Discussion :: GATE Exam Questions - Section 3 (Q.No.46)
|Koushik Pal said: (Mar 28, 2014)|
|e = v^2/225R.
So, e = 125^2/225*450.
|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.
= 0.07 answer.
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