# The Poisson Ratio and conservation of volume

This summer, I had the chance to work on modelling robot skin at Hanson Robotics and I’ve learned quite a bit of continuum mechanics in the process. But, one of the most important lessons I’ve learned so far turned out to be quite simple yet counter-intuitive.

Prior to reading any texts on elasticity, I would assume that volume was always conserved as I thought this would be useful for estimating elastic properties. In particular, I used this assumption when estimating the Poisson Ratio and Young’s Modulus of the rectangular sample I was modelling. However, this led to really bad estimates for the Poisson Ratio in particular which is defined to be the negative ratio of transverse to axial strain.

After reflecting on the source of my error, I wondered whether the the volume of an elastic material with a prismatic geometry wasn’t necessarily conserved when subject to small uni-axial strains.

I realised that this fact can be easily demonstrated for Poisson Ratios in the range $-1 \leq \nu < 0.5$:

1. We are given $\nu=-\frac{\Delta{W}/W_0}{\Delta{L}/L_0}=-\frac{\Delta{T}/T_0}{\Delta{L}/L_0}$ where $-1 \leq \nu < 0.5$

2. Let $\frac{\Delta{L}}{L_0}=\alpha > 0$ then $latex \begin{cases} \Delta{W}=-W_0*\nu*\alpha \\ \Delta{T}=-T_0*\nu*\alpha\\ \end{cases}$
3. $V'=L*T*W=(L_0+\alpha L_0)*T_0*(1-\nu \alpha)*W_0*(1-\nu \alpha)=L_0*T_0*W_0*(1+\alpha)*(1-\nu \alpha)^2=V*(1+\alpha)*(1-\nu \alpha)^2$

4. The equation in part 3 actually reduces to:

$\frac{\Delta{V}}{V}=(1+\alpha)*(1-\nu \alpha)^2-1$

Clearly, for $\nu \leq 0$, $\frac{V'}{V}>1$ so the volume has increased. And for $\nu >0$, the volume is constant only when $\nu = \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$. Since $\nu$ is supposed to be constant and $\alpha$ is allowed to vary we must conclude that the volume is not constant in this case either.

If I should trust my arguments then for any given non-negative poisson ratio, there is a single non-trivial value for $\alpha=\Delta{L}/L_0$ such that $V'=V$. The graph of the function $f(\alpha)= \frac{1}{\alpha}*(1-\sqrt{\frac{1}{1+\alpha}})$ is given below:

Now, we must deal with the case $\nu = \frac{1}{2}$ separately:

1. Let’s suppose, without loss of generality, that our material has the geometry of a cylinder. Then $V=\pi r^2 L$.
2. If its volume is constant(i.e. it’s incompressible), then: $\quad dV= 2\pi r L dr +\pi r^2 dL = 0$$\nu \equiv \frac{dr/r}{dL/L} = \frac{1}{2}$

So the moral of the story is that for most materials that we interact with in everyday life, their volume isn’t conserved when stretched or compressed unless they happen to have a poisson ratio of $\nu = \frac{1}{2}$.