Statically indeterminate shaft with torque

Method

The problem is statically indeterminate: only one equilibrium equation is available for two unknown reactions. Add a compatibility condition.

 //|A-----50mm-----C---40mm---B|//
 //|<---- 0.6 m --->|<-0.4 m->|//
                    | T0 = 2.5 kN m

Equations

1. Equilibrium: $T_A + T_B = T_0 = 2.5\times10^6$ N mm
2. Compatibility: the angle of twist of AC and CB must be equal at C (both ends are fixed and the section C rotates by one angle $\phi_C$)

$$\frac{T_A L_1}{GJ_1} = \frac{T_B L_2}{GJ_2}$$

Polar moments:

$$J_1 = \frac{\pi}{32}(50)^4 = 613.6\times10^3\ \text{mm}^4, \qquad J_2 = \frac{\pi}{32}(40)^4 = 251.3\times10^3\ \text{mm}^4$$

Torsional flexibility of each part, $L/J$: $\dfrac{600}{613.6\times10^3} = 9.778\times10^{-4}$ and $\dfrac{400}{251.3\times10^3} = 15.915\times10^{-4}$.

(a) Reactions

$$T_A = T_0\,\frac{L_2/J_2}{L_1/J_1 + L_2/J_2} = 2.5\times\frac{15.915}{9.778+15.915} = 1.549\ \text{kN m}$$ $$T_B = 2.5 - 1.549 = 0.951\ \text{kN m}$$

(The stiffer portion AC takes the larger share.)

(b) Maximum shear stresses

$$\tau_{AC} = \frac{16T_A}{\pi d_1^3} = \frac{16\times1.549\times10^6}{\pi\times50^3} = 63.1\ \text{MPa}$$ $$\tau_{CB} = \frac{16T_B}{\pi d_2^3} = \frac{16\times0.951\times10^6}{\pi\times40^3} = 75.7\ \text{MPa}$$

(c) Angle of twist of C

$$\phi_C = \frac{T_AL_1}{GJ_1} = \frac{1.549\times10^6\times600}{80000\times613.6\times10^3} = 0.0189\ \text{rad} = 1.08^\circ$$

Check using CB: $\dfrac{0.951\times10^6\times400}{80000\times251.3\times10^3} = 0.0189$ rad.

Answer: $T_A = 1.549$ kN m, $T_B = 0.951$ kN m; $\tau_{AC} = 63.1$ MPa, $\tau_{CB} = 75.7$ MPa; $\phi_C = 0.0189$ rad $= 1.08^\circ$.