CHAPTER
CHAPTER
STABILITY OF
STABILITY OF SLOPES
SLOPES
10.1 INTRODUCTION
10.1 INTRODUCTION
Slopes of earth are of two types Slopes of earth are of two types
1.1. NaturalNatural slopesslopes 2.
2. anan mademade slopesslopes
Natural slopes are those that exist in nature and are formed by natural causes. Such slopes Natural slopes are those that exist in nature and are formed by natural causes. Such slopes exist in hilly
exist in hilly areas.areas.The sides of cuttingThe sides of cutting s, ts, the slopes of embhe slopes of emb ankmankm ents constructed for roads, ents constructed for roads, railwayrailway lines, canals etc. and the slopes of earth dams constructed for storing water are examples of man lines, canals etc. and the slopes of earth dams constructed for storing water are examples of man made
made slopes. The slopes wslopes. The slopes w hether natural or artihether natural or artificial ficial may bmay b 1.1. InfiniteInfinite slopesslopes
2.
2. FiniteFinite slopesslopes The term
The term infiniteinfinite slope is used to designate a constant slope oslope is used to designate a constant slope o infiniteinfinite extent. The long slopeextent. The long slope of
of the face of a mothe face of a mo untain is an example of this type, whereasuntain is an example of this type, whereas finitefinite slopes are limited in extent. Theslopes are limited in extent. The slopes of
slopes of embemb ankmankm ents and earth dams are exaents and earth dams are examples ofmples of finitefinite slopes. Theslopes. The slopeslope length depends onlength depends on the height of the dam or embankment.
the height of the dam or embankment. Slope Stability:
Slope Stability: Slope stability is an extremely important consideration in the design andSlope stability is an extremely important consideration in the design and construction of earth dam
construction of earth dam s. The s. The stabilistability of a natural slope is alsty of a natural slope is also imo im portant. The portant. The results of a sloperesults of a slope failure
failure can often be catastrophic, invcan often be catastrophic, inv olving tolving the loss of consihe loss of considerablderable e property property and mand m any lives.any lives. Causes
Causesofof FailureFailure ofof Slopes:Slopes: hehe important factors that cause instabilityimportant factors that cause instability in in slopeslope anan leadlead toto failure
failure arar
1.1. Gravitational forceGravitational force 2.
2. ForceForce due todue to seepageseepage waterwater 3.
3. Erosion of the surface of slopes due toErosion of the surface of slopes due to flowingflowing waterwater
365 365
366
366 ChapterChapter 1010
4. The
4. The sudden loweringsudden loweringofof water adjacentwater adjacent to to slopeslope
5.
5. ForcesForces due todue toearthquakesearthquakes
The effect of all the forces listed above is to cause movement of soil from high point s The effect of all the forces listed above is to cause movement of soil from high point s to low
to low points.points. most importantmost important such forcessuch forces is theis the c o m p o n e n tc o m p o n e n t gravgrav ity that actity that act in thein the direction
direction probable motion.probable motion. various effectsvarious effects flowingflowing seeping waterseeping water rere generallygenerally recognized
recognized as very impoas very impo rtant in rtant in stabstab ility ility problemproblem s, but ofts, but often these effects have not en these effects have not beenbeen properly identified. It is a
properly identified. It is a factfact that the sthat the seepage oeepage o ccurring ccurring witwit hin a hin a soil mass causessoil mass causes seepageseepage
forces, which have mu
forces, which have mu ch greater effch greater effect ect than is than is comcom momo nly nly realized.realized. Erosion
Erosion on theon the surfacesurface f f slopeslope may be themay be the causecause of theof the removalremoval f f certain weightcertain weight soil,
soil, and mayand may thus leadthus lead to anto an increased stabilityincreased stability as far asas far asmass movementmass movement isis concerned.concerned. On theOn the other hand, erosion
other hand, erosion in thein the formform undercuttingundercuttingat the toe mayat the toe may increaseincreasethth heightheight of theof the slope,slope,
decrease
decrease thth lengthlength of theof the incipientincipient failurefailure surface, thusurface, thu s decreass decreasinging thth stability.stability. Whe
Whe n there is a lowering of the ground water or of a n there is a lowering of the ground water or of a freewater surface adfreewater surface ad jacent to the slope,jacent to the slope, fo
fo exampleexample in in sudden drawdownsudden drawdown of theof the water surfacewater surface in in reservoir therereservoir there is is decreasedecrease in thein the buoyancy
buoyancy of theof the soil whichsoil which is inis in effecteffect anan increaseincrease in thein the weight. This increaseweight. This increase inin weight causesweight causes increase in the shearing stresses that may or may not be in part counteracted by the increase in increase in the shearing stresses that may or may not be in part counteracted by the increase in
Component
Component ofof weightweight
Failure Failure surface surface
(a
(a InfiniteInfiniteslopeslope (b(b earthearthdada
Ground Ground water watertabletable Seepage
Seepage
parallel to parallel toslopeslope
(c
(c SeepageSeepagebelow abelow a naturalnaturalslopeslope
Lowering of water Lowering of water from
from levellevel toto
Earthquake Earthquake force force
(d) Sudden drawdown
(d) Sudden drawdown conditioncondition (e) Failure due to earthquake(e) Failure due to earthquake
Figure
366
366 ChapterChapter 1010
4. The
4. The sudden loweringsudden loweringofof water adjacentwater adjacent to to slopeslope
5.
5. ForcesForces due todue toearthquakesearthquakes
The effect of all the forces listed above is to cause movement of soil from high point s The effect of all the forces listed above is to cause movement of soil from high point s to low
to low points.points. most importantmost important such forcessuch forces is theis the c o m p o n e n tc o m p o n e n t gravgrav ity that actity that act in thein the direction
direction probable motion.probable motion. various effectsvarious effects flowingflowing seeping waterseeping water rere generallygenerally recognized
recognized as very impoas very impo rtant in rtant in stabstab ility ility problemproblem s, but ofts, but often these effects have not en these effects have not beenbeen properly identified. It is a
properly identified. It is a factfact that the sthat the seepage oeepage o ccurring ccurring witwit hin a hin a soil mass causessoil mass causes seepageseepage
forces, which have mu
forces, which have mu ch greater effch greater effect ect than is than is comcom momo nly nly realized.realized. Erosion
Erosion on theon the surfacesurface f f slopeslope may be themay be the causecause of theof the removalremoval f f certain weightcertain weight soil,
soil, and mayand may thus leadthus lead to anto an increased stabilityincreased stability as far asas far asmass movementmass movement isis concerned.concerned. On theOn the other hand, erosion
other hand, erosion in thein the formform undercuttingundercuttingat the toe mayat the toe may increaseincreasethth heightheight of theof the slope,slope,
decrease
decrease thth lengthlength of theof the incipientincipient failurefailure surface, thusurface, thu s decreass decreasinging thth stability.stability. Whe
Whe n there is a lowering of the ground water or of a n there is a lowering of the ground water or of a freewater surface adfreewater surface ad jacent to the slope,jacent to the slope, fo
fo exampleexample in in sudden drawdownsudden drawdown of theof the water surfacewater surface in in reservoir therereservoir there is is decreasedecrease in thein the buoyancy
buoyancy of theof the soil whichsoil which is inis in effecteffect anan increaseincrease in thein the weight. This increaseweight. This increase inin weight causesweight causes increase in the shearing stresses that may or may not be in part counteracted by the increase in increase in the shearing stresses that may or may not be in part counteracted by the increase in
Component
Component ofof weightweight
Failure Failure surface surface
(a
(a InfiniteInfiniteslopeslope (b(b earthearthdada
Ground Ground water watertabletable Seepage
Seepage
parallel to parallel toslopeslope
(c
(c SeepageSeepagebelow abelow a naturalnaturalslopeslope
Lowering of water Lowering of water from
from levellevel toto
Earthquake Earthquake force force
(d) Sudden drawdown
(d) Sudden drawdown conditioncondition (e) Failure due to earthquake(e) Failure due to earthquake
Figure
Stability
Stability of of SlopeSlope s s 367367
shearing strength, depending up
shearing strength, depending up on wheon whe ther or not ther or not thethe soilsoil is ablis able to undergo compression e to undergo compression which thewhich the load increase tends to cause. If a large mass of soil is saturated and is of low permeability, load increase tends to cause. If a large mass of soil is saturated and is of low permeability, practically
practically no volumno volum e changes we changes w ill ill be able to occur except at a sbe able to occur except at a slow rate, and in spite of low rate, and in spite of the increasethe increase of
of load the strength increase may be load the strength increase may be inappreciable.inappreciable.
Shear at constant volume may be accompanied by a decrease in the intergranular pressure Shear at constant volume may be accompanied by a decrease in the intergranular pressure an
an anan increaseincrease in thein the neutral pressure.neutral pressure. failurefailure may bemay be causedcaused byby suchsuch conditioncondition inin whichwhich thth entire soil mass passes into a state of liquefaction and
entire soil mass passes into a state of liquefaction and flowsflows like a liquid. like a liquid. A coA co ndition of this typendition of this type may be
may be developeddeveloped if theif the massmass ofof soilsoil isis subjectsubject toto vibration,vibration, fofo example,example, due todue to earthquake forces.earthquake forces. The various forces that act on slopes are illustrated in Fig.
The various forces that act on slopes are illustrated in Fig. 10.1.10.1.
10.2
10.2
G E N E R A L
G E N E R A L C O N S I D E R A T I O N S
C O N S I D E R A T I O N S
ASSUMPTIONS
ASSUMPTIONS
IN THE
IN THE
A N A L Y S I S
A N A L Y S I S
There are three distinct parts to an analysis of the stability of a slope. They are: There are three distinct parts to an analysis of the stability of a slope. They are:
Testing
Testing samplessamples toto determinedetermine thth cohesioncohesion angleangle internal frictioninternal friction IfIf the analysis is for a naturalthe analysis is for a natural slope,slope, it is esit is essentiasential thl th at the sample be unat the sample be un disturbdisturb ed. ed. In such impoIn such impo rtantrtant respect
respects s as rate of shear application and as rate of shear application and statstate of initial consolidate of initial consolidation, the condion, the cond ition of testing mustition of testing must represent as closel
represent as closely y as possible the most as possible the most unfavorabunfavorab le le conditions ever likely to occur in the actualconditions ever likely to occur in the actual slope.
slope. 2. The s
2. The s tudy of ittudy of items wems w hich are knohich are kno wn twn to enter but which cannot be o enter but which cannot be accoacco untedunted fo
fo in the computationsin the computations The m
The m ost important ost important of such itemof such item s is prs is progresogressive sive cracking wcracking w hich will stahich will start at the top of thert at the top of the slopeslope
where the
where the soilsoil is in tension, and aided by wis in tension, and aided by w ater prater pressure, mayessure, may progressprogress to considerable depth. Into considerable depth. In addition,
addition, there are the effects of the non-homogeneous nature of the typical soil and otherthere are the effects of the non-homogeneous nature of the typical soil and other variations
variations fromfrom the ideal conditions which must be assumed.the ideal conditions which must be assumed.
3.
3. ComputationComputation
IfIf a slope is toa slope is to failfail along a surface, all talong a surface, all the shearing strength mhe shearing strength m ust be overcome ust be overcome along that surfacealong that surface which
which thenthenbecomesbecomes surfacesurface rupture.rupture. suchsuch ABAB Fig. 10.1Fig. 10.1 representsrepresents infinite
infinite numbernumber ofof possible tracespossible traces onon whichwhich failurefailure might occur.might occur.
ItIt is assumed this assumed th at the problem is two at the problem is two dimensional, which theodimensional, which theo retically retically requires a long lengthrequires a long length of
of slope normslope norm al to al to the section. Howthe section. How ever, if ever, if the cross sectithe cross section invon inv estigated holds for a runnestigated holds for a runn ing ing lengthlength of
of roughly two orroughly two or moremore times the trace of the rupture, it is probable that the two dimensionaltimes the trace of the rupture, it is probable that the two dimensional casecase
he
he shearshear strength of soil is assstrength of soil is assumed umed to follto follow Cow C oulomb's oulomb's lawlaw = c' +
= c' + tantan 0"0" where,
where,
c'c' effective uniteffective unit cohesioncohesion effective
effective normal stress on the surface of rupture =normal stress on the surface of rupture = (cr(cr total normal stress
total normal stress on theon thesurfacesurface ofofrupturerupture
pore
porewater pressurewater pressure on theon thesurfacesurface ofofrupturerupture effective angle
effective angle ofof internal friction.internal friction. he
he itemitem ofof great importancegreat importance is theis the lossloss ofof shearing strength which many clays show whenshearing strength which many clays show when subjected to a large shearing strain. The stress-strain curves for such clays show the stress rising subjected to a large shearing strain. The stress-strain curves for such clays show the stress rising with
36
36 ChapterChapter 1010
value
value which may be much less than the maximum. Since a rupture surface tends t o developwhich may be much less than the maximum. Since a rupture surface tends t o develop progressively rather than with all the points at the same state of strain, it is generally the ultim ate progressively rather than with all the points at the same state of strain, it is generally the ultim ate value
value that should be used for tthat should be used for the shearing strength rhe shearing strength rather than tather than the mhe m aximaxim um value.um value.
10.3
10.3 F A C T O RF A C T O R S A F E T YS A F E T Y In
In stability analysis, two types of factors of safety are normally used. They arestability analysis, two types of factors of safety are normally used. They are 1.1. FactorFactor safety wsafety w ith respeith respectct toto shearing strength.shearing strength.
2.
2. Factor of safety with Factor of safety with respect to respect to cohecohe sion. This ision. This is s termed the factor of termed the factor of safety with safety with respect torespect to height.
height. Let, Let,
factor
factor safety wsafety w ith respith respectect toto strengthstrength
F,
F, factorfactor safety with respectsafety with respect toto cohesioncohesion factor of safety with respect to height factor of safety with respect to height F,
F, factorfactor safetysafety withwith respectrespect toto frictionfriction mobilized cohesion
mobilized cohesion mobilized angle
mobilized angle frictionfriction average value
average value mobilized shmobilized sh earing strengtearing strengt s s maximmaxim um um shearing shearing strengtstrength.h.
The factor of safety w
The factor of safety w ith ith respect to shearinrespect to shearin g strength,g strength, may be wmay be w rittritten asen as
j'j' tata < j ) '< j ) '
The shearing strength mobilized at each point on a failure surface may be written as The shearing strength mobilized at each point on a failure surface may be written as
_ _ LJLJ
r = c ; + < 7 ' t a n 0 ;
r = c ; + < 7 ' t a n 0 ;
(10.2)
(10.2)
., ., tanfitanfi where where TmTm ActuallyActually thth shearing resistshearing resistance ance (mobilized value(mobilized value shearing strength)shearing strength)doesdoes developdevelop to to like
like degreedegree at all points on an incipient failure surface. The shearing strains vary considerably andat all points on an incipient failure surface. The shearing strains vary considerably and th
th shearing stressshearing stress may be farmay be far from constant. Howeverfrom constant. However thth above expressionabove expression isis correctcorrect on theon the basisbasis averag
average coe co nditions.nditions.
IfIf the factors of safety with respect to cohesion and friction are different, we may write thethe factors of safety with respect to cohesion and friction are different, we may write the equation
equation of theof the mobilized shearing resistancemobilized shearing resistance
It will
It will shown latershown later thatthat F,F,dependsdepends on theon the heightheight of theof the slope.slope. From thisFrom this it may beit may be
conclu
conclu ded ded that tthat the factor of he factor of safety with respect to safety with respect to cohesion cohesion may be designated may be designated asas thth factorfactor of safetyof safety with respect
Stability
Stability SlopesSlopes the actual height, the criti
the actual height, the critical height being the mcal height being the m aximaxim um um height at which it is height at which it is possiblpossible e for a slfor a slope toope to stable.
stable. We mayWe may write fromwrite from (10.3)(10.3)
(1Q4) (1Q4)
where
where is arbitrarily taken equal to unity.is arbitrarily taken equal to unity.
Example
Example
10.110.1shearing strength parameters
shearing strength parameters f f soilsoil rere c'c' 26.126.1 kN/mkN/m = 15° = 15° 17.8 17.8 kN/mkN/m Calculate
Calculate thth factorfactor safetysafety (a(a with respectwith respect toto strength,strength, (b(b withwith respectrespect toto cohesioncohesion and (c)and (c) with
with respectrespect toto friction.friction. average intergran ular pressaverage intergranular pressureure tftf on theon the failure surfacefailure surface isis 102.5102.5 kN/mkN/m Solution
Solution th
th basisbasis of theof the given data,given data, thth average shearing strengthaverage shearing strength on theon the failure surfacefailure surface isis 26.7
26.7 102.5102.5 tan 15°tan 15° 26.7
26.7 102.5102.5 0.2680.268 54.254.2 and the
and the average valueaverage value mobilized shearing resistancemobilized shearing resistance isis T= T= 17 .8 +1 7.8 + 102.5102.5 tan 12°tan 12° 17.8 17.8 102.5102.5 0.2120.212 39.639.6 kN/mkN/m 39.6 17.80 39.6 17.80 tata 0.2120.212
above example shows
above example shows thth factorfactor ofof safety with respectsafety with respect toto shear strength,shear strength, isis 1.37,1.37, whereas the factors of safety with respect to cohesion and friction are different. Conside r two whereas the factors of safety with respect to cohesion and friction are different. Conside r two extreme
extreme cases:cases: When
When the factor of safety with resthe factor of safety with respect pect to cohesito cohesion on is is unity.unity. 2.
2. When When the fthe factor of safactor of safety with ety with respect to frespect to friction iriction is s unity.unity.
C a s e l C a s e l 102.50x0.268 102.50x0.268 12.90 12.90 Case Case 2.13 2.13 39.60 39.60 ————+102.50+102.50tata
37 Chapter 10 26.70 27.50
12.10
canhave anycombination of and between these two extremes cited above togive
the same mobilized shearing resistance of39.6 kN/m . Some of the combinationsof and are
given below.
Combination of an
1.00 1.26 1.37 1.50 2.20
2.12 1.50 1.37 1.26 1.00
UnderCase 2, thevalueof 2.20 when 1.0.Thefactorof safety 2.20 isdefined as the, actor of safety with respect to cohesion.
Example
10.2
hat will be the factors of safety with respect to average shearing strength, cohesion and internal friction of a soil, for which the shear strength parameters obtained from the laboratory tests are c' 32 kN/m an 18°; th expected parameters of mobilized shearing resistance ar
c' 21 kN/m an = 13° and the average effective pressure on the failure plane is 0 kN /m For the same value of mobilized shearing resistance determine the following:
Factor of safety w ith respect toheight;
2. Factor of safety with respect to friction when that with respect to cohesion is unity;an
3. Factor of safety w ith respect to streng th.
Solution
he available shear strength of the soilis
= 32 + 10 tan 18° = 32 + 35.8 = 67.8 kN/m The mobilized shearing resistance of the soil is
T = 2 1 110 tan 13° = 21 + 25.4 46.4 kN/m
67.8
Factor of safety with respect to average strength, —— 1-46 46.4
32
Factor of safety with respect to cohesion, = 1.52
_ tan 18° _ 0.3249
Factor of safety with respect to friction, F < t > TT TT 2309
Factor of safety with respect to height, (= will be at .0
., 110tanl8° 32
46.4 therefore, 3.0
1.0 46.4-35.8
Stability Slopes 371
32 110tanl8° 35.8
46.4 therefore, 2.49
1.0 46.4-32
Factor of safety with respect to strength is obtained when We may write 32 110 tan 18°
1.46
10.4
STABILITY
ANALYSIS
OF INFINITESLOPES
IN SANDAs an introduction to slope analysis, th problem f slope infinite extent is of interest. Imagine an infinite slope, as shown in Fig. 10.2, making an angle j8 with the horizontal. The soil is cohesionless completely hom ogeneous throu ghout. Then th stresses acting on any vertical plane in the soil are the same as those on any other vertical plane. stress at any point n plane
parallel to the surface at depth will be the same as at every point this plane.
Now consider vertical slice material ABCD having unit dimension normal to the page. forces acting this slice are its weight vertical reaction on the base of the slice, and two lateral forces acting on the sides. Since th slice is in equilibrium, th weight an reaction ar equal in magnitude an opposite in direction. They have common line action which passes through th center of the base lateral forces must equal opposite an their line action must parallel to the sloped surface.
normal an shear stresses plane re a' yzcos fi
where cr' effective normal stress,
effective un it weight of the sand.
If full resistance is mobilized plane th shear strength, s,of the soil Coulomb's la is
ta
when T= s, substituting fo tf have
or tan /3= tan
Chapter 10 (10.5a) Equation (10.5a) indicates that th maximum value of is limited to if the slope is to be stable. This condition holds true fo cohesionless soils whether th slope is completely dry or completely subm erged und er water.
he factor of safety of infinite slopes in sand may be written as
tanfi (10.5b)
10.5 STABILITY ANA LYS IS
INFINITESLOPES
IN C L A Y
Th vertical stress acting on plane (Fig. 10.3) where
yzcosfi
is represented in Fig. 10.3 in the stress diagram. normal stress on this plane is an th shearing stress is he line makes an angle (3with th cr-axis.
Mohr strength envelope is represented line whose equation is = c' +cr'tan^'
According to the envelope, th shearing strength is where th normal stress is When /3 is greater than the lines an meet. In this case the two lines meet at long as the shearing stress on a plane is less than the shearing strength on the plane, there is no danger of failure. Figure 10.3 indicates that at all depths at which th direct stress is less than there is no possibility of failure. However at particular depth at which th direct stress is th
O E
Stability Slopes
shearing strength and shearing stress values are equal as represented by failure is imm inent. This depth at which the shearing stress and shearing strength are equal is called the critical depth. depths greater than this critical value,Fig. 10.3indicates that the shearing stress is greater than the shearing strength but this is not possible. Therefore it may be concluded that the slope may be steeper than 0'as long as the depth of the slope is less than th critical depth.
Expression for the Stability of an Infinite Slope of Clay of Depth
Equation (10.2) gives th developed shearing stress
c' +(T'tan</>' (10.6)
Under conditions of no seepage and no pore pressure, the stress components on a plane at depth and parallel to the surface of the slope are
j' yHcos j3
Substituting these stress expressions in the equation above and sim plifyin g, we have
Y (tan tan
c'
/ ? ( t a n y t f - t a n ^ ) (10.7)
yti
where is the allowable height and the term c'Jy is a dimensionless expression called the stability n um ber and is designated as This dimension less num ber is proportional to the required cohesion and is inversely proportional to the allowable height. solution is for the case when seepage is occurring. If in Eq. (10.7)the factor of safety w ith respect to friction is unity , the stability number with respect to cohesion may be w ritten as
, where
stability number in Eq. (10.8) may be writtenas
where critical heigh t. From Eq. (10.9), we have
(10.10) indicates that the factor of safety w ith respect to cohesion, is the same as the factor of safety w ith respect to height
If there is seepage parallel to the ground surface throughout the entire mass, with the free water surface coinciding with the ground surface, the components of effective stresses on planes parallel to the surface of slopes at depth H are given as[Fig.10.4(a)].
Normal stress
374 Chapter 10
(a
(b
Figure 10.4 Analysis infinite slope
(a
with seepage flowthrough th
entiremass, and (b) with completely submerged slope.
the shea ring stres
sa sin /3
Now substituting Eqs (10.11a) and (10.l i b ) into equation
( l O . l l b )
simplifying, th stability expression obtained is -^2— 0- < / > ' „
sa sa
(10.12) before, if the f a c t o r safety with respect to friction is unity, th stability num ber which represents the coh esion m ay be w ritten as
F Y H c'sa Y 'sat
C/ tf
tan^--^-sa
(10.13) If the slope is completely submerged, and if there is no seepage as in Fig. 10.4(b), then (10.13) becomes
cos /?(tan ft ta < } > ' ) (10.14)
Stability Slopes
Example
10.3Findthe factorof safetyof a slope ofinfinite extent having slope angle 25°. The slope ismade of cohesionless soil with 0 = 30°.
Solution
Factor of safety
tan 30° 0.5774 tan/? tan 25° 0.4663
Example
10.4Analyzetheslope ofExample 10.3 if it ismadeofclay havingc' - 30kN/m 0' 20°, e =0.65and
2.7 and under th following conditions: (i when th soil is dry, (ii) when water seeps parallel
to the surfaceof the slope, and(iii) whenthe slope is submerged.
Solution
Fo e 0.65 an G 2.
27x^1= (2.7 0.65)x9.81
1 +0.65 / s a t 1 0.65
10.09 kN/m
(i For dry soil th stability number is
cos /?(tan/?- < j > ' ) when F,=l d
(cos25° (tan25° - tan20°) 0.084.
c' 30
Therefore, th critical height 22.25
16.05x0.084
(ii) For seepage parallel to thesurface of the slope [Eq. (10.13)]
c' 100Q
= cos 25° tan 5°-^-- ta 20° =0.2315 19.9
=6.51 19.9x0.2315
(iii) For the submerged slope [Eq. (10.14)] cos 25°(tan25° - tan20°) 0.084
Chapter
10.6 METHODS
STABILITY AN ALYSIS
SLOPES
FINITE
HEIGHT
stability slopes infinite extent ha been discussed in previous sections. more common problem is the one in whichth failure occurs curved surfaces. most widely used method analysis of homogeneous, isotropic, finite slopes is the Swedish method based on c ircular failure surfaces. Petterson (1955) first applied the circle method to the analysis of a soil failure in connection with th failure of quarry wall in Goeteberg, Sweden. Swedish National Comm ission, after studying large number failures, published report in 1922 showing that th lines of failure most such slides roughly approached th circumference f circle. failure circle migh t pass above the toe, through the toe or below it. By investigatin g the strength alon g the arc of a large number of such circles, it was possible to locate the circle which gave the lowest resistance to shear. This general method has been quite widely accepted as offering an approximately correct solution for the determination of the factor of safety of slope of an embankment and of its foundation. Developments in the method analysis have been made by Fellenius (1947), Terzaghi (1943), Gilboy (1934), Taylor (1937), Bishop (1955), and others, with th result that satisfactory an alysis of the stability of slopes, embankments an foundations means of the circle method is no longer an unduly tedious procedure.
There are other methods ofhistoric interest such as theCulmann method (1875) and the logarithmic spiral method. The Culmann method assumes that rupture will occur along plane. It is of interest only s classical solution, since actual failure surfaces re invariably curved. This method is approximately correct for steep slopes. The logarithmic spiral method was recommended by Rendulic (1935) with the rupture surface assuming the shape of logarithmic spiral. Though this method makes the problem statically determinate and gives
more accurate results, the greater length of time required for computation overbalances this accuracy.
Thereare several methods of stability analysis based on the circ ular arc surface of failure. fe of the methods are described below
Methods Analysis
The majority of the methods of analysis may be categorized as limit equilibrium methods. The basic assumption of the limit e quilibrium approach is that Coulomb's failure criterion is satisfied along th assumed failure surface. free body is taken from th slope an starting from know assumed values of the forces actin g upon the free body, the shear resistance of the soil necessary for equilibrium is calculated. This calculated shear resistance is then compared to the estimated or available shearstrength of the soil to give an indication of the factor safety.
Methods that consider only the whole free body are the (a) slope failure under undrained conditions, (b friction -circle metho d (Taylor, 1937, 1948) and (c) Taylor's stability number (1948).
Methods that divide th free body into many vertical slices an consider th equilibrium each slice are the Swedish circle method (Fellenius, 1927), Bishop method (1955), Bishop and Morgenstern method (1960) and Spencer method (1967). The majority of these methods are in chart form an cover wide varietyof conditions.
10.7 PLANE SUR FAC
FAILURE
Culmann (1875) assumed a plane surface of failure for the analysis of slopes which is mainly of interest because it serves as a test of the validity of the assumption of plane failure . In some cases
Stability Slopes 377
Forcetriangle Figure 10.5 Stability of slopes by Culmann method
The method asindicated above assumes that the critical surface of failureis plane surface passing through the toe of the dam asshown inFig. 10.5.
The forces th at act on the mass above trialfailure planeAC inclined at angle with the horizontal are shown in thefigure.The expression for the weight, and the total cohesion are respectively,
-yLHcosec /?sin(jtf- 0)
The use of the law of sines in the force triangle, shown in the figure, gives
C sm(6>-f)
cos^'
Substituting herein for and W, and rearranging we have
in which the subscript indicates that the stability number is for the trial plane at inclination 6. The most dangerous plane is obtained by setting the first derivative of the above equation with respect to equal to zero. This operation gives
where is the critical angle fo limiting equilibrium and the stability number fo limiting equilibrium may be written as
yH sin/? cos 0'
where is the critical height of the slope.
378 C h a p t e r10
If we write
tan
^' < > ~ t a n ^
where an are safety factors with respect to cohesion and friction respectively, Eq. (10.15) ay be modified for chosen values of an 0' as
sin/3co (/)' (10.16)
The critical angle for any assumed values of c' an 0' is
From Eq. (10.16), the allowable height of a slope is
Example
10.5Determine by Culmann's method the critical height of an embankment having a slope angle of 40° and the constructed soil having c' 630 psf, 0'= 20° and effective unit weight =1 14 lb/ft Find the allowable height of the embankment if F, .25.
Solution
4c'sin/?cos0' 4 x 630 x sin 40° cos 20°
H, 221 ft y[l-cos(0-4>')] 114(l-cos20°) or (> 1.25, 504lb/ft tan 20° tan 0.291, fa 16.23° ,, 4x504sin 40° cos 16.23° Allowable height, 128.7ft. 114[l-cos(40-16.23°)] 10.8
C I R C U L A R S U R F A C E S
FAILURE
investigations carried out in Sweden at the beginning this century have clearly confirmed that the surfaces of failu re of earth slopes resemble the shape of a circulararc.When soil slips along a circular surface, such a slide may be termed as a rotational slide. It involves downward and outward movement of a slice of earth as shown in Fig.10.6(a) and sliding occurs along the entire surface of contact between the slice and its base. The types of failure that normally occur may be classified as
Stability Slopes
2. Toe failure 3. Base failure
In slope failure , the arc of the rup ture surface meets the slope above the toe. This can happen when the slope an gle /3is quite high and the soilcloseto the toe possesses high stren gth. Toe failure occurs w hen the soil mass of the dam abo ve the base and below the base is homo gene ous. The base failure occurs particularly when the base anglej3 is low and the soil below the base is softer and more plastic than the soil above the base. The various modes of failure are shown in Fig. 10.6.
Rotational slide (a Rotational slide (b Slope failure (c) To failure (d Base failure
380 Chapter 10
10.9 FAILURE UNDER UNDRAINED CONDITIONS
(0
=
fully saturated clay slope fail under undrained conditions (0 = immediately after construction. T he stabil ity analy sis is based on the assum ption that the soil is homogene ou s and the potential failure surface is a circular arc. Two types of failures considered are
1. Slope failu re Base failure
The undrained shear strength of soil is assumed to be constant with depth. A trial failure circular surface with center radius is shown in Fig. 10.7(a) for a toe failure. The slope and the chord AB make angles /3 wi th the horizontal respectively. is the weight per unit
Firm base
(a) Toe failure (b Base failure
Figure 10.7 ritical circle positions for (a) slope failure (a fte r Fellenius, 192 7), (b) base failure 1> 50 40 20 1090 70 Values of (a 60 50 40° 30° 20° V a lu e s 10
Figure 10.8 (a) Relation between slope angle /3 and parameters a and fo
location critical circle
w h e n
/3 is g r e a t e r than 53°; (b relation b e t w e e n slopeangle and depth factor for various values of parameter (after Fellenius, 1 9 2 7 )
Stability of Slopes 381 length of the soil lying above th trial surface acting th rough th center of gravity of the mass. is the lever arm, is the length of the arc, the length of the chord an the mobilized cohesion for any assumed surface of failure.
We may express the factor of safety F^ as
(10.19) equilibrium of the soil mass lying above th assumed failure surface, we may write resisting moment actuating moment
he resisting moment Actuating moment,
Equation for the mobilized is
(10.20)
Now the factor of safety for the assumed trial arc of failure may be determined from q. (10.19). Thisis for one trial arc. The procedure has to be repeated for several trial arcs and the on that gives th least value is the critical circle.
If failure occurs along a toe circle, th center of the critical circle can be located by laying of the angles an 26as show n in Fig. 10.7(a). Values of an fo different slopeangles /3can be obtained from Fig. 10.8(a).
If there is base failure as shown in Fig. 10.7(b), th trial circle w ill be tangential to the firm base and as such th center of the critical circle lies on the vertical line passing throu gh midpoint on slope following equations may be written with reference to Fig. 10.7(b).
D
Depth factor, Distance factor, (10.21)
H
Values of can be estimated fo different values of an j8 by means of the chart Fig. 10.8(b).
Example 10.6
Calculate the factor of safety against shear failure along the slip circle shown in Fig. Ex. 10.6 Assume cohesion = 40 kN/m , angle of internal friction zero and the total unit weight of the soil = 20.0 kN/m
Solution
Draw the given slope ABCD as shown in F ig. Ex. 10.6. To locate the cen ter of rotation, extend the bisector line to cut the vertical line drawn from at point With as center an as radius, draw the desired slip circle.
Radius OC = R 36.5 m,Area BECFB = xEFxBC
- x 4 32.5 86.7
Therefore W 86.7 x 1 x 20 = 1734 kN
382 Chapter
36.5m
Figure. x. 10.6
From th figure have, x 15.2 m, an 9=
3.14
Length of arc =R0=36.5 33.8
180
length of a rc x cohesion radius 33.8x40x36.5 1734x15.2
10.10
FRICTION-CIRCLE
METHOD
Physical Concept of the Method
The principle of the method is explained with reference to the section through a dam shown in Fig. 10.9. A trial circle with center of rotation is shown in the figure. With center and radius
Friction circle
Trial circular f a i l u r e surface
Stability of Slopes 383
si 0", where is the radius of the trial circle, circle is drawn. line tangent to the inner circle must intersect th trial circle at an angle tf with Therefore, an vector representing intergranular pressure at obliquity to an element of the rupture ar m u s t tangent to the inner circle. This inner circle is called the friction circle or ^-circle. friction circle method of slope analysis is convenient approach fo both graphical an mathematical solutions. It is given this name because th characteristic assumption of the method refers to the 0-circle.
forces considered in the analysis ar
1. total weigh of the mass above th trial circle acting through th center of mass. center of mass may be determined by any one of the known methods.
2. The resultant boun dary neutral force The vector may be determined by a graphical method from flownet construction.
3. Theresultant in tergran ular force acting on theboundary. 4. The resultant cohesive force
Actuating
Forces
actuating forces may be considered to be the total weight and theresultant boundary force as shown in Fig. 10.10.
boundary neutral force always passes through th center rotation resultantof an designated as is shown in the figure.
Resultant Cohesive Force
Let the length of arc designated as th lengthof chord Let the arc length divided into number of small elements and let the mobilized cohesive force on these elements designated as , etc. as shown in Fig. 10.11. resultant of all these forces is shown the force polygon in the figure. The resultan t is A'B' wh ich is parallel and equal to the chord length
resultant of all the mobilized cohesional forces along the arc is therefore 'L
384 Chapter 10
(a Cohesive forces on trial arc (b) Polygon of forces
F i g u r e 1 0 . 1 1 R e s i s t a n t cohesive forces
maywrite
wherein c'=unit cohesion, factor of safety with respect to cohesion.
he line of action of may be determined by moment consideration. he moment of the total cohesion is expressed as
c' L R a c ' m c
where l moment arm. Therefore,
(10.22) It is seen that th line of action of vector is independent of the magnitude of c'
Resultant
Boundary
Intergranular Forceshe trial arc of the circle is divided into n u m b e r of small elements. et etc. be the intergranular forces acting on these elements as shown in Fig. 10.12. he friction circle is drawn with a radius of sin j/
where
lines action of the intergran ula r force etc. tangential to the friction circle make angle at the boundary. However, th vector sum of any two small forces has a line action through point missing tangency to the -circle y small a m o u n t . resultant of all granular forces must therefore miss tangency to the -circle
a m o u n t w h i c h is not considerable. Let the distance of the resultant of the granular force from th center of the circle designated si (a shown in Fig. 10.12).
Stability Slopes 385 KRsin<p'
Figure 10.12 Resultant of intergranular forces
magnitudeof dependsupon the typeof intergranular pressure distribution along the arc. The
most probable form of distribution is thesinusoidal distribution.
Thevariation of withrespect to thecentral angle a'isshowninFig. 10.13. Thefigure also gives relationships between an for a uniform stress distribution of effective normal stress along the arc of failure.
Thegraphical solution based on theconcepts explained above issimple inprinciple. For the
three forces Q, and ofFig. 10.14 to be inequilibrium, must pass through theintersection of
1.20 1.16 1.12 1.08 1.04 1.00 st Cent or e s s c ral angle anifo istrihrm> u t i o r or
tresssinusdistributi(oida 20 40 60 80
Central angle in degrees
386 Chapter 10
Figure 10.14 Force triangle
for the
friction-circlemethod
the know n lines of action of vectors The line of action of vector ust also be tange nt to the circle radius si . The value may be estimated by the use of curves given in Fig. 10.13, and the line action offeree may be drawn shown in Fig. 10.14. Since th lines action of all three forces and the m agn itude of force are know n, the magnitude of P and may-be obtained by the force parallelogra m c onstru ction that is indicated in the figure. The circle of radius of si rnis called th modified friction circle.
Determination
Factor
Safety
WithRespect to
StrengthFigure 10.15(a) is section f dam. is the trial failure arc. force th resultant is drawn as explained earlier. The line of action of is also draw n. Let the forces and C
(a Friction circle (b Factor safety
Figure 10.15 Graphical
method
determining
factor safety with respectto
Stability Slopes
meet at point D. An a rbitrary first trial using any reasonable value, which w ill be designated by 0'ml is given by the use of circle 1 or radiu in < j ) ' Subscript 1 is used for all other
quantities of the first trial. The f orce is then drawn through tange nt to circl 1. is parallel to chord and point 1 is the intersection of forces The mobilized cohesion is equal c' From this the mobilized cohesion c' is evaluate d. The factors of safe ty with respect to cohesion and friction are determined from the expressions
c' tanfl'
= — , and F*
These factors are the values used to plot point 1 in the graph in Fig. 10.15(b). Similarly other friction circles with radii in in 0'm3 etc. may be drawn and the procedure
repeated. Points , etc. re obtained as shown in Fig.10.15(b). The 45° line, representing F.,intersects the curve to give the factor of safety for this trial circle.
Several trial circles mu st be investigated in order to locate the critical circle, wh ich is the one having the minim um value of
Example
10.7em bank me nt has a slope of 2 (horizon tal) to 1 (vertical) with a height of 10 m. It is made of a soil having a cohesion of 30 kN/m , an angle of internal friction of 5° and a unit weight of 20 kN/m . Consider any slip circle passing through the toe. Use the friction circle method to find the factor of safety with respect to cohesion.
Solution
Refer to Fig. Ex. 10.7. Let FB be the slope and be the slip circle drawn with center nd radius R 20 m.
Length of chord B 27 Take as the midpoint then
AreaAKBFEA = areaAKBJA +area ABEA
-ABxJK -ABxEL
3
x 27 5.3 27 x 2.0 122.4 3
Therefore the weight of the soil mass = 122.4 x 1 x 20 =2448
It will act through point the centroid of the mass which can be taken as the mid point of FK
Now, 0 = 8 5 ° ,
31
Length of arc L RO = 20 x 85 x — 29.7 29.7
Moment arm of cohesion, / = 20 x = 22 m
From center at a distance draw the cohesive force vector C, which is parallel to the chord ow from the point of intersection of an draw a line tangent to the friction circle
388 Chapter 10 1 . 7 4 m
/ / = 1 0 m
Figure Ex. 10.7
drawn at with radius si = 20 sin 5° = .74 m . This line is the line of action of the third force
Draw triangle forces in which th magnitude and the direction fo is known only the directions of the other two forces are known.
Length gives th cohesive force C 520 kN Mobilized cohesion,
= 17.51 kN/m 29.7
Therefore th factor of safety w ith respec to cohesion, is
^=1.713
will .713 if the factor safety with respect to friction, F^ .0
tan5
Stability Slopes
The new radius of the friction circle is = R sin 0' = 20 x sin 3.3° 1.16
Th direction of changes and the modified triangle of force abd' gives, cohesive force C =lengthad 600 kN
C 600 Mobilised cohesino, c' - 20.2 kN/mr LJ Z*yI c' Therefore, = — = 1.5 20.2 10.1 T A Y L O R ' S STABILITY NUMBER
If th slope angle j8, height of embankment th effective unit weight of material y, angle of internal friction < / > ' ,an unit cohesion ar known, th factor of safety may be determined. In order
tomake unnecessary themore orless tedious stability determinations, Taylor (1937) conceived the
idea ofanalyzingthestability of alarge numberofslopes through wide range ofslope angles and
angles of internal friction, an then representing th results by an abstract number which he called
the "stability number".This number isdesignated as ^. Theexpression used is
From this thefactorof safety with respect tocohesion may be expressed as
Taylor published hi results in the form of curves which give th relationship between an
the slope angles /? for various values of 0' as shown in Fig. 10.16. These curves are for circles passing through th toe, although fo values of less than 53°, it has been found that th most dangerous circlepasses below thetoe. However, these curves may beused without serious error for
slopes down to fi 14°. The stability numbers are obtained for factors of safety with respect to
cohesion bykeeping thefactorofsafety with respect to friction equaltounity.
In slopes encountered inpractical problems, thedepthtowhichtherupture circle mayextend
isusually limited by ledgeorother underlying strong material asshowninFig. 10.17. Thestability number for thecase when 0"= 0 isgreatly dependent on the position of the ledge. The depth at
which theledge or strong material occurs may beexpressed interms of a depth factor whichis
defined as
rf ;| (10-25)
whereD - depth ofledge below the top of the embankment, H =heightofslope above thetoe.
or various values of and for the 0 = 0 case the chart in Fig. 10.17 gives the stability number fo various values of slope angle ft In this case th rupturecircle ay pass through th
toe or below the toe. The distance jc of the rupture circle from the toe at the toe level may be
•a Stability number, CD H° |_cu cr <° CD O) 0) V) Q) CT
(Q Stability num ber, N,
r-t-cn ~" r-+ Q) < ^~ CD -> cr CD -^
Stability Slopes
The chart in Fig. 10.17 shows the relationship between and If there is a ledge or other stronger material at the elevation of the toe, the depth factor for thiscase is unity.
Factor Safety with Respectto Strength
he developmentof the stability numberis based on the assumption that th factorof safety with respect to friction F,, is unity. The curves give directly the factor of safety with respect to cohesion only. If true factorof safety with respect to strengthis required, this factor should
apply equally to both cohesion and friction. The mobilized shear strength may therefore be expressed as
s c' a'tan (/)'
In the above expression, we may write
c' ta ( f > ' —=— or = — (approx.) (10.27)
5
c' and tf may be described as average values of mobilized cohesion and friction respectively.
Example
10.8he following particularsar givenfor anearth dam ofheight39 ft. The slope is submergedand the
slope anglej3 45°.
69 lb/ft
c' 55 lb/ft 0' 0°
Determine th factorof safety
Solution
Assumeas afirst trial .0
< t > ' 10 (approx.)
Fo (j)' 10°, an 45° the valueof from Fig. 10.16is 0.11, we may write From q. (10.23) substituting 55Q 2x69x# or =36. 3 ft 2x69x0.11 20 If F 1.9, = — 10.53° and 0.105 19
Chapter
.40ft 1.9x69x0.105
computed height 40 ft is almost equal to the given height 39 ft. The computed factor safety is therefore .9
Example
10.9
excavation is to be made in a soil deposit w ith a slope of 25° to the hor izontal and to a depth of 25 meters. The soil has the following properties:
c' 35kN/m = 15° and 7= 20kN/m
Determine th factor of safety of the slope assuming full friction is mobilized.
2. If the factor of safety with respect to cohesion is 1.5, what would be the factor safety with respect tofriction?
Solution
or = 15° and = 25°, Taylor's stability number chart gives stability number 0.03.
-233 0.03x20x25
2. 1.5, 0.047
xyxH 1.5x20x25
Fo A^ 0.047 an 25°, we have from Fig. 10.16, =
tan0' tan 15° 0.26
Therefore, 1.16
tan0 tan 13 0.231
Example
10.10
An embankment is to be made from a soil having c' 20 lb/ft = 18° and y= 21 lb/ft desired factor of safety with respect to cohesion as well as that with respect to friction is 1.5. Determine
he safe height if the desired slope is horizontal to vertical. 2. The safe slope angle if the desired height is 50 ft.
S o l u t i o n
0.325
tan 0' tan18° 0.325, tan 12.23°
1. Fo = 12.23° and (3 26.6° (i.e., 2 hor izon tal and 1 vertical) the chart gives 0.055
c' 20
Therefore, 0.055
Stability of Slopes 39 Therefore, 42 No w, = safe 1.5x121x0.055 42 42 ft 0.046 yH 1.5x121x50 Fo 0.046an 0' 12.23°, slope angle 23.5
10.12 TENSION CRA CK
If a dam is built of cohesive soil, tension cracks are usually present at the crest. The depth of such cracks may be computed from the equation
(10.28) where = depth of crack, ' unit cohesion, = unit weigh t of soil.
The effective length of any trial arc of failure is the difference between the total length of arc minus th depth crack as shown in Fig. 10.18.
10.13
STABILITY ANA LYSIS BY METHOD OF SLICES FOR
STEADY SEEPAGE
The stability analysis with steady seepage involves the development of the pore pressure head diagram along the chosen trial circle of failure. The simplest of the methods for know ing the pore pressure head at any point on the trial circle is by the use of flownets which is described below. Determination of Pore Pressure with Seepage
Figure 10.19 shows the section of a homogeneous dam w ith an arbitrarily chosen trial arc.There is steady seepage flow through the dam as represented by flow and equipotential lines. From the equipotential lines the pore pressure m ay be obtained at any point on the section. For example at point in Fig.10.19 th pressure head is h. Point is determined setting th radial distance
Tension crack
Effective length of trial arc of failure
394 Chapter 10 Trial circle 'R radius trial circle/' /s side Phreatic line Piezometer Pressure head at point a - h Discharge face Equipotential line ---- -' Pore pressure headdiagram -/
Figure 10.19 Determination of pore pressure with steady seepage
equal to A number of points obtained in the same manner as give the curved line through which is pore pressure head diagram.
Method
of Analysis (graphical method)
Figure 10.20(a) shows the section of a dam with an arbitrarily chosen trial arc.The center of rotation of the arc is 0. The pore pressure acting on the base of the arc as obtained from flow nets is shown in Fig.10.20(b).
When the soil forming the slope has to be analyzed under a condition where f u l l or partial
drainage takes place th analysis must take into account both cohesive frictional soil properties based on effective stresses. Since the effective stress acting across each elemental length of the assumed circular arc failure surface must be comp uted in this case, the method of slices is one of the convenient methods for this purpose. The method of analysis is as follow s.
The soil ma ss above the assum ed slip circle is divided in to a num ber of vertical slices of equ al width. The num ber of slices may be limited to a maxim um of eight to ten to facilitate computation. The forces u sed in the ana lysis acting on the slices are show n in Figs.10.20(a) (c). The forces are:
he weight W th slice.
2. The normal and tangential compo nents of the we ight acting on the base of the slice. They are designated resp ectively a an
3. The pore w ater pressure acting on the base of the slice.
4. The effective frictional and cohesive resistances acting on the base of the slice which is designated S.
he forces acting on the sides of the slices re statically indeterminate they depend on the
of the and we can
relative magnitudes.
In th conv ention al slice method analysis th lateral forces re assumed equal both sides of the slice. This assu mptio n is not strictly correct. The error due to this assum ption on the mass as
Stability Slopes 395
(a) Total norm al and tangential components
~--^ (b Pore-pressurediagram Trial failure surface «,/, Pore-pressure diagram
(c) Resisting forces on the base of slice (d) Graphical representation of all the forces Figure 10.20 Stability analysis of slope by the method of slices
39 Chapter 10 forces that re actually considered in the analysis re s h o w n in Fig. 10.20(c). various components may be determined as follows:
The weight, of a slice per unit length of dam may be com puted from W=yhb
where, total unit weight soil, h average height slice, b width slice. If the w idths of all slices are equal, and if the who le mass is homo geneous, the weigh
plotted s vector passing through th center f slice as in Fig. 10.20(a). may be mad e eq ual to the heigh t of the slice.
2. By constructing triangleABC, th weight can be resolved into normal component
a tangential component Sim ilar triangles can be constructed for all slices. The tan gential components of the weights cause th mass to slide down ward. The sum of all the weights cause the mass_ to slide down ward. The sum of all the tangential components may be expressed I.T.If the trial surface is curved upward near it lower end, th tangential component of the weight of the slice w ill act in the opposite direction along th curve. algebraic sum of should be considered.
3. The average pore pressure acting on the base of any slice of length may be found from th pore pressure diagram shown in Fig. 10.20(b) total pore pressure, on the baseof
ny slice is U=ul
effective norm al pressure N'acting on the base of any slice is N'=N- t/[Fig. 10.20(c)]
5. The frictional force acting on the base of any slice resisting the tendency of the slice to move d o w n w a r d is
(N tan
where 0'is the effective angle friction. Similarly th cohesive force opposing th movement of the slice and acting at the base of the slice is
where is the effective unit cohesion. total resisting force acting on the base of the slice is
= C c' (N ta
Figure 10.20(c) shows th resisting forces acting on the base f slice.
The sum of all the resisting forces ac ting on the base of each slice may be expressed as c'I,l ta I(W- /) c'L +ta 0' -
where £/ length of the curved surface.
The moments of the actuating and resisting forces about the point of rotation may be written as follows:
Actuating m o m e n t R~LT
Resisting moment R[c'L + tan £(jV factor safety may now be written as
Stability Slopes 397
he various components shown in Eq. (10.29) ca easily be represented graphically as shown in Fig. 10.20(d). line represents to suitable scale Z,(N - U). BC is drawn normalto at an equal to c'L + ta (N ). line drawn at an angle 0'to
gives the intercept BD on equal to tan 0'Z(N- ).The length BE on BC is equal to
IT. ow BC
BE (10.30)
Centers for Trial Circles Through Toe
Th factor of safety as computed an represented by Eq. (10.29) applies to one trial circle. This procedure is followedfor a numberof trial circles until one finds the one for whichth factorof safety is the lowest. This circle that gives the least is the one most likely to fail. The procedure is quite laborious. Th numberof trial circles may be minimizedif one follows th following method.
For any given slope angle /3(Fig. 10.21), th center of the first trial circle center may be determined as proposed Fellenius (1927). direction angles an may be taken from Table 10.1. For the centers of additional trial circles, the procedure is as follows:
Mark point whose position is as shown in Fig. 10.21. Join CO The centers of additional circles lie on the line extended. This method is applicable for a hom ogeneous c - < / > ) soil. hen th soil is purely cohesiv an homogeneous th direction angles given in Table 10.1 directly give th center for the critical circle.
Centers for Trial Circles Below Toe
Theoretically if the materialsof the dam and foundation ar entirely homo geneous, an practicable earth am slope ay have it criticalfailure surface below the toe of the slope. Fellenius found that th angle intersected at 0 in Fig. 10.22 fo this case is about 133.5°. find th center for the critical circle below th toe, th following procedure is suggested.
Locus of centers
of critical circles
Curve of factor of
39 Chapter 10
Figure 10.22 Centers trial circles fo base failure
Table 10.1 Direction angles a°
a°ofor
centers critical circlesSlope Slope angle Direction angles
0.6: : 1 . 5 : : : 5 : 60 45 33.8 26.6 1 8 . 3 1 1 . 3 29 28 26 25 25 25 40 37 35 35 35 37
Erect vertical at the midpoint of the slope. this vertical will be the center of the first trial circle. In locating th trial circle use an angle (133.5°) between the two radii which th circle intersects th surface of the embankment and the found ation. After th first trial circle been analyzed th center is some wh at move to theleft, th radius sho rtened and a new trial circle drawn
analyzed. Ad ditional center for the circles re spotted analyzed. Example 10.11
embankment is to be made f sandy clay having cohesion of 30 kN/m , angle internal friction of 20° and a unit weight of 18 kN/m . The slope and height of the emb ank ent are 1.6 :
respectively. Determ ine th factor safety using th trial circle given in Fig. Ex 10.11 by the method slices.
Solution
Consider the em ban km ent as shown in Fig. Ex.10 .11. The center of the trial circl is selected by taking = 26°an 35° from Table 10.1. The soil mass above the slip circle is divided into 13 slices of 2 m width each. he weight each slice er unit length of embankment is given W =
by where average heigh of the slice, b width of the slice, unit weight of the soil. The weight of each slice may be represented by a vector of height if an y, remain the same for the whole em ba nk ent. The vectors values were obtained g raphically. The he igh t vectors
Stability Slopes 39
Figure Ex. 10.11
ay be resolved into normal components and tangential components The values of and for the various slices are given below in a tabular form.
Values of al and/? Slice (m 1.8 5.5 9.5 10.6 11.0 10.2 (m 0.80 3.21 5.75 7.82 9.62 10.43 10.20 (m 1.72 4.50 5.30 5.50 4.82 3.72 2.31 Slice No. 10 11 12 13 (m) 9.3 8.2 6.8 5.2 .3 1.1 (m) 9.25 8.20 6.82 5.26 3.21 1.0 (m) 1.00 -0.20 -0.80 -1.30 -1.20 -0.50
The sum of these components and may be converted into forces ZN andIrrespectively by multiplying them as given belo
Sfc 81.57m, Ui 24.87m Therefore, ZN 81.57x 2 x 18 = 2937 kN 24.87 x2x 18 895kN Length of arc L =31.8 'L + tonfiZN 30x31.8 0.364x2937 Factor of safety 89 2.26
10.14
BISH OP'S SIMPLIFIED
METHOD
SLICES
Chapter
Bishop's method of slices (1955) i useful if a slope consists of several types of soil with different values and 0 and if the pore pressures in the slope re known or can be estimated. method of analysis is as follows:
Figure 10.23 gives section of an earth having sloping surface AB. ADC is an assumed trial circular failure surface w ith its center at The soil mass above the failure surface is divided into a num ber of slices. The forces acting on each slice are evaluated from limit equ ilibriu m of the slices. The eq uilib rium of the entire mass is determined by s um ma tion of the forces on each of the slices.
Consider for analysis a single slice abed (Fig. 10.23a) which is drawn to a larger scale in Fig. 10.23(b). forces acting this slice re
W weight of the slice
= total normal force on the failure surface
pore water pressure = ul on the failure su rface cd
shear resistance acting on the base of the slice normal forces on thevertical faces andad shear forces on the vertical faces be
6 the inclina tion of the failure surface cd to the horizontal
The system is statically indeterm inate. An approximate solution may be obtained by assuming that the resultant of nd is equal to that of an and their lines of action coincide. For equ ilibriu m of the system, the following equations hold true.
(a (b)
Stability of Slopes 40 N=Wcos6
(10.31) where tangential component of
The u nit stresses on the failu re surface of length , /, may be expressed as
Wcos6
normal stress,
sin0 (10.32)
shear stress,
equation fo shear strength, s, is ' c r't an ^' c' (cr-u)tan0' where rf effective normal stress
c' effective cohesion
ft effective angle of friction unit pore pressure
The shearing resistance to sliding on the base of the slice is si c' (Wcos 9 ul tan
where l U, the total pore pressure on the base of the slice (Fig 10.23b)
=
The total resisting force and the actuating force on the failure surface may be expressed as
Total resisting force is
[c7 ( W c o s 0 - M / ) t a n 0 ' ] (10.33)
Total actuating force is
Wsm0 (10.34)
The factor of safety is then given as
Eq (10.35)is the same as Eq. (10.29 obtained by the conventional method of analysis. Bishop (1955) suggests that the accuracy of the analysis can be improved by taking into account the forces the vertical faces of each slice. For the element in Fig. 10.23(b), may w rite an expression for all the forces acting in the vertical direction for the equ ilibrium condition
co&0 = W -T )-ulcos0- sin# (10.36)
If the slope is not on the verge of failure (F 1), the tangential force is equal to the shearing resistance on cd divided by
402 Chapter 10
c'
(10.37) where, N'=N-U,andU= l.
Substituting (10.37) into (10.36) solving N\ obtain
c' s i n < 9 co ta 0' sin .. (10.38) w h e r e , A T
For equilibrium of the mass above the failure surface, we have by taking moments about
Wsin0R (10.39)
By substitutingEqs.(10.37) (10.38) into (10.39) solving obtain expression forF
w h e r e , tan ( / > ' sin
(10.40) (10.41)
factor safety is present in Eq. (10.40) both sides. quantityAT
to be evaluated by means of successive approximation . Trial values of an that satisfy the equilibrium each slice, and the conditions
1.6 1.4 1.2
1.
f) co (sin ta )/
Note: is hen slop of failure arc is in the same quadrant as groun d slope.
0.6
-40 -30 -20 -10 0 10 20
Valuesof degrees 30 40
Stability Slopes 40
(E-E (r -T = 0
ar used. value ay then be computed first assuming an arbitrary value fo he value of ay then calculatedby making use of Eq. (10.40). If the calculated valueof differs appreciably from the assumed value, a second trial is made and the computation is repeated. Figure 10.24developed b y Janb u et al. (1956) helps to simplify the compu tation procedure
It is reported that an error about percent will occur if we assume Z(Tj tan0'= 0. But if we use the conventional method analysis u sing (10.35) th error introduced about
15 percent (Bishop, 1955).
10.15 BISHOP AND MORGENSTERN METHOD FOR SLOPE ANALYSI
Equation (10.40) developed based on Bishop's analysis of slopes, contains the term pore pressure u. The Bishop and Morgenstern method (1960) proposes the following equation for the evaluation of
yh (10.42)
where, u pore water pressure at any point on the assumed failu re surface unit weight of the soil
h the depth of the po int in the soil mass below the g round su rface
The pore pressure ratio is assumed to be constant throug hout the cross-section, which is called homogeneous pore pressure distribution. Figure 10.25shows th various parameters used in th analysis.
he factor of safety is defined as
nr,. (10.43)
where, m, n =stability coefficients.
an values may be obtained either from charts in Figs. B. 1 to B.6 or Tables B1 to B6 in Ap pendix B . The depth factor given in the charts or tables is as per Eq. (10.25), th at is DIH, where H height slope, an D depth firm stratum from the top of the slope. Bishop an Morgenstern (1960) limited their charts (o tables) to values c'ly H equal to 0.000, 0.025, an 0.050.
Center offailure surface
Failure surface
= unit weight of soil
/^^^^^^^^//^f^^^
Figure 10.25
Specifications
parameters
fo Bishop-Morgenstern
method
Chapter 10
Extension
of the
Bishopand Morgenstern Slope Stability Charts
As stated earlier, Bishop an Morgenstern (1960) charts tables cover values c'lyH equal to 0.000, 0.025, 0.050 only. These charts do notcover th values thatar normally encountered in natural slopes. Connor an Mitchell (1977) extended th work Bishop an Morgenstern to cover values c'lyH equal to 0.075 an 0.100 fo various values depth factors method employed is essentially th same that adopted by the earlier authors. extended values ar given in the form charts an tables from Figs. B.7 to B.14 an Tables B7 to 14 respectively in Appendix B.
Method Determining
1. Obtain th values an clyH
From the tables in ppend ix B, obtain the values of nd for the known values ofc/yH, 0 and /3 and for 1, 1.25 1.5.
3. Using Eq. (10.43), determ ine for each value of
4. The required value of is the lowest of the values obtained in step 3.
Example
10.12Figure 10.12gives typical section f homogeneous earth dam. soil parameters are: 30°,c' = 590 lb/ft an y lb/ft The dam has a slope :1 and a pore pressure ratio 0.5. Estimate the factor of safety by Bishop and Morgenstern method for a height of dam #=140
Solution
Height of H= 140ft
120x140 0.035
Given: 0' 30°,slope 4:1 and 0.5.
Since c'lyH 0.035, and 1.43fo H 140 ft, the for the dam lies between c'lyH 0.025 an 0.05an between 1.0 and 1.5. equation fo is
m-nr
Using the Tables in ppen dix B, the follow ing table can be prepared for the given values of c'lyH, 0, and
0'=30° c'= 590psf
120 pcf
/• =0.50 = 200 ft
Alluvium(same properties as above) Figure Ex. 10.12
