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Doubt on Fatigue Strength S-N diagrams

  1. Jan 7, 2014 #1
    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.
     
  2. jcsd
  3. Jan 7, 2014 #2

    nvn

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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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