Civil Engineering - Highway Engineering - Discussion

It's we have already given the design speed so we will use v2/127r.

e+f=v^2/gr, v in m/s.
e+f=v^2/127r, v in km/h.

For 75% of design speed f=0,as per irc.
e=(0.75v)^2/127r =v^2/225r.
e=75*75/(225*1000)=0.025=1:40.

For design of superelevation on road:-

(1)- design superelevation for 75% of design speed. (Hence s.e. resists 75% of centrifugal force rest is balanced by road friction)

e=v^2/225*R is correct.

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.

Super elevation = V^2 / 225 R.

= 75x75 / 225x1000.

= 0.025 = 1 in 40.

Super elevation = V^2 / 225 R.
= 75x75 / 225x1000.
= 0.025 = 1 in 40.

as per IRC e=v^/225.R.

Here, In question, it should be 1000m radius

Radius are not mentioned.