Topic: Strength of Materials
Two bolts A and B of same material and length are subjected to identical tensile load. If the elastic strain energy stored in bolt A is 16 times that of the bolt B and the mean diameter of bolt A is 13 mm, the mean diameter of bolt B (in mm) is
- A. 52
- B. 36
- C. 24
- D. 16
Answer: A. 52
Explanation: Strain energy $SE = \frac{P^{2}L}{2AE}$. Given $SE_{A} = 16 \cdot SE_{B}$. Since P, L, and E are the same for both bolts, $SE \propto \frac{1}{A}$. Therefore, $\frac{1}{A_{A}} = \frac{16}{A_{B}} \Rightarrow A_{B} = 16A_{A}$. Using area $A = \frac{\pi}{4}d^{2}$, we get $d_{B}^{2} = 16d_{A}^{2} \Rightarrow d_{B} = 4d_{A}$. Given $d_{A} = 13~mm$, $d_{B} = 4 \times 13 = 52~mm$.