GATE AG 2025 • Q55 NAT • 2 Marks

Section 2: Farm Machinery

Machine Design

Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts

Question Statement:

A solid round uniform-diameter shaft transmits 25 kW at 540 rpm. Allowable shear stress = 35 MPa. Maximum torque exceeds mean torque by 20% (bending neglected). Find the minimum shaft diameter (mm). (Rounded to 2 decimal places)

Official Answer Key:

41.50 to 43.50

Step-by-Step Solution & Explanation

Mean torque T_mean = P×60/(2πN) = 25000×60/(2π×540) = 442.10 N·m. Max torque T_max = 1.2×442.10 = 530.52 N·m. From torsion equation τ = 16T/(πd³): d = [16×T_max/(π×τ_allow)]^(1/3) = [16×530.52/(π×35×10⁶)]^(1/3) = 0.04258 m ≈ 42.58 mm.

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