Civil Engineering - Highway Engineering - Discussion

Use e= (0.75v)^2/gr.

It is right 1/40 by using the formula.

In this question clearly mentioned design for the road not for any particular vehicle.

In the design of road assume friction is zero and 75% velocity ie formula of e is v^2/225R.

Convert 75km/hr to m/ sec, which comes 20.85m/sec.
Take velocity as 75% of design speed.
Now, e= (0.75 * 20.85)^2/ 9.81X1000 = 0.024.
In ratio 1/1/0.024 = 1 in 40.

Here they have not given lateral friction, so we can take as e = V2/225*R.

Design superelevation is always taken as 75% of the Design Speed hence it is e = v^2/225R.

@Azanaw.

You will find a clear explanation in IRC 38 codebook. However, I will try to briefly summarise it to you.

Basically superelevation is provided to counteract the centrifugal force experienced by a vehicle when it's negotiating a curve. Now counteracting the full centrifugal force developed for a given design speed would require us to provide superelevation of more than 7%. But IRC restricts the maximum value of e= 7%.

So in practice for mixed traffic, most vehicles travel at less than design speed (where we have both slow and fast-moving vehicles) so as a compromise since we can\'t provide the actual value we design superelevation to resist the full centrifugal force developed at 75% design speed.

Eg: If the design speed of a National Highway is 80kmph. We design the superelevation to fully resist the centrifugal force developed due to 75% of 80kmph = 60 kmph.

This method of giving superelevation will not be inconvenient to fast-moving vehicles but would be of aid to slow-moving vehicles.

According to IRC 73-1990 Plate 1, the answer is 1:50.

Super elevation = V^2 / 225 R.

= 75x75 / 225x1000.

= 0.025 = 1 in 40.

Good explanation. Thanks all.