# 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.

## 1 Answer

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.

Since you're likely looking at polycrystalline metal plasticity, plastic incompressibility lies in the physical origin of plastic deformation: movement of dislocations through the atomic lattices. See wikipedia if you don't yet know what a dislocation is. In a single increment of plastic deformation, the atoms shift in such a way that the dislocation moves one atomic spacing through the lattice. Due to this, the overall volume of material remains the same at the end of the process as at the beginning. This is the motivation for the "plastic incompressibility" condition that you cite. In reality, there is a tiny bit of volume change in the transition state between the starting and ending configurations, but it occurs at the atomic scale and is not a consideration when you use continuum theories.

• Thanks for comments. Poission's ratio, ν, is a parameter in the stress-strain relationship for isotropic, linear elasticity only. Really? Poisson's ratio is defined for orthrotropic materials as well. A link: en.wikipedia.org/wiki/Orthotropic_material – 343_458 Jul 8 '15 at 22:59
• You should say Poisson's ratio is defined for linear elasticity. In fact, Poisson's ratio can be more than 0.5 for some orthrotropic materials. – 343_458 Jul 8 '15 at 22:59
• 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 $\mathbb{C}$. – Tyler Olsen Jul 8 '15 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/… – Tyler Olsen Jul 8 '15 at 23:52
• I understand that the ratio of strains is well defined; I just don't like the nomenclature when you get away from isotropy. I think we agree that the set of numbers exist for composites. – Tyler Olsen Jul 11 '15 at 22:58