Question

The maximum shear stress in a solid shaft of circular cross-section having diameter d subjected to a torque T is $\tau$. If the torque is increased by four times and the diameter of the shaft is increased by two times, the maximum shear stress in the shaft will be

• A. 2 $\tau$
• B. $\tau$
• C. $\tau$/2
• D. $\tau$/4

Answer 1

The correct option is C $\tau$/2

By torsion formula,

$\frac{\tau }{r}=\frac{T}{J}$

Where,
$\tau$ = shear stress
r = radius of shaft
T = Torsional moment
J = Polar moment of inertia
For solid circular shaft,

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

$\tau =\frac{T.r}{J}=\frac{T.\left(\frac{d}{2}\right)}{\left(\frac{\pi d^4}{32}\right)}$

$\Rightarrow \tau =\frac{16T}{\pi d^3}$

$Now,T_1=4T$

$and,d_1=2d$

${\tau }_1=\frac{16\left(4T\right)}{\pi (2d{)}^3}=\frac{1}{2}\left(\frac{16T}{\pi d^3}\right)$

${\tau }_1=\frac{\tau }{2}$