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Sonbull [250]
2 years ago
8

A 1.5 m x1.5 m square footing is supported by a soil deposit that contains a 16.5 m thick saturated clay layer followed by the b

edrock. The clay has μs = 0.50 and Es = 5,000 kN/m2 . The footing base is at 1.5 m below the ground surface. Determine the maximum vertical central column load so that the elastic settlement of the footing will not exceed 50.0 mm. If the square footing is replaced by a 1.2 m wide wall footing with all other conditions remaining the same.
Required:
What will be the elastic settlement under the same footing pressure?
Engineering
1 answer:
iren [92.7K]2 years ago
7 0

Answer:

somewhere around 34.2223 meters thick but that's what I am estimating.

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Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

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G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-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\\

The station is given as

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

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x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

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

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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