Civil Engineering - GATE Exam Questions - Discussion

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

[A]. 1.0
[B]. 0.05
[C]. 0.07
[D]. 0.154

Answer: Option D

Explanation:

No answer description available for this question.

Koushik Pal said: (Mar 28, 2014)  
e = v^2/225R.

So, e = 125^2/225*450.

= 0.154.

Berhane said: (Sep 5, 2014)  
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.

Shubham said: (Jul 18, 2016)  
But maximum superelevation should be 7%. It can't be more than 7%. So the answer should be 0.07.

Aashish Ahirwar said: (Jan 11, 2017)  
V^2/225R = (125)^2/225 * 450 = 0.154.

Vinay Singh said: (Aug 29, 2017)  
The answer should be .07. because it is max super elevation.

Sadulla said: (Nov 14, 2017)  
Yes @Vinay. It should be 0.07.

Priya said: (Dec 25, 2017)  
It's should be v2/225R.

Shubham Ramdasrao Umak said: (Jan 30, 2018)  
According to me, its answer is = 0.07.

Because whenever we are designing super elevation the maximum elevation is restricted to 0.07.

Ramyasree said: (Aug 21, 2018)  
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%.

Thonduru Radhika said: (Dec 8, 2018)  
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.

Ankit said: (Dec 19, 2019)  
e = .75v^2÷gR.
= .75x34.72^2÷(9.81x450).
= 0.07 answer.

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