Subtopic: Overload safety devices used in farm machinery
Question Statement:
An open belt drive system transmits 5 kW power using a flat belt of width and thickness as 80 mm and 5 mm, respectively. The driving shaft speed is 1500 rpm and the driven shaft speed is 500 rpm. The coefficient of friction between the belt and pulley is 0.20, and the wrap angle of the belt is 168°. If diameter of the smaller pulley is 200 mm, maximum stress induced in the belt will be _________ MPa.
Official Answer Key:
1.70 to 1.80
Step-by-Step Solution & Explanation
Speed ratio 1500:500 = 3:1, so the driver is the smaller pulley (D = 0.2 m, N = 1500 rpm).
Belt velocity: v = πDN/60 = π(0.2)(1500)/60 = 15.708 m s⁻¹.
Power: P = (T₁ − T₂)v ⇒ T₁ − T₂ = 5000/15.708 = 318.3 N.
Wrap angle θ = 168° = 2.9322 rad; μθ = 0.20 × 2.9322 = 0.5864; e^{μθ} = 1.7975.
T₁/T₂ = 1.7975 ⇒ T₂ = 0.5563 T₁ ⇒ T₁(1 − 0.5563) = 318.3 ⇒ T₁ = 717.4 N.
Belt cross-section A = 80 × 5 = 400 mm².
σ_max = T₁/A = 717.4/400 = 1.79 N mm⁻² = 1.79 MPa — within the official range 1.70–1.80.
Centrifugal tension is neglected as belt mass per unit length is not given.
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