A shaft is subjected to combined bending and torsional moments of 6 kN-m and 10 kN-m respectively. The equivalent torque will be equal to

Question

Question

This question was previously asked in

MPPSC AE Mechanical 2021 Official Paper

Answer 1

Concept:

The equivalent bending moment for a shaft which is subjected to the combined twisting moment (T) and bending moment (M) is given by the equation:

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

\({T_{eq}} = \left[ {\sqrt {{M^2} + {T^2}} } \right]\)

Calculation:

Given:

M = 6 kN-m and T = 10 kN-m

The equivalent torque is given by:

\({T_{eq}} = \left[ {\sqrt {{M^2} + {T^2}} } \right]\)

\({T_{eq}}=\sqrt {{{6}^2} + {{10}^2}} =\sqrt{136}\;kN-m\)