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 6 of 10.
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)
Aditya said:
1 decade ago
Guys can you please tell me the density of GSB, WMM, DBM and also BC?
Sushil said:
5 years ago
Super elevation = V^2 / 225 R.
= 75x75 / 225x1000.
= 0.025 = 1 in 40.
= 75x75 / 225x1000.
= 0.025 = 1 in 40.
Umesh said:
5 years ago
as per IRC e=v^/225.R.
Here, In question, it should be 1000m radius
Here, In question, it should be 1000m radius
Yugananth P said:
1 decade ago
e = V^2/225R.
= 75 * 75/(225*1000).
= 1/40 (or).
= 1 in 40.
= 75 * 75/(225*1000).
= 1/40 (or).
= 1 in 40.
Babugouda said:
8 years ago
Here,
e = v2/gr when v is in m/s.
e = v2 /127r when v is in kmph.
e = v2/gr when v is in m/s.
e = v2 /127r when v is in kmph.
Anmol said:
7 years ago
It's we have already given the design speed so we will use v2/127r.
Sonu said:
9 years ago
If f is not given so we apply e = v^2/225R . The answer is 1 in 40.
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)
Ram kumar said:
10 years ago
e = v^2/(127R) this formulas use.
And get the super elevation.
And get the super elevation.
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