Question

A mild steel bar of length $450mm$ tapers uniformly. The diameters at the ends are $36mm$ and $18mm$, respectively. An axial load of $12kN$ is applied on the bar. $E=2\times {10}^{5}N/m{m}^2$. The elongation of the bar will be

• A. $\frac{1}{3\pi }mm$
• B. $\frac{1}{6\pi }mm$
• C. $\frac{3\pi }{2}mm$
• D. $\frac{2}{3\pi }mm$

Answer 1

The correct option is B $\frac{1}{6\pi }mm$

Elongation
$\Delta =\frac{4PL}{\pi d_1{d}_{2}E}$
$=\frac{4\times 12\times {10}^3\times 450}{\pi \times 36\times 18\times 2\times {10}^5}=\frac{1}{6\pi }mm$