GATE2017 ME-2: 50

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A rod of length $20$ mm is stretched to make a rod of length $40$ mm. Subsequently, it is compressed to make a rod of final length $10$ mm. Consider the longitudinal tensile strain as positive and compressive strain as negative. The total true longitudinal strain in the rod is

1. $-0.5$
2. $-0.69$
3. $-0.75$
4. $-1.0$

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1 vote

True strain :

\begin{align*} \epsilon_t = \sum \left ( \frac{\Delta L}{L} \right ) = \int_{L_i}^{L_f} \frac{dl}{l} = \ln \frac{L_f}{L_i} \end{align*}

Now, here $L_f = 10 mm$  and  $L_i = 20 mm$

So, \begin{align*} \epsilon_t = \ln \frac{10}{20} = -0.6931 \end{align*}

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