This equation can be solved for the minimum permissible outside diameter $\mathrm{D}$:

$$\begin{gathered} \frac{\mathrm{\pi }}{16}{\mathrm{D}}^3=\frac{5 \mathrm{kN}-\mathrm{m}}{65 \mathrm{MPa}}=\frac{(5 \mathrm{kN}-\mathrm{m})(1,000 \mathrm{N}/\mathrm{kN})(1,000 \mathrm{mm}/\mathrm{m})}{65 \mathrm{N}/{\mathrm{mm}}^2} \\ =76,923.077 {\mathrm{mm}}^3 \end{gathered}$$

${\mathrm{D}}_{\min }=73.2 {\mathrm{mm}}^{}$