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 eo= 0.9 Y sat 115 pcf OCR =1.1 ROCK SAND W=5% 6 = 32° G=2.65 eo= 0.4 C=0.4 C = 0.05 C, = 0.2 ft/day Sult =D₁177 ft 59 months = D.109 ft 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.

Principles of Foundation Engineering (MindTap Course List)
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Chapter8: Mat Foundations
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Problem 8.7P
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PLEASE ONLY SOLVE PROBLEM 2, ONLY, I really want to undertsand the concept and theory behind this problem so please explain it with clear calculations and wording, thank you

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 C = 0.05
OCR =1.1
GENERAL
LOCAL
C, = 0.2 ft >/day
Sult =0,177 ft
59 months = 0.109 ft.
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.
Qult = 33,834 Hofft ²
gult = 4,744 Uofft 2
Ub
2
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 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 GENERAL LOCAL C, = 0.2 ft >/day Sult =0,177 ft 59 months = 0.109 ft. 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. Qult = 33,834 Hofft ² gult = 4,744 Uofft 2 Ub 2
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