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 2 of 10.

Etcherla karthik said:   2 years ago
e = v^2/225R is the correct formula.
(2)

Naresh said:   3 years ago
e = v^2/225R.
V in kmph, R in mtrs.
75×75/225 × 1000 = 1/40.
(25)

Anant chaudhari said:   3 years ago
Good explanation. Thanks all.

AKINDELE ADEGBOLA said:   3 years ago
I need clarity on super elevation, how do they arrive at 225, 227 if e+f=v2/gr. I need steps by steps calculations.

Anyone, please explain to me.
(3)

Siva gubbala said:   3 years ago
Given data :

R = 1000 m
V = 75 kmph=75x(1000/3600)=20.83m/s
e = (0.75xv'2/gr)
= ((0.75x20.83)'2/(9.81x1000))
= 0.025 = 1 in 40.
(11)

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

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.

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

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

Omkar said:   4 years ago
Yes @Yugananth

V^2/225R is the correct formula.


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