Why Abaqus give incorrect result in plane strain plasticity?!!

There is an incorrect result from Abaqus in plane strain plasticity simulation. In this short video, this  result is shown.

It would be pleased if anybody have any idea about why this happen ?!!

http://youtu.be/ERa4hVegtL4

Dear amin

Dear amin

I'm not an expert in abaqus, but your example is interesting, as you mentioned in your video the the strain is not physical....

Is there any contour of equivalaent platic strain?

Was your solution implicit or explicit?

because large substep explicit  gives wrong answeres.

Dear  aslan

Dear  aslan

As it is obviously shown in that simplest example, the reason is completely mathematical. We derived and developed continuum equations for plane strain plasticity and implemented in a Fortran code to resolve this fault.

For the Abaqus simulation, the implicit method was used. However, you can also decrease the time step, but you will never get the right results.

Dear amin, I think I found why abaqus gives you strange results. Since you have turned on nlgeom in your solution then your plastic formulation is not the same as small deformation theory.

If we neglet the plastic spin then the yield stress condition can be stated as folllows for large deformation theory:

$\sqrt[]{\left \| \frac{3}{2} \Xi^{dev}_{s}:\Xi^{dev}_{s} \right \|}-\sigma_{y_{0}}-R=0$

where $\Xi^{dev}_{s}$ is deviatoric part of symmetric mandel stress tensors and  G_y_0 is initial yield stress and R is isotropic hardening parameter. for isotropic materials the above equation can be reduced to:

$\sqrt[]{\left \| \frac{3}{2} \bar{\tau}:\bar{\tau} \right \|}-\sigma_{y_{0}}-R=0$

where $\bar{\tau}$ is spatial kirchhof stress tensor and the bar on it shows rotated kirchhof stress. since rotation has no effect on above formulation we can rewrite:

$\sqrt[]{\left \| \frac{3}{2} \tau:\tau \right \|}-\sigma_{y_{0}}-R=0$

and since the relation between Cauchy stress and Kirchhof stress is:

$\tau =J \sigma$

So it can be said:

$J|\sqrt[]{\left \| \frac{3}{2} \sigma:\sigma \right \|}-\sigma_{y_{0}}-R=0$

where $\sigma$ is Cauchy stress and J is jacobi of deformation gradient. The equivalent von Misses stress in abaqus is based on this formulation:

$\sqrt[]{\left \| \frac{3}{2} \sigma:\sigma \right \|}=S_{von}$

So the yield condition can be stated as follows:

$J S_{von}-\sigma_{y_{0}}-R=0$

The reason why you estimated unphysical large plastic strain is that you ignored J in your formulation.

We're not guess,

First of all, I want to thank you for your comment, but it's not reasonable. You just mentioned the simple relation of Mises plasticity theory. In computational plasticity, for computing the trial stresses and implementation, You can find clearer relation in detail (e.g. Simo and Hughes- Computational inelasticity).

Next, you must use it in FEM code to understand what is occurred in computation procedure.

It is worth to mention that the compatibility between stress and strain at This particular situation (Plane strain plasticity) in Abaqus is not True.

First apologize me for typing mistakes, all above stresses should be written in deviatoric parts. It was my first time to write equations in latex and html and it confused and distracted me a lot.
Thanks for your guide about trial stress and FEM application in large deformation plasticity, but I have written J2 large deformation plasticity code before, so please trust me ;-). The point is not the simplicity of yield condition, the point is that whether you have approximated the equivalent plastic strain correctly or not.

In your video you have written below equation to evaluate equivalent plastic strain:

3.16e8=2.5e8+(1e6)*eps

then you have concluded that eps=66 . Actually you mean:

S_von=sig_y_0+K*eps

where K is the isotropic hardening parameter. Your formulation is not reasonable since it is for small deformation theory. You have ignored Jacobian of deformation gradient which is wrong, You should approximate the equivalent plastic strain from this formulation which is better approximation for large deformation theory:

J*S_von=sig_y_0+K*eps

In small deformation theory J or Jacobian is nearly equal to one so the effect of nonlinear geometry is not clear.

For checking the correctness of Yield condition that I wrote, please check Simo and Hughes 9.2.14 and 8.2.1.

regarding to my last post,

regarding to my last post, You can check the abaqus stresses kind in plot result by simple tension example of a rod, if you check it correctly you will realize that abaqus shows cauchy stresses.

We all know

It seems that I had better consider it. However, considering J in this simple calculation does not make things OK (such as fictitious stresses).

We all know that the Abaqus give cauchy stresses as a result in current space. It should be pointed that output stresses must be according to the input strain-hardening, but it is not True for near rigid plastic material. You can check it out.