Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 3 (Q.No. 46)
46.
Design rate of super elevation for horizontal highway curve of radius 450 m for a mixed traffic condition, having a speed of 125 km/hour is
Discussion:
14 comments Page 1 of 2.
Koushik Pal said:
1 decade ago
e = v^2/225R.
So, e = 125^2/225*450.
= 0.154.
So, e = 125^2/225*450.
= 0.154.
Berhane said:
1 decade ago
For mixed traffic the supper elevation is calculated as follows:
e = v^2/225*R that means 125^2/225*450 = 15625/101250 = 0.15432098.
e = v^2/225*R that means 125^2/225*450 = 15625/101250 = 0.15432098.
Shubham said:
9 years ago
But maximum superelevation should be 7%. It can't be more than 7%. So the answer should be 0.07.
Aashish ahirwar said:
9 years ago
V^2/225R = (125)^2/225 * 450 = 0.154.
Vinay singh said:
8 years ago
The answer should be .07. because it is max super elevation.
Sadulla said:
8 years ago
Yes @Vinay. It should be 0.07.
Priya said:
8 years ago
It's should be v2/225R.
Shubham Ramdasrao Umak said:
8 years ago
According to me, its answer is = 0.07.
Because whenever we are designing super elevation the maximum elevation is restricted to 0.07.
Because whenever we are designing super elevation the maximum elevation is restricted to 0.07.
Ramyasree said:
7 years ago
The answer is 0.154 only because in the question he didn't asked as per IRC.
So as per IRC, it's is in between 7-10%.
So as per IRC, it's is in between 7-10%.
THONDURU RADHIKA said:
7 years ago
For mixed traffic, super elevation can be calculated.
It should be V^2 / 225R.
Since this value is less than 0.07, the super elevation of 0.154 may be adopted.
It should be V^2 / 225R.
Since this value is less than 0.07, the super elevation of 0.154 may be adopted.
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