Two footings rest in layer of sand 2.7m thick. The bottom of the footings are 0.9m below the ground surface. Beneath the clay layer is hard pan. The water table is at a depth of 1.8m below the ground surface.

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Solve this using the equation for approximation.
a. In footing A, compute the stress increase at the
center of the clay layer assuming that the pressure
beneath the footing is spread at an angle of 2 vertical
to 1 horizontal. (in kPa)
b. Determine the size of the footing B so that the
settlement in the clay layer is the same beneath
footing A and B. footing A is 1.5x1.5. (in m)
c. Determine the primary settlement for normally
consolidated clay beneath footing A. (in mm)
Transcribed Image Text:Solve this using the equation for approximation. a. In footing A, compute the stress increase at the center of the clay layer assuming that the pressure beneath the footing is spread at an angle of 2 vertical to 1 horizontal. (in kPa) b. Determine the size of the footing B so that the settlement in the clay layer is the same beneath footing A and B. footing A is 1.5x1.5. (in m) c. Determine the primary settlement for normally consolidated clay beneath footing A. (in mm)
Two footings rest in layer of sand 2.7m thick. The bottom
of the footings are 0.9m below the ground surface.
Beneath the clay layer is hard pan. The water table is at a
depth of 1.8m below the ground surface.
P-450 kN
P, = 900 kN
0.90 m
0.90 m
Y418,50 kN/m'
Sand
0.90 m
20.80 kN/m
Sand
y=18.80-kN/m
e=1.03, C,= 0.30
1.80 m
Clay
Solve this using the equation for approximation.
a. In footing A, compute the stress increase at the
Transcribed Image Text:Two footings rest in layer of sand 2.7m thick. The bottom of the footings are 0.9m below the ground surface. Beneath the clay layer is hard pan. The water table is at a depth of 1.8m below the ground surface. P-450 kN P, = 900 kN 0.90 m 0.90 m Y418,50 kN/m' Sand 0.90 m 20.80 kN/m Sand y=18.80-kN/m e=1.03, C,= 0.30 1.80 m Clay Solve this using the equation for approximation. a. In footing A, compute the stress increase at the
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