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GATE 2018 ME (ME2) – Question 41

Mechanics of Materials · Torsion of circular shafts · 2 marks · Multiple choice

A circular bar is clamped at P and Q. A torque T = 150 N·m acts 100 mm from P, with another 200 mm to Q. Find torsional reactions (TP, TQ) at the ends in N·m.

Diagram for GATE 2018 ME (ME2) question 41
  1. (50, 100)
  2. (75, 75)
  3. (100, 50)
  4. (120, 30)

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Correct answer: (C) (100, 50)

Explanation

Equilibrium requires $T_P+T_Q=150$ N·m. Zero relative end twist requires $T_P(100)/(GJ)=T_Q(200)/(GJ)$, hence $T_P=2T_Q$. Solving gives **(100, 50) N·m, C**.