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 1 of 10.
Naresh said:
3 years ago
e = v^2/225R.
V in kmph, R in mtrs.
75×75/225 × 1000 = 1/40.
V in kmph, R in mtrs.
75×75/225 × 1000 = 1/40.
(25)
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)
Aryan said:
2 years ago
e= v*v/(225 R).
=75*75/(225* 1000)
= 1/40.
0.025= 25/1000 = 1/40.
=75*75/(225* 1000)
= 1/40.
0.025= 25/1000 = 1/40.
(12)
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.
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)
Ada Jamaima said:
2 years ago
How and when to use v2/127R and v2/225R?
Please explain me in detail.
Please explain me in detail.
(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)
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.
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)
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)
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.
Anyone, please explain to me.
(3)
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.
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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