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Murrr4er [49]
2 years ago
10

A signalized intersection approach has three lanes with no exclusive left or right turning lanes. The approach has a 40-second g

reen out of a 75-second cycle. The yellow plus all-red intervals for phase total 4.0 seconds. If the start-up lost time is 2.3s/phase, the clearance lost time is 1.1s/phase, and thesaturation headway is 2.48 s/veh under prevailing conditions, what is the capacity of the intersection approach?
Engineering
1 answer:
ipn [44]2 years ago
6 0

Answer:

786.01

Explanation:

Calculate the saturation flow rate by using the following equation

S = 3600/h, where h is the saturation headway

S = 3600/2.48 = 1452 veh/hg

<u>Calculate the effective green time</u>

G(i) = G(i) + Y(i) – t(Li), where G(i) is the actual green time for movement, Y(i) is the sum of yellow and all red intervals for movement, and t(Li), is the total lost time for movement

G(i) = 40, Y(i) = 4, t(Li) = 2.3 + 1.1 = 3.4

G(i) = 40 + 4 – 3.4 = 40.6

<u>Calculate the capacity of an intersection approach by using the following equation</u>

C(i) = s(i) [g(i)/C], where s(i) the saturation flow rate, g(i) is the effective green time, and C is the cycle length

s(i) = 1452, g(i) = 40.6, C = 75

C(i) = 1452 x (40.6/75) = 786.01 veh/h/ln

The capacity of intersection approach is 786.01

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Replacing the equation for each condition:  

y = y_o + \frac{k}{\sqrt{(1 \times 10^{-2})}}\\\\\ \ \ \ \ \ \ 230 = yo + 10k\\\\ y = yo + \frac{k}{\sqrt{(6\times 10^{-3})}}\\\\275 = yo + 12.90k

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The equation is \sigma y = \sigma 0 + k y d^{\frac{1}{2}}

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75 = \sigma o+4 ky \\\\175 = \sigma o+12 ky\\\\ky = 12.5 MPa (mm)^{\frac{1}{2}} \\\\ \sigma 0 = 25 MPa\\\\d= 8.9 \times 10^{-3}\\\\d^{- \frac{1}{2}} =10.6 mm^{-\frac{1}{2}}\\

by putting the above value in the formula we get the \sigma y value that is= 157.5 MPa

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2 years ago
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