# Understanding incompressibility in continuum plasticity

I am a beginner in continuum plasticity and wondering physical meaning of incompressibility in continuum plasticity. Referring to MIT OCW link

the consequence of incompressibility condition is (eq 12.13 in the link)

$$\dot{\epsilon}_{11} + \dot{\epsilon}_{22} + \dot{\epsilon}_{33} = 0$$ or $$\dot{\epsilon}_{kk}=0$$ Where, $\dot{\epsilon}_{kk}$ denotes the summation of strain rates along Cartesian coordinate axes 1,2,3.

This also means that the Poisson's ratio $\nu$ would be 0.5. So, if I have a steel with $\nu = 0.3$ then is this ratio 0.5 during plastic deformation? I am very confused here, as Poisson's ratio is a material property and should not depend of deformation. Can someone explain this idea better? In fact the same idea is mentioned in the above link eqn. 12.15.

You have to be careful when you talk about plasticity, because you're not really allowed to carry over any of your elasticity parameters. Poission's ratio, $\nu$, is a parameter in the stress-strain relationship for isotropic, linear elasticity only. Poisson's ratio has no part in plasticity calculations.
• I'd contend that Poisson's ratio is only properly defined for isotropic materials, but you can generalize the concept for anisotropic materials to be the negative ratio of transverse strain to axial strain in any given set of directions. If you notice in the wiki article, they give $\nu_{ij}$ as a set of 9 "Poisson's ratios" instead of just a single number. It might be personal preference, but if you're going to allow that much freedom in your model, you should just call it generalized linear elasticity and use $\mathbb{C}$. Jul 8, 2015 at 23:48
• And that's an interesting point about the bounds on $\nu_{ij}$ for orthotropic materials. I'd never seen that before. Thanks! If you haven't seen it before, the derivation for the isotropic bounds on $\nu$ is pretty neat, and I've put an explanation on this site in the past: physics.stackexchange.com/questions/99077/… Jul 8, 2015 at 23:52