Subtopic: Measurement of force, stress, torque, speed, displacement and acceleration on machine elements – shafts, couplings, keys, bearings and knuckle joints
Question Statement:
A tractor PTO shaft, through a spur-gear reduction, drives a small grain auger at 540 rpm. The driving gear on the PTO shaft has 0.18 m pitch circle diameter. The auger requires 3.5 kW, transmitted through this gear. Normal force along the line of action between the meshing teeth is 760 N. Assume involute spur gears, single tooth contact, and neglect friction. The pressure angle (in degrees) is ________. (Rounded off to two decimal places) (Take π = 3.14)
Official Answer Key:
24.00 to 27.00
Step-by-Step Solution & Explanation
Step 1 — torque on the driving gear:
ω = 2πN/60 = 2 × 3.14 × 540/60 = 56.52 rad/s
T = P/ω = 3500/56.52 = 61.93 N·m
Step 2 — tangential (transmitted) force at pitch radius r = 0.18/2 = 0.09 m:
Fₜ = T/r = 61.93/0.09 = 688.1 N
Step 3 — pressure angle. In involute gearing, the resultant tooth force Fₙ acts along the line of action, inclined at the pressure angle φ to the common tangent at the pitch point:
Fₜ = Fₙ cos φ
cos φ = 688.1/760 = 0.90534
φ = cos⁻¹(0.90534) = 25.13°
(Sanity check: this is close to the standard 25° involute system.)
SECTION 3: FARM POWER
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