# Fatigue calculation using S-N curves.

Dear all,

I have a dummy question, but it seems there is no universally accepted answer. For a general structure subject to a cyclic loading with constant ampitude, we use FEA program to run stress analysis under peak and valley load point respectively. Now, we need to post process the FEA result to come up with the stress range that will be plugged into S-N curve to come up with a "N" value.

My question is, for a general structure, having complicated geometry and loadings, how will we calculate the stress range from valley to the peak? Here is what I have seen before used by various people and companies:

(1). By loading step operation that creates an artificial load step equal to the load step at peak minus load step at valley. Then use the first principal stress as the stress range. Load step operation in most commerical FEA package really starts from stress tensor operation to get a new stress tensor, from which the equivalent stress and principal stress are built.

(2). Doing the same as (1), but take equivalent stress of the aritificial load step as the stress range

(3). Use the principal stress at the load step at peak minus the principal stress at the load step at valley. Doing this could be tedious, and also hard to pick which principal stress (S1 or S3) at each load step we should use? Obviously it ignored the possible direction change from valley to peak.

(4). Doing the same as (2), but rather than having the operation with principal stress, they do operation using equivlent stress.

Can you guys elaborate?

Thanks & Regards,

William Lu

### Re: Fatigue calculation using S-N curves

William,

I have done various analysis for machine parts of mining equipment. For several years, I used what you wrote in (3). However, during the last years I changed to (4) because there are many fatigue books that recommend that approach (even basic mech. desing books such as Shigley, Mott, etc.). The von Mises approach has worked fine for me so far.

Best.

### Thanks, Alejandro,

Thanks, Alejandro,

I have had a typo in my original post for item (4), it should be:

(4). Doing the same as (3), but rather than having the operation with principal stress, they do operation using equivlent stress.

so if you still think you would change to (4), then question is for some cases, for example, you have a fully reversed bending cycle, from -M and +M, then for one hot spot, you would get the SAME equivalent stress. Therefore, using (4) will lead to ZERO stress range, which obviously does not make sense.

Another issue of using equivalent stress is that we are not able to find out the compression portion from a stress cycle. Usually we may need to discount the stress cycles that contains some compression part.

Thanks

William

### Re: Thanks, Alejandro

William,

The stress range cannot be zero. It is two times the alternating stress, hence a positive value.

If the compressive stress is relevant to your problem, I agree that the equivalent stress is not the right quantity. But most machine components made of steel behave equally in tension or compression (buckling is not a problem). On the other hand, a compressive stress (again assuming that other mechanism such as buckling does not govern the design) will always be beneficial from the fatigue point of view since there is evidence that compressive stress does not alter the fatigue life of the element under analysis. You can think of it as it will tend to close the small cracks that the fatigue cause. Furthermore, you can conservatively ignore the compressive stresses from your analysis.

Best.

### Re: Re: Thanks, Alejandro

I was talking about mean negative stress as I think that was you were refering to as "stress range". So, for the fully reversed case the mean stress value is zero. And compressive mean stress can be ignored from the analysis.

### Alejandro, like you said, the

Alejandro, like you said, the compression stress does not alter the fatigue life and therefore, we should give the credit to a stress range that includes some compression portion. For example, if your stress cycle from a loading cycle is -1ksi to +1ksi, then rather than using 2 ksi as stress range for fatigue calcualtion, we can use 1.2 ksi (based on DNV's method of using 60% of the full range).

Regaring the use of von Mises stress to calculate the stress range, what I am saying is, if we do that way, then for a fully reversed loading cycle, the equivalent stress will be 1ksi, 0, 1ksi. I am assuming that 1ksi is the von Mises stress I calculated at one loading point. Based on what you said above, our stress range should be 2*(1ksi - 0 ksi) = 2ksi. I do not know if we could use this in general, but I heard some other people they just use the amplitude of the stress cycle, i.e., 1ksi - 0 ksi = 1 ksi, rather than 2 ksi.

As said, another thing for using von Mises stress is how to give the credit to the cycles that may have some compression range. More specifically, for a structure under a 3D stress state, how can we tell if the hot spot is under tension or compression.

thanks

William.

### Re: Alejandro, like you said

I don't use the range. I have never seen someone using that. The theory expects you use the alternating stress (in the fully reversed case it coincides with the amplitude). So, I would use 1 ksi.

And for the compressive stress, I never give credits to it. So, I don't care about it. Perhaps we are in different reallities: I mostly work with equipment that is damaged after some use and so the owner of the equipment wants it to be repaired and doesn't want the equipment to be broken again. So, I redesign the component from a conservative point of view. Of course, if you are on the equipment manufacturer side you would want to give some credits to compressive stress as to reduce the weight of the equipment and hence its cost. 