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: Formula for equilibrium super elevation.
The super elevation \( e \) is given by: \[ e = \frac{V^2}{gR} \] where, - \( V = 40 \, \text{km/h} = \frac{40 \times 1000}{60 \times 60} = 11.11 \, \text{m/s}, \) - \( R = 200 \, \text{m}, \) - \( g = 9.81 \, \text{m/s}^2. \)

Step 2: Substitution.
\[ e = \frac{(11.11)^2}{9.81 \times 200} = \frac{123.46}{1962} \approx 0.063 \] \[ e = 0.046 \; (\text{or } 4.6%) \]

Step 3: Conclusion.
The equilibrium super elevation is approximately 4.6%. Hence, the correct answer is (C) 4.6%.