Stress life method is generally employed for high-cycle fatigue (HCF) applications. while HCF region is approximated by $\sigma=a(N)^b$ which gives a linear graph on log-log scale, for steels it starts from $0.9\sigma_{ut}$ up to $0.5\sigma_{ut}$. I presume that $0.9\sigma_{ut}> \sigma_{y}$ and in such a case how is my assumption of elastic strain and stress everywhere is valid. $\sigma_{ut}$ and $\sigma_{y}$ are the ultimate strength and yield strength of steel respectively.

  • $\begingroup$ Check out the Comet - a real example. $\endgroup$
    – Solar Mike
    Commented Jun 25, 2021 at 7:01
  • $\begingroup$ Please provide the source material for a better understand of your concerns. $\endgroup$
    – r13
    Commented Jun 25, 2021 at 16:44
  • $\begingroup$ I am not certain that there is an explicit assumption that material remains in the elastic region during fatigue testing. Could you provide the exact quote for that statement? If it states that explicitly, then your concerns are valid, but I don't think that's the case. $\endgroup$
    – NMech
    Commented Jun 25, 2021 at 17:39
  • $\begingroup$ If the loading causes nonlinear material behavior, you will not have high cycle fatigue by the usual definition of HCF - i.e. at least $10^4$ cycles to failure and often at least $10^7$. $\endgroup$
    – alephzero
    Commented Jun 26, 2021 at 1:07
  • $\begingroup$ I was referring to the following video youtu.be/Alvl5f8blgI .@7:20 the professor states the above assumption. $\endgroup$
    – huministic
    Commented Jun 26, 2021 at 14:40

1 Answer 1


The SN diagram below answers your question.

Note "High Cycle Fatigue (HCF)" failure is characterized by high repetitive cycles (N), usually equal to or greater than $10^4$, and low-stress (S) in the elastic zone.

enter image description here

  • $\begingroup$ But is not (0.7 to 0.9)σut > σy generally, which is still greater than elastic limit of the material? if so, shouldn't the material should be experiencing plastic strains and stress? $\endgroup$
    – huministic
    Commented Jun 26, 2021 at 14:49
  • $\begingroup$ I don't know what is the 0.7σut referred to, is that the lower bound of HCF? For the chart I've provided, when N = 10^4, the corresponding stress σa (590 Mpa) is approximately equal to 0.7σut. Assume σa is the breaking point and σy for that metal is 550 Mpa, then, it will fail in the high cycle fatigue mode and experience plastic deformation prior to breaking and, yes, the deformation after yield should be evaluated using the stress-strain curve in the plastic region. $\endgroup$
    – r13
    Commented Jun 26, 2021 at 15:39

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