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: State the torque balance condition.
The hydraulic system can only push the kingpin with a fixed maximum torque, and the tyre scrub resistance must not cross that ceiling: $T_{req} \le T_{avail}$.

Step 2: Compute the ceiling torque from the cylinder.
Piston area from a 32 mm bore: $A = \frac{3.14}{4}(0.032)^2 = 0.000804\ m^2$.
Force at the 8 MPa relief setting: $F = (8\times10^6)(0.000804) = 6430.7$ N.
Torque at the 60 mm arm: $T_{avail} = 6430.7 \times 0.060 = 385.84$ N.m.

Step 3: Write the resistance torque in terms of the offset $e$.
The tyre print is a circle of radius $r = b/2 = 0.14$ m carrying a uniform pressure.
For a print pivoting about an axis offset by $e$ from its own centre, the resistance combines the spin-in-place term $\frac{2}{3}\mu W r$ with an offset term, giving $T_{req} = \frac{\mu W}{3}(2r+e)$.

Step 4: Solve directly for $e$ at the limiting condition $T_{req}=T_{avail}$.
\[ e = \frac{3T_{avail}}{\mu W} - 2r \] $\mu W = 0.30 \times 12000 = 3600$ N, so $\dfrac{3 \times 385.84}{3600} = \dfrac{1157.5}{3600} = 0.3215$ m.
$e = 0.3215 - 2(0.14) = 0.3215 - 0.28 = 0.0415$ m.

Final Answer:
Converting to millimetres and rounding, the maximum kingpin offset allowed is \[ \boxed{42\ mm} \]