A road has a design speed of 20 m/s and the radius of the horizontal curve is equal to 250 m. Design the super-elevation that is needed on this road, with lateral friction coefficient value of 0.15. (Take g = 10 m/s².)
- A0.09
- B0.22
- C0.01Correct
- D0.07
Solution
Concept
On a horizontal curve, super-elevation plus side friction balances the centrifugal force: e + f = v2/(gR).
Formula
e + f = v2/(g R)
Step-by-step solution
1. Given v = 20 m/s, R = 250 m, f = 0.15, g = 10 m/s². 2. v2/(gR) = 400/(10 × 250) = 400/2500 = 0.16. 3. e + f = 0.16, so e = 0.16 − 0.15 = 0.01. 4. Required super-elevation e = 0.01.
Explanation
The total of super-elevation and friction must balance the centrifugal ratio v2/(gR) = 0.16. Friction gives 0.15, so super-elevation needs to give only 0.01.
Why the other options are wrong
A: 0.09 comes from the IRC first step e = (0.75v)2/(gR) without using f. B: 0.22 has no basis in either method. D: 0.07 is the IRC maximum super-elevation for plain areas, not the value asked here.
Shortcut
e = v2/(gR) − f = 0.16 − 0.15 = 0.01.
Common mistake
Using the speed in km/h with g, or applying the IRC 75% speed method when f is directly given.
Correct answer: C) 0.01