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elixir [45]
3 years ago
12

Using the AASHTO procedure, determine the thickness required for a base and a surface layer over existing subgrade. The structur

al number required for this pavement is 4.5. The available materials are a crushed stone base course with a material layer coefficient of 0.13 and an HMA surface with a material layer coefficient of 0.40. The drainage coefficient of the based course can be considered as 0.90.
Engineering
1 answer:
garri49 [273]3 years ago
7 0

Answer:

<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>

Explanation:

structural number = 4.5

stone base course material coefficient = 0.13

hma material layer coefficient = 0.40

drainage coefficient = 0.90

we will use layered analysis procedure to get thickness

D1 >= sN1/a1

when we cross multiply,

sN1 = a1D1 >=sN1

D2 >= -sN2-sN1/a2m2

sN2* + sN1* >= sN2

D3 >= sN3-(sN1*+sN2*)/a2m2

where a1,a2,a3 = layer coefficient

d1 d2 d3 = actual thickness

m2,m3 = coefficient of base

a1 = 0.4

a1 = 0.13

sN = 4.5

m2 = 0.9

D1 >= sN1/a1 = 4.5/0.4

= 11.25

thickness of surface = 12 inches

a1D1 = 0.4x12 = 4.8

we have value of sN2 = 5.5

(5.5 -4.8)/(0.13*0.9)

= 0.7/0.117

= 5.9829 inches

approximately 6 inches

so the pavement will have 12inch HMA surface and 6 inches granular base layer.

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3 years ago
Water is boiled in a pot covered with a loosely fitting lid at a location where the pressure is 85.4 kPa. A 2.61 kW resistance h
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4 0
2 years ago
An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
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-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

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St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

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St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

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Question 74
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z_{2} = 91.640\,m

7 0
3 years ago
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