ReviewofSoilMechanicsIl.DeterminationofcoefficientofconsolidationExample 1Inordertodeterminethepermeability,thevalue ofmymustbe calculated.Final void ratio, e1 = wiG,= 0.359 × 2.73 = 0.98Initialvoidratio,eo=ei+AeNowAHAeAe2.39>>>>△e = 0.35and eo= 1.33EU1H。1.98+e15.991+eo110.35eo - e1:7.0×10-4m2/kN=0.7m2/MNmyX2.332151+eo-%Coefficient of permeability0.45× 0.70x9.8k=cvmvw=60×140365×10=1.0×10-10m/21
21 Example 1 In order to determine the permeability, the value of mv must be calculated. Final void ratio, 𝑒1 = 𝑤1𝐺𝑠 = 0.359 × 2.73 = 0.98 Initial void ratio, 𝑒0 = 𝑒1 + ∆𝑒 Now Δ𝑒 1+𝑒0 = Δ𝐻 𝐻0 = 𝜀𝑣 ≫ Δ𝑒 1.98+Δ𝑒 = 2.39 15.99 ≫ Δ𝑒 = 0.35 and 𝑒0 = 1.33 𝑚𝑣 = 1 1 + 𝑒0 ∙ 𝑒0 − 𝑒1 𝜎1 ′ − 𝜎0 ′ = 1 2.33 × 0.35 215 = 7.0 × 10−4 m2/kN = 0.7 m2/MN 𝑘 = 𝑐𝑣𝑚𝑣𝛾𝑤 = 0.45 × 0.70 × 9.8 60 × 1440 × 365 × 103 = 1.0 × 10−10m/s Coefficient of permeability Review of Soil Mechanics II. Determination of coefficient of consolidation
ReviewofSoilMechanicsIll.Correctionforconstructionperiod>Previoussolution/chartissuddenlyappliedload> The load in real project is normally ramp loading (e.g. construction loading)linearincreaseandthenconstant> Howtofindasolutiontothisissueormakeacorrection?LoadActualTimeEffectiveconstructionperiod22
22 ➢ Previous solution/chart is suddenly applied load ➢ The load in real project is normally ramp loading (e.g. construction loading) - linear increase and then constant ➢ How to find a solution to this issue or make a correction? Review of Soil Mechanics III. Correction for construction period
Reviewof SoilMechanicsIll.Correction for construction periodLoadAssumedFor0<t≤tcPPSt,corr(t) = St(PtwhereSt()CS0teitTime512AttitTime2OtcFor t>tc:SSt,corr(t) = St(t -St,corr!2o-t.cor=u(tCorrected curveUiscalculated from"lnstantaneous"curveprevious solution/chartSct23
23 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 U is calculated from previous solution/chart 𝑃𝑡 ′ t t 𝑺t,corr t 𝑃 ′ 𝑺t,corr Review of Soil Mechanics III. Correction for construction period
ReviewofSoilMechanicsIll.CorrectionforconstructionperiodDegree of consolidationTwo sidesarefreeBottomsideis2ddrainageimpermeableCurve (2)Curve (1)Curve (3)Curve (1)Curve (1)Curve (1)(b)Half-closed layers(a)OpenlayersInitialvariationsofexcessporewaterpressure0100200.3020401U0.50(3)060T0.70U2forU<0.60,Tv4o9oforU>0.60,Ty=-0.933log(1-U)-0.0851000010.1000C,tT,d224Relationships between average degreeof consolidationand timefactor
24 Degree of consolidation Relationships between average degree of consolidation and time factor Two sides are free drainage Bottom side is impermeable Initial variations of excess pore water pressure 𝑓𝑜𝑟 𝑈 < 0.60, 𝑇𝑉= 𝜋 4 𝑈 2 𝑓𝑜𝑟 𝑈 > 0.60, 𝑇𝑉= −0.933 log 1 − 𝑈 − 0.085 Review of Soil Mechanics III. Correction for construction period
ReviewofSoilMechanicsIll.Correction for construction periodExample 2Alayerofclay8mthickliesbetweentwolayersofsand.Theuppersandlayerextendsfromground level to a depth of 4m, the water table beingat a depthof 2m.The lower sand layer isunderartesianpressure,thepiezometriclevelbeing6mabovegroundlevel.Fortheclaymy=0.94m?/MN and cy=1.4m?/year.As aresult of pumpingfrom the artesian layer thepiezometric level falls by 3m over a period of 2 years.Draw the time-settlement curve due toconsolidation of the clay for a period of 5 years from the start of pumping.Solution:13mg'=g-u2)16mAa'Ag'=-Au×29.43= 14.72kPa2Scf = m μs'H = 0.94 × 14.72 ×8 = 110mm4m-W.T.Sand4g=-4u=0Suddenly applied loading case:Layeris open, thus, d=4m, fort=5 years8mClayAo"=14.72kPa1.4 × 5CutT,=0.437=U=0.7342d2Da'SandZg'= -△u = -(-3 × 9.81)25=29.43kN/m2
25 Example 2 A layer of clay 8m thick lies between two layers of sand. The upper sand layer extends from ground level to a depth of 4m, the water table being at a depth of 2m. The lower sand layer is under artesian pressure, the piezometric level being 6m above ground level. For the clay mv = 0.94 m2 /MN and cv= 1.4 m2 /year. As a result of pumping from the artesian layer the piezometric level falls by 3m over a period of 2 years. Draw the time-settlement curve due to consolidation of the clay for a period of 5 years from the start of pumping. Solution: 𝜎 ′ = 𝜎 − 𝑢 Δ𝜎 ′ = −Δ𝑢 Δ𝜎lj ′ = 1 2 × 29.43 = 14.72𝑘𝑃𝑎 𝑠𝑐𝑓 = 𝑚 𝑣Δ𝜎lj ′𝐻 = 0.94 × 14.72 × 8 = 110𝑚𝑚 Suddenly applied loading case: Layer is open, thus, d=4m, for t=5 years 𝑇𝑣 = 𝑐 𝑣𝑡 𝑑 2 = 1.4 × 5 4 2 = 0.437 ⇒ 𝑈 = 0.73 Δ𝜎 ′ = −Δ𝑢 = 0 Δ𝜎 ′ = −Δ𝑢 = −(−3 × 9.81) = 29.43𝑘𝑁/𝑚2 Δ𝜎lj ′ = 14.72kPa Review of Soil Mechanics III. Correction for construction period