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2General Design

Procedure 2-1: General Vessel Formulas . . . . . . . . . . . . 38Procedure 2-2: External Pressure Design.. . . . . . . . . . . 42Procedure 2-3: Properties of Stiffening Rings . . . . . . 51Procedure 2-4: Code Case 2286.. . . . . . . . . . . . . . . . . . . . . . . 54Procedure 2-5: Design of Cones . . . . . . . . . . . . . . . . . . . . . . . 58Procedure 2-6: Design of ToriconicalTransitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67Procedure 2-7: Stresses in Heads Dueto Internal Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70Procedure 2-8: Design of Intermediate Heads . . . . . . 74Procedure 2-9: Design of Flat Heads . . . . . . . . . . . . . . . . . 76Procedure 2-10: Design of Large Openingsin Flat Heads .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81Procedure 2-11: Calculate MAP, MAWP, andTest Pressures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83Procedure 2-12: Nozzle Reinforcement.. . . . . . . . . . . . . . 85

Procedure 2-13 Find or Revise theCenter of Gravity of a Vessel . . . . . . . . . . . . . . . . . . . . . . . . . . 90Procedure 2-14: Minimum Design MetalTemperature (MDMT).. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90Procedure 2-15: Buckling of Thin WallCylindrical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95Procedure 2-16: Optimum VesselProportions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96Procedure 2-17: Estimating Weights of Vesselsand Vessel Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Procedure 2-18: Design of Jacketed Vessels. . . . . . . . 124Procedure 2-19: Forming Strains/FiberElongation.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

37

Procedure 2-1: General Vessel Formulas [1,2]

Notation

P ¼ internal pressure, psiDi, Do ¼ inside/outside diameter, in.

S ¼ allowable or calculated stress, psiE ¼ joint efficiencyL ¼ crown radius, in.

Ri, Ro ¼ inside/outside radius, in.K, M ¼ coefficients (See Note 3)

sx ¼ longitudinal stress, psisf ¼ circumferential stress, psiRm ¼ mean radius of shell, in.t ¼ thickness or thickness required of shell, head,

or cone, in.r ¼ knuckle radius, in.

Notes

1. Formulas are valid for:a. Pressures <3,000 psi.b. Cylindrical shells where t � 0.5 Ri or P � 0.385

SE. For thicker shells see Reference 1, Para. 1–2.c. Spherical shells and hemispherical heads where

t � 0.356 Ri or P � 0.665 SE. For thicker shellssee Reference 1, Para. 1–3.

2. All ellipsoidal and torispherical heads havinga minimum specified tensile strength greater than80,000 psi shall be designed using S¼ 20,000 psi atambient temperature and reduced by the ratio ofthe allowable stresses at design temperature andambient temperature where required.

3. Formulas for factors:

K ¼ 0:167

�2þ

�D2h

�2�

M ¼ 0:25

3þ

ffiffiffiLr

r !

Figure 2-1. General configuration and dimensional datafor vessel shells and heads.

38 Pressure Vessel Design Manual

Table 2-1General vessel formulas

Thickness, t Pressure, P Stress, S

Part Stress Formula I.D. O.D. I.D. O.D. I.D. O.D.

Shell

Longitudinal

[1, Section UG-27(c)(2)]

sx ¼ PRm

0:2t

PRi

2SEþ 0:4P

PRo

2SEþ 1:4P

2SEt

Ri � 0:4t

2SEt

Ro � 1:4t

PðRi � 0:4tÞ2Et

PðRo � 1:4tÞ2Et

Circumferential

[1, Section UG-27(c)(1);

Section 1-1 (a)(1)]

sf ¼ PRm

t

PRi

SE� 0:6P

PRo

SEþ 0:4P

SEt

Ri þ 0:6t

SEt

Ro � 0:4t

PðRi þ 0:6tÞEt

PðRo � 0:4tÞEt

Heads

Hemisphere

[1, Section 1-1 (a)(2);

Section UG-27(d)]

sx ¼ sf ¼ PRm

2t

PRi

2SE� 0:2P

PRo

2SEþ 0:8P

2SEt

Ri þ 0:2t

2SEt

Ro � 0:8t

PðRi þ 0:2tÞ2Et

PðRo � 0:8tÞ2Et

Ellipsoidal

[1, Section 1-4(c)]

See Procedure 2-7PDiK

2SE� 0:2P

PDoK

2SEþ 2PðK� 0:1Þ2SEt

KDi þ 0:2t

2SEt

KDo � 2tðK� 0:1Þ See Procedure

2-7

2:1 S.E.

[1, Section UG-32(d)]

See Procedure 2-7PDi

2SE� 0:2P

PDo

2SEþ 1:8P

2SEt

Di þ 0:2t

2SEt

Do � 1:8tSee Procedure

2-7

100%e6% Torispherical

[1, Section UG-32(e)]

See Procedure 2-70:885PLiSE� 0:1P

0:885PLoSEþ 0:8P

SEt

0:885Li þ 0:1t

SEt

0:885Lo � 0:8tSee Procedure

2-7

Torispherical L/r < 16.66

[1, Section 1-4(d)]

See Procedure 2-7PLiM

2SE� 0:2P

PLoM

2SEþ PðM� 0:2Þ2SEt

LiMþ 0:2t

2SEt

LoM� tðM� 0:2Þ See Procedure

2-7

Cone

Longitudinal sx ¼ PRm

2tcosf

PDi

4cosfðSEþ 0:4PÞPDo

4cosfðSEþ 1:4PÞ4SEtcosf

Di � 0:8tcosf4SEtcosf

Do � 2:8tcosf

PðDi � 0:8tcosfÞ4Etcosf

PðDo � 2:8tcosfÞ4Etcosf

Circumferential

[1, Section 1-4(e);

Section UG-32(g)]

sf ¼ PRm

tcosf

PDi

2cosfðSE� 0:6PÞPDo

2cosfðSEþ 0:4PÞ2SEtcosf

Di þ 1:2tcosf

2SEtcosf

Do � 0:8tcosf

PðDi þ 1:2tcosfÞ2Etcosf

PðDo � 0:8tcosfÞ2Etcosf

GeneralDesign

39

Figure 2-1a. Required shell thickness of cylindrical shell.


Figure 2-1a. (continued).

General Design 41

Procedure 2-2: External Pressure Design

Notation

A ¼ factor “A,” strain, from ASME Section II, PartD, Subpart 3, dimensionless

As ¼ cross-sectional area of stiffener, in.2

B ¼ factor “B,” allowable compressive stress, fromASME Section II, Part D, Subpart 3, psi

D ¼ inside diameter of cylinder, in.Do ¼ outside diameter of cylinder, in.DL ¼ outside diameter of the large end of cone, in.Ds ¼ outside diameter of small end of cone, in.E ¼ modulus of elasticity, psiI ¼ actual moment of inertia of stiffener, in.4

Is ¼ required moment of inertia of stiffener, in.4

I0s ¼ required moment of inertia of combined shell-ring cross section, in.4

L ¼ for cylindersdthe design length for externalpressure, including 1/3 the depth of heads, in.For conesdthe design length for externalpressure (see Figures 2-1b and 2-1c), in.

Le ¼ equivalent length of conical section, in.Ls ¼ length between stiffeners, in.

LT–T ¼ length of straight portion of shell, tangent totangent, in.

P ¼ design internal pressure, psiPa ¼ allowable external pressure, psiPx ¼ design external pressure, psiRo ¼ outside radius of spheres and hemispheres,

crown radius of torispherical heads, in.t ¼ thickness of cylinder, head or conical section,

in.te ¼ equivalent thickness of cone, in.a ¼ half apex angle of cone, degrees

Unlike vessels which are designed for internal pressurealone, there is no single formula, or unique design, whichfits the external pressure condition. Instead, there is a rangeof options available to the designer which can satisfy thesolution of the design. The thickness of the cylinder is onlyone part of the design. Other factors which affect the designare the length of cylinder and the use, size, and spacing ofstiffening rings. Designing vessels for external pressure isan iterative procedure. First, a design is selected with all ofthe variables included, then the design is checked todetermine if it is adequate. If inadequate, the procedure isrepeated until an acceptable design is reached.

Vessels subject to external pressure may fail at wellbelow the yield strength of the material. The geometry ofthe part is the critical factor rather than material strength.Failures can occur suddenly, by collapse of thecomponent.

External pressure can be caused in pressure vessels bya variety of conditions and circumstances. The designpressure may be less than atmospheric due to condensinggas or steam. Often refineries and chemical plants designall of their vessels for some amount of external pressure,regardless of the intended service, to allow for steamcleaning and the effects of the condensing steam. Othervessels are in vacuum service by nature of venturi devicesor connection to a vacuum pump. Vacuums can be pulledinadvertently by failure to vent a vessel during draining, orfrom improperly sized vents.

External pressure can also be created when vessels arejacketed or when components are within multichamberedvessels. Often these conditions can be many times greaterthan atmospheric pressure.

When vessels are designed for both internal andexternal pressure, it is common practice to first determinethe shell thickness required for the internal pressurecondition, then check that thickness for the maximumallowable external pressure. If the design is not adequatethen a decision is made to either bump up the shellthickness to the next thickness of plate available, or addstiffening rings to reduce the “L” dimension. If the optionof adding stiffening rings is selected, then the spacing canbe determined to suit the vessel configuration.

Neither increasing the shell thickness to remove stiff-ening rings nor using the thinnest shell with the maximumnumber of stiffeners is economical. The optimum solutionlies somewhere between these two extremes. Typically,the utilization of rings with a spacing of 2D for vesseldiameters up to about eight feet in diameter and a ringspacing of approximately “D” for diameters greater thaneight feet, provides an economical solution.

The design of the stiffeners themselves is also a trial anderror procedure. The first trial will be quite close if the oldAPI-ASME formula is used. The formula is as follows:

Is ¼ 0:16D3oPxLs

E

Stiffeners should never be located over circumferentialweld seams. If properly spaced they may also double as


insulation support rings. Vacuum stiffeners, if combinedwith other stiffening rings, such as cone reinforcementrings or saddle stiffeners on horizontal vessels, must bedesigned for the combined condition, not each indepen-dently. If at all possible, stiffeners should always clearshell nozzles. If unavoidable, special attention should begiven to the design of a boxed stiffener or connection tothe nozzle neck.

Design Procedure For Cylindrical Shells

Step 1: Assume a thickness if one is not alreadydetermined.

Step 2: Calculate dimensions “L” and “D.”Dimension “L”should include one-third the depth of the heads. Theoverall length of cylinder would be as follows for thevarious head types:

Step 3: Calculate L/Do and Do/t ratiosStep 4: Determine Factor “A” from ASME Code, Section

II, Part D, Subpart 3, Fig G: Geometric Chart forComponents Under External or Compressive Loadings(see Figure 2-1e).

Step 5: Using Factor “A” determined in step 4, enter theapplicable material chart from ASME Code, Section II,Part D, Subpart 3 at the appropriate temperature anddetermine Factor “B.”

Step 6: If Factor “A” falls to the left of the material line,then utilize the following equation to determine theallowable external pressure:

Pa ¼ 2AE3ðDo=tÞ

Step 7: For values of “A” falling on the material line of theapplicable material chart, the allowable external pres-sure should be computed as follows:

Pa ¼ 4B3ðDo=tÞ

Step 8: If the computed allowable external pressure is lessthan the design external pressure, then a decision mustbe made on how to proceed. Either (a) select a newthickness and start the procedure from the beginning or(b) elect to use stiffening rings to reduce the “L”

dimension. If stiffening rings are to be utilized, thenproceed with the following steps.

Step 9: Select a stiffener spacing based on the maximumlength of unstiffened shell (see Table 2-1a). The stiff-ener spacing can vary up to the maximum valueallowable for the assumed thickness. Determine thenumber of stiffeners necessary and the corresponding“L” dimension.

Step 10: Assume an approximate ring size based on thefollowing equation:

I ¼ 0:16D3oPxLs

E

Step 11: Compute Factor “B” from the following equationutilizing the area of the ring selected:

B ¼ 0:75PDo

tþ As=Ls

Step 12: Utilizing Factor “B” computed in step 11, find thecorresponding “A” Factor from the applicable materialcurve.

Step 13: Determine the required moment of inertia fromthe following equation. Note that Factor “A” is the onefound in step 12.

Is ¼�D2oLs�tþ As=Ls

A

14

Step 14: Compare the required moment of inertia, I, withthe actual moment of inertia of the selected member. Ifthe actual exceeds that which is required, the design isacceptable but may not be optimum. The optimizationprocess is an iterative process in which a new memberis selected, and steps 11 through 13 are repeated untilthe required size and actual size are approximatelyequal.

Notes

1. For conical sections where a< 22.5 degrees, designthe cone as a cylinder where Do ¼ DL and length isequal to L.

2. If a vessel is designed for less than 15 psi, andthe external pressure condition is not going to bestamped on the nameplate, the vessel does nothave to be designed for the external pressurecondition.

W/(2) hemi-heads L ¼ LT�T þ 0:333D

W/(2) 2:1 S.E. heads L ¼ LT�T þ 0:1666D

W/(2) 100% e 6% heads L ¼ LT�T þ 0:112D

General Design 43

For Case B, Le ¼ LFor Cases A, C, D, E:

Le ¼ 0:5

�1þ Ds

DL

�

te ¼ t cos a

DL=te¼

Le=DL¼

Figure 2-1b. External pressure cones 22 1/2� < a < 60�.


Design stiffener for large end of cone as cylinderwhere:

Do ¼ DL

t ¼ tL

Ls ¼ L1

2þ L2

2

Design stiffener for small end of cone as cylinderwhere:

Do ¼ Ds

t ¼ ts

Ls ¼ L3

2þ L2

2

�1þ Ds

DL

�

Figure 2-1c. Combined shell/cone design for stiffened shells.

Figure 2-1d. External pressure ~ spheres and heads.

General Design 45

Figure 2-1e. Geometric chart for components under external or compressive loadings (for all materials). (Reprinted bypermission from the ASME Code, Section VIII, Div. 1.)


Design Procedure For Spheres and Heads

Step 1: Assume a thickness and calculate Factor “A.”

A ¼ 0:125tRo

Step 2: Find Factor “B” from applicable material chart.

B ¼

Step 3: Compute Pa.

Figure 2-1f. Chart for determining shell thickness of components under external pressure when constructed of carbonor low-alloy steels (specified minimum yield strength 24,000psi to, but not including, 30,000psi). (Reprinted bypermission from the ASME Code, Section VIII, Div. 1.)

Figure 2-1g. Chart for determining shell thickness of components under external pressure when constructed of carbonor low-alloy steels (specified minimum yield strength 30,000 psi and over except materials within this range where otherspecific charts are referenced) and type 405 and type 410 stainless steels. (Reprinted by permission from the ASMECode, Section VIII, Div. 1.)

General Design 47

A to left of material line Pa ¼ 0:0625E

ðRo=tÞ2

A to right of material line Pa ¼ BtRo

Notes

1. As an alternative, the thickness required for 2:1 S.E.heads for external pressure may be computed fromthe formula for internal pressure where P ¼ 1.67 Pxand E ¼ 1.0.

Table 2-1aMaximum length of unstiffened shells

Thickness (in.)

Diameter (in.) 1 =

45=16

3 =

87=16

1 =

29=16

5 =

811=16

3 =

413=16

7 =

815=16 1 1

1=16 11 =

8 13=16

36 204

N42 168 280

313 N48 142 235 358

264 437 N54 122 203 306 437

228 377 N60 104 178 268 381

200 330 499 N66 91 157 238 336 458

174 293 442 626 N72 79 138 213 302 408 537

152 263 396 56t N78 70 124 193 273 369 483 616

136 237 359 508 686 N84 63 110 175 249 336 438 559

123 212 327 462 625 816 N90 57 99 157 228 308 402 510 637

112 190 300 424 573 748 N96 52 90 143 210 284 370 470 585 715

103 173 274 391 528 689 875 N102 48 82 130 190 263 343 435 540 661 795

94 160 249 363 490 639 810 1,005 N108 44 76 118 176 245 320 405 502 613 738 875

87 148 228 337 456 594 754 935 N114 42 70 109 162 223 299 379 469 571 687 816

79 138 211 311 426 555 705 874 1,064 N120 39 65 101 149 209 280 355 440 536 642 762 894

74 128 197 287 400 521 660 819 997 N126 37 61 95 138 195 263 334 414 504 603 715 839 974

69 120 184 266 374 490 621 770 938 1,124 N132 35 57 88 129 181 242 315 391 475 569 673 789 916 1,053

65 113 173 248 348 462 586 727 884 1,060 1,253 N138 33 54 83 121 169 228 297 369 449 538 636 744 864 994

62 106 163 234 325 437 555 687 836 1,002 1,185 N144 31 51 78 114 158 214 275 350 426 510 603 705 817 940 1,073

59 98 154 221 304 411 526 652 793 950 1,123 1,312 N150 49 74 107 148 201 261 332 405 485 573 669 774 891 1,017 1,152

92 146 209 286 385 499 619 753 902 1,066 1,246 1,442 N156 46 70 101 140 189 248 309 385 462 546 637 737 846 966 1,095

87 138 199 271 363 475 590 717 859 1,015 1,186 1,373 N162 44 67 96 133 178 233 294 367 440 520 608 703 806 919 1,042

83 131 189 258 342 448 562 684 819 968 1,131 1,309 1,509 N

1 =

45=16

3 =

87=16 ½ 9=16

5 =

811=16

3 =

413=16

7 =

815=16 1 1

1=16 11 =

8 13=16

Notes:

1. All values are in inches.

2. Values are for temperatures up to 500�F.3. Top value is for full vacuum, lower value is half vacuum.

4. Values are for carbon or low-alloy steel (Fy > 30,000 psi) based on Figure 2-1g.


Table 2-1bMoment of inertia of bar stiffeners

Thk t, in. Max, ht, in.

Height, h, in.

1 11 =

2 2 21 =

2 3 31 =

2 4 41 =

2 5 51 =

2 6 61 =

2 7 71 =

2 8

1 =

4 2 0.020 0.070 0.167

0.250 0.375 0.5

5=162.5 0.026 0.088 0.208 0.407

0.313 0.469 0.625 0.7813 =

8 3 0.031 0.105 0.25 0.488 0.844

0.375 0.563 0.75 0.938 1.125

7=163.5 0.123 0.292 0.570 0.984 1.563

0.656 0.875 1.094 1.313 1.531

½ 4 0.141 0.333 0.651 1.125 1.786 2.667

0.75 1.00 1.25 1.50 1.75 2.00

9=164.5 0.375 0.732 1.266 2.00 3.00 4.271

1.125 1.406 1.688 1.969 2.25 2.535 =

8 5 0.814 1.41 2.23 3.33 4.75 6.510

1.563 1.875 2.188 2.50 2.813 3.125

11=165.5 1.55 2.46 3.67 5.22 7.16 9.53

2.063 2.406 2.75 3.094 3.438 3.783 =

4 6 1.69 2.68 4.00 5.70 7.81 10.40 13.5

2.25 2.625 3.00 3.375 3.75 4.125 4.50

13=166.5 2.90 4.33 6.17 8.46 11.26 14.63 18.59

2.844 3.25 3.656 4.063 4.469 4.875 5.2817 =

8 7 4.67 6.64 9.11 12.13 15.75 20.02 25.01

3.50 3.94 4.375 4.813 5.25 5.688 6.125

1 8 5.33 7.59 10.42 13.86 18.00 22.89 28.58 35.16 42.67

4.00 4.50 5.00 5.50 6.00 6.50 7.00 7.50 8.00

Note: Upper value in table is the moment of inertia. Lower value is the area.

General Design 49

Moment of Inertia of Stiffening Rings

Figure 2-1h. Case 1: Bar-type stiffening ring.

Table 2-1cMoment of inertia of composite stiffeners

I1 ¼ t1H3

12

I2 ¼ wt3212

c ¼P

AnYnPA

I ¼ PAnY

2n þ

PI� C

PAnYn

Type H W t1 t2P

AP

I C I

1 3 3 0.375 0.5 2.63 0.87 2.50 2.84

2 3 4 0.5 0.5 3.50 1.17 2.50 3.80

3 4 4 0.375 0.5 3.50 2.04 3.28 6.45

4 4 5 0.5 0.625 5.13 2.77 3.41 9.28

5 4.5 5 0.5 0.5 4.75 3.85 3.57 11.25

6 5 4 0.5 0.625 5.00 5.29 3.91 15.12

7 5.5 4 0.5 0.5 4.75 6.97 4.01 17.39

8 6 5 0.5 0.625 6.13 9.10 4.69 25.92

9 6 6 0.625 0.625 7.50 11.37 4.66 31.82

10 5.5 6 0.875 0.875 10.01 12.47 4.42 37.98

11 6.5 6 0.75 0.75 9.38 17.37 4.99 48.14

12 7 6 0.625 0.75 8.88 18.07 5.46 51.60

13 8 6 0.75 1 12.00 32.50 6.25 93.25

14 8 6 1 1 14.00 43.16 5.93 112.47

Figure 2-1i. Case 2: T-type stiffening ring.


Procedure 2-3: Properties of Stiffening Rings

Notation

AC ¼ Cross sectional area of composite area, in2

Ar ¼ Cross sectional area of ring, in2

AS ¼ Cross sectional area of shell, in2

Dm ¼ Mean diameter of shell, inE ¼ Modulus of elasticity at design temperature,

PSIFy ¼ Minimum specified yield strength at design

temperature, PSIK8, K9 ¼ Zick’s coefficients

IC ¼ Moment of inertia of composite section, in4

P ¼ Internal pressure, PSIGPX ¼ External pressure, PSIGQ ¼ Load at saddle, Lbs

Rm ¼ Mean radius of shell, inr ¼ Inside radius of shell, inR ¼ Inside radius of shell in feet

S13 ¼ Circumferential stress in shell due to load Q,PSI

S14 ¼ Circumferential stress in ringdue to loadQ,PSIt ¼ Thickness, shell, inn ¼ Poisson’s ratio

sS ¼ Stress in shell due to internal pressure, PSIsT ¼ Stress in shell due to external pressure, PSI

Derivation of Formula for Influence of Shell withStiffening Ring

Length of shell acting with ring, L

:5 L ¼ ðRm tÞ1=2=�3�1� v21=4

For v ¼ :3;

:5 L ¼ ðRm tÞ1=2 = 1:285¼ :78ðRm tÞ1=2 ¼ :55 ðDm tÞ1=2

L ¼ 1:56 ðRm tÞ1=2 ¼ 1:1ðDm tÞ1=2

Lateral Buckling of Stiffening Rings per CC2286

Lateral buckling is dependent on stiffener geometry.The requirements for stiffener geometry per CC2286 areas follows;

Rm

L

t

.5 L

Figure 2-2. Stiffening Ring Dimensions.

Stiffening ring check for external pressure

Ls B ¼ 0:75PDo

t þ As=Ls

If B � 2,500 psi,

A ¼ 2B/E

If B > 2,500 psi,

determine A from

applicable material

charts

Moment of inertia w/o shell

ls ¼ D2o Lsðt þ As=LsÞA

14

Moment of inertia w/ shell

ls ¼ D2o Lsðt þ As=LsÞA

10:9

t

P

Do

As

E ¼ modulus of elasticity

From Ref. 1, Section UG-29.

Figure 2-3. (A), (B), (C). Stiffener Geometry Per CC2286

General Design 51

Case 1: Flat bar stiffener, flange of a tee stiffener, oroutstanding leg of an angle stiffener;

h1=t1 � :375�E=Fy

1=2Case 2: Web of tee stiffener or leg of angle stiffener

attached to the shell;

h2=t2 ��E=Fy

1=2

Shell Stresses Due to Internal or External Pressure on theRing Section

• Stress in shell due to external pressure, sSsS ¼ ðPX L RmÞ=AC

• Stress in shell due to internal pressure, sTsT ¼ ½ðP RmÞ=t�½AS=AC�

Horizontal Vessel: Shell Stresses at Internal or ExternalRing Section Due to Load Q

Case 1: External Ring Stiffener

• Stress in shell, S13

S13 ¼ ð�ÞðK8 QÞ=AC þ ðK9 Q r CÞ=IC• Stress in ring, S14

S14 ¼ ð�ÞðK8 QÞ=ACð�ÞðK9 Q r dÞ=IC• Combined stresses;

If S13 is negative;

��S13��sS < :5 Fy

If S13 is positive;

ðþÞS13ðþÞsT < 1:5 S

Rm

L

c

d

Figure 2-4. External stiffening ring dimensions.

Figure 2-3. (continued ).


Case 2: Internal Ring Stiffener

• Stress in shell, S13S13 ¼ ð�ÞðK8 QÞ=ACð�ÞðK9 Q r CÞ=IC

• Stress in ring, S14S14 ¼ ð�Þ ðK8 QÞ=ACðþÞðK9 Q r dÞ=IC

• Combined stresses;�� S13 �� sS < :5 Fy

Sample Problem Given

Q ¼ 267 kips

q ¼ 150�

Ar ¼ 1� 8 ¼ 8 in2

AS ¼ L t ¼ 12:93�:875

¼ 11:31 in2

AC ¼ 19:31 in2

S ¼ 20 KSI

Fy ¼ 32:6 KSI

IC ¼ 135:88 in4

C ¼ 2:27 in

d ¼ 6:61 in

K8 ¼ :3

K9 ¼ :032

r ¼ 78 in

P ¼ 175 PSIG

PX ¼ ð�Þ 15 PSIGRm ¼ 78:5625

t ¼ 0:875 in ðcorrÞ

L ¼ 1:56 ðRm tÞ1=2

¼ 1:56 ð78:5625 ð875ÞÞ1=2

¼ 12:93 in

Case 1: External Ring


S13 ¼ ð�Þ ðK8 QÞ=AC þ ðK9 Q r CÞ=IC¼ ð�Þ :3 ð267Þ=19:31þ ½:032 ð267Þ 78 ð2:27Þ�

=135:88

¼ ð�Þ 4:15 þ 11:13 ¼ ðþÞ 6:98 KSI

• Stress in ring, S14

S14 ¼ ð�Þ ðK8 QÞ=AC ð�Þ ðK9 Q r dÞ=IC¼ ð�Þ :3 ð267Þ=19:31 ð�Þ ½:032 ð267Þ 78

ð6:61Þ�=135:88¼ ð�Þ 4:15 ð�Þ 32:42 ¼ ð�Þ 36:57 KSI

• Stress in shell due to external pressure, sS

sS ¼ ðPX L RmÞ=AC

¼ ½ð�Þ 15 ð12:93Þ78:5625Þ�=19:31¼ ð�Þ7:89 KSI

• Stress in shell due to internal pressure, sT

sT ¼ ½ðP RmÞ=t� ½AS=AC�¼ ½175 ð78:5625Þ=:875� ½11:31=19:31�¼ ðþÞ 9:2 KSI

• Combined stresses;

Since S13 is positive;

ðþÞ S13 ðþÞ sT < 1:5 S

6:98þ 9:2 ¼ 16:18 KSI

<1:5 S ¼ 1:5 ð20Þ ¼ 30 KSI OK

Rm

L

C

d

Figure 2-5. Internal stiffening ring dimensions.

General Design 53

Case 2: Internal Ring (Same properties)


S13 ¼ ð�Þ ðK8 QÞ=AC ð�Þ ðK9 Q r CÞ=IC¼ ð�Þ 4:15 ð�Þ 11:13 ¼ ð�Þ 15:28 KSI

• Stress in ring, S14

S14 ¼ ð�Þ ðK8 QÞ=AC ðþÞ ðK9 Q r dÞ=IC¼ ð�Þ 4:15 ðþÞ 32:42 ¼ ðþÞ 28:27 KSI

• Combined stresses;�� S13 �� sS < :5 Fy

ð�Þ 15:28 ð�Þ 7:89 ¼ ð�Þ23:17 KSI<:5 Fy ¼ :5

�32:6

¼ 16:3 KSI

No Good!Conclusion: Use an external ring or increase the

properties of the internal ring.

Procedure 2-4: Code Case 2286 [1,8,21]

Nomenclature

A ¼ cross-sectional area of cylinder,p(Do – t)t, in2

AS ¼ cross-sectional area of a ring stiff-ener, in2

AF ¼ cross-sectional area of a large ringstiffener, in2

Di ¼ inside diameter of cylinder, in.Do ¼ inside diameter of cylinder, in.De ¼ outside diameter of assumed equiv-

alent cylinder for design of cones orconical sections, in.

DS ¼ outside diameter at small end ofcone, or conical section betweenlines of support, in.

DL ¼ outside diameter at large end ofcone, or conical section betweenlines of support, in.

E ¼ modulus of elasticity at designtemperature, ksi

fa ¼ longitudinal compressive membranestress from axial load, ksi

fb ¼ longitudinal compressive membranestress from bending moment, ksi

fh ¼ circumferential compressivemembranestress from external pressure, ksi

fq ¼ longitudinal compressive membranestress from pressure load on end ofcylinder, ksi

fv ¼ shear stress from applied loads, ksifx ¼ fa þ fq, ksi

Fba ¼ allowable longitudinal compressivemembrane stress from bendingmoment, ksi

Fca ¼ allowable longitudinal compressivemembrane stress from uniform axialcompression with lc > 0.15, ksi

Faha ¼ allowable longitudinal compressivemembrane stress from uniform axialcompression with hoop compressionwith 0.15 < lc < 1.2, ksi

Fbha ¼ allowable longitudinal compressivemembrane stress from bendingmoment with hoop compression, ksi

Fcha ¼ allowable longitudinal compressivemembrane stress from uniform axialcompression with hoop compressionfor 0.15 < lc < 1.2, ksi.

Fcha ¼ Faha when fq ¼ 0.Fhba ¼ allowable circumferential compres-

sive membrane stress from bendingmoment with hoop compression, ksi

Fhe ¼ elastic circumferential compressivemembrane failure stress fromexternal pressure, ksi

Fha ¼ allowable circumferential compres-sive membrane stress from externalpressure, ksi


Fhva ¼ allowable circumferential compres-sive membrane stress from shearload with hoop compression, ksi

Fhxa ¼ allowable circumferential compres-sive membrane stress from uniformaxial compression with hoopcompression, for lc � 0.15, ksi

Fva ¼ allowable shear stress from appliedshear load, ksi

Fve ¼ elastic shear failure stress fromapplied shear load, ksi

Fvha ¼ allowable shear stress from shearload with hoop compression, ksi

Fxa ¼ allowable compressive membranestress from axial compression, for lc� 0.15, ksi

Fxe ¼ elastic longitudinal compressivemembrane failure (local buckling)stress from hoop compression, ksi

Fxha ¼ allowable longitudinal compressivemembrane stress from uniform axialcompression with hoop compressionfor lc � 0.15, ksi.

FS ¼ stress reduction factor or designfactor (circumferential and longitu-dinal directions have separatevalues)

I ¼ moment of inertia of cylinder crosssection, p(Do – t)3t/8, in4

IS’ ¼ moment of inertia of ring stiffenerplus effective length of shell aboutits centroidal axis of combinedsection, in4

IS ¼ moment of inertia of ring stiffenerabout its centroidal axis, in4

K ¼ coefficient to approximate endconditions for lc calculation (2.1 forfree standing vessels supported atgrade)

L, L1, L2. ¼ distance of unstiffened vesselbetween lines of support, in.

LB, LB1, LB2. ¼ distance between bulkheads or largerings, in.

Lc ¼ axial length of cone or conicalsection, in.

LF ¼ one-half of the sum of the distances,LB, from one large ring to the next orline of support, in.

LS ¼ one-half of the sum of the distance,L, from one ring to the next of line ofsupport, in.

Lt ¼ overall length of vessel including 1/3depth of each head, in.

Lu ¼ laterally unsupported length fora free-standing vessel, for a skirt-supported vessel with no guide wires,the distance is from the top tangentline to the base of the skirt, in.

M ¼ bending moment across cylindercross section, in-kips

P ¼ design external pressure, ksiPa ¼ allowable external pressureQ ¼ uniform axial compression, kipsQp ¼ axial compression as a result of

external pressure, ksir ¼ radius of gyration of a cylinder,

r ¼ 0.25*(Do2 þ Di

2)1/2, in.R ¼ radius to centerline of shell, in.Rc ¼ radius of centroid of combined ring

stiffener and effective length of shell,Rc ¼ R þ Zc, in.

Ro ¼ radius to outside of shells ¼ section modulus of cylinder cross

section, p(Do – t)2t/4, in.t ¼ thickness of shell, less corrosion

allowance, in.tc ¼ thickness of cone, less corrosion

allowance, in.T/T ¼ tangent to tangent length, inV ¼ shear force, kipsZc ¼ radial distance fromcenterline of shell

to centroid of combined section ofring and effective length of shell, in.

Zc ¼ ASZSAS þ Let

ZS ¼ radial distance from centerline ofshell to centroid of ring stiffener(positive for outside rings), in.

lc ¼ slenderness factor for columnbuckling

lc ¼ KLupr

ffiffiffiffiffiffiffiffiffiffiffiffiFxaFS

E

r

In 1998, Code Case 2286 Alternative Rules for Deter-mining Allowable External Pressure and Compressive

General Design 55

Stresses for Cylinders, Cones, Spheres, and FormedHeadswas approved for Section VIII, Divisions 1 and 2 ofthe ASME Code. Currently, Code Case 2286 may beapplied only for Division 1, since the rules found in CodeCase 2286 were absorbed into Part 4 of Division 2.

Code Case 2286 not only has rules for cylinders,cones, spheres, and formed heads, but also for the sizingof cone-cylinder junction rings, stiffening rings forexternal pressure, tolerances, and reinforcement foropenings. When 2286 is used, it must be used for theentire vessel. Using 2286 will require a more rigorousanalysis than the vacuum chart method in Division 1 andmay provide a thinner shell, and fewer and/or smallerstiffening rings. In these cases, the overall cost of thevessel may be lowered as a result of the additionalanalysis and may be justified in many cases. An analogymay be used in comparing the use of Division 2 overDivision 1 in that more design and analysis of the vesselis required in Division 2, however as a result, lowerdesign margins are allowed. It should be noted that localjurisdictions may not always accept the use of 2286for all equipment though this should be evaluated ona case by case basis.

Code Case 2286 assumes that shells are axisym-metric, that for unstiffened vessels the shells are thesame thickness, and that for stiffened cylinders andcones the thickness between stiffeners is uniform.Additionally, capacity reduction factors (or knockdownfactors) are provided for general use, but are incorpo-rated in the allowable stress equations. Stress reductionfactors (FS) must be found for each direction of loadingso that the values of FS are determined independentlyfor both the longitudinal and circumferential directions.The stress reduction factors cover elastic and inelasticbuckling, and plastic collapse behavior for elements incompression.

There are several differences between Code Case 2286and the older vacuum chart method. One difference is theupper limit of temperatures that each method has. CodeCase 2286 has a limit of 800F for carbon steels found inUCS-23.1 and 800F for stainless steels in UHA-23,whereas the vacuum charts have higher limits for themore common carbon steels and stainless steels (exceptin a few cases). See Section II, Part D, MandatoryAppendix 3 for criteria and figures with temperaturelimits for various classes of materials under externalpressure and axial compression. It should be noted thatneither method has accounted for the effects of creep onbuckling. A complete list of what 2286 covers that the

vacuum chart method in Division 1 does not may befound in the 2286 document. A partial list of differencesis as follows:

Code Case

2286

Vacuum Chart

Method

Carbon steel (UCS-23.1)

temperature limits

800F 900F for most carbon

steels

Stainless steel (UHA-23)

temperature limits

800F 800F-1500F for most

stainless steels

Do / t upper limit 2,000 1,000

Slenderness ratio (KL/r) 200 None

Combined loads Yes No

Stiffener geometry

requirements

Yes No

In discussing the differences between the two proce-dures, most vessels in the refining industry have Do / tratios far less than 1,000, so the lower limits of thevacuum charts are usually not a concern. The limitationson slenderness ratios of 200 are from AISC specifica-tions and were based off of engineering judgment,economics in construction, and handling concerns. Theevaluation of combined load cases (such as externalpressure with uniform axial compression and bending) isalso something that is not considered in the vacuumchart method. The use of Code Case 2286 requires thecalculation of individual load cases first, and thendetermining if the combined load cases fulfill theacceptance criteria. This can become tedious sincecalculations are typically done for each weld plane.

Additionally, tolerances for cylindrical and conicalshells are provided when subjected to external pressure,and uniform axial compression and bending. If the toler-ances are exceeded, the allowable buckling stresses mustbe adjusted. In the case of vessels with large diameter overthickness ratios (approximately 300 or higher), it may beprudent to discuss shell tolerances with the fabricator toensure that if tolerances cannot be met, the reducedallowable buckling stresses can be used for checking thedesign of the vessel.

The procedure for designing a simple cylindrical vesselto Code Case 2286 or Section VIII, Division 2 is to firstestablish geometry as well as determine shell thicknessvalues to begin with. Typically, some thickness is estab-lished from, say, internal pressure and those thicknessvalues are used for starting the procedure in Code Case2286. External loads from wind and seismic designcriteria must be established to complete the procedure.


The allowable primary membrane compressive stress shallbe less than or equal to the maximum allowable tensilestress from Section II, Part D. The procedure is iterative.First, the allowable stresses are determined for eachloading condition without being combined, and then eachcombined case is evaluated.

Step 1: Determine the allowable external pressure.First, calculate Mx, Ch, and Fhe using the equations

supplied in the Code.Next, calculate Fic, which is the buckling stress that would

be found if FS ¼ 1 in the allowable stress equations.For external pressure, the equations are associated bythe Fhe/Fy ratios.

Then, calculate FS, where the equations are associatedwith the value of Fic as compared to Fy.

Calculate Fha with the calculated Fic and the calculated FSvalue. This is the allowable stress value to use in thethickness calculations, assuming no other loadings areoccurring.

Finally, determine the allowable pressure.If the allowable pressure is not sufficient for design

conditions, increase the thickness or add ringstiffeners.

Step 2: Determine the allowable longitudinal stress due toaxial compression.

First, calculate Mx, c(bar), Cx, and Fxe using the equationssupplied in the Code.

Next, calculate Fic, which is the buckling stress that wouldbe found if FS ¼ 1 in the allowable stress equations.For uniform axial compression, the equations areassociated by the Do/t ratios.


Calculate Fxa with the calculated Fic and the calculated FSvalue.

Calculate lc to see if the vessel is subject to columnbuckling. If it is not then the allowable longitudinalstress due to axial compression is Fxa.

If it is, then Fca must be calculated, which will bea reduced value of Fxa.

Finally, determine if the actual stress is less than or equal tothe allowable stress due to uniform axial compression.

Step 3: Determine the allowable longitudinal stress due tobending moment.

First, using the Do/t ratio and the g ¼ FyDo/ET ratio,determine the appropriate of the four equations to usefor Fba.

Next, calculate Fic, which is the buckling stress thatwould be found if FS ¼ 1 in the allowable stressequations.


Calculate Fba with the calculated Fic and the calculated FSvalue.

Finally, determine if the actual stress is less than or equalto the allowable stress due the bending moment.

Step 4: Determine the allowable in-plane shear stress dueto a shear force:

First, calculate av, Mx, Cv, Fve, and hv using the equationssupplied in the Code.

Next, calculate Fic, which is the buckling stress that wouldbe found if FS ¼ 1 in the allowable stress equations.


Calculate Fva with the calculated Fic and the calculated FSvalue.

Finally, determine if the actual stress is less than or equalto the allowable stress due the shear force. Note that theshear stress may be calculated at various angles aroundthe circumference.

Step 5: Determine the interaction of the stresses undercombined loads.

Step 6: Determine the size of the ring stiffener if stiffenersare needed. Ring stiffeners may be sized as either smallrings or large rings, as either ring is considered a line ofsupport.

There are several examples using the incorporatedCode Case 2286 in Division 2 that may be used asreference in the document ASME PTB-3 Section VIIIDivision 2 Example Problem Manual.

General Design 57

Procedure 2-5: Design of Cones

Notation

A ¼ factor A from ASMEAr ¼ Excess area available in junction without

a ring, in2

AS ¼ Cross sectional area of ring, in2

AeL ¼ Effective area of reinforcement, large end,in2

AeS ¼ Effective area of reinforcement, small end,in2

ArL ¼ Area of reinforcement required at largeend, in2

ArS ¼ Area of reinforcement required at smallend, in2

ATL ¼ Equivalent area of shell, cone, and ring atlarge end, in2

ATS ¼ Equivalent area of shell, cone, and ring atsmall end, in2

B ¼ Factor B from ASME, PSICa ¼ Corrosion Allowance, inDe ¼ OD of equivalent cylinder,in

E1, E2 ¼ Joint efficiency in shell or coneEs, Ec, Er ¼ Modulus of elasticity of shell, cone or ring,

psif1-4 ¼ Axial load per unit circumference excluding

pressure, Lbs/in-circFL ¼ Longitudinal load at large end if this is

a line of support, Lbs/inFS ¼ Longitudinal load at small end if this is

a line of support, Lbs/inI ¼ Available moment of inertia, ring only, in4

IS ¼ Required moment of inertia, in4

I’ ¼ Available moment of inertia of combinedshell-ring section, in4

IS’ ¼ Required moment of inertia of combinedshell-ring section, in4

K ¼ Ratio defined hereinL ¼ Length of cone along axis of cone, inLa ¼ Maximum distance along shell from

junction that can be included as part ofreinforcement, in

Lb ¼ Maximum distance along shell fromjunction to centroid of ring, in

LC ¼ Length of cone along shell, inLce ¼ Length of equivalent cylinder for external

pressure, in

LL ¼ Distance along shell at large end, fromjunction to first stiffening element, in

LS ¼ Distance along shell at small end, fromjunction to first stiffening element, in

M ¼ Factor at large end if this is a LOSM1, M2 ¼ Moment at Elev 1 or 2, in-Lbs

N ¼ Factor at small end if this is a LOSP ¼ Internal pressure, psigPx ¼ External pressure, psig

QL1-4 ¼ Axial load per unit circumference includingpressure, Lbs/in-circ

tS1, tS2 ¼ Thickness required, shell, intrc ¼ Thickness required, cone, inte ¼ Thickness, equivalent cylinder for large

end, int1-2 ¼ Thickness of shell, small/large end, intc ¼ thickness of cone, intrx ¼ Thickness required for cone external

pressure, inSs, Sc, SR ¼ Allowable stress of shell, cone and ring

respectively, psiW1, W2 ¼ Weights at Elev 1 or 2, Lbs

X, Y ¼ Factors for cones where a > 30�X1, Y1 ¼ ASME ratios defined herein

D ¼ From Tables 2-2, 2-3 and 2-4, degreessX ¼ Membrane longitudinal stress plus discon-

tinuity longitudinal stress due to bending,PSI

sf ¼ Membrane hoop stress plus averagediscontinuity hoop stress, PSI

LOS ¼ Line of Support

tct2t1

R

D

r

L

d

aα

Figure 2-6. Eccentric cone.


Geometry

FOR ECCENTRIC CONE.

a ¼ D� d

tan a ¼ a=L

L ¼ a tan a

FOR CONCENTRIC CONE.

a ¼ :5 ðD� dÞtan a ¼ a=L

L ¼ a tan a

LC ¼hL2 þ ðR� rÞ2

i1=2

1.0. Design Cone for Internal Pressure

1.1. Thickness Required• Required thickness of cone, Internal Pressure,a � 30�;Small End;

tC1 ¼ ðP rÞ=ðcos a ðSC E2 � :6PÞÞLarge End;

tC2 ¼ ðP RÞ=ðcos aðSC E2 � :6PÞÞ• Required thickness of Shell;

Small End;

tS1 ¼ ðP rÞ=ðSS E1 � :6PÞLarge End;

tS2 ¼ ðP RÞ=ðSSE1 � :6PÞ

1.2. Reinforcement Required• Values of X1 and Y1X1 ¼ Smaller of SS E1 or SC E2Y1 ¼ Greater of SS ES or SC EC

Large End

• P/X1 ¼D ¼ From Table 2-2If D < a then reinforcement is requiredIf D � a then no reinforcement is required

If reinforcement is required follow the followingsteps:

• Calculate ratio, K

K ¼ Y1=ðSR ERÞ > 1

K ¼ 1 if additional area of reinforcement is not required

• Area of reinforcement required at large end of cone,ArL

tc

t2

t1

R

Dr

L

d

aα

Figure 2-7. Concentric cone.

tc2

tc1

t1

ts

Ls

Lc L

t2

LL For intermediatestiffeners on cone

Minimum = √Rs(t1 – ts)

Figure 2-8. Dimensions of cone-cylinder intersections.

Table 2-2Values of D, degrees - large end,a £ 30 degrees, internal pressure

P/X1 D

.001 11

.002 15

.003 18

.004 21

.005 23

.006 25

.007 27

.008 28.5

.009 30

General Design 59

Case 1: External Loads are Not Included

ArL ¼ ��P R2 K

�2 X1

�1� �D=a tan a

Case 2: External Loads are Included

ArL ¼ ½ðK QL RÞ=X1� ½1� ðD=aÞ� tan a

From the worksheet, select the largest value of QL forthe large end in tension.

Reinforcement Available

• Effective area available at large end, Ael

Ael ¼�t2 � tS2

ðR t2Þ1=2þ

�tC � tr

� ðR tC=cos aÞ1=2• Area required at large end, Ar

Ar ¼ ArL � AeL

If Ar is negative, the design is adequate as is. If Ar is posi-tive, then either a ring must be added or the area ofsection increased. Ar is the net area required for the ringonly!

• If a ring is required, the maximum distance tocentroid of ring, Lb

Lb ¼ :25 ðR t2Þ1=2• If the shell is made thicker to provide more area forreinforcement, then the maximum length of shellincluded in the reinforcement shall be;

La ¼ ðR t2Þ1=2

Small End

• P/X1 ¼D ¼ From Table 2-3If D < a then reinforcement is requiredIf D � a then no reinforcement is required

If reinforcement is required follow the followingsteps:

Case 1: External Loads are Not Included:

ArS ¼ ��P r2 K

�2 X1

�1� �D=a tan a


ArS ¼ ½ðK QL rÞ=X1� ½1� ðD=aÞ� tan a

From the worksheet, select the largest value of QL forthe small end in tension.

Reinforcement Available

• Effective area available at large end, AeS

AeS ¼ :78�t1 � tS1

ðr t1Þ1=2þððtC � trÞ=cos aÞ1=2

• Area required at large end, Ar

Ar ¼ ArS � AeS



Lb ¼ :25 ðr t1Þ1=2

• If the shell is made thicker to provide more area forreinforcement, then the maximum length of shellincluded in the reinforcement shall be;

La ¼ ðr t1Þ1=2

Table 2-3Values of D, degrees - small end, a £ 30 degrees,

internal pressure

P/X1 D

.002 4

.005 6

.010 9

.02 12.5

.04 17.5

.08 24

.1 27

.125 30


2.0. Design Cone for External Pressure

2.1. Thickness Required

For a � 22.5�

• Design the cone as a cylinder where Do ¼ DL and thelength is equal to L.

• trx ¼ ________

For a > 22.5�

• Determine half apex angle, a

a ¼ arctan ½:5ðDiL � DiSÞ=L�• Length of equivalent cylinder, Lce

Lce ¼ L=cos a

• Diameter of equivalent cylinder, De

De ¼ :5 ½ðDL þ DSÞ=cos a�• Thickness of equivalent cylinder, tete ¼ tC cos a

• Calculate ratios;

Lce=De ¼De=te ¼

• Design the cone as a cylinder with the propertiescalculated above

• trx ¼ ________

2.2. Reinforcement Required

LARGE END

• Values of X1 and Y1

X1 ¼ Smaller of SS E1 or SC E2

Y1 ¼ Greater of SS ES or SC EC

• PX/X1 ¼D ¼ From Table 2-4If D < a then reinforcement is requiredIf D � a then no reinforcement is required

If reinforcement is required follow the following steps:

• Calculate ratio, K

K ¼ Y1=ðSR ERÞ > 1

K ¼ 1 if no additional area is required


ArL ¼ ��PX R2 K

�2 X1

�1� :25

�PX R

�D=a


ArL ¼ ½ðK QL R tan aÞ=X1� ½1� :25

ððPX R� QLÞ=QLÞ ðD=aÞ�

From the worksheet, select the largest value of QL forthe large end in compression.

Reinforcement Available• Effective area available at large end, AeL

AeL ¼ :55 ðDL t2Þ1=2�t2 þ tC=cos a

DS

Lctc

DIL

DL

DIS

α

Figure 2-9. Dimensions of concentric cone for externalpressure.

Table 2-4Values of D, degrees - large end, a £ 60

degrees, external pressure

PX/X1 D

0 0

.002 5

.005 7

.010 10

.02 15

.04 21

.08 29

.1 33

.125 37

.15 40

.2 47

.25 52

.3 57

.35 60

Note: If PX / X1 > .35 Use D ¼ 60

General Design 61


Ar ¼ ArL � AeL



Lb ¼ :25 ðR t2Þ1=2

• If the shell is made thicker to provide more area forreinforcement, then the maximum length of shellincluded in the reinforcement shall be;

La ¼ ðR t2Þ1=2

If the Large End is a Line of Support (LOS)1. Assume size of ring;

Size ¼ _____A ¼ ______I ¼ _______

2. Determine length of cylinder acting with cone, LLLL ¼ ______

3. Determine length of cone, LC


i1=24. Calculate equivalent area of cylinder, cone and

ring, ATL

ATL ¼ :5 ðLL t2Þ þ :5 ðLC tCÞ þ AS

5. Calculate Value M

M ¼ :5�� R tan a

þ :5 LL þ ��R2 � r2�

� �3 R tan a

6. Calculate Value FL

FL ¼ PX Mþ fn Tan a

Choose fn as worst value, compression7. Calculate Factor B

B ¼ :75 ½ðFL DLÞ=ATL�8. Find Factor A from applicable material curve or

calculate as follows;

A ¼ 2 B =En

Where En lesser of ES, EC or Er

9. Required moment of inertia, IS or IS’

IS ¼ ��A D2

L ATL�

14

IS’ ¼ ��A D2

L ATL�

10:9

10. Compare required values of I with actual;

I > ISI’ > IS’

SMALL END


Ars ¼ �PX r2 K Tan a

�2 X1


Ars ¼ ðK QL r Tan aÞ=X1

From the worksheet, select the largest value of QL forthe small end in compression.

Reinforcement Available• Effective area available at small end, AeS

AeS ¼ :55�t1 � tS

ðDS t1Þ1=2þ�tC � tr=cos a


Ar ¼ ArS � AeS



Lb ¼ :25 ðr t1Þ1=2• If the shell is made thicker to provide more area forreinforcement, then the maximum length of shellincluded in the reinforcement shall be;

La ¼ ðr t1Þ1=2

If the Small End is a Line of Support (LOS)1. Assume size of ring;

Size ¼ _____A ¼ ______I ¼ _______

2. Determine length of cylinder acting with cone, LSLS ¼ ______


3. Determine length of cone, LC


i1=2

4. Calculate equivalent area of cylinder, cone andring, ATS

ATS ¼ :5 ðLS t1Þ þ :5 ðLC tCÞ þ AS

5. Calculate Value N

N ¼ :5�r tan a

þ :5 LS þ��R2 � r2

�� 6 r tan a

6. Calculate Value FS

FS ¼ PX Nþ fn tan a

Choose fn as worst value, compression7. Calculate Factor B

B ¼ :75 ½ðFS DSÞ=ATS�8. Find Factor A from applicable material curve or

calculate as follows;

A ¼ 2 B=En

Where En lesser of ES, EC or Er9. Required moment of inertia, IS or IS’

IS ¼ ��A DS

2 ATS�

14

IS’ ¼ ��A DS

2 ATS�

10:9

10. Compare required values of I with actual;

I > IS

I’ > I’S

3.0. Design of Cones with Half Apex Angle, a, Between 30�and 60�, Internal Pressure Only!

Based on CC2150

LARGE END;

• Required thickness of shell;

tS2 ¼ ðP RÞ=ðSS E1 � :6PÞ• Required thickness of cone

tC2 ¼ ðP RÞ=ðcos a ðSC E2 � :6PÞÞCHECK STRESSES

• Determine ratio;

R=t2 ¼

• Determine values of factors, X and Y, from theappropriate figure. Use Fig 2-10 if shell and cone arethe same thickness. Use Fig 2-11 if the shell and coneare not the same thickness.X ¼Y ¼

• Calculate longitudinal stress, sX

sX ¼�P R=t2

h:5þ X ðR=t2Þ1=2

i

• Calculate circumferential stress, sf

sf ¼�P R=t2

h1� Y ðR=t2Þ1=2

i

• Allowable stresses;

sX < 3 SSsf < 1:5 SS

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

030 35 40

Half Apex Angle in Degrees

X or

Y

45 50 55 60

R/t=500 100

500

50

100

50

Y

X

Figure 2-10. X and Y for cone thickness¼ t (Use whenCone and Shell are the same thickness)

1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

030 35 40

Half Apex Angle in Degrees

X or

Y

45 50 55 60

R/t = 500 100

500

50

50Y

X

Figure 2-11. X and Y for cone thickness¼ t/cos a (Usewhen Cone and Shell are the not same thickness)

General Design 63

Sample Problem

a ¼ 45�P ¼ 500 psigD ¼ 120 inR ¼ 60.125 in CorrDT ¼ 600�FMatl: SA-516-70SS ¼ 19,400 PSIE ¼ 1Ca ¼ 0.125 inCode: ASME VIII-1

• Required thickness of Shell;

tS2 ¼ ðP RÞ=ðSS E1 � :6PÞ¼ ð500 ð60:125ÞÞ=ð19400� 300Þ¼ 1:57þ :125 ¼ 1:699

Use t2 ¼ 1.75 in (1.625 in Corr)

• Required thickness of cone

tC2 ¼ ðP RÞ=½cos a ðSC E� :6PÞ�¼ ð500 ð60:125ÞÞ=½cos 45 ð19400� 300Þ�¼ 2:22þ :125 ¼ 2:35

Use tC ¼ 2.375 in (2.25 in Corr)

CHECK STRESSES

• Determine ratio;

R=t2 ¼ 60:125=1:625 ¼ 37

• Determine values of factors, X and Y, from Fig 2-11X ¼ .53Y ¼ .24

• Calculate longitudinal stress, sX

sX ¼�P R=t2

h:5þ X ðR=t2Þ1=2

¼h500

�60:125

.1:625

ih:5þ :53 ð60:125=1:625Þ1=2

i¼ 68; 691 PSI

Allowable stress ¼ 3 SS ¼ 3 (19,400) ¼ 58,200 PSITherefore design is not acceptable!Increase shell thickness by 0.125 in and recalculate

stresses

• Calculate circumferential stress, sf

sf ¼�P R=t2

h1� Y ðR=t2Þ1=2

i

¼h500

�60:125

.1:625

ih1� :24 ð60:125=1:625Þ1=2

i

¼ ð�Þ 8; 507 PSI ðAcceptableÞ

Notes

1. For small vessels or horizontal vessels with coneswhere there are not significant external loads, thenthe external loads may be neglected.

2. Line of Support: This indicates a formal work line oran elevation on the vessel used for external pressurecalculations. It does not mean that the vessel issupported from that point.

3. The worksheet should be used to determine theloadings at the plane being investigated.

4. External Load Conditions;

INTERNAL PRESSURE:a: Large End: .5PR þ fn is in tension, use this valuein the procedure. If this value is (-), in compres-sion, then a stress analysis is required.

b: Small End: .5PR þ fn is in tension, use this valuein the procedure. If this value is (-), in compres-sion, then a stress analysis is required.

EXTERNAL PRESSURE:c: Large End: .5Px Rþ fn is in compression, use thisvalue in the procedure. If this value is (þ), intension, then a stress analysis is required.

d: Large End: .5Px Rþ fn is in compression, use thisvalue in the procedure. If this value is (þ), intension, then a stress analysis is required.


WM

ELEV 1

ELEV 2

R1

R2DL

f1 = +4M1–W1

f1 f2

f4f3

ds

W

M

ELEV 1

ELEV 2

R1

R2

DLf1 f2

f4f3ds

πds πds2 f1 = +

4M1–W1πDL πDL

2

f2 = –4M1–W1

πDL πDL2

f3 = +4M2–W2

πds πds2

f4 = –4M2–W2

πds πds2

f2 = –4M1–W1

πds πds2

f3 = +4M2–W2

πDL πDL2

f4 = –4M2–W2

πDL πDL2

General Design 65

f1 = +4M1–W1

πds πds2

f2 = –4M1–W1

πds πds2

f3 = +4M2–W2

πDL πDL2 f3 = +

4M2–W2πds πds

2

f1 = +4M1–W1

πDL πDL2

f2 = –4M1–W1

πDL πDL2

f4 = –4M2–W2

πDL πDL2 f4 = –

4M2–W2πds πds

2

WM

ELEV 1

ELEV 2

R1

R2DL

f1 f2

f4f3

ds

W

M

ELEV 1

ELEV 2

R1

R2

DLf1 f2

f4f3ds


Procedure 2-6: Design of Toriconical Transitions [1,3]

Notation

P ¼ internal pressure, psiS ¼ allowable stress, psiE ¼ joint efficiency

P1, P2 ¼ equivalent internal pressure, psif1, f2 ¼ longitudinal unit loads, lb/in.s1, s2 ¼ circumferential membrane stress, psi

a ¼ half apex angle, degm ¼ code correction factor for thickness of large

knucklePx ¼ external pressure, psi

M1, M2 ¼ longitudinal bending moment at elevation,in.-lb

W1, W2 ¼ dead weight at elevation, lb

Calculating Angle, a

Case 1

o > o0

Step 1:

sin f ¼ Rþ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2

p ¼f ¼ _____________

Step 2:

tan Q ¼ AB

Q ¼ _____________

Step 3:

a ¼ fþQ

a ¼ _____________

L ¼ cos fffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2

p

Figure 2-12. Dimensional data for a conical transition.

General Design 67

Case 2

o0 > oStep 1:

cos f ¼ Rþ rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2

p ¼

f ¼ _____________

Step 2:

tan Q ¼ AB

Q ¼ _____________

Step 3:

a ¼ 90�Q� f

a ¼_____________

L ¼ cos fffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiA2 þ B2

p

Dimensional Formulas

D1 ¼ D� 2ðR� R cos aÞD2 ¼ Dþ 2ðR� R cos aÞ

D0 ¼ D� 2R

�1� 1

cos a

�� 2‘ tan a

L1 ¼ D1

2 cos a

L2 ¼ D2

2 cos a

m ¼ 0:25

3þ

ffiffiffiffiffiL1

R

r !

Large End (Figure 2-13)

• Maximum longitudinal loads, f1(þ) tension; (–) compression

f1 ¼ �W1

pD1� 4M1

pD21

• Determine equivalent pressure, P1.

P1 ¼ Pþ 4f1D1

• Circumferential stress, D1.Compression:

s1 ¼ PL1

t� P1L1

t

�L1

2R

�

Figure 2-13. Dimensional data for the large end ofa conical transition.


• Circumferential stress at D1 without loads, s1.Compression:

s1 ¼ PL1

t

�1� L1

2R

�

• Thickness required knuckle, trk [1, section 1-4(d)].With loads:

trk ¼ P1L1m2SE� 0:2P1

Without loads:

trk ¼ PL1m2SE� 0:2P

• Thickness required cone, trc [1, section UG-32(g)].With loads:

trc ¼ P1D1

2 cos aðSE� 0:6P1ÞWithout loads:

trc ¼ PD1

2 cos aðSE� 0:6PÞ

Small End (Figure 2-14)

• Maximum longitudinal loads, f2.

(þ) tension; (–) compression

f2 ¼ �W2

pD2� 4M2

pD22

• Determine equivalent pressure, P2.

P2 ¼ Pþ 4f2D2

• Circumferential stress at D2.

Compression:

s2 ¼ PL2

tþ P2L2

t

�L2

2r

�

• Circumferential stress at D2 without loads, s2.

Compression:

s2 ¼ PL2

t

�1� L2

2r

�

• Thickness required cone, at D2, trc [1, section UG-32(g)].With loads:

trc ¼ P2D2

2 cos aðSE� 0:6P2ÞWithout loads:

trc ¼ PD2

2 cos aðSE� 0:6PÞ• Thickness required knuckle. There is no requirementfor thickness of the reverse knuckle at the small endof the cone. For convenience of fabrication it shouldbe made the same thickness as the cone.

Additional Formulas (Figure 2-15)

• Thickness required of cone at any diameter D’, tD’

tD’ ¼ PD’2 cos aðSE� 0:6PÞ

• Thickness required for external pressure [1, sectionUG-33(f)].

te ¼ t cos a

DL ¼ D2 þ 2te

Figure 2-14. Dimensional data for the small end ofa conical transition.

Figure 2-15. Dimensional data for cones due to externalpressure.

General Design 69

Ds ¼ D1 þ 2teL ¼ X� sin aðRþ tÞ � sin aðr � tÞLe ¼ L

2

�1þ DS

DL

�

Le

DL¼

DL

te¼

Using these values, use Figure 2-1e to determineFactor A.

• Allowable external pressure, Pa.

Pa ¼ 2AEteDL

; Pa > Px

where E ¼ modulus of elasticity at designtemperature.

Notes

1. Allowable stresses. The maximum stress is thecompressive stress at the tangency of the largeknuckle and the cone. Failure would occur in localyielding rather than buckling; therefore the allowablestress should be the same as required for cylinders.

Thus the allowable circumferential compressivestress should be the lesser of 2SE or Fy. Using a lowerallowable stress would require the knuckle radius tobe made very largedwell above code requirements.See Reference 3.

2. Toriconical sections are mandatory if anglea exceeds 30� unless the design complies with Para.1-5(e) of the ASME Code [1]. This paragraphrequires a discontinuity analysis of the cone-shelljuncture.

3. No reinforcing rings or added reinforcement isrequired at the intersections of cones and cylinders,providing a knuckle radius meeting ASME Coderequirements is used. The minimum knuckle radiusfor the large end is not less than the greater of 3tor 0.12(R þ t). The knuckle radius of the smallend (flare) has no minimum. (See [Reference 1,Figure UG-36]).

4. Toriconical transitions are advisable to avoid thehigh discontinuity stresses at the junctures for thefollowing conditions:a. High pressuredgreater than 300 psig.b. High temperaturedgreater than 450 or 500�F.c. Low temperaturedless than –20�F.d. Cyclic service (fatigue).

Procedure 2-7: Stresses in Heads Due to Internal Pressure [2,3]

Notation

L ¼ crown radius, in.r ¼ knuckle radius, in.h ¼ depth of head, in.

RL ¼ latitudinal radius of curvature, in.Rm ¼ meridional radius of curvature, in.sf ¼ latitudinal stress, psisx ¼ meridional stress, psiP ¼ internal pressure, psi

Formulas

Lengths of RL and Rm for ellipsoidal heads:

• At equator:

Rm ¼ h2

R

RL ¼ R

• At center of head:

Rm ¼ RL ¼ R2

h

Figure 2-16. Direction of stresses in a vessel head.


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