Hollow shaft stronger than solid shaft (same weight)

Statement: for the same material, same length and same weight (equal cross-sectional area), a hollow shaft can transmit more torque than a solid shaft, so it is stronger in torsion.

Let the allowable shear stress be $\tau$. Let the solid shaft have diameter $D_s$ and the hollow shaft external diameter $D$ and internal diameter $d$, with $k = d/D$.

Equal weight, material and length means equal area

$$\frac{\pi}{4}D_s^2 = \frac{\pi}{4}(D^2-d^2) \Rightarrow D_s^2 = D^2 - d^2 = D^2(1-k^2)$$

Torque capacity from the torsion formula $T = \dfrac{\tau J}{R}$

Solid shaft:

$$T_s = \frac{\pi}{16}\tau D_s^3$$

Hollow shaft:

$$T_h = \frac{\pi}{16}\tau\,\frac{D^4-d^4}{D} = \frac{\pi}{16}\tau\,\frac{(D^2-d^2)(D^2+d^2)}{D}$$

Ratio

Using $D_s^2 = D^2 - d^2$:

$$\frac{T_h}{T_s} = \frac{(D^2-d^2)(D^2+d^2)}{D\,D_s^3} = \frac{D^2+d^2}{D\,D_s} = \frac{D^2(1+k^2)}{D\cdot D\sqrt{1-k^2}} = \frac{1+k^2}{\sqrt{1-k^2}}$$

For any $0 < k < 1$:

$$(1+k^2)^2 - (1-k^2) = 3k^2 + k^4 > 0 \Rightarrow \frac{1+k^2}{\sqrt{1-k^2}} > 1$$

Therefore $T_h > T_s$: the hollow shaft carries a larger torque for the same weight.

Numerical illustration

(for $k = 0.6$: $\frac{1+0.36}{\sqrt{1-0.36}} = \frac{1.36}{0.8} = 1.70$).

Reason: in a solid shaft the material near the axis is lightly stressed because shear stress varies linearly from zero at the centre to $\tau$ at the surface. In the hollow shaft the same material is moved to the outside, where the lever arm and the stress are greatest, so it is used more efficiently.