Question

A solid shaft of diameter d and length L is fixed at both the ends. A torque, $T_0$ is applied at a distance L/4 from the left end as shown in the figure given below.


The maximum shear stress in the shaft is

• A. $\frac{16T_0}{\pi d^3}$
• B. $\frac{12T_0}{\pi d^3}$
• C. $\frac{8T_0}{\pi d^3}$
  
• D. $\frac{4T_0}{\pi d^3}$
  

Answer 1

The correct option is B $\frac{12T_0}{\pi d^3}$

$T_1$ + $T_2$ = $T_0$ .....(i)
Angle of twist for both shafts at section X-X will be same.
Therefore,

$\frac{T_{1}L}{4GJ}=\frac{T_2\times 3L}{4GJ}$

$\therefore T_1=3T_2$.....(ii)

Solving equation (i) and (ii)

$T_2=\frac{T_o}{4},T_1=\frac{3T_o}{4}$

Shear stress in 1

$=\frac{16T_1}{\pi d^3}=\frac{16}{\pi d^3}\times \frac{3T_0}{4}$

$=\frac{12T_0}{\pi d^3}$

Shear stress in 2

$=\frac{16T_2}{\pi d^3}=\frac{16}{\pi d^3}\times \frac{T_0}{4}=\frac{4T_0}{\pi d^3}$

Hence, maximum shear stress in shaft is
$\frac{12T_0}{\pi d^3}$