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 2 of 10.
Jack said:
9 years ago
A District road having the mixed traffic condition so speed is reduced by 25%.
Super elevation e = V^2 / 225 R.
e = 75 * 75/225 * 1000
= 0.025.
so for 1m,
1/0.025 = 40.
=> 1 in 40.
Super elevation e = V^2 / 225 R.
e = 75 * 75/225 * 1000
= 0.025.
so for 1m,
1/0.025 = 40.
=> 1 in 40.
Surafel Belayneh said:
7 years ago
A District road having the mixed traffic condition so speed is reduced by 25%.
Super elevation e = V^2/225 R.
e = 75 * 75/225 * 1000.
= 0.025.
So for 1m,
1/0.025 = 40.
=> 1 in 40.
Super elevation e = V^2/225 R.
e = 75 * 75/225 * 1000.
= 0.025.
So for 1m,
1/0.025 = 40.
=> 1 in 40.
Rahul said:
6 years ago
In this question clearly mentioned design for the road not for any particular vehicle.
In the design of road assume friction is zero and 75% velocity ie formula of e is v^2/225R.
In the design of road assume friction is zero and 75% velocity ie formula of e is v^2/225R.
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)
DJ.thakur said:
8 years ago
For design of superelevation on road:-
(1)- design superelevation for 75% of design speed. (Hence s.e. resists 75% of centrifugal force rest is balanced by road friction)
(1)- design superelevation for 75% of design speed. (Hence s.e. resists 75% of centrifugal force rest is balanced by road friction)
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)
Tabaamith said:
6 years ago
Convert 75km/hr to m/ sec, which comes 20.85m/sec.
Take velocity as 75% of design speed.
Now, e= (0.75 * 20.85)^2/ 9.81X1000 = 0.024.
In ratio 1/1/0.024 = 1 in 40.
Take velocity as 75% of design speed.
Now, e= (0.75 * 20.85)^2/ 9.81X1000 = 0.024.
In ratio 1/1/0.024 = 1 in 40.
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.
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)
Rehi said:
9 years ago
Actually, in the problem, it shows that horizontal curve is 1000m but in explanation, they take 1000m as a radius why don't they take it as length of the curve?
Shankar Meghwar said:
1 decade ago
R = V^2/225(e+fs).
Consider coefficient of friction as fs = 0.
Therefore R = V^2/225 e.
OR
e = V^2/225 R.
= 75*75/(225*1000).
= 0.025.
OR
= 1 in 40.
Consider coefficient of friction as fs = 0.
Therefore R = V^2/225 e.
OR
e = V^2/225 R.
= 75*75/(225*1000).
= 0.025.
OR
= 1 in 40.
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