Question

A mild steel rod tapers uniformly from $24mm$ dia to $12mm$ dia over its length of $400mm$. The rod when held vertical is subjected to an axial tensile load of $12kN$. $E=2\times {10}^{5}N/m{m}^2.$ The extension of the rod in $mm$ would be

• A. $\frac{3\pi }{2}$
• B. $\frac{2}{3\pi }$
• C. $\frac{\pi }{3}$
• D. $\frac{1}{3\pi }$

Answer 1

The correct option is D $\frac{1}{3\pi }$



$\Delta =\frac{4PL}{\pi E{d}_1{d}_2}=\frac{4\left(12000\right)400}{\pi (2\times {10}^5)24\left(12\right)}=\frac{1}{3\pi }$