Strain-strain concave upward-strain hardening, unlike steel.

[Mohamedsayed]

Hello

I am working on a material whose stress strain curve is strain hardening (concave upward), unlike the steel whose strain harening is concave downward.

the problem is when i calculate the equivalent plastic strain, it is negative. the following table represents the true stress-strain curve for the material that i am talking about.

0 0
143.4108527 0.1
405.1094891 0.2
637.9310345 0.25
768.8558235 0.27
1034.161491 0.3
1149.298597 0.31
1485 0.33250487

and this table represents the yield stress vs. Abs.plastic strain.
142.1994 0
169.8456962 -0.004242229
273.8821714 -0.032129522
445.4397035 -0.101488562
661.5232314 -0.208943057
912.2319088 -0.351022659
1183.168317 -0.515740219
1469.77675 -0.699728928
1485 -0.70950104

When i do that, i got error message that the value should be in ascending order.

Do you know how can i solve that problem.
Thanks

 

[Mohamedsayed]

I am posting the first and the second tables again because i noticed they are not clear. this table represents the stress-strain curve.

Stress strain
0 0
143.4108527 0.1
405.1094891 0.2
637.9310345 0.25
768.8558235 0.27
1034.161491 0.3
1149.298597 0.31
1485 0.33250487

This is the second table:
Yield stress Abs. plastic strain.
142.1994 0
169.8456962 -0.004242229
273.8821714 -0.032129522
445.4397035 -0.101488562
661.5232314 -0.208943057
912.2319088 -0.351022659
1183.168317 -0.515740219
1469.77675 -0.699728928
1485 -0.70950104

 

[IceBreakerSours]

I don't have background in plasticity but because of the negative signs, your plastic strains are in a descending order. Abaqus doesn't seem to like that.

Are you new to this forum? If so, please read these FAQ:

http://www.eng-tips.com/faqs.cfm?fid=376

http://www.eng-tips.com/faqs.cfm?fid=1083

 

[burningDiesel]

Question is, why did you put negatives?
why don't you just put:
143.4108527 0.0
405.1094891 0.1
637.9310345 0.15
768.8558235 0.17
1034.161491 0.2
1149.298597 0.21
1485 0.23

And do you have data between 0 and 0.1 strain?

 

[IceBreakerSours]

Precisely.

Also, you can't be serious about those decimal places! As if there is no uncertainty in the rest of the model! Come on. Two/three sig figs should be sufficient.

Are you new to this forum? If so, please read these FAQ:

http://www.eng-tips.com/faqs.cfm?fid=376

http://www.eng-tips.com/faqs.cfm?fid=1083

 

[mrgoldthorpe]

Your material obviously stiffens with increasing strain. Classical metal plasticity shows irrecoverable plastic strain beyond the yield point. It's an entirely different type of material behaviour.

So you can't use a classical metal plasticity law for this material. You need to use a more appropriate non-linear stress-strain law. I suggest you look at the various ABAQUS User Manuals for something appropriate.

(btw - there's no such thing as negative equivalent plastic strain.)

 

[Mohamedsayed]

Hello,
First of all, sorry about all numbers after decimal places, i just cut the numbers from the excel sheet.
I am not using metal plasticity, i am using Drucker Prager plasticity because i am working on rock not metal. In Drucker prager i need to define the yield stress as function of the plastic strain, i guess it is similar to metal plasticity, however i need to define other parameters such as Cohesion and friction angle of the rock.

The reason that i can not put the Abs. plastic strain in positive because the stress-strain curve concvaves upward unlike steel which concaves downward. Please see the excel sheet to see the stress-strain curve.

So if put Abs. Plastic strain in positive, the behavior of the material is different. Actually it is negative.

Please advice me.
Thanks

 

[Mohamedsayed]

I used the following equation to convert the true strain to true plastic strain which should be used in Abaqus.

e(plastic strain)=e(true strain)-(true stress/Young's Modulus).

If you use this equation, you will find that the equivalent plastic strain will be negative

for example:
e(plastic strain)=0.2 - (405/1434) = -0.82
etc.
Please help me, how can i overcome this negative plastic strain.

Thanks

 

[burningDiesel]

You should have stated you are modeling rock and using drucker-prager in your first post.
Furthermore, normal drucker prager wont do what you want.
Where/how did you get the test data?
Probably what you are seeing is an increase in stiffness because of an increase in density (because of an increased pressure).
The modified Drucker-Prager/Cap plasticity model should better suit your needs.

Also about your previous posts, you keep mentioning absolute strain, then why do you put the sign in front?
Maybe just taking the absolute would work, but I doubt it.

 

[Mohamedsayed]

Thanks Sdebock for your comment, and sorry about not mentioning that i am using linear Drucker prager plasticity model.

the Data that i sent before in the excel sheet is a published paper. and many people use it to calibrate their model, because it is very reliable.

So let me clear this time, because i was not clear in the previous posts. Modified Drucker-Prager/Cap can simulate material whose strain hardening curve as shown in the excel file i sent.

Thanks.

 

[MechIrl]

You're in a world of problems here....

First your stresses are similar to your elastic modulus so the additive decomposition (e=e_elastic+e_plastic) is a no-no. Second, your elastic strains seem to be large, so you can't use the built-in Abaqus plasticity models with a linear elastic material. Maybe the permanent-set model will do what you want, or else you're in UMAT territory.

 

[MechIrl]

Sorry, I was assuming the classical plasticity model. Maybe the pressure contribution in the DP model will keep the elastic strains small, but you definitely can't use e=e_elastic+e_plastic approach alone as you are doing for calibration.

 

[Mohamedsayed]

Thanks MechIrl,

although the first part of the stress strain curve is linear, but it does not represent an elastic behavior, i just assume that the first part is elastic so i can use elastic-plastic criterion in Abaqus.

This stress-strain curve represents the compaction of rock cuttings, so as you apply more load the voids decreases and stiffness increases.

I agree with you that we can not used (e = e (elastic) +e (plastic)