Brass-steel stepped shaft: max torque, rotation

Data: brass AB: $d = 80$ mm, $L = 1000$ mm, $\tau_{allow} = 88\ \text{N/mm}^2$, $G = 400\ \text{kN/mm}^2$; steel BC: $d = 60$ mm, $L = 1200$ mm, $\tau_{allow} = 110\ \text{N/mm}^2$, $G = 80\ \text{kN/mm}^2$. The same torque $T$ acts in both segments (applied at the free end C).

Maximum torque

$$T_{brass} = \frac{\pi}{16}\tau d^3 = \frac{\pi}{16}(88)(80)^3 = 8.8467\times10^{6}\ \text{N·mm} = 8.847\ \text{kN·m}$$ $$T_{steel} = \frac{\pi}{16}(110)(60)^3 = 4.6653\times10^{6}\ \text{N·mm} = 4.665\ \text{kN·m}$$

The smaller governs, so $T_{max} = 4.665$ kN·m (the steel segment reaches its limit first).

Rotation of the free end relative to the fixed end

$$\varphi_{AB} = \frac{TL}{GJ} = \frac{(4.6653\times10^{6})(1000)}{(400\times10^3)\frac{\pi}{32}(80)^4} = 0.00290\ \text{rad}$$ $$\varphi_{BC} = \frac{(4.6653\times10^{6})(1200)}{(80\times10^3)\frac{\pi}{32}(60)^4} = 0.05500\ \text{rad}$$ $$\varphi_C = \varphi_{AB} + \varphi_{BC} = 0.05790\ \text{rad} = 3.317^\circ$$

Answer: $T_{max} = 4.67$ kN·m; total rotation of C $= 0.0579$ rad $= 3.32^\circ$.