Nastran In-CAD non linear solution issue

[Anonymous]

I am trying to perform a limit load analysis.  The model is a test model and its set up in the following way

LGDISP=off

Elasto-plastic Material model

Isotropic hardening model

Tangent modulus = 0

Von Mises with a yield stress limit set to 20,000psi / Ultimate = 58,000psi

2x2x12in test block

The initial run was set with a load of 10,000 lbs and the solution failed to converge as expected and gave me a load factor of about 0.375 and a stress above the yield allowable as was expected (although the stress was higher than I expected for just one step over the limit load)

The second run, I set the load to 3700 lbs and the solution converged as expected, but the stress was still being reported above the yield allowable.  I repeated this process with loads slightly above and slightly below the reported limit load and each time, the result is that the reported stress is higher than the allowable yield that is set in the material properties. 

Why is the software reporting a stress higher than the yield when the material has no strength beyond the material set yield limit (20,000 psi)

[David_Kind]

@Anonymous, can you post a screenshot of your non-linear material definition?

[KubliJ]

Hi @Anonymous,

I am curious about your material setup too.

The reason is because you describe the material as elastic-plastic, which allows you to define the material properties after yield.  Which you seem to mention the ultimate strength.  Having a yield stress doesn't stop the material from stressing past that.  Just that we use a different material property after yield.

If you are looking for the elements to 'stop' stressing at yield strength, then you are looking to define a elastic-perfectly plastic setup.  You still use the elastic-plastic material model, but you set the new material strength post yield to a near zero value.  This will make the slope of the material strength relatively flat post yield.

Even with this setup, you will still see stresses slightly higher than the defined yield strength, but this has to do with where the stress is calculated in the element and where results are then interpolated and reported in the results.  This physical difference in location can cause the slight uptick in stress over the yield.

Also, if this is non linear, why do you have large displacement turned off?

Thanks,

James

[John_Holtz]

Hi @Anonymous

In some sense, the stress = force/area. So if you try to apply a force that causes a stress larger than "the yield stress", the software is going to create a stress larger than the yield stress. From the calculated stress at the applied load, it then checks if it follows the stress-strain curve. Now the question becomes this: how close to the stress strain curve are the results when it fails to converge? Is it within 10%? within 0.0001%? (I do not know the answer, but since it is a numerical solution, there could be some difference.)

 As James indicated, the location where the yield stress is calculated (at internal points known as gauss points) is different than the location where the stresses are shown (at the nodes). To make matters worse, the nodal stress is extrapolated. If the mesh is sufficiently fine, the two results are sufficiently close. This article goes into some of the details about this: Equivalent stress in a nonlinear analysis with Autodesk Nastran

[Anonymous]

for full disclosure, I am trying to get the model set up to run an analysis following the rules of ASME Sec VIII Div 2 part 5 limit load analysis.  This code requires we use a non linear analysis not for geometric non linearity, but for material non linearity so we can define the behavior in the plastic region and take advantage of the load step functions of the non linear analysis.  We typically would apply a unit load, say 10,000lbs, run the analysis and the solution would fail, but report the step that the non convergence occurred (the point when the model reached constant strain) and a load factor that is used to calculate the theoretical limit load.  The goal is to identify the point of global structural instability.  To do follow this procedure, we also need to apply small displacement theory, and have a tangent modulus of 0.  Near zero does not work for this procedure because it will allow the model to have strain capacity past the assigned material limit.  KubliJ is correct in using elastic-perfectly plastic, which I have done, but for the case of applying ASME code, a near perfectly plastic tan modulus does not serve the function of identifying global instability at the defined yield limit.  When I set the load just below the limit load, the stress produced what about 20% higher than the allowable stress and the same thing happens when I set the load slightly above the limit load except in this case, the solution fails to converge.  That's is where my question is...I have defined the perfectly plastic material properties (zero slope past the selected yield limit) but the model is still reporting stress....why?

[KubliJ]

Hi @Anonymous,

Are you expecting the stress to drop to zero when it hits yield? 

Thanks,

James

[Anonymous]

I am expecting the model to fail (inability to converge) and report the load that is associated with this failure in the form of a factor of the applied load. If the load selected results in a stress / strain point that’s on the curve (elastic region) I expect the model to reach a solution and report the resulting stresses ... which should be below the yield limit I set (this comes from the code allowable stresses)

[David_Kind]

@Anonymous try setting up the material as shown below (use an appropriate strain value in the second row for your particular material). This setup effectively creates a elastic-perfectly-plastic material model.

How do you plan on dealing with VIII-2 §5.3 (protection against local failure)? Are you going to export the stress values using a script? You can private message me about this if you wish.

[John_Holtz]

Hi @Anonymous

Please try @David_Kind 's suggestion, but I would expect it to produce the same result as your setup using the bilinear properties.

In some cases, you can replace the load with an enforced motion. This type of load is more stable, so the analysis should run to completion. The reaction force will show the load changing at the load limit as the part goes plastic. The calculated stress will still be different from the yield for the reasons mentioned in the article I referenced previously: the locations where the yield criteria is calculated is different from the location where the stress results are given in the contour plot. Because the stress in the contour plot is extrapolated, it can deviate from the yield stress. If my memory is correct, you can view the "Equivalent stress" result instead of the von Mises stress; the equivalent stress should follow the stress-strain curve.

[Anonymous]

In the past (using Autodesk Simulation) I was performing an elastic analysis and would export the stress tensors through the thickness for the linearization and use 5.3.2 for my protection against local failure check.  I'm trying to learn how to perform limit load and I haven't thought about how to get through this check yet.  I'm still trying to learn how to set up the model.  Any guidance you can provide on this would be appreciated.

[Anonymous]

I took a look at the equivalent stress plot and it was MUCH closer to the yield limit (around 10% over), but because this was a tester model, the mesh was not very fine.  I am in the process of running it with a much finer mesh to see just how close I can push it, but I think this is a good solution to the problem. 

[Anonymous]

I'm expecting to use 5.3.3, but not sure if its applicable to limit load.....

[David_Kind]

You're free to use either 5.3.2 or 5.3.3 (as far as I know). However, the problem will be extracting σ1, σ2, σ3, and εpep out of Nastran-In-CAD. The log files that it offers are more human readable than machine readable. You're going to need to spend some time writing a script to extract all the data. It's unfortunate that Nastran-In-CAD won't just product a .csv file with the data we need.

[Anonymous]

I ran the model with several different mesh densities and always got the same result...about 10% over when viewing the equivalent stress.  I need to have someone else review the file and provide feedback, I am not understanding how the calculation/reporting of the stress deviating from the material stress strain curve is acceptable from the standpoint of ASME Sec. VIII Div 2.  I have attached my test model.

Also, I downloaded the 2019 version today and will play with that one.  Its my understanding that this software has linearization capabilities.  This may help with the data extraction I need to do for the local failure check.  There also seems to be some kind of stand alone Nastran software...not sure yet what that is but ill be looking into it.

To all - please feel free to provide any comments or guidance

[David_Kind]

@Anonymous two comments:

1. Stress linearization is an obsolete technique that is not recommended (however it is permitted) for VIII-2.

2. The stress linearization tool did not make its way into NIC 2019. There is speculation that it will be available in 2019.1.

[John_Holtz]

Hi @Anonymous

The centroidal values (both the von Mises stress and equivalent stress) are slightly higher than 19999 psi stress. See the attached image for the settings.

The corner stresses, which are shown by default, are extrapolated from the centroidal values and that is why the corner stresses are higher. It looks like the nodes at the fixed end are especially susceptible to inaccuracies with the extrapolation. (But there are a few other nodes inward from the constraint that have corner stresses above the yield strength.)