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Sand: WT Ysat = 19 kN/m³ Ydry = 17 kN/m³ 3.5 m Sand: Ysat = 20 kN/m³ 4.5 m Ydry = 18 kN/m³ Ysat = 20 kN/m³ eo= 0.9 Clay: = Cc 0.36 Cr 0.12 = OCR = 1 6.0 m Sand: Ysat = 19 kN/m³ 2.0 m BEDROCK

Sand: WT Ysat = 19 kN/m³ Ydry = 17 kN/m³ 3.5 m Sand: Ysat = 20 kN/m³ 4.5 m Ydry = 18 kN/m³ Ysat = 20 kN/m³ eo= 0.9 Clay: = Cc 0.36 Cr 0.12 = OCR = 1 6.0 m Sand: Ysat = 19 kN/m³ 2.0 m BEDROCK

Structural Analysis
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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A surcharge of 250 kPa is applied over a large area above the soil profile shown below. The clay layer is initially normally consolidated and cv = 0.002 cm2/s. The water table is 3.5 m below the ground surface. Calculate the settlement of the clay layer in mm 40 days after the surcharge was applied. 

Hint: the final settlement at the end of primary consolidation is equal to 515 mm.

Sand: WT Ysat = 19 kN/m³ Ydry = 17 kN/m³ 3.5 m Sand: Ysat = 20 kN/m³ 4.5 m Ydry = 18 kN/m³ Ysat = 20 kN/m³ eo= 0.9 Clay: = Cc 0.36 Cr 0.12 = OCR = 1 6.0 m Sand: Ysat = 19 kN/m³ 2.0 m BEDROCK
Transcribed Image Text:Sand: WT Ysat = 19 kN/m³ Ydry = 17 kN/m³ 3.5 m Sand: Ysat = 20 kN/m³ 4.5 m Ydry = 18 kN/m³ Ysat = 20 kN/m³ eo= 0.9 Clay: = Cc 0.36 Cr 0.12 = OCR = 1 6.0 m Sand: Ysat = 19 kN/m³ 2.0 m BEDROCK
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