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.
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.
Arnab Chanda said:
1 decade ago
But it is not declared that 1000 m is the radius of the curve.
Muleta Totoba said:
1 decade ago
Right it is not declared but the term curve seems something to be the given figure 1000 is the radius.
Surya said:
1 decade ago
e = V^2/gR.
V = 75% of design speed.
V = 0.75*75.
v = V/3.2.
V = 75% of design speed.
V = 0.75*75.
v = V/3.2.
Jay mishra said:
1 decade ago
I think superelevation = v^2/127r.
Deepak mittal said:
1 decade ago
No it should not be v^2/127r. If the value of f then you can take the v^2/127r. Otherwise V^2/225R.
Biswa said:
1 decade ago
e = (0.75v)^2/Rg.
v in m/s.
Ans: 1 in 40.
v in m/s.
Ans: 1 in 40.
Reshma said:
1 decade ago
We have to design for 75% speed means super elevation for mixed traffic.
So,
e = (0.75V)^2/gR = V^2/225R.
So,
e = (0.75V)^2/gR = V^2/225R.
Anitha said:
1 decade ago
We have to design for 75% speed.
So, e = (0.75v)^2/g.r;
v = 75kmph = 75/3.6 = 20.833.
e = (0.75*20.833)^2/(9.81*1000) = 0.02488.
1/40 = 0.025.
So, e = (0.75v)^2/g.r;
v = 75kmph = 75/3.6 = 20.833.
e = (0.75*20.833)^2/(9.81*1000) = 0.02488.
1/40 = 0.025.
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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