${\tau }_{\mathit{max}}=\frac{{500×10}^{3}}{0.2218×{40×30}^{2}}=62.6\mathrm{N}-{\mathrm{mm}}^{2}$

#### Results

1. Angle of twist is $0.032\mathrm{rad}$

2. Maximum shear stress is $62.6\mathrm{N}-{\mathrm{mm}}^{2}$

### Determination of Allowable Torque

Problem 5: A solid shaft having its outer diameter of $100\mathrm{mm}$. Maximum shear stress of $80\mathrm{MPa}$ is applied on ...

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