CHAPTER STABILITY OF SLOPES
10.1
INTRODUCTION
Slopes of earth are of two types 1.
2.
Natural slopes an made slopes
Natural slopes are those that exist in nature and are formed by natural causes. Such slopes exist in hilly areas. The sides of cutting s, the the slopes of emb ankm ents constructed for roads, railway lines, canals etc. and the slopes of earth dams constructed for storing water are examples of man made slopes. The slopes w hether natural or artificial artificial may b 1.
Infinite slopes
2.
Finite slopes
The term infinite slope is used to designate a constant slope o infinite extent. The long slope the face of a mo untain is an example of this type, whereas finite slopes are limited in extent. The of slopes of emb ankm ents and earth dams are examples examples of finite slopes. The slope length depends on the height of the dam or embankment. Slope Stability: Slope stability is an extremely important consideration in the design and construction of earth dam s. The stability stability of a natural slope is also also im portant. The results of a slope failure can often be catastrophic, inv olving the the loss of considerabl considerablee property and m any lives. Causes of Failure of Slopes: he important factors that cause instability in slope an lead to failure ar 1.
Gravitational force
2. 3.
Force due to seepage water Erosion of the surface of slopes due to flowing water
365
C h a p t e r 10
366
4.
The sudden lowering of water adjacent to
5.
Forces due to earthquakes
slope
The effect of all the forces listed above is to cause movement of soil from high point s to low p o i n t s . most important such forces is the c o m p o n e n t g r a v i t y t h a t a c t in the direction probable motion. various effects flowing s e e p i n g w a t e r re g e n e r a l l y r e c o g n i z e d a s v e r y i m p o r t a n t i n s t a b i l i t y p r o b l e m s , b u t o f te te n t h e s e e f f e c t s h a v e n o t b e e n p r o p e r l y i d e n t i f i e d . I t i s a fact t h a t t h e se se e p a g e o c c u r r i n g w i t h i n a s o i l m a s s c a u s e s seepage f o r c e s , w h i c h h a v e m u c h g r e a t e r e f fe fe c t t h a n i s c o m m o n l y r e a l i z e d . Erosion on the s u r f a c e f slope may be the c a u s e of the r e m o v a l f certain weight soil, and may thus lead to an increased stability as far as m a s s m o v e m e n t is c o n c e r n e d . On the other hand, erosion in the form u n d e r c u t t i n g at the toe may increase th h e i g h t of the slope, decrease th l e n g t h of the i n c i p i e n t f a i l u r e s u r f a c e , t h u s d e c r e a si si n g th stability. Whe n there is a lowering of the ground water or of a freewater surface ad jacent to the slope, sudden drawdown of the water surface in reservoir there is decrease in the fo example in b u o y a n c y of the soil which is in effect an increase in the weight. This increase in weight causes increase in the shearing stresses that may or may not be in part counteracted by the increase in C o m p o n e n t of weight
Failure surface
(a Infinite slope
(b
earth da
Ground water table
Seepage parallel to slope
(c Seepage below a natural slope Lowering of water from level to Earthquake force
(d) Sudden drawdown condition
Figure 10.1
Forces
(e) Failure due to earthquake that act on
earth slopes
C h a p t e r 10
366
4.
The sudden lowering of water adjacent to
5.
Forces due to earthquakes
slope
The effect of all the forces listed above is to cause movement of soil from high point s to low p o i n t s . most important such forces is the c o m p o n e n t g r a v i t y t h a t a c t in the direction probable motion. various effects flowing s e e p i n g w a t e r re g e n e r a l l y r e c o g n i z e d a s v e r y i m p o r t a n t i n s t a b i l i t y p r o b l e m s , b u t o f te te n t h e s e e f f e c t s h a v e n o t b e e n p r o p e r l y i d e n t i f i e d . I t i s a fact t h a t t h e se se e p a g e o c c u r r i n g w i t h i n a s o i l m a s s c a u s e s seepage f o r c e s , w h i c h h a v e m u c h g r e a t e r e f fe fe c t t h a n i s c o m m o n l y r e a l i z e d . Erosion on the s u r f a c e f slope may be the c a u s e of the r e m o v a l f certain weight soil, and may thus lead to an increased stability as far as m a s s m o v e m e n t is c o n c e r n e d . On the other hand, erosion in the form u n d e r c u t t i n g at the toe may increase th h e i g h t of the slope, decrease th l e n g t h of the i n c i p i e n t f a i l u r e s u r f a c e , t h u s d e c r e a si si n g th stability. Whe n there is a lowering of the ground water or of a freewater surface ad jacent to the slope, sudden drawdown of the water surface in reservoir there is decrease in the fo example in b u o y a n c y of the soil which is in effect an increase in the weight. This increase in weight causes increase in the shearing stresses that may or may not be in part counteracted by the increase in C o m p o n e n t of weight
Failure surface
(a Infinite slope
(b
earth da
Ground water table
Seepage parallel to slope
(c Seepage below a natural slope Lowering of water from level to Earthquake force
(d) Sudden drawdown condition
Figure 10.1
Forces
(e) Failure due to earthquake that act on
earth slopes
Stability of Slope s
367
shearing strength, depending up on whe ther or not the soil is able able to undergo compression which the load increase tends to cause. If a large mass of soil is saturated and is of low permeability, slow rate, and in spite of the increase practically no volum e changes w ill be able to occur except at a slow of load the strength increase may be inappreciable. Shear at constant volume may be accompanied by a decrease in the intergranular pressure an an increase in the neutral pressure. failure may be caused by such condition in which th entire soil mass passes into a state of liquefaction and flows like a liquid. A co ndition of this type may be developed if the mass of soil is subject to vibration, fo example, due to earthquake forces. The various forces that act on slopes are illustrated in Fig. 10.1.
10.2 G E N E R A L C O N S I D E R A T I O N S ANALYSIS
ASSUMPTIONS IN THE
There are three distinct parts to an analysis of the stability of a slope. They are:
Testing
samples to determine th
cohesion
angle
internal friction
If the analysis is for a natural slope, it is essentia essentiall th at the sample be un disturb ed. In such impo rtant respects respects as rate of shear application and state state of initial consolidation, consolidation, the cond ition of testing must represent as closely closely as possible the most unfavorab le conditions ever likely to occur in the actual slope.
2. The s tudy of items items w hich are kno wn to to enter but which cannot be acco unted fo in the computations The m ost important of such item s is progres progressive sive cracking w hich will start start at the top of the slope where the soil is in tension, and aided by w ater pressure, pressure, may progress to considerable depth. In addition, there are the effects of the non-homogeneous nature of the typical soil and other variations from the ideal conditions which must be assumed. 3.
Computation
If a slope is to fail along a surface, all the the shearing strength m ust be overcome along that surface rupture. such AB Fig. 10.1 represents which then becomes surface infinite n u m b e r of possible traces on which failure might occur.
It is assumed th at the problem is two dimensional, which theo retically requires a long length section inv estigated holds for a runn ing length of slope norm al to the section. How ever, if the cross section of roughly two or more times the trace of the rupture, it is probable that the two dimensional case he shear strength of soil is assumed assumed to follow follow C oulomb's law = c' +
tan 0"
where, c'
effective unit cohesion effective normal stress on the surface of rupture = (cr
total normal stress on the surface of rupture pore water pressure on the surface of rupture effective angle of internal friction. he item of great importance is the loss of shearing strength which many clays show when subjected to a large shearing strain. The stress-strain curves for such clays show the stress rising with increasing strain to a maximum value, after which it decreases and approaches an ultimate
36
Chapter
10
value which 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 v a l u e t h a t s h o u l d b e u s e d f o r th th e s h e a r i n g s t r e n g t h ra ra t h e r t h a n th th e m a x i m u m v a l u e . FACTOR
10.3
SAFETY
In stability analysis, two types of factors of safety are normally used. They are 1.
Factor
s a f e t y w i t h r e s p ec ec t to s h e a r i n g s t r e n g t h .
2.
Factor of safety with respect to cohe sion. This is is termed the factor of safety with respect to height. Let, F,
factor
s a f e t y w i t h r e s pe pe c t to strength
factor
safety with respect to c o h e s i o n
factor of safety with respect to height F,
s a f e t y w i t h respect to friction
factor
mobilized cohesion mobilized angle average value s
friction mobilized sh earing strengt
m a x i m u m s h e a r i n g s t r e n g th th .
The factor of safety w ith respect to shearin g strength, j' ta
m a y b e w r i t te te n a s
The shearing strength mobilized at each point on a failure surface may be written as _
LJ
r=c;+<7'tan0; where
(10.2) .,
Tm
tanfi
A c t u a l l y th s h e a r i n g r e s i s ta ta n c e ( m o b i l i z e d v a l u e s h e a r i n g s t r e n g t h ) does develop to like degree at all points on an incipient failure surface. The shearing strains vary considerably and th shearing stress m a y b e f a r f r o m c o n s t a n t . H o w e v e r th above expression is correct on the basis a v e r a ge ge c o n d i t i o n s . If the factors of safety with respect to cohesion and friction are different, we may write the equation of the mobilized shearing resistance
It will shown later t h a t F, d e p e n d s on the h e i g h t of the slope. From this i t m a y b e conclu ded that the the factor of safety with respect to cohesion may be designated as th factor of safety with respect to height. This factor is denoted and it is the ratio between th critical height
Stability
Slopes
the actual height, the critical critical height being the m axim um height at which it is possible possible for a slope slope to stable. W e m a y write from (10.3) (1Q4)
where
is arbitrarily taken equal to unity.
Example
10.1
shearing strength parameters
f
soil re
26.1 k N / m
c'
= 15° 17.8 k N / m
Calculate th factor with respect to friction.
safety (a with respect to strength, (b w i t h respect to cohesion and (c) average intergran ular pressure pressure tf on the failure surface is 102.5 k N / m
Solution th basis of the given data, th average shearing strength on the failure surface is 26.7
102.5 tan 15°
26.7
102.5
and the average value
0.268
54.2
mobilized shearing resistance is
T= 1 7 . 8 + 102.5 tan 12° 17.8
102.5
39.6
0.212
39.6 k N / m 17.80
ta
0.212
above example shows th factor of safety with respect to shear strength, is 1.37, whereas the factors of safety with respect to cohesion and friction are different. Conside r two extreme cases: When the factor of safety with respect respect to cohesion cohesion is unity. 2.
When the factor factor of safety safety with respect to friction friction is is unity.
Casel
102.50x0.268
12.90
2.13
Case
39.60
—— +102.50
ta
37
C h a p t e r 10
26.70
27.50
12.10 can have any combination of
between these two extremes cited above to give
and
the same mobilized shearing resistance of 39.6 kN/m . Some of the combinations of
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
U n d e r Case 2, the value of 2.20 when as the, actor of safety with respect to cohesion.
1.0. The factor of safety
2.20 is defined
E x a m p l e 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 32 k N / m an c' 18°; th expected parameters of mobilized shearing resistance ar 21 kN/m an = 13° and the average effective pressure on the failure plane is 0 kN /m c' For the same value of mobilized shearing resistance determine the following: Factor of safety w ith respect to height; 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 soil is
= 32 +
10 tan 18° = 32 + 35.8 = 67.8 kN/m
The mobilized shearing resistance of the soil is T=21
110 tan 13° = 21 + 25.4
46.4 kN/m
67.8
Factor of safety with respect to average strength,
Factor of safety with respect to cohesion,
Factor of safety with respect to friction,
=
.,
46.4
110tanl8° 1.0
(=
therefore,
Factor of safety w ith respect to friction at F
46.4
32
_ tan 18° _ 0.3249
will be at 32
46.4-35.8
.0 is
1-46
1.52
TT
F
Factor of safety with respect to height,
——
TT
2309
.0
3.0
Stability
Slopes
46.4
371
32 1 1 0 t a n l 8 °
therefore,
1.0
Factor of safety with respect to strength 32
10.4
110 tan 18°
35.8
2.49
46.4-32
is obtained when
W e m a y write
1.46
STABILITY A N A L Y S I S
OF INFINITE
SLOPES
IN SAND
As 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 s a m e as at every point this plane. Now consider vertical slice material A B C D h a v i n g u n i t d i m e n s i o n n o r m a l 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 w e i g h t an reaction ar equal in m a g n i t u d e an opposite in direction. They have c o m m o n l i n e action which passes lateral forces must equal opposite an their line through th center of the base action must parallel to the sloped surface. re n o r m a l an shear stresses plane a'
yzcos fi
where cr'
effective normal stress, effective un it weight of the sand. plane If full resistance is mobilized
th shear strength, s, of the soil
is
ta w h e n T = s , substituting fo
Figure
10.2
tf
have
Stability analysis
infinite
slope in sand
C o u l o m b ' s la
C h a p t e r 10 or
tan /3 = tan
(10.5a)
if the slope is to be Equation (10.5a) indicates that th m a x i m u m v a l u e of is limited to 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
(10.5b)
tanfi
1 0 .5 Th
STABILITY ANA LYS IS
vertical stress
acting on plane
INFINITE
S L O P E S IN C L A Y
(Fig. 10.3) where
yzcosfi in Fig. 10.3 in the stress diagram. normal stress on this plane is is represented th shearing stress is he line makes an angle (3 with th cr-axis.
Mohr strength envelope is represented
line
an
whose equation is
= c' + cr'tan^'
According to the envelope, th shearing strength is where th normal stress is W h e n /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 w h i c h th direct stress is less than th there is no possibility of failure. However at particular depth at w h i c h th direct stress is
O
Figure 10.3
E
Stability a n a l y s i s of infinite s l o p e s in clay soils
Stability
Slopes
shearing strength and shearing stress values are equal as represented by failure i s i m m i n e n t . This depth at which the shearing stress and shearing strength are equal is called the critical depth. depths greater than this critical value, Fig. 10.3 indicates 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. E x p r e s s i o n for the Stability of an Infinite S l o p e of Clay of Depth
E q u a t i o n (10.2) gives th developed shearing stress (10.6)
c' +(T'tan'
depth
Under conditions of no seepage and no pore pressure, the stress components on a plane at 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 c'
yti
(tan
tan
/?(tanytf-tan^)
(10.7)
where is the allowable height and the term c'Jy is a dimensionless expression called the This dimension less num ber is proportional to the required stability n um ber and is designated as cohesion and is inversely proportional to the allowable height. solution is for the case w h e n seepage is occurring. If in Eq. (10.7) the factor of safety w ith respect to friction is unity , the stability n u m b e r with respect to cohesion may be w ritten as
where
,
stability number in Eq. (10.8) m a y b e w r i t t e n as
where
critical heigh t. From Eq. (10.9), we have
is the same as the (10.10) indicates that the factor of safety w ith respect to cohesion, 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 (lO.lla)
C h a p t e r 10
374
(a
(b
Analysis infinite slope (a with s e e p a g e flow t h r o u g h th m a s s , and (b) with completely submerged slope.
Figure 10.4
entire
the shea ring stres sin
sa
(lO.llb)
/3
N o w s u b s t i t u t i n g E q s (10. 11 a) and (10. l i b ) i n t o e q u a t i o n
s i m p l i f y i n g , th
stability e x p r e s s io n o b t a i n e d is
-^2—
'„
0-
sa
(10.12)
sa
b e f o r e , if the f a c t o r safety with respect to friction is unity, th s t a b i li t y n u m b e r w h i c h r e p r e s e n ts t h e c o h e s i o n m a y b e w r i t t e n a s C/
F c' Y
sa
H
Y 'sat
tf
tan^--^-
(10.13)
sa
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 where y,
ta
submerged u n i t w e i g h t o f t h e soil.
<}>')
(10.14)
Stability
Slopes
E x a m p l e 10.3 Find the factor of safety of a slope of infinite extent having of cohesionless soil with 0 = 30°.
slope angle
25°. The slope is made
Solution
Factor of safety
tan/?
tan 30°
0.5774
tan 25°
0.4663
E x a m p l e 10.4 Analyze the slope of Example 10.3 if it is made of clay having c' - 30 kN/m 0' 20°, e = 0.65 and 2.7 and under th following conditions: (i when th soil is dry, (ii) when water seeps parallel to the surface of the slope, and (iii) when the slope is submerged. Solution
Fo e
0.65 an
2.
G
(2.7
27x^1= /sat
1 +0.65
0.65)x9.81
1
0.65
10.09 kN/m (i For dry soil th
stability number cos /?(tan/?-
is when F,=l
')
d
(cos 25° Therefore, th
( t a n 25° - tan 20°)
0.084. c'
critical height
30
16.05x0.084
(ii) For seepage parallel to the surface of the slope [Eq. (10.13)] c'
100Q
= cos 25° tan 5°-^-- ta 20° =0.2315 19.9
19.9x0.2315
=6.51
(iii) For the submerged slope [Eq. (10.14)] cos 25° (tan 25° - tan 20°)
10.09x0.084
0.084
22.25
Chapter
10.6 METHODS HEIGHT
S T A B I L IT Y A N A L Y S I S
SLOPES
FINITE
infinite e x t e n t ha been discussed in previous sections. stability slopes more common problem is the one in w h i c h th 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 accepte d as offering an approximately correct solution for the determination of the factor of safety of slope of an e m b a n k m e n t and of its f o u n d a t i o n . D e v e lo p m e n t s 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 f o u n d a t i o n s m e a n s of the circle method is no longer an unduly tedious procedure. There are other methods of historic interest such as the Culmann method (1875) and the logarithmic spiral method. The C u l m a n n m e t h o d a s s u m e s t h a t r u p t u r e w i l l o c c u r a l o n g
p l a n e . It is of i n t e r e s t o n l y s classical solution, since actual failure surfaces re i n v a r i a b l y c u r v e d . 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. fe
There are several methods of stability analysis based on the circ ular arc surface of failure. 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 shear strength 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 variety of c o n d i t i o n s .
10.7
PLANE SUR FAC
FAILURE
C u l m a n n (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 this assumption is reasonable and in others it is questionable.
Stability
Slopes
377
Force triangle Stability of s l o p es by Culmann method
Figure 10.5
The method as indicated above assumes that the critical surface of failure is
plane surface
passing through the toe of the dam as shown in Fig. 10.5. The forces th at act on the mass above trial failure plane AC inclined at angle with the horizontal are shown in the figure. The expression for the weight, and the total cohesion are respectively, -yLH cosec /? 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
in which the subscript
and W, and rearranging we have
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 equilibrium may be written as
yH
where
limiting equilibrium and the stability number fo
sin/? cos 0' is the critical height of the slope.
limiting
(10.15)
C h a p t e r 10
378
If we write tan
^'
<>~tan^
where an are safety factors with respect to cohesion and friction respectively, Eq. (10.15) an 0' as ay be modified for chosen values of
sin/3 co ( / ) '
(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
10.5
Determine 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 w e i gh t = 1 1 4 lb/ft Find the allowable height of the embankment if F, .25.
Solution H,
4c'sin/?cos0'
4 x 630 x sin 40° cos20°
y[l-cos(0-4>')]
114(l-cos20°)
or
(>
,, Allowable height,
10.8
504 lb/ft
1.25,
tan
221 ft
tan 20°
0.291, fa
16.23°
4x504 sin 40° cos 16.23°
114[l-cos(40- 16.23°)]
CIRCULAR SURFACES
128.7 ft.
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 circular arc. 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 1.
Slope failure
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 w h e n the slope an gle /3 is quite high and the soil close to 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 angle j3 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
Figure 10.6
Types
failure
earth dams
C h a p t e r 10
380
1 0 .9
F A I L U R E U N D E R U N D R A I N E D C O N D I T IO N S (0
=
immediately after f u l l y saturated clay slope fail u n d e r u n d r a i n e d c o n d i t i o n s ( 0 = 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.
S l o pe f a i l u r e Base failure
The undrained shear strength circular surface with center and the chord AB m a k e a n g l e s /3
of soil is assumed to be constant with depth. A trial failure radius is s h o w n in Fig. 10.7(a) for a toe f a i l u r e . The slope wi th the horizontal respectively. is the weight per unit
Firm base
(b B a s e f a i l u r e
(a) Toe f a i l u r e
Figure 10.7
ritical circle positions for (a) slope fai lure (a fte r Fellenius, 192 7), (b) b a s e failure
50
1> 40
20
10
90
70 Values of
60
50
40°
30°
20°
10
V a lu e s
(a
Figure 10.8 ( a ) R e l a t i o n b e t w e e n s l o p e a n g l e /3 a n d p a r a m e t e r s a a n d 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 slope angle and depth factor for various values of parameter ( a f t e r 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 m o m e n t
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). This is 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 26 as show n in Fig. 10.7(a). Values of an fo different slope angles /3 can 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 following equations may be written with reference to Fig. 10.7(b). on slope D
Depth factor, Values of
H
Distance factor,
can be estimated fo
(10.21)
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 k N / m , angle of internal friction zero and the total unit weight of the soil = 20.0 kN/m Solution
Draw the given slope A B C D as shown in F ig. Ex. 10.6. To locate the cen ter of rotation, extend the bisector line With as center an to cut the vertical line drawn from at point as radius, draw the desired slip circle.
Radius
OC =R
36.5 m,Area B E C F B
- x4
32.5
=
xEFxBC
86.7
Therefore W 86.7 x 1 x 20 = 1734 kN acts through point which may be taken as the middle
Chapter
382
36.5m
Figure. From th figure
have,
Length of arc
=R0= 36.5
x
length of a rc x cohesion
15.2 m, an
radius
x.
10.6
9=
3.14
180
33.8
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
Figure 10.9
Principle
of friction circle method
Stability of Slopes
383
line tangent to the inner circle si 0", w h e r e is the radius of the trial circle, circle is d r a w n . must intersect th trial circle at an angle tf w i t h Therefore, an vector representing t a n g e n t to the i n n e r intergranular pressure at obliquity to an element of the rupture ar m u s t 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 m a s s . center of mass may be determined by any one of the known methods.
2.
3.
The resultant boun dary neutral force The vector may be determined by a graphical method from flownet construction. acting on the b o u n d a r y . The resultant in tergran ular force
4.
The resultant cohesive force
Actuating Forces
actuating forces m a y b e considered to be the total weight and the resultant boundary force s h as o w n in Fig. 10.10. boundary neutral force always passes through th center rotation resultant of an designated as is shown in the figure.
Resultant Cohesive Force designated as Let the length of arc th length of chord Let the arc length divided into n u m b e r of small elements and let the mobilized cohesive force on these elements designated as etc. as s h o w n in Fig. 10.11. resultant of all these forces is s h o w n , 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
Figure 10.10
Actuating forces
Chapter
384
(a
C o h e s i v e forces on
trial arc
Figure 1 0 . 1 1
10
(b) P o l y g o n of forces
Resistant cohesive forces
may w r i t e
wherein c'= u n i t cohesion,
he line of action of cohesion is expressed as c' L a R
where
c 'm
l
factor of safety with respect to cohesion.
may be d e t e r m i n e d by moment consideration.
he m o m e n t of the total
c
moment arm. Therefore, (10.22)
It is seen that th line of action of vector
Resultant
B o u n d a r y Intergranular
is independent of the m a g n i t u d e of
c'
Forces
etc. be the he trial arc of the circle is divided into n u m b e r of small elements. et intergranular forces acting on these elements as s h o w n in Fig. 10.12. he friction circle is d r a w n with a radius of sin j/ where lines a c t i o n of the i n t e r g r a n u l a r f o r c e etc. tangential to the f r i c t i o n at the b o u n d a r y . H o w e v e r , th v e c t o r s u m o f a n y t w o s m a l l circle make angle forces has a line action through point m i s s i n g t a n g e n c y to the small -circle y amount. r e s u l t a n t of all g r a n u l a r f o r c e s m u s t t h e r e f o r e m i s s t a n g e n c y to the -circle a m o u n t w h i c h is not c o n s i d e r a b l e . Let the d i s t a n c e of the r e s u l t a n t of the g r a n u l a r f o r c e si (a s h o w n in Fig. 10.12). designated from th c e n t e r of the c i r c l e
Stability
Slopes
385
KRsin
Figure
10.12
Resultant of intergranular forces
magnitude of depends upon the type of intergranular pressure distribution along the arc. The most probable form of distribution is the sinusoidal distribution. The variation of
with respect to the central angle a'is shown in Fig. 10.13. The figure also gives relationships between an for a uniform stress distribution of effective normal stress along the arc of failure. The graphical solution based on the concepts explained above is simple in principle. For the three forces Q, and of Fig. 10.14 to be in equilibrium, must pass through the intersection of
1.20 Cent ral angle 1.16
1.12
or anifo rm st e s s c istrih > u t i o r
1.08
1.04
1.00
or sinus oida tress distr ibuti( 20
40
60
Central angle
Figure 10.13
80
in degrees
Relationship between
and central angle a'
Chapter
386
Figure 10.14
10
Force triangle for the friction-circle method
the know n lines of action of vectors The line of action of vector ust also be tange nt to si . The v a l u e m a y b e estimated by the use of curves given in the circle radius Fig. 10.13, and the l i n e action o f f e r e e m a y b e d r a w n s h o w n in Fig. 10.14. Since th lines action of all three forces and the m agn itude of force a r e k n o w n , t h e m a g n i t u d e o f P a n d may-be obtained by the force parallelogra m c onstru ction that is indicated in the figure. The circle of radius of si is called th modified friction circle. rn Determination
Factor
S a f e t y With R e s p e c t to Strength
F i g u r e 10.15(a) is section f d a m . force is the trial failure arc. th r e s u l t a n t is drawn as explained earlier. The line of action of is also d r a w n . L e t t h e f o r c e s and C
(a Friction circle
Figure 1 0 . 1 5
Graphical method
(b
Factor
determining f a c t o r strength
safety
safety
with
respect to
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 i s t h e n d r a w n t h r o u g h 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' = — , and F*
tanfl'
T h e s e f a c t o r s are the v a l u e s u s e d to plot point 1 in the g r a p h in Fig. 10.15(b). S i m i l a r l y other friction circles with radii in in 0'm3 etc. may be drawn and the procedure repeated. Points , etc. re o b t a i n e d as s h o w n in Fig. 10.15(b). The 45° l i n e , r e p r e s e n t i n g for this trial circle. F., intersects the curve to give the factor of safety 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
10.7
em 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 radius R 20 m.
Length of chord Take
FB be the slope and
be the slip circle drawn with center
nd
27
B
as the midpoint
then
Area A K B F E A = area A K B J A + area A B E A -ABxEL
-ABxJK 3
3
x 27
5.3
27 x 2.0
122.4
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 ° ,
Length of arc
L
M o m e n t arm of cohesion, /
chord
31
RO = 20 x 85 x —
29.7
29.7 = 20 x = 22 m
From center at a distance draw the cohesive force vector C, which is parallel to the ow from the point of intersection of an draw a line tangent to the friction circle
Chapter
388
10
1.74m
//=10m
Figure
Ex.
10.7
d r a w n at w i t h radius si = 20 sin 5° = . 7 4 m . Th i s line is the line of action of the third force is k n o w n D r a w triangle forces in w h i c h th m a g n i t u d e and the direction fo only the directions of the other two forces are known. Length
gives th cohesive force C
520 kN
Mobilized cohesion,
29.7
= 17.51 k N / m
Therefore th factor of safety w ith respec to cohesion,
is
^=1.713
will If, F
.7 13 if the factor
1.5, then '
tan5
safety with respect to friction, F^
0.058 rad; or
3.34°
.0
Stability
Slopes
The new radius of the friction circle is
= R sin 0' = 20 x sin 3.3° Th
direction of
cohesive force
changes and the modified triangle of force abd' gives, C = length ad
C
Mobilised cohesino, c' Therefore,
1.16
LJ
c'
=—
20.2
-
600 kN
600 Z*yI
20.2 kN/mr
= 1.5
T A Y L O R ' S STABILITY NUMBER
10.1
th effective unit weight of material y, angle of If th slope angle j8, height of embankment internal friction < / > ' , an unit cohesion ar known, th factor of safety may be determined. In order to make unnecessary the more or less tedious stability determinations, Taylor (1937) conceived the idea of analyzing the stability of a large number of slopes through wide range of slope angles and angles of internal friction, an then representing th results by an abstract number which he called the "stability number". This number is designated as ^. The expression used is
From this the factor of safety with respect to cohesion 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 circle passes below the toe. However, these curves may be used without serious error for slopes down to fi 14°. The stability numbers are obtained for factors of safety with respect to cohesion by keeping the factor of safety with respect to friction equal to unity. In slopes encountered in practical problems, the depth to which the rupture circle may extend is usually limited by ledge or other underlying strong material as shown in Fig. 10.17. The stability number for the case when 0" = 0 is greatly dependent on the position of the ledge. The depth at which the ledge or strong material occurs may be expressed in terms of a depth factor which is defined as
rf
;|
(10-25)
where D - depth of ledge below the top of the embankment, H = height of slope above the toe. 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 rupture circle 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 expressed by distance factor which is defined as
Stability number,
•a
CD
O)
0) CD
H°
|_cu
V)
Q)
CT
cr <°
(Q
r-t-
~"
cn r-+
Q) <
^~ CD
->
cr
CD -^
S t a b i l it y n u m b e r , N,
Stability
Slopes
The chart in Fig. 10.17 shows the relationship between stronger material at the elevation of the toe, the depth factor
and
If there is a ledge or other
for this case is unity.
Safety with Respect to Strength
Factor
he development of the stability number is based on the assumption that th factor of safety with respect to friction F,, is unity. The curves give directly the factor of safety with respect to cohesion only. If true factor of safety respect strength required, this factor should to is with apply equally to both cohesion and friction. The mobilized shear strength may therefore be expressed as s
a' tan ( / ) '
c'
In the above expression, we may write
ta ( f > '
c'
—=—
or
= — (approx.)
(10 .27)
5
c' and tf
may be described as average values of mobilized cohesion and friction respectively.
E x a m p l e 10.8 he following particulars ar given for an earth dam of height 39 ft. The slope is submerged and the slope angle j3
45°.
69 lb/ft c'
55
0'
lb/ft
0°
Determine th
factor of safety
Solution
A s s u m e as a first trial 10
'
Fo
(j)'
From
.0
(approx.)
10°, an
45° the value of
q. (10.23)
from Fig. 10.16 is 0.1 1, we may write
substituting
55Q 2x69x# or
If F
2x69x0.11 1.9,
20
=—
19
=36. 3 ft
10.53° and
0.105
Chapter
1.9x69x0.105
.40ft
computed height 40 ft is almost equal to the given height 39 ft. The computed factor safety is therefore .9
E x a m p l e 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: 35kN/m
c'
= 15° and 7= 20
kN/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 with respect to friction?
safety
Solution or
= 15° and
= 25°, Taylor's stability number chart gives stability number
0.03x20x25
1.5,
2.
Fo A^
xyxH
0.047 an
Therefore,
0.03.
-233
1.5x20x25
25°, we have from Fig. 10.16, tan0'
tan 15°
0.26
ta n0
tan 13
0.231
0.047 =
1.16
E x a m p l e 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 w i t h respect to cohesion as well as that with respect to friction is 1.5. Determine he safe height if the desired slope is 2.
horizontal to
vertical.
The safe slope angle if the desired height is 50 ft.
Solution
tan 1.
0'
Fo
tan 18°
0.325,
= 12.23° and (3 Therefore, 0.055
tan
0.325
12.23°
26.6° (i.e., 2 hor izon tal and 1 vertical) the chart gives
20
c' yH
1.5
21
xH
0.055
Stability of Slopes
39
Therefore,
Now, Fo
10.12
=
42 safe
1.5x121x0.055
42 yH
0.046 an
1.5x121x50
0'
42 ft
0.046
12.23°, slope angle
23.5
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 m i n u s th depth crack as shown in Fig. 10.18.
S T A B I L I T Y A N A L Y S I S B Y M E T HO D O F S L IC E S F O R STEADY SEEPAGE 10.13
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
Figure 10.18 Tension crack in dams
built
cohesive soils
394
Chapter 10 Trial circle radius trial circle/' /s side 'R
Phreatic line Piezometer Pressure head at p o i n t a - h Discharge face
Equipotential line
---- -'
Pore pressure head diagram -/ Figure 10.19
equal to w h i c h is
Determination of pore pressure with steady seepage
A number of points obtained in the same manner as pore pressure head diagram.
give the curved line through
Method of Analysis (graphical m e t h o d )
Figure 10.20(a) shows the section of a dam w i t h 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 s h o w n 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 e f f e c t i v e 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 w i d t h . 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 w e i g h t W
th slice.
2.
The normal and tangential compo nents of the we ight They are designated resp ectively a an
acting on the base of the slice.
3. 4.
The pore w ater pressure acting on the base of the slice. 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. both sides In th conv ention al slice method a n a l y s i s th lateral forces re assumed equal of the slice. This assu mptio n is not strictly correct. The error due to this assum ption on the mass as w h o l e is a b o u t percent (Bishop, 1955)
Stability
Slopes
395
(a) Total norm al and tangential components
~--^
Trial failure surface «,/,
Pore-pressure diagram
(b Pore-pressure diagram
(c) Resisting forces on the base of slice Figure 10.20
(d) Graphical representation of all the forces
Stability analysis of slope by the method of slices
39
C h a p t e r 10
forces that re actually considered in the a n a l y s i s re s h o w n in Fig. 10.20(c). c o m p o n e n t s m a y b e d e t e r m i n e d a s fo l l o w s : The weight,
various
of a slice per u n i t l e n g t h o f d a m m a y b e c o m p u t e d from
W=yhb
where,
total u n i t w e i g h t
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 v e c t o r p a s s i n g t h r o u g h 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 c o n s t r u c t i n g t r i a n g l e A B C , th w e i g h t can be resolved into n o r m a l c o m p o n e n t a tangential component Sim ilar triangles can be constructed for all slices. The tan gential c o m p o n e n t s of the weights cause th m a s s to s l id e d o w n w a r d . The sum of all the weights cause the mass_ t o s l id e d o w n w a r d . T h e s u m o f a l l t h e ta n g e n t i a l c o m p o n e n t s m a y b e I.T. If the trial surface is c u r v e d u p w a r d n e a r it lower end, th tangential expressed c o m p o n e n t of the w e i g h t of the slice w ill act in the opposite direction along th c u r v e . algebraic sum of s h o u l d be considered.
3.
The average pore pressure acting on the base of any slice of length m a y b e f o u n d from on the base of th pore p r e s s u r e d i a g r a m s h o w n in Fig. 10.20(b) total pore pressure, ny slice is U=ul
e f f e c t iv e n o r m a l p r e s s u r e N' a c t i n g on the base of any slice is
N'=N- t / [ F i g . 10.20(c)] 5.
The frictional force m o v e d o w n w a r d is (N
acting on the base of any slice resisting the tendency of the slice to
tan
w h e r e 0' is the effective a n g l e friction. Similarly th cohesive force movement of the slice and acting at the base of the slice is where slice is
is the effective u n i t cohesion.
=C
c'
total resisting force
opposing th
acting on the base of the
ta
(N
Figure 10.20(c) s h o w s 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
w h e r e £/
ta
I(W- /)
c'L + ta
0'
-
l e n g t h of the curved surface.
The moments of the actuating and resisting forces about the point of rotation may be written as follows: Actuating moment
R~LT
Resisting moment
R[c'L + tan
factor
safety
£(jV
m a y n o w b e w r i t t e n as (10.29)
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 normal to 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
(10.30)
BE 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 followed for a number of trial circles until one finds the one for which th factor of safety is the lowest. This circle that gives the least is the one most likely to fail. The procedure is quite Th of may be laborious. number trial circles minimized if one follows th following method. For any given slope angle /3 (Fig. 10.21), th center of the first trial circle center may be an may be taken from determined as proposed Fellenius (1927). direction angles 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 materials of the dam and foundation ar entirely homo geneous, an practicable earth am slope ay have it critical failure 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
Figure 10.21
Curve of factor of
Location of centers of critical circle passing through toe of dam
Chapter
39
Figure 10.22
Centers
Direction angles
Table 10.1
Slope
a°
Slope angle
10
trial circles fo base failure
a ° o f o r centers Direction
critical circles
angles
0.6:
60
29
40
:
45
28
37
1.5:
33.8
26
35
:
26.6
25
35
:
18.3
25
35
5 :
11.3
25
37
Erect vertical at the m i d p o i n t 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 e m b a n k m e n t and the found ation. After th first trial circle been analyzed th center is some wh at move to the left, th radius sho rtened and a new trial circle drawn analyzed. Ad ditional center for the circles re spotted analyzed. Example 10.11
e m b a n k m e n t is to be m a d e f sandy clay having cohesion of 30 k N / m , angle internal friction of 20° and a unit weight of 18 k N / m . The slope and height of the emb ank ent are 1.6 : respectively. Determ ine th factor safety u s i n g 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 35° from Table 10.1. The soil mass above the slip circle is divided into 13 = 26° an W= slices of 2 m width each. he w e i g h t each slice er u n i t length of e m b a n k m e n t is given by where average heigh of the slice, b w i d t h 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
ay be resolved into normal components
Ex. 10.11
and tangential components
The values of
and
for the various slices are given below in a tabular form. Values of
(m
(m
(m
1.8
0.80
5.5
Slice
and
al
Slice No.
/? (m)
(m)
(m)
1.72
9.3
9.25
1.00
3.21
4.50
8.2
8.20
-0.20
5.75
5.30
10
6.8
6.82
-0.80
9.5
7.82
5.50
11
5.2
5.26
-1.30
10.6
9.62
4.82
12
3.21
-1.20
11.0
10.43
3.72
13
1.0
-0.50
10.2
10.20
2.31
The sum of these components
and
.3
1.1
may be converted into forces ZN and Irrespectively
by multipl ying them as given belo
Therefore,
Sfc
81.57m,
ZN
81.57 x 2 x 18 = 2937 kN 24.87 x2x
Length of arc Factor of safety
Ui
18
24.87m
895kN
L = 31.8
'L +
tonfiZN
30x31.8
0.364x2937
89
2.26
Chapter
10.14
BISH OP'S SIMPLIFIED M E T H O D
SLICES
Bishop's method of slices (1955) i useful if a slope consists of several types of soil with different values method and 0 and if the pore pressures in the slope re k n o w n or can be estimated. of analysis is as follows: Figure 10.23 gives section of an earth h a v i n g sloping surface AB. ADC is an a s s u m e d 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 p o r e water pressure = ul on the failure su rface cd shear resistance acting on the base of the slice normal forces on the vertical faces and ad 6
shear forces on the vertical faces be 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.
(b)
(a
Figure 10.23
Bishop's simplified
method
analysis
Stability of Slopes
40
N=Wcos6
(10.31)
tangential component of
where
The u nit stresses on the failu re surface of length , /, may be expressed as Wcos6
normal stress,
(10.32)
sin0
shear stress,
equation fo shear strength, s, is
cr'tan^'
' where rf
c'
(cr-u)tan0'
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
where
c'
(Wcos 9
ul tan
U, the total pore pressure on the base of the slice (Fig 10.23b)
l =
as
The total resisting force and the actuating force on the failure surface Total resisting force [c7
is (10.33)
(Wcos0-M/)tan0']
Total actuating force
may be expressed
is (10.34)
Wsm0
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-
(10.36)
sin#
If the slope is not on the verge of failure (F on cd divided by shearing resistance
1), the tangential force
is equal to the
Chapter
402
c'
10
(10.37)
where, N'=N-U,andU=
l.
(10.37) into
Substituting
c'
(10.36)
solving
obtain
N\
sin<9
ta 0' sin
co
(10.38)
..
where, AT For equilibrium of the mass above the failure surface, we have by taking moments about (10.39)
Wsin0R
(10.38) i n t o
B y s u b s t i t u t i n g Eqs. (10.37)
(10.39)
solving
obtain
expression
forF
(10.40) tan ( / > ' sin
where,
(10.41)
is p r e s e n t in Eq. (10.40) factor safety both sides. q u a n t i t y AT to be evaluated by means of successive approximation . Trial values of an that satisfy the equilibrium each slice, and the c o n d i t i o n s
1.6 Note:
1.4
h e n s l o p of f a i l u r e is arc is in the s a m e q u a d r a n t a s g r o u n d s l op e .
1.2
1.
0.6
-40
co
f)
-30
-20
Figure
10.24
-10
(sin
ta
0 10 20 V a l u e s of d e g r e e s
Values
fi
(after Janbu
)/
30
40
al., 1 9 5 6 )
Stability
Slopes
(E -E
40
(r -T
=0
ar used. ay then be c o m p u t e d he first assuming an arbitrary value fo value value of ay then differs calculated by making use of Eq. (10.40). If the calculated value of appreciably from the assumed value, a second trial is made and the computation is repeated. Figure 10.24 developed b y Janb u et al. (1956) helps to simplify the compu tation procedure tan0'= 0. But It is reported that an error about percent will occur if we assume Z ( T j 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
(10.42)
yh
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 10.25 shows th various parameters used in th analysis. he factor of safety is defined as (10.43)
nr,.
w h e r e , m, n = stability coefficients. an values may be obtained either from charts in Figs. B. 1 to B.6 or Tables B1 to B6 DIH, in Ap pendix B . The depth factor given in the charts or tables is as per Eq. (10.25), th at is firm stratum from the top of the slope. Bishop an w h e r e H height slope, an D depth Morgenstern (1960) limited their charts (o tables) to values c'ly H equal to 0.000, 0.025, an 0.050.
Center of failure surface
Failure surface
= unit weight of soil /^^^^^^^^//^f^^^
Figure
10.25
Specifications
parameters fo analysis
Bishop-Morgenstern method
Chapter 10
E x t e n s i o n of the Bishop a n d M o r g e n s t e r n S l o p e Stability C h a r t s As stated earlier, Bishop an M o r g e n s t e r n (1960) charts tables cover values c'lyH equal to 0.000, 0.025, 0.050 only. These charts do not cover th values that ar n o r m a l l y e n c o u n t e re d in natural slopes. C o n n o r an Mitchell (1977) extended th work Bishop an Morgenstern to cover values depth factors method c'lyH e q u a l to 0.075 an 0.100 fo various values employed is essentially th same that adopted by the earlier authors. extended values ar c h a r t s an tables from Figs. B.7 to B.14 an Tables B7 to 14 respectively in given in the form Appendix B. Method 1.
Determining
Obtain th v a l u e s From the tables in 0 and /3 and for
an
clyH
ppend ix B, obtain the values of 1, 1.25 1.5.
3.
Using Eq. (10.43), determ ine
4.
The required value of
nd
for the known values
ofc/yH,
for each value of
is the lowest of the values obtained in step 3.
E x a m p l e 10.12 Figure soil parameters are: 10.12 gives typical section f homogeneous earth dam. lb/ft The dam has a slope :1 and a pore pressure ratio 30°, c' = 590 lb/ft an y by Bishop and Morgenstern method for a height of dam 0.5. Estimate the factor of safety #=140 Solution
Height of
H=
140ft 0.035
120x140 G i v e n : 0' Since c'lyH 0.025 an 0.05 an
30°, slope 4:1 and
0.5.
0.035, and 140 ft, the 1.43 fo H b e t w e e n 1.0 and 1.5. e q u a t i o n fo
for the dam lies between c'lyH is
m-nr
Using the Tables in c'lyH, 0, and
ppen dix B, the follow ing table can be prepared for the given values of
0'=30° c' = 5 9 0 p s f 120 pcf /•
= 200 ft
=0.50 A l l u v i u m (same properties as above) Figure Ex. 10.12
Stability
Slopes
From Tables B2 and B3 for c'/yH =0.025
1.0
2.873
2.622
1.562
1.25
2.953
2.806
1.55
From Table B4, B5 nd B6 for c'ljH
Lowest
0.05 F,
1.0
3.261
2.693
1.915
1.25
3.221
2.819
1.812
1.50
3.443
3.120
1.883
Lowest
Hence 1.25 is the more critical depth factor. The value of fo c'lyH 0.035 lies between 1.55 (for c'lyH proportion F 1.655. 0.025) nd 1.812 (for c'lyH 0.05).
1 0 .1 6
MORGEN STERN METHOD
DRAWDOWN
ANALYSIS
RAPID
CONDITION
Rapid drawdown reservoir water level is one of the critical states in the design earth dams. Morgenstern (1963) developed th method a n a l y s i s fo rapid drawdown conditions based on the Bishop and Morgenstern method of slices. The purpose of this method is to compute the factor safety during rapid dra wdow n, which is reduced under no dissipation of pore w ater pressure. The assumptions made in the analysis are 1.
Simple slope of homogeneous material
2.
The dam rests on an impermeab le base
3.
The slope is completely submerged initially
4.
The pore pressure does Morgenstern used
dissipate during drawdown
pore pressure parameter 5
developed
Skempton (1954) which
states (10.45) where cr, h
y
total unit weight of soil or equal to twice the unit weight of w a t e r height of soil above the lower level of water after d r a w d o w n
The charts developed ta ke into account the drawdown ratio w hich is defined as (10.46) where
drawdown ratio //
h e i g h t of d r a w d o w n height of dam (Fig. 10.26) potential sliding circles mu st
t a n g e n t to the base of the section.
C h a p t e r 10
40
F u l l reservoir level
Drawdown /level
Figure 10.26
Dam section for draw down conditions
The stability charts are given in Figs. 10.27 to 10.29 covering a range of stability numbers c'/yH from 0.0125 to 0.050. he curves developed are for the values of 0'of 20°, 30°, and 40° for different values of
PL,
0.2
0.4
0.6 _0.8
D r a w d o w n ratio (a
/H
1.0
0
0.2
0.4
0.6
Drawdown ratio
_0.8 /H
:1
\
40 30 20°
0.2
0.4
0.6
D r a w d o w n ratio
Figure 10.27
_0.8 /H
1.0
0.2
0.4
0.6
(d ft
5:
Drawdown ratio
Drawdown stability chart fo c'/yH = 1963)
0.0125
_0.8 /H
1.0
(after Morgenstern,
Stability of Slopes
407
30° 20°
40°
30° 20
0.2
0.4
0.6
Drawdown ratio (a
_0.8 /H
1.0
0
0.2
0.4
0.6 _0.8
Drawdown rati
/H
1.0
(b ft = 3:1
ft = 2:1
i* >*
40 0°
30°
30
20°
20°
0.2
0.4
0.6
Drawdow n rati
_0.8 /H
1.0
0.2
(d
(c ft = 4 : 1
Figure 10.28
0.4
0.6 _0.8
D r a w d o w n r a ti o
D r a w d o w n stability chart fo c'lyH 1963)
/H
1.0
= 5:
0.025 (after Morgenstern,
Example 10.13 It is required to estimate the m inim um factor of safety for the comp lete draw dow n of the section s h o w n in Fig. x. 10.13 (Morgenstern, 1963)
.*._./:
Water level before drawdown
Water level after drawdown
Figure
Ex.
10.13
C h a p t e r 10
40
Solution From th data given in the Fig. —
31
x.
10.13
0.025
124.8x100
0.025, 0= 3:1,
From Fig. 10.28, fo
30°, an
'
H/H =
1,
1.20
It is evident than th critical circle is tangent to the base of the dam and no other level need investigated since this would only raise th effective value of resulting in higher factor of safety.
10.17
SPENCER METHOD
ANALYSIS
Spencer (1967) developed hi analysis based on the method of slices of Fellenius (1927) Bishop (1955). The analysis is in terms of effective stress and satisfies two equations of
40
30 20
.2
0.4
0.6
D r a w d o w n ratio
_0.8 /H
0.4 0.6 _0.8 0. D r a w d o w n r a ti o /H
1.0
40 n-
0.4 0.6 0. Drawdown rati
(c ft
Figure 10.29
_0.8 /H
4:
Drawdown stability c h a r t fo c'lyH
30 ^"•^
20
0.4 0.6 _0.8 0. D r a w d o w n r a t i o /H
(d ft
1.0
5:
0.05 (after Morgenstern, 1963)
Stability
40
Slopes
e q u i l i b r i u m , th first with respect to forces and the second with respect to m om ents. The interslice factor of safety forces ar a s s u m e d to be parallel as in Fig. 10.23. is expressed as
Shear strength available Shear strength mobilized The m obilized angle of shear resistance and other factors are expressed as (10.48)
pore pressure ratio,
nn 9) '
yh
c'
Stability factor,
(10.50)
The charts developed by Spencer for diff erent values of §' an The use o f these charts will be explained with worked out exam ples.
are given in Fig. 10.30.
E x a m p l e 10.14 Find th slope corresponding to factor safety of 1.5 for an e m b a n k m e n t 100 ft h i g h in whose properties are as follows: 0. c' = 870 Ib/sq ft y= 12 Ib/ft3, < / > ' 26°,
soil
Solution (by S p e n c e r ' s Method)
yt
ta
.,
1.5x120x100
tanf
0.488
0.325
1.5
Referring to Fig. 10.30c, of 0.048 is 3:1.
= 0 . 5 , th slope corresponding to
which
stability nu mb er
E x a m p l e 10.15 hat would be the change in strength on sudden draw down for a soil element at point P wh ich is s h o w n in Fig. 10.15? equipotential line passing throug h this element represents loss water head 1.2 m . T h e saturated u nit weight of the fill is 21 k N / m Solution
The data given are show n in Fig. Ex. 10.15. B e f o r e d r a w d o w n , he stresses at p o i n t are: 9.81 x 3 + 21 x 4
%
"o
(h
h'}
9.81(3 + 4
113 k N / m
1.2) = 57 kN/m
C h a p t e r 10
0.12
4:1
3:1
:1
4:1
3:1
:1
1.5:
0.10 0.08 ^0.06
\j
0.04 0.02
4
Figure 10.30
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 Slope a ngle/?, degrees
Stability charts (after Spencer, 1 9 6 7 )
Stability of Slopes
411
Figure Therefore tf
10.15
11 - 57 = 56 k N / m
After drawdown, o=
21 x 4 = 84 kN/m
sa
(h
h'
9.81(4
S4 27.5
1.2)
27.5 k N / m
56.5 k N / m
The change in strength is zero since the effective vertical stress does not change. Note: There is no change in strength due to sudden drawdown but the direction of forces of th seepage w ater changes from an inward direction before drawdow to an outward direction after drawdown and this is the main cause for the reduction in stability.
10.18 10.1
PROBLEMS Find the critical height of an infinite slope hav ing a slope ang le of 30°. T he slope is made of stiff clay having cohesion 20 kN/m angle internal friction 20°, void ratio 0.7 and specific gravity 2.7. Consider th following cases for the analysis. (a) the soil is dry. (b the water seeps parallel to the surface of the slope.
10.2
10.3
(c) the slope is submerged. infinite slope inclination with horizontal. underlain cohesive soil having 2.72 an e 0.52. There is thin weak layer 20 ft below an parallel to the slope (c 16°). Compute the factors of safety when (a) the 52 lb/ft slope is dry, and (b) ground water flows parallel to the slope at the slope level. An infinite slope is underlain with an overconsolidated clay hav ing c'
10 lb/ft
= 8°
an he slope is inclined at an angle of 10° to the horizontal. Seepage is sat = 120 lb/ft parallel to the surface and the ground water coincides with the surface. If the slope fails parallel to the surface along plane at depth of 12 f below th slope, determine th factor of safety. 10.4
A deep cut of 10 m depth is made in sandy clay for a road. The sides of the cut make an angle of 60° with the horizontal. The shear strength parameters of the soil are c' kN/m fi 25°, and 7= 18.5 kN/m If AC is the failure plane (Fig Prob. 10.4), estimate th factor of safety of the slope.
C h a p t e r 10
412
18.5kN/m
Figure Prob. 10.4
W
1050
Figure Prob. 10.5 10.5
A 40° slope is excavated to depth of 8 m in a deep layer of saturated clay ha ving strength parameters c 6 0 k N / m , 0 = 0, and y= 1 9 k N / m . Determine the factor of safety for the trial failure surface show n in Fig. Prob. 10.5.
10.6
excavation to depth of 8 m with slope of 1:1 clay having = 0. Determine the = 65 k N / m an passing thro ugh the toe of the cut and having a center weight of the saturated clay is 19 k N / m No tension
10.7
A 45° cut was m a d e in clayey silt with c 15 kN/m and y = 19.5 k N / m Site exploration revealed the presence of a soft clay stratum of 2 m thick having c 2 5 k N / m and 0 = 0 as shown in Fig. Prob. 10.7. Estimate the factor of safety of the slope for the assumed failure surface.
10.8
A cut was m ade in a homogeneous clay soil to a depth of 8 m as shown in Fig. Prob. 10.8. he total unit weight of the soil is 18 k N / m and its cohesive strength is 25 k N / m
as m a d e in deep layer of saturated factor of safety for a trial slip circle as show n in Fig. Prob. 10.6. The unit crack correction is required.
Stability of Slopes
Assuming
413
10.6
Figure Prob.
10.7
= 0 condition, determine the factor of safety with respect to a slip circle passing
through the toe. Consider
10.9
Figure Prob.
tension crack at the end of the slip circle on the top of the cut.
A deep cut of 10 m depth is made in natural soil for the construction of a road. The soil parameters are:
c' = 35
kN/m
= 15° and 7= 20 kN/m
Figure Prob.
10.8
Chapter
414
Figure Prob.
10.9
The sides of the cut make angles of 45° with the horizontal. Compute the factor of safety s h o w n in Fig. Prob. 10.9. u s i n g f r i c t i o n c i r c le m e t h o d for the f a i l u r e s u r f a c e
10.10
A n e m b a n k m e n t is to b e built to a height of 50 ft at an angle of 20° w ith the horizontal. The soil parameters are: c' 63 lb/ft = 18° and 7= 11 lb/ft Estimate th f o l l o w i n g ; 1. Factor of safety of the slope assuming full friction i s m o b i l i z e d .
2. Factor safety w ith respec to friction if the factor 1.5. Taylor's stability chart. 10.11
safety with respect to cohesion is
A cut w as made in natural soil for the construc tion of a railway line. The soil param eters are: c' 70 lb/ft 0' = 20° and 7= 110 lb/ft Determine the critical heigh t of the cut for a slope o f 30° with t h e h o r i z o n t a l b y m a k i n g u s e Taylo r's stability chart.
10.12
e m b a n k m e n t is to be constructed properties: c' 35 k N / m = 25° and
m a k i n g use of sandy clay having th f o l l o w i n g y= 19.5 k N / m
The he ight of the emba nkm ent is 20 m w ith a slope of 30° with the horizontal as show n in Fig. Prob. 10.12. Estimate the factor of safety by the method of slices for the trial circle s h o w n in the figure.
10.13 10.14
10.15 10.16
15°, If an e m b a n k m e n t of 10 m h e i g h t is to be m a d e from soil having c' 25 kN/m an 7 = 1 8 k N / m , w hat will be the safe angle of slope for a factor of safety of 1.5? e m b a r k m e n t is constructed for an earth dam of 80 ft h i g h at slope 3:1. properties of the soil used for the construction are: c 30°, 77 lb/ft estimated pore p ressuer ratio = 0 . 5 . D e t e r m i n e th factor safety 7 = 1 1 0 lb/ft Bishop Morgenstern method. safety fo 0' 20°. All the other data remain th For the Prob. 10.14, estimate th factor same.
For the Prob. 10.14, estimate th factor r e m a i n th s a m e .
safety for a slope of 2:1 w i t h all the oother data
Stability of Slopes
415
Figure Prob.
10.12
10.17
A cut of 25 m dopth is made in a compacted fill having shear strength parameters of = 25 kN/m an = 20°. The total unit weight of the material is 19 kN/m Th pore pressu er ratio has an average v alue of 0.3. The slope of the sides is 3:1. Estimate the factor of safety u s i n g th Bishop an Morgenstern method.
10.18
For th Prob. 10.17, estimate th factor of safety fo 0'= 30°, with all the other data remain th same.
10.19
For the Prob. 10.17, esatimate the factor of safety for a slope of 2: with all the other data remaining the same.
10.20
Estimate the minim um factor of safety for a complete draw dow n condition for the section of dam in Fig. Prob. 10.20. The full reservoir level of 15 m depth is reduced to zero after drawdown.
10.21
What is the safety factor if the reservoir level is brought down from 15 m to 5 m depth in th Prob. 10.20?
10.22
earth dam to be constructed at site has the following soil parameters: 20°. The height of of dam H 50 ft. 110 lb/ft an
c'
60
lb/ft
The pore pressure ratio 0.5. Determine the slope of the dam for a factor of safety of 1.5 using Spencer's method (1967).
c'
15 kN/m
'
y = 20 kN/m
Figure Prob. 10.20
C h a p t e r 10
416
45 ft
R
15ft
Figure Prob. 10.24 10.23
If the given pore pressure ratio is 0.25 in Prob. 10.22, wha t will be the slope of the dam?
10.24
An emb ankm ent has a slope of 1 .5 horizon tal to 1 vertical wit h a height of 25 feet. The soil parameters are: lb/ft
60
0'
20°,
and 7= 11
lb/ft
Determine th factor of safety using friction circle method for the failure surface in Fig. Prob. 10.24. 10.25
It is required to construct an emba nkme nt for a reservoir to a height of 20 2 horizontal to 1 vertical. The soil parameters are: = 4
kN/m
18°,
and 7=
shown
at a slope of
17.5 kN/m
Estimate the following: 1. Factor of safety of the slope assuming full friction is mobilized.
Factor of safety with respect to friction if the factor 1.5.
safety with respect to cohesion is
Use Taylor's st abilit y chart. 10.26
cutting of 40 ft depth is to be made for a road as shown in Fig. Prob. properties are: c'
= 500
15°,
lb/ft
and 7=
he soil
115 lb/ft
Estimate th factor of safety by the method 10.27
10.26.
slices for the trial circle show
in the figure.
An earth dam is to be construc ted for a reservior. The height of the dam is 60 ft. The properties of the soil used in the construction are: 00 lb/ft
0° =
20°, and 7= 11
lb/ft
an
ft
2:1.
Estimate th m i n i m u m f a c to r safety for the complete drawn from th full reservior level as shown in Fig. Prob. 10.27 by Morgenstern method. 10.28
W h a t is the factor of safety if the water level is brought down from 60 ft to 20 ft above th bed level of reservoir in Prob. 10.27?
Stability of Slopes
417
c' = 5001b/ft 0'=15°
y=1151b/ft
Figure Prob.
10.26
Full reservoir level
Figure Prob.
10.27
10.29 For the dam given in Prob. 10.27, determine th factor of safety for method.
= 0.5 by Spencer's
