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

Answer 1

Answer: (D)

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, $\sigma$ $=-\frac{\left(\Delta r/r\right)}{\left(\Delta L/L\right)}$
or $\frac{\Delta r}{r}$ $=-\frac{\sigma \Delta L}{L}$
On substituting this value of $\frac{\Delta r}{r}$ in Eq. $\left(i\right)$, we get
$\frac{\Delta V}{V}$ $=\frac{\Delta L}{L}\left(1-2\sigma \right)$ $\dots \left(ii\right)$
Young’s modulus, $Y=\frac{\left(F/\pi r^2\right)}{\Delta L/L}$or $\frac{\Delta L}{L}=\frac{F}{\pi r^{2}Y}$
On substituting this value of $\frac{\Delta L}{L}$ in Eq. $\left(ii\right)$, we get
$\frac{\Delta V}{V}$ $=\frac{F}{\pi r^{2}Y}$ $\left(1-2\sigma \right)$
$\frac{\Delta V}{\pi r^{2}L}$ $=\frac{F}{\pi r^{2}Y}$ $\left(1-2\sigma \right)$
$\Delta V$ $=\frac{FL}{Y}$ $\left(1-2\sigma \right)$