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.)

Concept:
∵ Design speed is given, so find superelevation for mixed traffic condition and in this condition different vehicles move at different speeds.

• For practical conditions, IRC recommends that superelevation should be provided to fully counteract the centrifugal force due to 75 % of design speed by neglecting lateral friction.

Following Design steps are involved into the design of super elevation :
Step 1: Calculate the superelevation corresponding to 75% of design speed and neglecting the role of lateral friction.

\(e_{equilibrium} = \frac{V^{2}}{225R}\)
If e_calculated < e_maximum, then provided e = equilibrium
if e_calculated > e , then take e = e_maximum (0.07)

Step 2: Provide e = e_maximum, and find the value of lateral friction 'f'
\(e + f = \frac{v^{2}}{127R}\)
if, f < 0.15, then provide e = e_maximum
if, f > 0.15, then fix f = 0.15

Step 3: Now take f = 0.15, e = e_maximum and find the actual velocity will be provided on the
highway
\(e + f = \frac{v_{a}^{2}}{127R}\)

\(e + 0.15 = \frac{v_{a}^{2}}{127R} \)
If V_design < V_actual
, then OK
If V_design > V_actual
, then restrict the speed by providing speed limits sign.
Calculation:
Given,
V = 20 m/s = 72 kmph

 R = 250 m
Step 1: \(e_{equilibrium} = \frac{V^{2}}{225R}\) 

\(e_{equilibrium} = \frac{72^{2}}{225 \times 250}\)= 0.09216

e_equilibrium > e_max(0.07), then go for step 2
 

Step 2: \(e + f = \frac{v^{2}}{127R}\) 

\(f = \frac{v^{2}}{127R}-e_{max} = \frac{72^{2}}{127\times 250}-0.07\)= 0.0932

Here, f < 0.15, then provide, e = e_max = 0.07
So Rate of super elevation is provide = 0.07