Question

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 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 \(\pi = 3.14\))

Hint:

Find the torque the hydraulic cylinder can deliver, then equate it to the tyre-scrub resistance torque, which combines a rotation-about-print-centre term with a kingpin-offset term.

Answer 1

Correct Answer: 42

Step 1: Find the torque the hydraulic cylinder can supply.
Cylinder bore diameter is 32 mm, so the piston area is
\[ A = \frac{\pi}{4}d^2 = \frac{3.14}{4}(0.032)^2 = 0.000804\ m^2 \] At the relief pressure of 8 MPa, the cylinder force is \(F = PA = 8 \times 10^6 \times 0.000804 = 6430.7\) N.
Through the 60 mm effective moment arm, the available kingpin torque is \(T_{avail} = 6430.7 \times 0.06 = 385.84\) N.m.

Step 2: Find the tyre print radius.
The tyre print is a circle of diameter \(b = 0.28\) m, so its radius is \(r = 0.14\) m.

Step 3: Set up the resistance torque about the kingpin.
With uniform tyre pressure, spinning a circular print about its own centre needs torque \(\tfrac{2}{3}\mu W r\) (this is the standard result for a uniformly loaded disc turning about its centre).
When the kingpin is offset by \(e\) from the print centre, the resistance grows with this offset too, and the two effects combine as
\[ T_{req} = \mu W \left( \frac{2r + e}{3} \right) \]

Step 4: Equate available and required torque, and solve for \(e\).
\(\mu W = 0.30 \times 12000 = 3600\) N.
\[ 385.84 = 3600 \left( \frac{2(0.14) + e}{3} \right) \] \[ \frac{2(0.14)+e}{3} = \frac{385.84}{3600} = 0.10718 \] \[ 0.28 + e = 0.3215 \implies e = 0.0415\ m = 41.5\ mm \]

Final Answer:
The maximum permissible kingpin offset is \[ \boxed{42\ mm} \]