SLOPE STABILITY - KSU · Methodology of Slope Stability Analysis It is a method to expresses the...

SLOPE STABILITY

Omitted parts:

Sections 15.13, 15.14,15.15

Chapter 15

TOPICS

Introduction

Types of slope movements

Concepts of Slope Stability Analysis

Factor of Safety

Stability of Infinite Slopes

Stability of Finite Slopes with Plane Failure Surface

o Culmann’s Method

Stability of Finite Slopes with Circular Failure Surface

o Mass Method

o Method of Slices

What is a Slope?An exposed ground surface that stands at an angle with the horizontal.

Why do we need slope stability?In geotechnical engineering, the topic stability of slopes deals with:

1. The engineering design of slopes of man-made slopes in advance

(a) Earth dams and embankments,(b) Excavated slopes,(c) Deep-seated failure of foundations and retaining walls.

2. The study of the stability of existing or natural slopes of earthworks andnatural slopes.

o In any case the ground not being level results in gravity components of theweight tending to move the soil from the high point to a lower level. Whenthe component of gravity is large enough, slope failure can occur, i.e. the soilmass slide downward.

o The stability of any soil slope depends on the shear strength of the soiltypically expressed by friction angle (f) and cohesion (c).

SLOPE STABILITY

TYPES OF SLOPE

A. Natural slope

• Hill sides

• Mountains

• River banks

B. Man-made slope

• Fill (Embankment)

• Earth dams

• Canal banks

• Excavation sides

• Trenches

• Highway Embankments

Slopes can be categorized into two groups:

• Some of these failure can cause dramatic impact on livesand environment.

Slope failures cost billions of $every year in some countries

Case histories of slope failure

Slope failures cost billions of $every year in some countries


Bolivia, 4 March 2003, 14 people killed, 400 houses buried


Brazil, January 2003, 8 people killed


LaConchita California Slump


Slides: Rotational (slump)

TOPICS

Introduction



Factor of Safety





o Mass Method

o Method of Slices

Falls Topples Slides Flows Creep Lateral spreads Complex

o Slope instability (movement) can be classified into sixdifferent types:

Types of Slope Movements

Falls

• Rapidly moving mass of material (rock or soil) that travels mostly through theair with little or no interaction between moving unit and another.

• As they fall, the mass will roll and bounce into the air with great force and thusshatter the material into smaller fragments.

• It typically occurs for rock faces and usually does not provide warning.• Analysis of this type of failure is very complex and rarely done.

• Gravitational effect and shear strength

Gravity has two components of forces:

T driving forces:

N resisting forces (because of friction)

T= W. sin b

N = W. cos bT

S = N tan f

the interface develop its

resistance from friction (f):

b

A = effective Base Area of sliding block

f = friction

N

S

In terms of stresses:

S/A = N/A tan f

tf = s tan f

or

Boulder

Falls

Topples

This is a forward rotation of soil and/or rock mass about an axis

below the center of gravity of mass being displaced.

Back-Scrap

Bulging at

Toe

Slides

Slides

A. Translational (planar)

o Movements occur along planar failure surfaces that may run more-or lessparallel to the slope. Movement is controlled by discontinuities or weakbedded planes.

A

Weak bedding

plane

(Planar)

Occur when soil of significantly

different strength is presented

Slides

B. Rotational (curved)

This is the downward movement of a soil mass occurring on an almost circular surface of rupture.

B

Back-Scrap

Bulging

Curved escarpment

(Slumps)C. Compound (curved)

Slides

Soil nails

Reinforcement

Slides

شدادات

Reinforcement

Possible failure

surface

Anchors

Slides

o The materials moves like aviscous fluid. The failure planehere does not have a specificshape.

Flows

It can take place in soil withhigh water content or in drysoils. However, this type offailure is common in the QUICKCLAYS, like in Norway.

Creep

• It is the very slow movement of slope material that occur over a

long period of time

• It is identified by bent post or trees.

o Lateral spreads usually occur on very gentle slopes or essentially flat terrain,especially where a stronger upper layer of rock or soil undergoes extensionand moves above an underlying softer, weaker layer.

weaker layer

Lateral spreads

Complex

1. Falls

2. Topples

3. Slides• A. Translational (planar)

• B. Rotational (slumps)

4. Flow

5. Creep

6. Lateral Spread

7. Complex

Complex movement is by a combination of

one or more of the other principal types of

movement.

Many slope movements are complex,

although one type of movement

generally dominates over the others at

certain areas or at a particular time.

1. Falls

2. Topples

3. Slides• Translational (planar)

• Rotational (curved)

4. Flows

5. Creep

6. Lateral spreads

7. Complex

Slide is the most

common mode of

slope failure, and it will

be our main focus in

this course

Types of Slope Failures

In general, there are six types of slope failures:

Types of Slide Failure Surfaces

• Failure of slopes generally occur along surfaces known as failure surfaces.The main types of surfaces are:

• Planar Surfaces: Occurs in frictional,non cohesive soils

• Rotational surfaces: Occurs in cohesive soils

Circular surface (homogeneous soil)

Non-circular surface (non-homogeneous soil)

Types of Slide Failure Surfaces

• Transitional Slip Surfaces:When there is a hard stratum at arelatively shallow depth

• Compound Slip Surfaces:When there is hard stratum at some depth that intersectswith the failure plane

Slid

es

Translational (planar)

InfiniteLong plane

failure surface

FinitePlane failure

surface

Rotational (curved)

Finite

Above the toe

Through the toe

Deep seated

Failure surface 1

2

3

Types of Failure Surfaces

Stability of infinite slopes

Stability of finite slopes with plane

failure surfaces

Stability of finite slopes with circular

failure surfaces

1

2

3


Types of Failure Surfaces Considered in this Course are

TOPICS

Introduction



Factor of Safety





o Mass Method

o Method of Slices

In general we need to check

The stability of a given existed slope

Determine the inclination angle for a slope that we want to

build with a given height

The height for a slope that we want to build with a given

inclination


Methodology of Slope Stability Analysis

It is a method to expresses the relationship between resisting forces anddriving forces

• Driving forces – forces which move earth materials downslope. Downslopecomponent of weight of material including vegetation, fill material, orbuildings.

• Resisting forces – forces which oppose movement. Resisting forces includestrength of material

• Failure occurs when the driving forces (component of thegravity) overcomes the resistance derived from the shearstrength of soil along the potential failure surface.

The analysis involves determining and comparing the shear stress developedalong the most likely rupture surface to the shear strength of soil.

Methodology of Slope Stability Analysis

1. Assume a probable failure surface.

2. Calculate the factor of safety by determining and comparing

the shear stress developed along the most likely rupture

surface to the shear strength of soil.

3. Repeat steps 1 and 2 to determine the most likely failure

surface. The most likely failure surface is the critical surface

that has a minimum factor of safety.

4. Based on the minimum FS, determine whether the slope is

safe or not.

Slope Stability Analysis Procedure

Methods of Slope Stability Analysis

o Limit equilibrium method

o Limit analysis method

o Numerical methods

We will consider only the limit equilibrium method, since it isthe oldest and the mostly used method in practice.

Assumptions of Stability Analysis

o The problem is considered in two-dimensions

o The failure mass moves as a rigid body

o The shear strength along the failure surface is isotropic

o The factor of safety is defined in terms of the average shear

stress and average shear strength along the failure surface

42

TOPICS

Introduction



Factor of Safety





o Mass Method

o Method of Slices

43

StressShear StrengthShear

Force DrivingForce Resisting safety ofFactor

d

fsF t

t

tf= Avg. Shear strength of soil

td= Avg. Shear stress developed along the failure surface

Factor of Safety

44

Fs is the ratio of resisting forces to the driving forces, or

• The most common analytical methods of slope stability use a

factor of safety FS with respect to the limit equilibrium condition,

FS < 1 unstable

FS ≈ 1 marginal

FS >> 1 stable

Generally, FS ≥ 1.5 is acceptable

for the design of a stable slope

average shear strength of the soil.

average shear stress

developed along the potential

failure surface.

Shear stress (driving movement)

Shear strength (resisting movement)

(developed)

(Available)

If factor safety Fs equal to or less than 1, the slope is

considered in a state of impending failure

Factor of Safety

45

Causes of slope failure

1. External causes

These which produce increase of shear stress, like steepeningor heightening of a slope, building on the top of the slope

2. Internal causes

These which cause failure without any change in externalconditions, like increase in pore water pressure.

Therefore, slopes fail due either to increase in stress orreduction in strength.

46

Where:c’ = cohesion f’ = angle of internal friction

= cohesion and angle of friction that develop along the potential failure surface

ddc f ,

ff FFFFF csc then When

Factor of safety with respect to cohesion

Factor of safety with respect to friction

When the factor of safety with respect to cohesion is equal to the

factor of safety with respect to friction, it gives the factor of safety

with respect to strength, or

Other aspects of factor of safety

Factor of Safety

47

TOPICS

Introduction



Factor of Safety





o Mass Method

o Method of Slices

48

What is an Infinite slope?

• Slope that extends for a relatively long distance and hasconsistent subsurface profile can be considered as infinite slope.

• Failure plane parallel to slope surface.

• Depth of the failure surface is small compared to the height ofthe slope.

• For the analysis, forces acting on a single slice of the sliding massalong the failure surface is considered and end effects isneglected.


49

Force parallel to the plane AB Ta = W sin b = g LH sin b

Infinite slope – no seepage

o we will evaluate the factor of safety against a possible slope failurealong a plane AB located at a depth H below the ground surface.

o Let us consider a slope element abcd that has a unit lengthperpendicular to the plane of the section shown.

o The forces, F, that act on the faces ab and cd are equal and opposite andmay be ignored.

The shear stress at the base of the slope element can be given by

The resistive shear stress is given by

(*)

50

The effective normal stress at the base of the slope element is given by

(**)

For Granular Soil (i.e., c = 0)b

f

tan

tan sF

This means that in case of infinite slope in sand, the value of Fs isindependent of the height H and the slope is stable as long as b < f’

(***)

Equating R.H.S. of Eqs. (*) and (**) gives


51

Case of Granular soil – Derivation From Simple Statics

Equilibrium of forces on a slice:

FSDriving Forces

Resisting Forces

L

L

Extra

52

The depth of plane along which critical equilibrium occurs isobtained by substituting Fs = 1 and H = Hcr into Eq. (***)

Critical Depth, Hcr


53

(*)

(**)

Seepage is assumed to be parallel to the slope

and that the ground water level coincides with

the ground surface.

The resistive shear stress developed at

the base of the element is given by

The shear stress at the base of

the slope element can be given

Infinite slope – with steady state seepage

54

Substituting Eq. (****) Into Eq. (***) and solving for Fs gives

Equating the right-hand sides of Eq. (*) and Eq. (**) yields

b

f

bbg tan

tan

tan cos H

c2

sF

No seepage

(***)

(****)

Recall

Infinite slope – with steady state seepage

55

EXAMPLE

56

EXAMPLE

57

EXAMPLE

58

• Cohesive Soils

With seepage No seepage

tan

'tan'

tan2cos

'

''

'tan'tan

b

f

g

g

bbg

ff

satHsat

csF

sF

cd

c

sFd

)'tan'tan(2cos

'

fgbgb

sat

ccrH

tan

'tan

tan2cos

'

''

'tan'tan

b

f

bbg

ff

H

csF

sF

cd

c

sFd

)'tan(tan2cos

1'

d

ccrH

fbbg


59


Granular Soils

With seepage No seepage

Independant of H Slope is stable as long as b < f

tan

'tan'

tan

'tan'

tan2cos

'0.0'

b

f

g

g

b

f

g

g

bbg

satsF

satHsat

csF

c

tan

'tan

tan

'tan

tan2cos

'0.0'

b

f

b

f

bbg

sF

H

csF

c

60

TOPICS

Introduction



Factor of Safety





o Mass Method

o Method of Slices

61

o For simplicity, when analyzing the stability of a finite slopein a homogeneous soil, we need to make an assumptionabout the general shape of the surface of potential failure.

o However, considerable evidence suggests that slopefailures usually occur on curved failure surfaces

,

o The simplest approach is to approximate the surface ofpotential failure as a plane.

o Hence most conventional stability analyses of slopes havebeen made by assuming that the curve of potential slidingis an arc of a circle.


62Plane Failure Surface

simple wedge

o Culmann’s analysis is based on the assumption that the failure of a slopeoccurs along a plane when the average shearing stress tending to cause theslip is more than the shear strength of the soil.

o Also, the most critical plane is the one thathas a minimum ratio of the averageshearing stress that tends to cause failureto the shear strength of soil.

o The method gives reasonablyaccurate results if the slope isvertical or nearly vertical.

o Culmann’s method assumes that the critical surface of failure is a planesurface passing through the toe.

Culmann’s Method

63

• The forces that act on the mass are shown in the figure, where trial failureplane AC is inclined at angle q with the horizontal.

• A slope of height H and that rises at an angle b is shown below.

The average shear stress on the

plane AC

Ta = W Sin q

Similar procedures as for infinite slope, only different

geometry. Also here we made optimization.

t (*)

Culmann’s Method

64

The average resistive shearing stress (Developed shear strength) developed

along the plane AC also may be expressed as

Culmann’s Method

Na

s’

td (**)

Equating the R.H.S of Eqs. (*) and (**) gives

(***)

65

Culmann’s Method

• To determine the critical failure plane, we must use theprinciple of maxima and minima (for Fs=1 and for given valuesof c’, f’, g, H, b) to find the critical angle q:

Critical failure plane

• The expression in Eq. (***) is derived for the trial failure planeAC.

• Substitution of the value of q = qcr into Eq. (***) yields

(****)

66

• For purely cohesive soils c 0 f = 0.

The maximum height of the slope for which critical equilibrium occurs can

be obtained by substituting iinto into Eq. (****)

Culmann’s Method

67

Culmann’s Method

• Steps for Solution

A. If Fs

is given; H is required

cos(1

cossin'c4 4.

)'d

-

'd H

sF

'tan '

dtan .3

sF

'c '

dc 2.

sFF

cF .1

d

f

f

ff

f

b

b

g

C

g

f‘ H

b

q

68

Culmann’s Method

• Steps for Solution

B. If H is given; Fs

is required

51 stepsRepeat .7

F another try F c

F If 6.

F c

Fs

F F c

F if Check 5.

'd

c

'c

cF .4

'd

)'d

- '

dc .3

sF

'tan '

dtan 2.

F Assume .1

cossin

cos(1

4

H

ff

ff

f

f

ff

f

b

bg

69

A cut is to be made in a soil having properties as shown in the

figure below.

If the failure surface is assumed to be finite plane, determine the

followings:

(a) The angle of the critical failure plane.

(b) The critical depth of the cut slope

(c) The safe (design) depth of the cut slope. Assume the factor of

safety (Fs=3)?

H

45o

g = 20 kN/m3

f’=15o

c’=50 kPa

Given equation:

EXAMPLE

70

(a) The angle of the critical failure plane q

can be calculated from:

(b) The critical depth of the cut slope canbe calculated from:

H g = 20 kN/m3

f’=15o

c’=50 kPa

45o

Key Solution

b 45o

f’ 15o

d

(c) The safe (design) depth of the cutslope.

where: c’d and f’d can be determined from:

71

TOPICS

Introduction



Factor of Safety





o Mass Method

o Method of Slices

72

Slid

es

Translational (planar)

InfiniteLong plane

failure surface

FinitePlane failure

surface

Rotational (curved)

Finite

Above the toe

Through the toe

Deep seated

Failure surface 1

2

3


73

Modes of Failure

Shallow slope failure

Finite Slopes with Circular Failure Surface

i. Slope failure

• Surface of sliding intersects the slope at or above its toe.

Under certain circumstances, a shallow slopefailure can occur.

1. The failure circle is referred to as a toe circle if itpasses through the toe of the slope

1. The failure circle is referred to as a slope circle ifit passes above the toe of the slope.

ii. Shallow failure

74

Firm Base

H

iii. Base failure

o The surface of sliding passesat some distance below thetoe of the slope.

o The circle is called themidpoint circle because itscenter lies on a vertical linedrawn through the midpoint ofthe slope.

o For b 53o always toe

o For b < 53o could be toe, slope, or midpoint and that depends on depthfunction D where:

Depth function:


75

• Summary

• Toe Circle all circles for soils with f > 3° & b > 53°

• Slope Circle always for D = 0 & b < 53°

• Midpoint Circle always for D > 4 & b < 53°


76

Various procedures of stability analysis may, in general, be divided into twomajor classes:

2. Method of slices

1. Mass procedure

Types of Stability Analysis Procedures

• In this case, the mass of the soil abovethe surface of sliding is taken as a unit.

• Most natural slopes and many man-made slopes consist of more than onsoil with different properties.

• This procedure is useful when the soilthat forms the slope is assumed to behomogeneous.

• In this case the use of mass procedureis inappropriate.

77

• In the method of slices procedure, the soil above the surface of sliding isdivided into a number of vertical parallel slices. The stability of eachslice is calculated separately.

b

2

1

1

2

WV

EE

V T

N'a

h

R

R

O

W

xa

• It is a general method that can be used for analyzing irregular slopes innon-homogeneous slopes in which the values of c’ and f’ are notconstant and pore water pressure can be taken into consideration.

Types of Stability Analysis Procedures

78

TOPICS

Introduction



Factor of Safety





o Mass Method

o Method of Slices

79

2. Slopes in Homogeneous clay Soil with c 0 , f = 0

Determining factor of safety using equilibrium equations (Case I)

Mdriving = Md = W1l1 – W2l2

W1 = (area of FCDEF) gW2 = (area of ABFEA) g

Mresisting = MR = cd (AED) (1) r= cd r2q

Mass Procedure

1.Slopes in purely cohesionless soil with c = 0, f 0

Failure generally does not take place in the form of a circle. So we will notgo into this analysis.

80

W1

W2

l1l2

Mdriving = Md = W1l1 – W2l2

Mresisting = MR = cd r2q

Mass Procedure

81

• The minimum value of the factor of safety thus obtained is the factor ofsafety against sliding for the slope, and the corresponding circle is thecritical circle.

• The potential curve of sliding, AED, was chosen arbitrarily.

• The critical surface is that for which the ratio of Cu to Cd is a minimum. Inother words, Cd is maximum.

• To find the critical surface for sliding, one must make a number of trialsfor different trial circles.

REMARKS

82

• Fellenius (1927) and Taylor (1937) have analytically solved forthe minimum factor of safety and critical circles.

• They expressed the developed cohesion as

H or

soil oft unit weighγ

slope ofheight H

number Stability m

Where

g

g

d

d

cm

mHc

• We then can calculate the min Fs as

• The critical height (i.e., Fs 1) of the slope can be evaluated by

substituting H = Hcr and cd = cu (full mobilization of the undrained

shear strength) into the preceding equation. Thus,


83

The results of analytical solution to obtain critical circles was representedgraphically as the variation of stability number, m , with slope angle b.

Toe, Midpoint or slope circles Toe slope

Firm Stratum

b

m is obtained from this chart depending on angle b


84

For a slope angle b > 53°, the critical circle is always a toe circle. The

location of the center of the critical toe circle may be found with the aid of

Figure 15.14.

For b < 53°, the critical circle may be a toe, slope, or midpoint circle,

depending on the location of the firm base under the slope. This is called

the depth function, which is defined as

Failure Circle


85

(radius)

Figure 15.13

Location of the center of the critical toe circle

The location of the center of the critical toe circle may be found with the aid of Figure 15.13

86

When the critical circle is a midpoint circle (i.e., the

failure surface is tangent to the firm base), its position

can be determined with the aid of Figure 15.14.

Figure 15.14

Firm base


87

The location of these circles can be

determined with the use of Figure 15.15

and Table 15.1.

Figure 15.15

Note that these critical toe circle are not necessarily the most critical circles that exist.

Critical toe circles for slopes

with b < 53°

88

How to use the stability chart? Given: b 60o, H, g, cu Required: min Fs

m = 0.195

1. Get m from chart

2. Calculate cd from

3. Calculate Fs

mHcd g

d

us

c

cF

89

Given: b 30o, H, g, cu, HD (depth to hard stratum) Required: min. Fs

m = 0.178

1. Calculate D = HD/H

2. Get m from the chart

3. Calculate cd from

4. Calculate Fs

mHcd g

d

us

c

cF

D = Distance from the top surface of slope to firm base

Height of the slope

Note that recent investigation put angle b at 58o instead of the 53o value.

How to use the previous chart?

90Rock layer

EXAMPLE

91

D=1.5m

SOLUTION

92

SOLUTION

93

The results of analytical solution to obtain critical circles was representedgraphically as the variation of stability number, m , with slope angle b.

Toe, Midpoint or slope circles Toe slope

Firm Stratum

b

m is obtained from this chart depending on angle b

Slopes in Homogeneous clay Soil with c 0 , f = 0

94

o Several trails must be made to obtain the most critical sliding surface,minimum factor of safety or along which the developed cohesion is a maximum

numberstabilitymfcd H

fqbag

,,, fqbag ,,,H fcd

o The maximum cohesion developed along the critical surface as

o Determination of the magnitude of described previously is based on atrial surface of sliding.

o The results of analytical solution to obtain minimum Fs was representedgraphically as the variation of stability number, m , with slope angle b forvarious values of f’ (Fig. 15.21).

o Solution to obtain the minimum Fs using this graph is performed by trial-and-error until Fs = Fc’=Ff’

o Since we know the magnitude and direction of W and the direction of Cd

and F we can draw the force polygon to get the magnitude of Cd.

o We can then calculate c’d from

REMARKS

Slopes in clay Soil with f = 0; Cu Increasing with Depth

EXAMPLE

98

o The Friction Circle method (or the f-Circle Method) is very useful forhomogenous slopes. The method isgenerally used when both cohesive andfrictional components are to be used.

Slopes in Homogeneous C’ – f’ Soils

o Here the situation is more complicatedthan for purely cohesive soils.

o The pore water pressure is assumed to be zero

o F—the resultant of the normal and frictionalforces along the surface of sliding. Forequilibrium, the line of action of F will passthrough the point of intersection of the lineof action of W and Cd.

99

fqbag ,,,H fcd

mHcd g

Friction Circle method

100

1. Assume fd (Generally start with = f’

i.e. full friction is mobilized)

2. Calculate

3. With fd and b Use Chart to get m

4. Calculate

5. Calculate

6. If Fc’ = Ff’ The overall factor of safety

Fs = Fc’ = Ff’

7. If Fc’≠ Ff’ reassume fd and repeat steps 2through 5 until Fc’ = Ff’

OrPlot the calculated points on Fc versus Fφ

coordinates and draw a curve through the points.[see next slide]. Then Draw a line through theorigin that represents Fs= Fc = Fφ

mHcd g

d

cc

cF

Procedures of graphical solutionGiven: H, b, g, c’, f’ Required: Fs

Taylor’s stability

number

101

Note: Similar to Culmann procedure for planar mechanism but here Cd is foundbased on m. In Culmann’s method Cd is found from analytical equation.

Procedures of graphical solutionGiven: H, b, g, c’, f’ Required: Fs

102

Hcr means that Fc’ = Ff’ =Fs = 1.0

1. For the given b and f’, use Chart to get m.

2.Calculate

m

cHcr

g

Given: b, g, C ’, f ’ Required: Hcr

Calculation of Critical Height

103

SUMMARY

f = 0

C f

Mass Procedure – Rotational mechanismneed only the use of Taylor’s chart.

104

• Example

• Given: cu

= 40 kN/m2

& g = 17.5 kN/m3

• Required:

1. Max. Depth

2. Radius r when Fs=1

3. Distance BC

• b = 60 °

> 53 °

from Fig.15.13 m= 0.195

60 °

muccrH 72.11

195.0*5.17 m

40

g

2

sin

H

2

ACDC

DCcr

2

ins

r

a

q

m 9.97 60cot - cot AE - AF EF BC

m 28.17 6.25)3in5)(3 (sin2

72.11

2

in sin2

o5.27 o53o06for 15.14 Fig.

a

qa

qab

crH

ss

crHr

andFrom

EXAMPLE

105

EXAMPLE

106

SOLUTION

107

SOLUTION

108

SOLUTION

109

1.42

SOLUTION

110

15 m

g = 16 kN/m3

C = 40 kN/m2

f = 15o

•Using Taylor’s stability chart determine the factor of safety for the slope shown in Fig.1.•For the same slope height, what slope angle must be used if a factor of safety of 1.5 is required?

50o

10 m

2nd Midterm Fall 1436-1437H QUESTION #2

111

fd Ff=tan f/tan fd m Cd = g H m Fc = C/Cd

15 1 0.092 14.7 2.70

10 1.52 0.116 18.6 2.20

7.5 2.0 0.125 20 2.0

a)

Fs = 2.0

b)

Cd = 40/1.5 =26.7 kpa

mHcd g26.7 = 16 X10 X m

m = 0.167

tan fd = tan f/1.5

fd

fd=10.1o

At m = 0.167 and fd=10.1o from chart b =75o

SOLUTION

112

Method of Slices

• Method of Slices

• Non-homogenous soils (mass procedure is not accurate)

• Soil mass is divided into several vertical Parallel slices

• The width of each slice need not be the same

• It is sometimes called the Swedish method

113

• It is a general method that can be used for analyzing irregular slopes innon-homogeneous slopes in which the values of c’ and f ’ are notconstant.

g1, c’1, f’1

g2, c’2, f’2

g3, c’3, f’3

Non-homogeneous Slope Irregular Slope

g, c’, f’b1

b2

Method of Slices

• Because the SWEDISH GEOTECHNCIAL COMMISION used this methodextensively, it is sometimes referred to as the SWEDISH Method.

• In mass procedure only the moment equilibrium is satisfied. Here attemptis made to satisfy force equilibrium.

114

• The soil mass above the trial slip surface is divided into several vertical parallelslices. The width of the slices need not to be the same (better to have it equal).

• The base of each slice is assumed to be a straight line.

• The inclination of the base to the horizontal is a.

• The height measured in the center line is h.

• The height measured in the center line is h.

• The accuracy of calculation increases if the number of slices is increased.

• The procedure requires that aseries of trial circles are chosenand analyzed in the quest forthe circle with the minimumfactor of safety.

Tr

Method of Slices

115

Method of Slices

• Forces acting on each slice

• Total weight wi=ghb

• Total normal force at the base Nr=s*L

• Shear force at the base Tr=t*L

• Total normal forces on the sides, Pn

and Pn+1

• Shear forces on the sides, Tn

and Tn+1

• 5 unknowns Tr ,P

n ,P

n+1 ,T

n ,T

n+1

• 3 equations SFx=0 , SF

y=0 ,SM=0

• System is statically indeterminate

• Assumptions must be made to solve the problem

• Different assumptions yield different methods

• Two Methods:

• Ordinary Method of Slices (Fellenius Method)

• Bishop’s Simplified Method of Slices

116

Method of Slices

For the whole sliding mass

sin*

)*tan**(

sin*

*

*

sF

sin*

*

sF

*

sin*

0*-sin*r*

0o

M

a

fsa

t

ta

tt

a

a

W

ln

lcFs

W

lf

Fs

lf

W

lf

ld

T

TW

rTW

S

S

S

S

S

SS

SS

S

117

Method of Slices

Slices of Method Simplified sBishop'•

Slices of MethodOrdinary

:Methods Two

N force of valuethe

findingin introduced are ionsapproximatbut exact isEquation

sin*

*tan*

*

sin*

)*tan**(

•

a

f

sa

fs

W

NlcFs

Nln

W

ln

lcFs

S

SS

SS

S

S

118

Fellenius’ Method

Assumption

For each slice, the resultant of the interslice forces is

zero.

The resultants of Pn

and Tn

are equal to the resultants

of Pn+1

and Tn+1

, also their lines of actions coincide.

Rn

Rn+1

Ordinary Method of Slices

119


n

sin*n

W

lu

c

sF

0 u

cc

:condition undrained For

n

sin*n

W

)tann

cos*n

Wn

l*c(

sF

nnr

n

cos*WN

0F )rT fromaway stay (to

aS

f

aS

faS

a

S

n

120

Steps for Ordinary Method of Slices

• Draw the slope to a scale

• Divide the sliding wedge to various slices

• Calculate wn

and an

for each slice

• an

is taken at the middle of the slice

• Calculate the terms in the equation

• Fill the following table

Slice# wn

an

sin an

cos an

ln

wn sin a

nw

n cos a

n

n

sin*n

W

)tann

cos*n

Wn

l*c(

sF

aS

faS

an

-ve

an

+ve

wn

wn


121

Find Fs

against sliding

Use the ordinary method of slices

EXAMPLE

122

Assumption

For each slice, the resultant of the interslice forces is

Horizontal.

i.e. Tn

=Tn+1

Bishop’s Simplified Method of Slices

123


n

s

rn

s

nnrn

s

r

s

nr

s

nn

s

nr

n

s

nndr

nrnrn

1n ny

sinF

tanNsin

F

lccos*NW

F

tanN

F

lcT

F

tanl

F

lcT

lF

tancl*T

sin*Tcos*NW

0F )P andP fromaway stay (to

af

a

a

f

fs

fs

t

aa

S

y

124


procedure error and Trail

F

sintancos

tanWcb

sinW

1F

cos

bl but

F

sintancos

F

sintancos

sinF

lcW

tancl

F

F

sintancos

sinF

lcW

N

s

nn

nn

nn

s

n

nn

s

nn

s

nn

n

s

nn

n

s

s

nn

n

s

nn

r

afa

f

aS

aafa

afa

a

fS

afa

a

S

125


Steps for Bishop’s Simplified Method of Slices

• Draw the slope to a scale

• Divide the sliding wedge to various slices

• Calculate wn

and an

for each slice

• an

is taken at the middle of the slice

• Calculate the terms in the equation

• Fill the following table

Slice# wn

an

sin an

cos an

bn

wn sin a

n

• Assume Fs

and plug it in the right-hand term of the equation

then calculate Fs

• Repeat the previous step until the assumed Fs = the

calculated Fs.

s

nn

nn

nn

s

F

sintancos

tanWcb

sinW

1F

afa

f

aSS

126


nn

)n(

nn

s

s

nn)n(

sinW

m

1)tanWcb(

F

F

sintancosm

aS

f

afa

a

a

S

)

tan incos

tan '(

sin

1

n

s

n

nn

nn

s

F

s

Wbc

WF

faa

f

a

127


• Example of specialized software: – Geo-Slope,

– Geo5,

– SVSlope

– Many others

128

Determine the safety factor for the given trial rupture surface shown in

Figure 3. Use Bishop's simplified method of slices with first trial factor of

safety Fs = 1.8 and make only one iteration. The following table can be

prepared; however, only needed cells can be generated “filled”.

Final Exam Fall 36-37 QUESTION #4

129

Fs = 1.8

Table 1. “Fill only necessary cell for this particular problem”

SliceNo.(1)

Widthbn

(m)(2)

Heighthl

(m)(3)

Heighth2

(m)(4)

AreaA

(m2)(5)

WeightWn

(kN/m)(6)

α(n)

(7)mα(n)

(8)

Wn sin a(kN/m)

(9)

1 22.4 70

2 294.4 54

3 38

4 435.2 24

5 390 12

6 268.8 0.0

7 66.58 -8

?

SOLUTION

130

Remarks on Method of Slices

o Bishop’s simplified method is probably the most widely used (but it hasto be incorporated into computer programs).

o The ordinary method of slices is presented in this chapter as a learningtool only. It is used rarely now because it is too conservative.

o It yields satisfactory results in most cases.

o Analyses by more refined methods involving consideration of the forces actingon the sides of slices show that the Simplified Bishop Method yields answersfor factors of safety which are very close to the correct answer.

o The Bishop Simplified Method yields factors of safety which are higherthan those obtained with the Ordinary Method of Slices.

o The two methods do not lead to the same critical circle.

o The Fs determined by this method is an underestimate (conservative) butthe error is unlikely to exceed 7% and in most cases is less than 2%.

131

Two Methods:


• Underestimate Fs

(too conservative)

• Error compared to accurate methods (5-20%)

• Rarely used


• The most widely used method

• Yields satisfactory results when applying computer

program

Remarks on Method of Slices

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