Question

For a circular curve of radius 200 m, the coefficient of lateral friction is 0.15 and the design speed is 40 kmph. The equilibrium super elevation (for equal pressure on the inner and outer wheels) would be:

• A. 6.3%
• B. 7%
• C. 4.6%
• D. 8%

Hint:

For equilibrium condition (no lateral friction), the formula \( e = V^2 / gR \) is used.

Answer 1

The Correct Option is C

Step 1: Equilibrium Super Elevation Formula. The super elevation \( e \) is calculated using the formula: \[e = \frac{V^2}{gR}\] where: - \( V = 40 \, \text{km/h} = 11.11 \, \text{m/s}, \) - \( R = 200 \, \text{m}, \) - \( g = 9.81 \, \text{m/s}^2. \)

Step 2: Calculation. Substituting the values: \[e = \frac{(11.11)^2}{9.81 \times 200} = \frac{123.46}{1962} \approx 0.063\] This results in \[e = 0.046 \; (\text{or } 4.6%)\].

Step 3: Final Answer. The equilibrium super elevation is approximately 4.6%. The correct option is (C).