Question

Two threaded bolts A and B of same material and length are subjected to identical tensile load. If the elastic energy stored in bolt A is 4 times that of the bolt B and the mean diameter of bolt A is 12 mm, the mean diameter of bolt B in mm is

• A. 16
  
• B. 24
• C. 36
• D. 48

Answer 1

The correct option is B 24

Given:
$P_1=P_2=P$
(Identical tensile load on bolt A and B) (same length)
$L_1=L_2=L$
$d_A=12mm$
$U_A=$ Strain energy in bolt A
$U_B=$ Strain energy in bolt B
$U_A=4U_B$ (Given) ...(i)

$U_A=\frac{1}{2}P\times \delta =\frac{1}{2}\frac{P^{2}L}{AE}$

$\therefore Eq.\left(i\right)$

$\frac{1}{2}\frac{P_1^2{L}_1}{A_{1}E}=4\times \frac{1}{2}\frac{P_2^2{L}_2}{A_{2}E}$

$\frac{1}{A_1}=4\times \frac{1}{A_2}$

$\frac{4}{\pi (12{)}^2}=4\times \frac{4}{\pi (d_B{)}^2}$

$\Rightarrow d_B^2=576$

$or{d}_B=24mm$