# 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. Commented May 14, 2019 at 17:07
• @John Custer No way. A Poisson ratio of 1/2 corresponds to an incompressible material. Commented May 14, 2019 at 22:57

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 $$\nu=1/2$$, $$\Delta V=0$$