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)

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

Answer 2

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

Volumetric strain

$= \frac{\Delta V}{V} = \frac{\Delta(\pi r^{2}L)}{\pi r^{2}L}$

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

$= \frac{\Delta L}{L} + \frac{2\Delta r}{r}$ …(i)

Poisson’s ratio,$\sigma = - \frac{(\Delta r/r)}{(\Delta L/L)}$

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

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

$\frac{\Delta V}{V} = \frac{\Delta L}{L}(1 - 2\sigma)$ …(ii)

Young’s modulus, $Y = \frac{(F/\pi r^{2})}{(\Delta L/L)}$

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

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

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

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

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