Calculate the total settlement under the retaining wall.(ɣ concrete = 24 kN/m3) +6.00 m0.00 m-10.00 m-18.00 mGWT6.00 mYsat 17.0 kN/m³=0.50 mYsat 18.0 kN/m³=N1,60 18=V₁ = 280 m/sc'=0 kPa and = 30°-No.200-5% (FC-5%)=Cu=50 kPa, c'=5 kPa and '= 25°Cr/(1+eo) = 0.02; Ce/(1+eo) = 0.2PI=24%, N1,60 = 6 Es= 10000 kPaV, 165 m/sYsat 21.0 kN/m³BACKFILLYn 20.0 kN/m³c'=0 kPa and '= 38°qu = 4 MPaVs = 845 m/s6.00 mEs= 30000 kPa0.50 mGWTVSOFT CLAYSILTY SANDWEATHERED ANDESITE

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+6.00 m 0.00 m -10.00 m -18.00 m GWT 6.00 m 0.50 m Ysat 17.0 kN/m³ = Ysat 18.0 kN/m³ = BACKFILL Yn 20.0 kN/m³ Ysat 21.0 kN/m³ = qu = 4 MPa V₁ = 845 m/s c'=0 kPa and = 38° 6.00 m Cu=50 kPa, c'=5 kPa and o'= 25° Cr/(1+eo) = 0.02; Ce/(1+eo) = 0.2 PI = 24%, N1,60 = 6 Es= 10000 kPa V, 165 m/s c'=0 kPa and = 30° -No.200-5% (FC-5%) N1,60 18 Es= 30000 kPa = V₁ = 280 m/s 0.50 m GWT V SOFT CLAY SILTY SAND WEATHERED ANDESITE

+6.00 m 0.00 m -10.00 m -18.00 m GWT 6.00 m 0.50 m Ysat 17.0 kN/m³ = Ysat 18.0 kN/m³ = BACKFILL Yn 20.0 kN/m³ Ysat 21.0 kN/m³ = qu = 4 MPa V₁ = 845 m/s c'=0 kPa and = 38° 6.00 m Cu=50 kPa, c'=5 kPa and o'= 25° Cr/(1+eo) = 0.02; Ce/(1+eo) = 0.2 PI = 24%, N1,60 = 6 Es= 10000 kPa V, 165 m/s c'=0 kPa and = 30° -No.200-5% (FC-5%) N1,60 18 Es= 30000 kPa = V₁ = 280 m/s 0.50 m GWT V SOFT CLAY SILTY SAND WEATHERED ANDESITE

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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