Fatigue analysis - Results do not match input data

[procenitelj]

Hello everyone,

I am new to fatigue analysis in Inventor Nastran and I came to a problem for what I cannot find a solution. @John_Holtz 

I am testing a piece of solid material (carcass of a conveyor belt). The piece was physically tested and these are the results:

Number of cycles to break	Stress in material (MPa)	Force (N)
1	200	80000
1000	120	48000
5000	100	40000
50000	80	32000
300000	70	28000
1000000	60	24000

I have set material parameters:

Load Scale factor VS time is set as:

When I run the test with 48.000 N, I get desired values, of about 1000 life cycles:

But when I run the analysis with the next level of force, 40.000 N, for which I expect 5000 life cycles, I get absurd results of 981567 life cycles:

Does anyone know where am I making a mistake?

[John_Holtz]

Hi @procenitelj

I assume the force and stress in your table are the maximum values when the load table multiplier is 1. Is that correct?

The "S" in the "S-N" curve is the stress amplitude (Sa), not the maximum stress (Smax). If the load were fully reversing, then "S" = Sa = Smax. You have a load that varies between 0.6 and 1.0 of the maximum stress, so "S"=Sa=Smax*(1-0.6)/2 = 0.2 times the maximum stress. If this understanding is correct, you need to plot the points 0.2*Smax versus N and recalculate the fatigue material properties.

John

[procenitelj]

Dear John,

Thank you for the fast reply. When I set the Load Scale Factor as following:

And leave old S-N data configuration:

I get proper results:

But when I use 0.2*Smax versus N:

with Load Scale factor 0,6 to 1, I do not get proper results:

Did I misunderstood you when u said: "You have a load that varies between 0.6 and 1.0 of the maximum stress, so "S"=Sa=Smax*(1-0.6)/2 = 0.2 times the maximum stress. If this understanding is correct, you need to plot the points 0.2*Smax versus N and recalculate the fatigue material properties." ?

[John_Holtz]

Hi @procenitelj 

Your understanding of what I said was correct. Sorry, but I think I misunderstood your original description I gave you wrong information. 😞 Let me know if this is the correct interpretation.

• The physical test was done using a standard rotating test specimen used for fatigue tests. (For some reason I was thinking the test numbers came from the actual conveyor belt which was cycling between 0.6 and 1.)
• In the standard fatigue test, the stress in the test specimen is fully reversing, so S = Sa = Smax, and your original material input was correct. (For fully reversing load, the stress amplitude Sa = 1/2(Smax - Smin = Smax.)
• The original question now make sense 🙂. Why does the analysis of the conveyor, with the load cycling between 0.6 and 1 times Smax, not give the calculated life from the S-N test data?

If the above is the correct interpretation, the answer is mostly the same! The "S" in the S-N curve is the stress amplitude. Since your stress amplitude Sa = 0.2* Smax, you need to read the S-N curve at that value.

For example, the load of 40000 N gives a stress of 100 MPa. Therefore the stress amplitude Sa = 100*0.2=20 MPa. Interpolating the S-N curve at 20 MPa indicates an infinite life, and that is approximately what Nastran calculated.

The full calculation is slightly more involved. The "S" on the S-N curve is the value Sa0 = Sa/(1-Sm/Su) where Sm is the mean stress. In your case, Sm = 1/2(60+100 MPa) = 80 MPa. Sa0=20/(1-80/120) = 60 MPa. Reading the S-N curve for 60 MPa indicates a life of 1 million cycles. Nastran calculated 982000 cycles.

In summary, the calculated stress to use for interpolating the S-N curve is the alternating stress with a correction due to the mean stress: Sa0

• Sa = alternating stress amplitude = 1/2*(Smax-Smin)
• Sm = mean stress = 1/2*(Smax+Smin)
• Su = ultimate stress (technically the stress at the first data point N0, Su)
• Sa0 = corrected alternating stress = Sa/(1-Sm/Su)

The attachment to this article gives more details about the calculations. See How to understand Fatigue analysis in Nastran | Inventor Nastran | Autodesk Knowledge Network.

John