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## Q. 10.9

Two steel shafts each of length L and outside diameter d are subjected to uniform torsion. The first shaft is solid, whereas the second one is hollow with inside diameter d/2. What is the ratio of the strain energies that they can absorb without exceeding a maximum allowable working shear stress, $\tau_w$?

## Verified Solution

Solid shaft: Strain energy due to torsion,

$U_1=\frac{T_1^2 L}{2 G J_1}$

But,

$T_1=\frac{\pi}{16} d^3 \tau_{ w } \quad \text { and } \quad J_1=\frac{\pi d^4}{32}$

Therefore,

$U_1=\frac{(\pi / 16)^2 d^6 \tau_{ w }^2 L}{G\left(\pi d^4 / 16\right)}=\frac{\pi}{16 G} d^2 L \tau_{ w }^2$              (1)

Hollow Shaft: Again,

$U_2=\frac{T_2^2 L}{2 G J_2}$

where

$\tau_{ w }=\frac{T_2\left\lgroup \frac{d}{2} \right\rgroup }{\frac{\pi}{32}\left\lgroup d^4-\frac{d^4}{2^4} \right\rgroup }=\frac{16^2 T_2}{15 \pi d^3}$

which gives

$T_2=\left\lgroup\frac{15 \pi}{16^2} \right\rgroup d^3 \tau_{ w } \quad \text { and } \quad J_2=\frac{\pi}{32}\left\lgroup d^4-\frac{d^4}{2^4} \right\rgroup =\frac{15 \pi}{(16)(32)} d^4$

Therefore,

$U_2=\frac{15^2 \pi^2}{16^4} \cdot \frac{d^6 \tau_{ w }^2 \cdot L}{15 \pi d^4} 16^2=\frac{15 \pi d^2 \tau_{ w }^2 L}{16^2 G}$             (2)

From Eqs. (1) and (2), we obtain $U_1 / U_2$ = 16/15.