2
$\begingroup$

I'm doing FEA of steel under high strain rates and using Elasto-ViscoPlastic material model, with Von-mises yield criterion along with Isotropic hardening. The strain rate sensitivity is addressed by Current yield Norton law.

Can anyone please outline how do I obtain the rate of plastic deformation, also called the accumulated plastic strain rate, dp/dt (stated in Current yield Norton law) from a uniaxial tensile stress-strain data?

$\endgroup$
7
  • $\begingroup$ Could you write out the Current Yield Norton Law in full? I've not worked with it before, but I might be able to figure out what you need. $\endgroup$ – Tyler Olsen Dec 5 '15 at 14:49
  • $\begingroup$ @TylerOlsen, I have uploaded it on imgur, hope it clarifies well - i.imgur.com/JiSOe7O.jpg $\endgroup$ – ShadowWarrior Dec 5 '15 at 17:18
  • 1
    $\begingroup$ You will need more information than the stress-strain curve to get $\dot{p}$ since it does not communicate any information about time. You will also need the strain vs time curve. In 1D, however, computing the time derivative of the accumulated plastic strain is easy, as it is simply the strain rate (once you have started yielding) if you assume an additive strain decomposition. $\endgroup$ – Tyler Olsen Dec 5 '15 at 19:37
  • $\begingroup$ The uniaxial test was done at a constant strain rate of 0.0001 1/s, up to a total strain of 0.15, of which 0.01 was elastic strain (so a linear strain vs time plot with 0.15 and 1500 seconds as extreme values). Loading was monotonic, so plastic strain would be 0.14. Do these info helps? Is the answer simply the strain rate? $\endgroup$ – ShadowWarrior Dec 5 '15 at 20:17
  • $\begingroup$ Yeah, in this case the answer is just the strain rate. From there, you should be able to determine what parameters fit your data (I assume that is the objective). $\endgroup$ – Tyler Olsen Dec 6 '15 at 19:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.