# Poisson ratio and volume change

Consider a rod of 1 m diameter and 2 m length. The Poisson's ratio of the material is 0.5. Assume the rod is stretched to 2.5 m length.

Now longitudinal strain is 0.5/2 = 0.25

lateral strain = longitudinal strain * Poisson's ratio $$= 0.25\cdot 0.5 = 0.125$$

Change in diameter is 0.125

Original volume $$= \pi (1\cdot 1)/4\cdot 2 = 1.57$$

Diameter is reduced to $$1-0.125 =0.875$$

New volume $$= \pi (0.875\cdot 0.875)/4 \cdot 2.5 = 1.502$$

The volume change is not zero when in comes to direct calculation. I know by $$E=3K(1-2\nu)$$, the volume change is zero, though.

Why is it so ?

Also, if you can give a valid explanation to it, Can this apply on a rectangular bar of square cross section?

• Only a Poisson ratio of 1/3 gives a strictly volume conserving material. – Jon Custer May 14 at 17:07
• @John Custer No way. A Poisson ratio of 1/2 corresponds to an incompressible material. – Chet Miller May 14 at 22:57

## 1 Answer

The Hookean elastic material is taken to be linear, which means that it applies strictly in the limit of small strains. For tiny strains, the volume change is zero (to linear terms in the principal strains) if Poisson ratio = 1/2. Try the same calculation for a case in which the change in length is 0.05 m and see what you get.

If the length is L and the diameter is D, the deformed length is $$L\left(1+\frac{\Delta L}{L}\right)$$ and the deformed diameter is $$D\left(1-\nu\frac{\Delta L}{L}\right)$$ So, to linear terms in $$\Delta L/L$$, the change in volume is $$\frac{\Delta V}{\left(\pi\frac{D^2}{4}L\right)}=(1-2\nu)\frac{\Delta L}{L}$$If $$\nu=1/2$$, $$\Delta V=0$$