Practice Problem 9-4

Practice Problem 9-4

Given: A certain steel (yield stress, s_y = 35 ksi) pressure vessel opperates at a pressure of 80 psi. The radius and thickness of the pressure vessel are 45" and 0.10", respectively.

Req'd: Based on the maximum distorsional energy theory (von Mises) what is the factor of safety for the pressure vessel??

Solution: Based on the data the hoop and longitudinal stresses are:

s_H = pR/t = 36 ksi and s_L = pR/2t = 18 ksi

The von Mises stress is then:

s_o = 1

2 s_x² s_xs_y + s_y² + 3t_y² ^1/2

s_o = 1

2 36² (36)(18) + 18² + 0 ^1/2

= 22.0

Therefore, the factor of safety is:

FS = s_y/s_o = (35 ksi)/(22.0 ksi) = 1.59