Subtopic: Power transmission systems – gear trains, differential, final drives and power take-off
Question Statement:
A 2WD tractor has to develop 30 kN gross tractive force at the ground. Rolling radius of the rear wheels is 0.73 m. Each rear wheel is driven by a simple planetary final-drive in which the sun gear (26 teeth) is input, the ring gear (78 teeth) is fixed, and the carrier is bolted to the rear-wheel hub. The sun-planet external mesh efficiency is 98.5%, while the planet-ring internal mesh efficiency is 99%. Losses upstream of the final-drive are neglected. The required sun-shaft input torque per rear wheel (in kN·m) is ________. (Rounded off to two decimal places)
Official Answer Key:
2.77 to 2.90
Step-by-Step Solution & Explanation
Step 1 — force and torque at one wheel. The 30 kN is shared by two rear wheels:
F_wheel = 30/2 = 15 kN
T_wheel = F × r = 15 × 0.73 = 10.95 kN·m
Step 2 — planetary speed ratio (sun input, ring fixed, carrier output). For a simple planetary set:
n_carrier/n_sun = Z_s/(Z_s + Z_r) = 26/(26 + 78) = 26/104 = 1/4
Since power = torque × speed, and the carrier turns at ¼ the sun speed, the ideal torque multiplication is 4 : 1.
Step 3 — ideal sun torque:
T_sun,ideal = 10.95/4 = 2.7375 kN·m
Step 4 — apply mesh efficiencies. Power flows through both meshes in series:
η_total = 0.985 × 0.99 = 0.97515
Step 5 — actual input torque required:
T_sun = 2.7375/0.97515 = 2.81 kN·m
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