The soil profile shown carries a uniformly distributed load of 75 kPa applied at the ground surface. Use Cs = 1/5 Cc. Determine the settlement of the clay layer due to primary consolidation considering the following cases: a. The clay is normally consolidated (0.242m) b. If Pc = 205 kPa (0.0484m) c. If OCR = 1.6219 (0.0764m) 75 kN/m2 2.4 m SAND ydry 14.75kN/m .GWT 4.6 m SAND ysat =17.45KN/m³ ysat = 20.62 kN/m³ e=0.9 LL - 50 5,2 m CLAY
The soil profile shown carries a uniformly distributed load of 75 kPa applied at the ground surface. Use Cs = 1/5 Cc. Determine the settlement of the clay layer due to primary consolidation considering the following cases: a. The clay is normally consolidated (0.242m) b. If Pc = 205 kPa (0.0484m) c. If OCR = 1.6219 (0.0764m) 75 kN/m2 2.4 m SAND ydry 14.75kN/m .GWT 4.6 m SAND ysat =17.45KN/m³ ysat = 20.62 kN/m³ e=0.9 LL - 50 5,2 m CLAY
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
The answer should be
A. 0.242m
B. 0.0484m
C. 0.0764 m

Transcribed Image Text:The soil profile shown carries a uniformly distributed load of 75 kPa applied at the ground
surface. Use Cs = 1/5 Cc.
Determine the settlement of the clay layer due to primary consolidation considering the
following cases:
a. The clay is normally consolidated (0.242m)
b. If Pc = 205 kPa (0.0484m)
If OCR = 1.6219 (0.0764m)
C.
75 kN/m
2.4 m
SAND
ydry 14.75kN/m
V GWT
4.6 m
SAND
ysat 17.45KN/m
ysat = 20.62 kN/m³
e =0.9
LL - 50
5,2m
CLAY
SETTLEMENT DUE TO SECONDARY CONSOLIDATION
Ss = C'a H log )
Eq. 14
Ca
1 + ep
Ae
C'a =
Eq. 15
Ca =
Eq. 16
log ()
where:
Ss = secondary consolidation settlement
Ca = secondary compression index
e, = void ratio at the end of primary consolidation
= e,-Ae
H = thickness of clay layer
t1, t2 = time
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