.A circular foundation 2m in diameter is shown in the figure below. A normallyconsolidated clay layer 5m thick is located below the foundation. Determine the consolidationsettlement of the clay. method 2:1(1) As one layer of clay of 5m thick:(2) Divide the clay layer into (5) sub-layers each of 1m thick:-Calculation of increase of stress below the center of each sub-layer Aa(i) :(3) Weighted average pressure increase (Simpson's rule):Circular foundationdiameter B = 2mG.S.1.0mSandq = 150 kN/m²y = 17 kN/m²0.5mW.T.Sand0.5mNormally consolidated clayY sat. = 18.5 kNm²C. =0.16, e, =0.855.0m

GivenCc=0.16 and e0=0.85 and load distribution 2:1 =H:V here n=2i Consolidation…

.A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 (1) As one layer of clay of 5m thick: (2) Divide the clay layer into (5) sub-layers each of lm thick: -Calculation of increase of stress below the center of each sub-layer Ag(i) :

.A circular foundation 2m in diameter is shown in the figure below. A normally consolidated clay layer 5m thick is located below the foundation. Determine the consolidation settlement of the clay. method 2:1 (1) As one layer of clay of 5m thick: (2) Divide the clay layer into (5) sub-layers each of lm thick: -Calculation of increase of stress below the center of each sub-layer Ag(i) :

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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