Question

A load of "W" kN is acting on a tyre having 160 mm nominal width. The effective friction coefficient of tyre and ground interaction is "\(\mu\)" and the kingpin offset is 10 mm. Assuming the tyre impression on ground as circle with diameter equal to nominal tyre width, the kingpin torque of the tyre in N.m will be

• A. \(W\mu\sqrt{3210}\)
• B. \(\frac{W\mu\sqrt{3210}}{1000}\)
• C. \(W\mu\sqrt{3300}\)
• D. \(\frac{W\mu\sqrt{3300}}{1000}\)

Hint:

The unit conversion factors cancel out beautifully:
\(1\text{ kN} \times 1\text{ mm} = 10^3\text{ N} \times 10^{-3}\text{ m} = 1\text{ N}\cdot\text{m}\).
Therefore, you can compute directly using raw values to get \(W\mu\sqrt{3300}\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The steering kingpin torque (scrub torque) consists of components resisting turning about the kingpin, dictated by both the kingpin offset and the pivoting resistance of the contact patch.

Step 2: Key Formula or Approach:
The combined effective radius (equivalent offset) \(r_{eq}\) taking both kingpin offset \(a\) and tyre contact diameter \(d\) into account is:
\[ r_{eq} = \sqrt{a^2 + \frac{d^2}{8}} \] The torque \(T\) is then:
\[ T = \mu \times W \times r_{eq} \]

Step 3: Detailed Explanation:
Given values:
- Kingpin offset (\(a\)) = \(10\text{ mm}\)
- Tyre nominal width (contact patch diameter \(d\)) = \(160\text{ mm}\)
Calculate the equivalent radius under the square root:
\[ r_{eq} = \sqrt{10^2 + \frac{160^2}{8}} \] \[ r_{eq} = \sqrt{100 + \frac{25600}{8}} \] \[ r_{eq} = \sqrt{100 + 3200} = \sqrt{3300}\text{ mm} \] Since \(W\) is given in kN and \(r_{eq}\) is in mm, convert both to SI base units (N and m):
\[ T = \mu \times (W \times 1000\text{ N}) \times \left( \frac{\sqrt{3300}}{1000}\text{ m} \right) \] The factor of 1000 cancels out:
\[ T = W \mu \sqrt{3300} \text{ N}\cdot\text{m} \]

Step 4: Final Answer:
The correct option is 3, which corresponds to \(W\mu\sqrt{3300}\).