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

Answer 1

Answer: (B)

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