Analysis of Superelevation Formula as Per I.R.C - Civil Engineering

Analysis of Superelevation

Let, W = Weight of the vehicle,
v = Speed of the vehicle in m/sec,
g = Acceleration due to gravity in m/sec²,
f = Coefficient of friction,
e = rate of superelevation (tanÎ¸),
F = Frictional force resistance due to centrifugal force,
P = Centrifugal force.
Now, P = Wv²/gR
 Analysis of Superelevation Formula as Per I.R.C

For equilibrium, resolving the forces parallel to the inclined plane, are
P CosÎ¸ = W SinÎ¸ + F
Or, P CosÎ¸ = W SinÎ¸ + f(W CosÎ¸ + P SinÎ¸)

After putting the value of P(P = Wv²/gR) we get,
Wv² CosÎ¸ / gR = W SinÎ¸ + f(W CosÎ¸ + Wv² SinÎ¸ / gR)
Or, v² CosÎ¸ / gR = SinÎ¸ + f(CosÎ¸ + v² SinÎ¸ / gR)

Now, both side divided by CosÎ¸, After that we get the following equation:
v² / gR = tanÎ¸ + f(1 + v² tanÎ¸ / gR)
Or, v² / gR = tanÎ¸ + f + fv² tanÎ¸ / gR
The term fv² tanÎ¸ / gR being too small and can be neglected.
Now,
v² / gR = tanÎ¸ + f
Or, v² / gR = e + f  ............(1)

Superelevation Formula

If the speed of the vehicle is given as V Km/h, then
v = V⨯1000 / 60⨯60 m/Sec
Or, v = 0.278V
Then the equation (1) becomes
(0.278V)² / 9.81 R = e + f
Or, e + f = V² / 127R
 Superelevation Formula