Question

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 sun gear (26 teeth) is input, 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)

Hint:

Find the wheel torque from the tractive force, get the planetary speed reduction ratio, then divide by both mesh efficiencies.

Answer 1

Correct Answer: 2.81

Step 1: Find the torque needed at each rear-wheel hub.
The total gross tractive force of 30 kN is shared equally by the two driven rear wheels, so each wheel must push with 15 kN.
Wheel (carrier) torque \(T_c = 15 \times 0.73 = 10.95\) kN.m.

Step 2: Find the speed reduction ratio of the planetary set.
With the ring gear fixed, sun as input and carrier as output, the standard epicyclic relation gives
\[ \frac{\omega_{sun}}{\omega_{carrier}} = \frac{Z_s + Z_r}{Z_s} = \frac{26+78}{26} = 4 \] So the sun shaft turns 4 times for every carrier turn, meaning ideal torque is stepped up 4 times from sun to carrier.

Step 3: Find the combined mesh efficiency.
Power passes through one external mesh (sun to planet) and one internal mesh (planet to ring, which reacts against the fixed ring).
\(\eta = 0.985 \times 0.99 = 0.97515\).

Step 4: Find the required sun-shaft torque.
Without losses the sun torque would be \(T_c/4 = 10.95/4 = 2.7375\) kN.m.
Since losses make the input work harder for the same output, divide by the efficiency:
\[ T_{sun} = \frac{T_c/4}{\eta} = \frac{2.7375}{0.97515} = 2.807 \]

Final Answer:
The required sun-shaft input torque per rear-wheel is \[ \boxed{2.81\ kN.m} \]