Fixed shaft: torsional reactions, rotations, max shear

Reading of the figure: shaft fixed at A and D; AB and CD are copper ($\phi 80$ mm), BC is steel ($\phi 60$ mm), each 1 m long. A torque $T_B = 20$ kN·m acts at B and $T_C = 4$ kN·m acts at C in the opposite sense. The problem is statically indeterminate (one equation of equilibrium, one of compatibility).

Polar moments

$$J_{80} = \frac{\pi}{32}(80)^4 = 4.0212\times10^{6}\ \text{mm}^4, \qquad J_{60} = \frac{\pi}{32}(60)^4 = 1.2723\times10^{6}\ \text{mm}^4$$

Compatibility

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

$$T_{AB} = T_A, \qquad T_{BC} = T_A + 20, \qquad T_{CD} = T_A + 16 \quad(\text{kN·m})$$

The fixed end D does not rotate relative to A, so the total twist is zero ($\varphi = \sum \frac{TL}{GJ} = 0$):

$$\frac{T_A L}{G_cJ_{80}} + \frac{(T_A+20)L}{G_sJ_{60}} + \frac{(T_A+16)L}{G_cJ_{80}} = 0$$

With $\dfrac{L}{G_cJ_{80}} = 5.6518\times10^{-9}$ and $\dfrac{L}{G_sJ_{60}} = 1.0076\times10^{-8}$ (per N·mm):

$$T_A = -\frac{20\,b + 16\,a}{2a + b} = -13.656\ \text{kN·m}$$

Torsional reactions

• At A: $|T_A| = 13.656$ kN·m, acting opposite to $T_B$.
• At D: $|T_D| = 2.344$ kN·m, acting opposite to $T_B$.

Check: $13.656 + 2.344 = 16.000 = 20 - 4$ kN·m.

Rotations of the joints (D is fixed, so its rotation is zero)

$$\varphi_B = \frac{|T_{AB}|L}{G_cJ_{80}} = 0.07718\ \text{rad} = 4.422^\circ$$ $$\varphi_C = \varphi_B - \frac{T_{BC}L}{G_sJ_{60}} = 0.01325\ \text{rad} = 0.759^\circ$$

Both are in the direction of the 20 kN·m torque, and $\varphi_D = 0$ (check: $\varphi_C - \frac{T_{CD}L}{G_cJ_{80}} = 0$).

Maximum shear stress ($\tau = \dfrac{16T}{\pi d^3}$)