SSC JE Mechanical Engineering 2018 Set 5 Paper

Question : 124 of 200

Marks: +1, -0

What is the equivalent torque (Nm) and bending moment (Nm) in a member subjected to the torque of magnitude 400 Nm and the bending moment of 300 Nm?

Solution:

Equivalent Torque is given by,

T_e=√{M^2+T^2}

T_e=(300^2+400^2)^{1\/2}

T_e = 500

Nm
Equivlent bending moment :

1/2[M+√{M^2+T^2]}

M_e=1/2[ 300+ 500]

M_e = 400 \Nm

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