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.
Etcherla karthik said:
2 years ago
e = v^2/225R is the correct formula.
(2)
Siva said:
3 years ago
Good platform. Very helpful for the competitive exams.
(2)
Just said:
6 years ago
75*75/(225*1000) = 75/(3*1000) = 25/1000 = 5/200 = 1/40.
(1)
Ramkishor Sahu said:
6 years ago
Superelevation is always.
e+f=v^2/127R.
But for the design of the superelevation we take,
e=V^2/225R.
So, Here velocity 75 kmph and horizontal curve 1000 meter given.
That means it is the radius of the horizontal curve.
Note:- Horizontal curve always given in the form of the radius.
Put these values to find the horizontal curve,
e = 75^2 / (225*1000).
e = 1/40.
Thank you.
e+f=v^2/127R.
But for the design of the superelevation we take,
e=V^2/225R.
So, Here velocity 75 kmph and horizontal curve 1000 meter given.
That means it is the radius of the horizontal curve.
Note:- Horizontal curve always given in the form of the radius.
Put these values to find the horizontal curve,
e = 75^2 / (225*1000).
e = 1/40.
Thank you.
(1)
Pramit Samanta said:
5 years ago
If mixed traffic or maximum value was mentioned then it has to be e + f = V2/225R. Otherwise for design purpose e + f = V2/127R.
(1)
Janardhan said:
8 years ago
e=V2/225R when V in KMPH,
e=v2/127R-f ,if f is given,
and v in m/s,other wise e=v2/127R if f is not given.
e=v2/127R-f ,if f is given,
and v in m/s,other wise e=v2/127R if f is not given.
(1)
Elavarasy R said:
4 years ago
v^2/225R.
= 75 x 75/227 x 1000.
= 0.025.
=1in40.
= 75 x 75/227 x 1000.
= 0.025.
=1in40.
(1)
Happy singh said:
4 years ago
Given data,
Speed = 75 kmp = 75*5/18 m/s = 20.83333333 m/s.
75% speed is the designed speed. So, 75% speed=20.83333333*0.75 m/s =15.625 m/s
Radius of curvature = 1000m.
Here 'g' is the gravitational acceleration and the value of 'g' is 9.8 m/s.
Formula of super elevation is (V2/gR) =(15.625*15.625/(9.8*1000)) =0.024912309%
0.024912309% of Super elevation means 1 in 40.1408.
So, the answer is 1 in 40 is correct.
Speed = 75 kmp = 75*5/18 m/s = 20.83333333 m/s.
75% speed is the designed speed. So, 75% speed=20.83333333*0.75 m/s =15.625 m/s
Radius of curvature = 1000m.
Here 'g' is the gravitational acceleration and the value of 'g' is 9.8 m/s.
Formula of super elevation is (V2/gR) =(15.625*15.625/(9.8*1000)) =0.024912309%
0.024912309% of Super elevation means 1 in 40.1408.
So, the answer is 1 in 40 is correct.
(1)
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)
KARA LALA said:
6 years ago
GSB - Granular Sab Base.
WMM - Wet Mix Macadam.
WMM - Wet Mix Macadam.
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