1 1 vote 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 $-0.5$ $-0.69$ $-0.75$ $-1.0$ Mechanics of Materials gateme-2017-set2 mechanics-of-materials applied-mechanics-and-design + – ♦Arjun 641 views answer comment Share Follow Add Sync Questions 0 reply Please log in or register to add a comment.
Best answer 1 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, $\begin{align*} \epsilon_t = \ln \frac{10}{20} = -0.6931 \end{align*}$ Answer B dd answered May 8, 2017 • selected May 8, 2017 by dd dd comment Share Follow 0 reply Please log in or register to add a comment.