Doubt on Fatigue Strength S-N diagrams

  • #1
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A standard way of plotting an S-N (fatigue strength vs no of cycles) graph is to start off by taking a value of Sm (for N=1000) cycles and joining it to the point depicting Se(Endurance Limit). Now we commonly take Sm= 0.9Sut for steels. Now I was wandering, before you actually get to use the component for 1000 cyces at 0.9 Sut the component will yield at around 0.5Sut(before you reach Sm), as Sy(steel)≈.5Sut. SO should yielding be the cause of failure for values of stress rather than fatigue?
Thank you in anticipation.
 

Answers and Replies

  • #2
nvn
Science Advisor
Homework Helper
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Abhishekdas: Yes, you are correct. I.e., if the allowable tensile stress (Sta) based on tensile yield strength (Sta = Sy/FSy) is less than the fatigue strength (Sn), then Sta governs over Sn, in which case you would use Sta, where FSy = yield factor of safety.
 

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