Hollow shaft 1670 kW at 4500 rpm: size

Data: $L = 1$ m, $P = 1670$ kW, $N = 4500$ rpm, $D = 1.75d$, $\tau_{max} = 210\ \text{MPa}$, $\theta \le 0.5^\circ$, $T_{max} = 1.3\,T_{mean}$, $G = 25.5\ \text{GPa}$. So $k = d/D = 1/1.75 = 0.5714$ and $1-k^4 = 0.8934$.

Torque

$$T_{mean} = \frac{60P}{2\pi N} = \frac{60(1670\times10^3)}{2\pi(4500)} = 3543.9\ \text{N·m}, \qquad T_{max} = 1.3T_{mean} = 4607.0\ \text{N·m}$$

Strength condition

$$T_{max} = \frac{\pi}{16}\tau D^3(1-k^4) \Rightarrow D^3 = \frac{16(4.6070\times10^{6})}{\pi(210)(0.8934)} = 1.2506\times10^{5}\ \text{mm}^3$$ $$D_1 = 50.01\ \text{mm}$$

Stiffness condition

$$\theta = \frac{T_{max}L}{GJ}, \quad J = \frac{\pi}{32}D^4(1-k^4), \quad \theta = 0.5^\circ = 0.00873\ \text{rad}$$ $$D^4 = \frac{32\,T_{max}L}{\pi G\theta(1-k^4)} = \frac{32(4.6070\times10^{6})(1000)}{\pi(25\,500)(0.00873)(0.8934)} = 2.3605\times10^{8}\ \text{mm}^4$$ $$D_2 = 123.95\ \text{mm}$$

Size of the shaft

The larger value governs (stiffness controls):

$$D = 123.95\ \text{mm}, \qquad d = \frac{D}{1.75} = 70.83\ \text{mm}$$

Answer: external diameter $\approx 124.0$ mm, internal diameter $\approx 70.8$ mm.