Question

A solid shaft is subjected to bending moment of 3.46 kN-m and a torsional moment of 11.5 kN-m. For this case, the equivalent bending moment and twisting moment are
1. 7.73 kN-m and 12.0 kN-m
2. 14.96 kN-m and 12.0 kN-m 
3. 7.73 kN-m and 8.04 kN-m
4. 14.96 kN-m and 8.04 kN-m

Answer 1

Correct Answer - Option 1 : 7.73 kN-m and 12.0 kN-m

Concept:

Equivalent moment when a shaft is subjected to bending and torsional moment is given by

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

Equivalent torque when a shaft is subjected to bending and torsional moment is given by

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

where, M and T are bending moment and torsional moment subjected on shaft respectively.

Calculation:

Given M = 3.46 kN-m and T = 11.5 kN-m

\({M_e} = \frac{1}{2}\left[ {3.46 + \sqrt {{{3.46}^2} + {{11.5}^2}} } \right] = 7.734\;kN - m\)

\({T_e} = \sqrt {{{3.46}^2} + {{11.5}^2}} = 12\;kN - m\;\)