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类型FREDLUND边坡稳定未来讲义课件.ppt

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    FREDLUND 稳定 未来 讲义 课件
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    1、So Why Change?WWWWWWWWNLimit EquilibriumMethod of AnalysisSm=t ta dldls sndlFinite Element Based Method of Analysisldt ta dlQUESTION:How can the Normal Stress at the base of a slice be most accurately computed?Consider the Free Body Diagrams used to calculate the Normal Stress?()msnsSFuFc=-+b bf fs

    2、sb btanFocus on SmbyxSmXREREXLSlip surface WhRN=s snb bb bfNL()()swasansmFuuFuFcSb bf fb bf fs sb bbtantan-+-+=Only new variable required for solving saturated-unsaturated soils problems is the shear force mobilized -+=NfWxtanRtantanuNR cFbwmf ff ff fb bb b -+=a aa af ff ff fb ba ab bsincostantantan

    3、cosNuNcFbwfPore-air pressures are assumed to be zero gauge()FFuFcXXWNbwLRtansincostansinsinf fa aa af fa ab ba ab b+-=s sxbaArea=Interslice normal force(E)width of slice,b bs sxt txys syDistance(m)Elevation(m)t txybaArea=Interslice shear force(X)Vertical sliceDistance(m)Elevation(m)=baxydyXt t=baxdy

    4、Es sMorgenstern&Price,1965Method Equilibrium Satisfied Assumptions Ordinary Moment,to base E and X=0 Bishops Simplified Vertical,Moment E is horizontal,X=0 Janbus Simplified Vertical,Horizontal E is horizontal,X=0,empirical correction factor,f0,accounts for interslice shear forces Janbus Generalized

    5、 Vertical,Horizontal E is located by an assumed line of thrust Spencer Vertical,Horizontal,Moment Resultant of E and X are of constant slope hWbb ba aN=s bN=s b nN NSmhWbb ba aN=s bN=s b nN NSmERELhWbb ba aN=s bN=s b nN NSmERELXRXLSpencersX=E f(x)f(x)is largest at mid-pointInflection points near cre

    6、st&toe()2/)(nnCKexfw w-=Wilson and Fredlund,1983X=E f(x)Unique function of“slope angle”for all slip surfacesUnique function of“slope angle”for all slip surfaces00.20.40.61.801.851.901.952.002.052.102.152.202.25l lJanbus GeneralizedSimplified BishopSpencerMorgenstern-Pricef(x)=constantOrdinary=1.928F

    7、fFmFredlund and Krahn1975Factor of safetyMoment and Force Limit Equilibrium Factors of SafetyFor a Circular typeslip surfaceMoment limit equilibrium analysisForce limit equilibrium analysisFredlund and Krahn,1975Lambda,l l Factor of safetyForce and MomentLimit equilibriumFactors of Safety for a plan

    8、ar toe slip surfaceForce limit equilibrium analysisMoment limit equilibrium analysisLambda,l l Factor of safetyKrahn 2003Force and MomentLimit equilibrium Factors of Safety for a composite slip surfaceMoment limit equilibrium analysisForce limit equilibrium analysisLambda,l l Factor of safetyFredlun

    9、d and Krahn 1975Force and MomentLimit equilibrium Factors of Safety for a“Sliding Block”type slip surfaceMoment limit equilibrium analysisForce limit equilibrium analysisLambda,l l Factor of safetyKrahn 2003Section Parallel to MovementSection Perpendicular to MovementParallelPerpendicularBaseX/EV/PI

    10、mprovement of Normal Stress ComputationsFredlund and Scoular1999Limit equilibrium and finite element normal stresses for a toe slip surfaceFrom limit equilibrium analysisFrom finite element analysisLimit equilibrium and finite element normal stresses for a deep-seated slip surfaceFrom finite element

    11、 analysisFrom limit equilibrium analysisLimit equilibrium and finite element normal stresses for an anchored slopeFrom finite element analysisFrom limit equilibrium analysisAssumption:The stresses computed from“switching-on”gravity are more reasonable than the stresses computed on a vertical slices

    12、snFinite Element Analysis for StressesLimit Equilibrium Analysiss snt tmMohr Circlet tmIMPORT:Acting Normal StressActuating Shear StressLimit Equilibrium AnalysisFinite Element Analysis for StressesStress LevelRezendiz 1972Zienkiewicz et al 1975Strength&Stress LevelAdikari and Cummins 1985Enhanced l

    13、imit methods(finite element analysiswith a limit equilibriumFinite Element Slope Stability MethodsDirect methods(finite element analysis only)Strength LevelKulhawy 1969F -Z=1313 D DD DLLfs ss ss ss s F=(c +tan)-c +tanA 1313 s s f fs ss ss ss ss s f fD D D DL Lf*F=(c +tan )K s s f ft tD D D DLLDefini

    14、tion of Factor of SafetyLoad increaseto failureStrength decreaseto failureanalysis)=mrFEMSSFb bf fs stan)u(cSwnr-+=xyx-Coordinatey-CoordinateSlip SurfaceFinite Element(r,s)srFictitious slice defined withthe Limit Equilibrium analysisCenter of the base of a slice(x,y)20406080100120204060800CrestPiezo

    15、metric LineToe21x-Coordinate (m)Note:Dry slope with&without piezometric liney-Coordinate (m)20406080100120204060800CrestToe21x -Coordinate (m)WaterPiezometric Liney -Coordinate (m)Note:Dry slope with&without piezometric line050100150200250300203040506070 x-Coordinate(m)Acting and restricting shear s

    16、tress(kPa)CrestToeShear StrengthShear ForcePoisson Ratio,m m=0.33012345672025303540455055606570 x-CoordinateFactor of SafetyCrestToeLocal F(m mLocal F(m m=0.33)Bishop Method,F=2.360=2.173Global Factors of SafetyBishop 2.360Janbu 2.173GLE(F.E.function)2.356Fs(m m=0.33 )2.342Fs(m m=0.48 )2.339Ordinary

    17、 2.226sJanbu Method,FsssFs=2.342Fs=2.339=0.48)0.00.51.01.52.02.50510152025Stability Number,g g Htan f f /c Factor of Safetyc =20kPac =10kPac=40kPaFs(m m =0.33)Fs(m m =0.48)Fs(GLE)2:1 Dry Slope 0.00.51.01.52.02.50.000.020.040.060.080.100.12Stability Coefficient,c /g g H Factor of Safety f f=30f f =10

    18、f f=202:1 Dry SlopesFs(m m =0.33)F(m m =0.48)Fs(GLE)s0.00.40.81.21.62.00.000.020.040.060.080.100.12Stability Coefficient,c/g g H Factor of safetyf f =30f f =20f f =102:1 Slope with piezometric lineFs(m m =0.33)Fs(m m =0.48)Fs(GLE)70102010060504030908070110506040102030 x-Coordinate (m)80GLE(F.E.funct

    19、ion)Fs(m m=0.33)Fs(m m=0.48)MethodXYRFactor of safetyGLE(F.E.Function)58.556.037.91.741Fs(m m=0.33)57.549.534.71.627Fs(m m=0.48)57.553.037.81.661Y-Coordinate(m)70102010060504030908070506040102030110 x -Coordinate (m)80Fs(m m=0.4 8 )Fs(m m=0.33)GLE(F.E.function)sMethodXYR Factor of safetyGLE(F.E Func

    20、tion.63.559.039.61.102Fs(m m=0.33)63.059.041.51.076F(m m=0.48)61.559.542.31.100y-Coordinate(m)0.00.51.01.52.02.53.03.50.000.020.040.060.080.100.12Stability Coefficient,c /g g HFactor of Safetyf f =20f f =102:1 Dry slope,one-half submergedf f =30Fs(m m =0.33)Fs(m m =0.48)Fs(GLE)0.00.51.01.52.02.50.00

    21、0.020.040.060.080.100.12Stability Coefficient ,c/H Factor of Safetyf f =30f f=10f f =202:1 Slope,one-half submergedg gFs(m m =0.33)Fs(m m =0.48)Fs(GLE)1020100605040309080701107050604010203080Fs (m=m=0.33)Fs(m m=0.48)GLE(F.E.Function)sMethodXYR Factor of safetyGLE(F.E.Function 58.058.5 40.22.303Fs(m

    22、m=0.33)52.550.5 31.82.259F (m m=0.48)51.551.5 31.02.273x-Coordinate(m)y-Coordinate(m)Local Factors of Safety can also be computed by the Enhanced Limit MethodImprovement on Shape and Location Ha and Fredlund2002Assumption:The stresses computed from“switching-on”gravity can be used to represent the s

    23、tress state in the soil massSmooth curveDiscrete form(1)(2)StageBState pointn+1AYi1Riii+1kjSijk.i i+1.Fs=(Shear Strength)/(Actuating Shear Stress)=BABAfsdLdLFt tt t =D DD D=niiiniiifsLLF11t tt tstage i+1stage ilijlijfttfijijjs sijtijqkijsijtElement(ij)Element(ij)R=Resisting Shear Strength:S=Actuatin

    24、g Shear StressFs=(Shear Strength)/(Actuating Shear Stress)=D D-=niiisfiLFG1)(t tt tdLFGsBAf)(t tt t-=-=niisiSFRG1)(S=Actuating Shear StressR=Resisting Shear Strength =D D=neijijijneijijiiil SLS11t tt t =D D=neijijfneijijifilRLRiji11t tt tijbijwaijaijneijijiluuucRtan)(tan)(1f f f fs s-+-+=the optimal

    25、 function obtained at point k of stage i+1,=the optimal function obtained at point j in stage i,and=the return function calculated when passing from the state point j in stage i to the state point k in stage i+1.where:Introduce an“optimal function”,H=Optimal FunctionG=Return Function=-=niisiSFRGG1mi

    26、n)(minmin)(jHi),()()(1kjGjHkHiii+=+)(1kHi+)(jHi),(kjGiAt the initial stage,(i=1):At the final stage,(i=n+1):where:=the number of state points in the final stage H=Optimal Function0)(1=jH 1.1 NPj=),()()(1kjGjHkHnnn+=+=+-=niisimnSFRGkH11).()(.1=n+1 NPk1+nNPDYNAMIC PROGRAMMING SOLUTION116487511114121H1

    27、(1)=092747H1(1)=13AH (2)=812310B5674STAGE NUMBER1234567d=(4,2)3G (1,2)=33105243252827224415532BATHE MINIMUM TRAVELLING TIME PROBLEMEntry point1InitialABpointYState point.i i+1.XBBn+1X.Stage No.Exit pointSiGrid elementboundarySearchingii+1kSearching gridjRiFinal pointjk5S6RSR3SR544RS223BR6SRS11AXYR11

    28、SR22SiSiRRnSn.Kinematical RestrictionRiii+1kjSiEliminated =0.33DYNPROG=1.02Enhanced=1.13Bishop;M-P=1.17Distance,mElevation,m=0.33DYNPROG=1.02Bishop;M-P=1.17Enhanced=1.13=0.33=0.48=0.33=0.48=0.33Distance,mElevation,mBishop;M-P=1.64Enhanced=1.62DYNPROG=1.49=0.33Enhanced=1.62Bishop;M-P=1.64DYNPROG=1.49

    29、Enhanced=1.10M-P=1.14DYNPROG=0.96Distance,mElevation,m=0.48DYNPROG=0.975Bishop=1.00ActualActualEnhanced=0.997=0.38Enhanced=1.02Bishop=1.00DYNPROG=0.997ActualDistance,mElevation,mActual40Distance(m)150510202530354550605565707580Medium layer20105Hard layer15Weak layerElevation(m)302535404550Hard layer:c=100 kPa,=30,=0,=20 kN/m,=0.35,E=100000 kPa.Medium layer:c=20 kPa,=30,=0,=15 kN/m,=0.33,E=15000 kPa.Weak layer:c=0 kPa,=10,=0,=18 kN/m,=0.45,E=2000 kPa.DYNPROGfoff(1.180)ffobfobogbggoom33mm3DYNPROG=1.18Distance,mElevation,mElevation,mDistance,mSlope/W=1.196

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