2.2 Slope Stability

2 2 STABILITY OF EARTH

CE 303 GEOTECHNICAL ENGINEERING - II

2.2. STABILITY OF EARTHSLOPES

by

Dr. T. Venkata Bharat, Ph.D.

Assistant ProfessorDepartment of Civil EngineeringIIT Guwahati Guwahati 781039IIT Guwahati, Guwahati-781039

METHODS OF STABILITY ANALYSIS| Limit Equilibrium Method (choice of analysis here!)

y based on equilibrium of forces

y requires knowledge of statics

y soil is considered to be on the verge of failure

| Limit Analysis based on Plasticity| Limit Analysis based on Plasticity

y based on equilibrium of stresses

y requires numerical methods

y generally, analysis is done using software packages such as Plaxis, Geostuido (Slope/w) etc.

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STABILITY OF INFINITE SLOPE| Infinite slopes have dimensions

that extend over great distances gas compared to their depth

| The assumption of an infinite length simplifies the analysis considerably.A i i f i fi i l i | A representative section of infinite slope is considered in the figure.

| In order to use LEM for the analysis the failure | In order to use LEM for the analysis, the failure mechanism should be postulated first.

| It is reasonable to assume that failure occurs on a 3plane parallel to slope.

STABILITY OF INFINITE SLOPE | A slice of soil is considered between the surface of

the slope and the assumed slip plane as shown in p p pfigure in the previous slide.

| Draw free-body diagram of forces acting on this slice and then formulate equilibrium equations.

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STABILITY OF INFINITE SLOPE | The factor of safety (F) of a slope is defined as the

ratio of the available shear strength of the soil (f) g ( f)to the minimum shear strength to maintain stability (mobilized strength, m

where fF=

tanf n =S

for Effective Stress Analysis (ESA)

for Total Stress

Case I: ESA without the effect of seepage forces

mf uS = Analysis (TSA)

=tantan

F

=at limit-

equilibrium5tan

STABILITY OF INFINITE SLOPE Case II: ESA with the effect of seepage forces (Js)

L t id d t ithi th lidi | Let us now consider groundwater within the sliding mass and assume that the seepage is parallel to the slope. The seepage force is given by

| from Statics| from Statics,

| At limit equilibrium,6

STABILITY OF INFINITE SLOPE Case III: TSA

Th h t th li l f TSA i| The shear stress on the slip plane for a TSA is

| The factor of safety F for TSA is given by:

| At limit equilibrium,

| Critical value of z occurs at:7

INFINITE SLOPES SALIENT POINTS| The maximum stable slope in a coarse-grained

soil, in the absence of seepage, is equal to the , p g , qfriction angle of the soil.

| The maximum stable slope in a coarse-grained soil, in presence of seepage, is roughly half of the friction angle of the soil.

| The critical slope angle in fine-grained soils is 45and the critical depth is equal to the depth of the tension cracks 2su/.tension cracks 2su/.

| Infinite slope mechanism is usually not observed for fine-grained soils. For such soils, rotational g ,failure mechanism is more common. 8

INFINITE SLOPE AN EXAMPLE| Dry sand is to be dumped from a truck on the side

of a roadway. The properties of the sand are = y p p 30, = 17 kN/m3 and sat = 17.5 kN/m3. Determine the maximum slope angle of the sand in (a) the dry state (b) the saturated state without (a) the dry state, (b) the saturated state, without seepage and (c) the saturated state if groundwater is present and seepage occurs parallel to the slope

d h f h l Wh i h f l towards the toe of the slope. What is the safe slope in the dry state for a factor of safety of 1.25?

(will be solved in the class)

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ROTATIONAL SLOPE FAILURE

| Slopes made up of homogenous fine-grained homogenous fine grained soils have been observed to fail through a rotational f il h ifailure mechanism.

| The failure surface is assumed to be circular (top assumed to be circular (top right) or noncircular (bottom right).

| The analysis also takes into account the presence of a

h i f i hi h phreatic surface within the sliding mass.

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STABILITY ANALYSIS OF A ROTATIONAL FAILURE

| A free-body diagram of the assumed circular mechanism would show the weight (W) of the soil g ( )within the sliding mass acting at the centre of mass.

| If seepage is present, the seepage forces (Js) would be present.

| The forces resisting the clockwise rotation of the sliding mass are the shear forces mobilized by the soil along the circular slip surface.the soil along the circular slip surface.

| We must now use statics to determine whether the disturbing moments created by W and Js exceed g y sthe restoring moment provided by the soil. 11

TSA:

Q (kN)

dQdw

c L ru ac L rFWd

=

u ac L rFWd Qd

= +u ac L rF

Wd P d= +

u ac L rFWd Qd P d

= + + 12QWd Qd+ w wWd P d+ Q w wWd Qd P d+ +Presence of load, Q Presence of crack Presence of load and crack

FRICTION-CIRCLE METHODConsidered forces:| Weight of soil mass in | Weight of soil mass in

failure zone, W

| Sum of cohesive forces ti ll l t h d acting parallel to chord

AB, Cm| The resultant of

frictional forces, R

| Factor of safety ti i b d equation is based on:

tanfm

cF F F

= = +13cF F F

such that F = Fc = F

FRICTION-CIRCLE METHOD| Important relations

( )Ch d L thcC ( )Arc Length tan | Procedure:

( )ChordLengthm ABc

CF

= ( )( )ArcLength

ChordLengthmAB

C

AB

L r= tan

=m F

y Assume a failure plane such as ABDAy Obtain the weight of soil mass, W, in the failure zone by

graphical techniquesy Find the direction (parallel to chord AB) and distance of Cm

from center, Oy Assume F and draw friction circle with radius rSinmy Find the direction of R (passes through intersection of W and

Cm, and runs tangent to -circle)y Draw force polygon and find the magnitude of Cmy Obtain Fc and compare with assumed Fy Change the value of F and repeat the procedure till Fc = F

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METHOD OF SLICES| One approach that is commonly used to analyze

rotational failure is to divide the sliding mass into gan arbitrary number of vertical slices and then sum the forces and moments of each slice.

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METHOD OF SLICES| Of course, the larger the number of slices, the

better the accuracy of our solution.y

| However, dividing the sliding mass into a number of vertical slices poses new problems.

| We now have to account for the internal or interfacial forces between two adjacent slices.

| Lets now attempt to draw a free-body diagram of an arbitrary vertical slice and examine the f i hi liforces acting on this slice.

16

FORCES ACTING ON A VERTICAL SLICE

17

METHOD OF SLICES KNOWN QUANTITIESQ

18

METHOD OF SLICES UNKNOWN QUANTITIESQ

19

METHOD OF SLICES| If there are n slices, we have to obtain the values of

6n-1 parameters.p| However, we only have 4n number of equations.| That leaves us with 2n 1 unknowns| That leaves us with 2n-1 unknowns.| Therefore, the problem is statically

indeterminateindeterminate.

| For example, if there are 10 slices, well have 6x10-1=59 unknowns but only 10x4=40 equations1 59 unknowns but only 10x4 40 equations.

| Therefore, in order to obtain a solution, we have to make certain simplifying assumptions or use an a e ce ta s p y g assu pt o s o use a iterative method 20

METHOD OF SLICES| Several solution methods have been developed

depending on the assumptions made about the p g punknown parameters and which equilibrium condition (force, moment or both) have been satisfiedsatisfied.

| Tables on the next two pages provide a summary of methods that have been proposedmethods that have been proposed.

| Computer programs (such as SLOPE/W or XSTABL) are available for all the methods listedXSTABL) are available for all the methods listedin the table.

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SWEDISH CIRCLE METHOD| Forces acting on a slice:y Weight of soil massWeight of soil mass

y Cohesive forces (C) in the opposite to the direction of

b bl d probable wedge movement

y Reaction (R) at the base inclined at to the normal, inclined at to the normal, assuming slippage is imminent

| Assumptions:y The interslice reaction forces are equal and opposite

Sh f h i li d b 22y Shear forces at the inter-slice are assumed to be zero

SWEDISH CIRCLE METHOD Factor of safety:

n

1

sec cos tann

j j jj

n

cb WF

=

+ =

1

sinj jj

W =

| It may be noted that the tangential component, Tj,

and base angle, j, may be negative for few slices

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SWEDISH CIRCLE METHOD | N and T curves:

NN A =

TT A = where AN and AT are areas of N- and T- diagrams, respectively

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BISHOPS SIMPLIFIED| The effect of forces acting on the

sides of the individual slices are taken into account

| Disregards the shear forces on the inter-slices (X1 = X2 = 0)

| Method satisfies moment equilibrium and vertical force equilibrium

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BISHOPS SIMPLIFIED | Factor of Safety:

1n ( )1

1tan

i

j j j jj

n

c b W ubm

FW

= + =

1

sinj jj

W =

where

A F b th th id it ti h i i d

( )1 tan tan cosm F = +| As F appears on both the sides, iterative approach is required

by assuming the F and finding the value. The assumed value is compared against the computed.

| The process is continues until both the values match 27

INTERSLICE FORCES| Interslice shear forces are required to calculate the

normal force at the base of each slice.| The interslice shear force (Xi) is computed as a

percentage of the interslice normal force (Ei) according to the following empirical equation proposed by g p q p p yMorgenstern and Price (1965):

where: = the percentage (in decimal form) of the function used and used, and f(x) = interslice force function representing the relative direction of the resultant interslice force

28

VARIOUS INTERSLICE FORCE FUNCTIONS

29

METHODS OF SLOPE STABILITY ANALYSIS

30

ASSUMPTIONS IN VARIOUS METHODS

31

COMPARISON OF DIFFERENT METHODS

32Provided in the Additional_read folder in CE303 dropbox link

DESIGN CHARTS| Slope stability analysis based on design charts is

useful y for preliminary analysisy for rapid means of checking the results of detailed

analysesanalysesy to compare alternates that can later be examined by

rigorous analysisy to determine the approximate value of the F as it allows

some quality control check for the subsequent computer-generated solutions

y To back-calculate strength values for failed slopes to aid in planning remedial measures

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TAYLORS CHARTS (1948)( )| Taylors charts provide the stability values in terms

of stability number, Sn using friction-circle y , n gmethod

F F F= =Condition:

| Analysis by these charts is valid for simple sections and homogeneous soils

cF F F= =

and homogeneous soils| In general, y failure surface passes through the toe when the slope is

steepy base failure (failure extends below toe) occurs when

either the slopes are flatter or/and firm stratum exists pbelow the toe 36

TAYLORS CHARTS (1948)( )

Fig. Conceptual section by Taylor

37

next figure Total Stress Analysis (TSA):

9

4

8

)

T

S

(

1

9 Stability number

C

H

A

R

T

In terms of F.S.

L

O

R

S

cd s

c cF

c N H = =

T

A

Y

L

38

c'' Analysis:9

4

8

)

T

S

(

1

9

In terms of F.S.c c

F = =

C

H

A

R

T

F=

cd s

Fc N H= =

L

O

R

S d

T

A

Y

L

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PROBLEM - 1

| Given a soil slope with height, H = 12 m, DH = 18 m, = 300 c = 58 kPa = 19 kN/m3 find = 300, c = 58 kPa, = 19 kN/m3, find y F of Sy The distance from toe to the point where critical circle

appears on the groundy F of S, if there are heavy loadings outside the toe.

(will be solved it in the class)

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PROBLEM - 2| Given a soil slope with height, H = 12 m, = 300, c

= 24 kPa, = 200, and = 19 kN/m3 What is the , , .factor of safety of the slope?

(will be solved it in the class)

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SPENCERS CHARTS (1967)( )| Based on solutions computed using Spencers

method, which satisfies complete equilibrium, p q| Charts are used to determine the required slope

angle for a preselected F of S| Solutions for three different pore pressure ratios,

ru: 0, 0.25, 0.5.y Pore water pressure ratio (ru) is the ratio of pore

water force on a slip surface to the total force due to weight of the soil and any external loadingweight of the soil and any external loading

| Assumption: firm stratum is at great depth below the slope

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67

)

Developed friction: t s

(

1

9

6

Developed friction: d = tan-1(tan/F)

s

c

h

a

r

t

n

c

e

r

s

S

p

e

43

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PROBLEM - 3| Given a slope with height H = 18 m, c = 9.6 kPa,

= 300, = 19.6 kN/m3, ru = 0.25, determine the , , u ,maximum slope angle for F of S of 1.5.

( ill b l d i h l )(will be solved in the class)

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PROBLEM FOR ASSIGNMENT - 1 | Given a soil slope with height, H = 12 m, = 300, c

= 24 kPa, = 200, and = 19 kN/m3 find the factor , , ,of safety of the slope using the following methods:y TSA based on moment equilibrium

y Friction-circle method

y Swedish circle method

y Bishops simplified method

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2.2 Slope Stability

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