The polar moment of inertia $\mathrm{J}$ for a hollow pipe or tube shaft is found from:

$\mathrm{J}=\frac{\mathrm{\pi }}{32}({\mathrm{D}}^4-{\mathrm{d}}^4)$

where $\mathrm{D}$ is the outside diameter and $\mathrm{d}$ is the inside diameter. For this shaft,

$$\begin{gathered} \mathrm{J}=\frac{\mathrm{\pi }}{32}\left[{(75 \mathrm{mm})}^4-{(65 \mathrm{mm})}^4\right] \\ =1,353,830.1 {\mathrm{mm}}^4 \end{gathered}$$