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Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

19


Module 5:

Lecture -1 on Stability of Slopes


Stability analysis of a slope and finding critical slipsurface;

Sudden Draw down condition, effective stress andtotal stress analysis;

Seismic displacements in marginally stable slopes;

Reliability based design of slopes,

Methods for enhancing stability of unstable slopes.

Contents


Contents of this lecture

Types of slopes

Failure types

Causes of slope failures

Analysis of slopes by using LE methods

Comparison

Concluding remarks


Application of shear strength theory

Earth pressure theories

Stability analysis of slopes

Infinite slopes

Slope that extends over along distance and theconditions remain identicalalong some surface orsurfaces for quite somedistance.

Finite slopes

Slope that connect land at oneelevation to land that is not faraway but is at differentelevation.

Can also exist in nature and man-made.


Slopes

Natural♦Hill side and valleys

♦Coastal and river cliffs

Man-made♦ Cuttings and embankments for

highways and rail roads

♦ Earth and ash pond dams

♦ Temporary excavations

♦ Waste heaps (landfill slopes)

♦ Landscaping for site development

Type of slopes


R

R

Types of slope failure

Circular

Non-circular


Translational slip




Compound slip

Rigid stratum


F

N

W

µsN

N

WF1

µkN

Block movement

As long as µsN > F --- block is said to be stationary

Resisting force FR : µsN

Disturbing force FD : F F1

D

R

FFFS =


Causes of slope failure

Gravity

Seepage

Earthquake

Erosion

Geological features

Construction activities


Typical slope failures


Courtesy: Geological natural hazardsSeptember 15, 2004



Landslide damage adjacent to a residential structure

Courtesy: North Carolina Geological Survey


Highway slope failure at Krishnabhir, Tribhuwan highway Nepal (Aryal, 2003)


(After Loher et al. 2002)

Typical sacrificial slope failure in highway embankment


Uttarakand (2013)


Landslide in Chongqing and Hong Kong

After Kwong et al. 2004


Effect of raising GWT


Aerial view of Waste slide on March 16, 1996 [USA]

Lateral displacements upto 275 m and vertical displacements upto 61 m

1.2 million m3 of waste

After Eid et al. (2000)


Aerial view of landfill on Feb. 6, 1996

Stark et al. (2000)


Slope FailureSlope failures depend on

• Soil type• Soil stratification• Ground water• Seepage• Slope geometry


Types of Slope Failure

Translational Slide• Failure of a slope along

a weak zone of a soil

• Sliding mass travelslong distances beforecoming to rest.

• Common in coarse-grained soils.

Thin layer of weak soil


Rotational slide• Common in homogenous fine-grained soil

• It has its point of rotation on an imaginary axis parallel to the slope

• There are three types of rotational failure:– Base slide– Toe slide– Slope slide

Types of Slope Failure


Base slide

• Occurs by an arc engulfing the whole slope.

• A soft soil layer resting on a stiff layer of soil is prone to base slide

Rotational slide


Toe slide

• The failure surfacepasses through thetoe of the slope.

Rotational slide


Slope slide

• The failure surface passes through the slope

Rotational slide


Flow slide• Occurs when internal and

external conditions force asoil to behave as a viscousfluid and flow down,spreading in all directions.

• Multiple failure surfaces occurand change continuously asflow proceeds.

• Occurs in dry and wet soils.


Block and wedge slide• Occurs when a soil mass is shattered along joints, seams,

fissures and weak zones by forces emanating from adjacentsoils.

• The shattered mass moves as blocks and wedges down theslopes.


FallsSimple detachment of rock mass from its parent body The process is only gravity governed.

Rock falls


Causes of Slope Failure

• Erosion

Water and windcontinuously erodeslopes.

Erosion changes thegeometry of the slopes,resulting in a slopefailure or a landslide. Steepening of slope by erosion


• Erosion:

Rivers and streamscontinuously scour theirbanks undermining theirnatural or man-madeslopes.

Scour by rivers and streams



• Rainfall:

Long periods of rainfallsaturate, soften, anderode soils.

Water enters intoexisting cracks and mayweaken underlying soillayers, leading to failure,(for example, mud slides)



• Earthquakes: Earthquakes induce dynamic forces especiallydynamic shear forces that reduce the shear strengthand stiffness of the soil.

Causes Of Slope Failure


• Earthquakes:Pore water pressures in saturated coarse-grainedsoils could rise to a value equal to the total meanstress and cause these soils to behave like viscousfluids. This phenomenon is known as dynamicliquefaction. Structures founded on these soils wouldcollapse.

The quickness in which the dynamic forces areinduced prevents even coarse–grained soils fromdraining the excess pore water pressures. Thus, failurein a seismic event often occurs under undrainedconditions.


• Geological features:

Many failures commonly result from unidentifiedgeological features.

Soil stratification



•External Loading:

Loads placed on thecrest of a slope add togravitational load andmay cause slopefailure.

Overloading at the crest of the slope



• Construction activities:Construction activities nearthe toe of an existing slopecan cause failure becauselateral resistance is removed.

Slope failures due toconstruction activities isdivided into two cases:

• Excavated slopes.• Fill slopes.

Excavation at toe of the slope


• Excavated slopes:

When excavation occurs, the total stresses arereduced and negative pore pressures aregenerated. With time the negative pore pressuresdissipate, causing a decrease in effective stressesand consequently lowering the shear strength of thesoil. If slope failures occur, they take place afterconstruction is completed.


Fill slopes:Fill slopes are common in embankmentconstruction. If the foundation soil is saturated, thenpositive pore water pressures are generated from theweight of the fill and the compaction process.

The effective stress decrease and consequentlyshear strength decreases. Slope failures in slope arelikely to occur during or immediately afterconstruction.


Factors that contribute to high shear stress:Factors contributing to instability of soil slopes

i) Removal of lateral supporta) Erosion – bank cutting by streams and riversb) Human agencies – cuts, canals, pits, etc.,

ii) Surchargea) Natural agencies – Weight of snow, ice and rain waterb) Human agencies – Fills, buildings, etc.,

iii) Transitory earth stresses – Earthquakesiv) Removal of underlying support

a) Sub aerial weathering – solutioning by ground waterb) Subterranean erosion – pipingc) Human agencies – mining

v) Lateral pressures – water in vertical cracks; freezing water in cracks; root wedging After Gray and Leiser (1982)


Factors contributing to instability of soil slopesFactors that contribute to low shear strengthi) Initial state

a) Composition – inherently weak materialsb) Texture – loose soils, metastable grain structuresc) Gross structure – faults, joining, bedding, planes, varying, etc.,

ii) Changes due to weathering and other physicochemical reactions

- Frost action and thermal expansion, Hydration of clay minerals,Drying and cracking, Leaching

iii) Changes in inter-granular forces due to pore water

- Seepage pressure of percolating ground water, loss in capillarytension upon saturation, buoyancy in saturated state.

ii) Changes in structure – Fissuring of pre-consolidated clays due torelease of lateral restraint; Grain structure collapse upon disturbance.

After Gray and Leiser (1982)


Slope Stability AnalysisGeneral Assumptions: The failure can be represented as a two dimensional

problem.

The sliding mass moves as a rigid body and thedeformations of the sliding mass has no significanteffects on the analysis.

The properties of soil mass are isotropic and shearresistance along failure surface remains sameindependent of the orientation of the failure surface.

The analysis is based on limit equilibrium method.


Infinite Slope Stability Analysis


Examples of slopes which can be infinite slopes

Ore or sand stock piling by dropping from a chute

Embankment formed by end dumping from a truck

Natural slopes formed in granular materials where thecritical failure mechanism is shallow sliding or surfaceravelling

Natural slopes formed in cohesive soils with great extent orweak cohesive material on ledge

Slopes in residual soils where a relatively thin layer ofweathered soil overlies a firmed soil or rock


Analysis of infinite slopesAssumptions:

Soil is homogenous.

The stress and soilproperties on every verticalplane are identical; On anyplane parallel to the slopestresses and soil properties areidentical.

⇔ Failure in such slope takes place due to sliding of the soilmass along a plane parallel to the slope at a certain depth.

Failure surface

b β

zhw


Analysis of infinite slopes

Weight of segment ABCD W = γzb(1)z

Db

W

B

C

A

N

T

Tangential stress τ down the slope

ββγββγτ cossin

cos/sin z

bzb

==

Normal stress σ within the segment

βγββγσ 2cos

cos/cos z

bzb

==

Pore water pressure u on the slip surface

( ) βγ 2coswwhzu −=


Analysis of infinite slopesNormal effective stress σ ′ =

Shearing strength τf at the base of segment

φστ ′′+′= tancf

For the general case:

ττ fFS =

( )( ) βγγγ

βγβγ2

22

cos

coscos

www

ww

hzzhzz

+−=

−−=

Factor of safety can bedefined as:

( )ββγ

γγγβφcossin

costan 2

zhzzcFS www +−′+′

=


Analysis of infinite slopesSpecial cases

For the critical case FS = 1 ⇔ β = φ′

zhc w ==′ ;0

⇔ FS of an infinite slope with a cohesion-lesssoil is independent of the depth of failureplane.β

φtantan ′

=FS

Case - A Dry Cohesion-less soil

Mohr failure envelope

β

φ′

σ

τ

(σ, τ)

(σf, τf)

For β < φ′ ⇒ τ < τf⇔ Slope is stable

(independent of depth of slope)

For β > φ′ ⇒ τ > τf Slope wouldhave already failed at all depths.

⇒ Slope is just stable


Analysis of infinite slopesCase – B

βγφγ

tantan ′′

=FS

0;0 ==′ whcSaturated cohesion-less slope

Factor of safety of a saturated cohesion-less slope is about ½ for a slope without saturation.

Case - C For a c - φ soil 0=wh

( )ββγγβφ

cossincostan 2

zzcFS′′+′

=


Analysis of infinite slopesCase - C For a c - φ soil 0=wh

( )ββγγβφ

cossincostan 2

zzcFS′′+′

=

Assuming FS = 1 z = hc

′′

−

′=

φγγβγ

β

tantan

sec2chc

β

φγγβ

γ 2sec

tantan

′′

−=

′

chc

Stability number

⇔ For c-φ soils there is a limiting depth for stability


Analysis of infinite slopes

For β1 < φ′ ⇒ τ < τf

For β2 > φ′ ⇒ τ = τf

The depth at which τ = τf is called the critical depth hc

Mohr failure envelope

β1

φ′

σ

τ

(σ, τ)

(σf, τf)

For this depth slope is just stable

β2

τ = τf

Case - C For a c - φ soil


20


Module 5:



Analysis of finite slopes Possible failure surfaces

Planar failure surface: Occurs along a specific plane or weakness

Excavations into stratified deposits (where strata dippingtoward the excavation); In earth dams along sloping cores ofweak material (not likely to occur in homogenous soils)Circular failure surface:

Soils exhibiting cohesion c or c and φ and no specific planes ofweakness or great strength.

Non circular failure surface

When the distribution of shearing resistance within an earthmass is non-uniform, failure can occur along surfaces morecomplex than a circle.


A finite slope with possible failure surfaces

1VnH


Typical patterns of rotational slides

a) Spoon-shaped b) cylindrical-shaped


Effective or Total stress parameters ?Short-term

• Low Permeability Soil, e.g. Clays:At the end of construction the soil is almost stillundrained. Hence total stress analysis making use ofundrained shear strength Cu is adopted.

• Free Draining Materials, e.g. sands/gravels:Drainage takes place immediately and hence effective stress parameters c′ and φ′ are used.

Long-termAfter a relatively long period of time, the fully drainedstage will have been reached, and hence effectivestress parameters c′ and φ′ are used.


The total stress strength is used for short–termconditions in clayey soils, whereas the effectivestress strength is used in long-term conditions inall kinds of soils, or any conditions where thepore pressure is known (Janbu, 1973)

Effective or Total stress parameters ?


Analysis of finite slopes

The Factor of Safety for finite slope depends on:

Assumed location of centre of rotation for the slip surface

Radius of failure surface

Type of failure (toe failure, base failure and slope failure).


Various definitionsof the factor ofsafety (FOS)


Review of Stability Analysis MethodsAll limit equilibrium methods utilize the Mohr-Coulombexpression to determine the shear strength τf along thesliding surface.

The available shear strength τf depends on the type of soil andthe effective normal stress, whereas the mobilized shear stress τdepends on the external forces acting on the soil mass.


Summary of LE methods


Section of unit width assumed for analysis


θ

dW

cu

R

R

Circular analysis – un-drained condition or φu = 0 analysis

Analysis in terms of totalstresses and applies to theshort-term condition for acutting or embankmentassuming soil profile tocomprise fully saturated clay.

D

R

MMFS =

( )Wd

RLcFS Au=

Demerit: Determination of W and d


Calculation of FS

2211 dWdWM D +=

LRcM uR =

( )2211 dWdWLRcFS u

+=

Section of unit width assumed for analysis


Un-drained analysis stability charts –Taylor’s method

2θ

dW

c

R

R

β DH

H

α

Nomenclature used fordeveloping stability numbers

D = depth factor


Un-drained analysis stability charts – Taylor’s method

WdcLRFS =

W = f (γ, H, geometry of failure surface)

⇔ Geometry of failure surface can be characterized by the three angles α, β, and θ

(1)

Rewriting (1): ( )θβαγ ,,HfcFSc

r ==

cr = required cohesion to just maintain a stable slope andf (α, β, θ) is pure number, designated as the Stability number Ns

HcN r

s γ=Taylor’s Stability number

FS is lowest Factor of safetyobtained from circular arcanalysis.


Taylor’s curves

β > 53°

Slope inclination β

β [°] Ns

60 0.191

65 0.199

70 0.208

75 0.219

80 0.232

85 0.246

90 0.261

For φ = 0 soils



This is because for such steep slopes, the critical failure surfacepasses through the toe of the slope and does not go below thetoe.

γcHc

85.3=Critical height

For β < 53° Ns = f (β, D/H)

For β > 53° Ns = f (β) [all critical slip circles pass through toe]

For gentle slopes, the critical failure surface goes below toe andalways restricted above strong layer (hence depends on itslocation).

For a vertical cut (β = 90°) Ns = 0.26 (short-term condition)

Obtained from Taylorstability number



Position of the critical slip circle ( for FS = 1) may be limited by two factors:

a) The depth of stratum in which sliding can occur

b) The possible distance from the toe of the rupture surface to the toe of the slope.

Prof. B V S Viswanadham, Department of Civil Engineering, IIT BombayDepth factor D/H

Stability numbers for homogeneous simple slopes for φ = 0

After Taylor (1948)

For example:

By knowing D/H and β -- Nsand n can be obtained fromthis chart

n = 0.65

For D/H = 1; β > 53°n = 0


Un-drained analysis stability charts – Taylor’s methodImportant points: It is necessary to ignore possibility of tensioncracks – otherwise geometrically similar failuresurfaces do not occur on slopes having differentheights --- hc = f (c and γ) and is not proportional to H.

Taylor’s stability numbers were determined froman analysis of total stress only.

Taylor’s method is practically restricted toproblems involving un-drained saturated clays orto much common cases where the pore pressureis everywhere zero.


γucz 2

0 =

Pw

Vertical tension crack in cohesive soils

( )lzWd

RLcFS

w

cAu

+

=2

021 γ

l


The ordinary method of slicesIn this method, the potential failure surface is assumedto be a circular arc with centre O and radius r.

The soil mass (ABCD) above a trial surface (AC) isdivided by vertical planes into a series of slices of widthb.

The base of each slice is assumed to be a straight line.

The factor of safety (FS) is defined as the ratio of theavailable shear strength τf to the shear strength τm whichmust be mobilized to maintain a condition of limitingequilibrium.


The Ordinary method (OM) satisfies the momentequilibrium for a circular slip surface, but neglects boththe interslice normal and shear forces. The advantageof this method is its simplicity in solving the FOS, sincethe equation does not require an iteration process.

The ordinary method of slices


rsinα

r

r

h

α

b

A B

CD

l

α

X2

X1

E1

E2

α

FBD of slice i

The method of slices

OLA = length of arc AC



m

fFSττ

=The FS is taken to be the same for eachslice, implying that there must be mutualsupport between slides. i.e. forces must actbetween slices.

1. Total weight of slice W = γbh

2. Total normal force N = σl ( includes N′ = σ′l and U = ul)u = PWP at the centre of the base and l is the length of the base.

3. The shear force on the base, T = τml

4. Total normal forces on sides E1 and E2

5. The shear forces on the sides, X1 and X2


The method of slices Considering moments about O, the sum of themoments of the shear forces T on the failure arc ACmust be equal the moment of the weight of the soilmass ABCD.

∑ ∑= αsinWrTr

( )∑ ∑= ατ

sinWlFS

f

Using ( ) lFSlT f

m

ττ ==

∑∑=

ατsinW

lFS f


The method of slicesFor an analysis in terms of effective stress:

∑∑ ′′+′

=αφσ

sin)tan(

Wlc

FS

∑∑ ′′+′

=α

φsin

tanW

NLcFS a

Equation (1) is exact but approximations areintroduced in determining the forces N′.

(1)


The Fellenius (or Swedish ) SolutionIt is assumed that for each slice the resultant of theinterslice forces is zero.

The solution involves resolving the forces on each slicenormal to the base i.e. N′ = Wcosα - ul

∑∑ −′+′

=α

αφsin

)cos(tanW

ulWLcFS a

Rewriting Equation (1):


r

r

α4

A

CD

12

34

567

α1

α3

-α5

The Fellenius (or Swedish ) method of slices

The components of Wcosα andWsinα can be determinedgraphically for each slice.

For an analysis in terms of totalstress the parameters cu and φuare used and the value of u = 0

∑∑+

=α

αφsin

)cos(tanW

WLcFS uau

∑=

αsinWLcFS auFor φu = 0


In this solution it is assumed that the resultant forces onthe sides of the slices are horizontal. i.e X1 – X2 = 0For equilibrium the shear force on the base of any slice is:

( )φ′′+′= tan1 NlcFS

T

Resolving forces in the vertical direction:αφααα sintansincoscos ′

′+

′++′=

FSN

FSlculNW

After some rearrangement and using l = b secα:

∑∑

′+

′−+′=)/tan(tan1

sec]tan)([sin

1FS

ubWbcW

FSφα

αφα

Bishop simplified Method (BSM)


Bishop (1955) also showed how non-zero values of theresultant forces (X1-X2) could be introduced into theanalysis but refinement has only a marginal effect onthe factor of safety.

The pore water pressure can be related to the total fillpressure at any point by means of dimensionless porepressure ratio ru = u/γh .

For any slice, ru = u/W/b

∑∑

′+


sec]tan)1([sin

1FS

rWbcW

FS u φααφ

α

By rewriting:



Bishop simplified Method (BSM)Bishop’s simplified method (BSM) considers theinterslice normal forces but neglects the intersliceshear forces. It further satisfies vertical forceequilibrium to determine the effective base normalforce (N’).


Janbu’s simplified methodJanbu’s simplified method (JSM) is based on acomposite slip surface (i.e. non-circular) and the FOSis determined by horizontal force equilibrium. As inBSM, the method considers interslice normalforces (E) but neglects the shear forces (T).


21


Module 5:



Slope Stability Analysis Methods


The ordinary method of slicesIn this method, the potential failure surface is assumedto be a circular arc with centre O and radius r.

The soil mass (ABCD) above a trial surface (AC) isdivided by vertical planes into a series of slices of widthb.

The base of each slice is assumed to be a straight line.

The factor of safety (FS) is defined as the ratio of theavailable shear strength τf to the shear strength τm whichmust be mobilized to maintain a condition of limitingequilibrium.


The Ordinary method (OM) satisfies the momentequilibrium for a circular slip surface, but neglects boththe interslice normal and shear forces. The advantageof this method is its simplicity in solving the FOS, sincethe equation does not require an iteration process.

The ordinary method of slices


rsinα

r

r

h

α

b

A B

CD

l

α

X2

X1

E1

E2

α

FBD of slice i


OLA = length of arc AC



m

fFSττ

=The FS is taken to be the same for eachslice, implying that there must be mutualsupport between slides. i.e. forces must actbetween slices.

1. Total weight of slice W = γbh

2. Total normal force N = σl ( includes N′ = σ′l and U = ul)u = PWP at the centre of the base and l is the length of the base.

3. The shear force on the base, T = τml

4. Total normal forces on sides E1 and E2

5. The shear forces on the sides, X1 and X2


The method of slices Considering moments about O, the sum of themoments of the shear forces T on the failure arc ACmust be equal the moment of the weight of the soilmass ABCD.

∑ ∑= αsinWrTr

( )∑ ∑= ατ

sinWlFS

f

Using ( ) lFSlT f

m

ττ ==

∑∑=

ατsinW

lFS f


The method of slicesFor an analysis in terms of effective stress:

∑∑ ′′+′

=αφσ

sin)tan(

Wlc

FS

∑∑ ′′+′

=α

φsin

tanW

NLcFS a

Equation (1) is exact but approximations areintroduced in determining the forces N′.

(1)


The Fellenius (or Swedish ) SolutionIt is assumed that for each slice the resultant of theinterslice forces is zero.

The solution involves resolving the forces on each slicenormal to the base i.e. N′ = Wcosα - ul

∑∑ −′+′

=α

αφsin

)cos(tanW

ulWLcFS a

Rewriting Equation (1):


r

r

α4

A

CD

12

34

567

α1

α3

-α5

The Fellenius (or Swedish ) method of slices

The components of Wcosα andWsinα can be determinedgraphically for each slice.

For an analysis in terms of totalstress the parameters cu and φuare used and the value of u = 0

∑∑+

=α

αφsin

)cos(tanW

WLcFS uau

∑=

αsinWLcFS auFor φu = 0


In this solution it is assumed that the resultant forces onthe sides of the slices are horizontal. i.e X1 – X2 = 0For equilibrium the shear force on the base of any slice is:

( )φ′′+′= tan1 NlcFS

T

Resolving forces in the vertical direction:αφααα sintansincoscos ′

′+

′++′=

FSN

FSlculNW

After some rearrangement and using l = b secα:

∑∑

′+


sec]tan)([sin

1FS

ubWbcW

FSφα

αφα



Bishop (1955) also showed how non-zero values of theresultant forces (X1-X2) could be introduced into theanalysis but refinement has only a marginal effect onthe factor of safety.

The pore water pressure can be related to the total fillpressure at any point by means of dimensionless porepressure ratio ru = u/γh .

For any slice, ru = u/W/b

∑∑

′+


sec]tan)1([sin

1FS

rWbcW

FS u φααφ

α

By rewriting:



Bishop simplified Method (BSM)Bishop’s simplified method (BSM) considers theinterslice normal forces but neglects the intersliceshear forces. It further satisfies vertical forceequilibrium to determine the effective base normalforce (N’).


Janbu’s simplified methodJanbu’s simplified method (JSM) is based on acomposite slip surface (i.e. non-circular) and the FOSis determined by horizontal force equilibrium. As inBSM, the method considers interslice normalforces (E) but neglects the shear forces (T).


Morgenstern-Price method (M-PM)The Morgenstern-Price method (M-PM) also satisfiesboth force and moment equilibriums and assumes theinterslice force function.


Spencer’s method

Spencer’s method (SM) is the same as M-PMexcept the assumption made for interslice forces. Aconstant inclination is assumed for interslice forcesand the FOS is computed for both equilibriums(Spencer 1967)


A 45 slope is excavated to a depth of 8m in a deeplayer of saturated clay of unit weight 19 kN/m3: therelevant shear strength parameters are cu = 65 kN/m2

and φu = 0. Determine the factor of safety for the trialfailure surface specified in Figure. The cross-sectionalarea ABCD is 70m2.

Example 1

After Craig (2004)


Figure for Example 1

Example 1


Solution for Example 1

This is the factor of safety for the trial failure surfaceselected and is not necessarily minimum factor ofsafety.



The minimum factor of safety can be estimated by using FS = cu/NsγH.

Using Taylor’s chart for Ns vs Slope inclination β, For β= 45° and assuming that D is large, the value of Ns is0.18.


Taylor’s curves

β > 53°

Slope inclination β

β [°] Ns

60 0.191

65 0.199

70 0.208

75 0.219

80 0.232

85 0.246

90 0.261

For φ = 0 soils


Example 2

Using the Fellenius method of slices, determine thefactor of safety, in terms of effective stress, of the slopeshown in Figure for the given failure surface usingpeak strength parameters c′ = 10 kPa and φ′ = 29°. Theunit weight of the soil above and below the watertable is 20 kN/m3.

After Craig (2004)


Table giving computations (After Craig 2004)


Example 3

2 m 2 m 2 m 2 m 2 m 2 m 1 m

Fellenius method of slices

b =

l = bsecα


Slice W(kN)

c′(kPa)

tanφ′ l (m) N=Wcosα

T=Wsinα

c′l(kN)

ul(kN)

N-ul(kN)

(N-ul) tanφ′

1 27.7 8 0.466 2.3 24.5 -13 18.3 4.8 19.7 9.2

2 96.5 8 0.466 2.1 93.9 -22.5 16.5 14.8 79.1 36.9

3 148 8 0.466 2 148 0 16 22.2 125.8 58.6

4 188.7 8 0.466 2.1 183 44.4 16.5 29 154 72.8

5 199.8 8 0.466 2.3 176.1 94 18.3 34.2 141.9 61.1

6 148 15 0.364 2.8 105.5 103.9 42 31.4 74.1 27

7 37 15 0.364 2 16.5 32.6 30.4 11.4 5.1 1.9

Fellenius method of slices ∑∑∑ −′+′

=T

ulNlcFS

)(tanφ

∑T = 239.4

∑c′ l = 158

∑ = 267.5FS = (158+267.5)/239.4

= 1.78



Using the Bishop method of slices, determine thefactor of safety in terms of effective stress for the slopedetailed in Figure for the specified failure surface. Thevalue of ru is 0.20 and the unit weight of the soil is 20kN/m3 and the shear strength parameters are c′ = 0kN/m2 and φ′= 33°

Example 3


Example 3


Comparison of Slope stability analysis methods


Comparison of LE methods

Grid and radius option used to search for circular CSS

Entry and exit option used to search for circular CSS


After Lambe and Whitman, 1969)Schematic diagram slope cross-section

Slope material Properties Value

Unit wt (kN/m3) 19.64

Cohesion (kPa) 4.31

Friction angle (0) 32



Slice 11 - Ordinary Method

36.661

13.699

29.689

Slope stability analysis (Geo-slope 2012) Slice free body diagram

Ordinary method of slices

1.161


Slice 11 - Bishop Method

36.661

14.727

34.614

40.322

32.668


1.289



Slice 11 - Janbu Method

36.661

15.325

34.2

40.322

32.668

Slope stability analysis (Geo-slope 2012)

Slice free body diagram

Janbu’s simplified method

1.2221.222


Slice 19 - Morgenstern-Price Method

22.893

7.9803

18.608

34.425

17.394

31.108

14.466


Morgenstern-Price method (M-PM)


Ordinary method of slices (OSM)

Bishop’s simplified method (BSM)

Geo-slope 2012 1.161 1.289

Lambe and Whitman (1969)

1.17 1.3

Comparison of factor of safety


Aryal (2003)

PLAXIS


22


Module 5:



Sudden drawdown


Determination of most critical slip surfaceCriteria for most critical slip surface = Minimum factor

of safety

Trial and error approach involves following parameters

a)Center of rotation of the slip surfaceb)Radius of slip surfacec)Distance of intercept of slip surface from the toed)Minimum factor of safety achieved


Fellenius (1935) proposed empirical approach for cohesive soils (φu = 0)

Slope ratio α Ψ1 : 0.58 29° 40°1 : 1 28° 37°1 : 1.5 26° 35°1 : 2 25° 35°1 : 3 25° 35°1 : 5 25° 37°

α

Ψ

H

β

O1

Draw line through corners of slope at angle α and Ψas per in table.

O1 will be center of rotation for slip circle.


Jumikis (1962) extended the method for c’- φ’ soil

Possible locations of centers for c’- φ’ soil

P

α

Ψ

H

β

H

4.5 H

O1 Center of rotation of critical circle is assumed to lie on PO1line. Point P is at distance H below the toe in vertical direction and 4.5 H away from toe in horizontal direction



Grid and radius option used to search for circular CSS

Entry and exit option used to search for circular CSS


After Lambe and Whitman, 1969)Schematic diagram slope cross-section

Slope material Properties Value

Unit wt (kN/m3) 19.64

Cohesion (kPa) 4.31

Friction angle (0) 32



Slice 11 - Bishop Method

36.661

14.727

34.614

40.322

32.668


1.289



Slices data (Bishop’s method) for Lambe and Whitman problem

0

5

10

15

20

25

30

35

40

0 5 10 15

Distance from toe of the slope (m)

Nor

mal

stre

ss a

t the

bas

e of

slic

es (k

Pa)

0

5

10

15

20

25

0 5 10 15

Distance from toe of the slope (m)

Shea

r stre

ss m

obili

sed

(kPa

)


Impenetrable strata

Embankment

FOS with FEM = 1.29

Finite element modeling with help of Plaxis 2D

Impenetrable strata

Embankment

Possible failure surfaces

Slope stability analysis Lambe and whitman problem


Method of analysis Factor of safetyLimit Equilibrium

Ordinary method of slices 1.161Bishops method 1.289Janbu’s method 1.222Morgenstern-Price method 1.306

Finite EquilibriumStrength reduction factor 1.29

Comparison of FOS in LEM and FEM


Aryal (2003)

PLAXIS


Development of phreatic surfaces within the slope

u/γh


Comparison of Phreatic surfaces measured and computed from SEEP/W

β = 63.43°


Variation of FS with u/γh


A cutting 9 m deep is to be excavated in a saturatedclay of unit weight 19 kN/m3. The design shear strengthparameters are cu = 30 kN/m2 and φu = 0°. A hardstratum underlies the clay at a depth of 11 m belowground level. Using Taylor’s stability method, determinethe slope angle at which failure would occur. What isthe allowable slope angle if a factor of safety of 1.2 isspecified.

Example 4 for Practice


Example 5 for Practice

For the given failure surface, determine the factor ofsafety in terms of effective stress for the slope detailedin Figure, using the Fellenius method of slices. The unitweight of the soil is 21 kN/m3 and the characteristicshear strength parameters are c′ = 8 kN/m2 and φ′= 32°.


After Craig (2004)


Rapid Drawdown Condition


After the reservoir or dam has been full for sometime, conditions of steady seepage becomeestablished through the dam with the soil below thetop flow line in the fully saturated state. This conditionmust be analysed in terms of effective stress withvalues of pore pressure being determined from theflow net.

Values of ru up to 0.45 are possible in homogeneousdams but much lower values can be achieved indams having internal drainage. The factor of safety forthis condition should be at least 1.5.

Steady state Seepage


After a condition of steady seepage has becomeestablished, a drawdown of the reservoir level willresult in a change in the pore water pressuredistribution.

If the permeability of the soil is low, a drawdownperiod measured in weeks may be ‘rapid’ in relation todissipation time and the change in pore water pressurecan be assumed to take place under undrainedconditions.

Rapid drawdown


Slope stability analysis in drawdown condition

Typical variations in water level during drawdownResponse of slope to rapid drawdown


'o w wu (h h h )= γ + −

Pore water pressure before drawdown at a point P on a potential failure surface is given by

Change in total major principal stress = Total or Partial removal of water above the slope on P.


And the change in pore water pressure is then given by

1 w wh∆σ = −γ

1u B∆ = ∆σ

w wB h= γ

Therefore the pore water pressure at P immediately after rapid drawdown is:

ou u u= + ∆

( ) 'w w(h h 1 B h )= γ + − −


Hence, pore water pressure ratio

usat

urh

=γ

'w w

usat

h hr 1 (1 B)h h

γ= + − − γ

For a decrease in total stresses, the value of B isslightly greater than 1. An upper bound value of rucould be obtained by assuming B = 1 and neglectingh0.

Typical values of ru immediately after drawdown arewithin the range 0.3–0.4. A minimum factor of safety of1.2 may be acceptable after rapid drawdown.


The pore water pressure distribution after drawdownin soils of high permeability decreases as pore waterdrains out of the soil above the drawdown level.

The saturation line moves downwards at a ratedepending on the permeability of the soil.

A series of flow nets can be drawn for differentpositions of the saturation line and values of porewater pressure obtained. The factor of safety can thusbe determined, using an effective stress analysis, forany position of the saturation line.


Typical flow net in case of drawdown (After Craig, 2004)


Pore pressure ratio (ru) can be used for stability analysis as explained by Bishop and Morgenstern (1960)This method is based on “effective stress method”. It involves following five parameters:

i) Slope angle, ii) Depth factor, iii) angle of shearing resistance (φ’), iv) non-dimensional parameter (c’/ γH), and v) pore pressure ratio (ru).

Factor of safety can be computed by using charts provided by Bishop-Morgenstern (1960).


Submerged slope of height 7m and slope of 1 V: 3H

Schematic diagram of lope (After Berilgen, 2007)

DH

Drawdown rate (R) = D/H

Seepage and stability analysis for drawdown condition

R1 = 1 m/day (rapid drawdown)R2 = 0.1 m/day (slow drawdown)


Steady state seepage analysis (constant hydraulic boundaries i.e. total head)

Transient seepage analysis (varying hydraulic boundaries i.e. total head)

Stability analysis (consideration of driving forces for failure

i.e. body forces, pore water pressure, etc.)


Property ValueUnit weight (kN/m3) 20Coefficient of permeability (m/sec) 10-6 and 10-8

Cohesion (kPa) 10Internal friction angle (degree) 20

Four cases were studied considering two drawdown rates and two types of soil.


Flow paths during drawdown phenomena

Drawdown

Pore pressure contours at the steady state condition

Steady state seepage analysis using SEEP/W


40

45

50

55

60

65

70

75

80

0 5 10 15 20 25 30 35

Time (days)

Pore

wat

er p

ress

ure

(kPa

)

Depletion of phreatic surfaces

P1

Drawdown rate R1 = 1 m/day

Variation of pore water pressure at point “P1”

Pore water pressure dissipation with time

Transient seepage analysis using SEEP/W


1

1.25

1.5

1.75

2

2.25

2.5

2.75

0.1 1 10 100

Time (days)

Min

imum

fact

or o

f saf

ety

R = 1 m/day; k = 10-6 m/sec

Critical failure surface at the end of drawdown

Slope stability analysis using SLOPE/W

Factor of safety decreases as drawdown progresses

Critical FOS = 1.497


Effect of drawdown rate

Transient seepage analysis for R = 1 m/day

Transient seepage analysis for R

Transient seepage analysis for R = 0.1 m/day

More amount of depletion of phreatic surface

At the end of drawdown

At the end of drawdown

K = 10-6 m/sec

K = 10-6 m/sec


20

30

40

50

60

70

80

90

0.1 1 10 100 1000

Time (days)

Pore

wat

er p

ress

ure

(kPa

)

R = 1 m/dayR = 0.1 m/day

Variations of pore water pressure with time at the point “P1”

Higher dissipation of pore water pressure in case of slow drawdown


0

0.5

1

1.5

2

2.5

3

3.5

4

0.1 1 10 100 1000Time (days)

Min

imum

fact

or o

f saf

ety

R = 1 m/day; k = 10-6 m/secR = 0.1 m/day; k = 10-6 m/sec

Higher factor of safety due to dissipation of pore water pressure

Variations of factor of safety with seepage time


Effect of coefficient of permeability of soil

Transient seepage analysis for k = 1x 10-6 m/sec

Transient seepage analysis for k = 1x 10-8 m/sec

R = 1m/day

R = 1m/day

Depletion of phreatic surface is marginal for soils with k


20

30

40

50

60

70

80

90

0.1 1 10 100

Time (days)

Pore

wat

er p

ress

ure

(kPa

)

k = 10-6 m/seck = 10-8 m/sec

Dissipation of pore water pressure is less for soils with low k

Variations of pore water pressure with time at the point “P1”


0

0.5

1

1.5

2

2.5

3

3.5

4

0.1 1 10 100Time (days)

Min

imum

fact

or o

f saf

ety

R = 1 m/day; k = 10-6 m/secR = 1 m/day; k = 10-8 m/sec Critical FOS = 1

Higher FOS for soils having high coefficient of permeability

Variations of factor of safety with seepage time


Total stress analysis

Requirement Comment

Total stresses in soil mass Common to both methods

Strength of soil when subjected to changes in total stress similar to stress changes in field

Accuracy is doubtful, since strength depends upon induced pore pressures


Effective stress AnalysisRequirement CommentTotal stresses in soil mass Common to both methodsStrength parameters of soil in relation with effective stress

considerable accuracy, since this is insensitive to test condition

Determination of changes in external loads

Accuracy depends on measurement of pore water pressure


Slopes subject to rainfall


Slope instability is a common problem in manyparts of the world, and cause thousands of deathsand severe infrastructural damage each year.

Rainfall has been identified as a major cause fortriggering landslides and slope failure.

The mechanism leading to slope failure is that thepore water pressure starts increasing when waterinfiltrates the unsaturated soil.

The problem becomes severe if the fill material haslow- permeability, and cannot dissipate the porewater pressure generated due to rainfall.


To investigate the effect of rainfall on slope stability, alimit equilibrium analysis was carried out by usingSLOPE/W, a product of Geostudio (2012) software.

Two slope configurations (45° and 63° inclination)were selected, and were subjected to rainfall ofvarious intensities (2mm/hr-80 mm/hr) for 24 hrs.

Phreatic surfaces were fed into SLOPE/W, and stabilityanalyses were performed at the onset of rainfall,during rainfall, and upto 24 hours after rainfall.


Slope configuration selected (45° inclination)

Applied rainfall intensity

Water table position


Soil parameters used in SLOPE/W(FOS was computed by Bishop’s modified method of slices)


Note: Slope stability reduces with increasing intensities of rainfall

Effect of rainfall intensity on Slope stability

0.8

1

1.2

1.4

1.6

1.8

2

0 10 20 30 40 50 60

Fact

or o

f saf

ety

Time (hours)

2 mm/hr9 mm/hr22 mm/hr36 mm/hr80 mm/hrLimiting factor of safety

Slope inclination: 45°

Rainfall stopped


Note: Steeper slopes have lower initial FOS, and the effect of rainfall onsuch slopes is more devastating as compared to flatter ones.

Effect of rainfall intensity on Slope stability

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

0 10 20 30 40 50 60

Fact

or o

f saf

ety

Time (hours)

2 mm/hr

9 mm/hr

22 mm/hr

36 mm/hr

80 mm/hr

Limiting factor of safety

Slope inclination: 63°

Rainfall stopped

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