Question

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

• A. $$\sqrt{136}kN-m$$
• B. $$16kN-m$$
• C. $$\sqrt{16}kN-m$$
• D. $$8kN-m$$

Answer 1

Answer:

$$\sqrt{136}kN-m$$

Given: M = 6kN - m and T = 10kN - m The equivalent torque is given by: $T_{eq} =[ \sqrt {M^ 2 +T^ 2} ] \Rightarrow T_eq = \sqrt {6^ 2 +10^ 2} = \sqrt {136} kN-m$