Subtopic: Overload safety devices used in farm machinery
Question Statement:
A tractor PTO runs at a constant speed of 540 rpm to drive a rotary tiller via a flange coupling having a shear pin parallel to the shaft. The shear pin axis is located at 60 mm from the PTO shaft axis. The allowable shear stress of the pin material is 200 MPa. For overload safety, it must fail at 150% of the rated torque to protect the gearbox. The tiller requires 40 kW under normal load at rated PTO speed. Neglecting bending, stress concentration effects and other losses, the pin diameter, in mm, is nearest to (Take π = 3.14)
Options:
(A) 7.5
(B) 10.6
(C) 6.1
(D) 8.6
Official Answer Key:
B
Step-by-Step Solution & Explanation
Step 1 — rated torque:
ω = 2πN/60 = 2 × 3.14 × 540/60 = 56.52 rad/s
T_rated = P/ω = 40,000/56.52 = 707.7 N·m
Step 2 — design (failure) torque at 150% overload:
T_fail = 1.5 × 707.7 = 1061.6 N·m
Step 3 — shear force carried by the pin at radius 60 mm:
F = T_fail/r = 1061.6/0.06 = 17,693 N
Step 4 — required cross-sectional area (pin in single shear, since it is parallel to the shaft through the flange):
A = F/τ_allow = 17,693/(200 × 10⁶) = 8.847 × 10⁻⁵ m² = 88.47 mm²
Step 5 — pin diameter:
d = √(4A/π) = √(4 × 88.47/3.14) = √112.7 = 10.62 mm ≈ 10.6 mm
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