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Deformation of Tapered Bars

A mild steel ...

Question

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

\frac{3 π}{2}

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\frac{2}{3 π}

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\frac{π}{3}

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\frac{1}{3 π}

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Solution

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

$Δ = \frac{4 P L}{π E d_{1} d_{2}} = \frac{4 (12000) 400}{π (2 \times 10^{5}) 24 (12)} = \frac{1}{3 π}$

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