lesson 06-chapter 6 slope stability

SOILS AND FOUNDATIONSSOILS AND FOUNDATIONS

Testing

Experience

Theory

Lesson 06Lesson 06Chapter 6 Chapter 6 –– Slope StabilitySlope Stability

TopicsTopics

ggTopic 1 (Section 6.0 Topic 1 (Section 6.0 –– 6.8)6.8)-- Stability analysis of slopesStability analysis of slopes

ggTopic 2 (Section 6.9)Topic 2 (Section 6.9)-- Improving the stability of embankmentsImproving the stability of embankments

SLOPE STABILITYSLOPE STABILITY

Lesson 06 Lesson 06 -- Topic 1Topic 1Stability analysis of slopesStability analysis of slopes

Section 6.0 Section 6.0 –– 6.86.8

Learning OutcomesLearning Outcomes

ggAt the end of this session, the participant will At the end of this session, the participant will be able to:be able to:-- Recall modes of slope failureRecall modes of slope failure-- Explain effects of water on slope stabilityExplain effects of water on slope stability-- Discuss slope stability circular and block Discuss slope stability circular and block

analysesanalyses-- Compute safety factor by chart solutionCompute safety factor by chart solution

Stability ProblemsStability Problems

Embankment Fill

Firm Soil

ggShallow Shallow translational translational failure (Infinite failure (Infinite Slope)Slope)

ggCircular Circular FailureFailure

Stability ProblemsStability Problems

ggSliding Sliding block block failurefailure

ggLateral Lateral squeezesqueeze-- Lesson 7Lesson 7

Effect of Water on Slope StabilityEffect of Water on Slope Stability

ggFrictional SoilsFrictional Soils-- Below Water Table, Buoyancy Reduces Shearing Below Water Table, Buoyancy Reduces Shearing

ResistanceResistanceggClaysClays

-- Cohesive Strength Decreases as Water Content Cohesive Strength Decreases as Water Content IncreasesIncreasesCohesive Strength

Water Content

Effect of Water on Slope Stability Effect of Water on Slope Stability (Cont’d)(Cont’d)ggFills on Clays and SiltsFills on Clays and Silts

-- Soil Consolidates as Water is Squeezed Out Soil Consolidates as Water is Squeezed Out --Factor of Safety Increases With TimeFactor of Safety Increases With Time

ggCuts in ClayCuts in Clay-- Soil Absorbs Water When Overburden Pressure Soil Absorbs Water When Overburden Pressure

Removed Removed -- Factor of Safety Decreases With TimeFactor of Safety Decreases With Time

Effect of Water on Slope Stability Effect of Water on Slope Stability (Cont’d)(Cont’d)ggShalesShales, , ClaystonesClaystones, Siltstones, Etc., Siltstones, Etc.

-- Weak Rock Materials “Slake” When Exposed to Weak Rock Materials “Slake” When Exposed to Water Water -- Embankments Undergo Internal Embankments Undergo Internal Settlement or FailureSettlement or Failure

Design Factor of SafetyDesign Factor of Safety

ggMinimum FS = 1.25 for highway side slopesMinimum FS = 1.25 for highway side slopes

ggUse FS = 1.3 to 1.5 for critical slopes such as Use FS = 1.3 to 1.5 for critical slopes such as end slopes under abutments, slopes end slopes under abutments, slopes containing footings, major retaining containing footings, major retaining structuresstructures

Design Factor of SafetyDesign Factor of Safety

ggSelection of FS depends on:Selection of FS depends on:

-- Method of stability analysisMethod of stability analysis

-- Method used to determine shear strengthMethod used to determine shear strength

-- Degree of confidence in reliability of subsurface Degree of confidence in reliability of subsurface datadata

-- Consequences of failureConsequences of failure

-- Criticality of the applicationCriticality of the application

Infinite Slope AnalysisInfinite Slope Analysis

ggSlope that extends for a relatively long Slope that extends for a relatively long distance and has consistent subsurface distance and has consistent subsurface profile can be considered as infinite slopeprofile can be considered as infinite slope

ggFailure plane parallel to slope surfaceFailure plane parallel to slope surface

Embankment Fill

Firm Soil

Infinite Slope Analysis in Dry SandsInfinite Slope Analysis in Dry Sands

T

W

b

h

SlopeSurface

FailureSurface

S

N

β

W

S

N β

Force Polygon

N

W = γ b h

N = W cos β

T = W sin β

S = N tan φ

βφ

βφ

= tan

tan = sinW

N tanTS = FS

ggFS is independent of slope height hFS is independent of slope height hggFS is a function of only FS is a function of only φφ and and ββ

cc--φφ soils with Watersoils with Water

h cos2β

Seepage Flow

Failure Surface

Slope Surface

h

β

h W

N'+U Pore Water Force U = γwbh cosβ

b

T

ββγ

φβγγ cos sin h

'tan ) ( cos ) - (h + c' = FS

sat

2 wsat

βφ

γγ

tan'tan ' = FS

sat

For c' = 0

Circular Arc FailureCircular Arc Failure

Resistance Force

Circular Arc FailureCircular Arc Failure

WLForceWeightSLStrengthShearTotalFS

××

=MomentgOverturnin

MomentResistingFS =

Simple RuleSimple Rule--ofof--Thumb for FSThumb for FS

ggOnly for preliminary “Only for preliminary “guestimateguestimate” for FS” for FS

c c = cohesion of foundation soil= cohesion of foundation soilγγFillFill = unit weight of fill= unit weight of fillHHFillFill = Height of fill= Height of fill

ggNo waterNo water

FillHFill

c6FS×γ

≅

What is the FS for following case?What is the FS for following case?

1.69ft)pcf)(30(130

psf)(6)(1100FS ==

Circular Arc Stability Analysis(ORDINARY METHOD OF SLICES)

O

Firm

Soft

FirmSlipSurface

Fill

RCircle Radius R

Step by Step ProcedureStep by Step Procedure

1.1. Draw crossDraw cross--section to natural scalesection to natural scale2.2. Select failure surfaceSelect failure surface3.3. Divide the failure mass into 10Divide the failure mass into 10--15 slices 15 slices

using suggestions on Page 6using suggestions on Page 6--1414

Forces on a SliceForces on a SliceWithout WaterWithout Water With WaterWith Water

Failure Mass Divided into SlicesFailure Mass Divided into Slices

16

O

R

R

12345678

9

11

12131415

2:1

10

α=+60°

+54°

+51°+43°

+34°

+25°+16°

+9°

+1°

−7°

−15°−2

4°−32°

−42°

−49°−53°

Note that slices 1 through 9 have positive α angles and contribute to the driving force. Slices 10 through 16 have negative α angles and reduce the net driving force.

Step by Step ProcedureStep by Step Procedure4.4. Compute total weight ( WCompute total weight ( WTT ) of each slice) of each slice5.5. Compute frictional resisting force for each slice Compute frictional resisting force for each slice

N TanN Tanφφ -- ulul6.6. Compute cohesive resisting force for each sliceCompute cohesive resisting force for each slice

ClCl7.7. Compute tangential driving force (T) for each Compute tangential driving force (T) for each

sliceslice8.8. Sum resisting and driving forces for ALL slices Sum resisting and driving forces for ALL slices

and compute FSand compute FS

T1ctanN

ForcesDrivingForcesResistingFS

∑∑+φ∑

=∑

∑=

Example of One Slice w/o WaterExample of One Slice w/o Water

Assume: Assume: gg γγ totaltotal = 120 = 120 pcfpcf, slice height = 10, slice height = 10’’

slice width = 10slice width = 10’’, , φφ = 25= 25°°, , αα = 20= 20°°, l =11, l =11’’, C = 200 , C = 200 psfpsf. .

ggFind: Resisting and Driving Forces Find: Resisting and Driving Forces

Compute Slice Weight and Normal Compute Slice Weight and Normal ForceForce

W W TT = = γγ totaltotal x slice area (x 1x slice area (x 1’’ thick)thick)= 120 = 120 pcfpcf x 10x 10’’ x 10x 10’’= 12,000 lbs= 12,000 lbs

N N = W= WTT coscos αα -- ulul= 12,000 lbs x = 12,000 lbs x coscos 2020°°= 11,276 lbs= 11,276 lbs

Compute Resisting and Driving Compute Resisting and Driving ForcesForcesN Tan N Tan φφ = 11276 x Tan 25= 11276 x Tan 25°°

= 5,258 lbs = 5,258 lbs

ClCl = 200 = 200 psfpsf x 11x 11’’ x 1x 1’’= 2,200 lbs= 2,200 lbs

T = T = WWtt Sin Sin αα= 12,000 lbs x Sin 20° = 12,000 lbs x Sin 20° = 4,104 lbs= 4,104 lbs

Group ExerciseGroup Exercise

ggAssuming the water is 5’ above the slice Assuming the water is 5’ above the slice base, which of the force components change base, which of the force components change in this exercise?in this exercise?

SolutionSolution

ggThe water will affect the normal force, NThe water will affect the normal force, N

N = WN = WTT Cos Cos αα -- ulul= 12,000 lbs x Cos 20= 12,000 lbs x Cos 20°° -- 5 x 62.4 x 115 x 62.4 x 11= 11,276 lbs = 11,276 lbs –– 3,432 lbs3,432 lbs= 7,844 lbs= 7,844 lbs

(N=11,276 lbs for original water level)(N=11,276 lbs for original water level)

Tabular Form for CalculationsTabular Form for Calculations

ggFigure 6Figure 6--1111ggFigure 6Figure 6--1212

Recommended Stability MethodsRecommended Stability Methods

ggLimit equilibrium methodsLimit equilibrium methods-- Summation of moments, vertical and horizontal Summation of moments, vertical and horizontal

forcesforces

ggOrdinary Method of Slices (OMS) ignores Ordinary Method of Slices (OMS) ignores both shear and normal both shear and normal intersliceinterslice forces and forces and considers only moment equilibriumconsiders only moment equilibrium


ggVariations of OMS are Bishop method, Variations of OMS are Bishop method, Simplified Simplified JanbuJanbu method, Spencer method, method, Spencer method, etc.etc.

ggBishop methodBishop method-- Also known as Simplified Bishop methodAlso known as Simplified Bishop method-- Includes Includes intersliceinterslice normal forcesnormal forces-- Neglects Neglects intersliceinterslice shear forcesshear forces-- Satisfies only moment equilibriumSatisfies only moment equilibrium


ggSimplified Simplified JanbuJanbu methodmethod-- Includes Includes intersliceinterslice normal forcesnormal forces-- Neglects Neglects intersliceinterslice shear forcesshear forces-- Satisfies only horizontal force equilibriumSatisfies only horizontal force equilibrium

ggSpencer methodSpencer method-- Includes both normal and shear Includes both normal and shear intersliceinterslice forcesforces-- Considers moment equilibriumConsiders moment equilibrium-- More accurate than other methodsMore accurate than other methods


ggOMS is conservative and gives OMS is conservative and gives unrealistically lower FS than Bishop or other unrealistically lower FS than Bishop or other refined methodsrefined methods

ggFor purely cohesive soils, OMS and Bishop For purely cohesive soils, OMS and Bishop method give identical resultsmethod give identical results

ggFor frictional soils, Bishop method should be For frictional soils, Bishop method should be used as a minimumused as a minimum

ggRecommendation: Use Bishop, Simplified Recommendation: Use Bishop, Simplified JanbuJanbu or Spenceror Spencer

Slope Stability Guidelines for DesignSlope Stability Guidelines for Design

ggTable 6Table 6--11

ggComputer analysis is nowComputer analysis is now--aa--days commonly days commonly performed by use of slope stability softwareperformed by use of slope stability software-- XSTABL, UTEXAS, XSTABL, UTEXAS, ReSSAReSSA, SLOPE/W, etc, SLOPE/W, etc

Remarks on Safety FactorRemarks on Safety Factor

ggMinimum FS = 1.25 using OMSMinimum FS = 1.25 using OMS

ggUse FS = 1.3 to 1.5 for critical slopes such as Use FS = 1.3 to 1.5 for critical slopes such as end slopes under abutments, slopes end slopes under abutments, slopes containing footings, major retaining containing footings, major retaining structuresstructures

ggUse FS = 1.5 for cut slopes in fineUse FS = 1.5 for cut slopes in fine--grained grained soils which can lose strength with timesoils which can lose strength with time

Critical Failure SurfaceCritical Failure Surface

Critical Failure SurfaceCritical Failure Surface

ggCheck multiple failure surfaces and compare Check multiple failure surfaces and compare the lowest safety factorsthe lowest safety factors

ggSearch all areas of slope to find the lowest Search all areas of slope to find the lowest safety factorsafety factor

ggBe careful of “secondary” features such as Be careful of “secondary” features such as thin weak layersthin weak layers

ggEvaluate all loading and unloading Evaluate all loading and unloading conditions, e.g., rapid drawdownconditions, e.g., rapid drawdown

ggUse stability charts to develop a “feel” for Use stability charts to develop a “feel” for the safety factorthe safety factor

Stability ChartsStability Charts

ggAssumptionsAssumptions-- TwoTwo--dimensional limit equilibrium analysisdimensional limit equilibrium analysis-- Simple homogeneous slopesSimple homogeneous slopes-- Circular slip surfaces onlyCircular slip surfaces only

ggUseful for preliminary analysis prior to Useful for preliminary analysis prior to computer analysis to develop a “feel” for computer analysis to develop a “feel” for safety factorsafety factor

Taylor’s Stability Taylor’s Stability ChartsChartsggStability NumberStability Number

gg In terms of In terms of FFcc

ggFS = FS = FFcc = F= Fφφ

See Figure 6-15

Slope Angle, β

HFcN

cs γ

′=

HNcFs

c γ′

=

Taylor’s Stability Taylor’s Stability ChartsChartsggChart for Chart for φφ′′=0=0

conditions and conditions and ββ<54<54ºº

β = 53ºFor β > 53º, use Figure 6-14

Determination of Factor of SafetyDetermination of Factor of Safety

FStan

FSc

FSφ′

σ ′+′

=τ ′

ddd tanc φ′σ ′+′=τ ′

FStantan;

FScc;

FS dddφ′

=φ′′

=′τ ′=τ ′

FSdφ ′

=φ ′

Example 6Example 6--11

gg3030--ft high slope ft high slope ggSlope angle, Slope angle, ββ = 30= 30ººggTotal unit weight, Total unit weight, γγ = 120 = 120 pcfpcfggEffective cohesion, cEffective cohesion, c′′ = 500 = 500 psfpsfggEffective friction angle, Effective friction angle, φφ′′=20=20ºº

ggDetermine the Factor of Safety, FSDetermine the Factor of Safety, FS

Example ComputationExample Computationgg Assume FS = 1.6Assume FS = 1.6gg FS = FS = FFcc = F= Fφφ

gg For For φφ′′dd=12.5=12.5ºº and and ββ = 30= 30ºº, the stability factor, N, the stability factor, Nss, is , is 0.06. Thus,0.06. Thus,

oo

d 5.126.1

20FS

==φ ′

=φ ′

)H()pcf120()6.1(psf50006.0 =

ft30ft4.43)06.0()pcf120()6.1(

psf500H >==

Taylor’s Stability ChartsTaylor’s Stability Charts

Slope Angle, β

0.06

0.075

Example ComputationExample Computation

gg Since 43.4 ft > 30 ft, the actual FS is higher than 1.6.Since 43.4 ft > 30 ft, the actual FS is higher than 1.6.gg Assume FS=1.9Assume FS=1.9

gg FS = FS = FFcc = F= Fφφ= 1.9 => => N= 1.9 => => Nss ≈≈ 0.0750.075

gg Computed H is close to actual height of 30 ft Computed H is close to actual height of 30 ft -- Therefore, FS Therefore, FS ≈≈ 1.91.9

o5.109.1

o20FSd ==φ′

=φ′

ft2.29)075.0()pcf120()9.1(

psf500H ==

Janbu’sJanbu’s Stability ChartsStability Charts

ggAccount for:Account for:-- Surcharge loading at top of slopeSurcharge loading at top of slope-- SubmergenceSubmergence-- Tension cracksTension cracks-- SeepageSeepage

ggSection 6.6.3Section 6.6.3

Sliding Block Failure TypesSliding Block Failure Types

Shallow Weak Soil Layer

Firm Soil

Fill

Fill

Firm Soil

Thin Seam Weak Clay

Firm Soil

Clay

Lens of Silt or SandFill

Impermeable Clay

w/o Frictional Resistance

Clay

1

2

3

After SlideAfter Slide

C SR 42 OregonC SR 42 OregonL

SandstoneSandstone

FillFill 18’18’12’12’

Silty ClaySilty Clay24’24’

SlidingSlidingBlockBlockAnalysisAnalysis

ActiveActiveWedgeWedge

CentralCentralBlockBlock

PassivePassiveWedgeWedge

PP

PP

LL

aa

pp

cLcL

FillFill

SandSand

SoftSoftClayClaySeamSeam SandSand










PP

PP

LL

aa

pp

cLcL

FillFill

SandSand


PP

PP

LL

aa

pp

cLcL

FillFill

SandSand


PPaa = Active = Active DrivingDriving Force = Force = ½½ γγ HH22KKaa

PPpp = Passive = Passive ResistingResisting Force = Force = ½½ γγ HH22KKpp

cLcL = = ResistingResisting Force Due To Clay CohesionForce Due To Clay Cohesion

aPcLpP

ForcesDriving ForcesResistingFS

+==

Example 6Example 6--33ggFind the Safety Factor for the 20Find the Safety Factor for the 20′′ high high

embankment by the simple sliding block embankment by the simple sliding block method using method using RankineRankine pressure coefficients, pressure coefficients, for the slope shown belowfor the slope shown below

11’’

1010′′

Firm Material Firm Material

Soft Clay C = 400psfSoft Clay C = 400psf

2020′′

22

11

γγT T = 110 = 110 pcfpcfφφ = 30= 30°°

γγTT = 110 = 110 pcfpcfφφ = 30= 30°°

11’’

1010′′

Firm Material Firm Material

Soft Clay C = 400psfSoft Clay C = 400psf

2020′′

22

11

22

11

γγT T = 110 = 110 pcfpcfφφ = 30= 30°°

γγTT = 110 = 110 pcfpcfφφ = 30= 30°°

Example 6Example 6--33

ggAdd solutionAdd solution

Student Exercise 2Student Exercise 2gg Using a Using a RankineRankine sliding block analysis, determine sliding block analysis, determine

the safety factor against sliding for the the safety factor against sliding for the embankment and assumed failure surface shownembankment and assumed failure surface shown

4545°° + + φφ/2/2

4545°° -- φφ/2/2

3030°°

1122

OGSOGS

C = 250 C = 250 psfpsfSoft Clay Soft Clay

1616′′

55′′

1010′′

3030′′

Sand Sand γγ ′′ = 60 = 60 pcfpcfφφ = 30= 30°°

Sand Sand γγ = 120 = 120 pcfpcf φφ = 30= 30°°

Sand Fill Sand Fill γγ = 120 = 120 pcfpcfφφ = 30= 30°° OGSOGS

4545°° + + φφ/2/2

4545°° -- φφ/2/2

3030°°

1122

OGSOGS


1616′′

55′′

1010′′

3030′′

Sand Sand γγ ′′ = 60 = 60 pcfpcfφφ = 30= 30°°

Sand Sand γγ = 120 = 120 pcfpcf φφ = 30= 30°°

Sand Fill Sand Fill γγ = 120 = 120 pcfpcfφφ = 30= 30°° OGSOGS

SolutionSolution

←==

←==γ=

→==γ=

=°+°=φ−°=

=°−°=φ−°=

Kips15)ft1)(ft60)(ksf250.0(cL

Kips18)ft1)(0.3(2)ft10)(kcf120.0(21pK2H2

1pP

Kips32)ft1)(33.0(2)ft40)(kcf120.0(21aK2H2

1aP)ft.per (

0.3)23045(2tan)245(2tanKp

33.0)23045(2tan)245(2tanaK

Kips32Kips15Kips18

PacLpP

ForcesDriving ForcesResistingFS +

=+

=Σ

Σ=

F.S. = 1.03 F.S. = 1.03 –– TOO LOW!!TOO LOW!!

Student ExerciseStudent ExerciseggSame as previous exercise except that water Same as previous exercise except that water

table rises of 10 ft to OGStable rises of 10 ft to OGS

SolutionSolution

gg 10 ft rise in water lowers the FS from 1.03 to 0.7710 ft rise in water lowers the FS from 1.03 to 0.77

77.0Kips31

Kips15Kips9

aPcLpP

FS

Kips15)'1)('60)(ksf250.0(cL

PreviousKips18Kips9)3(2)'10)(kcf060.0(21pK2Hb2

1pP

Kips31Kips13Kips18aTotalP

Kips13)'1)('10(2

)ksf4.1ksf2.1(aSandP

)footper(ksf4.1)33.0)('10)(kcf060.0(ksf2.12aP

Kips18)'1)(21)('30)(Ksf2.1(aFillP

)footper(ksf2.1)33.0)('30)(kcf120.0(1aK1H11aP

=+

=+

=

==

←<<==γ=

→=+=

→=+

=

=+=

→==

==γ=

Use of Computer ProgramsUse of Computer Programs

ggSeveral methods for stability analysisSeveral methods for stability analysis-- Consideration of Consideration of intersliceinterslice forces, irregular forces, irregular

failure surfaces, seismic forces, external forces, failure surfaces, seismic forces, external forces, tieback forces, tieback forces, piezometricpiezometric level, heterogeneous level, heterogeneous soil systems, etc.soil systems, etc.

ggUser friendly input and outputUser friendly input and outputggUser documented and verified programUser documented and verified programggXSTABL, UTEXAS, SLOPE/W, XSTABL, UTEXAS, SLOPE/W, ReSSAReSSA

Be Careful with Computer ProgramsBe Careful with Computer Programs

ggPlace emphasis where it belongsPlace emphasis where it belongs-- InvestigationInvestigation-- SamplingSampling-- TestingTesting-- Development of soil profileDevelopment of soil profile-- Design soil strengthsDesign soil strengths-- Water table locationWater table location

ggGarbage in Garbage in –– Garbage outGarbage out


ggAt the end of this session, the participant will At the end of this session, the participant will be able to:be able to:-- Recall modes of slope failureRecall modes of slope failure-- Explain effects of water on slope stabilityExplain effects of water on slope stability-- Discuss slope stability circular and block Discuss slope stability circular and block

analysesanalyses-- Compute safety factor by chart solutionCompute safety factor by chart solution

Any Questions?Any Questions?

THE ROAD TOUNDERSTANDING

SOILSAND

FOUNDATIONS

SLOPE STABILITYSLOPE STABILITY

Lesson 06 Lesson 06 -- Topic 2Topic 2Improving the stability of embankmentsImproving the stability of embankments

Section 6.9Section 6.9


ggAt the end of this session, the participant will At the end of this session, the participant will be able to:be able to:-- Recall methods for stabilizing fill slopesRecall methods for stabilizing fill slopes-- Describe reinforced soil slopesDescribe reinforced soil slopes-- List techniques to improve cut slopesList techniques to improve cut slopes

Mitigating Slope Stability ProblemsMitigating Slope Stability Problems

ggTable 6Table 6--2 2

Solutions to Slope Stability Solutions to Slope Stability ProblemsProblemsggChange alignmentChange alignmentggLower gradeLower gradeggCounterweight Counterweight bermbermggExcavate and replace weak soilExcavate and replace weak soil

Solutions to Slope Stability Solutions to Slope Stability Problems (Cont’d)Problems (Cont’d)ggDisplace weak soilDisplace weak soilggStage construct fillStage construct fillggLightweight fillLightweight fillggGround improvementGround improvementggReinforcement of embankment soilsReinforcement of embankment soils

Reduce Grade Reduce Grade

Firm

Fill

Effect: Reduces Driving Weight

Slip Surface

Firm

Soft

Foundation Overstressed Reduced Load

Counterweight Berm Counterweight Berm

Effect: Provides Resisting Weight

Fill

Soft

Firm

Berm

Additional Resisting Weight

Slip Surface

Excavate and Replace Weak Soil Excavate and Replace Weak Soil

Firm

Effect: Stronger Soil Resists Sliding

Soft

FillFill

Shear KeyGranular Fill

Direction of Work

Water Table Water Table Embankment

Soft, Weak Compressible

Soil

Mud Wave

Firm Bottom

Good Material Replacing

Displaced Poor Material

Rolling Surcharge Desired

Grade

Displacement of Weak SoilsDisplacement of Weak Soils

STAGE 2

STAGE 1

WEAK SOIL

FILL

Fill Stage Construction Fill Stage Construction

Slip Surface

Firm

Soft

Effect: Reduce Driving Weight

Granular Fill

Lightweight Fill Lightweight Fill Lightweight Fill

Examples of Lightweight Fill Examples of Lightweight Fill MaterialsMaterialsggWood FiberWood FiberggShredded TiresShredded TiresggEPSEPS

Cut Slope StabilityCut Slope Stability

ggDeepDeep--Seated Failure (clays)Seated Failure (clays)ggShallow Surface “Sloughs” in Saturated Shallow Surface “Sloughs” in Saturated

Slopes of Clay, Silt and/or Fine SandSlopes of Clay, Silt and/or Fine Sand

Ground ImprovementGround Improvement

gg GroutingGroutinggg Vertical Wick DrainsVertical Wick Drainsgg Stone ColumnsStone Columnsgg Vibro CompactionVibro Compactiongg Dynamic CompactionDynamic Compaction

gg Soil Mixing Soil Mixing gg Soil NailingSoil Nailinggg Reinforced Soil SlopesReinforced Soil Slopesgg MicropilesMicropiles

Ground ImprovementGround ImprovementStone ColumnsStone Columns

Reinforced Soil SlopesReinforced Soil Slopes

Preliminary Design of RSSPreliminary Design of RSS

ggFigure 6Figure 6--2828

Solutions to Cut Slope Stability Solutions to Cut Slope Stability ProblemsProblemsTable 6Table 6--33ggFlatten or Bench SlopeFlatten or Bench SlopeggBench slopeBench slopeggButtress ToeButtress ToeggLower Water TableLower Water TableggReinforcement (e.g., soil nail, biotechnical)Reinforcement (e.g., soil nail, biotechnical)

Cut Slope StabilityCut Slope Stability

Undrained Clay in Cut Gradually WeakensAnd May Fail Long After Construction

Undrained Clay in Cut Gradually WeakensAnd May Fail Long After Construction

Before CutBefore Cut After CutAfter Cut FailureFailure

SeepageSeepage

ToeToeToeToe

Slip Surface

Slip SurfaceSwellingSwellingClay SoilClay Soil

Water Table

Water Table

Water Table

Water Table

Minimum Recommended Safety Factor = 1.50Minimum Recommended Safety Factor = 1.50

Cut slopes may deteriorate with time as a Cut slopes may deteriorate with time as a result of natural drainage conditions that result of natural drainage conditions that embankments do not experienceembankments do not experience

Factor of Safety for Cut SlopesFactor of Safety for Cut Slopes


ggAt the end of this session, the participant will At the end of this session, the participant will be able to:be able to:-- Recall methods for stabilizing fill slopesRecall methods for stabilizing fill slopes-- Describe reinforced soil slopesDescribe reinforced soil slopes-- List techniques to improve cut slopesList techniques to improve cut slopes



SOILSAND

FOUNDATIONS

Interstate 0 Interstate 0 –– Apple FreewayApple FreewayNote: Scale shown in Station FormNote: Scale shown in Station Form

Baseline Stationing

Baseline Stationing

S.B. Apple Frwy

N.B. Apple Frwy

Proposed Toe of SlopeProposed Toe of Slope

Existing Ground SurfaceExisting Ground Surface

12

Proposed Final GradeProposed Final GradeProposed AbutmentProposed Abutment

Interstate 0Interstate 0

9090 9191 9292 9393

Apple Freeway Apple Freeway ExerciseExerciseggAppendix AAppendix A

-- Section A.5Section A.5

Subsurface Explorations

Terrain reconnaissance Site inspection Subsurface borings

Basic Soil Properties Visual description

Classification tests Soil Profile

Laboratory Testing Po diagram

Test request Consolidation results Strength results

Slope

Stability

Design soil profile Circular arc analysis Sliding block analysis Lateral squeeze analysis

Approach Roadway Settlement

Design soil profile Magnitude of settlement Rate of settlement Surcharge Vertical drains

Spread Footing Design

Design soil profile Pier bearing capacity Pier settlement Abutment settlement Surcharge Vertical drains

Driven Pile Design Design soil profile

Static analysis – pier Pipe pile H – pile Static analysis – abutment Pipe pile H – pile Driving resistance Lateral movement - abutment

Construction Monitoring

Wave equation Hammer approval Embankment instrumentation

Design Soil Profile (East Approach Design Soil Profile (East Approach Embankment)Embankment)

Fill

7' Sand

35' Clay

DenseGravel

10'

25'

2:1

= 110 pcf = 36°

= 130 pcf= 40°

= 125 pcf= 0

C = 0

C = 1100 psf

C = 0

30'30'

5'5'

= 130 = 130 pcfpcf= 43= 43°°

C = 0C = 0

33′′ OrganicOrganic

γγ = 90 = 90 pcfpcf

w = 120%w = 120% Fill

7' Sand

35' Clay

DenseGravel

10'

25'

2:1

= 110 pcf = 36°

= 130 pcf= 40°

= 125 pcf= 0

C = 0

C = 1100 psf

C = 0

30'30'

5'5'

= 130 = 130 pcfpcf= 43= 43°°

C = 0C = 0



w = 120%w = 120% 30'30'

5'5'

= 130 = 130 pcfpcf= 43= 43°°

C = 0C = 0



w = 120%w = 120%

Compute FS Against Circular Arc Compute FS Against Circular Arc Failure Failure –– Rule of Thumb AnalysisRule of Thumb Analysis

FillFill HC6.)S.F(SafetyofFactor

×=

γ

Soft Clay Soft Clay

γγFillFill = 130 = 130 pcfpcf

Bedrock Bedrock

3030’’

C = 1100 C = 1100 psfpsf

69.1)30)(130()1100)(6(.S.F ==

FillFill HC6.)S.F(SafetyofFactor

×=

γ

Soft Clay Soft Clay


Bedrock Bedrock

3030’’



Bedrock Bedrock

3030’’

C = 1100 C = 1100 psfpsf

69.1)30)(130()1100)(6(.S.F ==

Compute FS Against Circular Arc Compute FS Against Circular Arc Failure Failure –– Normal Method Normal Method (Hand Solution)(Hand Solution)

gg For deep clay subsoil, the “critical” (min FS) failure surface For deep clay subsoil, the “critical” (min FS) failure surface will generally pass deep into the weakest clay layer. The will generally pass deep into the weakest clay layer. The center of the circle usually lies above the fill slopecenter of the circle usually lies above the fill slope

FillFill33'33'

7'7' SandSand

35'35' ClayClay

DenseDenseGravelGravel

10'10'

25'25'

2:12:1RR

RROO

FillFill33'33'

7'7' SandSand

35'35' ClayClay

DenseDenseGravelGravel

10'10'

25'25'

2:12:1RR

RROO


gg Note that slices 1 through 9 have positive Note that slices 1 through 9 have positive αα angles and angles and contribute to the driving forces. Slices 10 through 16 have contribute to the driving forces. Slices 10 through 16 have negative negative αα angles and reduce the net driving forcesangles and reduce the net driving forces

O

R

R

33'

7'

35'25'

10'

Fill

Sand

Clay

Dense Gravel

1234567

89

1112131415

16

2:1

10

α=+60°+54°+51°+43°+34°

+25°+16°

+9°

+1°

−7°

−15°

−24°−3

2°−42°

−49°−53°

α = 0°

u ul

u ul


Fill33'

7' Sand

35' Clay

DenseGravel

10'

25'

2:1R

ROF.S. = 1.36Normal

Fill33'

7' Sand

35' Clay

DenseGravel

10'

25'

2:1R

ROF.S. = 1.36Normal

Comparison of Factors of SafetyComparison of Factors of Safety

gg FS = 1.36 Normal method (hand solution)FS = 1.36 Normal method (hand solution)gg FS = 1.63 Bishop method (computer program)FS = 1.63 Bishop method (computer program)gg Remember that Normal method is very conservative when Remember that Normal method is very conservative when

the soil profile has friction soil and the Bishop method is the soil profile has friction soil and the Bishop method is more theoretically correct.more theoretically correct.

Fill33'

7' Sand

35' Clay

DenseGravel

10'

25'

2:1R

RO

Critical circle Fill33'

7' Sand

35' Clay

DenseGravel

10'

25'

2:1R

RO

Critical circleCritical circle

Sliding Block AnalysisSliding Block AnalysisEast Approach EmbankmentEast Approach Embankmentgg Estimate FS (assume failure surface as shown)Estimate FS (assume failure surface as shown)

Fill33'

7' Sand

35’Clay

DenseGravel

10'

25'

2:1

= 110 pcf = 36

= 130 pcf= 40o

= 125 pcf= 0

C = 0

C = 1100 psf

C=0

P

CLL = 60'

P

Assumed FailureSurface

ActiveWedge

PassiveWedge

CentralBlock

PPAA Fill33'

7' Sand

35’Clay

DenseGravel

10'

25'

2:1

= 110 pcf = 36

= 130 pcf= 40o

= 125 pcf= 0

C = 0

C = 1100 psf

C=0

P

CLL = 60'

P

Assumed FailureSurface

ActiveWedge

PassiveWedge

CentralBlock

PPAAPPAA

Sliding Block AnalysisSliding Block AnalysisEast Approach EmbankmentEast Approach Embankment

gg Circular Arc Failure More CriticalCircular Arc Failure More Critical

Fill33'

7' Sand

35' Clay

DenseGravel

10'

25'

2:1

= 125 pcfC = 1100 psf

PA = 24 K

CL = 66 K

ActiveWedge

PassiveWedge

CentralBlock

24 K

PP = 18 K

Compute F.S. :

F.S. =Horiz. Resisting Forces

Horiz. Driving Forces

18 K + 66 K =84 K24 K = 3.5 (O.K.)

Fill33'

7' Sand

35' Clay

DenseGravel

10'

25'

2:1

= 125 pcfC = 1100 psf

PA = 24 K

CL = 66 K

ActiveWedge

PassiveWedge

CentralBlock

24 K

PP = 18 K

Compute F.S. :

F.S. =Horiz. Resisting Forces

Horiz. Driving Forces

18 K + 66 K =84 K24 K = 3.5 (O.K.)

Summary of Embankment Slope Summary of Embankment Slope StabilityStabilityggDesign soil profileDesign soil profile

-- Soil layer unit weights and strengths estimatedSoil layer unit weights and strengths estimatedggCircular arc analysisCircular arc analysis

-- Approach embankment slope stability safety Approach embankment slope stability safety factor of 1.63 against circular failurefactor of 1.63 against circular failure

ggSliding block analysisSliding block analysis-- Approach embankment slope stability safety Approach embankment slope stability safety

factor of 3.5 against sliding failurefactor of 3.5 against sliding failure



SOILSAND

FOUNDATIONS

lesson 06-chapter 6 slope stability

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