Chapter 10 Slope Stability

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



,, 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 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 ' 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