2 2 STABILITY OF EARTH

CE 303 GEOTECHNICAL ENGINEERING - II

2.2. STABILITY OF EARTH–SLOPES

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!)

based on equilibrium of forces

requires knowledge of statics

soil is considered to be on the verge of failure

Limit Analysis based on PlasticityLimit Analysis based on Plasticity

based on equilibrium of stresses

requires numerical methods

generally, analysis is done using software packages

such as Plaxis, Geostuido (Slope/w) etc.2

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.

4

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 (τf) g ( f)to the minimum shear strength to maintain stability (mobilized strength, τm

where fFττ

=tanf nτ σ φ′ ′=

S

for Effective Stress Analysis (ESA)

for Total Stress

Case – I: ESA without the effect of seepage forces

mτf uSτ = Analysis (TSA)

α φ′=tantan

Fφα

′=

at limit-equilibrium

5tanα

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 Staticsfrom Statics,

At limit equilibrium,6

STABILITY OF INFINITE SLOPE …Case – III: TSA

Th h t th li l f TSA iThe 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 45°and the critical depth is equal to the depth of the tension cracks 2su/γ.tension cracks 2su/γ.

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 φ’ = y p p φ30°, γ = 17 kN/m3 and γsat = 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.

10

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

φu = 0 Analysisφu y

Q (kN)

dQdw

c L ru ac L rF

Wd=

u ac L rF

Wd Qd=

+u ac L r

FWd P d

=+

u ac L rF

Wd 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 METHOD

Considered 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

τ σ φτ′ ′

= = +13cF F Fφ

such that F = Fc = Fφ

FRICTION-CIRCLE METHOD…Important relations

( )Ch d L thc

C ( )Arc Length tanφ′

Procedure:

( )Chord Lengthm ABc

CF

= × ( )( )Arc Length

Chord Lengthm

ABC

AB

L r= × tanFφ

φψ =

Assume a failure plane such as ABDAObtain the weight of soil mass, W, in the failure zone by graphical techniquesFind the direction (parallel to chord AB) and distance of Cm from center, OAssume Fφ and draw friction circle with radius r×SinψFind the direction of R (passes through intersection of W and Cm, and runs tangent to φ-circle)Draw force polygon and find the magnitude of Cm

Obtain Fc and compare with assumed Fφ

Change the value of Fφ and repeat the procedure till Fc = Fφ

14

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.

15

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.

Let’s 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 unknownsThat leaves us with 2n-1 unknowns.

Therefore, the problem is statically indeterminateindeterminate.

For example, if there are 10 slices, we’ll 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.

21

SWEDISH CIRCLE METHOD

Forces acting on a slice:Weight of soil massWeight of soil mass

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

b bl d probable wedge movement

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

Assumptions:The interslice reaction forces are equal and opposite

Sh f h i li d b 22Shear 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

α α φ=

⎡ ⎤+⎣ ⎦=

⎡ ⎤

∑

∑1

sinj jj

W α=

⎡ ⎤⎣ ⎦∑

It may be noted that the tangential component, Tj, and base angle, αj, may be negative for few slices

23

SWEDISH CIRCLE METHOD …

N and T curves:

NN A γ= ×∑

TT A γ= ×∑

where AN and AT are areas of N- and T- diagrams, respectively24

RIGOROUS METHODS

Bishop's Simplified

J b ' Si lifi d Janbu's Simplified

Janbu's Generalized

Spencer

Morgenstern-Price

General Limit Equilibrium (GLE)

Corps of Engineers Corps of Engineers

Lowe-Karafiath25

BISHOP’S 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

26

BISHOP’S SIMPLIFIED …Factor of Safety:

1n

⎡ ⎤( )1

1tan

i

j j j jj

n

c b W ubm

FW

α

φ=

⎡ ⎤′ ′+ −⎣ ⎦=

⎡ ⎤

∑

∑1

sinj jj

W α=

⎡ ⎤⎣ ⎦∑

where

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

( )1 tan tan cosm Fα φ α α′= +

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:λ = 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

for preliminary analysisfor rapid means of checking the results of detailed analysesanalysesto compare alternates that can later be examined by rigorous analysisto determine the approximate value of the F as it allows some quality control check for the subsequent computer-generated solutionsTo back-calculate strength values for failed slopes to aid in planning remedial measures

33

DESIGN CHARTS…Taylor’s chart (1948)Bishop & Morgenstern (1960)Bishop & Morgenstern (1960)Spencer (1967)Janbu (1968)( )Hunter & Schuster (1968)Chen & Giger (1971)O’Connor & Mitchell (1977)Cousins (1978)Ch l & S ( 98 )Charles & Soares (1984)Barnes (1991)

34

DESIGN CHARTS…Taylor’s chart (1948)Bishop & Morgenstern (1960)Bishop & Morgenstern (1960)Spencer (1967)Janbu (1968)( )Hunter & Schuster (1968)Chen & Giger (1971)O’Connor & Mitchell (1977)Cousins (1978)Ch l & S ( 98 )Charles & Soares (1984)Barnes (1991)

35

TAYLOR’S CHARTS (1948)( )Taylor’s 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 Fφ= =

and homogeneous soilsIn general,

failure surface passes through the toe when the slope is steepbase failure (failure extends below toe) occurs when either the slopes are flatter or/and firm stratum exists pbelow the toe 36

TAYLOR’S CHARTS (1948)…( )

Fig. Conceptual section by Taylor

37

next figure φ = 0 Analysis:94

8)TS

(19 Stability number

CH

AR

T

In terms of F.S.′ ′

LOR

’S

cd s

c cF

c N Hγ′ ′

= =

TAYL

38

c'−φ' Analysis:94

8)TS

(19

In terms of F.S.c c

F′ ′

= =

CH

AR

T

Fφφ′

=

cd s

Fc N Hγ

= =

LOR

’S dφ φ

TAYL

39

PROBLEM - 1

Given a soil slope with height, H = 12 m, DH = 18 m, β= 300 c = 58 kPa γ = 19 kN/m3 find = 300, c = 58 kPa, γ = 19 kN/m3, find

F of SThe distance from toe to the point where critical circle appears on the groundF of S, if there are heavy loadings outside the toe.

(solve it during the tutorial class)

40

PROBLEM - 2Given a soil slope with height, H = 12 m, β = 300, c’ = 24 kPa, φ’ = 200, and γ = 19 kN/m3 What is the , φ , γ .factor of safety of the slope?

(solve it during the tutorial class)

41

SPENCER’S CHARTS (1967)( )Based on solutions computed using Spencer’s 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.

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

42

67)

Developed friction: ts (1

96

Developed friction: φd = tan-1(tanφ/F)

s ch

art

ncer

’sSp

e

43

43

PROBLEM - 3Given a slope with height H = 18 m, c’ = 9.6 kPa, φ’= 300, γ = 19.6 kN/m3, ru = 0.25, determine the , γ , u ,maximum slope angle β for F of S of 1.5.

( l i d i h i l l )(solve it during the tutorial class)

44

PROBLEM FOR ASSIGNMENT - 1 Given a soil slope with height, H = 12 m, β = 300, c’ = 24 kPa, φ’ = 200, and γ = 19 kN/m3 find the factor , φ , γ ,of safety of the slope using the following methods:

φu = 0 analysis

Friction-circle method

Swedish circle method

Bishop’s simplified method

45