Von Mises Stress Strain Discrepancy 3

[jcc2000]

I just completed a non-linear analysis. The following are the inputs for the material:
E=200GPa
v=0.27
Von Mises Criteria, Initial Yield 248MPa
Plasticity Modulus 500MPa

The output stress and strains do not match the bilinear stress strain curve. For example I am getting 400MPa of Von Mises stress and 0.002000 Von Mises Strain. If I am 152MPa of stress past yield, shouldn't the strain be 152/500 = 0.300000?

 

[GBor]

Oh the many questions to ask:

What software are you using?
Where is the stress located?
What type of deflections are you getting?
What is this a model of?
How are parts connected or is it a single part?
Are you looking at total strain, or directional?

Can you provide MUCH more detail?

Thanks,

 

[EdR]

"If I am 152MPa of stress past yield, shouldn't the strain be 152/500 = 0.300000?"


Only if you are looking at a case of uniaxial stress....

Ed.R.

 

[johnhors]

Don't stress-strain curves usually plot uniaxial stress against strain, not Von Mises?

And anybody know what "Von Mises Strain" is ?

 

[meshparts]

I agree with EdR and Johnhors.

But supposing jcc2000 has the approximately the case of uniaxial stress, then I would have a correction for the equation given by jcc2000.

So jcc2000 gives us the following Data:

E_e=200e9; % Elasticity Modulus
E_p=500e6; % Plasticity Modulus
sigma_y=248e6; % Yield Stress
sigma=400; % Computed Stress
eps=2e-3; % Computed Strain

The yield strain can be computed as:

eps_y=sigma_y/E_e; % Yield Strain

Using the provided date the yield strain is equal:
eps_y=1.2e-3;

Since jcc2000 is using a bilinear Stress-Strain relationship, the equation in the plasticity domain is:

sigma_th=E_p*(eps-eps_y)+sigma_y;

where sigma_th is the theoretically computed stress. So sigma_th can be computed now:

sigma_th=248.38e9; % Means 248.38 GPA

The computed value of stress (400 GPa) is much more grater than the theoretical value (248.38).

So either my calculations are wrong or in my opinion the computed value of equivalent stress should be much smaller, in a case of almost uniaxial stress of course...

Regards
Alex

 

[meshparts]

Correction:

sigma_th=248.38e6; % Means 248.38 MPA

Regards
Alex

 

[crisb]

As JohnHors has pointed out there is no such thing as von mises strain. It is misleading for your program to offer such a quantity and it looks like it is simply dividing the von Mises stress by the elastic modulus.
Maybe its time to get another FEA program?

 

[hondaknight1]

Hmmm crisb with all due respect I dunno if I agree with you completely. Whilst it is obvious by inspection that the "von mises strain" quoted above does equal the calculated von mises stress divided by E. But there is a term that is used in FEA, fracture mechanics, and materials engineering called the "equivalent von mises strain", have a look in the Abaqus, Ansys, Algor, Diana and many other user manuals for instance, and I can put my hand on a few text books here in my office that define it, and I still ahve lecture notes from my degree course in MAterials Science and Engineering which also define it. Its normally derived from the complete strain tensor in a similar way that the equivalent von mises stress is derived from the stress tensor. A link to a definition is here

http://eng.snu.ac.kr/service/on-line-documents/diana/Analys/node362.html

Its true it may not have been defined by Richard Von Mises himself, but I think it has become common parlance.

So I read johnhors post as a question not of wether it existed but wether anyone here could define it?

The result above could well be coincidental and the FEA package may be calculating it correctly, we cannot really comment as we do not know the model..

 

[meshparts]

Crisb,

what jcc2000 means, is the "equivalent strain". The equivalent strain is defined similar to the "equivalent stress" or "von Mises stress". The equation for the equivalent strain is very similar to the equation for equivalent stress.

Unfortunately I can't advise you to any solid literature in which the term is described. One link is the Chapter "19.12.1. Physical Interpretation of Equivalent Strain" of the ANSYS Theory Reference Manual.

By searching in Internet I have found, that the concept of "equivalent strain" is being also used in Abaqus.

Regards
Alex

 

[rb1957]

pretty soon we'll get back to that old saw ... "there isn't anything called von mises stress, von mises defines a failure criteria".

looking at the numbers posted it looks to me as tho' the program as applies the elastic modulus ... 400MPa/200GPa = 0.002.

but anyways, doesn't von mises determine an equivalent uni-axial stress for a given complex stress state, to be compared with the material strength (from uni-axial tests) ?

 

[crisb]

Effective stress, equivalent stress, and von Mises stress are widely used, and understood to be the same thing. I am not sure that the same applies to the various descriptions of strain used by different programs. Maybe the only safe course is to check the definition of the variable with the software vendor.

For a nonlinear analysis with nonlinear material there is also the possibilty of unloading and a change in the yield stress, so it extra important to have the correct variables when postprocessing. Effective strain, effective plastic strain, accumulated effective strains will usually be different and probably called different things in different programs.

Its possible that the answer given by the OP was correct if there was an unloading cycle and the output is an "instantaneous" strain or something of the sort!

 

[hondaknight1]

rb157 with all due respect but your statement "there isn't anything called von mises stress, von mises defines a failure criteria" is just not true!

There is a stress you can calculate called the von mises stress, which tells you the equivlent uniaxial stress, so you can then compare that stress to the uniaxial yield stress. Without scanning and sending you a page from my trusty text book try looking at this website :

http://en.wikipedia.org/wiki/Von_Mises_yield_criterion

It tells you all about it..

 

[meshparts]

rb1957,

I will give you a star, but not for your first phrase but for the second one:

"looking at the numbers posted it looks to me as tho' the program as applies the elastic modulus ... 400MPa/200GPa = 0.002"

This could be the solution for the problem of jcc2000. Hopefully he/she can confirm your statement.

Secondly: What program are you using anyway, jcc2000? Have you defined the nonlinear analysis?

Regards
Alex

 

[hondaknight1]

mihaiupb and RB157 I have a feeling both of you you must have not read any of crisb or my posts, because he said this "simply dividing the von Mises stress by the elastic modulus" which is exactly what rb157 said and I agreed with him...

 

[meshparts]

You are right, hondaknight1, I have not read your post because we both have answered at the same time (12 Aug 08 7:27) and my browser shows my post after yours. And I have looked for new posts just after my last reply... Sorry, just a technical problem...

And you are also right, crisp deserve a star too, though I cannot agree with all his statements...

Alex

 

[jcc2000]

Thanks for the posts. HondaKnight1 your link to the formula for equivalent stain was especially helpful. I pulled some nodes from my model for max, min, mid and von mises stresses and strains (see below). Then applied the formula for stress and strain. The handcalcs matched pretty closely to the model. I understand now the discrepancy between the von mises stress and von mises strain.

However I am now puzzled about the high stresses to the low strains, for both max, min and von mises. My failure criteria is von mises with an initial yield of 248MPa. How can the stresses be that high 366MPa von mises node 3 (exceeds tensile of A36) strain 0.00184194 (Not close to tensile).

I think the core issue for me is I'm not sure what happens when the stresses (max, min, von mises?) exceed the initial yield stress? The math should answer that, but I don't know what that is.

In reference to some of the other posts:
The .002 strain and 400MPa stress was just coincidental that it jives with the elastic modulus (See the table belwo for some examples). Good eyes though.
I am using the following program:
NEiNastran/FEMAP
Single load (no cycling)


Strain Stress
Node Mpa
Max Prin 1 0.00155839 283.2418
2 0.000778489 65.01225
3 0.00113718 100.9485
4 0.000894366 85.2513

Mid Prin 1 -0.00031123 -7.470545
2 0.000257073 -13.52022
3 0.000296915 -12.61719
4 0.000333001 0.887024

Min Prin 1 -0.0008067 -84.26731
2 -0.0015611 -279.9757
3 -0.0019398 -308.149 Hand Calcs
4 -0.0017462 -311.7101 Strain Stress
Mpa
Von Mises 1 0.00144031 335.7741 0.001455154 335.763575
2 0.0014214 313.884 0.001439714 313.1952724
3 0.00184194 366.9315 0.001851875 365.7847599
4 0.00160649 362.2747 0.001624807 362.2241337

 

[crisb]

Just a thought.
If you are using 3D elements it might be worth checking unsmoothed values at integration points. I think you will find that these are more consistent.
Smoothed values and values at nodes will probably be extropolated from the stress or strain at the integration points, so are not "exact".

 

[rb1957]

now i'm puzzled ... no 400 MPa, maybe a different element ??

more puzzling (on the face of it) are the elastic stress/strain ratios (not equal to 200GPa) ...
65/.0007785 = 83.5 GPa
101/.001137 = 88.8 GPa
(near enough given the accuracy of the input)
nowhere near 200 GPa

another thought would be to print the intermediate steps ... the algorithim should determine the onset of yield (at some %age of your applied load) and then slowly step up towards your required load.

maybe having a nearly perfectly elastic/plastic material (E = 200GPa, Ep = 0.5GPa) is a problem for the code ??

 

[hondaknight1]

If its not the smoothing thing I think I know the answer to this, I am just trying to work out how to explain it...

 

[jcc2000]

Turns out the root cause was too coarse of a mesh. I plotted only the intergraton points and the stresses were in line with the stress-strain curve. I refined the mesh and the stresses started to approach 248MPa.

In the original run, the nodal stresses and strains were off by as much as 47.5%. Using an intergration point of 248MPa/.00124strain which is just strating to yield being off %47.5 on strain only gives 0.00183. For this analysis, hardly anything to be concerned about. For stress %47.5 gives 366MPa which raises a red flag.

The fact that the principals that are fully elastic don't line up with the elastic modulus I would attribute to the model being multi-axial loaded and with nodal averaging.

Thanks for the help.