Civil Engineering - Highway Engineering - Discussion

Discussion Forum : Highway Engineering - Section 1 (Q.No. 1)
1.
A district road with a bituminous pavement has a horizontal curve of 1000 m for a design speed of 75 km ph. The super-elevation is
1 in 40
1 in 50
1 in 60
1 in 70
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
95 comments Page 7 of 10.

Rahul said:   6 years ago
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.

Tabaamith said:   6 years ago
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.

KARA LALA said:   6 years ago
GSB - Granular Sab Base.

WMM - Wet Mix Macadam.

Arjun said:   6 years ago
Design superelevation is always taken as 75% of the Design Speed hence it is e = v^2/225R.

Mohammed said:   6 years ago
@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.

Ramkishor Sahu said:   6 years ago
Superelevation is always.

e+f=v^2/127R.

But for the design of the superelevation we take,

e=V^2/225R.

So, Here velocity 75 kmph and horizontal curve 1000 meter given.

That means it is the radius of the horizontal curve.

Note:- Horizontal curve always given in the form of the radius.

Put these values to find the horizontal curve,

e = 75^2 / (225*1000).

e = 1/40.

Thank you.
(1)

Jeet said:   6 years ago
Super elevation = V^2 / 225 R.

= 75x75 / 225x1000.

= 0.025 = 1 in 40.

Just said:   6 years ago
75*75/(225*1000) = 75/(3*1000) = 25/1000 = 5/200 = 1/40.
(1)

Pramit Samanta said:   5 years ago
If mixed traffic or maximum value was mentioned then it has to be e + f = V2/225R. Otherwise for design purpose e + f = V2/127R.
(1)

Sushil said:   5 years ago
Super elevation = V^2 / 225 R.
= 75x75 / 225x1000.
= 0.025 = 1 in 40.


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