Subtopic: Steering and hydraulic control systems used in tractors
Question Statement:
A stationary Ackerman-steer tractor on a dry, clean concrete surface carries 12 kN vertical load on each front steered wheel of 0.28 m nominal tyre width (b). Assume uniform pressure distribution on the circular tyre-print area having diameter b. Effective friction coefficient between tyre and surface is 0.30.
The steering is individually power-assisted by a single-rod hydraulic cylinder, acting through a pitman arm resulting in a 60 mm effective moment arm. The hydraulic relief valve is set to 8 MPa, and the cylinder bore is 32 mm (ignore rod area). Neglect other losses. The maximum kingpin offset (in mm), which does not cause the maximum allowable kingpin torque to be exceeded, is ________. (Rounded off to the nearest integer) (Take π = 3.14)
Official Answer Key:
37 to 45
Step-by-Step Solution & Explanation
Part 1 — maximum kingpin torque the hydraulics can deliver.
Cylinder piston area: A = (π/4)d² = (3.14/4)(0.032)² = 8.038 × 10⁻⁴ m²
Cylinder force at relief setting: F = p·A = 8 × 10⁶ × 8.038 × 10⁻⁴ = 6431 N
Torque about the kingpin: T_max = F × moment arm = 6431 × 0.060 = 385.8 N·m
Part 2 — resisting torque of a stationary steered tyre.
For a tyre print of diameter b under uniform pressure, turning about a kingpin axis offset a distance P from the print centre, the friction-resisting torque (Taborek’s relation) is:
T = μW √(P² + b²/8)
The b²/8 term is the “spin” or scrub torque that exists even at zero offset; the P² term adds the effect of the offset.
Part 3 — solve for the offset.
μW = 0.30 × 12,000 = 3600 N
385.8 = 3600 × √(P² + 0.28²/8)
√(P² + 0.0098) = 385.8/3600 = 0.10718
P² + 0.0098 = 0.011487
P² = 0.001687
P = 0.0411 m ≈ 41 mm
SECTION 4: SOIL AND WATER CONSERVATION ENGINEERING
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