For the tall vessel described in Example 16.10, what are the longitudinal stresses when the vessel is designed for earthquake zone 2 using the Uniform Building Code?

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Accepted Answer

It is necessary to determine the lateral earthquake forces and the overturning moment at the support line. Using Eq. 16.1, the lateral earthquake forces may be determined.

V = ZIKCSW (16.1)

[latex]Z=\frac{3}{8}[/latex] for zone 2

I = 1.0

K = 2.0

W = 607,000 lb

h = 1400 in.

[latex]w=\frac{W}{h}=\frac{607,000}{1400}=433.6 ext { lb /in. }[/latex]

E = [latex]26.38 imes 10^{6} \ psi[/latex]

[latex]\frac{d+t}{t}=\frac{120+0.75}{0.75}=161[/latex]

[latex]I_{x}=0.049\left(121.5^{4}-120^{4} ight)=517,600 ext { in. }^{4}[/latex]

[latex]T=0.0908 \sqrt{\frac{(433.6)(1400)^{4}}{\left(26.38 imes 10^{6} ight)(517,600)}}=1.0029[/latex]

[latex]C=\frac{1}{15 \sqrt{(1.0029)}}=0.0666[/latex]

KC = (2.0)(0.0666) = 0.1331

CS = (0.12)(1.5) = 0.18 ; maximum is CS = 0.14

KCS = (2.0)(0.14) = 0.28 ; maximum KCS = 0.3

V = (3/8)(1)(.28)(607,000) = 63,740 lb

[latex]F_{t}=0.07(1.0592)(63,740)=4,730[/latex] lb but not more than 0.25 V = 0.25 (63,740) = 15,940 lb; [latex]F_{t}[/latex]=4730 lb

[latex]F_{x}=V-F_{t}=63,740-4730=59,010[/latex] lb

[latex]M_{e}=(59,010)(1400)(2 / 3)+(4730)(1400)=61,698,000 ext { in.- lb }[/latex]

Using Eq. 16.27, determine total compressive loading as

[latex]P_{v}=\frac{W}{\pi D_{m}}+\frac{4 W e}{\pi D_{m}^{2}}+\frac{4 M}{\pi D_{m}^{2}}[/latex] (16.27)

[latex]P_{v}=\frac{(117,820)}{\pi(120.75)}+\frac{(4)(61,698,000)}{\pi(120.75)^{2}}=310.6+5387.7=5698.3[/latex]

[latex]P_{o}=15.0[/latex]

[latex]\alpha=\frac{P_{v}}{P_{o} D_{o}}=\frac{(5698.3)}{(15)(121.5)}=3.1266[/latex]

m = 0.1401

n = 3.0

[latex]P_{o}^{\prime}=\frac{9-1+0.1401+0.1401(3.1266)}{9-1+0.1401}(15)=15.8 \ psi[/latex]

Because t = 0.75 in., MAWP = 23.9 psi at 550° F, the shell is still acceptable with the earthquake loading. When [latex]\alpha[/latex] is larger than 1.0, the vessel is checked as a cantilever beam. The axial unit stress due to [latex]P_{o}[/latex] is

[latex]\frac{P_{o} D_{m}}{4}=\frac{(15)(120.75)}{4}=452.8 ext{ lb / in.}[/latex]

Axial stress is (452.8 + 5698.3 /(0.75) = 8,200 psi.

Following the method given in UG-23(b) of the ASME Code, VIII-1, the allowable axial compressive stress on a cylinder is determined as follows:

determine [latex]A=\frac{0.125}{R_{o} / t}=0.125 / 81=0.0015[/latex]

At 550°F, the value of B from Fig. 5-UCS-28.1 of the ASME Code, VIII-1, is 8700 psi. Consequently, the actual compressive stress of 8200 psi is less than the allowable compressive stress of 8700 psi, and the plate thickness of 0.75 in. is satisfactory.

Question 16.11: For the tall vessel described in Example 16.10, what are the...

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