Slope Stability

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SLOPE STABILITY

Chapter 15

Omitted parts: Sections 15.13, 15.14,15.15 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 TOPICS

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 and natural slopes. o In any case the ground not being level results in gravity components of the weight tending to move the soil from the high point to a lower level. When the component of gravity is large enough, slope failure can occur, i.e. the soil mass slide downward. o The stability of any soil slope depends on the shear strength of the soil typically expressed by friction angle (f) and cohesion (c). TYPES OF SLOPE

Slopes can be categorized into two groups:

A. Natural slope • Hill sides • Mountains • River banks

B. Man-made slope • Fill (Embankment) • Earth dams • Canal banks • Excavation sides • Trenches • Highway Embankments Case histories of slope failure

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

Slope failures cost billions of $ every year in some countries Case histories of slope failure

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

Slope failures cost billions of $ every year in some countries Case histories of slope failure

Brazil, January 2003, 8 people killed Case histories of slope failure

LaConchita California Slump Case histories of slope failure Case histories of slope failure Case histories of slope failure

Slides: Rotational (slump) Case histories of slope failure TOPICS

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

 Falls  Topples  Slides  Flows  Creep  Lateral spreads  Complex Falls

• Rapidly moving mass of material (rock or soil) that travels mostly through the air 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 thus shatter 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. Falls

• Gravitational effect and shear strength

Gravity has two components of forces: T driving forces: T= W. sin b

Boulder N resisting forces (because of friction) N = W. cos b N T the interface develop its resistance from friction (f): S f = friction S = N tan f

In terms of stresses: S/A = N/A tan f or b tf = s tan f A = effective Base Area of sliding block 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. Slides

o Movements occur along planar failure surfaces that may run more-or less parallel to the slope. Movement is controlled by discontinuities or weak bedded planes. Back-Scrap A Slides A. Translational (planar)

Bulging at Toe Weak bedding plane Occur when soil of significantly different strength is presented (Planar) Slides

B. Rotational (curved) This is the downward movement of a soil mass occurring on an almost circular surface of rupture.

Back-Scrap

Bulging

Curved escarpment C. Compound (curved) (Slumps) Slides Slides

Reinforcement

Soil nails Slides

Reinforcement

Anchors شدادات

Possible failure surface Flows o The materials moves like a viscous fluid. The failure plane here does not have a specific shape.

It can take place in soil with high water content or in dry soils. However, this type of failure is common in the QUICK CLAYS, like in Norway. Flows 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. Lateral spreads o Lateral spreads usually occur on very gentle slopes or essentially flat terrain, especially where a stronger upper layer of rock or soil undergoes extension and moves above an underlying softer, weaker layer.

weaker layer Complex

Complex movement is by a combination of one or more of the other principal types of movement. 1. Falls 2. Topples 3. Slides • A. Translational (planar) • B. Rotational (slumps) 4. Flow 5. Creep 6. Lateral Spread 7. Complex Many slope movements are complex, although one type of movement generally dominates over the others at certain areas or at a particular time. Types of Slope Failures

In general, there are six types of slope failures:

1. Falls 2. Topples Slide is the most 3. Slides common mode of • Translational (planar) slope failure, and it will • Rotational (curved) be our main focus in this course 4. Flows 5. Creep 6. Lateral spreads 7. Complex 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 Non-circular surface (homogeneous soil) (non-homogeneous soil) Types of Slide Failure Surfaces

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

• Transitional Slip Surfaces: When there is a hard stratum at a relatively shallow depth Slides Rotational Translational (curved) (planar) Types of FailureSurfaces Typesof Failure surface Finite Finite Infinite Abovetoe the Deep Deep seated Through thetoe Plane failurePlane failure surface surface Long Long plane 1 3 2 Types of Failure Surfaces

Types of Failure Surfaces Considered in this Course are 1

Stability of infinite slopes

2 Stability of finite slopes with plane failure surfaces

3 Stability of finite slopes with circular failure surfaces TOPICS

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 and driving forces

• Driving forces – forces which move earth materials downslope. Downslope component of weight of material including vegetation, fill material, or buildings.

• Resisting forces – forces which oppose movement. Resisting forces include strength of material

• Failure occurs when the driving forces (component of the gravity) overcomes the resistance derived from the shear strength of soil along the potential failure surface. Methodology of Slope Stability Analysis

The analysis involves determining and comparing the shear stress developed along the most likely rupture surface to the shear strength of soil. Slope Stability Analysis Procedure

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. 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 is the 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 TOPICS

42 Factor of Safety

Factor of safety Resisting Force Driving Force

 Shear Strength Shear Stress

t t = Avg. Shear strength of soil F  f f s t t = Avg. Shear stress developed along the failure surface d d

43 Factor of Safety

• The most common analytical methods of slope stability use a

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

Fs is the ratio of resisting forces to the driving forces, or Shear strength (resisting movement) (Available) average shear strength of the soil.

Shear stress (driving movement) average shear stress (developed) developed along the potential failure surface.

FS < 1  unstable Generally, FS ≥ 1.5 is acceptable F ≈ 1  marginal S for the design of a stable slope FS >> 1  stable

If factor safety Fs equal to or less than 1, the slope is considered in a state of impending failure 44 Causes of slope failure

1. External causes These which produce increase of shear stress, like steepening or heightening of a slope, building on the top of the slope

2. Internal causes These which cause failure without any change in external conditions, like increase in pore water pressure.

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

45 Factor of Safety

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

cd ,fd = cohesion and angle of friction that develop along the potential failure surface Other aspects of factor of safety

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

When Fc  Ff then Fs  Fc  Ff 46 TOPICS

47 Stability of Infinite Slopes

What is an Infinite slope?

• Slope that extends for a relatively long distance and has consistent 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 of the slope. • For the analysis, forces acting on a single slice of the sliding mass along the failure surface is considered and end effects is neglected.

48 Infinite slope – no seepage o we will evaluate the factor of safety against a possible slope failure along a plane AB located at a depth H below the ground surface. o Let us consider a slope element abcd that has a unit length perpendicular to the plane of the section shown. o The forces, F, that act on the faces ab and cd are equal and opposite and may be ignored. The shear stress at the base of the slope element can be given by

Force parallel to the plane AB

Ta = W sin b = g LH sin b (*)

The resistive shear stress is given by

49 Infinite slope – no seepage

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

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

(***)

tan f F  For Granular Soil (i.e., c = 0) s tan b

This means that in case of infinite slope in sand, the value of Fs is independent of the height H and the slope is stable as long as b < f’ 50 Case of Granular soil – Derivation From Simple Statics Extra

L Equilibrium of forces on a slice:

FS  Resisting Forces Driving Forces

51 Infinite slope – no seepage

Critical Depth, Hcr

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

52 Infinite slope – with steady state seepage

Seepage is assumed to be parallel to the slope and that the ground water level coincides with the ground surface.

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

(*)

The resistive shear stress developed at the base of the element is given by

(**) 53 Infinite slope – with steady state seepage

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

(***)

Recall

(****)

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

c tan f F   s g H cos2b tan b tan b

No seepage

54 EXAMPLE

55 EXAMPLE

56 EXAMPLE

57 Stability of Infinite Slopes

• Cohesive Soils With seepage No seepage

tanf ' tanf' c' c' tanf ' c ' tanf ' c ' d d d d Fs Fs Fs Fs c' g ' tanf' c' tanf' Fs   Fs   g H cos2 b tanb g tanb 2 tanb sat sat gH cos b tanb

c' c' 1 Hcr  Hcr  cos2 b(g tanb g 'tanf') g cos2 b(tanb tanf ') sat d

58 Stability of Infinite Slopes

Granular Soils With seepage No seepage

c'0.0 c'0.0 c' g ' tanf' c' tanf' Fs   F   g H cos2 b tanb g tanb s 2 sat sat gH cos b tanb tanb g ' tanf' tanf' Fs  F  g tanb s sat tanb

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

59 TOPICS

60 Stability of Finite Slopes with Plane Failure Surface

o For simplicity, when analyzing the stability of a finite slope in a homogeneous soil, we need to make an assumption about the general shape of the surface of potential failure. o The simplest approach is to approximate the surface of potential failure as a plane. o However, considerable evidence suggests that slope failures usually occur on curved failure surfaces o Hence most conventional stability analyses of slopes have been made by assuming that the curve of potential, sliding is an arc of a circle.

61 Culmann’s Method o Culmann’s method assumes that the critical surface of failure is a plane surface passing through the toe. o Culmann’s analysis is based on the assumption that the failure of a slope occurs along a plane when the average shearing stress tending to cause the slip is more than the shear strength of the soil. o Also, the most critical plane is the one that simple wedge has a minimum ratio of the average shearing stress that tends to cause failure to the shear strength of soil.

o The method gives reasonably accurate results if the slope is vertical or nearly vertical. Plane Failure Surface 62 Culmann’s Method

• A slope of height H and that rises at an angle b is shown below. • The forces that act on the mass are shown in the figure, where trial failure plane AC is inclined at angle q with the horizontal. Similar procedures as for infinite slope, only different geometry. Also here we made optimization.

The average shear stress on the plane AC

= Ta W Sin q

t (*)

63 Culmann’s Method

The average resistive shearing stress (Developed shear strength) developed along the plane AC also may be expressed as

Na s’ td (**)

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

(***) 64 Culmann’s Method

Critical failure plane • The expression in Eq. (***) is derived for the trial failure plane AC. • To determine the critical failure plane, we must use the

principle of maxima and minima (for Fs=1 and for given values of c’, f’, g, H, b) to find the critical angle q:

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

(****)

65 Culmann’s Method

The maximum height of the slope for which critical equilibrium occurs can be obtained by substituting iinto into Eq. (****)

• For purely cohesive soils c  0 f = 0.

66 Culmann’s Method

• Steps for Solution

A. If Fs is given; H is required 1. F  F  F c f s c' 2. c '  d F C s H g tanf' f‘ 3. tanf '  d F s b   4c ' sin b cosf ' q 4. H  d  d  g  1 cos( b -f ' )  d 

67 Culmann’s Method

• Steps for Solution

B. If H is given; Fs is required

1. Assume F f tanf' 2. tanf '  d F s gH  1 cos( b -f ' ) 3. c '   d  d 4  sin b cosf '   d  c' 4. F  c c ' d 5. Check if F  F  F  F  F c f s c f 6. If F  F  try another F c f f 7. Repeat steps 1 5 68 EXAMPLE

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)?

Given equation:

H g = 20 kN/m3 f’=15o c’=50 kPa 45o 69 Key Solution

(a) The angle of the critical failure plane q b  45o can be calculated from: f’ 15o (b) The critical depth of the cut slope can be calculated from:

H g = 20 kN/m3 (c) The safe (design) depth of the cut f’=15o slope. c’=50 kPa 45o

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

70 TOPICS

71 Types of Failure Surfaces

Failure surface 1

Long plane Infinite failure surface Translational (planar) 2 Plane failure Finite

surface Slides 3 Above the toe Rotational Finite Through the toe (curved) Deep seated

72 Finite Slopes with Circular Failure Surface

Modes of Failure i. Slope failure • Surface of sliding intersects the slope at or above its toe. 1. The failure circle is referred to as a toe circle if it passes through the toe of the slope

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

ii. Shallow failure Under certain circumstances, a shallow slope failure can occur. Shallow slope failure 73 Finite Slopes with Circular Failure Surface iii. Base failure o The surface of sliding passes at some distance below the toe of the slope.

o The circle is called the H midpoint circle because its center lies on a vertical line drawn through the midpoint of the slope. Firm Base o For b  53o always toe o For b < 53o could be toe, slope, or midpoint and that depends on depth function D where: Depth function:

74 Finite Slopes with Circular Failure Surface

• 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°

75 Types of Stability Analysis Procedures

Various procedures of stability analysis may, in general, be divided into two major classes: 1. Mass procedure • In this case, the mass of the soil above the surface of sliding is taken as a unit.

• This procedure is useful when the soil that forms the slope is assumed to be homogeneous.

2. Method of slices • Most natural slopes and many man- made slopes consist of more than on soil with different properties.

• In this case the use of mass procedure is inappropriate. 76 Types of Stability Analysis Procedures

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

• It is a general method that can be used for analyzing irregular slopes in non-homogeneous slopes in which the values of c’ and f’ are not constant and pore water pressure can be taken into consideration. O b

R V1 a x W R

E1 h E2 W

V2 T

77 a N' TOPICS

78 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 not go into this analysis. 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) g W2 = (area of ABFEA) g

Mresisting = MR = cd (AED) (1) r 2 = cd r q

79 Mass Procedure

Mdriving = Md = W1l1 – W2l2

2 Mresisting = MR = cd r q

W1 l2 l1

80 REMARKS

• 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. In other words, Cd is maximum.

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

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

81 Finite Slopes with Circular Failure Surface

• Fellenius (1927) and Taylor (1937) have analytically solved for the minimum factor of safety and critical circles. • They expressed the developed cohesion as cd  g H m Where m  Stability number • We then can calculate the min Fs as H  height of slope γ  unit weigh t of soil c or m  d g H

• 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,

82 Finite Slopes with Circular Failure Surface

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

Toe slope Toe, Midpoint or slope circles

Firm Stratum

m is obtained from this chart depending on angle b 83 Finite Slopes with Circular Failure Surface

Failure Circle

 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

84 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

(radius)

Figure 15.13

85 Finite Slopes with Circular Failure Surface

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

86 Critical toe circles for slopes with b < 53°

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

m = 0.195

1. Get m from chart

2. Calculate cd from

cd  g H m

3. Calculate Fs

cu Fs  cd

88 How to use the previous chart?

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

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

m = 0.178

1. Calculate D = HD/H 2. Get m from the chart

3. Calculate cd from cd  g H m

cu 4. Calculate Fs Fs  cd

o o Note that recent investigation put angle b at 58 instead of the 53 value. 89 EXAMPLE

Rock layer 90 SOLUTION

D=1.5m

91 SOLUTION

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

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

Toe slope Toe, Midpoint or slope circles

Firm Stratum

m is obtained from this chart depending on angle b 93 REMARKS

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 o Determination of the magnitude of described previously is based on a trial surface of sliding. 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 o The maximum cohesion developed along the critical surface as c c  g Hf a,b,q,f d  f a, b,q,f  m  stability number d g H o The results of analytical solution to obtain minimum Fs was represented graphically as the variation of stability number, m , with slope angle b for various 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’ 94 Slopes in clay Soil with f = 0; Cu Increasing with Depth Slopes in clay Soil with f = 0; Cu Increasing with Depth EXAMPLE Slopes in Homogeneous C’ – f’ Soils o Here the situation is more complicated than for purely cohesive soils. o The Friction Circle method (or the f- Circle Method) is very useful for homogenous slopes. The method is generally used when both cohesive and frictional components are to be used.

o The pore water pressure is assumed to be zero o F—the resultant of the normal and frictional forces along the surface of sliding. For equilibrium, the line of action of F will pass through the point of intersection of the line

of action of W and Cd. 98 Friction Circle method

cd  g Hf a,b,q,f cd  g H m 99 Procedures of graphical solution

Given: H, b, g, c’, f’ Required: Fs

1. Assume f (Generally start with = f’ d Taylor’s stability i.e. full friction is mobilized) number 2. Calculate

3. With fd and b Use Chart to get m

4. Calculate cd  g H m c 5. Calculate Fc  cd 6. If Fc’ = Ff’ The overall factor of safety

Fs = Fc’ = Ff’

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

Plot the calculated points on Fc versus Fφ coordinates and draw a curve through the points. [see next slide]. Then Draw a line through the 100 origin that represents Fs= Fc = Fφ Procedures of graphical solution

Given: H, b, g, c’, f’ Required: Fs

Note: Similar to Culmann procedure for planar mechanism but here Cd is found based on m. In Culmann’s method Cd is found from analytical equation. 101 Calculation of Critical Height

Given: b, g, C ’, f ’ Required: Hcr

Hcr means that Fc’ = Ff’ =Fs = 1.0 1. For the given b and f’, use Chart to get m. 2.Calculate

c H  cr g m

102 SUMMARY

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

f = 0

C f 103 EXAMPLE

• Example 2 3 • Given: cu = 40 kN/m & g = 17.5 kN/m • Required: 1. Max. Depth

2. Radius r when Fs=1 3. Distance BC • b = 60 ° > 53 ° from Fig.15.13 m= 0.195

cu 40 Hcr   11.72 m g m 17.5*0.195 °  H  60  cr  DC AC sin a r  DC     q sin 2 2 2 From Fig. 15.14 for b  60oa 35oand q  72.5o H r  cr  11.72 17.28 m q 2sin a sin 2(sin 35)(sin36.25) 2 BC  EF  AF - AE  Hcr cot a - cot 60 9.97 m 104 EXAMPLE

105 SOLUTION

106 SOLUTION

107 SOLUTION

108 SOLUTION

1.42

109 2nd Midterm Fall 1436-1437H QUESTION #2

•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?

10 m g = 16 kN/m3 C = 40 kN/m2 15 m f = 15o

50o

110 SOLUTION

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

Fs = 2.0

b) fd

Cd = 40/1.5 =26.7 kpa

cd  g H m 26.7 = 16 X10 X m m = 0.167

o tan fd = tan f/1.5 fd=10.1

o o 111 At m = 0.167 and fd=10.1 from chart b =75 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

112 Method of Slices

• It is a general method that can be used for analyzing irregular slopes in non-homogeneous slopes in which the values of c’ and f ’ are not constant.

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

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

g1, c’1, b2 f’1 g2, c’2, g, c’, f’ f’2 b1 g3, c’3, f’3

Non-homogeneous Slope Irregular Slope 113 Method of Slices

• The soil mass above the trial slip surface is divided into several vertical parallel slices. The width of the slices need not to be the same (better to have it equal). • The accuracy of calculation increases if the number of slices is increased. • 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 procedure requires that a series of trial circles are chosen and analyzed in the quest for the circle with the minimum factor of safety.

114 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 ,Pn ,Pn+1 ,Tn ,Tn+1

• 3 equations SFx=0 , SFy=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 115 Method of Slices

For the whole sliding mass S Mo  0 SW *r*sina -ST *r  0 SW *sina  ST t f T t *l  *l d F s t f SW *sina  *l F s St *l f Fs  SW *sina S(c*l s *tanf *l) Fs  n SW *sina 116 Method of Slices

S(c*l s *tanf *l) Fs  n SW *sina Ssn *l  SN Sc*l  tanf *SN Fs  SW *sina Equation is exact but approximations are introduced in finding the value of force N

Two Methods: •Ordinary Method of Slices •Bishop's Simplified Method of Slices

117 Ordinary Method of Slices

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+1

118 Ordinary Method of Slices

SFn  0 (to stay away from Tr )

Nr  Wn * cosa n n S ( c* l W * cosa tanf ) F  n n n s SW * sina n n For undrained condition: c  cu f  0 c l F  u s SW * sina n n

119 Ordinary Method of Slices

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 • a n is taken at the middle of the slice wn wn • Calculate the terms in the equation S(c* l W * cosa tanf ) F  n n n s an SW * sina an n n -ve +ve

• Fill the following table

Slice# wn an sin an cos an ln wn sin an wn cos an

120 EXAMPLE

Find Fs against sliding Use the ordinary method of slices

121 Bishop’s Simplified Method of Slices

Assumption For each slice, the resultant of the interslice forces is Horizontal.

i.e. Tn =Tn+1

122 Bishop’s Simplified Method of Slices

SFy  0 (to stay away from Pn and Pn1 ) y Wn  Nr * cosa n  Tr * sina n  c  s tanf   n  Tr  t d * ln   ln  Fs 

cln s nln tanf Tr   Fs Fs

cln Nr tanf Tr   Fs Fs

cln Nr tanf Wn  Nr * cosa n  sina n  sina n Fs Fs

123 Bishop’s Simplified Method of Slices

cln Wn  sina n Fs Nr  tanf sina n cosa n  Fs  cl  W  n sina  n F n  Scl  tanf s  n tanf sina cosa  n   n F  b F   s  but l  n s tanf sina n n cosa n cosa n  Fs 1 cb W tanf F  n n s S tanf sina SWn sina n n cosa n  Fs Trail and error procedure 124 Bishop’s Simplified Method of Slices

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 1 cb W tanf F  n n s S tanf sina SWn sina n n cosa n  Fs

• Fill the following table

Slice# wn an sin an cos an bn wn sin an

• 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. 125 Bishop’s Simplified Method of Slices

tanf sina n ma ( n )  cosa n  Fs 1 ( cbn Wn tanf ) ma ( n ) Fs  S SWn sina n

1 c' b W tanf F  ( n n ) s  sin a tanf Wn sin a n n cosa n  Fs

126 Bishop’s Simplified Method of Slices

• Example of specialized software: – Geo-Slope, – Geo5, – SVSlope – Many others

127 Final Exam Fall 36-37 QUESTION #4

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”.

128 SOLUTION

Fs = 1.8 Table 1. “Fill only necessary cell for this particular problem” Width Height Height Area Weight Slice W sin a b h h A W α m n No. n l 2 n (n) α(n) (kN/m) (m) (m) (m) (m2) (kN/m) (7) (8) (1) (9) (2) (3) (4) (5) (6) 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

129 Remarks on Method of Slices o Bishop’s simplified method is probably the most widely used (but it has to be incorporated into computer programs). o It yields satisfactory results in most cases. o The Fs determined by this method is an underestimate (conservative) but the error is unlikely to exceed 7% and in most cases is less than 2%. o The ordinary method of slices is presented in this chapter as a learning tool only. It is used rarely now because it is too conservative. o The Bishop Simplified Method yields factors of safety which are higher than those obtained with the Ordinary Method of Slices. o The two methods do not lead to the same critical circle. o Analyses by more refined methods involving consideration of the forces acting on the sides of slices show that the Simplified Bishop Method yields answers

for factors of safety which are very close to the correct answer. 130 Remarks on Method of Slices

Two Methods: Ordinary Method of Slices

• Underestimate Fs (too conservative) • Error compared to accurate methods (5-20%) • Rarely used

Bishop’s Simplified Method of Slices • The most widely used method • Yields satisfactory results when applying computer program

131