Question

A uniform rod of length $L$ and area of cross-section $A$ is subjected to tensile load $F$. If $\sigma$ be Poisson's ratio and $Y$ be the Young's modulus of the material of the rod, then find the volumetric strain produced in rod.

• A. $\frac{F}{A Y}(1+2 \sigma)$
• B. $\frac{F}{A Y}(1-2 \sigma)$
• C. Zero
• D. None of these

Answer 1

Answer: (B)

$Y=\frac{F / A}{\Delta L / A}$
$\Rightarrow Y=\frac{\Delta L}{L}=\frac{F}{A Y}$ ......(i)
Now, by definition of Poisson's ratio
$\sigma=\frac{\Delta r / r}{\Delta L / L}$
$\Rightarrow \frac{\Delta r}{r}=-\frac{\sigma L}{L}=-\frac{\sigma F}{A Y} $ [by using Eq. (i)]
$\Rightarrow \frac{\Delta r}{r}=-\frac{\sigma F}{A Y}$ .......(ii)
Now $V=\pi r^{2} L$
$\Rightarrow \frac{\Delta V}{V}=\frac{2 \Delta r}{r}+\frac{\Delta L}{L}$
By using Eqs. (i) and (ii)
$\frac{\Delta V}{V}=2\left(\frac{-\sigma F}{A Y}\right)+\frac{F}{A Y}$
$\Rightarrow \frac{\Delta V}{V}=\frac{F}{A Y}[1-2 \sigma]$