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.

Haylee Ethio said:   9 years ago
Well,

We Have,
R=1000m
V= 75Km/h
Required?
Superelevation(e)
e=v^2/225R
= 75^2/225(1000)
=1/40 is the Correct one.

Danjah said:   9 years ago
here 1000 m is the length of curve, not radius.

Yohannis said:   9 years ago
The question stated that 75km/hr is design speed, thus simply we can calculate e=(75^2)/225R. b/c 'f' is not given.

There fore, 1:40 is the correct answer.
If 'f' were given, we use (e+f) = (75^2)/127R.

Titan said:   9 years ago
Why is everyone assuming R=1000?

I think the curve length is 1000, not R. And the angle is not given.

Sonu said:   9 years ago
If f is not given so we apply e = v^2/225R . The answer is 1 in 40.

Jayrajsinh said:   9 years ago
Practically super elevation e = v^2/225R.

So answer 1 in 40 is correct.

Mukeah senani said:   9 years ago
E = V^2/225R -> this method is Right.

Happy said:   9 years ago
But, here v^2/225 is the highest unit of super-elevation.

Joel said:   9 years ago
The provided answer 1 in 40 is correct.

Kruti nayak said:   9 years ago
v^2/225R is correct.
v^2/127R when f = 0.


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