Question 8.5: A cylindrical shell with length 18 ft and outside diameter 6...
A cylindrical shell with length 18 ft and outside diameter 6 ft is constructed of carbon steel with a yield stress of 38,000 psi. Determine the thickness needed to resist an external pressure of 300 psi.
Let t =1.25 in.
\frac{L}{D_{ o }}=3.0, \quad \frac{D_{ o }}{t}=57.6From Figure 8.10, the factorA=0.00095. From Eq. (8.6) with E_{0}=29 \times 10^{6} psi,
P=\frac{2 A E_{0}}{3\left(D_{ o } / t\right)} (8.6)
P_{ a }=\frac{2 \times 0.00095 \times 29 \times 10^{6}}{3 \times 57.6}=320 \text { psi }Now check for the inelastic region. From Figure 8.11, factor B=12,000 psi in the plastic region. Hence, the first of Eq. (8.6) cannot be used. From the second of Eq. (8.6),
P=\frac{4 \times 12,000}{3 \times 57.6}=278 psi inadequate
Try t =1.375 in. Then,
\frac{D_{ o }}{t} = 52.4 and A = 0.0011.
From Figure 8.11,
B = 12,400 psi,
P=\frac{4 \times 12,400}{3 \times 52.4} = 316 psi acceptable
Use t = 1.375 in.
Figure 8.11 and Eqs. (8.6) and (8.7) are applicable only when D_{ o } / t is less than or equal to 1000. For ratios above 1000, many engineers use the following equation, which was developed by the US Experimental Model Basin:
P_{2}=\frac{2 \sigma}{D_{ o } / t}\left(1-\frac{1}{D_{ o } / t}\right) (8.7)
P_{ cr }=\frac{2.42 E_{0}}{\left(1-\mu^{2}\right)^{3 / 4}} \frac{\left(t / D_{ o }\right)^{2.5}}{\left[L / D_{ o }-0.45\left(t / D_{ o }\right)^{1 / 2}\right]}A minimum DF of 3.0 is to be applied to this equation in order to obtain the allowable external pressure.
