# Fatigue Problems w/ alternating stresses



## navyasw02 (Oct 7, 2010)

Ok so every time I do a fatigue problem, I do it completely differently from the way the solution has it. I tried #19 in Lindeburg's Practice problems which is as follows:

Circular tube, .5in OD, .03 in wall thickness, alternating stresses -3in-lbf and 30in-lbf. 42kpsi Sy, 72kpsi Sult, 24k Se, no stress concentrations. The problem asks for factor of safety for infinite life fatigue.

Well, I tried this method - I cracked open my Shigley's to page 339, set N = 10^6 (infinite life region) and used the good old N = ((Sf/n)/a)^(1/b) equation. It says if determining finite life with a factor of safety n, use sigma a for the Sf/n term and sigma a = Se/n with n being the factor of safety. I figured that by using 10^6 for infinite life, that would work the same way. I got roughly 1.6 for my factor of safety which is one of the answers listed. I figured Se was already .5* Su, why would Lindeburg possibly have us spend way more than 6 minutes to calculate the additional factors so the Se was probably good enough for the final value of Su. WRONG! Then I went back and recalculated Se using .59 as the ke factor and got an even lower factor of safety.

Turns out the solution wanted me to use eqn 50.32 from the MERM and I dont understand why that one is being used over the fatigue life method outlined in Shigley's.

Any ideas? I'm guessing that when in doubt, fall back on the MERM equations?


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## Lily (Oct 10, 2010)

I am not used to shigley, I got it recently because I realized that some problems do not have a reference in the MERM. What I usually do is look in the MERM first then if I can't find anything related to the problem I look for it in shigley.


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