I'm simulating a tube end forming process using Ls-dyna. The tube is made of AL6061-T6. The equivalent plastic strain (EPS) obtained by Ls-dyna exceeds 0.8 (which is the failure strain according to the stress-strain curve). Is that mean the process will fail??. I made the process experimentally, and it works. But on the simulation, the EPS is high (range:1.5 to 2).
Should I compare the In-plane strain to the FLD of AL?
If so, how can I get that FLD?
Dear Amr, the maximum plastic strain EPS that you can have is a function of the stress state. The value that you have in the monoaxial test is just one point of the so called fracture locus. This is basically a curve in which the maximum plastic strain at fracture is plotted as a function of the stress state (trough the triaxiality) and, sometimes, also as function of the load angle.
The shape of this surface is governed by model such as the Rice&Tracey one ore the Johnson-Cook.
For the majority of materials, such data are available in literature.
For furter explanation see my references.
Franco
PS: in the attachment the fracture locus of an alluminium A357 T6 alloy
Dear Franco,
Thank you so much for your reply. It's really helpful.
To make sure that I got it right.
You mean that the max EPS obtained from the FE-solver (Ls-dyna) is due to a state of stress that differs from that in case of mono-axial test, and consequently, we can not simply compare the max EPS from Ls-dyna with the 0.8 (the failure strain), and we should rather use a curve like the one you attached. Am I right?
If you please give more details on the attached curve.
I know the Y-axis is the failure EPS.
Does X-axis represent the triaxiality factor?
Can you explain more about triaxialilty factor?
What is the relation between Johnson-Cook and the attached curve?
Thanks in advance. @ Franco Concli
Dear Amr Shaaban
Do you have the static tensile/compressive, shear test results of AL6061-T6 material? If you have these results, you may reference to fracture strain. Based on this strain, you can select to critical equivalent plastic strain in your numerical analysis.
Amr Shaaban
" You mean that the max EPS obtained from the FE-solver (Ls-dyna) is due to a state of stress that differs from that in case of mono-axial test, and consequently, we can not simply compare the max EPS from Ls-dyna with the 0.8 (the failure strain), and we should rather use a curve like the one you attached. Am I right? "
Yes exactly!
Does X-axis represent the triaxiality factor?
Yes X-axis is the triaxiality! Is defined as the ratio between the idrostatic stress and the vM equivalent stress.
What is the relation between Johnson-Cook and the attached curve?
Johnson-Cook is the general form of the fracture locus. The curve is a simplified version that neglects the effect of the strain rate and the temperature.+
Franco
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