Non-linear geometry analysis

  • Last Post 02 December 2018
  • Topic Is Solved
Schwarz18 posted this 01 December 2018

Hi all,

I'm opening this thread in order to have some clarification about a non linear geometry analysis I'm running on APDL. The model is very simple, it's characterized by a 1500x3000 mm grid with a 40x40 mm pitch of beam188 elements, with translational constraints on the external borders and borders every 1000 mm, so that we have along the 1500 dimension 4 borders constrained, while 2 borders along 3000 dimension all constrained. I used a circular section with 2 mm radius for the beam188 elements. The load applied is at the center of my grid, on a 200x200 mm area, with a value of 1400 N, equally distributed. Since my model undergoes large displacement, I had to switch the analysis to a non-linear geometry solution. I used, as non linear material model, the BIHP (bilinear isotropic hardening plasticity) with a value of 235 MPa as YS and 2000 as Et. The isotropic part would be the classical of a steel, so that Ex would be equal to 210000 MPa and poisson's ratio of 0.3. Now the things I noticed running different analysis are really strange. Indeed, with the same load condition and Et, if I change the value of YS (let's say, 235 then 300 then 400 and so on), the von mises and equivalent plastic stress that I have as output will increase as well. This is a thing that I can't explain, since the load I'm applying every time is the same, what I'm changing are the mechanical properties of my plasticity field. If during my first analysis, with YS at 235 MPa, I have a VM stress of 270 MPa, I don't expect it to increase during my second run, with a YS of 300 MPa, to about 340 MPa! The displacements are the same during the two runs, what change are the stress and strains, that's why I can't make a good correlation between my FE model and what I'm expecting to happen. Could someone give me an explanation about it? Thanks in advance for any replies!

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peteroznewman posted this 01 December 2018

Please post an image of the grid so it is more easily visualized for the community.

Schwarz18 posted this 01 December 2018

Hi peter,

thanks for pointing that out!


That is the grid used for the analysis. The constraints applied on the borders as I said, are ux, uy and uz. In order to have the 1400 N total load, I divided it by 25, the number of keypoints that cover a 200x200 mm area.

peteroznewman posted this 01 December 2018

Plot the Plastic Strain using a consistent Legend for color between the two plots. In one plot, set the material YS value to 235 MPa and in the second plot, use a material YS value of 400 MPa.  This may help to explain why you get what you got.

jj77 posted this 01 December 2018

It is normal what you are seeing.


Run it without nonlinear material, thus only nonlinear geometry, and very likely you will see that the stress (not sure what you look at in Beams, perhaps VonMises (VM) or very often total fibre stress, which is axial + the two bending stresses) will be above 300 MPa.


Thus (with nonlinear material and geometry) if you yield at 235 MPa (perfectly plastic curve or with some small hardening modulus), then the largest stress (VM) should be around there, while if you yield at 300 MPa, then the largest stresses (VM) should be around there about (300 MPa) again.


This is as expected.

Schwarz18 posted this 02 December 2018

Hi jj77, thanks for your reply

Well, the value of the VM stress, considering same load conditions and boundary ones, should not be the same in the two cases?

I mean, if I have two different materials, one with a YS of 235 MPa and the other with a YS of, let's say, 300 MPa, applying the same conditions mentioned above, the results obtained shouldn't be consistent?

If during the first run I got a VM of 270 MPa, entering the plastic zone, I expect having the same stress in the second run but, this time, without entering the plastic zone since my YS is higher.

Is that correct or am I missing some point here?

Schwarz18 posted this 02 December 2018

Meanwhile I made some hand calculations based on the curve I have from the assumptions above and I found out something interesting.


The value I mentioned of the stresses (270 MPa and 340 MPa) were the VM ones and I found out they are simply referred to the values I would obtain from an elastic point of view. Indeed dividing the values ansys gave me as output by the elastic modulus, I would obtain the elastic part of the stress, named as elastic von mises stress. In other words in an elastic-plastic analysis with some hardening these values are, I think, of no use. The real values I should look at, instead, are the equivalent/von mises plastic strain and plastic equivalent stress.


Now said, my doubt still remains. With a YS of 235 MPa, I obtain a plastic equivalent stress of 237 MPa, while with a YS of 300 MPa, I obtain a plastic equivalent stress of 300.7 MPa, with the same load and boundary conditions.


How is this coherent with the argument of my post above?

jj77 posted this 02 December 2018

Vm stress or total fibre is used for elastoplast iterations.

Run it as I said with linear elastic material no plasticity and tell us the largest vm stress.

Schwarz18 posted this 02 December 2018

The max VM stress obtained from the analysis you mentioned is around 426 MPa.


So you were right about it! You say it's normal having different values of plastic equivalent stress, changing YS. But I still can't understand this point.


Because I think about what would happen in reality... I got some yielding with a S235JR, with that applied load, then I can think of using a different steel with an higher yielding stress, for example a S275JR. Why, then, my material would reach yield as well? Shouldn't I have, instead, a material that has a maximum stress around 240 MPa (like in the first case), using only a different steel?

jj77 posted this 02 December 2018

That is it. As explained my first point.

Easy to think abut it. Simple uniaxial test with 500 Mpa applied to it. So if you have yield at 300 Mpa then your max vm stress is caped at 300. While if you change yield thus material to yield at 400 then the largest vm will be 400. Finally if you changed again to 1000 Mpa then of course it does not yield and largest vm is 500 Mpa. That is it for me. Enjoy the weekend.

Schwarz18 posted this 02 December 2018

I think I got the point now


In other words ansys is telling me that if I apply 1400 N on a 200x200 mm area, with a S235JR I would reach yield zone, same as using S275JR and so on. If, instead, I use a material that has a YS > 430 MPa I would have a maximum stress around that value, however without ever reaching plasticity.


This makes sense. I was thinking, instead, that my output would be, every time, the real stress obtained, as you said, without plasticity. I was not thinking about the cap value set by choosing different YS.


Thanks for your replies and have a good WE as well!