Applying Stress-Strain curve in Abaqus 4

[yaston4]

Hi,

I would really like to clear this up I have some stress stain curve data as follows:

stress (MPa) %
0 0
567 0.0008
600 0.021
640 0.053
690 0.11
680 0.1022

Can I apply this data directly into Abaqus? It seems that the first values can not be zero. I have also defined a elastic modulus and Poisson's ratio.

Thanks

 

[yaston4]

Apologise the raw data is as follows:

stress (MPa) %
0 0
567 0.08
600 2.1
640 5.3
690 11
680 10.22

 

[burningDiesel]

Easiest is to define linear elastic material with E mod from
0 0
567 0.08
Then define plastic behavior with curve from:
600 0.021
640 0.053
690 0.11
680 0.1022

But why does your strain drop at the last value?

 

[TGS4]

sdebock has given you the (mostly) correct advice. The one small correction that I would add is that your data is in total strain. You need to separate plastic strain from elastic strain. Input plastic strain in the plastic material definition, not total strain.

Another piece of advice, make sure that you are inputting true stress-true strain curves, and not engineering stress-engineering strain curves. ABAQUS is designed to input the former, not the latter.

 

[yaston4]

Is there any standard or guidance on seperating the total, elastic and plastic strains? I have found some information saying that:

total strain : sigma / E
plastic strain : .002
elastic strain = total - plastic

Also for the true stress/trus strain I have found some information here, is this the correct procedure for converting engineering to true data?

http://www.drd.com/techsupport/eng_true_strain.htm

Thank-you for all you input and help.

P.S. sdebock, the drop off at the end is the fracture point, it would probably be better to leave this piece of data out.

 

[burningDiesel]

I indeed was too quick to reply regarding the plastic strains.
And yes, leave out the fracture point, and website seems to give the correct formulas.

 

[Drej]

It's not always necessary to input true stress-strain data, as this depends on the magnitude of strain seen in the analysis. For most engineering steels (for example) and where the response is of the order of less than say 5%, it is reasonable to use engineering stress-strain data.


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[yaston4]

After doing the calcs from "engineering" to "true" the values seem similar, but not exactly the same.

Still need to clear one last thing about TGS4's comments

You need to separate plastic strain from elastic strain. Input plastic strain in the plastic material definition, not total strain.

Can anyone shed some more advice on this and how this should be done in good practice?

Thanks

 

[MechIrl]

You need to determine a yield stress for you material beyond which you are assuming plastic behaviour. The easiest option is then to determine the strain at the yield stress based on your assumed Young's modulus. This is the elastic strain, which can then be subtracted from your total strain values to give plastic strains. Obviously the elastic strain will usually be quite small.

 

[yaston4]

Thanks MechIrl,

So these equations hold true?

total strain : sigma / E (at yeild point)
plastic strain : .002
elastic strain = total - plastic

 

[MechIrl]

Not for the method I have suggested. You should bear in mind that when you are modelling the stress strain behavior with the linear elastic-plastic law in abaqus you are assuming an initial linear elastic behavior, immediately followed by plastic behavior. Of course this doesn't really happen in many metals (I assume you're modelling a metal) as around the yield strength material can become non-linear elastic and have a gradual transition to plastic behavior. One approach is to use something like the 0.2% offset yield strength as your transition point (I was purposely vague in my last answer regarding how to get Yield Stress as in some cases the 0.2% offset strength may not be appropriate).

So, you will have:
elastic strain = Yield Stress/E
plastic stain = total strain (the strain in your raw data) - elastic strain

Clearly the elastic strain is not necessarily exactly 0.002 if using the offset method to get Yield Stress.