Torsional rigidity; hollow shaft vs solid shaft strength

Torsional rigidity is the torque required to produce a twist of one radian per unit length of the shaft; it equals the product $GJ$ of the modulus of rigidity and the polar moment of inertia (from $\dfrac{T}{\varphi} = \dfrac{GJ}{L}$). Unit: N·mm² (or N·m²).

Torque of the hollow shaft

Data: $D = 100$ mm, $d = 70$ mm, $\tau = 80\ \text{MN/m}^2 = 80\ \text{N/mm}^2$, $N = 300$ rpm (the speed is not needed for the torque).

$$J = \frac{\pi}{32}(100^4 - 70^4) = 7.4603\times10^{6}\ \text{mm}^4$$ $$T_h = \frac{\tau J}{R} = \frac{80\times7.4603\times10^{6}}{50} = 1.1936\times10^{7}\ \text{N·mm} = 11.936\ \text{kN·m}$$

(Power at 300 rpm would be $P = \frac{2\pi N T}{60} = 375.0$ kW.)

Comparison with a solid shaft of the same weight

Same material, weight and length means equal area:

$$\frac{\pi}{4}D_s^2 = \frac{\pi}{4}(100^2 - 70^2) \Rightarrow D_s = \sqrt{5100} = 71.41\ \text{mm}$$ $$T_s = \frac{\pi}{16}\tau D_s^3 = \frac{\pi}{16}(80)(71.41)^3 = 5.7210\times10^{6}\ \text{N·mm} = 5.721\ \text{kN·m}$$ $$\frac{T_h}{T_s} = \frac{11.936}{5.721} = 2.086$$

Answer: torque of the hollow shaft $= 11.94$ kN·m; it is 2.09 times as strong as the solid shaft of equal weight (solid shaft torque $= 5.72$ kN·m).