5.00-aoThe root time methodInitialcompressionD-asTheoreticalcurve4.50UC0.9)BeerneeBA4.001.0L主固结10VT.PrimaryXao-asconsolidationro =ao-af1.15X10as-ag0as-a100ea3.50rp =rp =9 ao-afao-afrs = 1 - (ro + rp)0.848d23.00Cua1octoSecondaryconpression次固结/蠕燮(creep)22.50142101246Vtgo80t9otEOP16Vf (min)
X 1.15X t90 主固結 次固結/蠕變(creep) 𝑟0 = 𝑎0 − 𝑎𝑠 𝑎0 − 𝑎𝑓 𝑟𝑝 = 𝑎𝑠 − 𝑎100 𝑎0 − 𝑎𝑓 ; 𝑟𝑝 = 10 9 𝑎𝑠 − 𝑎90 𝑎0 − 𝑎𝑓 𝑟𝑠 = 1 − (𝑟0 + 𝑟𝑝) 𝑡𝐸𝑂𝑃 The root time method 16 𝑐𝑣 = 0.848𝑑 2 𝑡90
Reviewof SoilMechanicsIl.DeterminationofcoefficientofconsolidationDeterminationofcoefficientofconsolidationInpractical projects,the horizontalcoefficient ofconsolidation(ch)is ofinterest,Rowe Cell is normally adopted for determining Ch.LVDTtovacuumlinesettiementdrainagerodtodiaphragmpressuresystemairbleediaphregmtopressurvolumecontrollerrim drainperipheralfilterpapersinteredporousbronzediscSandClayperipheralrubbermembraneconnectingrodwellCh = (2~4)cukn = (2~4)ku17
17 Determination of coefficient of consolidation Clay Sand well In practical projects, the horizontal coefficient of consolidation (ch ) is of interest, Rowe Cell is normally adopted for determining ch . 𝑐ℎ = 2~4 𝑐𝑣 𝑘ℎ = (2~4)𝑘𝑣 Review of Soil Mechanics II. Determination of coefficient of consolidation
ReviewofSoilMechanicsIl.DeterminationofcoefficientofconsolidationRowe cell testforhorizontal hydraulicconductivitydeterminationClaysampleSandwell18
18 Review of Soil Mechanics II. Determination of coefficient of consolidation Sand well Clay sample Rowe cell test for horizontal hydraulic conductivity determination
Reviewof SoilMechanicsIlDeterminationofcoefficientofconsolidationExample 1Thefollowingcompressionreadingsweretakenduringanoedometertestona saturated clayspecimen(G=2.73)whentheappliedpressurewas increasedfrom214to429kN/m2:00.250.512.25491625Time (min)4.534.015.004.674.624.414.283.753.49Gauge (mm)648136491002004001440Time (min)3.062.842.762.613.283.153.002.96Gauge (mm)After1440minthethicknessofthespecimenwas13.60mmandthewatercontentwas35.9%Determinethecoefficientofconsolidationfromboththelogtimeandtheroottimeplotsandthevaluesofthethreecompressionratios.Determinealsothevalueofthecoefficientofpermeability.Solution:Totalchangeinthicknessduringincrement=5.00-2.61=2.39mm?6Averagethicknessduringincrement=13.60+2.39/2=14.80mm15Lengthofdrainagepath, d=14.80/2=7.40mmConventionaloedometercell19
19 Example 1 The following compression readings were taken during an oedometer test on a saturated clay specimen (Gs=2.73) when the applied pressure was increased from 214 to 429 kN/m2 : Time (min) 0 0.25 0.5 1 2.25 4 9 16 25 Gauge (mm) 5.00 4.67 4.62 4.53 4.41 4.28 4.01 3.75 3.49 Time (min) 36 49 64 81 100 200 400 1440 Gauge (mm) 3.28 3.15 3.06 3.00 2.96 2.84 2.76 2.61 After 1440 min the thickness of the specimen was 13.60 mm and the water content was 35.9%. Determine the coefficient of consolidation from both the log time and the root time plots and the values of the three compression ratios. Determine also the value of the coefficient of permeability. Solution: Total change in thickness during increment = 5.00-2.61 = 2.39 mm Average thickness during increment = 13.60+2.39/2 = 14.80 mm Length of drainage path, d= 14.80/2=7.40 mm Review of Soil Mechanics II. Determination of coefficient of consolidation
ReviewofSoilMechanicsIlDetermination ofcoefficient ofconsolidationExample 1Fromthelogtimeplottso = 12.5 min0.196d20.196×7.4021440×3650.45m2/year+Cu10612.5tso5.00-4.790.088ro5.00-2.614.79-2.98rs= 1- (0.088 +0.757)= 0.1550.757rp5.00-2.61From the root time plot, tg = 7.30, and thereforetgo=53.3min0.848d20.848 × 7.4021440×365= 0.46 m2/yearXC.10653.3too5.00-4.810.080ro5.00-2.6110(4.81- 3.12)r,= 1- (0.080 + 0.785)=0.1350.785rp9(5.00 - 2.61)20
20 Example 1 𝑡50 = 12.5 min From the log time plot 𝑐𝑣 = 0.196𝑑2 𝑡50 = 0.196 × 7.402 12.5 × 1440 × 365 106 = 0.45 m2/year 𝑟0 = 5.00 − 4.79 5.00 − 2.61 = 0.088 𝑟𝑝 = 4.79 − 2.98 5.00 − 2.61 = 0.757 𝑟𝑠 = 1 − 0.088 + 0.757 = 0.155 From the root time plot, 𝑡90 = 7.30, and therefore 𝑡90 = 53.3 min 𝑐𝑣 = 0.848𝑑2 𝑡90 = 0.848 × 7.402 53.3 × 1440 × 365 106 = 0.46 m2/year 𝑟0 = 5.00 − 4.81 5.00 − 2.61 = 0.080 𝑟𝑝 = 10(4.81 − 3.12) 9(5.00 − 2.61) = 0.785 𝑟𝑠 = 1 − 0.080 + 0.785 = 0.135 Review of Soil Mechanics II. Determination of coefficient of consolidation