Subtopic: Design and selection of machine elements – gears, pulleys, chains and sprockets and belts
Question Statement:
In a single V-belt drive having 30° groove angle, the cross-sectional area of the belt is 250 mm² and the mass density of the belt material is 1200 kg m⁻³. The maximum stress bearing capacity of the belt material is 6000 kN m⁻². Coefficient of friction between the belt and the pulley is 0.2 and the angle of lap on the driving pulley is 150°. The maximum power transmitted through the V-belt in kW is
Options:
(A) 5.6
(B) 10.5
(C) 17.7
(D) 35.4
Official Answer Key:
D
Step-by-Step Solution & Explanation
Maximum allowable tension T_max = σ_max·A = 6000×10³ × 250×10⁻⁶ = 1500 N. Maximum power occurs when centrifugal tension Tc = T_max/3 = 500 N. Belt velocity at this condition: Tc = m·v² (m = ρA = 1200×250×10⁻⁶ = 0.30 kg/m). v = √(Tc/m) = √(500/0.30) = 40.82 m/s. T1 = T_max − Tc = 1500−500 = 1000 N. For a V-belt: T1/T2 = e^(μθ/sinβ), β = half groove angle = 15°. T1/T2 = e^(0.2×2.618/sin15°) = e^(0.5236/0.2588) = e^2.023 = 7.56. T2 = 1000/7.56 = 132.3 N. Power = (T1−T2)·v = (1000−132.3)×40.82 = 35,425 W ≈ 35.4 kW, matching option (D).
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