$\mathrm{D}=40 \mathrm{mm}$

and

$$\begin{gathered} \mathrm{d}=\mathrm{D}-2\mathrm{t} \\ =40 \mathrm{mm}-2(5 \mathrm{mm}) \\ =30 \mathrm{mm} \end{gathered}$$

The polar moment of inertia $\mathrm{J}$ for the hollow shaft is computed as:

$$\begin{gathered} \mathrm{J}=\frac{\mathrm{\pi }}{32}({\mathrm{D}}^4-{\mathrm{d}}^4) \\ =\frac{\mathrm{\pi }}{32}\left[{(40 \mathrm{mm})}^4-{(30 \mathrm{mm})}^4\right]=171,805.85 {\mathrm{mm}}^4 \end{gathered}$$