When a shaft is subjected to a bending moment M and a twisting moment T, then the equivalent twisting moment is equal to 

Explanation:

Circular cross-section when subjected to pure bending develops normal stress which is given by:

\({σ _b} = \frac{M}{I}{y_{max}} \Rightarrow \frac{M}{{\frac{{\pi {d^4}}}{{64}}}} \times \frac{d}{2} = \frac{{32M}}{{\pi {d^3}}}\)

Circular cross-section when subjected to pure twisting moment develops shear stress which is given by:

\({τ _t} = \frac{T}{J}{r_{max}} \Rightarrow \frac{T}{{\frac{{\pi {d^4}}}{{32}}}} \times \frac{d}{2} = \frac{{16T}}{{\pi {d^3}}}\)

The combined effect of bending and torsion produces principal stress which is given by:

\({σ _{1,2}} = \frac{{{σ _x} + {σ _y}}}{2} \pm \sqrt {{{\left\{ {\frac{{{σ _x} - {σ _y}}}{2}} \right\}}^2} + {τ ^2}} \)

\({σ _{1,2}} = \left\{ {\frac{{\frac{{32M}}{{\pi {d^3}}}}}{2}} \right\} \pm \sqrt {{{\left\{ {\frac{{\frac{{32M}}{{\pi {d^3}}}}}{2}} \right\}}^2} + {{\left\{ {\frac{{16T}}{{\pi {d^3}}}} \right\}}^2}} \)

\({σ _1} = \frac{{16}}{{\pi {d^3}}}\left\{ {M + \sqrt {{M^2} + {T^2}} } \right\}\)

\({σ _2} = \frac{{16}}{{\pi {d^3}}}\left\{ {M - \sqrt {{M^2} + {T^2}} } \right\}\)

Maximum shear stress:

\({τ _{max}} = \frac{{{σ _1} - {σ _2}}}{2} \Rightarrow \frac{{16}}{{\pi {d^3}}}\left\{ {\sqrt {{M^2} + {T^2}} } \right\}\)

Equivalent Twisting TM:

It is the twisting moment Teq that alone produces maximum shear stress equal to the maximum shear stress produce due to combined bending and torsion.

Let Teq be equivalent TM.

\(τ = \frac{{16{T_{eq}}}}{{\pi {d^3}}}\)

As per the definition of Teq τ  = τmax

\(\frac{{16{T_{eq}}}}{{\pi {d^3}}}= \frac{{16}}{{\pi {d^3}}}\left\{ {\sqrt {{M^2} + {T^2}} } \right\}\)

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

Additional Information

The effect of torsion and bending can be combined in such a way that single action either bending or torsion may result in normal stress or shear stress. 

Equivalent Bending BM:

It is the BM that alone produces maximum normal stress equal to the maximum normal stress produce due to combined bending and torsion.

Let Meq be equivalent BM.

\(σ = \frac{{32{M_{eq}}}}{{\pi {d^3}}}\)

As per the definition of equivalent BM σ = σ1

\(\frac{{32{M_{eq}}}}{{\pi {d^3}}}=\frac{{16}}{{\pi {d^3}}}\left\{ {M + \sqrt {{M^2} + {T^2}} } \right\}\)

\(\therefore{M_{eq}} = \frac{1}{2}\left\{ {M + \sqrt {{M^2} + {T^2}} } \right\}\)