retagged by
1 votes
1 votes

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. $-1.0$
retagged by

1 Answer

Best answer
1 votes
1 votes

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

selected by
Answer:

Related questions

0 answers
0 votes