Geotechnical Engineering II CE 481

Geotechnical Engineering II CE 481
4. Slope Stability Chapter 15 Omitted parts: Sections 15.9, 15.14, 15.15, 15.16, 15.17 Figures 15.24, 15.25, 15.26, 15.27 and related texts

Contents 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 Culmann’s Method Stability of Finite Slopes with Circular Failure Surface Mass Method Method of Slices

Introduction 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: 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.

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. 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 Slopes Slopes can be categorized into two groups:-
Natural slope Hill sides Mountains River banks 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

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

Central America??

Brazil, January 2003, 8 people killed

Human activities disturb large volumes of earth materials in construction of buildings, transportation routes, dams and reservoirs, canals, and communications systems, and thus have been a major factor in increases in damages due to slope failures.

LaConchita California Slump

Slides: Rotational (slump)

Falls Topples Slides Flows Creep Lateral spreads Complex
Types of Slope Movements Slope instability (movement) can be classified into six different types: Falls Topples Slides Flows Creep Lateral spreads Complex

I. 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.

A = effective Base Area of sliding block
Gravitational effect and shear strength Gravity has two components of forces: T driving forces: N resisting forces (because of friction); T= W. sin b Boulder N = W. cos b W T N 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

II. Topples This is a forward rotation of soil and/or rock mass about an axis below the center of gravity of mass being displaced.

III. Slides 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 Slides A. Translational (planar) A Occur when soil of significantly different strength is presented Bulging at Toe Weak bedding plane (Planar)

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)

Reinforcement Soil nails

Reinforcement Possible failure surface Anchors شدادات

IV. Flows 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.

V. 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.

VI. Lateral spreads 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

VII. Complex Falls Topples Slides Flow Creep Complex
Complex movement is by a combination of one or more of the other principal types of movement. Falls Topples Slides A. Translational (planar) B. Rotational (slumps) Flow Creep Complex Many slope movements are complex, although one type of movement generally dominates over the others at certain areas or at a particular time.

Mass Movements: Five Main Types
To view this animation, click “View” and then “Slide Show” on the top navigation bar. 31

Factors Affecting Stability of Slopes

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.

Concept of Instability
Instability occurs when the shear stresses t that cause movement (e.g. gravitational forces W) overcome the internal shear strength tf of the soil (cohesion C and friction f between the soil grains) along potential plane of failure. W Accordingly, the factors causing instability may be divided to: Factors causing increased Shear Stresses t Factors causing a reduction in Shear Strength tf tf (f, c) t tf t (f, c) tf (f, c)

Types of Slope Failures Considered in this Course
In general, there are six types of slope failures: Falls Topples Slides Translational (planar) Rotational (curved) Flows Creep Lateral spreads Complex Slide is the most common mode of slope failure, and it will be our main focus in this course

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)

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

Translational (planar) Long plane failure surface
Types of Failure Surfaces Considered in this Course Failure surface 1 Slides Translational (planar) Infinite Long plane failure surface Finite Plane failure surface Rotational (curved) Above the toe Through the toe Deep seated 2 3

Types of slope failures considered in this course
1 Stability of infinite slopes 2 Stability of finite slopes with plane failure surfaces 3 Stability of finite slopes with circular failure surfaces

Concepts of Slope Stability Analysis
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.

  d f 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 Assume a probable failure surface 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. 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. Based on the minimum FS, determine whether the slope is safe or not.

Methods of Slope Stability Analysis
Limit equilibrium Limit analysis 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 The problem is considered in two-dimensions The failure mass moves as a rigid body The shear strength along the failure surface is isotropic The factor of safety is defined in terms of the average shear stress and average shear strength along the failure surface

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) (developed) average shear stress developed along the potential failure surface. FS < 1  unstable FS ≈ 1  marginal FS >> 1  stable W (f, c) td tf H Generally, FS ≥ 1.5 is acceptable for the design of a stable slope td If factor safety Fs equal to or less than 1, the slope is considered in a state of impending failure.

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

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.

I. Infinite slope – no seepage
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. Let us consider a slope element abcd that has a unit length perpendicular to the plane of the section shown. 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

For Granular Soil (i.e., c = 0)
The effective normal stress at the base of the slope element is given by …….(##) Equating R.H.S. of Eqs. (#) and (##) gives …….(###) For Granular Soil (i.e., c = 0) Critical Depth, Hcr The depth of plane along which critical equilibrium occurs is obtained by substituting Fs = 1 and H = Hcr into Eq. (###). 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’

Extra Case of Granular soil – Derivation From Simple Statics
Equilibrium of forces on a slice: L Driving Forces Resisting Forces

II. 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 (**)

Equating the right-hand sides of Eq. (*) and Eq. (**) yields
(***) Recall (****) Substituting Eq. (****) Into Eq. (***) and solving for Fs gives No seepage

General Case of Seepage
EXTRA ($) 𝝉 𝒅 = 𝒄 ′ 𝑭 𝒔 +(𝝈−𝒖) 𝒕𝒂𝒏 𝝋′ 𝑭 𝒔 𝝈 ′ = 𝑵 𝒂 𝑨𝒓𝒆𝒂 𝒐𝒇 𝒃𝒂𝒔𝒆 = 𝜸𝑯𝑳. 𝒄𝒐𝒔 𝟐 𝜷 𝑳 𝑪𝒐𝒔𝜷 =𝐴=𝜸𝑯. 𝒄𝒐𝒔 𝟐 𝜷 𝝉 𝒅 = 𝒄 ′ 𝑭 𝒔 +(𝜸𝑯. 𝒄𝒐𝒔 𝟐 𝜷 −𝒖) 𝒕𝒂𝒏 𝝋′ 𝑭 𝒔 ($$) Equating R.H.S of ($) and ($$) and rearranging gives The challenge here is to evaluate the value of u. We need to go to seepage problems 𝑭𝑺= 𝒄 ′ 𝜸.𝑯 .𝐜𝐨𝐬 𝜷 .𝐬𝐢𝐧 𝜷 +(𝟏− 𝒖 𝜸.𝑯. 𝒄𝒐𝒔 𝟐 𝜷 ) 𝒕𝒂𝒏 𝝋′ 𝒕𝒂𝒏 𝜷

Extra Summary of Stability of Infinite Slopes H
𝑭𝑺= 𝒄 ′ 𝜸 .𝑯 .𝐜𝐨𝐬 𝜷 .𝐬𝐢𝐧 𝜷 +(𝟏− 𝒖 𝜸.𝑯. 𝒄𝒐𝒔 𝟐 𝜷 ) 𝒕𝒂𝒏 𝝋′ 𝒕𝒂𝒏 𝜷 H With water seepage: 𝒖= 𝜸.𝑯. 𝒄𝒐𝒔 𝟐 𝜷 (The ground water level coincides with the ground surface) 𝑭𝑺= 𝒄 ′ 𝜸.𝒉 .𝐜𝐨𝐬 𝜷 .𝐬𝐢𝐧 𝜷 + 𝒕𝒂𝒏 𝝋′ 𝒕𝒂𝒏 𝜷 This indicates that in an infinite slope in sand, the value of Fs is independent of the height H and the slope is stable as long as b < f’. If no water (u = 0): 𝑭𝑺= 𝒕𝒂𝒏 𝝋′ 𝒕𝒂𝒏 𝜷 If no water (u = 0) and soil is granular (C’=0) :

GS (ground surface)

From Eq

Concepts of Slope Stability Analysis Factor of Safety Stability of Infinite Slopes Stability of Finite Slopes with Plane Failure Surface Culmann’s Method Stability of Finite Slopes with Circular Failure Surface Mass Procedure Method of Slices

Stability of Finite Slopes with Plane Failure Surface
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. The simplest approach is to approximate the surface of potential failure as a plane. However, considerable evidence suggests that slope failures usually occur on curved failure surfaces Hence most conventional stability analyses of slopes have been made by assuming that the curve of potential sliding is an arc of a circle. ,

Culmann’s Method Culmann’s method assumes that the critical surface of failure is a plane surface passing through the toe. 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. Also, the most critical plane is the one that has a minimum ratio of the average shearing stress that tends to cause failure to the shear strength of soil. Plane Failure Surface simple wedge The method gives reasonably accurate results if the slope is vertical or nearly vertical.

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 (&) 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 (&&&)

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 (&&&&)

For purely cohesive soils c  0  = 0.
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  = 0. 𝜽 𝒄𝒓 = 𝜷 𝟐 𝑯 𝒄𝒓 = 𝟒 𝒄 ′ 𝜸 𝐬𝐢𝐧 𝜷 𝟏− 𝐜𝐨𝐬 𝜷

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 = 20 kN/m3 f’=15o c’=50 kPa 45o

(a) The angle of the critical failure plane q can be calculated from:
Key Solution: (a) The angle of the critical failure plane q can be calculated from: (b) The critical depth of the cut slope can be calculated from: b = 45o f’ =15o H = 20 kN/m3 f’=15o c’=50 kPa (c) The safe (design) depth of the cut slope. 45o d where: c’d and f’d can be determined from:

Possible Cases I. Given b, c’ and f’ required critical height
Directly from II. Given b, Fs, c’ and f’ required height for the given factor of safety Solve for H III. Given H, b, q, c’ and f’ required factor of safety Assume Ff tan fd = tan f/Ff Find Cd Fc = C/Cd Repeat until Fc = Ff

Solution 2nd Midterm Fall 1436-1437H (QUESTION #1)
A cutting is to be excavated in a clay stratum at an angle that is equal to 60o. The relevant parameters are g= 17 kN/m3, c = 20 kN/m2 and f = 25o. Assume the failure mechanism to be planar (i.e. Cullman’s method) and determine:- The height of the excavation that will have a factor of safety of 1.5 against sliding. The critical height of the excavation. Repeat (b) if the soil is purely cohesive. Compare the answer with the result of part (b) and comment. Solution Mother equation

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

iii. Base failure The surface of sliding passes at some distance below the toe of the slope. The circle is called the midpoint circle because its center lies on a vertical line drawn through the midpoint of the slope. H Firm Base For   53o always toe For  < 53o could be toe, slope, or midpoint and that depends on depth function D where: Depth function:

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.

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 2 1 W V E T N' a h R a x R W

I. 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 = cd r2q 𝑭 𝒔 = 𝒓 𝟐 𝜽 𝒄 𝒖 𝒘 𝟏 𝒍 𝟏 − 𝒘 𝟐 𝒍 𝟐

Mdriving = Md = W1l1 – W2l2 Mresisting = MR = cd r2q W1 l2 l1 W2 𝑭 𝒔 = 𝒓 𝟐 𝜽 𝒄 𝒖 𝒘 𝟏 𝒍 𝟏 − 𝒘 𝟐 𝒍 𝟐 MR = cu r2q

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.

Fellenius (1927) and Taylor (1937) have analytically solved for the minimum factor of safety and critical circles. They expressed the developed cohesion as We then can calculate the min Fs as 𝑭 𝒔 = 𝑪 𝒖 𝜸𝑯𝒎 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, 𝑯 𝒄𝒓 = 𝑪 𝒖 𝛄𝒎

m is obtained from this chart depending on angle b.
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 b Firm Stratum m is obtained from this chart depending on angle b.

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 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

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 15.14 (radius) Figure 15.14

When the critical circle is a midpoint circle (i. e
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 Figure 15.15 Firm base

Critical toe circles for slopes with b < 53°
The location of these circles can be determined with the use of Figure and Table 15.1. Figure 15.16 Note that these critical toe circle are not necessarily the most critical circles that exist.

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

How to use the previous chart?
Given: b =30o, 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 Calculate D = HD/H Get m from the chart Calculate cd from 4. Calculate Fs Note that recent investigation put angle b at 58o instead of the 53o value.

Rock layer

D=1.5m

𝑭 𝒔 = 𝑪 𝒖 𝜸𝑯𝒎

2. 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 b Firm Stratum m is obtained from this chart depending on angle b.

3. Slopes in Homogeneous C’ – f’ Soils
Here the situation is more complicated than for purely cohesive soils. 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. 𝑨𝑪 is a trial circular arc that passes through the toe of the slope, and O is the center of the circle. The pore water pressure is assumed to be zero 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.

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. We can then calculate c’d from Determination of the magnitude of described previously is based on a trial surface of sliding. 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). The maximum cohesion developed along the critical surface as 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 ). Solution to obtain the minimum Fs using this graph is performed by trial-and-error until Fs = Fc’=Ff’

Procedures of graphical solution
Taylor’s stability number Given: H, b, g, c’, f’ Required: Fs Assume fd (Generally start with = f’) i.e. full friction is mobilized) Calculate With fd and b Use Chart to get m If Fc’ = Ff’ The overall factor of safety Fs = Fc’ = Ff’ If Fc’≠ Ff’ reassume fd and repeat steps 2 through 5 until Fc’ = Ff’ Or 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 origin that represents Fs= Fc = Fφ

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.

Hcr means that Fc’ = Ff’ =Fs = 1.0
Calculation of Critical Height Given: b, g, C ’, f ’ Required: Hcr Hcr means that Fc’ = Ff’ =Fs = 1.0 For the given b and f’, use Chart to get m. Calculate

SUMMARY Mass Procedure – Rotational mechanism need only the use of Taylor’s chart. f = 0 C- f

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 f = 15o 15 m 50o

Solution a) Fs = 2.0 fd b) Cd = 40/1.5 =26.7 kpa 26.7 = 16 X10 X m
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 Fs = 2.0 fd b) Cd = 40/1.5 =26.7 kpa 26.7 = 16 X10 X m m = 0.167 tan fd = tan f/1.5 fd=10.1o At m = and fd=10.1o from chart b =75o

Non-homogeneous Slope Irregular Slope
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, f’1 b2 g2, c’2, f’2 g, c’, f’ b1 g3, c’3, f’3 Non-homogeneous Slope Irregular Slope

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

Extra Analysis Using the Method of Slices
Taking moment about O, the sum of the moments of the shear forces Ti on the failure arc must equal the moment of the soil mass ABC For equilibrium of each slice S Mo = 0 Wn r Sin an = Tr r Wn sin an = Tr Still here we do not have the free body diagram. In away we have mass procedure for each slice. When we come to evaluate Nr we will have free body diagram for each slice . DLn is approximately equal to (bn)/(cos an), where bn the width of the nth slice. The value of an may be either positive or negative. The value of an is positive when the slope of the arc is in the same quadrant as the ground slope. For clarity Tr and Nr we drop the subscript n

Extra For Equilibrium of all slices
r sinan This is exact but approximations are introduced in determining the force Nr. bn To get Nr we have to look at the Forces on each slice. Wn = weight of slice = g h bn Nr= normal force at the base Tr = shear force at base Pn, Pn+1 = Normal force on sides of slice Tn, Tn+1 = Shear forces on sides of slice 6 No. of Unknowns  Nr, Tr, Pn, Pn+1, Tn, Tn+1 3 equilibrium equations  SFx =0, SFy =0, S M=0 h The system is statically indeterminate. Assumptions must be made to solve the problem. Different assumptions yield different methods.

Extra Therefore, some assumptions must be made to achieve a statically determinate solution. The most common procedures to render the analysis determinate have involved an assumption regarding the interslice forces. Varying assumptions regarding the interslice forces have lead to several different methods of slices, one exception is the Ordinary or Fellenius Method, which simply ignores the interslice forces. This method can lead to substantial error and is no longer commonly used in practice. The various limit equilibrium methods not only use different assumptions to make the number of equations equal to the number of unknowns but also differ with regard to which equilibrium equations are satisfied.

Slope Analysis Methods: Rotational Failure Surface
Extra Slope Analysis Methods: Rotational Failure Surface “OMS” (“Modified Bishop’s”) The U.S. Army Corps of Engineers’ Modified Swedish Methods Nonvellier Lowe –kara fiath Methods that satisfy static equilibrium fully are referred to as complete equilibrium methods.

Ordinary Method of Slices (OMS) (Fellenius’ Method)
This method is also referred to as "Fellenius' Method" and the "Swedish Circle Method". The method ignores both shear and normal interslice forces and considers only moment equilibrium. This can be put as: The resultant of Pn and Tn = the resultant of Pn+1 and Tn+1 The line of action of the resultant coincide. Therefore, the interslice forces cancel each other. From SFvertical = 0 an Y’ Nr = Wn cos an (*) (**) Substituting Eq. (*) into Eq. (**)

Soil Properties g C’ f’ xn hn an bn

Steps for solution using Ordinary Method of Slices
Draw the slope to a scale Divide the soil above the sliding surface into slices (choose 5 to 8 slices) Measure bn, xn, hn from the drawing. Fill the following table: Fs = sum of col 12/sum of col 8 Col 1 Col 2 Col 3 Col 4 Col 5 Col 6 Col 7 Col 8 Col 9 Col 10 Col 11 Col 12 Slice # bn hn g Wn xn an Wn sina DLn c’ f’ c’ DLn + Wn cosan tan f’ Sin an = xn/r DLn= (bn)/(cos an) To find the minimum factor of safety—that is, the factor of safety for the critical circle—one must make several trials by changing the center of the trial circle.

Assume there is no seepage in the slope.

Note that the value of an may be either positive or negative
Note that the value of an may be either positive or negative. The value of an is positive when the slope of the arc is in the same quadrant as the ground slope.

Nr = Wn cos an Ordinary Method of Slices (OMS) (Fellenius’ Method)
From SFY’= 0 Y’

Bishop’s Simplified Method of Slices (BSMS)
In 1955, Bishop proposed a more refined solution to the ordinary method of slices. In this method, the effect of forces on the sides of each slice are accounted for to some degree. Assumption Tn = Tn+1 Recall To get Nr , S Fvertical = 0 ( to avoid Pn & Pn+1 ) Recall for OMS Nr = Wn cos an

Nr From the above equations, we can see that Fs is on both sides of the equation (indeterminate problem). Therefore, a trial-and-error procedures needs to be adopted to find the value of Fs.

Note: In the textbook he set
Then using the force polygon to express Nr and at the end assuming DT = 0. In our case from the beginning we put DT = 0 and we directly use S Fy =0.

2. Repeat step 1 until Fs (RHS) = Fs (LHS)
Procedures 1. we assume Fs and insert in RHS of the equation and calculate Fs in the LHS of the equation. 2. Repeat step 1 until Fs (RHS) = Fs (LHS) As in the ordinary method of slices, a number of failure surfaces must be investigated so that we can find the critical surface that provides the minimum factor of safety. Mass Method vs. Method of slices in search for the minimum factor of safety.

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

Notes Bishop’s simplified method is probably the most widely used (but it has to be incorporated into computer programs). It yields satisfactory results in most cases. 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%. 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. The Bishop Simplified Method yields factors of safety which are higher than those obtained with the Ordinary Method of Slices. The two methods do not lead to the same critical circle. 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.

Stability Analysis by Method of Slices for Steady-state Seepage
The fundamentals of the ordinary and Bishop’s simplified method of slices were presented above, assuming the pore water pressure u =0. However, for steady-state seepage through slopes, as is the situation in many practical cases, the pore water pressure must be considered when effective shear strength parameters are used. The factor of safety becomes: For the ordinary method of slices For Bishop’s simplified method of slices un is the average pore water pressure at the base of the slice:

Slopes in Homogeneous Clay and Layered Soil with f = 0 Mass procedure
X1 W1 X2 W2

Slopes With Layered Soil – method of slices
The method of slices can be extended to slopes with layered soil. The general procedure of stability analysis is the same. Only when the factor of safety is calculated, the values of f’ and c’ will not be the same for all slices.

Loaded Slopes P x + P . X

Slopes in Homogeneous Clay Soil with f = 0 and tension crack
X1 W1

Berm for increasing stability

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”.

Fs = 1.8 Table 1. “Fill only necessary cell for this particular problem” Slice No. (1) Width bn (m) (2) Height hl (3) h2 (4) Area A (m2) (5) Weight Wn (kN/m) (6) α(n) (7) mα(n) (8) Wn sin a (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 ?

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Geotechnical Engineering II CE 481

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Geotechnical Engineering II CE 481

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