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 8 of 10.

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

Here, In question, it should be 1000m radius

Elavarasy R said:   4 years ago
v^2/225R.
= 75 x 75/227 x 1000.
= 0.025.
=1in40.
(1)

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.

Happy singh said:   4 years ago
Given data,

Speed = 75 kmp = 75*5/18 m/s = 20.83333333 m/s.
75% speed is the designed speed. So, 75% speed=20.83333333*0.75 m/s =15.625 m/s
Radius of curvature = 1000m.

Here 'g' is the gravitational acceleration and the value of 'g' is 9.8 m/s.

Formula of super elevation is (V2/gR) =(15.625*15.625/(9.8*1000)) =0.024912309%
0.024912309% of Super elevation means 1 in 40.1408.
So, the answer is 1 in 40 is correct.
(1)

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.

Mahesh said:   4 years ago
Here they have not given lateral friction, so we can take as e = V2/225*R.

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.

Siva said:   3 years ago
Good platform. Very helpful for the competitive exams.
(2)


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