Hollow shaft 40 kW at 240 rpm: diameters

Data: $P = 40$ kW, $N = 240$ rpm, $\tau_{max} = 100\ \text{MPa}$, $D = 2d$ (so $k = d/D = 0.5$ and $1-k^4 = 0.9375$).

Torque

$$T = \frac{60P}{2\pi N} = \frac{60(40\times10^3)}{2\pi(240)} = 1591.5\ \text{N·m} = 1.5915\times10^{6}\ \text{N·mm}$$

Diameters

$$T = \frac{\pi}{16}\tau\,D^3(1-k^4) \Rightarrow D^3 = \frac{16T}{\pi\tau(1-k^4)} = \frac{16(1.5915\times10^{6})}{\pi(100)(0.9375)} = 8.6461\times10^{4}\ \text{mm}^3$$ $$D = 44.22\ \text{mm}, \qquad d = \frac{D}{2} = 22.11\ \text{mm}$$

Answer: external diameter $\approx 44.2$ mm, internal diameter $\approx 22.1$ mm. (In practice adopt 45 mm and 22.5 mm.)