A typical computer output may look like the solid line shown in Fig. 8.1. The FE line must be divided
into a membrane stress and bending stress as shown. These values can then be compared to the allowable
stress given in Table 8.4.
A typical output of a detailed stress analysis consists of three normal stresses, IT" ITI, ITh, and three shearing
stresses 'Trj, Trh. 'Tlh' From these six stresses, the designer can obtain three principal stresses (Tj, (Tz, CT) by
using the classical equation
(T
= (aj
+
(Tj)/2 ± [(O'i - uY/4
+
'T~PI2
min
'Thlcknoss: I, (trOll
Oulslde
Membrane.Plus·Bendlng
plates :l: 8m1l'
• • •
SCL
lAembran.
(PIma'
FIG. 8.1
LINEARIZING STRESS DISTRIBUTION
(8.1)
Analysis of Components in Vill-2 241
The maximum stress intensity defined in VIII-2 is the absolute value of the larger of the following values
The maximum stress intensity is compared with allowable values in Table 8.4.
Example 8.3
Problem
The forces and bending moments in sections a-a and b-b due to design pressure in Example 8.1 were
calculated from the classical theory of plates and shells as
Section a-a
Membrane force in the radial direction = 2602.3 lb
Membrane force in the hoop direction = 0.0
Bending moment in the radial direction = 89,052.0 in.-lb
Bending moment in the hoop direction = 89,052.0 in.-lb
Section b-b
Bending moment in the axial direction = 5988.0 in.-lb
Bending moment in the tangential direction = 1796.4 in.-Ib
Shearing force in the radial direction = 2602.3 lb
Membrane force in the axial direction = 9600 lb
Membrane force in the boop direction = 0 (assunting the shell cannot grow radially at this location. A
more accurate solution of this problem can be obtained by taking
into consideration the inward deflection of point b-b due to the
edge rotation. The value of this deflection can be taken as the
edge rotation times half the flat cover thickness.)
Deterntine the stress values at sections a-a and b-b in accordance with the VIIl-2 procedures and compare
them with the allowable stresses.
Solution
Section a-a
The membrane stress is
P« = forcelt = 2602.3/6.375
~
410 psi
~
13,150 psi
The bending stress is
P,
~
6MIi'
~
6 x 89052/6.375'
242
Chapter 8
From Table 8.4,
~
Allowable Pm
Allowable P;
+ P,
~
20,000 psi> 410 psi
30,000 psi> 13,560 psi (13150
+ 410)
Section b-b
The axial membrane stress at design pressure is
~
foreelt = 960011.0 = 9600 psi
The axial membrane stress at operating pressure is
P«
~
(operating Pldesign P)(foreelt)
=
8400 psi
~
(700/800)(9600/1.0)
The axial bending stress at operating pressure is
Q = (operating Pldesign P)(6Mlf)
~
~ (7001800)(6 X 5988.0/1.0')
31,440 psi
From Table 8.4,
Allowable P;
Allowable Pm
+
Q
~
~
20,000 psi> 9600 psi
60,000 psi> 39,840 psi (8400
+ 31,440)
The hoop membrane stress at design pressure is
Pm
~
foreelt
= 0/1.0
~
0 psi
The hoop membrane stress at operating pressure is
P;
~
(operating Pldesign P)(forcelt)
~
0 psi
~
(700/800)(0/1.0)
The hoop bending stress at operating pressure is
Q
~
(operating Pldesign P)(6Mlf)
= 9430
psi
~ (700/800)(6 X 1796.4/1.0')
Analysis of Components in VIII-2 243
From Table 8.4,
Allowable P; = 20,000 psi> 0 psi
Allowable Pm + Q = 60,000 psi> 9430 psi (9430 + 0)
Example 8.4
Problem
A finite element (FE) analysis was performed on a flat head-to-shell junction, shown in Fig. E8A. Three
different loading conditions were calculated. They were pressure, mechanical.vandthermal loading. The
results of the FE stress output are shown in Table E8.4. Assume the operating and design pressures are
the same and all initial stress values are equal to zero. Assume the allowable stress value to be 14 ksi,
Calculate the primary membrane stress and the secondary stress at the junction.
AXISYMMETRIC FINITE ELEMENT MODEL
7
r- i f
CL
IrL
~V
% .S tress Classification Line
rscu
r
h
L..
FIG. E8.4
MODEL OF A FINITE ELEMENT LAYOUT IN A FLAT HEAD-TO-SHELL JUNCTION
244
Chapter 8
TABLE E8.4
SUMMARY OF FINITE ELEMENT OUTPUT
seL Membrane Stress, psi
Loads
<T,
<T,
<T,
0'"
Pr
Me
Th
-900
400
200
6200
1000
100
11400
500
100
100
-600
-800
COMBINATION 1
Pr + Me
-500
7200
11900
-500
COMBINATION 2
Pr+Me+Th
-300
7300
12000
-1300
Pressure
Mechanical
Thermal
(not applicable; not used)
act, Membrane Plus Bending Stress, psi
<T,
Pressure
Mechanical
Thermal
<T,
0',
0'"
Pr
Me
Th
-2000
400
200
2100
1500
9200
11000
-700
3700
100
-600
-800
COMBINATION 3
Pr + Me
-1600
3600
10300
-500
COMBINATION 4
Pr+Me+Th
-1400
12800
14000
-1300
SeL Peak Stress, psi
0',
Pressure
Pr
Me
Th
Mechanical
Thermal
Pr + Me
COMBINATION 5
COMBINATION 6
Pr
+
Me
+
Th
0',
<T,
0'"
0
850
-200
-100
2400
2000
0
-1250
1900
100
-1350
700
850
2300
-1250
-1250
650
4300
650
-550
NOTES:
COMBINATIONS computed for each component of stress. Calculations of Principal Stresses and Stress Intensities made
AFTER combinations.
r = in-plane component in radial direction
I = In-plane component in longitudinal direction
h = out-of-plane component in hoop direction
Solution
Primary Membrane Stress
Table 8.4 indicates that primary membrane stress is produced by mechanical loads ouiy. Thus, in Table
E8.4 under the Membrane Stress part, ouly the pressure, mechanical, or a combination of pressure and
mechanical are to be used. Thermal stresses are ignored in this case. The FE results indicate that there is
a shearing stress in the r.l plane. Thus the two principal stresses, "1 and "2, in this plane are calculated
from Eq. (8.1), while the third principal stress is "h. The three principal stresses become
Pressure, psi
Mechanical, psi
Pressure plus mechanical, psi
0',
0',
<T.
6200
1370
7230
-900
30
-530
11,400
500
ll,900
Analysis of Components in VIII-2
245
And the maximum stress intensity values are given by
Maximum Stress Intensity, psi
Pressure
Mechanical
Pressure plus mechanical
12,300
1340
12,430
Allowable Pm = 14,000 psi> 12,430 psi
Secondary Stress
Table. SA indicates that secondary stress is produced bymechanical and thermal loads, The. FE results
indicate that there is a shearing stress in the r.l plane. Thus the two principal stresses, 0"1 and 0"2, in this
plane are calculated from Eq. (8'\), while the third principal stress is CJh' The three principal stresses become
0",
Pressure, psi
Mechanical, psi
Pressure plus mechanical, psi
Pressure plus mechanical plus thermal, psi
2100
1760
3650
12,920
~540
140
-1650
~1520
0",
11,000
~700
10,300
14,000
And the maximum stress intensity values are given by
Maximum Stress Intensity, psi
Pressure
Mechanical
Pressure plus mechanical
Pressure plus mechanical plus thermal
11,540
2460
11,950
15,520
Allowable Pm + Q = 42,000 psi> 15,520
8.5 FATIGUE EVALUATION
When a fatigue evaluation is required in accordance with AD-160 of VIIl-2 or by the user or a qualified
engineer, it shall be performed in accordance with the requirements of Appendix 5 of VIIl-2, The number
of cycles are evaluated from fatigue charts such as the one shown iu Fig. 8,2 for carbon steel.
In most applications the process cycle is easily determined. Each cycle consists of a start-up condition
in pressure and temperature, a steady state, and then shutdown of pressure and temperature. This is illustrated
in Fig. 8.3a. More complex cycles often occur where there is reversal of stress, as shown in Fig, 8,3b, On
occasion, complicated cycles occur, such as the one shown in Fig. 8.3c. For each cycle the designer
determines the maximum stress range, which is the algebraic difference between the maximum and minimum
stress intensities in a cycle. The alternating stress, which is half the maximum stress range, is then obtained.
With this value, the fatigue chart is then used to obtain the number of permissible cycles for each stress
range. A Cumulative Usage Factor is then determined for each type of cycle considered by the designer.
Example 8.5
Problem
Use the Peak stress values given in Example 8.4 to determine the fatigue life at the location indicated in
Fig. E8A, Use Fig. 8.2 for a fatigue chart,
246 Chapter 8
,
00
ll-
I'
J
!- Nom
(1)E~30>:lO"p8i.
l2f lll19rPOlate for UTS 8(1-1' S bi-
(3) Tabla 5-110.1 eontBi", tfbul-'ed vallHll and II form",l. lor lin IIl;CUrat.. lnlllrpolsllon of ttl_ eurve •.
i-
,
'......
~L
S., psi
Fo' UTS" SO k'i
~
~-
.......
s
""" ~
Fo' UTS 115·130 ~.i
.....
I-~ .....
.....
iI
"
"
I
I
I I
I
I
I
I I I "
•
I
I
-..... -..... --"" I
I
I
I";"
1I1I
00'
Numberof Cycles
FIG. 8.2
FATIGUE CURVES FOR CARBON, LOW ALLOY, 4XX, HIGH ALLOY, AND HIGH STRENGTH
STEELS FOR TEMPERATURES NOT EXCEEDING 700°F (ASME VIII-2)
Solution
Table 804 indicates that the peak stress must be combined with the membrane aud bending stresses to
determine fatigue life.
Peak Plus Secondary Stress
From Table E8A, we combine the peak stresses for pressure, mechanical, and thermal conditions with those
of membrane plus bending stresses. This gives peak plus membrane plus bending stress, as shown below:
0',
Pressure, psi
Mechanical, psi
Pressure plus mechanical, psi
Pressure plus mechanical plus thermal, psi
-2000
1250
-750
-950
11,000
-1950
9050
10950
2000
3900
5900
7900
200
-1950
-1750
-1050
The three principal stresses become
Pressure, psi
Mechanical, psi
Pressure plus mechanical, psi
Pressure plus mechanical plus thermal, psi
0',
+0',
0'.
2010
4930
--2010
220
-280
-1070
11,000
-1950
9050
10,950
6400
8020
Analysis of Components in VIII-2
P.T
-'::'
I C . = ~
_ Trme
(0)
P.T
-,T,me
~_ _--..J.-
(b)
P.T
f-
~-----_r--:T,me
(e)
FIG. 8.3
CYCLIC CURVES
247
248 Chapter 8
And the maximum stress intensity values are giveu by
Maximum Stress Intensity, psi
Pressure
Mechanical
Pressure plus mechanical
Pressure plus mechanical plus thermal
13,010
6880
9330
12,020
The maximum alternating stress is
S"
=
13010/2
= 6500 psi
From Fig. 8.2, with S" equal to 6500 psi, the maximum uumber of cycles is
> 1,000,000.
REFERENCES
API, 1996, American Petroleum Institute, Recommended Rules for Design and Construction of Large,
Welded, Low-Pressure Storage Tanks, API 620, Washington, D.C., API.
ASCE, 1998, American Society of Civil Eugineers, Minimum Design Loads for Buildings and Other
Structures, ASCE 7-98, New York, ASCE.
ASME, 2001a, American Society of Mechanical Engineers, Boiler and Pressure Vessel Code, Section
VIII, Division I, Pressure Vessels, New York, ASME.
ASME, 2001b, American Society of Mechanical Eugineers, Boiler and Pressure Vessel Code, Section
VIII, Division 2, Alternative Rules for Pressure Vessels, New York, ASME.
ASME, 1999, American Society of Mechanical Engineers, B31.3, Chemical Plant and Petroleum Refinery
Piping, New York, ASME.
ASME, 1969, American Society of Mechanical Engineers, Criteria of the ASME Boiler and Pressure
Vessel Code for Design by Analysis in Sections III and VIII, Division 2, New York, ASME.
ASME, 1968, American Society of Mechanical Engineers, Section Ylll-Division 2 of the ASME Boiler
and Pressure Vessel Code-Guide to Alternative Rules for Pressure Vessels, New York, ASME.
Beer, F. P., and Johnson, Jr., E. R., 1992, Mechanics of Materials, New York, McGraw Hill.
Bergman, E. 0., 1955, "The Design of Vertical Vessels Subjected to Applied Forces," Transactions of
the ASME, New York, ASME.
Flugge, W., 1967, Stresses in the Shells, New York, Springer-Verlag,
Gilbert, N., and Polani, J. R., May 1979, Stability Design Criterion for Vessels Subjected to Concurrent
External Pressure and Longitudinal Compressive Loads, New York, ASME.
ICBO, 1997, International Conference of Building Officials, Uniform Building Code, Whittier, CA, ICBO.
Jawad, M. H., 1994, Theory and Design of Plate and Shell Structures, New York, Chapman and Hall.
Jawad, M. H., and Farr, J. R., 1989, Structural Analysis and Design of Process Equipment, New York,
John Wiley & Sons.
G+ W Taylor-Bonney, Bulletin 502: Modem Flange Design, 7th Edition, Southfield, MI, G+ W.
Peterson, R. E., 1974, Stress Concentration Factors, New York, John Wiley & Sons.
Prager, W., and Hodge, P. G., 1965, Theory of Perfectly Plastic Solids, New York, John Wiley & Sons.
Shield, R. T., and Drucker, D. C., June 1961, "Design of Thin-Walled Torispherical and Toriconical
Pressure Vessel Heads," Journal of Applied Mechanics, New York, ASME.
TEMA, 1999, Tubular Exchanger Mannfacturers Association, Inc., Standards of Tubular Exchanger
Manufacturers Association, 8tb ed., Tarrytown, NY, TEMA.
249
250 References
Waters, E. 0., Wesstrom, D. B., Rossheim, D. B., and Williams, F. S. G., 1937, "Formulas for Stresses
in Bolted Flanged Connections," Transactions of the ASME, New York, ASME.
Zick, L. P., and Germain, A. R., May 1963. "Circumferential Stresses in Pressure Vessel Shells of
Revolution," Journal of Engineering for Industry, New York, ASME.
ApPENDIX A
GUIDE TO
VIII.. 1
REQUIREMENTS
251
252 Appendix A
1----------------------------------------------------------------------------------------------1
i
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QUICK REFERENCE GUIDE
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'
ASME PRESSURE VESSEL CODE (SECTION VIII, DIVISION 1)
,
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fUll fACE CA$KtT,APPX. '-62-'
SP+lERICALlY OISH£O COVERS,
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no. 2-4. A!'PX. S,
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STUOS "BOI.TS.\l0-12.
UCS-l0. LlNI"12.IJHA-12
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UO-54, 1J0-82
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UC·82. APPX, 0
LONOIHJOlNAl .OrNTS. 1JW-33,
IJW-3. UW-3S. U\II·9
tru, TALE 'lOl.ES, \l0-2~, UCI.-25
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no. U'I'/-9, U\II'13, I'IC. UW-13.1
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UW-15. flC. UW-16.1.1JW-15
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COVER.APPX, 1-6
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COlolPRESsrON lImG, AWX, 1·~. 1'6
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UW·15.FLC. UW·15.1, FlO. UW·16.2
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nLLET WEWS. UW-18, UW-35. TAaLE !JW'12
KNUCKLE RAOllIS.UG·n. UCS·79
FLUID OPENINGS. l!G·J6. flG,UG-J6
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UCS-lD. UCS-ll.UNI'-12. UNF-1J
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UCS-56. TA!lLE UCS'S6, UCS-19l<j1
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t.rn4'l. lIalVICE UW-21~1, UClJ'2, UCI-2 ~ UC'99(9)(4)
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MIIlliftW.S lIC-, IffRU UG-IS, oc-e.
UCl'l1 " UW-5,
TA!lLES NF-l T11RU NF-S
ua-rr.
,
THE PURPOSE OF THIS GUIDE IS TO ILLUSTRATE SOME OF THE lYPES OF PRESSURE VESSEL
CONSTRUCTION WHICH ARE PROVIDED UNDER SECTION VIII,DIVISION " OF THE ASME CODE
ISSUED IN JULY 2001,AND TO FURNISH DIRECT REFERENCE TO THE APPUCABLE RULE IN THE
CODE. IN THE EVENT OF A DISCREPANCY, THE RULES IN THE CURRENT EDITION OF THE CODE
SHALL GOVERN.
P.O.BOX 451
ST,lOUIS. MISSOURI
TEL (314) 621-0000
CORPORA1E COMMUNICATtoNS
FAX 421-7704
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FOR STA¥S. lm-14.UW-19. FIG.
UW-19.2. STAY(O SUilFACES, 1)(;·47
,
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ApPENDIXB
MATERIAL DESIGNATION
TABLE B.l
CARBON STEEL PLATE
Nominal Composition
Current
Formerly
C-SI
C-SI
C-SI
SA-515 Gr. 60
SA-515 Gr. 65
SA-515 Gr. 70
A-201 Gr. B
A-212 Gr. A
A-212 Gr. B
C-SI
C-Mn-SI
C-Mn-SI
C-Mn-SI
SA-516
SA-516
SA-516
SA-516
A~201
1/2 Cr-1/5 Mo-V
3/4 NI-1/2 Cr-1/2 Mo-V
SA-517 Gr. B
SA-517 Gr. F
Gr. 55
Gr. 60
Gr. 65
Gr. 70
Gr.
A-201 Gr.
A~212 Gr.
A-212 Gr.
A fine
B fine
A fine
8 fine
Yield Stress,
ksi
Tensile Stress,
ksi
32
35
38
60
65
70
30
32
35
38
55
60
65
70
100
100
115
115
grain
grain
grain
grain
T1A
T-1
NOTE: Old "Fire Box" quality steel more closely corresponds to modern-day pressure vessel steels, while "Flange" quality
contains more impurities and may not be as homogeneous,
TABLE B.2
CHROME-MOLY STEEL PLATE SPECiFICATIONS, SA-387
Nominal Composition
1/2 Cr-1/2 Mo
1 Cr-1/2 Mo
1 1/4 Cr-1/2 Mo-SI
2 1/4 Cr-1 Mo
3 Cr-1 Mo
5 Cr-1/2 Mo
Current
Formerly
2
12
11
22
21
5
A
B
C
D
E
(SA·357)
Yield Stress, ksi
Ci. 1 ~
Ci. 1 ~
Ci. 1 ~
Ci. 1 ~
Ci. 1 ~
CI. 1 ~
33,
33,
35,
30,
30,
30,
Ci. 2 ~
Ci. 2 ~
Ci. 2 ~
ci 2 ~
Ci. 2 ~
CI. 2 ~
45
40
45
45
45
45
Tensile Stress, ksi
Ci.
Ci.
Ci.
Ci.
Ci.
CI.
1 ~
1 ~
1 ~
1 ~
1 ~
1 ~
55, Ci. 2 ~
55, Ci. 2 ~
60, Ci. 2 ~
60, Ci. 2 ~
50, Ci. 2 ~
60, CI. 2 ~
70
65
75
75
75
75
NOTE: Each grade of SA-387 is available in two classes of tensile strength levels, which depend on heat treatment. Class
1 is the lower-strength material and generally is the material that has been annealed. Class 2 is the higher-strength
material that has been normalized and tempered or quenched and tempered.
TABLE B.3
CHROME-MOLY STEEL FORGING SPECIFICATIONS, SA-182
Nominal Composition
1/2 Cr·1/2 Mo
1 Cr·1/2 Mo
1 Cr-1/2 Mo
11/4Cr·1/2Mo
11/4Cr·1/2Mo
2114 Cr-1 Mo
2 1/4 Cr·1 Mo
3 Cr·1 Mo
5 Cr·1/2 Mo
5 Cr-1/2 Mo
Current
Formerly
Yield Stress, ksi
Tensile Stress, ksi
Gr. F2
Gr. F12, CI. 1
Gr. F12, CI. 2
Gr.F11,CI.1
Gr. F11, CI. 2
Gr. F22, CI. 3
Gr. F22, CI. 1
Gr. F21
Gr. F5
Gr. F5a
Gr. F2
Gr. F12b
Gr. F12
Gr.F11b
Gr. F11
Gr. F22
Gr. F22a
Gr. F21
Gr. F5
Gr. F5a
40
30
40
30
40
45
30
45
40
65
70
60
70
60
70
75
60
75
70
90
NOTE: Several Chrome-Moly forgings have the same nominal composition but vary in strength and/or chemical requirements.
253
254
Appendix B
TABLE B.4
CHROME-MOLY STEEL FORGING SPECIFICATIONS, SA·336
Nominal Composition
1 Cr-1/2 Mo
11/4Cr-1/2Mo
11/4Cr-1/2Mo
11/4Cr-1/2Mo
2 1/4 Cr-1 Mo
21/4 Cr-1 Mo
3 Cr-1 Mo
3 Cr-1 Mo
5 Cr-1/2 Mo
5 Cr'1/2 Mo
Current
Formerly
Yield Stress, ksi
Tensile Stress, ksi
Gr. F12
Gr.F11,CI.1
Gr.F11,CI.2
Gr.F11,CI.3
Gr. F22, CI. 1
Gr. F22, CI. 3
Gr. F21, CI. 1
Gr. F21, CI. 3
Gr. F5
Gr. F5a
CI. F12
CI.F11b
CI.F11
CI.F11.
CI. F22a
CI. F22
CI. F21a
CI. F21
CI. F5
CI. F5a
40
30
40
45
30
45
30
45
36
50
70
60
70
75
60
75
60
75
60
80
NOTE:Several Chrcrne-Molyfcrqinqs havethesamenominal composition but vary instrength and/or chemical requirements.
TABLE B.5
QUENCH & TEMPERED CARBON AND ALLOY STEEL FORGINGS, SA-50B
Nominal Composition
C-Si
C-Mn-Si
3/4 Ni-1/2 Mo-1/3 Cr-V
3/4 Ni-1/2 Mo-Cr-V
31/2 Ni-1/2 Mo-1 3/4 Cr-V
31/2 Ni-1/2 Mo-1 3/4 Cr-V
31/2 Ni-1/2 Mo-1 3/4 Cr-V
Current
Formerly
Yield Stress,
kst
Tensile Stress,
ksi
Gr. 1
Gr.1A
Gr. 2, CI.
Gr. 3, CI.
Gr. 4, CI. 1
Gr. 4, CI. 2
Gr. 4, CI. 3
CI. 1
CI. 1.
CI. 2
CI. 3
CI. 4
CI. 4.
CI. 4b
36
36
50
50
85
100
70
70
70
80
80
105
115
90
ApPENDIXC
JOINT EFFICIENCY FACTORS
JOINT EFFICIENCY FACTORS
SECTION VIII DIVISION 1
C
B
A
B
D
SEAMLESS
1+--1 ELLIPSOIDAL
OA
TORISPHERICAL
A TYPE 1 BUTT WELD FULL RT
B TYPE 1 BUTT WELD FULL RT
SEE UW·11 (a)(5)(a)/UW·12(a)/TABLE UW·12(a)
E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=1.00 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR)
E=1.00 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD FULL RT
B TYPE 1 BUTT WELD SPOT RT
SEE UW·11 (a)(5)(a)&(b)/UW·12(a),(b)&(d)/TABLE UW·12(a)&(b)
E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=1.00 SEAMLESS HeAD CALCULATIONS (QUALITY FACTOR)
E=O.85 LONGITUDINAL STRESS CALCULATIONS
FIG. C.1
255
256 Pcppendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
c
A
A
D
SEAMLESS
+Ii-! HEMISPHERICAL
A TYPE 1 BUTT WELD FULL RT
SEE UW·11 (a)(5)(a)/UW·12(a)1TABLE UW·12(a)
E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=1.00 HEAD CALCULATIONS (JOINT EFFICIENCY)
E=1.00 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD IN SHELL FULL RT
A TYPE 2 BUTT WELD HEAD TO SHELL FULL RT
SEE UW·11 (a)(5)(a)/UW·12(a)1TABLE UW·12(a)
E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.90 HEAD CALCULATIONS (JOINT EFFICIENCY)
E=O.90 LONGITUDINAL STRESS CALCULATIONS
FIG. C.2
Joint Efficiency Factors
257
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
A
A
D
A
SEAMLESS
+f-l HEMISPHERICAL
A TYPE 1 BUn WELD IN SHELL FULL RT
A TYPE 2 Bun WELD HEAD TO SHELL SPOT RT
SEE UW-11 (a}(5)(a)&(b)/UW·12(a),(b)&(d)/TABLE UW-12(a)&(b)
E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.80 HEAD CALCULATIONS (JOINT EFFICIENCY)
E=O.80 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 Bun WELD IN SHELL NO RT
A TYPE 2 Bun WELD HEAD TO SHELL NO RT
SEE UW-12(c) lit (d)/TABLE UW-12(c)
E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.65 HEAD CALCULATIONS (JOINT EFFICIENCY)
E=O.65 LONGITUDINAL STRESS CALCULATIONS
FIG. C.3
258 Appendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
A
A
D
A
SEAMLESS
~0...-4 HEMISPHERICAL
A TYPE 1 BUTT WELD IN SHELL SPOT RT
A TYPE 2 BUTT WELD HEAD TO SHELL SPOT RT
SEE UW-11 (a)(5)(a)&(b)/UW-12(b)&(d)/TABLE UW-12(b)
E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.80 HEAD CALCULATIONS (JOINT EFFICIENCY)
E=O.80 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD IN SHELL SPOT RT
A TYPE 2 BUTT WELD HEAD TO SHELL NO RT
SEE UW-11 (a)(5)(a)&(b)/UW-12(~),(c)&(d)/T ABLE UW-12(b)&(c)
E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.65 HEAD CALCULATIONS (JOINT EFFICIENCY)
E=O.65 LONGITUDINAL STRESS CALCULATIONS
FIG. C.4
Joint Efficiency Factors
259
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
B
B
D
SEAMLESS
+lH ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUrr WEI..Il SPOT FIT
B TYPE 1 BUrr WEI..Il SPOT FIT
SEE UW·11 (a)(5)(a)&(b)/UW.12(b)&(d)/TABLE UW-12(b)
E:O.85 SHELL CALCULATIONS (JOINT EFFICIENCy)
E:1.00 HEAD. CALCULATIONS (QUALITY FACTOR)
E:O.85 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUrr WELD SPOT FIT
B TYPE 1 BUrr WELD NO FIT
SEE UW·11 (a)(5)(a)&(b)/UW-12(b),(c)&(d)/TABLE. UW-12(b)lIt(c)
E:O.85 SHELL CALCULATIONS (JOINT EFFICIENCy)
E:O.85 HEAD CALCULATIONS (QUALITY FACTOR)
E:O.70 LONGITUDINAL.. STRESS CALCULATIONS
FIG. C.5
260 Appendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
B
A
D
B
SEAMLESS
<lH1---l 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=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=I.OO HEAD CALCULATIONS (QUALITY FACTOR)
E=O.80 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD SPOT RT
B TYPE 2 BUTT WELD NOT RT
SEE UW-11 (a)(5)(a)&(b)/UW-12(b)(c)&(d)/TABLE UW-12(b)&(c)
E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.85 HEAD CALCULATIONS (QUALITY FACTOR)
E=O.65 LONGITUDINAL STRESS CALCULATIONS
FIG. e.6
Joint Efficiency Factors
261
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
B
A
D
B
SEAMLESS
......f--l ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 2 BUTT WELD NO RT
B TYPE 3 BUTT WELD NO RT
SEE UW·11(a)(5)(a)&(b)/UW·12(c)&(d)ITABLE UW·12(c)
E=O.65 SHelL CALCULATIONS (JOINT EFFICIENCY)
E=O.85 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR)
E=O.60 THE LONGITUDINAL STRESS CALCULATIONS
A TYPE 2 BUTT WELD SPOT RT
B TYPE 3 BUTT WELD FULL OR SPOT RT
SEE UW·11(a)(5)(a)&(b)/UW·12(a),(b)&(d)ITABLE UW-12(a)&(b)
E=O.80 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.85 SEAMLESS HEAD CALCULATIONS (QUALITY FACTOR)
E=O.60 LONGITUDINAL STRESS CALCULATIONS
FIG. C.7
262 Appendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
B
A
D
B
SEAMLESS
.~I-j ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUTT WELD NO RT
B TYPE 1 BUTT WELD NO RT
SEE UW-12(c)&(d)/TABLE UW-12(c)
E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.85 HEAD CALCULATIONS (QUALITY FACTOR)
E=O.70 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD NO RT
B TYPE 1 BUTT WELD SPOT RT
SEE UW-11(a)(5)(a)&(b)/UW-12(b),{c)&(d)/TABLE UW-12(b)&(c)
E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E=O.85 LONGITUDINAL STRESS CALCULATIONS
FIG. C.B
Joint Efficiency Factors
263
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
A
B
D
B
SEAMLESS
......--i ELLIPSOIDAL
OR
TORISPHERICAL
A TYPE 1 BUTT WELD NO AT
B TYPE 6 SINGLE FILLET WELD
SEE UW-11 (a)(5)(a)&{b)/UW·12{d)/TABLE UW-12{C)
E=O.70 SHELL CALCULATIONS (JOINT EFFICIENCy)
E=O.85 HEAD CALCULATIONS (QUALITY FACTOR)
E=O.45 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD SPOT AT
B TYPE 6 SINGLE FILLET WELD
SEE UW-11 (a){5)(a)&{b)/UW·12{b)&{d)/Table UW-12{b)&(c)
E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCy)
E=O.85 HEAD CALCULATIONS (QUALITY FACTOR)
E=O.45 LONGITUDINAL STRESS CALCULATIONS
B TYPE 4 DOUBLE FILLET WELD SAME AS B TYPE 6 EXCEPT
E=O.55 LONGITUDINAL STRESS CALCULATIONS
. FIG. e.9
264 Appendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
c
B
B
D
SEAMLESS
+1-1 ELLIPSOIDAL
SEAMLESS SHELL
OR
TORISPHERICAL
SEAMLESS SHELL
B TYPE 1 BUTT WELD FULL RT
SEE UW-11(a)(5)(a)/UW-12(a)&(d)/TABLE UW-12(a)
E=1.00 SHELL CALCULATIONS (QUALITY FACTOR)
E=1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E=1.00 LONGITUDINAL STRESS CALCULATIONS
SEAMLESS SHELL
B TYPE 1 BUTT WELD SPOT RT
SEE UW·11 (a)(5)(a)&(b)/UW-12(b)&(d)/TABLE UW-12(c)
E=1.00 SHELL CALCULATIONS (QUALITY FACTOR)
E=1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E=O.85 LONGITUDINAL STRESS CALCULATIONS
FIG. C.10
Joint Efficiency Factors
265
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
B
B
D
SEAMLESS
4-1---4 ELLIPSOIDAL
SEAMLESS SHELL
OR
TORISPHERICAL
SEAMLESS SHEL.L.
B TYPE 6 SINGL.E FIL.L.ET WELD
SEE UW·11 (a)(5)(a)!UW·12(d)/TABL.E UW-12(c)
E=O.85 SHELL CAL.CULATIONS (QUAL.ITY FACTOR)
E=O.85 HEAD CAL.CUL.ATIONS (QUAL.ITY FACTOR)
E=O.45 L.ONGITUDINAL. STRESS CAL.CUL.ATIONS
B TYPE 4 DOUBL.E FIL.L.ET WEL.D SAME AS B TYPE 6 EXCEPT
E=O.55 LONGITUDINAL STRESS CAL.CULATIONS
NOTE: INTENT IN INTERPRETATIONS·VOLUME 20·JUL.Y,1987
REVISIONS APPEAR IN THE 12/31/87 ADDENDA
FIG. C.11
266 Appendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
A
B
B
D
~~
SEAMLESS
ELLIPSOIDAL
OR
TORISPHERICAL
A ERW PIPE
B TYPE 1 BUTT WEL.O SPOT RT
SEE UW·11 (a)(5)(a)&(b)/UW.12(b)&(d)/TABLE UW.12(b)
E=1.DO SHELL. CALCULATIONS (QUALITY FACTOR), NOTE 1
E=1.00 HEAD CALCULATIONS (QUALITY FACTOR)
E=0.85 LONGITUDINAL STRESS CALCULATIONS. NOTE 2
A ERW PIPE
B TYPE 1 BUTT WEL.O NO RT
SEE UW·11 (a)(5)/UW.12(c)&(d)/TABLE UW.12(c)
E=0.85 SHELL. CALCULATIONS (QUALITY FACTOR), NOTE 1
E=O.85 HEAD CALCULATIONS (QUALITY FACTOR)
E=0.70 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
NOTE 1: THE QUALITY FACTOR SHOWN IN THE SHELL CALCULATIONS
IS IN ADDITION TO THE CATEGORY A JOINT FACTOR (E=0.85)
INCl.UDED IN THE ALLOWABLE STRESS VALUE OBTAINED FROM
THE APPLICABLE STRESS TABLE.
NOTE 2: DIVIDE THE Al.l.OWABLE STRESS VAl.UE OBTAINED
FROM THE APPLICABl.E STRESS TABLE BY 0.85 BEFORE APPLYING
A JOINT EFFICIENCY IN THE LONGITUDINAl. STRESS CALCUl.ATION
(eg, SEE NOTE 26, TABLE UCS.23)
FIG. C.12
Joint Efficiency Factors
267
JOiNT EFFICiENCY FACTORS
SECTION VIII, DIVISION 1
C
B
A
D
SEAMLESS
EL.lIPSOIDAL.
OR
TORISPHERICAL.
D TYPE 1 BUTT WELD
SEE UW·11(a)(5) FULL RT IF PART IS DESIGNED E:1.00
RT NOT REQUIRED FOR tr or trn CALCULATIONS E:1.00
E:EFFICIENCY IF OPENING THROUGH CATEGORY A WELD
D FULL OR PARTIAL PENETRATION CORNER WELD
RT NOT REQUIRED E:1.00
RT NOT REQUIRED FOR tr or trn CALCULATIONS E:1.00
E:EFFICIENCY OF THE BUTT WELD PENETRATED
FIG. C.13
268
Appendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
B
A
o
B
SEAMLESS
+f-l ELLIPSOIDAL
OR
TORISPHERICAL
B&C TYPES I OR 2 BUTT WELD IN NOZZLES/COMMUNICATING CHAMBERS
RT NOT REQUIRED TO CALCULATE trn
RT MAY BE REQUIRED FOR:
SERVICE RESTRICTION
MATERIAL THICKNESS
USERIDESIGNATED AGENT
FIG. C.14
Joint Efficiency Factors
269
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
C
•
B
po
~C
1
D
.
A
C-'
•
FLAT HEAD
C IS PARTIAL OR FULL PENETRATION CORNER JOINT FOR
ALL CASES
E FOR C IS NOT ESTABLISHED BY CODE RULES
A
A
A
A
A
TYPE 1 BUTT WELD FULL RT - E=1.00
TYPE 1 BUTT WELD SPOT RT - E=O.85
TYPE 1 BUTT WELD NO RT - E=O.70
ERW BUTT WELD-E=1.00 (JOINT EFFICIENCY IN STRESS VALUE)
SEAMLESS· E=1.00
NOTE: FLAT HEAD FORMULA HAS BUILT-IN STRESS MULTIPILER
REGARDLESS OF TYPE OF JOINT OR EXAMINATION. A FACTOR
FOR E WILL APPLY IN THE FLAT HEAD FORMULA IF A CATEGORY
A JOINT DOES EXIST IN THE HEAD
FIG. C.15
270 Appendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
•
C
~C
t
D
B
'-
A
•
C-'
TUBE SHEET
OR
FLANGE
A TYPE 1 BUTT WELD FULL RT
C TYPE 1 BUTT WELD FULL RT
E=1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=1.00 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD FULL RT
C TYPE 1 BUTT WELD SPOT RT
E.1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E.. O.85 LONGITUDINAL STRESS CALCULATIONS
FIG. C.16
Joint Efficiency Factors
JOINT EFFICIENCY FACTORS
SECTION VIII. DIVISION 1
c
•
~c
1D
B
.
A
C
TUBE SHEET
OR
FLANGE
A TYPE 1 BUTT WELD FULL RT
C TYPE 1 BUTT WELD NO RT
E=O.85 SHELL CALCULATIONS (QUALITY FACTOR)
E=O.70 LONGITUDINAL STRESS CALCULATIONS
A TYPE 1 BUTT WELD SPOT RT
C TYPE 1 BUTT WELD NO RT
E=O.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E=O.70 LONGITUDINAL STRESS CALCULATIONS
FIG. C.17
271
272 J\ppendix C
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
c
•
~C
1D
B
•
A
C
-
TUBE SHEET
OR
FLANGE
A TYPE 1 BUTT WELD FULL RT
C WELD FULL OR PARTIAL PENETRATION CORNER JOINT
E:1.00 SHELL CALCULATIONS (JOINT EFFICIENCY)
E FOR C IS NOT ESTABLISHED BY CODE RULES
A TYPE 1 BUTT WELD SPOT RT
C WELD FULL OR PARTIAL PENETRATION CORNER JOINT
E:O.85 SHELL CALCULATIONS (JOINT EFFICIENCY)
E FOR C IS NOT ESTABLISHED BY CODE RULES
A TYPE 1 BUTT WELD NO RT
C WELD FULL OR PARTIAL PENETRATION CORNER JOINT
E:O.70 SHELL CALCULATIONS (JOINT EFFICIENCY)
E FOR C IS NOT ESTABLISHED BY CODE RULES
FIG. C.18
Joint Efficiency Factors
273
JOINT EFFICIENCY FACTORS
SECTiON Viii, DIViSION 1
c
.
El
~C
T
0
.
A
TUBE SHEET
OR
C.....
F!..ANGE
A ERW PIPE
C TYPE 1 BUTT WELD SPOT RT
E=1.00 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E=0.85 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
A ERW PIPE
C TYPE 1 BUTT WELD NO RT
E=0.85 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E=0.70 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
A ERW PIPE
C TYPE 2 BUTT WELD NO RT
E=0.B5 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E=0.65 LONGITUDINAL STRESS CALCULATIONS, NOTE 2
A ERW PIPE
C FULL OR PARTIAL PENETRATION CORNER JOINT
E=1.00 SHELL CALCULATIONS (QUALITY FACTOR), NOTE 1
E FOR C IS NOT ESTABLISHED BY CODE RULES, NOTE 2
NOTE 1: THE QUALITY FACTOR SHOWN IN THE SHELL CALCULATIONS IS IN ADDITION
TO THE CATEGORY A JOINT FACTOR (E=O.B5) INCLUDED IN THE ALLOWABLE
STRESS VALUE OBTAINED FROM THE APPLICABLE STRESS TABLE.
NOTE 2: DIVIDE THE ALLOWABLE STRESS VALUE OBTAINED FROM THE APPLICABLE
STRESS TABLE BY 0.B5 BEFORE APPLYING A JOINT EFFICIENCY IN THE
LONGITUDINAL STRESS CALCULATIONS (eg. SEE NOTE 26, TABLE UCS-23).
FIG. C.19
274 Appendix C
EXAMPLE CALCULATION FOR
JOINT EFFICIENCY FACTORS
SECTION VIII, DIVISION 1
c
B
A
D
B
DESIGN SPECIFICATIONS
175PSI AT 600F
60F AT 175PSI
SHELL MATERIAL SA-414A
SHELL DIAMETER 40"
S=12,800 FROM SECTION II PART D-TABLE 1
CATEGORY A TYPE 1 NO RT
E=0.70 FROM TABLE UW-12 COLUMN (c)
ELLIPSOIDAL HEAD MATERIAL SA-285A
HEAD SKIRT DIAMETER 40"
S=12,300 FROM SECTION II PART D-TABLE 1
CATEGORY B TYPE 1 SPOT RT
JOINT EFFICIENCY E=0.85 FROM FIG. UW-12 (d)
QUALITY FACTOR E=1.0 FROM UW-12(d)
NOZZLE MATERIAL SA-53A ERW
NOZZLE DIAMETER 14"
S=11,700 FROM SECTION II PART D-TABLE 1
CATEGORY B TYPE 1 NO RT
JOINT EFFICIENCY E=0.85 FACTORED IN STRESS VALUE
QUALITY FACTOR E=0.85 FROM UW-12 (d) & (e)
NOZZLE FLANGE SA-105 300 LB SLIP ON B16.5
FIG. C.20.E
~_~
SEAMLESS
ELLIPSOIDAL
Joint Efficiency Factors
CIRCUMFERENTIAL STRESS IN SHELL UG-27(c)(1)
1= PRISE-0.6P
1= 175 X20/12,800X 0.70-0.6 X 175
t =3,500/8855
1= 0.395" USE 0.500"
GO TO FIG. UCS-66
SA-414A IS IN CURVE B
60F IS ABOVE CURVE - NO IMPACT TEST REQUIRED
NOT NECESARY TO CALCULATE LONGITUDINAL STRESS IN SHELL
CATEGORY B SPOT RT ·E=0.85 IF CALCULATED
ELLIPSOIDAL HEAD UG-32(d)
1= PD/2SE-0.2P
t= 175 x40/2 X 12,300X 1.0-0.2x 175
t 7,000/24,565
t=0.285 USE 0.375
GO TO FIG. UCS-66
SA-285A IS IN CURVE B
60F IS ABOVE CURVE· NO IMPACT TEST REQUIRED
=
CIRCUMFERENTIAL STRESS IN NOZZLE UG-27(c)
t= PRISE-0.6P
t= 175 x7/11 ,700 X 0.85-0.6 X 175
1= 1 ,225/9480
1=0.129° USE 0.210°
GO TO FIG. UCS-66
SA-53A IS IN CURVE A
60F IS ABOVE CURVE· NO IMPACT TEST REQUIRED
FIG. C.20.E (CONTD.)
275
ApPENDIXD
FLANGE CALCULATION SHEETS
FLANGE CALCULATION SHEETS
Blank fill-in calculation sheets are given for the following types of flanges:
Sheet D.l - Ring flange with ring-type gasket
Sheet D.2 - Slip-on or lap-joint flange with ring-type gasket
Sheet D.3 - Welding neck flange with ring-type gasket
Sheet D.4 - Reverse welding neck flange with ring-type gasket
Sheet D.S - Slip-on flange with full-face gasket
Sheet D.6 - Welding neck flange with full-face gasket
1
2
DESIGN CONomONS
3
fA.CE
GA.SKET
N_
b
G
!.oIling MOhHial
4
COl',olion "110","01\0
LOAO AND BOLT CALCULATIONS
Duign Temp.. Sf.
W .. z
AIm. Temp., Si.
Hp
= 2brGmP =
Delign Temp., Sb
H
.....G'lrP/4 -
AIm. Temp., S.
W .. l
=
CONDITION
LOA.O
5
He -W.,
H-
Hr-H-Hp-
b.Cy
Hp
+
H
=
x
lEVER ARM
ho -
SIC -
he -
.5(C
M.OMENT
Mp
B)
Gl
hI - .S! ho
=
Me -
+ j'cl
Hpho -
Helle -
Mr =Hrf,r
=
M.
Hc-W
he - .SIC -
6
Gl
M.
-
SHAPE CONSUNTS
J(
11 boll,poclI''i1 u.ceedl 2<;1
M. ond MiJ
7
=
A(B
+ to mulhply _ I &011 ,pacing
~"
20 +,
In I "quollOf'lI by
OPERATING
~-
1=
GR~TER
,== ~ ~:~
SEATING
t
f
= 'J ~
•
. ._ _.....
Cc>mputed
0.·'
1
C!welted
Numb.,
)
FIG. 0.1
RING FLANGE WITH RING-TYPE GASKET
277
278 Appendix D
0..
1
Desir
2
DESIGN CONDITIONS
"0.__,
FAa
GASKET
3
l'
N
•
r....p.toIVf.
0
flaftge MOI.rio!
,
lolling MOlerlol
•
w.o,ion
Flange
II
~.
4
Al1owon<.
Bolling
W.,
AI.... Temp., 51.
H.
Oelign T.mp~ 51>
H
Aim. remp., S.
W.,
CONDITION
5
OJnraJ.ing
8
S,.
S"
S,.
s"
LOAD
.
S"
S,.
,.
S,.
+H
LEVER
X
Ho
W.,
H
ho
.5(C
Ho -
h,
.sre + "1 + j'GI
Hc=W
R+91
=
he
~
.SIC -
Gl
-
M,
M.
~
"f'
-
gl/g.
Rodigl fig., 51
tJmG/A/1
h.
a
t-
I
F,
"
U h.g.1
V,
STRESS FORMULA FACTORS
1 ~9O'
A
"
W
~g,"
-+
~
It + 1
= .4/3 te +
, ==
-r
x
tljd
,+,
(flo_KG/
••
I
«rt
KGI
II
II
OPERATING SEATING.
+ /,
If boll fpO(i"O .XU.df 20
multiply
11I0 o"d 1II(l in ebeve equolionf by:
,~
1--8'
t"
-C'
I
b:4=
'801t$
~
I
Ho
~R"
-
e=h,
-/8g.
7
== "'0Y/t":- zs•
.sts, + SrI
::.,~" .SIS.'! + 5.] or
-l>
h/I>.
F,
U
Sealing
mol}"",)
IH
-
V,
tong. Hub, SH
1.
~
K and HUB FACTORS
V
STRESS CALCULATION
i
~
Z
::••;;.srs, + 5.} or.5 (SH + 5r)
Tong. fig., Sf
Hlh,
M,
,
lS,
I1php
Me - Hchc
r
Rodiol Fig., S~ - tim. /'io.P
m. YIII
,1,\.0
6
m./)..gll
Tong. FI"., 5r _
MOMENT
~
01
STRESS CAlCULATION-operating
long. Hub, Sit
'RM
-
hI)
9~'
1.5
w- .5(A... + At.}S.
GlrP/4
__ N,
::-:t W.J/S. or W .. ljSb
A.
....
_ 2bll"Gmf'_
~$I"I.4
All_.. b1.
$I....
1.5
CALCULATIONS
brGy
No
Ht_H
SeQllnll
ecu
tOAD AND
Oesign remp_, Sf.
He
H,
G'
Comput.d
Oat.
C....ck.d
Numb..
FIG. D.2
SLlp·ON OR LAp·JOINT FLANGE WITH RING·TYPE GASKET
&olt ,pod""
2Q
+'
Flange Calculation Sheets
1
2
DESIGN CONI)fTIONS
3
FAa
GASKET
D.Jion "e.wr., ,
N
D.tiOn r.llIJ:lerolure
b
G
,
Flonoe Motorial
•
loltinO Material
Cotlo,ior\ Allowonco
4
..
Dctsign romp., 5#.,
W.. ~ _ bll'Gy
AIm. Temp., S,.
H,
= 2bll'GmP =
H
_ Gl ll'PI 4
W.,
=H,.
h
I'
,J
Flong,
Do,;on
lohmg
Tf!mp~
51.
Atm. Tomp., S,
CONDITION
5
lOAD AND BOLT CALCULATIONS
lOAD
s..:.ol;n'il
H, _
1.5 S,.
hG -
.5(C -Gl
+ 0'
Long. Hub, Sa _ Im,/).,;,'
S,.
:;'~, ".S(SH
frno/}.Ql"
So.
Rodiol FIg., S.
PIno/At'
S,.
Tong. flg., S'l' _frio Y-/t"-ZS.
S"
:;":t .SIStI + S.l or ,S!SIl + 5,}
t
h'
t"
I
=
-
M.
-
Mo
=
-
f
-
, _ FII>,
F' 1
~'-R'
.../ ~
~ :-ho
,-91<
=
yu h"g.l
_I.
=
4/3 I, + I
,
-r - alT
= Il/d
,
-
+1
-'Y+~
=
-
=
-
_ Mol B _ Operaling_
"
==
Mo/B
== 'ealing =
+
If bolt spacing eJl:eeeds 20
f, mllltiply
m. and mo in above equalionl by:
I
I
;Bolfs
Hr
,I
Boll IpaeinQ
20
+I
.
HO
t~ +
-
STRESS FORMULA FACTORS
,
"~
me.
B'
d
.fBi.
7
-
=
-
V
gl/g.
...
-
hlh,
F
-
,
noting
Mr_Hrhr
K AND HUB FACTORS
U
-n~'
..E'
-
z
STRESS CALCULATION
-
MG - HanG -
T
+ Sd or.s ISH + Sri -
!.orlg. Hub, Sa
+hcl
MOMENT
Mo - Hphp
K _ AlB
-
Radial Fig., Sa _
9 AI,_obl,
$I, ...
=
=
6
STRESS CALCULATION-Op'l'Otillg
S,.
1.5 S,.
11:+ .50'
hG = .51C - G)
13m../).,t"
Tong. Fig., ST _m,Y/t" -IS._
S,.
_
=
lEVER ARM
bQ _
h, -.S(R
=
Hc=W
8 .."o.....bI.
50,...
A~lS,
+H=
-
He;; -W.. l-H
Hr _H
-
W _ .S{A.. +
-
X
fip.~ffB~PI.l~
Operating
.
A.. _ :;'.:" W",~/S. Of W .. l/S~_
-
G'
ComplIted
Dote
Checked
Numbe.
I-f<;
FIG. 0.3
WELDING NECK FLANGE WITH RING·TYPE GASKET
279
280 Appendix D
1
2
DESIGN CONDmONS
GASKET
fACE
DeliVR P'.,wre, ,.
H
0
O.sign Temp.raW••
G
.,
RCl""_ MOlorial
lolling Mot.rig!
4
COHo,ion Allowonc;.
5,.
il
!j
FIORile
W., - bTCy
A.
A. _
01uign Temp., S.
H
== G'TPjJ =
W _ .5{)...
Aim. Te..,p., S.
W., _ lip
CONDITION
LOAD
Hp,..,.d J ,. /
O/Hroting
:;0;;'
_ 2bll"GmP_
A'm. remp., Sf.
tolling
5
LOAD AND BOLT CALCULATIONS
H.
Design Temp.,
+
LEv ER ARM
hp,=·,5(c:.+9'
Ha_W.. 1 H_
0,
Hr_H_Hp_
hT
:::::
.5(C
GJ
.5(C -
B
8 Al'o-ob!<o
:lot...,
1.5
t
G)
MOMENT
~
BI_
211~
=
MIJ _ Hpllo -
-
M,
STRESS CAlCUlATION-operating
tong. Hub, 5/1 _fmol>'QI 1
5,.
Rodicll fjg., S.
5"
5,.
Tong. Fig., S-r_ m.YRIt' -ZSR(O.67Ie+I}/f3_
.lJ1....
1,5 St.
=
Tang_ fIg, 5T (AT 8') = ~;
"to.
2K' (1
(K'
[ V
STRESS CALCULAnON
+ ';. tel
Radial Fig., SR
f3ma/At'
S,.
S,.
Tong. Fig., 5T
maYl<ll'
I) A
~
a.
S"
Tong. Flg.• $or(AT8)=_ y
U,
(K'
"
H
1) A
"f
W
~"o{-h(;j
I
t=
~"'~
7
~
1<---",
U~ h
1_ '-,
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282 Appendix D
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DESIGN CONDITIONS
3
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ApPENDIXE
CONVERSION FACTORS
CONVERSION OF U.S. CUSTOMARY UNITS TO SI UNITS!
Multiple U.S. Customary Units
inches
feet
square iuches (in.')
square feet (ft")
U.S. gallons
cubic feet (ft.'}
pounds, mass
pouuds, force
psi, pressure
bars
Btu
ksi-x/in., fracture toughness
OF
By Factor
0.0254
0.3048
0.0006452
0.092903
0.003785
0.02832
0.4536
4.448
6,894.8
100,000.
1,005.056
1.099 X 10'
t, = (t, - 32)/1.8
'Por other conversions, see ASTM E 380.
283
To Get SI Units
meters
meters
m-
m'
m'
m'
kilograms
Newtons
Pascals
Pascals
Joules
Pa-x/rn
°C
INDEX
Categories, welded joint, 5, 6
Charpy impact-test requirements, 16
A
A value, 43-44
Allowable stress, 2-3, 5, 6
ASME Boiler and Pressure Vessel Code,
Section. VIII, v,seealso Section VIII
Assignment of materials to CUI\!es, 10
Axial buckling stress, critical, 39
Axial compression, cylindrical shells
under, 36-42
Chrome-moly steel forging specifications
SA-182,253
SA-336,254
Chrbrile~m()lysteelplatespecificatibns,
253
Circular flat plate, 101
Circumferential membrane stress, 28
Circumferential stress, 34, 70
Closure design details, 175-178
Cold temperatures, 24
Compensation, inherent, openings with,
136
Component analysis in VIII-2, 233-248
Compression
axial, cylindrical shells under, 36-42
cone-to-shell junction at large end of
cone in, 75-76
B
B value, 43-44
Beams on Elastic Foundation, 134
Bellows-type expansion joints, 230, 231
Bending moment, 41
Bending stress, 41-42
ligament efficiency for multi-diameter
openings for, 193-197
Blind flanges, 106-107
Boiler and Pressure Vessel Code, Section
VIII, ASME, v, see also Section VIII
Bolt loads, 106
Bolt sizing, 106
Bolted flanges, 105-106
connections with ring type gaskets,
108-124
Bolted flat head, 12
Bolted flat plates and covers, 105-106
Bolting, flat plates and covers with,
106-107
Bolting rings, 124
Braced and stayed construction, 169-173
Braced and stayed surfaces, 169-172
Brittle fracture, 9-13
Buckling, of cylindrical shells, 42-43
Buckling equation, 38
Buckling stress, critical axial, 39
Butt joints, 6
Butt welded, 5, 7
Butt welded components, II
cone-to-shell junction at small end of
cone in, 86-87
Compressive stress, 70
Cone-to-cylinder junction
large end of
inherent reinforcement for, 96
values of Q for, 96
small end of
inherent reinforcement for, 98
values of Q for, 99
Cone-to-shell junction
large, 80-81, 89-92
at large end of cone
in compression, 75-76
in tension, 85-86
small, 81-82, 92-94
at small end of cone
in compression, 86-87
in tension, 76-77
Conical sections
external pressure on, 85-94
internal pressure on, 74-85
VIII-l,74-94
VIII-2, 95-99
Conical shells, 74
Conversion factors, 283
Comer joint, 12
Corner welded, 5, 7
Corrosion allowance, 31, 32
c
C value, 102
Carbon steel plate specifications, 253
Cast Ductile Iron, rules of Part VCD for,
23-24
Cast Iron, rules of Part VCI for, 23
285
286 INDEX
Covers
flat, see Flat plates and covers
spherically dished, 124-131
Creep, 2
Creep rate, 3
Critical axial buckling stress, 39
Critical strain, lowest, 43
Crown radius of ellipsoidal heads, 67
Curves, assignment of materials to, 10
Cylinders
effective length of, 44
elliptical, 56
Cylindrical shells, 27_56
under axial compression, 36-42
buckling of, 42-43
equations, VIII-2, 53-54
external pressure on, 42-53
hoop stress in, 29
lines of support of, under external
pressure, 45
mitered, 54-55
openings in, 137
under tensile forces, 27-36
thick, 33-36, 53
thin, 27-33, 53
D
Design rules, 1
Design temperatures, 2, 4
Dimpled and embossed assemblies,
welded stays for, 170-172
Double full fillet lap joint, 6
E
E (Joint Efficiency Factors), 3-6, 255-275
Earthquake forces, 27
Edge moment, 101
Effective length of cylinders, 44
EJMA (Expansion Joint Manufacturers
Association) Standard, 201
Elastic foundation theory, 134, 213
Elasticity, modulus of, 38, 206
Ellipsoidal heads
crown radius of, 67
pressure on concave side of, 65-66
pressure on convex side of, 67-68
VIII-l,65-68
VIII-2,72-74
Elliptical cylinders, 56
Elliptical shells, 55-56
Empirical equations, 47
Enameled vessels, 23
EPC (External Pressure Charts), 37-38
Excess area, 77
Expansion Joint Manufacturers
Association (EJMA) Standard, 201
Expansion joints, 230, 231
External pressure
on conical sections, 85-94
on cylindrical shells, 42-53
on ellipsoidal heads, 67
on torispherical heads, 71
lines of support of cylindrical shells
under, 45
reinforced opening design for, 134
in spherical shells, 61-64
External Pressure Charts (EPC), 37~38
F
F-factor, 137, 138
Factor 1.1, 2
Fatigue, stress concentration factors used
in, 239
Fatigue curves, 20, 246
Fatigue evaluation, 245-248
Fatigue requirements, 19-22
F&D heads (Flanged and Dished heads),
68
Figures, list of, xiii-xv
Fillet welded,S, 7
Fixed tubesheets, 213-230
design equations for, 217-225
details for, 214, 215
Fixity factor, 209
Flange calculation sheets, 277-282
Flange connections, bolted, with ring type
gaskets, 108-124
Flange rings, 124
Flanged and Dished heads (F&D heads),
68
Flanged and flued expansion joints, 230,
231
Flanges, 12, 77
blind, 106-107
bolted, see Bolted flanges
flat-face, 124
full-face gasket, 123
integral, 109, 110, 125, 126
lap-joint, see Lap-joint flanges
loose, 109, 125
optional, 109, 125
reverse, 118-123
ring, 109
slip-on, see Slip-on flanges
standard, 109-118
welding neck, see Welding neck flanges
Flat-face flanges, 124
287
Flat head
bolted, 12
integral, 12
Flat plates and covers, 101-108
bolted, 105-106
with bolting, 106-107
circular, with bolting, 106-107
circular integral, 101-104
integral, 101-105
multiple openings in rims of, 109
noncircular, with bolting, 107
noncircular integral, 104-105
openings in, 107-108
unstayed, 103
Flues, 77
Forces
earthquake, 27
meridional, 64
tensile, see Tensile forces
Fracture, brittle, 9-13
Full-face gasket
slip-on flange with, 281
welding neck flange with, 282
Full-face gasket flanges, 123
G
Gasket crushout, check for, 106
Gasket design requirements, 105
Gaskets
full face, see Full-face gasket
ring type, see Ring-type gasket
Geometric parameters, 217
Glass-lined vessels, 23
Group numbers, 2
H
Half-pipe jackets, 181-186
maximum allowable internal pressure
in, 181-182
minimum thickness of, 182-184
Head configurations, 57
Heads
conical, 74
ellipsoidal, see Ellipsoidal heads
hemispherical, see Hemispherical heads
spherically dished, 124
tori spherical, see Torispherical heads
toriconical, 77
Heat exchangers
configurations of, 202
design of, 201-231
design rules for components of, 201
example, 8
tubesheet, see Tubesheets
If-tube, see Uetube exchangers
Hemispherical heads, 57
pressure on concave side of, 57-61
pressure on convex side of, 61-64
thickness, 60-61, 63-64
VIII-I,57-64
VIlI-2, 64-65
High alloy steels, without impact testing,
minimum design metal temperatures
in, 18
Hillside position, 146
Hoop stress, 4
basic equation for, 55
in cylindrical shells, 29
Hydrostatic, term, 22
Hydrostatic test, 22
for VIII-I, 23
for VIII-2, 24-25
I
Impact energy, minimum, 16
Impact-test exemption curves, 15
Impact-test requirements, Charpy, 16
Impact testing
high alloy steels without, minimum
design metal temperatures in, 18
reduction of MDMT without, 17
Inertia, moment of, calculating, 87-88
Inherent compensation, openings with,
136
Inside radius, 58
Integral flanges, 109, 110, 125
Integral flat head, 12
Internal pressure, 27
on conical sections, 74-85
maximum allowable, in half-pipe
jackets, 181-182
reinforced opening design for, 134
in cylindrical shells, 27
in spherical shells, 57-61
in eliiptical shells, 55
in torispherical shells, 68
J
Jacket closure bars, 4
Jacket penetrations, 179-181
Jacketed vessels, 173-181
design of closure member for, 175-179
openings in, 179
types of, 174
welded stays for, 170-171
Jackets
half-pipe, see Half-pipe jackets
288 INDEX
spiral, 182
Joint categories, welded, 5, 6
Joint Efficiency Factors (E), 3, 4, 6,
255-275
Mitered cylindrical shells, 54-55
Modulus of elasticity, 38, 206
Moment of inertia, calculating, 87-88
N
K
K factor, 181-186
Knuckle, thickness required for, 77
Knuckle radius, 65
L
l\ factor, 207 ~208
Lame's equation, 33
Lap-joint flanges, 109
with ring-type gasket, 278
Layered vessels, 30-31
Ligament efficiency
for constant-diameter openings, 192
for multi-diameter openings
for bending stress, 193-197
for membrane stresses, 192-193
Ligament efficiency method, 133
Ligament efficiency rules, VIII-I, 164-167
Loads, 1
bolt, 106
radial, 52
shear, 50
vacuum, 47
wind,27
Longitudinal stress, 34
Loose flanges, 109, 125
M
Mandatory rules, 1
Material designation, 253-254
Materials to curves, assignment of, 10
MAWP (maximum allowable working
pressure), 23, 24
Maximum allowable working pressure
(MAWP), 23, 24
MDMT, see Minimum Design Metal
Temperature
Membrane stress, 24
circumferential, 28
ligament efficiency for multi-diameter
openings for, 192-193
in spherical shells, 58
Meridional forces, 64
Meridional stress, 70
Minimum Design Metal Temperature
(MDMT),11
in high alloy steels without impact
testing, 18
reduction of, without impact testing, 17
Non-Mandatory rules, 1
Noncircular cross section, vessels of,
186-199
Noncircular flat plates, 104, 107
Nozzle connections, 54
Nozzle design, alternative rules for,
157-164
Nozzle nomenclature, 154
Nozzle reinforcement, 140-151
of series of openings, 165-165
o
Obround cross section, vessels of, 183,
191
Openings, 133-168
code bases for acceptability of, 133
constant-diameter, ligament efficiency
for, 192
in cylindrical shells, 137
exceeding size limits, 151-153
in flat plates and covers, 107-108
with inherent compensation, 136
in jacketed vessels, 179
multi-diameter, ligament efficiency for,
see Ligament efficiency for multidiameter openings
multiple, in rims of flat heads or covers,
109
nozzle reinforcement of series of,
165-166
reinforced, see Reinforced openings
in spherical shells, 139
terms and definitions for, 134
in vessels of noncircular cross section,
187,192-197
Optional flanges, 109
Outside radius, 58
p
P numbers, 2
Part Dcn for Cast Ductile Iron, rules of,
23-24
Part DCI for Cast Iron, rules of, 23
Part DLT rules, 24
Peak stress, 133, 235
Penetrations, jacket, 179-181
Plate diameter
opening diameter does not exceed half,
107-108
289
opening diameter exceeds half, 108
Plates, flat, see Flat plates and covers
Plates on elastic foundation, 213
Pneumatic, term, 22
Pneumatic test, 22
for VIII-\' 23
for VIII-2, 24
Poisson's ratio. effective, 206
Pressure
on concave side
of ellipsoidal heads, 65-66
of hemispherical heads, 57-61
of torispherical heads, 68-71
on convex side
of ellipsoidal heads, 67-68
of hemispherical heads, 61-64
of torispherical heads, 71-72
external, see External pressure
internal, see Internal pressure
maximum, for thickness, 27
Pressure-area procedure, 77
Pressure boundary, 133
Pressure test requirements
for VIII-I, 23-24
for VIII-2, 24-25
Pressure testing, 22-25
Pressure vessels, 22
Primary stress, 133, 233, 234
Process cyclic curves, 247
Q
Q factor, 95
Quality factor, 255, 259-266, 271, 273
Quality Factors, 3
Quench and tempered carbon and alloy
steel forging specifications, 254
R
Radial loads, 52
Radius-to-depth ratio, 65
Reactor, example, 14
Rectangular cross section, vessels of,
188-190, 197-199
Reducers, 54
Reduction factor, 209
References, 249-250
Reinforced openings
area of reinforcement available for,
140-151
area of reinforcement required for,
137-139
general requirements, 134-136
limits of reinforcement for, 140
rules
VIII-l,136-153
VIII-2, 153-164
shape and size of, 137
Reinforcement, nozzle, see Nozzle
reinforcement
Reinforcement limits, 134-136
Reinforcement plate, welded connection
with,l1
Reverse flanges, 118-123
Reverse welding neck flange with ringtype gasket, 280
Ring flanges, 109
with ring-type gasket, 277
Ring girders, 4
Ring-type gaskets
bolted flange connections with, 108-124
reverse welding neck flange with, 280
ring flange with, 277
slip-on or lap-joint flange with, 278
welding neck flanges with, 279
Rings, stiffening, see Stiffening rings
Rupture, 2, 3
s
Secondary stress, 233
Section VIII
background information, 1-25
Divisions 1 and 2, 1
Section VIII-1, pressure test requirements
for, 23-24
Section VIIl-2, pressure test requirements
for, 24-25
Sections, conical, see Conical sections
Shear loads, 50
Shearing stress, 52
Shells
conical, see Conical sections
cylindrical, see Cylindrical shells
elliptical, 55-56
spherical, see Spherical shells
51 units, conversion of U.S. customary
units to, 283
Single full fillet lap joint, 6
Single-welded butt joints, 6
Slip-on flanges, 109
with full-face gasket, 281
with ring-type gasket, 278
Special components, VIII-I, 169-199
Spherical radius, 65
Spher-ical shells
external pressure in, 61-64
internal pressure in, 57~61
large, 58
membrane stress in, 58
openings in, 139
290 L'lDEX
thick,58
VIII-I, 57-64
VIII-2, 64-65
Spherically dished covers, 124-131
Spiral jackets, 182
Standard flanges, 109-118
Stays
and staybolts. 172-173
welded, see Welded stays
Stiffening rings, 48-50
attachment of, 50-53
designing, 48-50
Strain, 37
critical, lowest, 43
Stress
allowable, 2-5
bending, 41-42
circumferential, 34, 70
classification of, 237-238
combinations of, 240-245
compressive, 70
critical axial buckling, 39
hoop, see Hoop stress
longitudinal, 34
membrane, see Membrane stress
meridional, 70
peak, 133, 235
primary, 133,233,234
secondary, 233
shearing, 52
tensile, 2
thermal, 233, 235
Stress categories, 233-239
and limits, 236
Stress concentration, 239
Stress multipliers, 3
Stress rupture, 3
Stress-strain diagrams, 37-38
Structural discontinuity, 234
T
Tables, list of, xvii
TEMA (Tubular Exchanger Manufacturers
Association) standards, 201
Temperatures
cold,24
design, 2, 4
metal, minimum design, see Minimum
Design Metal Temperature
Tensile forces, cylindrical shells under,
27-36
Tensile strength, 3
Tensile stress, 2
Tension
cone-to-shell junction at large end of
cone in, 85-86
cone-to-shell junction at small end of
cone in, 76-77
Thermal stress, 233, 235
Thickness
details governing, used for toughness,
11-13
hemispherical heads, 60-61, 63-64
maximum pressure for, 27
minimum, of half-pipe jackets, 182-186
required for knuckle, 77
Threaded-end stay construction, special
limitations for, 170
Threaded-end stays, special limitations
for, 172
Torispherical heads
pressure on concave side of, 68-71
pressure on convex side of, 71-72
shallow, 68
VIII-I, 68-72
VIIl-2,72-74
Toughness, governing thickness details
used for, 11-13
Toughness rules, 13
Transition sections, 54, 57
conical, see Conical sections
Tube patterns, 206
Tubesheet attachments, 4
Tubesheet design
rules for, 201
in U-tube exchangers, 201-213
Tubesheets, 12
fixed, see Fixed tubesheets
integral, design equations for, 209-213
simply supported, design equations for,
207
types of, 201
Tubular Exchanger Manufacturers
Association (TEMA) standards, 201
u
U-tube exchangers, tubesheet design in,
201-213
Unified Numbering System (UNS), 2
UNS (Unified Numbering System), 2
Unstayed flat plates and covers, 103
U.S. customary units to SI units,
conversion of, 283
v
Vacuum loads, 47
Venting, 23
291
Vessels
jacketed, see Jacketed vessels
layered, 30-31
of noncircular cross section, 186-199
of obround cross section, 187-191
of rectangular cross section, 188-190,
197-199
VIII-l requirements, guide to, 252
w
Weld efficiencies, 3
Welded attachments, 13
Welded connection with reinforcement
plate, 11
Welded-in stay construction, special
limitations for, 170, 171
Welded joint categories, 5, 6
Welded stays
for dimpled and embossed assemblies,
170-172
for jacketed vessels, 170, 171
Welding neck flanges, 114, 115
witb full-face gasket, 281
with ring-type gasket, 279
Wind loads, 27
Wind moment, 40
y
Yield strength, 3
interpolation between, 16
z
Z factor, 104