Given: A cylindrical steel tank (pressure vessel) has length L = 3.5 ft, radius R = 10 in. and thickness t = 0.10 in. The maximum stress that the steel can withstand is sf = 50 ksi. Assume E = 30 Msi and n = 0.3
Req'd: Determine (a) the maximum allowable working pressure pw if the Factor of Safety is 2.0, and (b) the change in radius at this pressure.
Sol'n: If the failure stress of the steel is 50 ksi and
the Factor of Safety is 2.0, the allowable (maximum working) stress is:
sw = sf /(FS) = (50 ksi)/2 = 25 ksi
In a cylindrical vessel, the hoop stress is twice the longitudinal stress, so the hoop stress controls the design. Therefore, the hoop stress is limited to 25 ksi:
pw = 250 psi
The change in radius is:
