Chapter 10 Slope Stability h4kn

Figure 10.12

Resultant of intergranular forces

magnitude of K depends upon the type of intergranular pressure distribution along the arc. The most probable form of distribution is the sinusoidal distribution. The variation of K with respect to the central angle a'is shown in Fig. 10.13. The figure also gives relationships between of and K 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, C and P of Fig. 10.14 to be in equilibrium, P must through the intersection of

1.20

ox 1.16

£

1.12

J Cent ral angle

71

/ /

/ For ianifo rm

str essc istrih>utiori —S / /

1.08 j/ /

1.04

/ s<

/ 1.00

20^

^40

/

' /

y

/

/

/

For sinus oida

s tress

distributi(^n

^ 60

80

100

120

Central angle a ' in degrees Figure 10.13

Relationship between K and central angle a'

386

Chapter 10

Figure 10.14

Force triangle for the friction-circle method

the known lines of action of vectors Q and C. The line of action of vector P must also be tangent to the circle of radius KR sin 0' . The value of K may be estimated by the use of curves given in Fig. 10.13, and the line of action offeree P may be drawn as shown in Fig. 10.14. Since the lines of action of all three forces and the magnitude of force Q are known, the magnitude of P and C may-be obtained by the force parallelogram construction that is indicated in the figure. The circle of radius of KR sin 0'rn is called the modified j jfriction circle. T

Determination of Factor of Safety With Respect to Strength Figure 10.15(a) is a section of a dam. AB is the trial failure arc. The force Q, the resultant of W and U is drawn as explained earlier. The line of action of C is also drawn. Let the forces Q and C

D (a) Friction circle

Figure 10.15

(b) Factor of safety

Graphical method of determining factor of safety with respect to strength

Stability of Slopes

387

meet at point D. An arbitrary first trial using any reasonable $m value, which will be designated by 0'ml is given by the use of circle 1 or radius KR sin <j)'ml. Subscript 1 is used for all other quantities of the first trial. The force Pl is then drawn through D tangent to circle 1. Cl is parallel to chord and point 1 is the intersection of forces C{ and Pr The mobilized cohesion is equal c'm]Lc. From this the mobilized cohesion c'ml is evaluated. The factors of safety with respect to cohesion and friction are determined from the expressions c' F' = ——, and F*,

tanfl'

These factors are the values used to plot point 1 in the graph in Fig. 10.15(b). Similarly other friction circles with radii KR sin <j/m2, KR sin 0'm3. etc. may be drawn and the procedure repeated. Points 2, 3 etc. are obtained as shown in Fig. 10.15(b). The 45° line, representing Fc = F., intersects the curve to give the factor of safety Fs for this trial circle. Several trial circles must be investigated in order to locate the critical circle, which is the one having the minimum value of F5. Example 10.7 An embankment has a slope of 2 (horizontal) to 1 (vertical) with a height of 10 m. It is made of a soil having a cohesion of 30 kN/m2, an angle of internal friction of 5° and a unit weight of 20 kN/m3. Consider any slip circle ing 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 EFB be the slope and AKB be the slip circle drawn with center O and radius R = 20 m. Length of chord AB = Lc = 27 m Take J as the midpoint of AB, then Area AKBFEA = area AKBJA + area ABEA = -ABxJK + -ABxEL 3 2 = - x 27 x 5.3 + - x 27 x 2.0 = 122.4 m2 3 2 Therefore the weight of the soil mass = 122.4 x 1 x 20 = 2448 kN

It will act through point G, the centroid of the mass which can be taken as the mid point of FK.

Now, 0=85°, 314 Length of arc AKB = L = RO = 20 x 85 x — = 29.7 m 6 180

L 29.7 Moment arm of cohesion, / = R— = 20 x —— = 22 m L

c

21

From center O, at a distance /fl, draw the cohesive force vector C, which is parallel to the chord AB. Now from the point of intersection of C and W, draw a line tangent to the friction circle

Chapter 10

388

1.74m

//=10m

Figure Ex. 10.7

drawn at 0 with a radius of R sin 0' = 20 sin 5° = 1 .74 m. This line is the line of action of the third force F. Draw a triangle of forces in which the magnitude and the direction for W is known and only the directions of the other two forces C and F are known. Length ad gives the cohesive force C = 520 kN Mobilized cohesion, c'm

= - = — = 17.51 kN/m 2 L 29.7

Therefore the factor of safety with respect to cohesion, Fc, is F =11 = ^=1.713 FC will be 1 .7 13 if the factor of safety with respect to friction, F^ - 1 .0 If, F = 1.5, then 0' =

tan5 c = 0.058 rad; or 0' = 3.34° F.

Stability of Slopes

389

The new radius of the friction circle is

r{ = R sin 0'm = 20 x sin 3.3° = 1.16 m. The direction of F changes and the modified triangle of force abd' gives, cohesive force = C = length ad' = 600 kN

C 600 Mobilised cohesino, c'm = ~— - 20.2 kN/mr LJ

c'

Z*yI /

30

Therefore, Fc = — = = 1.5 c' 20.2

10.1 1

TAYLOR'S STABILITY NUMBER

If the slope angle j8, height of embankment H, the effective unit weight of material y, angle of internal friction ', and unit cohesion c' are known, the 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 a wide range of slope angles and angles of internal friction, and then representing the results by an abstract number which he called the "stability number". This number is designated as A^. The expression used is

From this the factor of safety with respect to cohesion may be expressed as

F

-=7

<10-24>

Taylor published his results in the form of curves which give the relationship between Ns and the slope angles /? for various values of 0' as shown in Fig. 10.16. These curves are for circles ing through the toe, although for values of 13 less than 53°, it has been found that the most dangerous circle es 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 (FJ 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 Ns 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 of a depth factor nd which is defined as » r f =;|

(10-25)

where D - depth of ledge below the top of the embankment, H = height of slope above the toe. For various values of nd and for the 0 = 0 case the chart in Fig. 10.17 gives the stability number NS for various values of slope angle ft. In this case the rupture circle may through the toe or below the toe. The distance jc of the rupture circle from the toe at the toe level may be expressed by a distance factor n which is defined as

Stability number, Ns

•a C CD _j O

O) 0)

o"

CD

H° |_cu

V)*

o cra

Q) CT

-* CD

<° o ^^

(Q C

Stability number, N,.

Q) -. <.

Q) O

—f> -i >

r-t-

CD cn ~" w

—J Q)

r-+ 0)

<. E

o ^~ CD C

co 3 ->J cr

'** CD -^

cn

oee

Stability of Slopes

391

The chart in Fig. 10.17 shows the relationship between nd and nx. If there is a ledge or other stronger material at the elevation of the toe, the depth factor nd for this case is unity. Factor of Safety with Respect to Strength The development of the stability number is based on the assumption that the factor of safety with respect to friction F,, is unity. The curves give directly the factor of safety Fc with respect to cohesion only. If a true factor of safety Fs with respect to strength is required, this factor should apply equally to both cohesion and friction. The mobilized shear strength may therefore be expressed as

s

c'

a' tan (/)'

In the above expression, we may write — = c'm,

tan (f>'m = —=— ,

S

or #, = — (approx.)

5

(10.27)

S

c'm and tf m may be described as average values of mobilized cohesion and friction respectively.

Example 10.8 The following particulars are given for an earth dam of height 39 ft. The slope is submerged and the slope angle j3 = 45°. Yb = 69 lb/ft3 c' = 550 lb/ft2

0' = 20° Determine the factor of safety FS. Solution Assume as a first trial Fs = 2.0

20 'm = Y = 10° (approx.) For (j)'m = 10°, and (3 = 45° the value of Ns from Fig. 10.16 is 0.1 1, we may write c' From Eq. (10.23) N = - , substituting

55Q 2x69x# or H =

— =36.23 ft 2x69x0.11

20 19

If F5 = 1.9, $ = — = 10.53° and N = 0.105

*

392

Chapter 10

1.9x69x0.105

.40ft

The computed height 40 ft is almost equal to the given height 39 ft. The computed factor of safety is therefore 1 .9.

Example 10.9 An excavation is to be made in a soil deposit with a slope of 25° to the horizontal and to a depth of 25 meters. The soil has the following properties: c'= 35kN/m2, 0' = 15° and 7= 20 kN/m3 1 . Determine the 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 of safety with respect to friction? Solution 1 . For 0' = 15° and (3 = 25°, Taylor's stability number chart gives stability number Ns = 0.03.

0.03x20x25 2.

-233

For F = 1.5, JN = --- - -—- = 0.047 FcxyxH 1.5x20x25 For A^ = 0.047 and (3 = 25°, we have from Fig. 10.16, 0'm = 13 tan0' tan 15° 0.268 Therefore, F, = -— = - = - = 1.16 0 tan0 tan 13° 0.231

Example 10.10 An embankment is to be made from a soil having c' = 420 lb/ft2, 0' = 18° and y= 121 lb/ft3. The desired factor of safety with respect to cohesion as well as that with respect to friction is 1.5. Determine 1 . The safe height if the desired slope is 2 horizontal to 1 vertical. 2. The safe slope angle if the desired height is 50 ft.

Solution , 0.325 tan 0' = tan 18° = 0.325, 0'm - tan ' — - = 12.23° 1.

For 0' = 12.23° and (3 = 26.6° (i.e., 2 horizontal and 1 vertical) the chart gives Ns = 0.055 Therefore, 0.055 =

c' FcyH

420 1.5 x 121 xH

393

Stability of Slopes

Therefore, #safe. =

2.

Now, NS = • FcyH

420 = 42 ft 1.5x121x0.055

420 = 0.046 1.5x121x50

For N = 0.046 and 0' = 12.23°, slope angle P = 23.5C

10.12

TENSION CRACKS

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)

r

where z0 = depth of crack, c' = unit cohesion, y = unit weight of soil. The effective length of any trial arc of failure is the difference between the total length of arc minus the depth of crack as shown in Fig. 10.18.

10.13 STABILITY ANALYSIS BY METHOD OF SLICES FOR STEADY SEEPAGE The stability analysis with steady seepage involves the development of the pore pressure head diagram along the chosen trial circle of failure. The simplest of the methods for knowing 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 with 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 may be obtained at any point on the section. For example at point a in Fig. 10.19 the pressure head is h. Point c is determined by setting the radial distance ac

Tension crack

Effective length of

trial arc of failure

Figure 10.18 Tension crack in dams built of cohesive soils

394

Chapter 10

Trial circle - ' 'R = radius / of trial circle/' d/s side / Phreatic line Piezometer Pressure head at point a - h Discharge face

\- Equipotential line

x

r ---- -'

Pore pressure head diagram -/ Figure 10.19

Determination of pore pressure with steady seepage

equal to h. A number of points obtained in the same manner as c give the curved line through c which is a pore pressure head diagram. Method of Analysis (graphical method) Figure 10.20(a) shows the section of a dam with an arbitrarily chosen trial arc. The center of rotation of the arc is 0. The pore pressure acting on the base of the arc as obtained from flow nets is shown in Fig. 10.20(b). When the soil forming the slope has to be analyzed under a condition where full or partial drainage takes place the analysis must take into both cohesive and frictional soil properties based on effective stresses. Since the effective stress acting across each elemental length of the assumed circular arc failure surface must be computed in this case, the method of slices is one of the convenient methods for this purpose. The method of analysis is as follows. The soil mass above the assumed slip circle is divided into a number of vertical slices of equal width. The number of slices may be limited to a maximum of eight to ten to facilitate computation. The forces used in the analysis acting on the slices are shown in Figs. 10.20(a) and (c). The forces are: 1 . The weight W of the slice. 2. The normal and tangential components of the weight W acting on the base of the slice. They are designated respectively as N and T. 3. The pore water pressure U acting on the base of the slice. 4. The effective frictional and cohesive resistances acting on the base of the slice which is designated as S. The forces acting on the sides of the slices are statically indeterminate as they depend on the stress deformation properties of the material, and we can make only gross assumptions about their relative magnitudes. In the conventional slice method of analysis the lateral forces are assumed equal on both sides of the slice. This assumption is not strictly correct. The error due to this assumption on the mass as a whole is about 15 percent (Bishop, 1955).

Stability of Slopes

395

(a) Total normal and tangential components

B ~--^ C

Trial failure surface

f\l

/ 7"

U} = «,/,

Pore-pressure diagram U2 = M2/2 U3 = M3/3

(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

396

Chapter 10

The forces that are actually considered in the analysis are shown in Fig. 10.20(c). The various components may be determined as follows: 1 . The weight, W, of a slice per unit length of dam may be computed from W=yhb where, y = total unit weight of soil, h = average height of slice, b - width of slice. If the widths of all slices are equal, and if the whole mass is homogeneous, the weight W can be plotted as a vector AB ing through the center of a slice as in Fig. 10.20(a). AB may be made equal to the height of the slice. 2. By constructing triangle ABC, the weight can be resolved into a normal component N and a tangential component T. Similar triangles can be constructed for all slices. The tangential components of the weights cause the mass to slide downward. The sum of all the weights cause the mass_ to slide downward. The sum of all the tangential components may be expressed as T= I.T. If the trial surface is curved upward near its lower end, the tangential component of the weight of the slice will act in the opposite direction along the curve. The algebraic sum of T should be considered. 3. The average pore pressure u acting on the base of any slice of length / may be found from the pore pressure diagram shown in Fig. 10.20(b). The total pore pressure, U, on the base of any slice is U=ul 4. The effective normal pressure N' acting on the base of any slice is N'=N- t/[Fig. 10.20(c)] 5. The frictional force Ff acting on the base of any slice resisting the tendency of the slice to move downward is

F = (N - U) tan 0' where 0' is the effective angle of friction. Similarly the cohesive force C" opposing the movement of the slice and acting at the base of the slice is where c is the effective unit cohesion. The total resisting force S acting on the base of the slice is

S = C + F' = c'l + (N - U) tan 0' Figure 10.20(c) shows the resisting forces acting on the base of a slice. The sum of all the resisting forces acting on the base of each slice may be expressed as Ss = c'I,l + tan 0' I(W- £/) = c'L + tan 0' X(N - U) where £/ = L = length 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 0' £(jV - U)] The factor of safety F? may now be written as (10.29)

397

Stability of Slopes

The various components shown in Eq. (10.29) can easily be represented graphically as shown in Fig. 10.20(d). The line AB represents to a suitable scale Z,(N - U). BC is drawn normal to AB at B and equal to c'L + tan 0' Z(N - U). The line AD drawn at an angle 0'to AB gives the intercept BD on BC equal to tan 0'Z(N- U). The length BE on BC is equal to IT. Now

F =

BC BE

(10.30)

Centers for Trial Circles Through Toe

The factor of safety Fs as computed and 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 the factor of safety is the lowest. This circle that gives the least Fs is the one most likely to fail. The procedure is quite laborious. The number of trial circles may be minimized if one follows the following method. For any given slope angle /3 (Fig. 10.21), the center of the first trial circle center O may be determined as proposed by Fellenius (1927). The direction angles aA and aB may be taken from Table 10.1. For the centers of additional trial circles, the procedure is as follows: Mark point C whose position is as shown in Fig. 10.21. CO. The centers of additional circles lie on the line CO extended. This method is applicable for a homogeneous (c - ) soil. When the soil is purely cohesive and homogeneous the direction angles given in Table 10.1 directly give the center for the critical circle. Centers for Trial Circles Below Toe

Theoretically if the materials of the dam and foundation are entirely homogeneous, any practicable earth dam slope may have its critical failure surface below the toe of the slope. Fellenius found that the angle intersected at 0 in Fig. 10.22 for this case is about 133.5°. To find the center for the critical circle below the toe, the following procedure is suggested.

Locus of centers of critical circles

Figure 10.21

Curve of factor of safety

Location of centers of critical circle ing through toe of dam

398

Chapter 10

Figure 10.22 Table 10.1

Centers of trial circles for base failure

Direction angles a°A and a°ofor centers of critical circles

Slope

0.6: 1 1 :1 1.5: 1 2: 1 3: 1 5: 1

Slope angle

60 45 33.8 26.6 18.3 11.3

Direction angles

29 28 26 25 25 25

40 37 35 35 35 37

Erect a vertical at the midpoint M of the slope. On this vertical will be the center O of the first trial circle. In locating the trial circle use an angle (133.5°) between the two radii at which the circle intersects the surface of the embankment and the foundation. After the first trial circle has been analyzed the center is some what moved to the left, the radius shortened and a new trial circle drawn and analyzed. Additional centers for the circles are spotted and analyzed. Example 10.11 An embankment is to be made of a sandy clay having a cohesion of 30 kN/m2, angle of internal friction of 20° and a unit weight of 18 kN/m3. The slope and height of the embankment are 1.6 : 1 and 10m respectively. Determine the factor of safety by using the trial circle given in Fig. Ex. 10.11 by the method of slices. Solution Consider the embankment as shown in Fig. Ex.10.11. The center of the trial circle O is selected by taking aA = 26° and aB = 35° from Table 10.1. The soil mass above the slip circle is divided into 13 slices of 2 m width each. The weight of each slice per unit length of embankment is given by W = haby;, where ha = average height of the slice, b = width of the slice, yt = unit weight of the soil. The weight of each slice may be represented by a vector of height ha if b and y, remain the same for the whole embankment. The vectors values were obtained graphically. The height vectors

Stability of Slopes

399

Figure Ex. 10.11

may be resolved into normal components hn and tangential components h{. The values of ha, hn and ht for the various slices are given below in a tabular form. Values of hoal /hvn and /?,r Slice No.

1 2 3 4 5 6 7

ha(m)

hn(m)

ht(m]

1.8 5.5 7.8 9.5 10.6 11.0 10.2

0.80 3.21 5.75 7.82 9.62 10.43 10.20

1.72 4.50 5.30 5.50 4.82 3.72 2.31

Slice No. ha(m) 8 9 10 11 12 13

9.3 8.2 6.8 5.2 3.3 1.1

hn(m)

9.25 8.20 6.82 5.26 3.21 1.0

ht(m)

1.00 -0.20 -0.80 -1.30 -1.20 -0.50

The sum of these components hn and ht may be converted into forces ZN and Irrespectively by multiplying them as given below Sfcn = 81.57m, Therefore,

Uit = 24.87m

ZN = 81.57 x 2 x 18 = 2937 kN Zr = 24.87 x2x 18 = 895kN

Length of arc = L = 31.8 m Factor of safety =

'L + tonfiZN 30x31.8 + 0.364x2937 = 2.26 895

Chapter 10

400

10.14

BISHOP'S SIMPLIFIED METHOD OF SLICES

Bishop's method of slices (1955) is useful if a slope consists of several types of soil with different values of c and 0 and if the pore pressures u in the slope are known or can be estimated. The method of analysis is as follows: Figure 10.23 gives a section of an earth dam having a sloping surface AB. ADC is an assumed trial circular failure surface with its center at O. The soil mass above the failure surface is divided into a number of slices. The forces acting on each slice are evaluated from limit equilibrium of the slices. The equilibrium of the entire mass is determined by summation 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). The forces acting on this slice are W = weight of the slice N = total normal force on the failure surface cd U = pore water pressure = ul on the failure surface cd FR = shear resistance acting on the base of the slice Er E2 - normal forces on the vertical faces be and ad Tr T2 = shear forces on the vertical faces be and ad 6 = the inclination of the failure surface cd to the horizontal The system is statically indeterminate. An approximate solution may be obtained by assuming that the resultant of £, and T^ is equal to that of E2 and T2, and their lines of action coincide. For equilibrium of the system, the following equations hold true.

O

(a)

Figure 10.23

(b)

Bishop's simplified method of analysis

Stability of Slopes

401

N=Wcos6

(10.31)

where F( = tangential component of W The unit stresses on the failure surface of length, /, may be expressed as Wcos6 normal stress,
Wsin0 rn = -

(10.32)

The equation for shear strength, s, is s = c' + cr'tan^' = c' + (cr-u)tan0' where rf = effective normal stress c' - effective cohesion (ft = effective angle of friction u = unit pore pressure The shearing resistance to sliding on the base of the slice is si = c'l + (Wcos 9 - ul) tan $ where ul = U, the total pore pressure on the base of the slice (Fig 10.23b) d -= rFR At

The total resisting force and the actuating force on the failure surface ADC may be expressed as

Total resisting force FR is FR=

[c7 + (Wcos0-M/)tan0']

(10.33)

Total actuating force Ft is Ft =

Wsm0

(10.34)

The factor of safety Fs is then given as

F

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 the forces E and Ton the vertical faces of each slice. For the element in Fig. 10.23(b), we may write an expression for all the forces acting in the vertical direction for the equilibrium condition as N' co&0 = W + (T^ -T2)-ulcos0- FR sin#

(10.36)

If the slope is not on the verge of failure (Fs > 1), the tangential force Ft is equal to the shearing resistance FR on cd divided by Fg.

402

Chapter 10

c'l (10.37)

where, N'=N-U,andU= ul. Substituting Eq. (10.37) into Eq. (10.36) and solving for N\ we obtain

c'l — sin<9 F tan 0' sin 6 cos <9 + F..

(10.38)

where, AT= T{ - Tr For equilibrium of the mass above the failure surface, we have by taking moments about O Wsin0R =

FRR

(10.39)

By substituting Eqs. (10.37) and (10.38) into Eq. (10.39) and solving we obtain an expression forF as F

(10.40) tan (/>' sin 9 F

where,

(10.41)

The factor of safety FS is present in Eq. (10.40) on both sides. The quantity AT= T{ - T2 has to be evaluated by means of successive approximation . Trial values of E^ and Tl that satisfy the equilibrium of each slice, and the conditions

1.6 1.4

i

i

i

Note: 0 is + when slope of failure arc is in the same quadrant as ground slope.

1.2 1.0

0.6

-40

mf) = cos 6 + (sin 6 tan d) )/F

-30

-20

-10

0 10 20 Values of 6 degrees

30

40

Figure 10.24 Values of mfi (after Janbu et al., 1956)

Stability of Slopes

(El-E2) = Q and

403

(r l -T 2 ) = 0

are used. The value of Fs may then be computed by first assuming an arbitrary value for Fs. The value of Fs may then be calculated by making use of Eq. (10.40). If the calculated value of Fv differs appreciably from the assumed value, a second trial is made and the computation is repeated. Figure 10.24 developed by Janbu et al. (1956) helps to simplify the computation procedure. It is reported that an error of about 1 percent will occur if we assume Z(Tj - T"2) tan0'= 0. But if we use the conventional method of analysis using Eq. (10.35) the error introduced is about 15 percent (Bishop, 1955). 10.15

BISHOP AND MORGENSTERN METHOD FOR SLOPE ANALYSIS

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 u (10.42)

yh

where, u = pore water pressure at any point on the assumed failure surface Y= unit weight of the soil h = the depth of the point in the soil mass below the ground surface The pore pressure ratio ru is assumed to be constant throughout the cross-section, which is called a homogeneous pore pressure distribution. Figure 10.25 shows the various parameters used in the analysis. The factor of safety F is defined as F_ = m - nr,.

(10.43)

where, m, n = stability coefficients. The m and n values may be obtained either from charts in Figs. B. 1 to B.6 or Tables B1 to B6 in Appendix B. The depth factor given in the charts or tables is as per Eq. (10.25), that is nd = DIH, where H = height of slope, and D = depth of firm stratum from the top of the slope. Bishop and Morgenstern (1960) limited their charts (or tables) to values of c'ly H equal to 0.000, 0.025, and 0.050.

Center of failure surface

Failure surface

y = unit weight of soil /^^^^^^^^//^f^^^

Figure 10.25 Specifications of parameters for Bishop-Morgenstern method of analysis

Chapter 10

404

Extension of the Bishop and Morgenstern Slope Stability Charts As stated earlier, Bishop and Morgenstern (1960) charts or tables cover values of c'lyH equal to 0.000, 0.025, and 0.050 only. These charts do not cover the values that are normally encountered in natural slopes. O' Connor and Mitchell (1977) extended the work of Bishop and Morgenstern to cover values of c'lyH equal to 0.075 and 0.100 for various values of depth factors nd. The method employed is essentially the same as that adopted by the earlier authors. The extended values are given in the form of charts and tables from Figs. B.7 to B.14 and Tables B7 to B14 respectively in Appendix B. Method of Determining Fs 1. Obtain the values of ru and clyH 2. From the tables in Appendix B, obtain the values of m and n for the known values ofc/yH, 0 and /3, and for nd - 0, 1, 1.25 and 1.5. 3. Using Eq. (10.43), determine Fs for each value of nd. 4. The required value of Fs is the lowest of the values obtained in step 3. Example 10.12 Figure Ex. 10.12 gives a typical section of a homogeneous earth dam. The soil parameters are: 0' = 30°, c' = 590 lb/ft2, and y = 120 lb/ft3. The dam has a slope 4:1 and a pore pressure ratio ru = 0.5. Estimate the factor of safety Fs by Bishop and Morgenstern method for a height of dam #=140 ft. Solution

Height of dam H= 140ft c'

590 = 0.035 120x140

Given: 0' = 30°, slope 4:1 and ru = 0.5. Since c'lyH = 0.035, and nd = 1.43 for H = 140 ft, the Fs for the dam lies between c'lyH 0.025 and 0.05 and nd between 1.0 and 1.5. The equation for Fs is = m-nr

Using the Tables in Appendix B, the following table can be prepared for the given values of c'lyH, 0, and /3.

0'=30° c' = 590psf y - 120 pcf /•„ =0.50

D = 200 ft

Alluvium (same properties as above) Figure Ex. 10.12

Stability of Slopes

405

From Tables B2 and B3 for c'/yH =0.025

n

d

m

n

1.0 1.25

2.873 2.953

2.622 2.806

F, 1.562 1.55

Lowest

From Table B4, B5 and B6 for c'ljH - 0.05 n

d 1.0 1.25 1.50

m

n

3.261 3.221 3.443

2.693 2.819 3.120

F, 1.915 1.812 1.883

Lowest

Hence nd = 1.25 is the more critical depth factor. The value of Fs for c'lyH = 0.035 lies between 1.55 (for c'lyH = 0.025) and 1.812 (for c'lyH = 0.05). By proportion F = 1.655.

10.16 MORGENSTERN METHOD OF ANALYSIS FOR RAPID DRAWDOWN CONDITION Rapid drawdown of reservoir water level is one of the critical states in the design of earth dams. Morgenstern (1963) developed the method of analysis for rapid drawdown conditions based on the Bishop and Morgenstern method of slices. The purpose of this method is to compute the factor of safety during rapid drawdown, which is reduced under no dissipation of pore water pressure. The assumptions made in the analysis are 1. 2. 3. 4.

Simple slope of homogeneous material The dam rests on an impermeable base The slope is completely submerged initially The pore pressure does not dissipate during drawdown Morgenstern used the pore pressure parameter 5 as developed by Skempton (1954) which

states 5=—

(10.45)

where cr, = y h j- total unit weight of soil or equal to twice the unit weight of water h = height of soil above the lower level of water after drawdown The charts developed take into the drawdown ratio which is defined as (10.46) where Rd = drawdown ratio // = height of drawdown H = height of dam (Fig. 10.26) All the potential sliding circles must be tangent to the base of the section.

406

Chapter 10 Full reservoir level

"

Drawdown /level

Figure 10.26

H

Dam section for drawdown 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. The curves developed are for the values of 0'of 20°, 30°, and 40° for different values of B.

PL,

0.2 0.4 0.6 _0.8 Drawdown ratio H/H

1.0

0

0.2 0.4 0.6 _0.8 Drawdown ratio H/H

(a) 0 = 2:1

\

40° 30° 20°

0.2 0.4 0.6 _0.8 Drawdown ratio H/H

1.0

30°

UH

20°

0

0.2 0.4 0.6 _0.8 Drawdown ratio H/H

(c) ft = 4:1

Figure 10.28

1.0

(d) 0 = 5:

Drawdown stability chart for c'lyH = 0.025 (after Morgenstern, 1963)

Example 10.13 It is required to estimate the minimum factor of safety for the complete drawdown of the section shown in Fig. Ex. 10.13 (Morgenstern, 1963)

.*._./:

Water level before drawdown

Water level after drawdown

Figure Ex. 10.13

Chapter 10

408

Solution From the data given in the Fig. Ex. 10.13

312 N =— = = 0.025 yH. 124.8x100 From Fig. 10.28, for W = 0.025, 0= 3:1, ' = 30°, and H/H = 1, Fs = 1.20

It is evident than the critical circle is tangent to the base of the dam and no other level need be investigated since this would only raise the effective value of NS resulting in a higher factor of safety.

10.17

SPENCER METHOD OF ANALYSIS

Spencer (1967) developed his analysis based on the method of slices of Fellenius (1927) and Bishop (1955). The analysis is in of effective stress and satisfies two equations of

X

40° 30° 20° 0.2 0.4 0.6 _0.8 Drawdown ratio H/H

0.2 0.4 0.6 _0.8 Drawdown ratio H/H

1.0

k

\ .

\

\l X \X n- 2

0

0.2 0.4 0.6 _0.8 Drawdown ratio H/H (c) ft = 4:1

Figure 10.29

0

\

X X

\X

xx^ — X. ^"•^ »^_

——

E^M

0.2 0.4 0.6 _0.8 Drawdown ratio H/H

40° 30° 20°

1.0

(d) ft = 5:1

Drawdown stability chart for c'lyH = 0.05 (after Morgenstern, 1963)

Stability of Slopes

409

equilibrium, the first with respect to forces and the second with respect to moments. The interslice forces are assumed to be parallel as in Fig. 10.23. The factor of safety Ff is expressed as

F5 =

Shear strength available Shear strength mobilized

CIO 47) ' '

The mobilized angle of shear resistance and other factors are expressed as (10.48)

u pore pressure ratio, r = — yh

n n 49) ^ ' '

c' Stability factor, NS=——

(10.50)

The charts developed by Spencer for different values of Ns, §'m and ru are given in Fig. 10.30. The use of these charts will be explained with worked out examples.

Example 10.14 Find the slope corresponding to a factor of safety of 1.5 for an embankment 100 ft high in a soil whose properties are as follows: c' = 870 Ib/sq ft, y= 120 Ib/ft3, ' = 26°, ru = 0.5 Solution (by Spencer's Method) N = ^L = 5 Fsytl

870 1.5x120x100

., tanf 0.488 _ „ _ tan 0 = -— = - = 0.325 F 1.5 t

Referring to Fig. 10.30c, for which r =0.5, the slope corresponding to a stability number of 0.048 is 3:1.

Example 10.15 What would be the change in strength on sudden drawdown for a soil element at point P which is shown in Fig. Ex. 10.15? The equipotential line ing through this element represents loss of water head of 1.2 m. The saturated unit weight of the fill is 21 kN/m3. Solution

The data given are shown in Fig. Ex. 10.15. Before drawdown, The stresses at point P are: % = /A + nA = 9.81 x 3 + 21 x 4 = 113 kN/m2 "o = Yw (hw + hc- h'} = 9.81(3 + 4 - 1.2) = 57 kN/m2

Chapter 10

410

0.12

4:1

3:1

2:1

4:1

3:1

2:1

1.5:

0.10 0.08

?L

^0.06 \j

0.04 0.02

2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 Slope angle/?, degrees

Figure 10.30

Stability charts (after Spencer, 1967)

411

Stability of Slopes

Figure Ex. 10.15 Therefore tf0 = (JQ - UQ = 113 - 57 = 56 kN/m2 After drawdown, o= ysathc = 21 x 4 = 84 kN/m2 u = yw (hc - h'} = 9.81(4 - 1.2) = 27.5 kN/m2 of = a-u = S4-27.5 = 56.5 kN/m2 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 the seepage water changes from an inward direction before drawdown to an outward direction after drawdown and this is the main cause for the reduction in stability.

10.18

PROBLEMS

10.1 Find the critical height of an infinite slope having a slope angle of 30°. The slope is made of stiff clay having a cohesion 20 kN/m2, angle of internal friction 20°, void ratio 0.7 and specific gravity 2.7. Consider the following cases for the analysis. (a) the soil is dry. (b) the water seeps parallel to the surface of the slope. (c) the slope is submerged. 10.2 An infinite slope has an inclination of 26° with the horizontal. It is underlain by a firm cohesive soil having Gs = 2.72 and e = 0.52. There is a thin weak layer 20 ft below and parallel to the slope (c' - 525 lb/ft2, 0' = 16°). Compute the factors of safety when (a) the slope is dry, and (b) ground water flows parallel to the slope at the slope level. 10.3 An infinite slope is underlain with an overconsolidated clay having c' - 210 lb/ft2, 0' = 8° and ysat = 120 lb/ft3. The slope is inclined at an angle of 10° to the horizontal. Seepage is parallel to the surface and the ground water coincides with the surface. If the slope fails parallel to the surface along a plane at a depth of 12 ft below the slope, determine the 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' - 20 kN/m2, fi = 25°, and 7= 18.5 kN/m3. If AC is the failure plane (Fig Prob. 10.4), estimate the factor of safety of the slope.

412

Chapter 10

y = 18.5kN/m 3

Figure Prob. 10.4

W = 1050 kN

Figure Prob. 10.5 10.5

A 40° slope is excavated to a depth of 8 m in a deep layer of saturated clay having strength parameters c = 60 kN/m2, 0 = 0, and y= 19 kN/m3. Determine the factor of safety for the trial failure surface shown in Fig. Prob. 10.5. 10.6 An excavation to a depth of 8 m with a slope of 1:1 was made in a deep layer of saturated clay having cu = 65 kN/m 2 and 0M = 0. Determine the factor of safety for a trial slip circle ing through the toe of the cut and having a center as shown in Fig. Prob. 10.6. The unit weight of the saturated clay is 19 kN/m3. No tension crack correction is required. 10.7 A 45° cut was made in a clayey silt with c = 15 kN/m2, 0 = 0 and y = 19.5 kN/m3. Site exploration revealed the presence of a soft clay stratum of 2 m thick having c = 25 kN/m2 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 made in a homogeneous clay soil to a depth of 8 m as shown in Fig. Prob. 10.8. The total unit weight of the soil is 18 kN/m3, and its cohesive strength is 25 kN/m2.

Stability of Slopes

413

<§)

Figure Prob. 10.6

Figure Prob. 10.7

10.9

Assuming a 0 = 0 condition, determine the factor of safety with respect to a slip circle ing through the toe. Consider a 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/m2, 0' = 15° and 7= 20 kN/m3.

Figure Prob. 10.8

414

Chapter 10

Figure Prob.

10.10

10.11

10.12

10.13 10.14

10.15 10.16

10.9

The sides of the cut make angles of 45° with the horizontal. Compute the factor of safety using friction circle method for the failure surface AC shown in Fig. Prob. 10.9. An embankment is to be built to a height of 50 ft at an angle of 20° with the horizontal. The soil parameters are: c' - 630 lb/ft2, 0' = 18° and 7= 115 lb/ft3. Estimate the following; 1. Factor of safety of the slope assuming full friction is mobilized. 2. Factor of safety with respect to friction if the factor of safety with respect to cohesion is 1.5. Use Taylor's stability chart. A cut was made in natural soil for the construction of a railway line. The soil parameters are: c' = 700 lb/ft 2 , 0' = 20° and 7= 110 lb/ft3. Determine the critical height of the cut for a slope of 30° with the horizontal by making use of Taylor's stability chart. An embankment is to be constructed by making use of sandy clay having the following properties: c' = 35 kN/m2, 0' = 25° and y= 19.5 kN/m3. The height of the embankment is 20 m with a slope of 30° with the horizontal as shown in Fig. Prob. 10.12. Estimate the factor of safety by the method of slices for the trial circle shown in the figure. If an embankment of 10 m height is to be made from a soil having c' = 25 kN/m2, 0' = 15°, and 7=18 kN/m3, what will be the safe angle of slope for a factor of safety of 1.5? An embarkment is constructed for an earth dam of 80 ft high at a slope of 3:1. The properties of the soil used for the construction are: c - 770 lb/ft2, 0' = 30°, and 7=110 lb/ft3. The estimated pore pressuer ratio r =0.5. Determine the factor of safety by Bishop and Morgenstern method. For the Prob. 10.14, estimate the factor of safety for 0' = 20°. All the other data remain the same. For the Prob. 10.14, estimate the factor of safety for a slope of 2:1 with all the oother data remain the same.

415

Stability of Slopes

Figure Prob. 10.12 10.17 A cut of 25 m dopth is made in a compacted fill having shear strength parameters of c = 25 kN/m2, and 0' = 20°. The total unit weight of the material is 19 kN/m3. The pore pressuer ratio has an average value of 0.3. The slope of the sides is 3:1. Estimate the factor of safety using the Bishop and Morgenstern method. 10.18 For the Prob. 10.17, estimate the factor of safety for 0'= 30°, with all the other data remain the same. 10.19 For the Prob. 10.17, esatimate the factor of safety for a slope of 2:1 with all the other data remaining the same. 10.20 Estimate the minimum factor of safety for a complete drawdown 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 the Prob. 10.20? 10.22 An earth dam to be constructed at a site has the following soil parameters: c'= 600 lb/ft2, y = 110 lb/ft3, and 0' = 20°. The height of of dam H = 50 ft. The pore pressure ratio ru = 0.5. Determine the slope of the dam for a factor of safety of 1.5 using Spencer's method (1967).

c' = 15 kN/m2

' = 30° y = 20 kN/m3

Figure Prob. 10.20

416

Chapter 10 O

R = 45 ft 15ft

Figure Prob. 10.24 10.23 10.24

If the given pore pressure ratio is 0.25 in Prob. 10.22, what will be the slope of the dam? An embankment has a slope of 1.5 horizontal to 1 vertical with a height of 25 feet. The soil parameters are: c - 600 lb/ft2, 0' = 20°, and 7= 110 lb/ft3. Determine the factor of safety using friction circle method for the failure surface AC shown in Fig. Prob. 10.24. 10.25 It is required to construct an embankment for a reservoir to a height of 20 m at a slope of 2 horizontal to 1 vertical. The soil parameters are: c = 40 kN/m2, f = 18°, and 7= 17.5 kN/m3. Estimate the following: 1. Factor of safety of the slope assuming full friction is mobilized. 2. Factor of safety with respect to friction if the factor of safety with respect to cohesion is 1.5. Use Taylor's stability chart. 10.26 A cutting of 40 ft depth is to be made for a road as shown in Fig. Prob. 10.26. The soil properties are: c' = 500 lb/ft2, 0' = 15°, and 7= 115 lb/ft3. Estimate the factor of safety by the method of slices for the trial circle shown in the figure. 10.27 An earth dam is to be constructed for a reservior. The height of the dam is 60 ft. The properties of the soil used in the construction are: c = 400 lb/ft2, 0° = 20°, and 7= 115 lb/ft 3 , and ft = 2:1. Estimate the minimum factor of safety for the complete drawn from the full reservior level as shown in Fig. Prob. 10.27 by Morgenstern method. 10.28 What is the factor of safety if the water level is brought down from 60 ft to 20 ft above the bed level of reservoir in Prob. 10.27?

417

Stability of Slopes

c' = 5001b/ft2 0'=15° y=1151b/ft 3

Figure Prob.

10.26

Full reservoir level

1

Figure Prob.

10.27

10.29 For the dam given in Prob. 10.27, determine the factor of safety for r « = 0.5 by Spencer's method.