Fixed shaft: end couples, diameter, section of no twist

Reading of the figure: shaft of uniform diameter fixed at A and D; $T_B = 15$ kN·m at B and $T_C = 25$ kN·m at C (opposite sense to $T_B$); $AB = 1.5$ m, $BC = 1.2$ m, $CD = 1.5$ m; $\tau_{max} = 50\ \text{MN/m}^2$.

End fixing couples

Let $T_A$ be the reaction at A (positive in the sense of $T_B$). The torques in the segments are

$$T_{AB} = T_A, \qquad T_{BC} = T_A + 15, \qquad T_{CD} = T_A + 15 - 25 = T_A - 10 \quad(\text{kN·m})$$

The ends are fixed, so the total twist is zero. $G$ and $J$ are constant:

$$T_A(1.5) + (T_A+15)(1.2) + (T_A-10)(1.5) = 0 \Rightarrow 4.2\,T_A + 3 = 0 \Rightarrow T_A = -0.714\ \text{kN·m}$$

The negative sign means the reaction at A acts opposite to $T_B$. Also $T_D = 10.714$ kN·m from equilibrium ($T_A + T_D + 15 - 25 = 0$ with the signs above).

Answer (end couples): $|T_A| = 0.714$ kN·m, acting opposite to $T_B$; $|T_D| = 10.714$ kN·m, acting in the same sense as $T_B$.

Diameter

The maximum torque is $T_{BC} = 14.286$ kN·m.

$$d^3 = \frac{16T_{max}}{\pi\tau} = \frac{16(1.4286\times10^{7})}{\pi(50)} = 1.4551\times10^{6}\ \text{mm}^3 \Rightarrow d = 113.32\ \text{mm}$$

Section with no angular twist

The rotation of a section relative to A is $\varphi = \sum \dfrac{TL}{GJ}$. At B (taking $GJ = 1$ units, kN·m²): $\varphi_B = 0.714\times1.5 = 1.071$. In BC the twist changes at the rate $T_{BC} = 14.286$ in the opposite direction, so it reaches zero at a distance

$$x = \frac{1.071}{14.286} = 0.075\ \text{m from B}$$

Answer: the section $\approx 0.075$ m from B (i.e. 1.575 m from A, inside BC) suffers no angular twist; $d \approx 113.3$ mm.