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)
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)
Anant chaudhari said:
3 years ago
Good explanation. Thanks all.
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)
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)
Siva said:
3 years ago
Good platform. Very helpful for the competitive exams.
(2)
MILI said:
3 years ago
You are absolutely right @Anitha.
We cannot use the formula e=v^2/225 directly, as the question does not mention Hill road.
We cannot use the formula e=v^2/225 directly, as the question does not mention Hill road.
Mahesh said:
4 years ago
Here they have not given lateral friction, so we can take as e = V2/225*R.
Anupama patra said:
4 years ago
I think the superelevation is v^2/127R and this question not mention the radius are 1000.
Omkar said:
4 years ago
Yes @Yugananth
V^2/225R is the correct formula.
V^2/225R is the correct formula.
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