Find the value of strain energy of a metallic bar with cross-sectional area of 2 cm\(^2\) and gauge length of 10 cm if it stretches 0.002 cm under a load of 10 kN.
DSSSB JE - 2023,
Answer: A
For a gradually applied load the strain energy stored is the work done by the load, equal to half the load times the extension:
\[ \begin{aligned} U = \tfrac{1}{2}P\,\delta \\ &= \tfrac{1}{2} \times 10000\ \text{N} \times 0.002\ \text{cm} \\ &= 10\ \text{N-cm} \end{aligned} \]The area and gauge length are not needed once the load and the extension are given. The options 12, 14 and 16 N-cm are larger than the value this relation gives, and would correspond to larger loads or extensions than those in the question.
Please log in or register to report this question.