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 5 of 10.

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

Umesh said:   5 years ago
as per IRC e=v^/225.R.

Here, In question, it should be 1000m radius

Himanshu kumar said:   8 years ago
But there speed in kilometer and horigental curve in meter so how it is possible? Please explain.

Dr. ANUJ said:   4 years ago
This will as per the design steps given by IRC for heterogeneous traffic. Step one is, you have to find the value of e (superelevation) neglecting friction for 75% of design speed.

If the value so obtained is less than 0.07, the design is over and that value is taken as the designed value of superelevation.

Jahidul Islam said:   4 years ago
V^2\225R is correct.

MILI said:   3 years ago
You are absolutely right @Anitha.

We cannot use the formula e=v^2/225 directly, as the question does not mention Hill road.

Omkar said:   4 years ago
Yes @Yugananth

V^2/225R is the correct formula.

Anupama patra said:   4 years ago
I think the superelevation is v^2/127R and this question not mention the radius are 1000.

Aditya said:   1 decade ago
Guys can you please tell me the density of GSB, WMM, DBM and also BC?

Jack said:   9 years ago
A District road having the mixed traffic condition so speed is reduced by 25%.

Super elevation e = V^2 / 225 R.
e = 75 * 75/225 * 1000
= 0.025.

so for 1m,
1/0.025 = 40.
=> 1 in 40.


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