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 9 of 10.

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)

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)

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

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

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

Ada Jamaima said:   2 years ago
How and when to use v2/127R and v2/225R?

Please explain me in detail.
(10)

Chittaranjan sankhua said:   2 years ago
Given Option is wrong.

According to 1st, we have to convert the speed in m/s.
Then the e = v2/127r and e = v2/225r is for mix traffic.
(4)

Ovuta Stanley said:   2 years ago
@All.

With this formula V2/225R for supper elevation where V is 75kmph and R is 1000m. How do I get 1 in 40 if after dividing I'm getting 0.025? Please explain me.
(1)

Aryan said:   2 years ago
e= v*v/(225 R).
=75*75/(225* 1000)
= 1/40.

0.025= 25/1000 = 1/40.
(12)

Aryalabi said:   2 years ago
The value of super elevation is calculated for 75 % of the design speed.
So, e=[(0.75V)^2]/[g*R].
put v = 75*1000/3600 = 20.83 m/s.
R=1000.
g=9.81 we get,
e=0.024 i.e. 1 in 40.
(3)


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