1. For the circular foundation below, determine: a) the ultimate settlement and b) the settlement after 9 months, due to consolidation of the clay layer. Base your calculations on the values at the midpoint of the clay layer. F=200,000 lb 4 ft 6 ft 6 ft 14 ft 10 ft CLAY SAND W=5% = 32° G=2.65 eo= 0.4 eo= 0.9 C = 0.4 Y sat 115 pcf = 0.05 OCR =1.1 ROCK C, = 0.2 ft z/day Sult =D₁177 ft 59 months = D.109 ft
1. For the circular foundation below, determine: a) the ultimate settlement and b) the settlement after 9 months, due to consolidation of the clay layer. Base your calculations on the values at the midpoint of the clay layer. F=200,000 lb 4 ft 6 ft 6 ft 14 ft 10 ft CLAY SAND W=5% = 32° G=2.65 eo= 0.4 eo= 0.9 C = 0.4 Y sat 115 pcf = 0.05 OCR =1.1 ROCK C, = 0.2 ft z/day Sult =D₁177 ft 59 months = D.109 ft
Principles of Foundation Engineering (MindTap Course List)
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ISBN:9781305081550
Author:Braja M. Das
Publisher:Braja M. Das
Chapter8: Mat Foundations
Section: Chapter Questions
Problem 8.7P
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Please solve with step by step solutions so I can understand the prioblem and theory
![1. For the circular foundation below, determine: a) the ultimate settlement and b) the
settlement after 9 months, due to consolidation of the clay layer. Base your calculations on the
values at the midpoint of the clay layer.
F=200,000 lb
4 ft
6 ft
6 ft
14 ft
All
10 ft
CLAY
SAND
w=5%
= 32°
G = 2.65
eo= 0.4
eo= 0.9
C=0.4
Y sat 115 pcf C = 0.05
OCR=1.1
C = 0.2 ft 2/day
Sult = 0,177 ft
59 months = 0.109,
ROCK
2.
Assume that the footing in problem 1 is of a 6 ft radius. Calculate the ultimate bearing
capacity for the assumptions of general and local shear failures.
2
GENERAL gult = 33,834 Lofft ²
gult = 4,744 Ubift?
2
LOCAL](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78a1cdca-9a2c-4803-aedc-ba4e0e310e87%2F3bffa8cd-8375-4e68-a168-76882cfb7e41%2Fky7flxp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. For the circular foundation below, determine: a) the ultimate settlement and b) the
settlement after 9 months, due to consolidation of the clay layer. Base your calculations on the
values at the midpoint of the clay layer.
F=200,000 lb
4 ft
6 ft
6 ft
14 ft
All
10 ft
CLAY
SAND
w=5%
= 32°
G = 2.65
eo= 0.4
eo= 0.9
C=0.4
Y sat 115 pcf C = 0.05
OCR=1.1
C = 0.2 ft 2/day
Sult = 0,177 ft
59 months = 0.109,
ROCK
2.
Assume that the footing in problem 1 is of a 6 ft radius. Calculate the ultimate bearing
capacity for the assumptions of general and local shear failures.
2
GENERAL gult = 33,834 Lofft ²
gult = 4,744 Ubift?
2
LOCAL
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