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sq
2005-09-15, 13:27
Assume the following:

I have an effectively incompressible elastomer such that if in uniaxial tension or compression, the first principal stretch lambda(1) = lambda, then lambda(2) = lambda(3) = lambda^(-1/2), obviously.

Deviatoric stress is expressed as a derivative of the energy function:

t(i) = lambda(i) * d(W)/d lambda(i) + p

Where p is indeterminate. For example, for Gaussian statistical theory, W = G/2 *(lambda(1)^2+lambda(2)^2+lambda(3)^2-3), which yields

t(i) = G*lambda(i)^2-p

Now, here's the trick: If we use p to set t(2) = t(3) = 0, which holds for uniaxial tension, then

t(1) = G*(lambda^2-1/lambda), the familiar result.

However, in the numerical implementation (such as in a VUMAT), we do not know for a given stress field how to properly set p. If p is set to a hydrostatic pressure stress as a function of J (as is often done), then the stress field for the uniaxial case illustrated above will be offset by (t2 - t1).

Does anybody have insight into how this discrepancy may be resolved?

Jorgen
2005-09-18, 18:18
Good question. Here are a few comments:
:arrow: Explicit finite element simulations cannot really use an incompressible material model, since that would give an infinite wave speed.
:arrow: You can use an incompressible material model with implicit simulations, however.
:arrow: As you mentioned, most FE programs get around the undetermined pressure issue by adding compressibility through a term like [TeX:7bc7a1313f]\kappa(J-1)[/TeX:7bc7a1313f].

I am not quite sure I follow your statement on (t2-t1). Can you explain more?

- Jorgen

sq
2005-09-20, 07:22
Thanks for replying, Jorgen, it's always great to hear from you.

Okay, what I mean is that for even the approximate case of isovolumetric uniaxial stretch, the off-axis orthogonal stress components do not resolve to zero. Unless I am missing something, which is entirely possible.

Jorgen
2005-09-22, 18:20
Hi sq,

Is that you Steve Q.? Good to hear from you!

I have to admit that I still don't quite follow your question. Let's see, isovolumetric deformation, that sounds like incompressibility. In this case, as I think you mentioned, the off-axis stresses cannot simply be determined from the equilibrium equations but also require the boundary conditions in terms of the pressure p. In most cases when I deal with elastomers I add some compressibility to make the numerical predictions easier. In my experience, the purely incompressible solution can be obtained in a numerical simulation by gradually increasing the bulk modulus.

- Jorgen

sq
2005-09-23, 06:07
Yes, it's me! :)

Actually, I think you followed my question better than how well I phrased it! My concern (which, actually, you've answered now) was caused by my wanting to associate the results of properly applying the boundary conditions with the results of the constitutive model (in evaluating my own VUMAT). Oops. :oops:

P.S.,

What a great site! I've really enjoyed reading the posts, and hoping I can contribute here and there.