Question

A bar of circular cross section is clamped at ends P and Q as shown in the figure. A torsional moment T = 150 Nm is applied at a distance of 100 mm from end P. The torsional reactions $(T_P,T_Q)$ in Nm at the ends P and Q respectively are

• A. (50, 100)
• B. (75, 75)
• C. (100, 50)
• D. (120, 30)

Answer 1

The correct option is C (100, 50)

$T_P+T_Q=T$
${\theta }_{PR}+{\theta }_{RQ}=0$
$\Rightarrow \frac{T_{P}L}{G{I}_P}+\frac{(T_P-T)2L}{G{I}_P}=0$
$\Rightarrow T_P+2T_P=2T$
$\Rightarrow T_P=\frac{2T}{3}=\frac{2}{3}\times 150=100Nm$
$\therefore$ From equation (i),
$\Rightarrow T_Q=\frac{T}{3}=\frac{150}{3}=50N-m$