Reviewof SoilMechanicsIll.CorrectionforconstructionperiodExample 2m(2)Dam4m-W.T.SandThis clay layer is a“"aquitard"layer8.mClaySandThis sand layer is an"artesian" layer with higherAaSandwater pressureabovegroundsurface26
26 Example 2 Review of Soil Mechanics III. Correction for construction period
Reviewof SoilMechanicsIll.Correction for construction periodExample 2For0<t=1.42≤tc=2years1.42Load× 14.71.42Assumed2St.corr(1.42) = St(214.7PPt=28.6×0.71=20.31mm0TimeIt=1.421.421.42=0.26×110mm=28.6mm22t/2=1.42/2=0.71t=1.42Time1.42t1.4 ×Cu22St,corr(t)=0.062142d242=0.26=26%UGCorrected curveTvUt(years)Sc (mm)110.100.0080.09"lnstantaneous"curve220.200.0310.35set0.300.79330.0701.42440.400.126For0<t=1.42≤tc=2years0.502.21550.1961.42Pt(t)x14.7 = 10.44kPa3.22660.600.28520.735.00800.437P=14.7kPa27
27 Example 2 U Tv t (years) sc (mm) 0.10 0.008 0.09 11 0.20 0.031 0.35 22 0.30 0.070 0.79 33 0.40 0.126 1.42 44 0.50 0.196 2.21 55 0.60 0.285 3.22 66 0.73 0.437 5.00 80 𝐹𝑜𝑟 0 < 𝑡 = 1.42 ≤ 𝑡𝑐 = 2𝑦𝑒𝑎𝑟𝑠 𝑠𝑡,𝑐𝑜𝑟𝑟(1.42) = 𝑠𝑡( 1.42 2 ) 1.42 2 × 14.7 14.7 = 28.6 × 0.71 = 20.31𝑚𝑚 ∵ 𝑠𝑡( 1.42 2 ) = 𝑈( 1.42 2 )𝑠𝑓 = 0.26 × 110𝑚𝑚 = 28.6𝑚𝑚 𝑇𝑣( 𝑡 2 ) = 𝑐𝑣 𝑡 2 𝑑 2 = 1.4 × 1.42 2 4 2 = 0.0621 𝑈( 𝑡 2 ) = 𝑈( 1.42 2 ) = 0.26 = 26% 𝐹𝑜𝑟 0 < 𝑡 = 1.42 ≤ 𝑡𝑐 = 2𝑦𝑒𝑎𝑟𝑠 𝑃𝑡 ′ (𝑡) = 1.42 2 × 14.7 = 10.44𝑘𝑃𝑎 𝑃′ = 14.7𝑘𝑃𝑎 Review of Soil Mechanics III. Correction for construction period
ReviewofSoilMechanicsIll.CorrectionforconstructionperiodExample 3An8mdepthofsandoverliesa6mlayerofclay,belowwhichisanimpermeablestratum;thewatertable is2mbelowthe surfaceof the sand.Over aperiod of 1year, a3m depth of fill(unit weight 20kN/m3)is tobedumped onthe surface over an extensivearea.The saturatedunitweightofthesandis19kN/m3andthatoftheclayis20kN/m3abovethewatertabletheunitweightofthesandis17kN/m3.Fortheclay,therelationshipbetweenvoidratioandeffectivestress(units:kN/m3)canberepresentedbytheequationae = 0.88- 0.32log100and the coefficient of consolidation is 1.26 m2/year(a)Calculate thefinal settlementof the area dueto consolidation of the clay and the settlementafter aperiod of 3years fromthe start of dumping(b) If a very thin layer of sand, freely draining, existed 1.5m above the bottom of the clay layer,whatwouldbethevaluesofthefinaland3-yearsettlements?Solution:(a) One layersand = Ysat,sand -Yw = 19 - 9.8 = 9.2kN/m3;Yclay = 20 - 9.8 = 10.2kN /m328
28 Example 3 An 8 m depth of sand overlies a 6m layer of clay, below which is an impermeable stratum; the water table is 2m below the surface of the sand. Over a period of 1 year, a 3m depth of fill (unit weight 20 kN/m3 ) is to be dumped on the surface over an extensive area. The saturated unit weight of the sand is 19 kN/m3 and that of the clay is 20 kN/m3 ; above the water table the unit weight of the sand is 17 kN/m3 . For the clay, the relationship between void ratio and effective stress (units: kN/m3 ) can be represented by the equation 𝑒 = 0.88 − 0.32 log 𝜎 ′ 100 and the coefficient of consolidation is 1.26 m2 /year. (a) Calculate the final settlement of the area due to consolidation of the clay and the settlement after a period of 3 years from the start of dumping. (b) If a very thin layer of sand, freely draining, existed 1.5m above the bottom of the clay layer, what would be the values of the final and 3-year settlements? Solution: (a) One layer 𝛾𝑠𝑎𝑛𝑑 ′ = 𝛾𝑠𝑎𝑡,𝑠𝑎𝑛𝑑 − 𝛾𝑤 = 19 − 9.8 = 9.2𝑘𝑁/𝑚3 ; 𝛾𝑐𝑙𝑎𝑦 ′ = 20 − 9.8 = 10.2𝑘𝑁/𝑚3 Review of Soil Mechanics III. Correction for construction period
ReviewofSoilMechanicsIll.Correction for construction periodeo -e1Example 3C=0.32eo - e1 = C. log(o'/o);Colog(oi/o)Aeeo-eAEu1+eo1+eoW.T% = (17 ×2) +(9.2 × 6) + (10.2 ×3) = 119.8kN/m28mSandi=119.8+(3×20)=179.8kN/m2119.8-e=0.88-0.32log0.855eo=0.88-0.320g100ClayCcg16 me。Od-6mHSf=AHHlog1+eo1+eo0.32179.8x6000=182mm10gImpermeable1+0.855119.8(a)PttFor<t≤tc: St,corr(t)=where0St(2)ptsForwhere St(tU(tt>tc:St.corr(t)= St(tGood equations to use!211.26 × 2.5tc：0.875=U=0.3355years;TVt2.62d222LU(t=0.335x182=61mm(t)=St(tar29
29 Example 3 Cc=0.32 𝑒0 − 𝑒1 = 𝐶𝑐 log( 𝜎1 ′ /𝜎0 ′ ); 𝐶𝑐 = 𝑒0 − 𝑒1 log( 𝜎1 ′ /𝜎0 ′ ) Δ𝜀𝑣 = Δ𝑒 1 + 𝑒0 = 𝑒𝑜 − 𝑒 1 + 𝑒0 𝜎0 ′ = (17 × 2) + (9.2 × 6) + (10.2 × 3) = 119.8𝑘𝑁/𝑚2 ; 𝑠𝑓 = Δ𝜀𝑣𝐻 = 𝑒𝑜 − 𝑒 1 + 𝑒0 𝐻 = 𝐶𝑐 1 + 𝑒0 log 𝜎1 ′ 𝜎0 ′ 𝐻 = 0.32 1+0.855 log 179.8 119.8 × 6000 = 182𝑚𝑚 𝑒 = 0.88 − 0.32 log( 𝜎 ′ 100) ⇒ 𝑒0 = 0.88 − 0.32𝑜𝑔( 119.8 100 ) = 0.855 𝜎1 ′ = 119.8 + (3 × 20) = 179.8𝑘𝑁/𝑚2 𝐹𝑜𝑟 0 < 𝑡 ≤ 𝑡𝑐 : 𝑠𝑡,𝑐𝑜𝑟𝑟(𝑡) = 𝑠𝑡( 𝑡 2 ) 𝑃𝑡 ′ 𝑃 ′ 𝑤ℎ𝑒𝑟𝑒 𝑠𝑡( 𝑡 2 ) = 𝑈( 𝑡 2 )𝑠𝑓 𝐹𝑜𝑟 𝑡 > 𝑡𝑐 : 𝑠𝑡,𝑐𝑜𝑟𝑟(𝑡) = 𝑠𝑡(𝑡 − 𝑡𝑐 2 ) 𝑤ℎ𝑒𝑟𝑒 𝑠𝑡(𝑡 − 𝑡𝑐 2 ) = 𝑈(𝑡 − 𝑡𝑐 2 )𝑠𝑓 Good equations to use ! 𝑠𝑡,𝑐𝑜𝑟𝑟(𝑡) = 𝑠𝑡 (𝑡 − 𝑡𝑐 2 ) = 𝑈(𝑡 − 𝑡𝑐 2 )𝑠𝑓 = 0.335 × 182 = 61𝑚𝑚 𝑡 − 𝑡𝑐 2 = 3 − 1 2 = 2 − 5𝑦𝑒𝑎𝑟𝑠; 𝑇𝑣 = 𝑐𝑣 𝑡 − 𝑡𝑐 2 𝑑 2 = 1.26 × 2.5 6 2 = 0.875 ⇒ 𝑈 = 0.335 Review of Soil Mechanics III. Correction for construction period
Reviewof SoilMechanicsIll. Correction for construction periodExample 3(b) Two layerMethod 1: Settlement is proportional to the thickness H4.5W.TFor Layer H, of 4.5m:x182=136.5mmSf1=6Bm1.5Sandx182=45.5mmFor Layer H, of 1.5m:Sf26c(t-)1.26 × 2.5T,1.40U2=0.97山d21.524.5(1)d=2.25mSc2 = U2Sf2 = 0.97 × 45.5 = 44,1mm(2)1.5mId=1.5m(t-2)(b)1.26 × 2.5T,=0.622 →U, = 0.825d22.252Sc1 = USf1 = 0.825× 136.5 _ 112.6mmSc= Sc1+ Sc2= 112.6+44.1 =156.7mm30
30 Example 3 (b) Two layer Method 1: Settlement is proportional to the thickness H For Layer H1 of 4.5m: For Layer H2 of 1.5m: 𝑠𝑐 = 𝑠𝑐1 + 𝑠𝑐2 = 112.6 + 44.1 = 156.7𝑚𝑚 𝑠𝑓1 = 4.5 6 × 182 = 136.5𝑚𝑚 𝑠𝑓2 = 1.5 6 × 182 = 45.5𝑚𝑚 𝑇𝑣 = 𝑐𝑣 𝑡 − 𝑡𝑐 2 𝑑 2 = 1.26 × 2.5 1.5 2 = 1.40 ⇒ 𝑈2 = 0.97 𝑠𝑐2 = 𝑈2 s𝑓2 = 0. 97 × 45. 5 = 44. 1mm 𝑇𝑣 = 𝑐𝑣 𝑡 − 𝑡𝑐 2 𝑑 2 = 1.26 × 2.5 2.252 = 0.622 ⇒ 𝑈1 = 0.825 𝑠𝑐1 = 𝑈1 s𝑓1 = 0. 825 × 136. 5 = 112. 6mm Review of Soil Mechanics III. Correction for construction period