PEEQ and Maximum Principal plastic Strain

Good afternoon,

Based on a simple cube under compression and a True Stress-Strain Isotropic Hardening curve my goal is to understand properly the meaning of PEEQ in ccx and to obtain (later) the value of the “Maximum principal plastic strain”. inp attached.

I have run a push analysis (load driven) up to the maximum VM stress defined in the curve 541.6MPa. An additional 0.1 MPa makes it fail which is a good sign of agreement.

After the analysis is finished, I have plotted the INPUT True Stress-Strain curve against the RESULTS VM against (PEEQ + σprop/E Strain) obtaining a very good agreement.

I have extracted the Plastic Strain Tensor SDV2-7 + SDV1 from the .frd as reference.

I have computed the equivalent plastic strain according to the manual formula at the last step in three different ways and found that there is a difference compared with SDV1.

Strain tensor last step t =541.6:

[[ 6.4610e-02 4.9700e-18 -1.1700e-18]
[ 4.9700e-18 6.4610e-02 7.7000e-18]
[-1.1700e-18 7.7000e-18 -1.6243e-01]]

Frobenius Norm: 0.152167 .As expected from the manual (ccx 2.19 formula 332)
Check using Tensor dot: 0.152167
Check using Double contraction tra(DDT): 0.152167

Ccx SDV1 (equivalent plastic strain): 0.1391 < 0.152167

¿Where does this difference come from? Refinement doesn’t seem to change anything.
¿Is ccx using a different definition of the norm (2-norm?¿?) or maybe is the PEEQ evaluated at the Gauss points and later extrapolated ?


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I think that the formula holds only for small strains which is not the case for last step…try checking it at 0.030 or so…Bathe in his book sets a cutoff of about 4% strain so it can be called “small deformation”

Thanks for your help Juan,

I have proceed as suggested. Attached is the deviation computed as (SDV1-epeq)/ SDV1 *100
It increases linearly up to a 9% approximately. I really don’t know how to proceed now.

¿Am I misunderstanding the formula to compute SDV1 or SDV (2 to 7) is not the Plastic strain tensor I should use to compute it?

My goal was to obtain the maximum principal plastic strain. PEEQ was a previous verification step to be sure I was extracting those values from the right tensor.

Something to pay attention.

This problem (.inp) doesn’t have shear strain.

The strain tensor is almost diagonal. SDV2 , SDV3 and SDV4 are virtually the Principal Plastic Strains.

Principal Strain 1= Principal Strain Elastic part+ SDV2

Principal Strain 2= Principal Strain Elastic part + SDV3

Principal Strain 3= Principal Strain Elastic part + SDV4

Principal Strain Elastic part is negligible compared to the Plastic part.

I have plot:

Principal Strain 1 versus SDV2

Principal Strain 2 versus SDV3

Principal Strain 3 versus SDV4

I would expect the graphs to almost overlap.

That’s true for Strain 1 and Strain 2 but the Strain 3 shows something that I think it is very suspicious.

Principal Plastic strain 3 is bigger than the corresponding Principal Strain 3. The part can’t be bigger than the total.

but the manual says that SDV variables are NOT the plastic strain tensor (approximate IF small strains), the formula for strains is
but you equal eps_p to SDV and that’s contrary to the manual. You should calculate eps_p=eps-eps_e and then you get the plastic strain tensor and can calculate derived quantities.

Hi and thanks again for your help,

There are new things to me, and theory is rough without some help.

The formula you show I understand is valid for NLGEOM=OFF where the total strain rate tensor is additive decomposed into an elastic and a plastic part.
Now I understand there is a limit of validity for that approximation to small strains. The graph agrees very well in that sense with your announced 4%.

¿Where could I find or how could I compute the elastic part I need to substract?
I would like to replicate the PEEQ value (although I assume that its value could be outside the range of validity)

By other hand I have switch the NLGEOM=ON. That means a nonlinear strain tensor is used. Now there is not an additive decomposition but Multiplicative.
I was expecting an extension of the validity range to larger strains but I get the same deviation when comparing SDV1 by hand calculation and strange to me SDV4 is still > Principal Strain 3.

Hi ,

I 'm going through the manual in detail. Apart of it I have look at other source and I’m guessing if the manual has not lost some dots on top of the strains to denote their derivative nature.

I’m referring to the starting point where the differential equations are stated. ¿Isn’t the additive decomposition of the elastic and plastic strains referring to the elastic and plastic strain rates?.

¿Souldn’t the formula shown on the manual sqrt(2/3)*||ep|| refer to the equivalent plastic strain rate?. Seems there is an additional integral over the time to get the equivalent plastic strain.

I think that SDV2-7 are the components of the plastic strain rate tensor, and the manual is missing the dot on top of e. I guess now I understand better why SDV1 is called accumulated (integral) equivalent plastic strain.

-According to the manual equivalent plastic strain PEEQ is a logarithmic/true strain.
-According to formula (388) manual ccx 2.20 that equivalent plastic strain is build from the infinitesimal plastic strain tensor ep . That suggest ep is a logarithmic strain tensor too. ?¿?¿
-According to formula (382) manual ccx 2.20 ep is shown as part of the total strain tensor which in this case is a Lagrangian strain tensor ?¿?


Seems the plastic strain tensor is indistinctly added in two different frameworks.
I would kindly ask if someone could help me to clarify about ccx Output for SDV from 2 to 7 (infinitesimal plastic strain tensor). ¿Is it a Lagrangian , Eulerian or Logarithmic strain tensor?

Thanks in advance.