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## 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 (tanθ),

F = Frictional force resistance due to centrifugal force,

P = Centrifugal force.

Now, **P = Wv²/gR**

For equilibrium, resolving the forces parallel to the inclined plane, are**P Cosθ = W Sinθ + F**

Or, **P Cosθ = W Sinθ + f(W Cosθ + P Sinθ)**

After putting the value of P(P = Wv²/gR) we get,**Wv² Cosθ / gR = W Sinθ + f(W Cosθ + Wv² Sinθ / gR)****Or, v² Cosθ / gR = Sinθ + f(Cosθ + v² Sinθ / gR)**

Now, both side divided by Cosθ, After that we get the following equation:**v² / gR = tanθ + f(1 + v² tanθ / gR)****Or, v² / gR = tanθ + f + fv² tanθ / gR**

The term fv² tanθ / gR being too small and can be neglected.

Now,**v² / gR = tanθ + f****Or, v² / gR = e + f …………(1)**

### Formula of Superelevation

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**

**Formula Of equilibrium Superelevation**

**If we provide e = V² / 127R, **then the value of ‘f’ will be equal to zero and the lateral friction will not be considered. This superelevation is called equilibrium superelevation.

Equilibrium superelevation (e) = **V² / 127R**

**Design Formula of Superelevation as Per IRC**

As per I.R.C, superelevation is calculated at three fourth the design speed

e = (0.75V)**²** / 127R. Or, e = V**²** / 225R

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