The mechanism of faliure in this problem will be yielding, so we have to consider yield strength of the material to find factor of safety. Generally, yield strength of aisi L6 tool steel is 55114.3 lbf/in2 due to qunched and tempered it will increase upto 130534 lbf/in2.
Since we are talking about static faliure, load on the pipe will be,
(Load) P = 15000 + 5000 = 20000lbf
(Area of pipe) A = (162-152)/4 = 24.35in2
(Maximum permissible stress) = P/A = 20000/24.35 = 821.35lbf/in2
Factor of safety = (yield strength)/ = 130534/821.35 = 159
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