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
Discussion:
95 comments Page 5 of 10.
Sarath chandra said:
10 months ago
When shall we need to consider the speed in terms of metres per second? Please explain to me.
Manoj kumawat said:
1 decade ago
According IRC : The super-elevation is calculated for 75% of design speed.
So e = V^2/225R.
So e = V^2/225R.
Yash said:
7 years ago
Already design speed is given, then why it converted again to design speed? Please tell me.
Arjun said:
6 years ago
Design superelevation is always taken as 75% of the Design Speed hence it is e = v^2/225R.
Anupama patra said:
4 years ago
I think the superelevation is v^2/127R and this question not mention the radius are 1000.
Osi said:
9 years ago
But we have studied it in the formula = v^2/126R. I am confusing which one is correct.
IES AKP said:
10 years ago
1000 m is radius, it can't be length because there is no relation to provide length.
Mahesh said:
4 years ago
Here they have not given lateral friction, so we can take as e = V2/225*R.
Jayrajsinh said:
9 years ago
Practically super elevation e = v^2/225R.
So answer 1 in 40 is correct.
So answer 1 in 40 is correct.
Jeet said:
6 years ago
Super elevation = V^2 / 225 R.
= 75x75 / 225x1000.
= 0.025 = 1 in 40.
= 75x75 / 225x1000.
= 0.025 = 1 in 40.
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