Hollow versus solid shaft for power

Torque to be transmitted

$$T = \frac{60P}{2\pi N} = \frac{60\times150\times10^3}{2\pi\times300} = 4774.6\ \text{N m} = 4.775\times10^6\ \text{N mm}$$

Solid shaft

$$T = \frac{\pi}{16}\tau d^3 \;\Rightarrow\; d^3 = \frac{16\times4.775\times10^6}{\pi\times60} = 405.3\times10^3$$ $$d = 74.0\ \text{mm} \;\rightarrow\; \text{provide } 75\ \text{mm}$$

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

$$T = \frac{\pi}{16}\tau\,\frac{d_o^4 - d_i^4}{d_o} = \frac{\pi}{16}\tau\,d_o^3(1-k^4)$$ $$1 - k^4 = 1 - 0.6^4 = 0.8704$$ $$d_o^3 = \frac{16\times4.775\times10^6}{\pi\times60\times0.8704} = 465.7\times10^3 \;\Rightarrow\; d_o = 77.5\ \text{mm}$$ $$d_i = 0.6\times77.5 = 46.5\ \text{mm}$$

Provide $d_o = 78$ mm, $d_i = 47$ mm.

Weight comparison

For the same length and material, weight is proportional to cross-sectional area:

$$\frac{A_{hollow}}{A_{solid}} = \frac{d_o^2(1-k^2)}{d^2} = \frac{77.5^2(1-0.36)}{74.0^2} = \frac{3843.6}{5476} = 0.702$$ $$\text{Saving} = 1 - 0.702 = 29.8\ \%$$

The hollow shaft is more economical because the material near the axis, where shear stress is small, is removed.

Answer: hollow shaft $d_o = 77.5$ mm (78 mm), $d_i = 46.5$ mm (47 mm); solid shaft $d = 74.0$ mm (75 mm); weight saving $\approx 29.8\ \%$.