PEEQ and Maximum Principal plastic Strain

[Disla]

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 ?

Thanks.

DNV-RP-C208_Materials_Test_SDV.inp

[JuanP74]

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”

[Disla]

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.

[Disla]

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.

[JuanP74]

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.

[Disla]

@JuanP74

approximate IF small strains

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.

[Disla]

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.

[Disla]

-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.

[vicmw]

@Disla did you ever make sense of this? It’s all pretty confusing to me but I thought I could reproduce PEEQ using the regular strain output (E) on a single element unit cube with axis-aligned strain like this:

1. Convert the Lagrangian strain tensor from CCX to Logarithmic strain. Shear is zero.

el_xx = ln(sqrt(e_xx * 2 + 1))

el_yy = ln(sqrt(e_yy * 2 + 1))

el_zz = ln(sqrt(e_zz * 2 + 1))

2. Use that logarithmic strain in the von Mises formula to obtain equivalent strain.

equivalent strain = sqrt( ((el_xx - el_yy)^2 + (el_yy - el_zz)^2 + (el_zz - el_xx)^2) / 2 )

With negligible elastic strain, I think that should equal PEEQ. It is roughly proportional to PEEQ but off by a factor of about 2 for some reason I can’t understand.

[Disla]

Hi Victor,

The formula for the equivalent Von Misses Strain involves a coefficient of 2/9 not 1/2 as Von Misses equivalent Stress. (Additional 1/(1+0.5) in front)

It is valid in the small strain deformation regime. It disregards the volumetric change (It focuses in the deviatoric part of the Strain Tensor). If I recall properly ccx Nonlinear algorithm is isochoric so if I’m not wrong our Strain Tensor is deviatoric itself.

Mecway custom formula should be :

sqrt( (( log(sqrt(2e.xx+1))-log(sqrt(2e.yy+1)) )^2+(log(sqrt(2e.yy+1))-log(sqrt(2e.zz+1)))^2+(log(sqrt(2e.zz+1))-log(sqrt(2e.xx+1)))^2)*2/9)

[vicmw]

Scaling it by 2/3 (= 1/1+0.5) doesn’t get agreement.

I don’t understand the small strain requirement. Isn’t all plastic strain not “small”? Regardless, the curves don’t seem to be converging towards each other as strain gets smaller. I did it again with nu=0.499, approximately no change in volume and the results are about the same:

[Disla]

Hi Victor I’m using this inp as reference.

For validation purposes of the formula the material definition has a zero Yield Stress. I do that just in this particular case because I’m interested in comparing PEEQ with the formula so I get rid of any elastic strain. The elastic part of the strain tensor is zero.

*PLASTIC
0,0
500000000,1

This is the abstract of the expected Strain measures for this model. I’m imposing a maximum enginieering strain of 1 (imposed tensile uniaxial displacement of 1 mm) same as you. Make sure you are imposing a tensile stress state not compressive. As all my strains are plastic this is going quite far but …I expect PEEQ=Logarithmic Strain=0.69315 in this particular set up (All strain is Plastic).

Results. Maximum deviation is 0.23% for the whole range.

** Generated by Mecway 30
*NODE
1,0,0,0
2,0,0.001,0
3,0,0.001,0.001
4,0.001,0,0
5,0.001,0,0.001
6,0,0,0.001
7,0.001,0.001,0.001
8,0.001,0.001,0
*ELEMENT,TYPE=C3D8
1,1,4,8,2,6,5,7,3
*ELSET,ELSET=MATERIALTEST
1
*ELSET,ELSET=EALL
1
*BOUNDARY
1,1,,0
1,2,,0
1,3,,0
2,3,,0
2,1,,0
3,1,,0
4,3,,0
4,2,,0
5,2,,0
6,1,,0
6,2,,0
8,3,,0
*AMPLITUDE,NAME=A_1
0,0
1,-0.001
*material, name=steel
*ELASTIC,
210000000000.0,0.3
*PLASTIC
0,0
500000000,1
*density
7850

** material assignment to bodies
*solid section, elset=Eall, material=steel
*STEP,NLGEOM=YES,INC=1000,AMPLITUDE=STEP
*STATIC
0.001,1,0,0.01
*BOUNDARY,AMPLITUDE=A_1
3,3,,-1
*BOUNDARY,AMPLITUDE=A_1
5,3,,-1
*BOUNDARY,AMPLITUDE=A_1
6,3,,-1
*BOUNDARY,AMPLITUDE=A_1
7,3,,-1
*NODE FILE,GLOBAL=YES
U
*EL FILE
PEEQ,E,S,NOE
*END STEP

[vicmw]

Wow that’s much better agreement. I’ll try to reproduce it. Is “EQUIVALENT STRAIN FORMULA” the one from your previous post?

sqrt( (( log(sqrt(2e.xx+1))-log(sqrt(2e.yy+1)) )^2+(log(sqrt(2e.yy+1))-log(sqrt(2e.zz+1)))^2+(log(sqrt(2e.zz+1))-log(sqrt(2e.xx+1)))^2)*2/9)

[Disla]

Yes it is. Copy Paste

sqrt( (( log(sqrt(2*e.xx+1))-log(sqrt(2*e.yy+1))  )^2+(log(sqrt(2*e.yy+1))-log(sqrt(2*e.zz+1)))^2+(log(sqrt(2*e.zz+1))-log(sqrt(2*e.xx+1)))^2)*2/9)

Deviation:

(sqrt( (( log(sqrt(2*e.xx+1))-log(sqrt(2*e.yy+1))  )^2+(log(sqrt(2*e.yy+1))-log(sqrt(2*e.zz+1)))^2+(log(sqrt(2*e.zz+1))-log(sqrt(2*e.xx+1)))^2)*2/9)-peeq)/peeq*100

[Disla]

Regarding why it is only valid for small strain deformation regime note my plastic flow is linear. That’s why the agreemen is so good in the validation file. It is intentional. For the general case scenario, where the plastic flow is non linear you need to integrate .

To visualize and simplify the problem , you can imagine the plastic Strain as the lenght of the curve. The more it deviates from a straight line the larger it will be and more deviation you will find with PEEQ if you use the aproximation it represents to use that formula.

[vicmw]

Got it. I had too much elastic strain, now my results match yours.

It seems like the rate of change of plastic strain doesn’t have to be linear (is that what you meant?) but it does have to be monotonic. If you reverse direction, PEEQ continues increasing monotonically while this formula decreases by the same amount. So this integral you mention, is that just the time integral of the equivalent plastic strain formula applied to the differential strain? Then an increase or a decrease would look identical.

[Disla]

Hi Victor,

@vicmw

reverse direction

“Reverse” word could be missread here. “Reverse” in a displacement driven test means imposing a new displacement path with it’s corresponding generation of new plastic strain (regardless the path ovelaps the previous one). Equivalent Plastic Strain should always be monotonically increasing.

@vicmw

If you reverse direction, PEEQ continues increasing monotonically while this formula decreases by the same amount

The strain values we see in the GUI e.xx ,e.yy and e.zz used for the formula disregard where the model comes from or where it has pass through. It shows the actual Strain state based on initial lenght. That’s why the formula returns a final zero value with reversal.

@vicmw

So this integral you mention, is that just the time integral of the equivalent plastic strain formula applied to the differential strain?

The only way I saw to solve it with a straight formula was with a straight line path. Maybe that condition could be relaxed as you say. Have not tested. (I think that would not work for a Strain *rate dependant plasticity )

For a more general case including sign reversal on strains I would add all the small increments of equivalent plastic strain along the arbitrary path. In discrete form:

Note all the increments are added in absolute value. Not sure if this is very rigurous or one need to compute it directly inisde the formula for each single component as (e.xxi - e.xx i-1 …….)

That can’t be written as a unique custom formula because it uses the previous state. Maybe you can find the way with some script.
With Excel I have try a simple case and there is a good agreement even with the displacement reversal. I’m Using a 1/2 sinus wave function BC to stretch and return to the initial position.

Regards

[Disla]

This is the inp.

I have introduce some variation to find how flexible is all the exposed above.

Sinus amplitude for BC to have sign reversal.

Nonlinear Stress Strain Curve.

Overal time step =2.

PEEQ RESULT Directly from ccx.

Results WITH FORMULA ADDING INDIVIDUAL INCREMENTS Excel resume:

** Generated by Mecway 30
*NODE
1,0,0,0
2,0,0.001,0
3,0,0.001,0.001
4,0.001,0,0
5,0.001,0,0.001
6,0,0,0.001
7,0.001,0.001,0.001
8,0.001,0.001,0
*ELEMENT,TYPE=C3D8
1,1,4,8,2,6,5,7,3
*ELSET,ELSET=MATERIALTEST
1
*ELSET,ELSET=EALL
1
*SURFACE,NAME=DISP_DRIVEN
1,S2
*BOUNDARY
1,1,,0
1,2,,0
1,3,,0
2,3,,0
4,3,,0
4,2,,0
8,3,,0
*AMPLITUDE,NAME=A_1
0,0
0.061400307,-9.629791909631E-05
0.106400532,-0.0001663565430627
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1.938609693,-9.628228412304E-05
2.00001,1.570796326712E-08
*material, name=steel
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19404.8,0.02
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150553.6,0.04
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249980595.2,0.98
249997537.35,0.99
250000000,1
*density
7850

** material assignment to bodies
*solid section, elset=Eall, material=steel
*STEP,NLGEOM=YES,INC=1000,AMPLITUDE=STEP
*STATIC,SOLVER=PARDISO
0.01,2.00001,0,0.01
*BOUNDARY,AMPLITUDE=A_1
3,3,,-1
*BOUNDARY,AMPLITUDE=A_1
5,3,,-1
*BOUNDARY,AMPLITUDE=A_1
6,3,,-1
*BOUNDARY,AMPLITUDE=A_1
7,3,,-1
*NODE FILE,GLOBAL=YES
U,RF
*EL FILE
PEEQ,E,S,NOE
*END STEP

[vicmw]

Cool, seems like it matches well. I guess for a general tensor, there could be a difference between

sum ( | eps_i - eps_i-1 | )

and

sum ( eps(e_i - e_i-1) )

where eps() is the von Mises function, but either way, it’s pretty reassuring that it makes sense now

Thanks a lot.

[Disla]

You are welcome.

An important comment here.

We have simplified the problem as much as possible for validation of the formula. In this case we can just use e.xx , e.yy and e.zz because all shear strains are zero in a pure tensile test.

That formula should be using the three principal strains e1, e2 and e3 to be more general and consider the plastic strain generated by the shear components too.