Hollow shaft against solid shaft

Torque

$$T = \frac{60P}{2\pi n} = \frac{60 \times 90\,000}{2\pi \times 250} = 3437.7\ \text{N.m}$$

Solid shaft

$$\begin{aligned} T &= \frac{\pi}{16}\tau d^3 \\ d^3 &= \frac{16 \times 3437.7\times10^3}{\pi \times 40} = 4.377\times10^5\ \text{mm}^3 \\ d &= 75.9\ \text{mm} \end{aligned}$$

Hollow shaft ($k = d_i/d_o = 0.6$)

For the same torque and stress,

$$\frac{\pi}{16}\tau d_o^3(1 - k^4) = T$$ $$d_o^3 = \frac{d^3}{1 - k^4} = \frac{4.377\times10^5}{1 - 0.6^4} = \frac{4.377\times10^5}{0.8704} = 5.029\times10^5\ \text{mm}^3$$ $$d_o = 79.5\ \text{mm}, \qquad d_i = 0.6 \times 79.5 = 47.7\ \text{mm}$$

Mass saving

Mass is proportional to cross-sectional area (same length and material):

$$\frac{A_h}{A_s} = \frac{d_o^2(1 - k^2)}{d^2} = \frac{79.5^2 \times 0.64}{75.9^2} = 0.702$$

Saving $= 1 - 0.702 = 0.298$, about 30 percent.

Hollow shafts save material because the inner core carries little shear stress. The outside diameter is only about 5 percent larger, but hollow shafts are costlier to make.

Answer: Solid d = 75.9 mm; hollow $d_o$ = 79.5 mm, $d_i$ = 47.7 mm; mass saving about 29.8 percent.