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