Question

A steel rod of $10\text{mm}$ diameter is loaded with a force of $20\text{kN}$. Change in diameter of the rod is $3.82\times {10}^{-3}\text{mm}$ . Given, Young's modulus for steel is $200\text{GPa}$, find its Poisson's ratio.

• A. $0.30$
• B. $0.28$
• C. $0.21$
• D. $0.88$

Answer 1

The correct option is A $0.30$

We know that ,
Young's modulus $\left(Y\right)=\frac{\text{longitudinal stress}}{\text{longitudinal strain}}$
$Y=\frac{\left(F/A\right)}{\text{longitudinal strain}\left(\Delta L/L\right)}$
$\Rightarrow$ Longitudinal strain $\frac{\Delta L}{L}=\frac{F}{\frac{\pi }{4}{d}^2\times Y}$
From the data given in the question we can say that,
$\frac{\Delta L}{L}=\frac{20\times {10}^3}{\frac{\pi }{4}\times {10}^2\times {10}^{-6}\times 200\times {10}^9}$
$=1.27\times {10}^{-3}$

To find Poisson's ratio, we have to find the value of lateral strain.
From the definition of lateral strain, we can say that,
Lateral strain $=\frac{\text{change in diameter (}d\text{)}}{\text{original diameter (}D\text{)}}=\frac{3.82\times {10}^{-3}}{10}=3.82\times {10}^{-4}$
Poisson's ratio $\left(\sigma \right)=\frac{\text{lateral strain}}{\text{longitudinal strain}}$
$\therefore \left(\sigma \right)=\frac{3.82\times {10}^{-4}}{1.27\times {10}^{-3}}$
or $\left(\sigma \right)=0.30$
Thus, option (a) is the correct answer.