Fixed shaft, two torques: couples, diameter

Data: length $= 11.25$ m, fixed at both ends. Viewed from A: $T_C = 30$ kN·m clockwise at 4.5 m from A, $T_D = 37.5$ kN·m anticlockwise at 7.5 m from A. Segments: $L_1 = 4.5$ m (A to C), $L_2 = 3.0$ m (C to D), $L_3 = 3.75$ m (D to B). $\tau_{max} = 60\ \text{N/mm}^2$.

End fixing couples

Take clockwise (viewed from A) as positive. Let $T_A$ be the end couple at A. The torques in the segments are

$$T_1 = T_A, \qquad T_2 = T_A + 30, \qquad T_3 = T_A + 30 - 37.5 = T_A - 7.5 \quad(\text{kN·m})$$

For a shaft fixed at both ends the total twist is zero ($G$, $J$ constant):

$$T_A(4.5) + (T_A+30)(3.0) + (T_A-7.5)(3.75) = 0$$ $$11.25\,T_A + 90 - 28.125 = 0 \Rightarrow T_A = -5.50\ \text{kN·m}$$

The negative sign means the couple at A is anticlockwise. From overall equilibrium, $T_A + 30 - 37.5 + T_B = 0$:

$$T_B = 13.00\ \text{kN·m}\ \text{(clockwise)}$$

End couples: at A, $5.50$ kN·m anticlockwise; at B, $13.00$ kN·m clockwise (viewed from A). Check: $13.00 - 5.50 = 7.50 = 37.5 - 30$.

Diameter

The maximum torque is in segment CD: $T_{max} = 24.50$ kN·m.

$$d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(2.4500\times10^{7})}{\pi(60)} = 2.0796\times10^{6}\ \text{mm}^3$$ $$d = 127.64\ \text{mm}$$

Answer: end couples $5.5$ kN·m (A, anticlockwise) and $13$ kN·m (B, clockwise); diameter $\approx 127.6$ mm.