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Applied Geotechnics

LANDFILL STABILIYA/Prof Hadi Khabbaz

Email: hadi.khabbaz@uts.edu.auRoom 2.511B

• Introduction

• Important factors in design of landfills

OUTLINE

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• Geotechnical Failure

• Sliding failure

• Slope failure

Text Book

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Qian, Koerner and Gray

(2002)

GEOENVIRONMENTAL ENGINEERING

Site Remediation, Waste Containment, and Emerging Waste

Management Technologies

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By: Sharma and ReddyHari D. Sharma, GeoSyntec Consultants - Walnut

Creek, CaliforniaKrishna R. Reddy, University of Illinois - Chicago, IL

ISBN: 978-0-471-21599-8John Wiley ©2004

Introduction to Environmental

Engineeringth

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4th Edition

M. L. Davis and D.A. Cornwell

McGraw Hill company

Tri-Cities LandfillFremont, San Francisco Bay Area, California

6

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Bottom and lateral side liners system Leachate collection and removal systemGas collection and control system

Landfill Components

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Final cover systemStormwater management systemGroundwater management systemGas monitoring system

Schematic Diagram of a MSW Landfill

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Double Liner and Leachate Collection System

9 10

11 12

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13

Leachate collection sumpNatural rock foundation

Membrane liner

Municipal Solid Waste Landfill

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Hazardous Waste Landfill

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17 18

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To be addressed in the preliminary design report:Landfill geometry and configurationLiner SystemLeachate collection and removal systemFinal cover system

Conceptual Design of a Landfill

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Final cover systemSurface drainage systemMonitoring programs for site and groundwater conditionsGas collection systemFinal use of the landfill property

Taken from Sharma and Reddy 2004

Major steps in designLandfill footprint layoutSub-base gradingC ll l t d filli

Landfill Construction

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Cell layout and fillingTemporary cover selectionFinal cover gradingFinal cover selection

Protect environment from waste and pollutionPrevent groundwater contamination

* Landfill location with respect to water table

Limit leachate

Important Factors in Design

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* Prevent water infiltration:

During construction and operationAfter closure

Prevent structural failure* Failure of liners

* Failure of side slopes

Landfill Envelope Encapsulates the waste and

isolates it from surrounding environment.Main Components are:

Li S t

Landfill Envelope

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Liner SystemLeachate collection and removal systemLimit rainfall collectionGas collection and control system generationFinal cover system

General: the site must be geologically, hydrologically & environmentally suitable

Location with respect to water tableAbove ground landfill

Landfill Location

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Above ground landfill* When water table is close to the ground level

* Little or no excavation

Below and above ground landfill* When water table is deep

* Depth of excavation depends on the depth of water table

and the natural clay layer

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Landfill LocationGeotechnical Considerations

Soil conditions and topographySoil is required for daily cover and final cap coverTopography dictates types of landfill and extent of cover

Faults areasLandfills must not be located close to an active fault

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Landfills must not be located close to an active faultA minimum of 60 m distance to faults is normally required

Seismic impact zoneAreas with >10% probability of horizontal acceleration of 0.1g in

250 yearsMay result failure of side slopes in above ground

landfillsUnstable areas

Poor foundation conditions or susceptible to mass movement

Geotechnical FailureSlope failure

Side slope failure

Ground failure

Waste failure

Sliding failure

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Sliding failure

Lining failure during construction

Cover failure after completion

SettlementAfter construction

Long time settlement

Lining crack

Slope StabilitySliding failure of lining and cover systems

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Slope Stability

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Slope Stability

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Slope Stability

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Slope Stability

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Slope Stability

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Crown

Main scarpToe bulge

Minor scarp

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Toe

p

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τ = c’ + σ’ tan φ’Failure criterion:W

b

T

• No water, effective stress analysis:

d

α

N

T = W (sin α)– Disturbing force:

– Resisting force (at the onset of failure):Tf = ?

Sliding Failure

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Tf = ∫ τf . dAτf = c + σ tan φFailure criterion:

Tf = C + N tan φ’

N = ∫ σ’.dAC = ∫ c’.dA = c’ . b/cos α

Tf

N

F = Tf / T

N

= W (cos α)

– Factor of safety:

b

W

T

N

• Effect of water:

d

α

U b d

τf = c’ + σ’ tan φ’Failure criterion:

Tf = ∫ τf . dAN’ = ∫ σ’.dA = N - U

At the onset of failure:

Sliding Failure

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N

Tfdw

U = γw b dw cos α

Tf = C + N’ tan φ’C = ∫ c’.dA = c’ . b/cos αN = γt b d cos α

TTF f=

T = W sin α = γt b d sin α

ααγφ′αγγ+′

=cos sin d

tan cos)d - d(c

t

2wwt

U

N’

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b

W

T

N

• Special case, cohesionless soils: d

α

⎞⎛

– With water:

ααγφ′αγγ+′

=cos sin d

tan cos)d - d(cF

t

2wwt

Sliding Failure

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N

Tfdwα

φ′⎟⎟⎠

⎞⎜⎜⎝

⎛γ

γ−=tan

tand d 1F ww

– Without water:

αφ′

=tan

tanF

– For dry slopes the angle of repose is equal to φ.U

N’

Sliding FailureSelection of Parameters:

Sliding of clayey soil over sandy soilObtain 2 safety factors using properties of 2 soils separately and use the minimum of the 2 safety factors.

Sliding of soil over geomembrane

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Sliding of soil over geomembraneUse the strength parameters of geomembrane Adhesion between soil and geomembrane: caInterface friction angle between soil and geomembrane: δ Failure criterion τf = ca + σ’ tan δ

During construction, the weight of the construction equipment needs to be included.

Consider a long slope with a uniform soil cover of 300 mm thickness. The soil has a unit weight of 18 kN/m3, friction angle of 35°, and zero cohesion (i.e., it is sand). The cover soil is placed directly on a geomembrane, which has an interface friction angle of 25° and zero adhesion.

Example 1

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• What is the factor of safety against sliding failure at a slope angle of 3(H)-to-1(V)?

• What is the factor of safety against sliding failure if the soil is saturated?

Use δ here.

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Why?

If the soil is not saturated, the slope is stable.

Conclusions for Example 1:

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However, if the soil is saturated, for instance due to leachate, the slope fails and slides down from top of the membrane.

Consider a long slope with a uniform soil cover of 300 mm thickness. The soil has a unit weight of 18 kN/m3, friction angle of 0°, and cohesion of 50 kPa (i.e., it is clay). The cover soil is placed directly on a sand drain which has a friction

Example 2

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directly on a sand drain, which has a friction angle of 30° and zero cohesion.

• What is the factor of safety against sliding failure at a slope angle of 3(H)-to-1(V)?

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73.130tan3tantanF1 ==

αφ′

= o

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L = 20 m

Tension crack

Given: As shown in the following Figure Find: The Factor of Safety against Sliding

Example 3

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L = 20 m

β=10 o

15 m

DrainSandstone

c′=18 kPa φ′=25 γ =18 kN/m3

10 m

Clay

Solution:

L = 20 m

15 γw Pw2

Pw1W

Tβ=10 o

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Uplift

15 γw

10 γw

10 γw

Uplift

Tf

AB

β−+βφ′β−−β+′

=cos)PP(sinW

tan)sin)PP(cosW()AB(cF2w1w

2w1w

TtanNC

TTF f φ′′+′

==

β= cos/20AB

Calculate the factor of safety against sliding in the following infinite slope when the tension crack is full of water.

Quiz Question

Tension crackTension crack

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L = 20 m

β=12 o

14 m

DrainSandstone

C′=18 kPa φ′=30 γ =19 kN/m3

9 m

Clay

L = 20 m

β=12 o

14 m

DrainSandstone

C′=18 kPa φ′=30 γ =19 kN/m3

9 m

Clay

L = 20 m

Uplift 9 γw

14 γw Pw2

Pw1W

Tf

A

β=12 o

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pmrkN320,5191420W =××=

14 γw

9 γw

UpliftA

B

2/14P w2

1w γ×=

2/9P w2

2w γ×=pmrkN5.563PP 2w1w =−

TtanNC

TTF f φ′′+′

== β−+βφ′β−−β+′

=cos)PP(sinW

tan)sin)PP(cosW()AB(c2w1w

2w1w

oo

oooo

12cos5.56312sin532030tan)12sin5.56312cos320,5()12cos/20(18F

+−+=

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99.13.16578.3304F ==

OK2F ≈

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• A common mode of slope failure is a rotational slip along an approximately circular failure surface.

Circular Failure Mechanism

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Shallow failure Deep-seated failure

• Stability analysis:

Circular Slip

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O

x

bPdPsPxWRLF

2w1w

f

−−+τ=

p

q

P

s

momentDisturbingmomentsistingReF =

A failure mechanism with minimum F must be found.

Circular Slip

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WPw1

Pw2R

Lτf

x

• Undrained stability analysis:

O x

Circular Slip

RLF fτ=

52τf

R

L

W

xWF =

Example 4

A 12 m deep excavation in a clay, which has an undrained shearing strength of cu = 40 kPa (φu = 0) and a unit weight of 20 kN/m3.

(a) Calculate the area of the sliding section of the soil.

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(b) Calculate the factor of safety of the slope against sliding.

W

100o

7m

24m

12m

6m

12m

Slip circle

Point of Rotation

20o

100o

24m

12m

6m

12m

Point of Rotation

20o

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W 7m

Slip circle

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55 56

100o

24m

12m

Point of Rotation

20o

A 6m-berm is excavated to increase the factor of safety, of the slope. Calculate the factor of safety of the slope after the berm is excavated.

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W7m

24m6m

12m

Slip circle

Berm

6m

2b m3666A =×=

)xAxA(40205

xW40205F

bbtttnetnet −γ==

m936xb =+=

With a Berm

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118.1)9367289(20

40205F >=×−×

=

But still is less than 1.3.

What do you suggest to increase the factor of safety?

Based on a total stress analysis, in which case, the factor of safety

against the slope failure would be the least? Why?(Assume other design parameters are the same)

a.cu = 35 kPa, γ = 20 kN/m3,

b.cu = 35 kPa, γ = 18 kN/m3,

30 kP 20 kN/ 3

QUIZ

c.cu = 30 kPa, γ = 20 kN/m3,

d.cu = 30 kPa, γ = 18 kN/m3,

e.Information is not enough to compare these cases.

Slip circle

cu, γ

• A generalized method that may be used for:Total or effective stress analyses with φ ≠ 0;Any geometry or loading;Any pore pressure state;

i

Method of Slices

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Any soil or rock layering.i

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• A generalized method that may be used for:– Total or effective stress analyses with φ ≠ 0;

i

Method of Slices

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– Any geometry or loading;– Any pore pressure state;– Any soil or rock layering.

i

• Forces on a slice:

E’i+1

E’i X

Xi

bi

W

Ui

Ui+1

R

x

αi

Statically indeterminate;Solutions based onsimplified assumptions:

Method of Slices

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i Xi+1Li

N’i

Tf

Ubi

i

Li = Arc length of slice base = R sinαx

simplified assumptions:• Ordinary slices or

Swedish method;• Bishop’s method;• Simplified Bishop’s

method;• …

A circular base slide is considered in clayey soil consisting of two layers (the top clay layer and the bottom clay layer).

Assume the values of variables shown on the figure are given.

R

OPoint of rotation

(Centre of circle) X

P1d1

α3

P2d2

α1 Top clay: cu1α2R

OPoint of rotation

(Centre of circle) X

P1d1

α3

P2d2

α1 Top clay: cu1α2

QUIZ

The total weight of the sliding section is W and the distance between the centre of mass to the point of rotation is X.

Accordingly, provide a proper formula for the factor of safety of this slope against sliding in the undrained condition (φu = 0).

Cross Section of a Circular Base SlideNot to Scale

W

dw Bottom clay: cu2

Top clay

Cross Section of a Circular Base SlideNot to Scale

W

dw Bottom clay: cu2

Top clay

)dx(PWx)(RcRc

F11

211u1u2

32

++α+α+α

=

• A number of chart based solutions exist for simple geometries.

• For undrained (total stress) analysis of slopes charts produced by Taylor are often used.

The charts are based on the analysis of circular failure surfaces, and

Graphical Methods

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y ,assume that soil strength is given by a Mohr-Coulomb analysis.Tension cracks are not considered.The charts can be used for drained (effective stress) analysis of dry slopes.

• The stability of homogeneous slopes is a function of: Slope geometry;“Stability number”.

• The factor of safety for undrained frictionless slopes is a function of:– Undrained cohesion, cu;– Inverse of slope height, H;– Inverse of unit weight of soil, γ. cN u=

H cF u

γ∝

Stability Number

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• A dimensionless number is defined as: FH N

γ=

• Slopes of similar stability number and similar geometry have the same safety factor.

c1 , γ1

i1H1

c2 , γ2

i2H2i1= i2

N1=N2 F1=F2

0.15

0.20

0.25

mbe

r, c/

γHF

φ = 0o, D=∞Taylor’s

Chart 1

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0.00

0.05

0.10

Sta

bilit

y nu

0 10 20 30 40 50 60 70 80 90

Slope angle (degree)

H DH

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Case 1: The most dangerous of the circles passing through the toe, represented by full lines in chart. Where full lines do not appear, this case is not appreciably different from case 2.Case 2: Critical circles passing below the toe,

Taylor’s Charts

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represented by long dashed lines in chart. Where long dashed lines do not appear, the critical circle passes through the toe.Case 3: Surface of ledge or a strong stratum at the elevation of the toe (D=1), represented by short dashed lines in chart.

53o

num

ber,

c/γH

F

0.14

0.15

0.16

0.17

0.18

0.19Slope angle

Taylor’s

Chart 2

φ = 0

68

1 2 3 4Depth factor, D

Sta

bilit

y n

0.09

0.10

0.11

0.12

0.13

DHH

DHH

nH

Example 5Find the safety factor of the slope given below:

6m γ = 20 kN/m3

25kP

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60oφu = 10ocu = 25kPa

0.15

0.20

0.25m

ber,

c/γH

F

φ = 0o, D=∞Taylor’s

Chart 1

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0.00

0.05

0.10

Sta

bilit

y nu

0 10 20 30 40 50 60 70 80 90

Slope angle (degree)

OK5.149.114.0620

25HNcF ≈=

××=

γ=

(a) Using Chart 1: N ≈ 0.14

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Type of Failure: Toe Failure

Example 5Find the safety factor of the slope given below:

H=8m γ = 20 kN/m3

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H=8m30o

γ = 20 kN/m3

φu = 0ocu = 36 kPaDH

Check for D=1 and D=2

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0.15

0.20

0.25

mbe

r, c/

γHF

φ = 0o, D=∞Taylor’s

Chart 1

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0.00

0.05

0.10

Sta

bilit

y nu

0 10 20 30 40 50 60 70 80 90

Slope angle (degree)

53o

num

ber,

c/γH

F

0.14

0.15

0.16

0.17

0.18

0.19Slope angle

Taylor’s

Chart 2

φ = 0

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1 2 3 4Depth factor, D

Sta

bilit

y n

0.09

0.10

0.11

0.12

0.13

DHH

DHH

nH

(b) H = 8m, D = 1, Using Chart 1: N = 0.132

φ = 0°, Using Chart 2: N = 0.132 n < 0

75

OK7.1132.0820

36HNcF =

××=

γ=

53o

num

ber,

c/γH

F

0.14

0.15

0.16

0.17

0.18

0.19Slope angle

Taylor’s

Chart 2

φ=0

76

1 2 3 4Depth factor, D

Sta

bilit

y n

0.09

0.10

0.11

0.12

0.13

DHH

DHH

nH

(c) Chart 1: cannot be usedφu = 0°, Using Chart 2: N = 0.172 n ≈ 1.3

31.1172.0820

36F =××

=

m4.1083.1nH =×=

77

Unsafe zone for building construction 30°

10.4 m

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Important Factors

Interface shear strength between various geosynthetics, geomembranes, soil and solid waste layers

Pore pressures acting on liner

Excavated side height and slope

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Groundwater level

Waste filling height and slope

for waste failure the strength properties of waste can be used in slope stability analysis.

Difficult to determine as the friction angle varies between 30° and 60°

• Frictional soils– Below water table, buoyancy reduces shearing resistance;

• Clays– Cohesive strength decreases as moisture content

increases.

Effect of Water

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• Fills on clays.– Soil consolidates as water is squeezed out - factor of

safety increases with time, short term critical.• Cuts in clay

– Soil absorbs water when overburden pressure removed - factor of safety decreases with time, long term critical.

Any Questions?

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Thank you for your attention