122
Cylindrical and Spherical Parts Subjected to Internal and External Pressure
Chapter 7
The calculated maximum allowable external pressure of 217 psi is much greater than the required 30 psi. Therefore assume a thinner section and redo calculations. Assume t = 0.25 inch Ro
=
A =
120 in. + 0.25 in. = 60.25 inch 2
(0.125) (0.25 in.) = 0.00052 60.25 in.
From Figure CS−2, B = 7,500 psi. Therefore: Pa
=
7,500 psi (0.25 in.) 60.25 in.
= 31 psi
The calculated maximum allowable external pressure of 31 psi is just slightly greater than the required value of 30 psi. Therefore, the minimum required thickness is 0.25 inch.
UG 29 Stiffening Rings for Cylindrical Shells Under External Pressure The Code provides two methods for sizing stiffening rings. Method (a) is based on the stiffening ring providing all additional stiffening, and method (b) is based on a combination of the stiffening ring and a part of the shell providing the additional stiffening. Both methods require that one assume an initial size and shape for the ring. (a) Stiffening ring alone: 1) Determine B, where:
B = 0.75 (
PDo t + A S /L S
)
A S = cross-sectional area of stiffening ring, square inches. LS = distance between support lines on both sides of the stiffening ring, inches. Do = outside diameter of shell, inches. 2) Enter appropriate external pressure chart in Section II, Part D and determine Factor A that corresponds to the calculated Factor B. 3) Determine the required moment of inertia of the stiffening ring only, IS:
A D2o L S t + S IS
=
LS
A
14
CASTI Guidebook to ASME Section VIII Div. 1 – Pressure Vessels – Third Edition
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Chapter 7
123
4) Determine actual moment of inertia of ring only, I, in.4 5) I must be equal to or greater than IS (b) The stiffening ring/shell combination: 1) Determine B, where:
B = 0.75 (
PDo t + A S /LS
)
A S = cross-sectional area of stiffening ring, square inches. LS = distance between support lines on both sides of the stiffening ring, inches. Do = outside diameter of shell, inches. 2) Enter appropriate external pressure chart in Section II, Part D and determine Factor A that corresponds to the calculated Factor B. 3) Determine the required moment of inertia of the ring/shell combination, I S', in.4
A D2o L S t + S IS
=
LS
14 A D2o L S t + S
I ' S
=
A
LS
A
10.9
4) Determine the moment of inertia of the combined ring and shell section acting together, I', inch4. The length of shell used in the calculation shall not be greater than 1.10 (D ot)¹⁄₂. No overlap of contribution is allowed. 5) I' shall be equal to or greater than I S'. Note that the formulas for B are identical. If the value of B falls below the left end of the material/temperature curve, then A is calculated using A = 2B/E where E is the modulus of elasticity. If different materials are used for the shell and stiffening ring, then use the external pressure chart that gives the lowest value of A.
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Cylindrical and Spherical Parts Subjected to Internal and External Pressure
Chapter 7
General Requirements Stiffening rings must extend completely around the circumference of the cylinder. All joints and connections between rings must maintain the required moment of inertia of the combined ring-shell section. Inside and outside rings are allowed and stiffening rings may be continued on the opposite surface. See Figure 7.10.
Gap (not to exceed 8 times the thickness of the shell plate) This section shall have moment of inertia required for ring unless requirements of butt weld UG-29 (c) are met. gap
see UG-29 (c) shell
A
E
web of stiffener flange of stiffener
butt weld
gap in ring for drainage J
This section shall have moment of inertia required for ring.
F
strut member
D
K
section J-K
Length of any gap in unsupported shell not to exceed length of arc shown in Fig. UG-29.2
B butt weld in ring unstiffened cylinder Type of construction when gap is greater than length of arc shown in Fig. UG-29.2
at least 120 deg.
C
support
This section shall have moment of inertia required for ring.
K
Figure 7.10 Various arrangements of stiffening rings for cylindrical vessels subjected to external pressure. Fig. UG 29.1.
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125
Limited gaps in stiffening rings adjacent to the shell are allowed. The allowed gap length depends on L/Do and Do /t and is given by Figure 7.11. Gaps greater than the allowed length must be reinforced on the opposite surface or comply with additional requirements given in UG−29. The additional requirements are: 1. 2. 3.
limit the unsupported shell arc to 90 degrees, stagger unsupported shell arcs in adjacent stiffening rings, and increase the unsupported shell length L to the distance between alternate rings or at heads to the second ring from the head.
Baffles, trays, tray supports, or other internal structures perpendicular to the axis of the longitudinal axis and welded to the cylinder may function as stiffeners provided they satisfy all the requirements in paragraphs UG−29 and UG−30.
2000 o D 0 3 o 0 D 0. o 5 = 3 D 0 c . 0 r o 0 A D 0 4 5 = 0 . 4 r c 0 = . A 0 r c = A
1000 800
,
600 500 400
s
300
s
k
n
e
s
D
t/
o D 5 5 o 0 . D o 0 5 D = 6 o 0 5 D . r c 7 5 0 A 0 . = 8 o 0 0 D . r c = 0 0 A c 1 r = . o A 0 c r D = A 5 c 2 r 1 . o A 0 D = 0 o c 5 r D 1 . A o 5 0 D 7 = 1 0 o . 0 c 0 D r 2 . A 0 = 0 5 = r c 2 . o c A 0 r D A = 0 0 r c 3 . A 0 = r c A
r c A
200
O
u
st
di
e
d
ai
m
e
te
r
ht
ic
100 80 60 50 40 30
o D 0 9 3 . 0 = r c A
20 10 0.01
0.02
0.04 0.06
0.10
0.2
Design length
0.4 0.6
1.0
2
3 4 5 6
8 10
20
outside diameter, L/D
Figure 7.11 Maximum arc of shell left unsupported because of gap in stiffening ring of cylindrical shell under external pressure. Fig. UG 29.2.
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Cylindrical and Spherical Parts Subjected to Internal and External Pressure
Chapter 7
UG 30 Attachment of Stiffening Rings To insure that stiffening rings are effective, UG −30 provides minimum welding requirements. Stiffeners may be welded or brazed to the shell. The rings may be attached with continuous, intermittent, or a combination of continuous and intermittent welds or brazes. See Figure 7.12 for some acceptable details.
S
2 in. min.
2 in. min.
24t max.
2 in. min. S
tw
tw w Staggered intermittent weld
In-line intermittent weld
Continuous fillet weld one side, intermittent other side
S < 8t external stiffeners S < 12t internal stiffeners
stiffener
tw
tw
t
shell
w (b)
(a)
(c)
tw continuous full penetration weld
t
w (d)
(e)
Figure 7.12 Some acceptable methods of attaching stiffening rings. Fig. UG 30. CASTI Guidebook to ASME Section VIII Div. 1 – Pressure Vessels – Third Edition
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Cylindrical and Spherical Parts Subjected to Internal and External Pressure
127
Example 7.6 Minimum Thickness of a Thick Cylinder Under External Pressure A forged 7.5 inch inside diameter PI vessel is exposed to an external design pressure of 1,700 psi at 700°F. The material is SA-182 Type F321. The unsupported length is 18 feet. Find the minimum required thickness for the vessel.
Solution: Table 1A of Section II, Part D gives an allowable tensile stress of 13,000 psi for SA-182 Type F321 material. The applicable external pressure chart in Section II, part D is HA-2. The solution procedure is shown in Figure 7.9. A) Calculate Do /t and L/Do ratios. Assume t = 1.0 inch Do = 7.5 in. + 2 (1.0 in.) Do = 9.5 inch Do /t = 9.5 in./1 in. Do /t = 9.5 L/Do = (18 ft) (12 in./ft)/9.5 in. L/Do = 22.7 Do /t is less than 10 but greater than 4. Therefore, use the procedure outlined in Figure 7.9. From Figure G of Section II, Part D, with Do /t = 9.5 and L/Do = 22.7, A = 0.014. Chart HA-2 gives a factor B value of 11,800 psi for A = 0.014 and 700°F. B) Calculate Pa1 using the formula in Step 6 of Figure 7.9.
Pa1
= 11,800 psi
Pa1
= 1,708 psi
2.167 − 0.0833 9.5
C) Calculate Pa2 using the formula in Step 10 of Figure 7.9. The allowable tensile stress for the material is 13,000 psi. Therefore: S1 = 2 (13,000 psi) = 26,000 psi.
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Cylindrical and Spherical Parts Subjected to Internal and External Pressure
Chapter 7
B* is one-half the yield strength at temperature and is equal to the maximum value of B for the material at 700°F. B* = 13,100 psi S2 = 1.8 (13,100) psi S2 = 23,580 psi S is the smaller of S1 and S2 and equals 23,580 psi, therefore,
Pa2 Pa2
=
=
2 (23,580 psi) 1 1− 9.5 9.5 4,442 psi
D) The maximum allowable external pressure is the smaller of Pa1 and Pa2. Therefore, Pa for the 1 inch thick section is 1,708 psi, which is greater than the required 1,700 psi.
Example 7.7 Design of Shell and Stiffening Rings The lower cylindrical part of the vessel shown in Figure 7.3 has a 96 inch inside diameter and Type 1 butt welds. The corrosion allowance is 0.125 inch. The design temperature is 800°F and the internal design pressure is 200 psi. The external design pressure is 15 psi. The material is SA −516 Grade 70 and full radiography will be performed. The minimum unsupported length of the section L is 360 inches. Find the minimum required thickness and the maximum allowable external pressure of the section.
Solution: The allowable stress in Part D of Section II, Table 1A, for SA −516 Grade 70 at 800°F is 12,000 psi. Table UW−12 gives a joint efficiency of 1.0 for f ully radiographed Type 1 joints. A) Determine the thickness t required for internal pressure.
t=
PR SE - 0.6P
t=
200 psi (96 in./2) 12,000 psi (1) − 0.6 (200 psi)
t = 0.81 inch t < R/2 and P < 0.385SE. Therefore, 0.81 inch is the minimum required thickness for internal pressure.
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treq = t + 0.125 in. treq = 0.81 in. + 0.125 in. treq = 0.94 inch. Use 1 inch plate. B) Determine the maximum allowable external pressure for the section with t = 0.81 inch and ts = 1.0 inch. Calculate ratios L/Do and Do /t where Do is the outside diameter and t is 0.81 inch. Do = 96 in. + 2 x ts Do = 96 in. + 2 (1 in.) = 98.0 inch Do /t = 98.0 in./0.81 in. Do /t = 121.0. Do /t is greater than 10. Therefore, use procedure in UG−28(c)(1) which is diagrammed in Figure 7.8 L/Do = 360 in./98 in. L/Do = 3.70 From Figure G of Section II, Part D, with L/Do and Do /t A = 0.00026 Table 1A of Section II, Part D lists External Pressure Chart No. CS-2 for SA −516 Grade 70 material. Chart CS-2 gives a B factor of 3,200 psi for A = 0.00026 at 800°F. Therefore using the formula in Step 6 of Figure 7.8:
Pa
=
4 (3,200) psi 3 (121.0)
Pa = 35.3 psi Pa is much greater than the required 15 psi. The value may be reduced by decreasing the thickness or increasing L. The thickness is required for internal pressure. Therefore, increase L and redo the procedure. Assume L = 612 inches
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Cylindrical and Spherical Parts Subjected to Internal and External Pressure
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L/Do = 612 in./98 in. L/Do = 6.24 A = 0.00014. A is to the left of the temperature line on Chart CS−2 E = 22,800,000 psi The allowable external pressure must be calculated using the formula in Step 7 of Figure 7.8.
Pa
=
2 (0.00014) (22,800,000 psi) 3 (121.0)
Pa = 17.6 psi Pa is greater than the required 15 psi. Therefore use the one inch shell plate and space stiffeners at 612 inches. C) Size the stiffening rings assuming rings act alone. Use the procedure from UG−29. Determine B. Assume a 1 x 8 inch bar of SA −516 Grade 70 material. A S = 8 in.2 LS = 612 inches t = 0.810 inch P = 15 psi DO = 98 inches
B = 0.75
(15) (98) (0.81 + 8/612)
B = 1,339 psi B is to the left of the curve on Figure CS-2. Therefore A must be calculated using the following equation:
A =
2B E
=
(2) (1,339 psi) 22.8 x 106 psi
= 0.0001175
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131
Determine the required moment of inertia for the stiffener.
D2 L t + IS
=
A S L
A
14
612 in. (98 in.)2 0.81 in. + IS
=
8 in.2 (0.0001175) 612 in.
14
IS = 40.6 inch4 Determine the actual moment of inertia of the 1 inch x 8 inch bar.
Figure 7.13 Stiffening ring.
I=
bd3 12
=
(1 in.)(8 in.) 3 12
=
42.7 inch 4
I > IS. Therefore a 1 inch x 8 inch bar may be used. Attach per Fig. UG- 30. D) Size stiffening ring assuming ring and shell act together. Note that B does not change. Therefore calculate the required moment of inertia for the combined section assuming a 1 inch x 8 inch bar.
D2L t + IS
=
A S L
A
10.9 CASTI Guidebook to ASME Section VIII Div. 1 – Pressure Vessels – Third Edition
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Cylindrical and Spherical Parts Subjected to Internal and External Pressure
(98 in.)2 612 in. 0.81 in. + IS
=
8 in.2 612 in.
Chapter 7
(0.0001175 )
10.9
IS = 52.15 inch4 Allowed length of shell contribution is 1.1 (Dt)¹⁄₂ or 9.8 inches. I for the 1 inch x 8 inch combination section is 120.4 in. 4, which is much larger than required Is. Therefore, assume a 1 inch x 6 inch ring. The IS for the 6 inch ring combination is 51.94 in.4. The centroid of the composite section (Figure 7.14) consisting of 6 inch ring and shell is 1.871 inches from the ID of the shell.
Figure 7.14 6 inch Stiffening ring and portion of shell. Calculate the actual moment of inertia of the combined 6 inch ring and shell combination. I = (1 in.) (6 in.)3
1 12
+ (9.80 in.) (0.81 in.) (1.466 in.)
2
+ (6 in.) (1 in.) (1.939 in.)
2
I = 57.62 inch4 I > IS therefore, use a 6 inch x 1 inch ring. Attach per Fig. UG−30. Note that a smaller ring is required in part D. If the stiffeners are designed as a combination ring and shell section, then paragraph UG −29 does not contain additional requirements other than calculating the moment of inertia of the combined section. The material savings will often outweight the additional design cost.
CASTI Guidebook to ASME Section VIII Div. 1 – Pressure Vessels – Third Edition
