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S-N curve is drawn from the results of fatigue test and fatigue test is conducted under completely reversed bending stress condition on a standard specimen. Is there any reason for conducting the test under completely reversed bending stress condition and why specifically "reversed stress condition" is used? I mean it can be a repeated stress condition or some other cyclic stress.

#Update

enter image description here Figure source- Material science by Callister

So, for the same stress amplitude increasing the mean stress reduces the fatigue life.

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  • $\begingroup$ normally you would test in the direction of the stress. A part that was non-uniform would react differently in different orientations. $\endgroup$
    – Tiger Guy
    Commented Mar 20, 2023 at 15:17

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Complete reversal is the most basic type of cyclic loading. You can describe it just using the amplitude, because the mean stress is zero. S-N curves obtained in this way can also be used for situations where the mean stress is not zero, in which case you have to tweak the amplitude, which can be done for example using Goodman rule.

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  • $\begingroup$ Ok..is there any specific reason for conducting the test under reversed "bending", I mean it can be done under twisting or reversed axial loading? $\endgroup$
    – MechaTrex
    Commented Mar 21, 2023 at 5:09
  • $\begingroup$ You have to choose one of these as a baseline anyway and I presume the reversed bending will be the closest to most real situations where fatigue is of concern. Another consideration might be complexity of the testing apparatus. $\endgroup$
    – Tomáš Létal
    Commented Mar 21, 2023 at 20:36
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If you ever recall needing to cut a wire by bending it repeatedly you will probably get the idea why full reversed is a much harsher condition that only one sided.

The fully reversed condition is the most harsh (for a given stress) and it will make a specimen fail sooner (usually faster that half the time of the one side test). In fatigue tests that is a big incentive, since testing a million cycles even at 10 Hz takes about a full day of testing. So there is an incentive of using the harshest (repeatable) condition to speed up the process.

UPDATE ====

Below is the Goodman diagram (Source wikipedia). It associates mean stress $\sigma_m$ and alternating stress $\sigma_a$.

enter image description here

In most cases, the different curves involve the followign factors $\frac{\sigma_m}{\sigma_b}$ and $\frac{\sigma_a}{\sigma_w}$. E.g.:

$$\frac{\sigma_m}{\sigma_b} + \frac{\sigma_a}{\sigma_w}= \frac{1}{n} \tag{eq.1}$$ where:

  • $\sigma_b$ is the ultimate tensile stress
  • $\sigma_w$ is the fatigue limit.

As you may know the $\sigma_b$ (UTS) is always greater than $\sigma_w$ (Fatigue). Therefore:

  • the mean stress has grow to the ultimate tensile stress in order to have failure.
  • On the other hand the alternating stress can only go up to the fatigue limit (which is lower).

As a result, the sum of the absolute values is minimal when mean stress is zero $(|\sigma_m|=0)$ and alternating stress is maximum $(|\sigma_m|=\sigma_w )$.

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  • $\begingroup$ Ok.. I get that "full reversed is a much harsher condition than only one sided". But regarding "fully reversed condition is the most harsh" I have the following doubt. S-N curve is between stress amplitude and cycles on log basis. So, If I consider a stress amplitude of 400 Mpa, and consider 2 conditions, (1) fully reversed stress of +-400 Mpa and (2) cyclic stress of +500 Mpa and -300 Mpa. $\endgroup$
    – MechaTrex
    Commented Mar 21, 2023 at 19:12
  • $\begingroup$ Then how can I conclude that specimen will fail sooner in (1).. because if I compare (1) and (2) then stress amplitude is same 400 Mpa but in (2) mean stress of 100 Mpa will also be present. So, how can I conclude that fully reversed condition is most harsh? $\endgroup$
    – MechaTrex
    Commented Mar 21, 2023 at 19:12
  • $\begingroup$ I would disagree with full reversed being much harsher condition than one sided. For the same amplitude, positive mean stress should be harsher and negative mean stress should be less harsh. "Full reversed" is just a convenient baseline. $\endgroup$
    – Tomáš Létal
    Commented Mar 21, 2023 at 20:39
  • $\begingroup$ Have a look at the Gooman relation/diagram. It associates the life of a material for the mean stress and the amplitude of the alternating stress. For the maximum mean stress, no alternating stress is possible. For mean stress zero you have the maximum value of alternating stress. On that stress note the slope, and how its calculated, it should tell you that the alternating stress is more harsh condition. $\endgroup$
    – NMech
    Commented Mar 21, 2023 at 20:43
  • $\begingroup$ @mechatrex also please consider that for fatigue purposes there are two represenations. The min-max ($\sigma_{min}, \sigma_{max}$) and the mean-alternating $(\sigma_{m}, \sigma_a)$. For the example you are giving (like you said), with $\sigma_{max} = 400$ and $(\sigma_{min} = 0)$, then $\sigma_m = 200$ and $\sigma_a=200$. However, for the fully reversed, $\sigma_m = 0$ and $\sigma_a=400$, and if you take note on the Goodman relation you will understand that it is worse. $\endgroup$
    – NMech
    Commented Mar 21, 2023 at 21:02

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