stress is greater than yield but plastic strain is zero?

[feadude]

I am doing a plastic analysis of cantilevered curved steel beam of I shape. Post processing results are: stress in tension flange (smaller radius) is 135 ksi , total engineering strain is 0.00387 , and plastic strain is zero.

my input (engineering) stress strain data are:
point 1 (yield) epsilon=.0043333 sigma=130 ksi
point 2 (uts) epsilon=.055 sigma=140 ksi
I am using von mises plastic yield function, isotropic plastic hardening rule, static, material nonlinear FEA, IDEAS NX 10
If stress is greater than yield, shouldn't plastic strain be greater than zero?
thank you

 

[corus]

It depends on which stress you are referring to. A stress component can be above yield but the Von Mises stress can be less.

corus

 

[feadude]

I am referring to von mises stress, it is 135 ksi and it is higher than yield of 130 ksi. Maximum principal stress is about 84 ksi and minimum principal stress is about -82 ksi.
If I plug these values in von mises formula I get 143.76 ksi which is more than 135 ksi. I think this is because I am using the stress contour values from the color bar and they could be slightly off.
thank you corus

 

[GregLocock]

well, the strain is less than the elastic limit, so something seems odd.



Cheers

Greg Locock

 

[corus]

It should also be remebered that the yield criteria may be met only at the element gauss points. Your results probably refer to the extrapolated nodal values, obtained from the gauss point. Thus the element gauss point stress may still be within yield but your nodal value may not.

corus

 

[feadude]

yes they are nodal values extrapolated from integration points. IDEAS does not give you the option of outputting integration point stresses for post processing but you can list it into a list file but that will be just a bunch of numbers that will not be suitable for presentation. Do you know if other FEA softwares give the option of outputting the integration point stresses into postprocessor. But I still do not understand when nodal stresses were extrapolated from integration points , should it not nodal strains been extrapolated from integration point strains and they should show a plastic strain value other than zero, after all stresses are calculated from strains?
thank you corus

 

[jetmaker]

feadude,

Let me start by saying that I do not use IDEAS, I generally use PATRAN w/ ABAQUS for my non-linear problems, so my questions/comments will be general in nature.

1) Are you sure that you are suppose to be entering ENGINEERING stress/strain data? In ABAQUS, you enter only the plastic component of the TRUE stress/strain profile. In NASTRAN, it is the whole TRUE stress/strain profile.

2) in the post-processor, can you turn off the element averaging feature? In some cases, the results for an averaged element do not agree. Using unaveraged data is the raw form and provides a more "TRUE" value for the element.

3) have you gone and checked the text data file to see what the corresponding stress/strain is for a given element?

Regards,

jetmaker

 

[jetmaker]

oh.. just another quick addition,

Regardless of what strees/strain values you enter, it should follow the curve within reason. Since you are getting conflicting results as to what side of the yield point you are on, you might want to increase your load by 1.5 factor and see whether you are following the curve at all.

Another quick question... you mention that you are running a static analysis with non-linear material FEA. Can you do that? In NASTRAN and ABAQUS, one has to run a non-linear static analysis using a plastic constituent model. Just wanting to make sure that is something that was not overlooked.

Thanks,

jetmaker

 

[feadude]

jetmaker:
1. yes. IDEAS uses engineering stress/strain data for material nonlinear and true stress/strain data for geometric nonlinear.

2.von mises averaged stress = 135 ksi
von mises unaveraged stress = 145 ksi

3.text data does not give me corresponding stress/strain for a given element.

4. I increased the load by 1.5 factor and I am following the curve for this higher load .

5. I am doing material non-linear static analysis with von mises plastic yield function.

Thank you jetmaker

 

[jetmaker]

feadude,

You said you've increased the load by a factor of 1.5 and the results seem to be following the curve. Can you provide the stress/strain values for that point for Von Mises and principal stress/strains relations?

Do you enter the Young's Modulus for material anywhere? If so, what is the value that you are using?

Is it possible to better define the stress/strain curve behaviour in IDEAS than just using the 2 entered points? If so, have you tried better defining the region in which the yield occurs?

Later,
jetmaker

 

[feadude]

jetmaker,
1. Factor of 1.5 is ouside of uts=140 ksi and at this factor beam has more than collapsed. I only made that run to see if I get plastic strain which I did. The area that I am interested is for von mises to be between yield=130 ksi and ultimate=140 ksi.

2.I am not entering E. I am using the IDEAS default value of E = 30000 ksi.

3. In IDEAS you can enter more points for stress strain data if you are using isotropic hardening rule but for kinematic hardening you can only use bilinear curve (two points). I am using Ziegler Prager kinematic hardening rule because my application is cyclic loading and isotropic hardening is not proper for cyclic loading.

Thank you Jetmaker

 

[jetmaker]

feadude,

Your problem has me stumpped. As far as I can tell, you have a working plastic material model by the fact that you have plastic strains when your load is high (1.5 factor).

Only thing I can think of is to check the input deck non-linear material card to make sure it reads something like the following:

0 0.00 0.00
1 0.004333 130000
2 0.055 140000

The 3 points will define your bilinear curve.

Also, IDEAS should be writing an ASCII file with all the analysis output somewhere. All FEA programs that I have dealt with produce such a file which can then be read in by post-processors. Finding this file will help tremendously in trouble shooting the results cause they report Gausian or centroidal values. This may be a feature that you have to turn on in the analysis preference menus.

Have you contacted SDRC about this? It could be a bug.

Keep me informed on the results. I am very curious.

jetmaker.

 

[feadude]

jetmaker,
My input points are:
1 0.0043333 130000
2 0.055 140000
The zero point is entered 0.0 0.0 as default by IDEAS.
Since I get plastic strains for higher loads (factor of 1.5) and this particular steel has a tensile so close to yield (140 is close to 130), the only thing that comes to mind right at this time is what "corus" stated above:

"It should also be remebered that the yield criteria may be met only at the element gauss points. Your results probably refer to the extrapolated nodal values, obtained from the gauss point. Thus the element gauss point stress may still be within yield but your nodal value may not. "

thank you jetmaker
thank you corus

 

[jetmaker]

feadude,

Agreed, that's why looking for the ASCII output files is important. The data there is raw, and will be reported directly at the Gausian points. There you should be able to directly correlate strain to stress values.

Hope it works out,

jetmaker

 

[nagi]

One more thing feadude,

If you have the unaveraged nodal values you can obtain the integration point values. For a 4-noded 2D element the stress (or any other quantity) at an integration point is 0.6220*value at closest node + 0.0446*value at farthest node + 0.16667*value at each of the 2 remaining nodes.

As Jetmaker and Corus mentioned the stress-strain curve is only accurately satisfied at the integration points. For large stress gradients this effect could be quite large. Here's a numerical example. If you have just one 4-noded 2D element and a stress of 1000 at one of its integration points and 0 at the other 3, then the "nodal" stresses will be 1866.0 at the closest node, 144.0 at the furthest node and -500 at the other 2 nodes! Assume now that the yield stress was 120, then all plastic strains will be zero even though you have a 1866 stress at one node!

There is another issue (less important though) when it comes to von-Mises and other "equivalent" nodal stresses. Different nodal v.Mises stress values will result if you first interpolate all the stress components to the nodes and then use the v.Mises formula, OR directly interpolate v.Mises stresses from integration points.

Nagi

 

[feadude]

Naji
I am using solid parabolic tetrahedral elements. How do I get integration point values from unaveraged nodal values? I plotted element stresses (unaveraged) not contour stresses .There are only 3 elements over a small area on the ouside surface of flange that have yielded but the surrounding elements 3 faces has yielded but the fourth face has not. Inside surface of flange has not yielded.

thank you Naji

 

[nagi]

Feadude,

I don't recall the values for the quadratic tet, but is should be relatively straightfoward (to go from nodal values to integration point values). You find the r,s,t coordinates of the integration point and substitute them in the shape function for each node. You then multiply the number that you get with the unaveraged stress value and add them up. That should give you the stress at the integration point.

That should always give you a good estimate. It will only be exact however if the FE code actually used the reverse of that interpolation to get the nodal stresses. In many case involving quadratic elements however, they do some simplifications.

For all practical purposes you won't have to do such a calculation. The main point I believe was to show that nodal stresses could be off-limit because of the fact that they are extrapolated from integration point values. The amount of "error" in these values is proportional to the variation of the quantity within the element.

Nagi