Stepped shaft: angle of twist at free end

Reading of the figure: fixed at the left; segment 1: $d = 110$ mm, $L = 3$ m, with a 10 kN·m torque at its right end; segment 2: $d = 60$ mm, $L = 2$ m, with a 5 kN·m torque at the free end, both in the same sense. $G = 0.9\times10^5\ \text{N/mm}^2$.

Internal torques

• Segment 2: $T_2 = 5$ kN·m
• Segment 1: $T_1 = 10 + 5 = 15$ kN·m

Polar moments

$$J_1 = \frac{\pi}{32}(110)^4 = 1.4374\times10^{7}\ \text{mm}^4, \qquad J_2 = \frac{\pi}{32}(60)^4 = 1.2723\times10^{6}\ \text{mm}^4$$

Angle of twist

$$\varphi_1 = \frac{T_1L_1}{GJ_1} = \frac{(15\times10^6)(3000)}{(0.9\times10^5)(1.4374\times10^{7})} = 0.03479\ \text{rad}$$ $$\varphi_2 = \frac{T_2L_2}{GJ_2} = \frac{(5\times10^6)(2000)}{(0.9\times10^5)(1.2723\times10^{6})} = 0.08733\ \text{rad}$$ $$\varphi = \varphi_1 + \varphi_2 = 0.12211\ \text{rad} = 6.997^\circ$$

Answer: angle of twist at the free end $= 0.1221$ rad $= 7.00^\circ$.