A circular shaft of diameter d and length l is subjected to a torque T and a bending moment M. The ratio of maximum shear stress to bending stress is:

Explanation:

In some applications the shaft is simultaneously subjected to bending moment M and Torque T.

From the simple bending theory equation:

\(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

If σ_b is the maximum bending stresses due to bending moment M on shaft:

\({\sigma _b} = \frac{{32M}}{{\pi {d^3}}}\)

Torsion equation:

\(\frac{T}{J} = \frac{\tau }{r} = \frac{{G\theta }}{L}\)

The maximum shear stress developed on the surface of the shaft due to twisting moment T:

\(\tau = \frac{{16T}}{{\pi {d^3}}}\)

The ratio of the maximum shear stress to maximum bending stress developed in a shaft is

\(\frac{\tau }{\sigma } = \frac{{\left( {\frac{{16T}}{{\pi {D^3}}}} \right)}}{{\left( {\frac{{32M}}{{\pi {D^3}}}} \right)}} = \frac{T}{{2M}}\)

Additional Information

Equivalent Bending Moment:

\({M_e} = \frac{1}{2}\left[ {M + \sqrt {{M^2} + {T^2}} } \right]\)

Equivalent Torque:

\({T_e} = \sqrt {{M^2} + {T^2}}\)