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Calculate the consolidation settlement in clay layer depending on an increment in stress level of AP=150KN/m² for the given figure. Notes: Pre-consolidation pressure, Po=200KN/m2 Y=17KN/m3, y2=20KN/m3, y3=19KN/m3, Ywater 10kN/m3, Initial void ratio, eo=0.8, Swelling&compression curve parameters, Cc=0.25, c,=0 05

Calculate the consolidation settlement in clay layer depending on an increment in stress level of AP=150KN/m² for the given figure. Notes: Pre-consolidation pressure, Po=200KN/m2 Y=17KN/m3, y2=20KN/m3, y3=19KN/m3, Ywater 10kN/m3, Initial void ratio, eo=0.8, Swelling&compression curve parameters, Cc=0.25, c,=0 05

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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AP=150KN/m2 Calculate the consolidation settlement in clay layer depending on an increment in stress level of AP=150KN/m² for the given 3.5m Sand 71 V Underground water level 3.5m Sand 72 figure. 3.5m Notes: Rock Pre-consolidation pressure, Po=200KN/m² Y=17KN/m3, y2=20KN/m3, y3=19KN/m3, Ywater 10kN/m3, Initial void ratio, eo=0.8, Swelling&compression curve parameters, Co=0.25, c,=0 05
Transcribed Image Text:AP=150KN/m2 Calculate the consolidation settlement in clay layer depending on an increment in stress level of AP=150KN/m² for the given 3.5m Sand 71 V Underground water level 3.5m Sand 72 figure. 3.5m Notes: Rock Pre-consolidation pressure, Po=200KN/m² Y=17KN/m3, y2=20KN/m3, y3=19KN/m3, Ywater 10kN/m3, Initial void ratio, eo=0.8, Swelling&compression curve parameters, Co=0.25, c,=0 05
Equations to be used in analyses: K. = tan| 45- K, = tan| 45+- Ye = Yza -Ym Q =Qa * FS %3D S= M lo I =1.3cN, + qN, +0.4YBN, tano max ahw 4 Na Ne = (Na – 1)Coto Kpy = 3tan?|45 + %3D 12cos2 (플+ 용) Ny 2 Cos?0 tano Ae-c*log(P./Po')+ c*log((Po'+ 4P )/P.') ốc=del(1+eo)*He B' B' F =1+0.2tan? 45+ L' F =1+0.2 f=0°) F =F, = 1- L' 90 B' F = F =1+0.1 L' - (if o>10°) F =F, =1 (if 6=0°) F = 1- %3D !! D, D, F =1+0.2- tan 45 +- B (if o>10°) Fa =1+0.2 (f 6=0°) Cy = (1/0.01* o, )es В F = F =1+0.1 45+ (if >10°) Fng = F =1 (if 6 = 0*) C; = ER/ 60 N0 = N CCyC,C,C, %3D
Transcribed Image Text:Equations to be used in analyses: K. = tan| 45- K, = tan| 45+- Ye = Yza -Ym Q =Qa * FS %3D S= M lo I =1.3cN, + qN, +0.4YBN, tano max ahw 4 Na Ne = (Na – 1)Coto Kpy = 3tan?|45 + %3D 12cos2 (플+ 용) Ny 2 Cos?0 tano Ae-c*log(P./Po')+ c*log((Po'+ 4P )/P.') ốc=del(1+eo)*He B' B' F =1+0.2tan? 45+ L' F =1+0.2 f=0°) F =F, = 1- L' 90 B' F = F =1+0.1 L' - (if o>10°) F =F, =1 (if 6=0°) F = 1- %3D !! D, D, F =1+0.2- tan 45 +- B (if o>10°) Fa =1+0.2 (f 6=0°) Cy = (1/0.01* o, )es В F = F =1+0.1 45+ (if >10°) Fng = F =1 (if 6 = 0*) C; = ER/ 60 N0 = N CCyC,C,C, %3D
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