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
Discussion:
95 comments Page 10 of 10.
Kishor bist said:
2 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.
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.
(6)
Ramesh said:
2 years ago
Concept:
The superelevation is given by:
e = (V ^ 2)/(225R) Where,
V = speed (in kmph), and R = Radius (in m).
Calculation:
Given: V = 75 kmph, and R = 1000m,
e = ((75) ^ 2)/(225 * 1000) = 1/40.
So, the super elevation is 1/40.
The superelevation is given by:
e = (V ^ 2)/(225R) Where,
V = speed (in kmph), and R = Radius (in m).
Calculation:
Given: V = 75 kmph, and R = 1000m,
e = ((75) ^ 2)/(225 * 1000) = 1/40.
So, the super elevation is 1/40.
(16)
Timirbaran Daata said:
1 year ago
According to IRC 73-1990 Plate 1, the answer is 1:50.
Sarath chandra said:
10 months ago
When shall we need to consider the speed in terms of metres per second? Please explain to me.
Taba Tallum said:
3 months ago
Actually, E =v^2/ (225R) is for Design purposes, considering 75% of the design speed.
The correct equation for calculating the superelevation should be e=V^2/ (127R).
The correct equation for calculating the superelevation should be e=V^2/ (127R).
(3)
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