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)


$T_P+T_Q=T$ ...(i)
$Q_{PR}+Q_{RQ}=0$

$\frac{T_P\cdot L}{G{I}_P}+\frac{(T_P-T)2L}{G{I}_P}=0$

$T_P+2T_P=2T$

$T_P=\frac{2T}{3}=100Nm$

From equation (i), we get,

$T_Q=\frac{T}{3}=50Nm$