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.
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.
Shashank saklani said:
3 years ago
.07 is the design value as per IRC but the question does not say e(design ) as per IRC.
So the value will be according to the formula i.e, v2/225r.
So the value will be according to the formula i.e, 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.
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.
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%.
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.
Vinay singh said:
8 years ago
The answer should be .07. because it is max super elevation.
Ankit said:
6 years ago
e = .75v^2÷gR.
= .75x34.72^2÷(9.81x450).
= 0.07 answer.
= .75x34.72^2÷(9.81x450).
= 0.07 answer.
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.
Bogale Emiru said:
3 years ago
Anyone, please explain the Calculation part.
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