Determine the ultimate consolidation settlement of the entire deposit. Where the distance of A is 3m and the distance of B is 2.25. 7. A 4.0 m thick fill is to be made of a soil with a Proctor maximum dry unit weight of 19.4 kN/mand an optimum moisture content of 13.5%. This fill will be compacted at optimum moisturecontent to an average relative compaction of 91%. The underlying soils are as shown in Figure6. Consolidation tests were performed at points A and B, with the following results:0.kPa)75100SampleC.0.590.190.141.901210.37The silty sand is normally consolidated. Using hand computations, determine the ultimateconsolidation settlement of the entire deposit due to the weight of this 4m fill.3-0mY = 14,0 KN/mSitty sand (SM)Dr =45%4-5m17-5 KN/m²>-13-5 KN/mSoft onganicclayce)om4.5m8-15-0 KNmMedium ClayCCH)15-0mglacial tillFigure 6

Thicknes of clay fill = 4.0 mMaximum dry unit weight(γd-max) = 19.4 kN/m3Optimum moisture content = 13.5%Relative compaction = 91%For sample A,Compression Index (Cc) = 0.59Recompression Index (Cr) =0.19Initial void ratio (eo) = 1.9Overconsolidated pressure (σ'c) = 75For sample B,Compression Index (Cc) = 0.37Recompression Index (Cr) =0.14Initial void ratio (eo) = 1.21Overconsolidated pressure (σ'c) = 100Relative Compaction = γdγd-maxFinal Settement due to consolidation,Sf = CrH1+eolog10σc σo+CcH1+eolog10σo+∆σ σcHere,σo = overburden pressureH = Height of soil strata∆σ = Increase in soil pressure Pressure from soil fillRelative Compaction =γdγd-maxγd = 0.91×19.4γd = 17.65 kN/m3For bulk unit weight(γ),γd = γ1+w17.65 = γ1+0.135γ = 20.03 kN/m3∆σ =20.03 ×4 = 80.14 kN/m2Settlement of soft organic clay,H =6 mσo=14×3+17.5×4.5+13.5×3 -9.81×(3+4.5+3)σo=161.25 - 103.0 σo =58.25 kN/m2Sf = CrH1+eolog10σc σo+CcH1+eolog10σo+∆σ σcSf = 0.19×61+1.9log107558.25+0.59×61+1.9log1058.25+80.1475Sf = 0.043+0.325Sf =0.367 mSettlement of medium clay,H =6 mσo=14×3+17.5×4.5+13.5×6+15×2.25 -9.81×(3+4.5+6+2.25)σo=235.5 - 154.51 σo=81 kN/m2Sf = CrH1+eolog10σc σo+CcH1+eolog10σo+∆σ σcSf = 0.19×61+1.9log1010081+0.59×61+1.9log1081+80.14100Sf = 0.036+0.253Sf =0.289 mSo total settlement of soil deposit = 0.367 + 0.289 = 0.656 m