$$\begin{gathered} \mathrm{J}=\frac{\mathrm{\pi }}{32}{\mathrm{d}}^4 \\ =\frac{\mathrm{\pi }}{32}{(25 \mathrm{mm})}^4 =38,350 {\mathrm{mm}}^4 \end{gathered}$$

Using the internal torques ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$, compute the shear stress in each shaft segment.

${\mathrm{\tau }}_1=\frac{{\mathrm{T}}_1\mathrm{c}}{\mathrm{J}}=\frac{(400 \mathrm{N}-\mathrm{m})(12.5 \mathrm{mm})(1,000 \mathrm{mm/m})}{38,350 {\mathrm{mm}}^4}=\mathbf{130}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{MPa}$

${\mathrm{\tau }}_2=\frac{{\mathrm{T}}_2\mathrm{c}}{\mathrm{J}}=\frac{(-200 \mathrm{N}-\mathrm{m})(12.5 \mathrm{mm})(1,000 \mathrm{mm/m})}{38,350 {\mathrm{mm}}^4}=\mathbf{-}\mathbf{65}\mathbf{.}\mathbf{2}\mathbf{ }\mathbf{MPa}$