A valley curve has a descending gradient of 1 in 40 followed by an ascending gradient of 1 in 50. The length of the valley curve required for a design speed of 80 km/hour for comfort condition is

37 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

58 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

116 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

199 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $58 m$
$L = 2 {[\frac{N v^{3}}{C}]}^{\frac{1}{2}}$

$= 0.38 [N V^{3}]^{1 / 2}$
$[since C = 0.6 m / s^{2}]$

$N = \frac{1}{40} - (- \frac{1}{50}) = \frac{1}{40} + \frac{1}{50} = 0.045$

$∴ L = 0.38 [0.045 \times 80^{3}]^{1 / 2}$

$= 57.67 m ≃ 58 m$

Suggest Corrections

Similar questions

Q. In a valley curve, a descending gradient of 1 in 60 is meeting an ascending gradient of 1 in 75. The total length of valley curve is ________ m. Assume design speed as 75 kmph and rate of change of centrifugal accleration is 0.6

m / s^{3}

Q. A descending gradient of 1 in 20 meets an ascending gradient of 1 in 30. What will be length(in m) of a valley curve to fulfill comfort criterion for a design speed of 80 kmph.
[Assume allowable rate of change of centrifugal acceleration 0.6

m / s e c^{3}

and answer correct upto one decimal place].

An ascending gradient of 1 in 45 meets a descending gradient of 1 in 60. A summit curve has to be designed gradient for design speed of 80 km/h so as to provide a safe overtaking sight distance of 230 m. Estimate the length of the summit curve.

Q. A descending gradient of 4% meets an ascending grade of 1 in 40 where a valley curve of length 200 m is to be formed.The distance of the lowest point on the valley curve from its first tangent point will be

Q. An ascending gradient of 1 in 40 meets a descending gradient of 1 in 60. What will be the length of summit curve for a stopping sight distance of 100 m?

Join BYJU'S Learning Program

A valley curve has a descending gradient of 1 in 40 followed by an ascending gradient of 1 in 50. The length of the valley curve required for a design speed of 80 km/hour for comfort condition is

The correct option is B 58mL=2[Nv3C]12 =0.38[NV3]1/2 [since C=0.6m/s2] N=140−(−150)=140+150=0.045 ∴L=0.38[0.045×803]1/2 =57.67m≃58m

The correct option is B $58 m$
$L = 2 {[\frac{N v^{3}}{C}]}^{\frac{1}{2}}$

$= 0.38 [N V^{3}]^{1 / 2}$
$[since C = 0.6 m / s^{2}]$

$N = \frac{1}{40} - (- \frac{1}{50}) = \frac{1}{40} + \frac{1}{50} = 0.045$

$∴ L = 0.38 [0.045 \times 80^{3}]^{1 / 2}$

$= 57.67 m ≃ 58 m$