Question

A metal cylinder of length L is subjected to a uniform compressive force F as shown in the figure. The material of the cylinder has Young's modulus Y and Poisson’s ratio $\sigma$ . The change in volume of the cylinder is

• A. $\frac{\sigma FL}{Y}$
• B. $\frac{(1-\sigma )FL}{Y}$
• C. $\frac{(1+2\sigma )FL}{Y}$
• D. $\frac{(1-2\sigma )FL}{Y}$

Answer 1

The correct option is D

$\frac{(1-2\sigma )FL}{Y}$

Volume of the cylinder, $V=\pi r^{2}L$

Volumetric strain $=\frac{\Delta V}{V}=\frac{\Delta \left(\pi r^{2}L\right)}{\pi r^{2}L}$

$\frac{\Delta V}{V}=\frac{\pi r^2\Delta L+2\pi rL\Delta R}{\pi r^{2}L}$

$=\frac{\Delta L}{L}+\frac{2\Delta r}{r}\dots \left(i\right)$

Poisson's ratio, $\sigma =-\frac{\left(\frac{\Delta r}{r}\right)}{\left(\frac{\Delta L}{L}\right)}$

or $\frac{\Delta r}{r}=-\frac{\sigma \Delta L}{L}$

on substituting this value of $\frac{\Delta r}{r}$ in Eq.(i), we get

$\frac{\Delta V}{V}=\frac{F}{\pi r^{2}Y}$

On substituting this value of $\frac{\Delta L}{L}$ in Eq.(ii), we get

$\frac{\Delta V}{V}=\frac{F}{\pi r^{2}Y}(1-2\sigma )$

$\frac{\Delta V}{\pi r^{2}L}=\frac{F}{\pi r^{2}Y}(1-2\sigma )$

$\Delta V=\frac{(1-2\sigma )FL}{Y}$