Question

A stepped circular shaft is fixed at A and C as shown in the below figure. The diameter of the shaft along BC is twice that of as along AB. The torsional rigidity of AB is GJ. The torque required for unit twist at B is

• A. $\frac{2GJ}{l}$
• B. $\frac{5GJ}{l}$
• C. $\frac{9GJ}{l}$
• D. $\frac{17GJ}{l}$

Answer 1

The correct option is D $\frac{17GJ}{l}$

${\theta }_{AC}=0$

$\Rightarrow {\theta }_{AB}+{\theta }_{BC}=0$

$\Rightarrow \frac{T_{A}l}{GJ}+\frac{(T_A-T)l}{G{J}^{\prime }}=0$

$J^{\prime }=\frac{\pi (2d{)}^4}{32}\left(i\right)$

$=16.\frac{\pi d^4}{32}=16J$

From equation (i)

$\Rightarrow \frac{T_{A}l}{GJ}+\frac{(T_A-T)l}{16GJ}=0$

$\Rightarrow 16T_A+T_A-T=0$

$\Rightarrow 17T_A=T$

$\Rightarrow T_A=\frac{T}{17}$

${\theta }_{AB}={\theta }_B-{\theta }_A$

$=\frac{T_{A}l}{GJ}=\frac{Tl}{17GJ}$

$\Rightarrow {\theta }_B=\frac{Tl}{17GJ}(As{\theta }_A=0)$

$\Rightarrow \frac{T}{{\theta }_B}=\frac{17GJ}{l}$

$\therefore$ Torque required for unit twist at B

$=\frac{17GJ}{l}$