Question

A bending moment of 400 kNm and a twisting moment of 300 kNm is applied on a circular solid shaft. If direct stress based on maximum principal stress theory is σ and shear stress based on maximum shear stress theory is τ, then \(\frac{\sigma }{\tau }\) is:

1. \(\frac{5}{9}\)
2. \(\frac{5}{3}\)
3. \(\frac{3}{5}\)
4. \(\frac{9}{5}\)

Answer 1

Correct Answer - Option 4 : \(\frac{9}{5}\)

Concept:

According to maximum principal stress theory:

For solid shaft,

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

σ_max = \(\frac{{32{M_e}}}{{\pi {d^3}}}\) = \(\frac{{16}}{{\pi {d^3}}}\left[ {{M_b} + \sqrt {{{\left( {{M_b}} \right)}^2} + {{\left( {{M_t}} \right)}^2}} } \right]\)

According to maximum shear stress theory:

For solid shaft,

T_e = \(\sqrt {\left( {{M_b}} \right) + \left( {{M_t}} \right)} \)

τ_max = \(\frac{{16{T_e}}}{{\pi {d^3}}}\) = \(\frac{{16}}{{\pi {d^3}}}\left[ {\sqrt {{{\left( {{M_b}} \right)}^2} + {{\left( {{M_t}} \right)}^2}} } \right]\)

Where, 

σmax = Direct stress, τ_max = Shear stress, Me = Equivalent moment, T_e = Equivalent twisting moment, Mb = Bending moment, Mt = Twisting moment, d = Diameter of circular shaft.

Calculation:

Given:

Mb = 400 kNm, Mt = 300 kNm 

\(\frac{\sigma }{\tau } = \frac{{\frac{{16}}{{\pi {d^3}}}\left[ {{M_b} + \;\sqrt {{{\left( {{M_b}} \right)}^2} + {{\left( {{M_t}} \right)}^2}} } \right]}}{{\frac{{16}}{{\pi {d^3}}}\left[ {\sqrt {{{\left( {{M_b}} \right)}^2} + {{\left( {{M_t}} \right)}^2}} } \right]}}\)

= \(\frac{{\left[ {400 + \sqrt {{{400}^2} + {{300}^2}} } \right]}}{{\sqrt {{{400}^2} + {{300}^2}} }}\)

= \(\frac{9}{5}\)

 

The maximum Shear stress theory is applicable for ductile materials.