T=\frac{P}{\omega} \Rightarrow F=\frac{T}{(D/2)}
S_{sy}=0.577 S_{y}
L_{shear} = \frac{Fn}{bS_{sy}}, L_{crush} = \frac{2Fn}{hS_{y}}
According to Budynas (p. 390-394), shaft diameter determine standard sizes for width, height, and key depth. The designer chooses an appropriate key length to carry the torsional load. Failure of the key can be by direct shear, or by bearing stress. |