A surface load of 60 kPa is applied on the ground surface over a large area. The soil profile consists of a sand layer 2 m thick, the top of which is the ground surface, overlying a 4 m thick layer of clay. An impermeable boundary is located at the base of the clay layer. The water table is 1 m below the ground surface. If the preconsolidation pressure for a sample of soil from the mid-point of the clay layer is 60 kPa, calculate the consolidation sețtlement of the clay layer. The properties of the soil section are: sand: Pdry = 1.6 Mg/m, Psat = 1.9 Mg/m³, clay: psat = 1.65 Mg/m’, eo = 1.5, C = 0.6, C, = 0.1.

Principles of Foundation Engineering (MindTap Course List)
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Author:Braja M. Das
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Chapter16: Soil Improvement And Ground Modification
Section: Chapter Questions
Problem 16.9P
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Q4
A surface load of 60 kPa is applied on the ground surface over a large area. The soil
profile consists of a sand layer 2 m thick, the top of which is the ground surface, overlying
a 4 m thick layer of clay. An impermeable boundary is located at the base of the clay
layer. The water table is 1 m below the ground surface. If the preconsolidation pressure
for a sample of soil from the mid-point of the clay layer is 60 kPa, calculate the
consolidation sețtlement of the clay layer. The properties of the soil section are: sand:
Pdry = 1.6 Mg/m", Psat = 1.9 Mg/m°, clay: psat= 1.65 Mg/m³, eo = 1.5, Ce = 0.6, C,= 0.1.
Hint: Ppc= [Ho/(1+einit)]x[C,log(p'/ơʻinit) + Clog(oʻfin/p')]
Transcribed Image Text:Q4 A surface load of 60 kPa is applied on the ground surface over a large area. The soil profile consists of a sand layer 2 m thick, the top of which is the ground surface, overlying a 4 m thick layer of clay. An impermeable boundary is located at the base of the clay layer. The water table is 1 m below the ground surface. If the preconsolidation pressure for a sample of soil from the mid-point of the clay layer is 60 kPa, calculate the consolidation sețtlement of the clay layer. The properties of the soil section are: sand: Pdry = 1.6 Mg/m", Psat = 1.9 Mg/m°, clay: psat= 1.65 Mg/m³, eo = 1.5, Ce = 0.6, C,= 0.1. Hint: Ppc= [Ho/(1+einit)]x[C,log(p'/ơʻinit) + Clog(oʻfin/p')]
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