Civil Engineering - Highway Engineering - Discussion

5. 

The total length of a valley formed by two gradients - 3% and + 2% curve between the two tangent points to provide a rate of change of centrifugal acceleration 0.6 m/sec2, for a design speed 100 km ph, is

[A]. 16.0 m
[B]. 42.3 m
[C]. 84.6 m
[D]. none of these.

Answer: Option C

Explanation:

No answer description available for this question.

Venkatesh said: (Sep 10, 2013)  
Here N = 0.03+0.02 = 0.05.

L = 0.38(NV^3)^0.5.

L = 84.76MTS APPROX.

Ram Kumar said: (Nov 18, 2013)  
Length is 84.16 means, then what s the answer? option A or option C.

Alik Biswas said: (Jan 21, 2014)  
We Know, L= 2*(N*v^3/c)^0.5 (Length of valley curve).

Here, N = 2%-(-3%) (Algebraic difference in grades).
= 5% Or 0.05.

v = 100 km/hr or 27.778 m/sec (Design Speed).
c = 0.6 m/sec3 (Centrifugal acceleration).

Therefore, L= 2*(0.05*27.778^3/0.6)^0.5 = 84.53 m.

Say,84.60 m ([C] is the correct answer.

Twizere Kagunga Elie said: (Sep 24, 2014)  
I don't understand well if N is load or if is median.

Nitin said: (Jan 2, 2015)  
N is sum of upward & downward gradient.

Manikanth said: (Jan 23, 2015)  
Lenth of valley curve for comfort condition is 2(Nv^3/c)^0.5.

Tarun said: (May 26, 2015)  
V = 1km/hr*5/18 is m/s.

Selvi said: (Aug 12, 2015)  
L = 0.378(NV3)^1/2.
= 0.378(.05*100*100*100)^.5.
= 86.4 m.

Rishabh said: (Oct 4, 2015)  
L is the length of valley curve.

And this formula is applicable in this case only 2(Nv^3/c)^0.5.

Harpreet Singh said: (Dec 26, 2015)  
I don't understand. Please explain.

Nikhil said: (Jan 3, 2016)  
What is mean by V?

Santhi said: (Jan 27, 2016)  
Here v is the design speed of the vehicle.

Ayebare Junior Jeezy said: (Apr 27, 2016)  
If V is the speed =100km/hr.
Difference in grades = A = (m - n).
= (-3 - 2) = 5% of acceleration
= 0.6.

But V =100 x 1000/60 x 60 = 27.8m/s.
L = 2*(aV^3)^0.5.
=((2)*0.05 x 27.8^3/0.6)^0.5).
= 84.53m.

Elsadig Ahmed Ali said: (Jul 14, 2016)  
V = 1000 km/hr = 27.8 m/s.
N = 5%.
C = .6.

L = 2 * ((N * V^3) / C)^.5,
L = 2 * ((.05 * 27.8^3) / .6)^.5,
L = 2 * ((.05 * 21484.952) / .6)^.5,
L = 2 * (1074.2 / .6)^.5,
L = 2 * (1790.3)^.5,
L = 2 * 42.3.
L = 84.6.

Samm said: (Aug 21, 2016)  
But N is not a difference in grade, it is deviation angle in radians. Please clarify this.

Ravindra said: (Aug 21, 2016)  
@Alik Biswas, Well. Here, 'c' means what?

Elsadig Ahmed Ali said: (Sep 10, 2016)  
@Samm.
N = |n2 - n1| = |-3 - 2| =|-3 + -2| = |-5| = 5.

@Ravindra.
C = centrifugal acceleration = 0.6 m/s^2.

Sarang Mote said: (Sep 25, 2016)  
L = 2x(NV3/c)"0.5.
L = 2*((0.05 * 27.78 * 27.78 * 27.78)/0.6)"0.5.
L = 84.53m.

Adikeri said: (Dec 1, 2016)  
Not understanding. Explain in detail.

Doha said: (May 3, 2017)  
L = (A*V^2)/1300a,
L = (5 * 100^2)/1300 * 0.6,
L = 64.1m.

Dame Abebe said: (May 22, 2017)  
Yes, L =84.60 so option "C" is the answer.

Sandi said: (Jun 7, 2017)  
L = 2X(NV3/C)^.5,
N = .05,
V = 27.78,
L = 84.498.

Vamsi said: (Aug 19, 2017)  
Answer is C.

David Bielstein said: (Jan 29, 2018)  
I thought the centrifugal acceleration at the beginning of a parabolic curve would be 0 m/sec^2, increasing to the maximum acceleration at the point directly above or below the PI, and then back to 0.

Is 0.6 m/sec^2 assumed to be the maximum instantaneous centrifugal acceleration?

Shahbaz Bashir said: (Jan 4, 2019)  
N is the difference between raising gradient and falling gradient i.e N = n1 - n2 and also falling gradient = negative (-) similarly raising gradient =positive (+).

Shaguftasultana said: (Jul 2, 2019)  
N=|-3-2|=5,
L=2* {(Nv^3)/C},
=2* 5*(100/3.6)/0.6,
= 84.57.

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