Fatigue results on Ansys Mechanical 4

[TeoAlfa]

Hello to all.
I am evaluating a project in terms of fatigue.
This is a large steel structure which is designed to take 40 tons of load.
I have done an analysis of 50ton just to be evaluate what will be the results in case the user loads the structure more than the specified maximum load.
Maximum stress occurs on a pin, which is close to 1000Mpa. Pin material is C45E steel with min yield stress of 370Mpa and maximum tensile strength of 780Mpa.
This pin obviously will fail on the first application of the load.

I then started to run a fatigue analysis of the whole structure.
Fatigue analysis input data:
Fatigue Strength Factor (Kf) = 0.8
Loading: Zero based
Analysis Type: Strain Life (the structure is designed to be low cycle loading)
Main Stress Theory: Smith-Watson-Topper (chosen because it's the most conservative option)
Stress Component: Equivalent von-Mises

The result i got was around 6.000 cycles on the pin where maximum stress occured.

My question is: How reliable this result could be? In first sight it seems impossible that a pin can withstand 6.000 cycles at a stress level close to 3 times the material yield stress!
Thanks in advance for every reply!

 

[L_K]

I would first think through the modelling of the pin
(element type, contacts, material model, meshing).
Did you make some assumptions (for example linear model)
which might affect pin stresses?

 

[TeoAlfa]

I have used high quality and density hex mesh for the pin.
The contact with the bushing is "no separation" as this is dictated by the available computing resources. With my current workstation capabilities, nonlinear contact options will result in days of analysis time.
My analysis is linear static. Do you think i should insert some nonlinear controls? And what would you recommend?

 

[L_K]

I think nonlinear analysis is needed if there is a risk for static failure. This depends
a lot on the particular structure, loading etc. Could you share some pictures of the model?
You could make a smaller model of the detail to solve faster.

Regarding strain-life fatigue analysis: Did you define a new material for the steel
pin material with the correct strain-life parameters?

 

[Petr Vymlatil]

The explanation why you can obtain positive number of cycles with stress higher than tensile stress might be elastic–plastic correction method (Neuber’s rule). This method is a part of the strain-life approach in ANSYS fatigue tool. You can find more information here on slide 16 here or here.

Keep in mind, that strain-life is based on stresses and strains, which are calculated by Neuber's rule or plastic calculation. In other words you cannot reach stresses higher strain tensile stress.

I would also recommend a nonlinear calculation with the stress-strain diagram, which corresponds to your strain-life parameters. In this way you can check the amplitude of strains and estimate the number of cycles from your strain-life curve.

Still, check your inputs carefully.


Petr Vymlatil (

http://www.designtec.info)

 

[TeoAlfa]

@ L K
I attach some images from the analysis. I hope it helps.

I will try to decompose the area of interest and run a nonlinear analysis.
And yes, i have defined the full strain-life parameters of the material.
Strength coefficient: 920Mpa
Strength exponent: -0.106
Ductility coefficient: 0.213
Strength exponent: -0.47
Cyclic strength coefficient: 1000Mpa
Cyclic strain hardening exponent: 0.2

 

[TeoAlfa]

@ Petr Vymlatil

If i understand correctly, my fatigue analysis results are not correct just because the stress and strain i got from the linear static analysis are not correct either because the pin is outside the tensile strength limit.

 

[Petr Vymlatil]

@ TheaAlfa
No I did not want to say, that your fatigue analysis results are not correct, because the pin is outside the tensile strength limit.

You should not use the tensile strength limit as strict criterion. Keep in mind that in reality have the material plastic behavior and the stresses will redistribute due to plasticity.

Your assessment procedure should look like this:
I. static assessment
- perform linear static analysis
- in your case are the stresses high above the yield strength. The basic question is, if you reached the yield strength in a critical cross-section completely. Your plots with contour 250 MPa are not very useful, use your yield strength (370 MPa). If not, there is chance for redistribution and you should perform a nonlinear calculation (with stress-strain curve). If yes, change your design.
- perform a nonlinear calculation (with stress-strain curve) and look for maximum plastic strain. If the maximum plastic strain is relatively small or acceptable for you there is chance that your design stands also cyclic load (low cycle fatigue <10e4 cycles). If not, change your design.
II. low cycle fatigue
- perform linear calculation and ANSYS will use Neuber's rule and calculate the strain amplitudes from your linear calculation. Keep in mind that Nueber's rule should be used only if the regions with stresses above yield are small (most likely not the case in your current design according the figures you shared).

However, low cycle fatigue assessment is not trivial. I would also recommend better software for low-cycle fatigue analysis (for example FEMFAT).



Petr Vymlatil (

http://www.designtec.info)

 

[L_K]

In my opinion, based on the stress results, bilinear plasticity model and frictional contacts
should be used to model the detail. One thing to consider in the subsequent strain-life analysis
is the choice of stress component.

 

[TeoAlfa]

@ Petr Vymlatil + L K

Many thanks for your replies.
I post again some images with corrected stress plot (370MPa).
I think the situation is quite serious as the whole circumference of the pin exceeds yield strength.
Section images also attached.
What do you think?

Would it be better to choose another steel alloy with higher yield strength or i will have more serious fatigue problems due to the brittle nature (nucleation etc) of that steel?

 

[Petr Vymlatil]

@TeoAlfa fist I would recommend to follow advice from L_K - use nonlinear frictional contacts and nonlinear material model and check the amount of plasticity

Petr Vymlatil (

http://www.designtec.info)

 

[TeoAlfa]

Hi again,

I have run a nonlinear analysis (large deflections on) but i could not use nonlinear contacts between the pin and bush because the angle of the acting force causes the structure to rotate so no convergence. I will try again though.
So what we have is "large deflections on", and again bonded contacts.
As for the material model i have entered full strain-life parameters, as well as bilinear isotropic hardening data. See attached image from Engineering Data.

http://oi64.tinypic.com/fcs6qp.jpg

The loading case was 2 steps loading, zero force at the start, linear increase of the force up until 1sec, and then releasing the force up to 2sec, see attached image.

http://oi64.tinypic.com/25qchox.jpg

Finally I got some interesting results.
The maximum stress on the pin was just over yield (around 380Mpa), compared to more than 1000Mpa in the linear analysis.

Maximum total strain on the pin was 0.048.

And here is the stress-strain diagram at the node of maximum stress on the pin.
It seems that it's perfectly linear, although not returning fully to zero.

What conclusion would you come up with?
Thank you in advance!!

 

[L_K]

In first picture, yield strength is defined as 250 MPa. Should it be 380 MPa instead?

 

[TeoAlfa]

Yes, you are right.... i just copied the data from structural steel... How can i find the tangent modulus for the material?

 

[L_K]

For example in Eurocode 3, for strain hardening plasticity model tangent modulus is defined as E/100.

 

[TeoAlfa]

Thank you, i will run the analysis with the correct data and come back with the results.

 

[TeoAlfa]

So, i am back with the final results.

Material model

http://oi65.tinypic.com/2ln9z47.jpg

Load case. Here i made a bit of a change, i added one second after the release of the force to give more time to the pin to come to its final steady state.

http://oi67.tinypic.com/2ia5x8i.jpg

Here is the maximum stress, significantly higher than the previous analysis. 382Mpa/471Mpa.
I cannot figure out why it resulted that different as what i changed were the only the bilinear isotropic hardening data.

Maximum plastic strain. Now it's significantly less than previously.

And here is the stress-strain diagram at the node of max plastic strain.

As i am new to nonlinear analysis, i would appreciate any help. Thanks!!

 

[TeoAlfa]

Based on these results, i attach a picture of the fatigue life.
Analysis Type: Strain Life
Mean Stress Theory: SWT
Stress component: Max Shear (this gives the worst result)

Fatigue life of 100.000 cycles is perfectly acceptable to my application, but something tells me that this is not a reliable result.
Could you please review this?

https://i.imgur.com/q3KbSZK.jpg

 

[L_K]

Static analysis:
Stress increases over yield strength because of strain hardening.
After load step 2 (unloading) only plastic strain remains.

What is interesting that max stress is achieved at 70% load, then it
decreases, and again increases to 100% load. It is hard to say what leads
to this behaviour without seeing the model fully.

In my opinion, increasing the length of load step 2 does not really make sense.
Remember this is static analysis. Equilibrium is the same at t=2 and t=3.
Results should be identical. Actually you will get all the information you need
from load step 1.

Fatigue analysis:
If you think results are not reliable, review your material parameters.
You could read the result (cycles) directly from the EN-graph, now that
you know the strain for one cycle.

 

[TeoAlfa]

One thought about the weird stress diagram. The stress results as shown above are for the whole pin, not for a specific node. As such, i suppose that maximum stress occurred at different nodes during the loading time period. So maybe due to stress redistribution, we had let's say point X with stress A at time T, and stress A-1 at time T+1.
Let's take a look at the stress strain diagram of a certain node where maximum plastic strain occurs. Isn't that weird too (see image below) ?
How could you explain that stress decreases so much (29Mpa) at 1.4sec and then increases again after 1.5sec.
And why stress remains after unloading? Is it remaining internal stresses due to plastic deformation?

http://oi66.tinypic.com/16aqfdt.jpg

I also attach an image of the model. This is a small part of the whole assembly which i isolated to run the nonlinear analysis.

http://oi68.tinypic.com/r09icm.jpg