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

Answer: (D)

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

Volume of the cylinder, $V=\pi r^{2}L$
Volumetric strain $=\frac{\Delta V}{V}=\frac{\pi r^2\Delta L+2\pi rL\Delta r}{\pi r^{2}L}=\frac{\Delta L}{L}+\frac{2\Delta r}{r}....\left(i\right)$
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 (i), we get
$\frac{\Delta V}{V}=\frac{\Delta L}{L}(1-2\sigma )...\left(ii\right)$
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 (ii), we get
$\frac{\Delta V}{V}=\frac{F}{\pi r^{2}Y}(1-2\sigma )$

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

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