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Deformation of Tapered Bars

A mild steel ...

Question

A mild steel rod tapers uniformly from $30 m m$ diameter to $12 m m$ diameter in a length of $300 m m$ . The rod is subjected to an axial load of $12 k N$ . $E = 2 \times 10^{5} N / m m^{2}$ . What is the extension of the rod in mm?

\frac{4 π}{5}

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\frac{4}{5 π}

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\frac{π}{5}

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\frac{1}{5 π}

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Solution

The correct option is D $\frac{1}{5 π}$
Total extension in the rod will be
$Δ L = \frac{4 P L}{π E D_{1} D_{2}}$
$= \frac{4 \times 12 \times 10^{3} \times 300}{π \times 2 \times 10^{5} \times 12 \times 30} = \frac{1}{5 π}$

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