Practice Problem 9-4 | Back | Index | Next |
Given: A certain steel (yield stress, S_{y} = 35 ksi) pressure vessel operates at a pressure of 300 psi. The radius and thickness of the pressure vessel are 40 in. and 0.50 in., respectively. Req'd: Based on the maximum distortion energy theory (von Mises), determine the factor of safety for the pressure vessel. |
s_{H} = pR/t = 24 ksi and s_{L} = pR/2t = 12 ksi
The von Mises stress is:
Therefore, the Factor of Safety is:
F.S. = S_{y}/s_{o} = (35 ksi)/(20.8 ksi) = 1.6827 = F.S. = 1.68
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