Stepped shaft: maximum torque at C

Data: $\tau_{allow} = 50\ \text{MPa}$; segment AC: $d_1 = 20$ mm, $L_1 = 0.7$ m; segment CB: $d_2 = 30$ mm, $L_2 = 0.5$ m; same material; torque $T$ applied at C; both ends fixed.

Distribution of torque

Let $T_A$ and $T_B$ be the end reactions: $T_A + T_B = T$ (equilibrium). The rotation of C calculated from either end must be equal:

$$\frac{T_AL_1}{GJ_1} = \frac{T_BL_2}{GJ_2} \Rightarrow \frac{T_A}{T_B} = \frac{L_2/J_2}{L_1/J_1} = \frac{0.5/30^4}{0.7/20^4} = 0.1411$$

So $T_A = 0.1236\,T$ and $T_B = 0.8764\,T$.

Limiting torque

Stress in each segment is $\tau = \dfrac{16T_i}{\pi d^3}$.

• AC reaches 50 MPa when $T_A = \dfrac{\pi(50)(20)^3}{16} = 78.54$ N·m. Then $T = T_A/0.1236 = 635.2$ N·m.
• CB reaches 50 MPa when $T_B = \dfrac{\pi(50)(30)^3}{16} = 265.07$ N·m. Then $T = T_B/0.8764 = 302.5$ N·m.

The smaller value governs (segment CB is stressed to the limit first):

$$T_{max} = 302.5\ \text{N·m}$$

Shear stress at A

With this torque, $T_A = 37399.9$ N·mm and $T_B = 265071.9$ N·mm, so

$$\tau_A = \frac{16T_A}{\pi(20)^3} = 23.81\ \text{MPa}, \qquad \tau_B = \frac{16T_B}{\pi(30)^3} = 50.00\ \text{MPa}$$

Answer: maximum torque at C $= 302.5$ N·m ($\approx 0.302$ kN·m); shear stress at A (segment AC) $= 23.8$ MPa.