Question 16.7: A vertical vessel with a cylindrical shell and hemispherical...
A vertical vessel with a cylindrical shell and hemispherical heads is installed inside a building. The shell has 5 ft inside diameter and 0.5 in. nominal thickness, and is 30 ft from tangent to tangent. The vessel contains a fluid at 35 lbs/ft^{3}. Determine the total longitudinal stress in the cylindrical shell above and below the support lines, which is at the lower shell-to-head junction. SE=15,000 psi.
Assume that the internal pressure is set by Eq. (16.17) for a value of SE=15,000 psi. Rearranging the terms gives
P=\frac{ SE t}{R+0.6 t}=\frac{15,000(0.5)}{30+0.6(0.5)} = 245 psi.
The dead load of vessel above the support line is
Shell =\pi\left(30.5^{2}-30^{2}\right)(360)(490 / 1728)
= 9700 lb.
Upper head =(4 / 3) \pi\left(30.5^{3}-30^{3}\right) \times(490 / 1728)(1 / 2) =815 lb
Total weight = 9700 + 815 = 10,515 lb.
The longitudinal stress using Eq. (16.19) is
\sigma_{ L }=+P\left(\frac{R}{2 t}-0.2\right) \pm \frac{W}{\pi D_{ m } t} (16.19)
\sigma_{ L }=+245\left(\frac{30}{2 \times 0.5}-0.2\right)-\frac{10,515}{\pi(60.5)(0.5)}\sigma_{ L }=+7300-110 psi
\sigma_{ L }=\left\{\begin{array}{ll} 7190 \text { psi tension } & \text { with internal pressure} \\ 110 \text { psi compression } & \text { without internal pressure } \end{array}\right.
The dead load of vessel and contents below the support line is
Fluid in shell =\pi(30)^{2}(360)(35 / 1728)
= 20,620 lb.
Fluid in heads =(4 / 3) \pi(30)^{3}(35 / 1728)
= 2290 lb.
Weight of lower head = 815 lb.
Total weight = 20,620 + 2290 + 815
= 23,725 lb.
The longitudinal stress, using Eq. (16.19), is
\sigma_{ L }=+245\left(\frac{30}{2 \times 0.5}-0.2\right)+\frac{23,725}{\pi(60.5)(0.5)}
\sigma_{ L } = + 7300 + 250 = 7550 psi tension with internal pressure.