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slope stability 2 - 1998 - 1999


									ENV-2E1Y Fluvial Geomorphology 2004 - 2005

Multiple Landlsides at Yuen Mo Village, Kowloon East during the rain storm of 29 - 31st May 1982 when over 530 mm of rain fell. The collapse occurred in the late morning of 30 th May and most of the huts in the village were destroyed or severely damaged. Three people were killed. At 16:15, the site was inspected by Emergency Duty Officer, N. K. Tovey who had previously inspected 4 other landslides in neighbouring villages, each one of which involved deaths. All remaining huts were condemed by Dr N.K. Tovey and a permanent evacuation order on all 120 inhabitants of Yuen Mo was issued. From that time Yuen Mo Village ceased to exist.

Slopes and related topics Section 5 Slope Stability

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

Slope Stability and Related Topics 5. Slope Stability
5.1. Introduction
The stability of slopes and whether or not massive failure in the form of landslides occurs is dependent on several factors as were described in the introduction to this course. These may be summarised as:the geometry of the slope including the geometric configuration of the varying strata - determined by surveying methods, water flow within the slope - analysed using techniques covered in section (2) of this course, the material properties of the differing strata, including the unit weight angle of friction and cohesion, which are in turn dependent on the previous consolidation history of the soil, additional loading by man. There are several methods by which the stability of a slope may be analysed, many are valid only under certain conditions. There is also a group of more general solutions which can be applicable in all cases, but sometimes it is difficult to find a solution even with the aid of a computer. Assuming we know the geometry of the slope and the underlying strata, the relevant material properties, and we also understand the water flow, then all methods of analysis begin with postulating a failure mechanism. It is essential that we correctly identify the most critical mechanism, and this usually is a matter of experience. In the past, some slopes have been analysed and given a clean bill of health, but as a less than critical failure mechanism was identified failures have occurred on "theoretically" stable slopes sometimes with potentially disastrous consequences (e.g. the Tsing Yi, Hong Kong failures above the PEPCO oil storage depot following the rainstorm of 29th - 31st May 1982).

5.2. Types of failure
Failures in slopes may:1) be straight lines (particularly so in granular media) 2) approximate to arcs of circles 3) approximate to logarithmic spirals 4) be a combination of straight lines, arcs of circles, and/or logarithmic spirals. Examples are shown in Fig. 5.1

Fig. 5.1 Examples of different methods for analysing stability of slopes. a) example applicable for a purely cohesive soil where slip surface is a circle; c) Infinite slope where slope is of approximately constant slope over a significant distance which is much greater than depth to bedrock. The failure surface is also parallel to this slope. Water flow can be included if it is parallel to slope as can changes in strata;

b) example applicable for soil with both cohesion and friction, but water flow must be absent. Failure is a straight line; 55

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

d) general case for slope of general shape. Slip surface may be of any form and may be composite including arcs or circles and straight line portions. In the example shown, there is a straight line section which is along the bedrock plane. Water flow can be incorporated as can variation in strata and the presence of tension cracks.

Strain is defined as a non-dimensional ratio which is the displacement during shearing over the original length of the sample. In a triaxial test, the sample is usually 75 mm long and a deformation of about 1 - 2 mm is needed to achieve peak strength in a dense test representing 1 - 3% strain. For loose samples, the deformation will be around 10 mm in a sample of comparable length. In a slope the soil mass is large, and it is not possible for the whole slope to deform (with the associated volume change instantaneously. Near the base of the slope, the material can expand and bulge slightly and allow small strains along the potential failure plane as shown by the shaded region in Fig. 5.2. The corresponding point on the stress - strain diagram is shown at point A on the rising part of the curve. Further around the failure zone, the points B and C have low amount of strain on the stress - strain plot, and the mobilised shear strength is thus small.

5.3. Progressive failure.
Most slope failures occur during or immediately after periods of heavy rain when the water table is high. Thus slope failures on the North Norfolk Coast are more common in winter during times of high water table. Equally, movement of the Mam Tor Landslide in Derbyshire occurs during the winter months and usually only if more than 400 mm of rain falls in the critical period. In Hong Kong, landslides are rare in the winter months from November to March, and are very common in the summer months (May - August) and over 500 landslides occurring in a single day have been reported. Slopes may be triggered by rainfall, and may catastrophically fail if the rainstorm is prolonged (e.g. Po Shan Road, Hong Kong, 1972), but not infrequently, the failure is progressive with small amounts of movement until eventually the failure is catastrophic in a particular event (e.g. Aberfan, Tsing Yi). After massive and catastrophic failure, continued movement may take place (e.g. Mam Tor).

Fig. 5.3 Failure is now more advanced. The most highly stressed region has just passed peak shear strength, while at B, the strength is approaching peak. Region C is still relatively lightly stresses. Bulging at toe might noticeable in aerial photographs and might be visible to naked eye.

Fig. 5.2

Region of high stress in a slope prior to failure.

The stress-strain diagram indicates the approximate states of stress at points along the potential failure zone. Slight bulging of the toe would be discernible with accurate survey measurements. Unlike materials such as steel which show relatively little deformation before failure, soils deform by a considerable amount before the peak shear strength is achieve in the case of dense sand or over consolidated clays. For loose sands and normally consolidated clays, the deformation before the ultimate strength is reached is large. 56

Once the lower part of the slope has deformed, the next part can deform (see Fig. 5.3. Here, the lowest part of the failure zone is at the peak shear strength, while moderate strengths have been mobilised further along the failure arc. The stress points corresponding to points B and C have moved further up the curve. Finally (Fig. 5.4), after further deformation, bulging should become very evident at the base of the slope while the stress at the point A will now be less than previously as it has past the peak strength while the stress at B is now at peak and that at C is rising rapidly. All along the failure arc, the mobilising shear stress will vary and it is the integrated value of the strength along the whole failure surface which will determine

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

whether or not failure will occur. Such a failure which develops in this fashion is called a progressive failure. Of importance is the fact that there will be a time delay (albeit quite short in some cases) from the start of the failure to the time of catastrophic failure. Normally, evidence of failure may be detected from bulging of the toe (early stage) and the development of tension cracks at the top (later stage), and finally a settlement of the crest immediately prior to failure.

failure envelope on the Mohr - Coulomb envelope is a constant irrespective of normal stress (i.e.  = 0). This method can be used for any slope profile, but is more suited to simple shapes. Water flow must be absent, i.e. excess water pressures, although the method may be use for slopes which are entirely submerged. Only single strata must be present. 2) those methods where the potential failure surface approximates to a straight line. It is valid for solids with both friction and cohesion, but there must be no water flow. The analysis is possible for irregular shaped surfaces, but it is more usually used for simple shapes. It is not really suited if there is more than one stratum. 3) those methods for the analysis of slopes which are approximately infinite in extent compared to the depth of the soil material. The angle of the slope is approximately constant over a large distance. Differing strata may be present, but only parallel to the surface. The method can deal with water flow provided that it is parallel to the surface. Both frictional and cohesive materials may be present. This method of analysis is known as the Infinite Slope Method. 4) those methods which are applicable to the analysis of general slope stability. They are valid for varying ground water flow conditions, for various modes of failure (straight - line, arcs of circles or various combinations), for slopes with varying strata which may or may not be parallel to the failure surface or slope surface, and for slopes in which tension cracks have developed from desiccation of the surface layers. These methods are collectively known as the Method of Slices, and there are several variants depending on the extent of approximations made. Generally speaking all assumptions are SAFE ASSUMPTIONS in that they underestimate the stability of the slope. In the case of the Infinite slope method, the failure surface will always be parallel to the surface, and for a single stratum it can be shown (see section 5.7), that the stability is unaffected by the depth of the potential failure surface. For all other methods of analysis, the method first assumes a failure surface of appropriate shape and analyses the stability to obtain a factor of safety (Fs) 57

Fig. 5.4.

Whole of potential failure surface is now highly stressed with region A well beyond peak strength at the residual strength, region B at peak strength and region C approaching peak strength. Noticeable bulging at the toe which should be seen by naked eye. Failure is imminent.

Though the circumstances leading up to the Aberfan disaster on October 21st 1966 were contributory to the disaster. There were many signs in the months and years before that a potential disaster that a disaster might occur, the consequences of the disaster could have been avoided even at a late stage. Two people who were working at the top of the waste tip about 30 minutes before the disaster noted tension cracks and a settlement at the top. In vain they attempted to raise the alarm, but vandals had removed the wires of the communication telephones.

5.4. Methods of analysis
There are many methods available for studying the stability of slopes, and for some there are several variants. In this course we shall consider 4 basic methods:1) those methods for relatively shallow slopes in normally consolidated or lightly over consolidated materials and in which the soil material may be considered to be purely cohesive ( and undrained) situations (i.e. the

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

Fs 

inherent strength of the soil strength required for stability

If the computed factor of safety is less than unity, the slope is clearly unstable and likely to fail. If the factor of safety is greater than unity we cannot assume that the slope is stable as we may not have chosen the most critical mode of failure (i.e. failure surface). It is thus necessary to repeat the calculations with a different failure surface until the most critical one is found. Usually, experience will give a guidance as to what types of failure are likely to be more critical than others. In the case of the straight line failures (other than the infinite slope cases), it is possible to assess the stability on a number of different failure planes each one at a different angle to the horizontal. A graph such as Fig. 5.5 is now plotted with the angle of the potential failure surface as the X - axis, and the computed Factor of Safety as the Y - axis. The value of Fs will be high for shallow angles, and fall as increases. After a critical angle, the value of s will rise again as continues to increase. In this example, the critical value of Fs is clearly the minimum of the curve. A similar approach may be used for the purely cohesive failures. In this case, the it is the radius of curvature of the failure arc which is used as the independent variable on the X - axis.

This cohesive force will be the resisting force, while the weight of the slope acting through the centre of gravity of the potential siding segment will be the mobilising force. In this case since the cohesive force and the weight do not act at a point we must also consider the moments of the forces in assessments of equilibrium, i.e. since the centre of gravity is not below the centre of the circle of failure, the weight will act as a pendulum and attempt to cause the segment to rotate. This tendency to move is counteracted by the cohesive force. The assess the moments of the two key forces we need to determine the distances of their respective lines of action from the centre of the circle in a direction at right angles to the force. The weight acts at a distance X from the point O, while the cohesion acts along a tangent (i.e. at the radius of the slip surface).

Fig. 5.6

Failure of a slope in a purely cohesive medium. This failure surface is an arc of a circle. It is necessary to determine the distances X, R and L for analysis.

the factor of safety = is given by:= Fig. 5.5. Variation of factor of safety with orientation of failure zone to horizontal in a straight line failure.

Restoring Moment ------------------------------Mobilising Moment



Though this is a simple method of analysis there are some practical problems which require some ingenuity to solve. A question similar to this was set as an exam question a few years ago. One problem is to determine the weight of the sliding mass:- this is equal to the area of the slice multiplied by the unit weight. The area could be computed by drawing the sliding mass on graph paper and counting squares, or it can be derived from geometry (rather more complex!). In the case in question, since this was a 40 minutes question (rather than the 60 minute question now set), the area of the wedge was given. 58


Method of Analysis - I - purely cohesive failures.

Fig. 5.6 illustrates this type of failure. The failure surface is an arc of a circle with centre at point O. There is no frictional component in the resistance to failure and pure cohesion is developed along the failure arc. If the cohesion is c kPa, then the total cohesive force will be c .  . 1 (the 1 comes from unit distance at right angles to the plane of the paper).

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005 b)

Section 5

This still left the question of the position of the centre of gravity, the position of the centre of the sliding circle, and the length of the sliding arc to determine. In the question, candidates were given a cardboard template of the exact shape of the sliding wedge, and a drawing pin on which to balance to wedge. So the solution to the question began with attempting to balance the wedge on the drawing pin. Once the approximate centre of gravity had been found, the template was pricked through so that the diagram on the question paper could be marked with the centre of gravity. First join ends of slip circle, then through middle draw line at right angles. Point D is determined from equation 5.2, and hence the centre of the circle can be found.

use a ruler and approximately rotate it around following the shape of the curve and read off the length. adopt the method used by one candidate who pulled out a strand of her hair and laid this to follow the circle. Finally, the relevant length was measured on a ruler!.


5.6 Method of Analysis - II - Straight Line Failures
Fig. 5.8 shows this type of failure and the key forces to be considered. To solve such a problem we need the weight of the sliding wedge. Once again this is derived from the area on a scale diagram (i.e. weight is area multiplied by unit weight). In all examples, the wedge is triangular, and the area may be obtained either from:area = 0.5 x base x height (i.e. 0.5 b h in this case).

or area = 0.5 a b sin C where a and b are the lengths of two sides of the triangle and C is the included angle. In this example, all the forces pass through a single point (i.e. directly below the centre of gravity of the wedge and on the sliding surface). Though reference is made to the centre of gravity, we do not actually have to determine its position in this method of analysis. The weight acting downwards will have two components, one acting down the slope, and the other acting perpendicular to the slope. If the slope is just stable then the frictional force will just balance this component of the weight acting down the slope and attempting to cause failure. Equally, the component of the weight acting at right angles will balanced by the normal force N.

Fig. 5.7 Geometric construction to find centre of circle accurately when using first method of analysis.

The centre of the sliding circle can be estimated by trial and error using a pair of compasses, or alternatively a geometric theorem can be used. The extreme ends of the slip circle are connected by a line which is then bisected at right angles (Fig. 5.7), such that AY = BY. These two distances can be measured from the diagram as can the distance CY. Finally, the geometric theorem states that DY x CY = AY x BY ......................5.2

Hence it is possible to determine the distance DY and then the diameter of the circle CD. Finally, the distance CD is halved (i.e. OC = OD) to get the centre of the circle The length of arc may be determined in one of three ways:a) join the lines from the extreme ends of the slip circle to the centre of the circle and measure the angle. Convert this angle to radians () and multiply by the radius of the circle. i.e.


= R.

Fig. 5.8 Straight line failure of a slope. The key forces are the weight (W), the normal

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005 5.7.1 Dry Cohesionless Slope

Section 5

force (N), and the fractional force (F) which all act through a single point. The governing equations are thus (by resolving parallel and perpendicular to the slope):-

Four forces act on a block of slope (Fig. 5.9): W = weight of block of depth Z R = Reaction upwards X1 & X2 are lateral forces on block. It is readily shown that X1 and X2 have a common line of action [from Moments], and resolution parallel to the slope shows that they are equal and opposite. The effect of this is that we can effectively disregard these in subsequent analysis. Hence but

F = W sin  and N = W cos  ........................5.3
For our factor of safety we note that:inherent shear strength of soil ------------------------------------------------force which would just resist failure (i..e mobilising force)

Fs =

the mobilising force =

F = W sin 

W = R W =  . bzcos 
| area of block

Also the restraining force will be the inherent strength of the soil which can be derived from the Mohr Coulomb envelope, i.e.



N tan  ............................5.4

Note we MUST use forces here, and so we are dealing with the cohesive force (not the stress as defined by the intrinsic property of the material).

Thus the factor of safety may be specified as:-

Fs 

c L  W cos  tan  W sin 


For those who are Mathematically inclined:-

Fig. 5.9 Diagram of a typical slice on an infinite slope showing relevant forces which act. The reaction force may be further split into two components one parallel and the other perpendicular to the slope.

Note: with this geometry, it is possible to define the weight in terms of the slope angle  and the failure angle , and also the length (L) of the failure surface in terms of  and  Equation 5.5 can thus be generalised, and it is possible to differentiate equation 5.5 to get

Resolving perpendicular to the slope:-

Rcos  = N = W cos 

= dzbcos2 

dFs d

N is the normal force so the normal stress = dzcos2. Resolving parallel to the slope

and hence determine the

minimum factor of safety from all possible failure angles directly. If you are mathematically inclined you may wish to check this out.

Rsin  = F = Wsin 

= dzcos sin .

This may be plotted on a Mohr Diagram, remembering that for cohesionless materials c = 0.

5.7 Method of Analysis - III - Infinite Slope Method
This method of analysis assumes that the slope is of infinite extent, and that any water seepage is parallel to the slope. There are several cases which can be considered, some of which will be treated in detail below. 60

The angle the point subtends at the origin () is known as the angle of obliquity,

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

Fig. 5.10 Mohr - Coulomb Diagram for dry case stability. The point A represents the stress point on a plane at depth z below the surface.

  z cos. sin   1  i.e.   tan 1  d   z cos 2    tan (tan  )   d  
i.e. in this case the angle of obliquity slope angle  .

Fig. 5.11 Evaluation of the pore water pressure in the full seepage case. The pore water pressure at A is the vertical distance AB. Thus pore water pressure equals But AC = so.

() equals the

w x AB.

z cos 

If  is increased the point moves along the dashed line which is part of a circle which also passes through the origin. Failure would occur if, or when .

AB = AC cos  = z cos2 w z cos2.

 reaches the value  =  = ,

and pore water pressure =

Hence in this case failure occurs when

However, the pore water pressure only affects normal stresses, so new normal stress equals:-

i.e. the slope angle equals the angle of internal friction. This latter is sometimes known as the angle of repose and is normally in the region of 30o - 40o for granular media. Its value is higher for denser sands and lower for loose sands. 5.7.2. Fully Wet Cohesionless Slope - Full Seepage Along any equipotential, there is no change in excess pore water pressure. Thus the pressure at A can be evaluated in static head terms alone if the vertical distance AB corresponding to the equipotential AC is used (Fig. 5.11).

 ' =  - u =  z cos2  -  wz cos2 
[Note: We must also change the unit weight from its dry value to its saturated value.] Thus  ' = ( - w)z cos2 or 'z cos2, apart from the change in unit weight, the shear stress remains the same.

Fig. 5.12 Mohr - Coulomb Diagram for full seepage case. The stress point A has moved to point A' as a result of the excess pore water pressure. This stress point is closer to the 61

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

failure line and the slope is thus less stable. The locus of the point A' is now an ellipse.

the imposed shear stress will be:-

(d + z) cos.sin 
The effect of the positive pore water pressure is to bring the stress point closer to the failure line, and thus the slope is less stable. Once again failure occurs when

 (  z  d)cos. sin   z   d  and tan     ( z  d   z )cos 2    ( z  d   z ) tan  w w  

 
For typical values of z and d = 1m and the same values for the unit weights as before:-

  zcos. sin     and tan    2  (    )zcos   (    ) tan  w w  
Typical values: - for a sand and Hence or


tan  

= 20 kNm-3 w = 10 kNm-3. = 2 tan  2 .

20 x 1  20 x 1 40 tan   tan  ( 20 x 1  20 x 1  10 x 1) 30

Failure once again occurs when  =


tan  

Thus failure occurs when  is approximately  /2 and the slope approximately half as stable as in the dry case. 5.7.3. Case with dry top layer and full seepage below The next case to consider is the one where there is full seepage below a given depth d and material is dry above. We shall neglect the effect of capillary action at the interface (this has the effect of improving the stability so by neglecting it will be a SAFE APPROXIMATION).

In this case  is approximately 0.75, in other words the surface dry layer has increased the stability again 5.7.4 Case with damp soil - i.e. negative pore water pressures throughout. If the whole section is damp with negative pore water pressures throughout, then the stress point in the Mohr - Coulomb Diagram (Fig. 5.10) will move towards the right making the slope more stable (i.e. the reverse direction to that with full seepage..

5.7.5 Case with soil with cohesion. If the material forming the slope has cohesion as well as frictional properties, then some interesting things happen. If we return to the first case, we note that the slope is table provided that the stress point lies below the Mohr - Coulomb line. If cohesion is present, then the failure line is displaced upwards, and the slope becomes more stable. Indeed it is quite possible for a slope to plot as a stress point shown as B (Fig. 5.14 ) and be stable. However, the question of stability must be qualified since it was noted that failure on an infinite slope was independent of depth: i.e. at greater depth both the normal and frictional components increased at a constant proportion so that the angle of obliquity remained constant, and hence the factor of safety. In the case of material with cohesion and the stress point lying above the purely friction envelope, the above statement is no longer true. Thus as the depth increases, the stress point will move along the line of constant obliquity and will eventually cross the failure envelope when failure will occur. The conclusion that can be drawn from this is that infinite slopes with material having cohesion are stable but only up to a given normal stress. 62

Fig. 5.13 case with full seepage with a dry layer of soil above. The dry layer increases the normal stress faster than the shear stress and hence the slope becomes more stable.

In this example we would consider a potential failure in the full seepage region (which is the more critical region). the total normal stress will now be:-

( d + z) cos2
the water pressure will be:

w cos2

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

What does this mean in practice?

when dealing with Mohr's Circles. However, the result is simple, the maximum depth that will remain stable is given by:depth of stable vertical slope =

2c 


Since c is approximately 20 - 40 kPa for many soils, and is approximately 20 kN m-3, the maximum depth for a stable vertical slope will be 1 - 2 m.


An Infinite Slope which is completely submerged

Fig. 5.14 Mohr - Coulomb Diagram for case with soil having cohesion. Slope is more stable and slope angles steeper than the angle of friction are possible but only for a limited slope height.

An infinite slope which is completely submerged will not have any water seepage (except in extreme conditions), and analysis will show that the stability is identical to that in the dry case. It is suggested that you check this out for yourself. Remember in this case you should use the submerged unit weight to work out both the normal stress and shear stress and that excess pore water pressures are now zero. 5.7.7 Depth of Tension Cracks Tension cracks will appear in cohesive materials when desiccated. This will appear on flat surfaces, the most classic examples are the cracking that occurs on reservoir floors when the water level falls in summer. From the above, we can predict the maximum depth of such cracks. If the cracks deepen, then minor failures will occur and debris will fill the base of the crack such that the maximum depth will never be much greater than that predicted for long. The development of tension cracks can also be of importance in the stability of finite slopes as will be considered in section 5.8. The maximum depth of such cracks can be predicted from the relationship in equation 5 6 and used in more detailed analysis of the stability of slopes.

The normal stress is dependent only on the angle of the slope, the depth to the failure surface, and the unit weight of the material. Since for a given slope it will only be the depth that can vary, it is clear that any cohesion present will allow slopes of an angle steeper than the angle of friction only for a limited depth. We may estimate this critical depth by drawing the diagram as shown in Fig,. 5.14 and reading off the normal stress () on the X - axis Hence the critical depth is given by:-

 z  cos2 
It is also possible to work out what z is directly without plotting the graph by trigonometry. Thus the vertical distance on the stress plot of the critical points is given by two relationships - one from the Mohr - Coulomb line and the other from the shear stress i.e.:-

5.7.8 The factor of safety for Infinite Slope Analysis.

from Mohr Coulomb. . . . . . . c   tan  from shear stress . . . . . . . . . . . .  z cos  sin  i. e. z  ( c   tan  )  cos  sin 

A consequence of the above observation is to note that we can readily dig vertical trenches in clay when it is not possible to do so in dry sand. A question that immediately arises is what is the maximum depth for which a vertical slope in a cohesive material will be stable? This analysis is beyond the scope of this course but will be covered indirectly in the Seismology Course 63

Fig. 5.15 Determination of factor of safety in the dry cohesionless case.

N. K. Tovey

ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

The factor of safety may be readily determined when using the infinite slope method of analysis. It is obtained directly from the Mohr - Coulomb Diagram. Thus Fig. 5.15 is a revised form of Fig. 5.10. The point A represents the stress point for a slope of angle  while the point C represents the point on the failure envelope directly above the stress point. Thus the factor of safety is

AB / BC. i.e.

Fs 

BC  z cos2  . tan  tan    AB  z cos  . sin  tan 
Fig. 5.16 General slope with slices drawn. Note that one slice boundary coincides with the intersection of the water table with the slip circle. Both the actual slip circle and the surface are approximated to a series of straight lines. For analysis purposes, it is convenient to number the slices separately.

Thus the factor of safety may be determined either graphically using the Mohr - Coulomb Diagram or entirely numerically using the above equation. The factor of safety in all other cases may be similarly estimated.

5.8 General Solution to stability of Slopes
5.8.1 Introduction The general method for analysing the stability of slopes was developed by Fellenius and is usually referred to by the method of slices. There are several variants on the method with fewer or more simplifying assumptions. The Swedish Slice Method has the most such assumptions, but all are SAFE ASSUMPTIONS, i.e. they underestimate the factor of safety. All the methods begin by dividing the slope into a number of vertical slices and analysing the stability of each. The overall stability is then given by a summation of the stability effects on all the slices. As with many problems in Geotechnics, there is a trade off between accuracy and time as to the number of slices chosen. Typically, the number chosen should not be less than 5, and numbers more than about 8 get tedious to compute. There are some computer methods to analyse the stability of slopes, but these require careful data input as the whole of the slope geometry must be specified, and there can be problems with convergence if simple rules are not followed. The method begins by drawing a realistic slip circle and then dividing the area of the slope above this circle into several slices. The slices may be of varying width, but the general rule is that the height no slice should be more than about 2 - 2.5 its width, nor should it be less than 0.4 times the width. Further, it simplifies calculations, if slice boundaries are arranged to coincide where different strata cut the slip circle and also where the water table intersects the slip surface. Fig. 5.17 Details of forces acting on a single slice. The effect of tension cracks at the crest can be included as can the effects of varying shear strength properties. Once the factor of safety has been estimated along a single slip circle, further potential slip circles are analysed to find the most critical one (or group of critical slip circles.

5.8.2 Method of Analysis A typical slope is shown in Fig. 5.16 with the various slices included. In addition the approximate straight lines to the various curved sections of both the slip circle and the surface as shown. In Fig. 5.17 the details of the forces acting on each slice are shown. These are (in the case without water):1) the weight acting vertically downwards (W) 2) the normal force acting orthogonal to the local slip surface (N) 3) the frictional force acting parallel to the local slip circle. (F) 64

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Section 5

4) vertical inter-slice forces arising from friction between slices (X1 and X2). 5) horizontal inter-slice reaction forces between adjacent slices (E1 and E2). The inter-slice forces were encountered previously in the Infinite Slope Method of Analysis, but in that case it was easy to show that such forces on either side of each slice were equal and opposite and would thus cancel each other out. In the general method, this is not the case as the slices are of different sizes. However, overall there must be no net horizontal force or vertical force arising from these inter-slice forces and various approximations are possible. A) The first approximation is to assume that the vertical forces on each slice balance (this is more likely than the horizontal ones). Overall they must do so, but not necessarily for a particularly slice. By making this simplifying assumption to lead to an underestimate of the factor of safety and thus it is a SAFE ASSUMPTION. The computer method which you will use in the practical uses two methods, a rigorous method which does not make the approximation and a less rigorous one which does. In execution using the rigorous method, it is necessary to first guess what the imbalance of the two forces on the sides are. The calculation is then completed at which time a revised estimated can be determined from the calculations. If the revised estimate and the initial guess are close then the computed factor of safety is accepted. If the two are not close, then a second trial is made using the computed values from the first set of calculations. The processes is repeated through several iterations until convergence is achieved. Unfortunately, if the slices are illconditioned (i.e. they do not conform to the rules laid out above), convergence will not occur, and the solution can become unstable. B) The second approximation is to assume that both the horizontal and vertical forces on the sides of the slices cancel out. This is a more severe approximation, but is a SAFE APPROXIMATION once again.

the area of each slice (so that we can work out the weight, the orientation of the local angle of the slip circle to the horizontal ().

The analysis is most conveniently conducted in tabular form such as that shown on the next page. a) We begin by measuring the local angle of the slip circle and enter this in column 1 of the table. NOTE: some of the local angles of the slip circle will be negative in deep seated failures such as that shown in Fig. 5.16 (e.g. 1). b) We now need to estimate the weight of each slice and we do this as in other methods for slope stability by determining the area of the slice. If there is a single stratum, the we determine the area of the whole slice whereas if there are several strata we must sub-divide the slice so that the area of each stratum in each slice can be determined. Since the unit - weight of most materials is approximately the same, this refinement is not always necessary. The estimated values of area are entered into column [2]. The shape of slices may be approximated to either:-

triangles [ area = 0.5 x base x height. It is usually convenient to use the vertical side of the slice as the base of the triangle and the distance from this slice edge to the opposite apex as the height in these estimates]. trapezia [ area is mean height of parallel sides multiplied by distance between them]. c) The weight of the slice is then the area multiplied by the unit weight. In the case of multiple strata, the each separate stratum can be approximated to a trapezium, and the resultant weight is the summation of all areas multiplied by the respective unit weights. We have three basic forces controlling the equilibrium of each slice [if we use approximation B], and as with previous examples, the mobilising force will be W sin  while the normal force will be W cos  We compute this values from the data in columns [1] and [3] and enter the results in columns [5] and [4] respectively. The cumulative effects of all slices may then be determined by summing the values in both columns [4] and [5].


5.8.3 Analysis of Slopes without water seepage In most of the discussion regarding the Method of Slices we shall assume that both the horizontal and vertical inter-slice forces cancel. We will return briefly to deal with the more accurate solution using approximation A above later. Basically the analysis follows exactly the same form as that for a straight line failure. (We may incorporate the effects of water pressure later). We treat each slice separately and from each we will need to determine:f) 65



some of the values in column [5] will be negative!!!

Finally, we require (for the dry case), information on the cohesive force. To obtain this which measure the length

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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

of the segment on the slip circle in each slice (column [6] ), and multiply these values by the cohesion to give the values in column [7]. Finally we sum all values in column [7]. Equation 5.5 which was used for determining the stability of a single wedge is then modified to incorporate all slices by a summation on each part of the equation:i.e. Fs 

column [8] from which the pore water force U may be determined (column [9]).

 c    W cos  tan   W sin 


This factor of safety will give the situation in the dry case ( i.e. without water seepage).

Fig. 5.18 Flowlines and equipotentials in ground water flow in a slope NOTE: we must work in terms of forces not stresses (i.e. we multiply the pore water pressure by the area of the base of the slice).

5.8.4 Analysis of Slopes with water seepage The analysis of slope stability in the wet case proceeds in the same fashion as the dry case, and once again , it is most convenient to use a tabular form for analysing the slope. Columns [1] - [7] are completed as in the dry case. We need to assess the effects of water seepage. There is a more exact method involving the construction of a flow net and an approximate method. The approximate method will over estimate the water pressure and hence once again it will be a SAFE APPROXIMATION. We first need to sketch the water table (Fig. 5.18), then remembering that with the phreatic surface (water table), the positions of the intersections of the equipotential drops with the water table will be spaced at points to given equal vertical head drops. This gives us the start of the flownet which we can then complete by sketching curvilinear squares in the normal way. Once we have done this we can then estimate the water pressure at any point on the slip circle by following the relevant equipotential line from the point on the slip circle up to the water table and then evaluating this vertical head drop. (Fig. 5.19). This distance (h) will be less than the direct vertical distance (h') and so will over-estimate pore water pressures. The water table is shown steeply curved in the diagram for clarity, but normally the water table is flatter and the difference between the two estimates is much less [typically being around 10% or less. The values for the head of water (hw) , whether accurate or approximate, are entered for each slice in Space for additional notes

Fig. 5.19 Estimation of true water pressure approximate pressure (h').

(h) and

The approximation will always overestimate true value and will thus be a safe approximation. In the example shown the water table is steeply curved and this accentuates the difference between the two values. Normally the water table is less steep and the two values usually differ by less than 10%. Finally, we modify the Normal force to account for the pore water pressure and enter the value in column [10]. Column [10] is effectively column [4] - column [9] The factor of safety in the wet case is then given by:-

Fs 

 c    (W cos   U ) tan   W sin 



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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

Cohesion (c) = Slice No  kPa Angle of Friction (  ) = Area of Slice Weight of Slice (W)

Unit Weight ( ) W sin 

= c

kN m-3

W cos 




U =hw  gw

N= W cos -U

Slice No










1 2 3 4 5 6 7 8 9 10  

1 2 3 4 5 6 7 8 9 10

c   ( W cos  U )tan  Fs( wet)    W sin  

c   W cos )tan  Fs(dry )    W sin 

= ========


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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

5.8.5. Modification to allow for inter-slice horizontal forces - Approximation A. Equations 5.6 and 5.7 give the factors of safety in both the dry and the wet case for the situation where the interslices forces are assumed to cancel out. Bishop proposed methods to avoid this approximation although the full analysis with no approximations is difficult to execute in practice, needs a computer program, and then can be beset with convergence problems if the correct procedure is not rigidly followed. However, the simplified method of Bishop does allow us to obtain a more accurate estimate by removing the restriction that the horizontal forces on each slice must balance. The full derivation of the formula is beyond the scope of this course, but it is sufficient to note that if we do know what the factor of safety is it is possible to estimate what the horizontal inter-slice forces are and include them in our analysis. This begs a question though. How do we know what the factor of safety is until we have estimated it using an appropriate formula? The formula to use is:-

and impose additional lateral stresses on the slope which will make the slope less stable (see Fig. 5.20).

Fig. 5.20

Slope with a tension crack arising from desiccation. The presence of the crack reduces the area of the sliding surface and hence stability is reduced. Water filling crack will impose lateral stress on the sides of the crack causing a further reduction in stability.

Fs 

 (cb W tan ).M  W sin

Analysis of the slope in the dry case will follow the same procedure as described above except that the failure surface terminates at the base of the crack. When water is present the analysis is more involved an beyond the scope of this course 5.8.7 Effect of water filled tension crack on a straight line failure

 tan .tan   where M  cos 1   Fs  

Fig . 5.21 shows a potential straight line failure surface with a tension crack both dry and filled. It is relatively straightforward to analyse the reduction in strength in these two cases as shown below. Analysis of slope stability in the general case with dry cracks is easily accommodated in the analysis, but the effect of water filled cracks is more complex and beyond the scope of this course except in general descriptive terms. Equation 5.5 for the stability of a straight line failure is:-

and bn is the width of the nth slice (not the length parallel to the slip circle) The term M involves Fs so the way we proceed is to first make a guess that the factor of safety is unity (i.e. the slope is just stable), we work through the analysis and obtain an estimate for Fs which is general will be different from our original guess. We now replace the computed value of Fs as the denominator to the expression for M and repeat. We will find that the difference in the two values on the second interation is much less and we continue until both the guess and the computed value are closely the same (usually agreement to 1% is sufficient), and this is then the correct value of the factor of safety. 5.8.6 Effect of Tension Cracks - general description. Tension cracks may, in dry weather, open to a depth of 2c/ (see section 5.7.7) and will affect the stability of a slope in two ways. Firstly, there will be a reduction in the length of the slip surface which has inherent strength, and secondly immediately after a drought, water can fill such a crack 68

Fs 

c L  W cos  tan  .....................5.5 W sin 

Fig. 5.21 Straight Line failure with water filled tension crack.

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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

Incorporating these modifications changes the factor of safety to:1 c L'  ( W' cos  .  w d 2 . sin  )tan  ….5.11 2 Fs  1 ( W' sin   .  w d 2 . cos ) 2 5.8.8. Effect of Man Made Loading Man-made (or natural loading) on a slope can affect its stability. It can be incorporated by adding the load to the appropriate slices before analysis. The positioning of the load also affects the stability. In some cases it will increase the stability, in other cases it will decrease. Generally speaking, loads near the base of the slope will tend to improve stability and those near the crest will reduce the stability, but this is not always the case as it also depends on the position of the water table.

Fig. 5.22 The water pressure distribution with depth for the tension crack in Fig. 5.21 With a dry tension crack, the equation is essentially the same except the length L' is now the relevant length, and the weight W' will be reduced to omit the small wedge to the right of the tension crack. The formula now becomes:-

An example of a slope with a crack at the top
Straight Line failure with crack at top which can become water filled

Fs 

c L'  W' cos tan  W' sin 



[based on Exam question in January 1996] The geometry is shown in Fig. 5.23

In the case of the crack filled with water, the water pressure will be hydrostatic and thus the total water force from the crack acting horizontally will be:

1  wd2 2

Height of slope = 10 m, angle of slope = 60o, c = 15 kPa;  = 19o;  = 16 kN m-3 Critical failure surface will be about (  +  ) /2, i.e. about 40o We need the length of the failure surface (L), and also the weight of the sliding mass. L = H / sin 

Note that because the pressure is hydrostatic it has a triangular distraction and increases linearly wt. depth, and the above expression is effectively the pressure at mean depth multiplied by the depth. This force acts horizontally and thus perpendicular to the slope it will have a component of:-

This is computed in column [2] of Table 5.2  Column [3] then computes the area of the wedge and column [4] the total weight using the unit weight of 16 kN m-3. the cohesive resistance is computed in colum [5] column [6] computes the Normal Force W cos  tan  = 0.3444 W cos  the Shear Force (W sin ) is evaluated in column [7] the factor of safety is calculated as ([5] + [6]) / [7] the procedure is repeated for other failure angles around 40o.

1 .  w d2 . sin  2
( W' . cos 

acting in a      

direction outwards from the slope and will thus reduce the effective normal force from W'.cos to

1 .  w d2 . sin  ) 2

At the same time there will also be a component of the water force acting parallel to and downwards along the slope equal to:-

1 .  w d 2 . cos 2

So the mobilising force increases from to

W'.sin 

The factor of safety is now plotted against failure surface angle (Fig.5.24), or from the Data Sheets

) s o c . 2 d w  .

1   nis . 'W ( 2

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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

Fs 

c . L  W. cos . tan  W. sin 

Fig. 5.23. Geometry of the Slope showing tension crack. This question appeared in the 1996 Exam Paper. failure surface angle [1] 35 40 45 38 failure surface length [2] 17.4 15.6 14.1 16.2 area of wedge [3] 42.5 30.7 21.0 35.1 weight of wedge [4] 680 491 336 562 cohesion x L [5] 261 234 212 243 Normal Force [6] 198 129 82 152 Shear Force [7] 390 316 238 346 Factor of Safety [8] 1.177 1.148 1.235 1.142

Table 5.2 Analysis of factor of Safety at various Failure Angles without tension crack. Fig. 5.24 Plot of Factor of safety against failure surface angle Having obtained values for 35o, 400, and 450, it is clear that the minimum factor of safety representing the critical failure occurs between 35o and 40o, so a further angle of 38o is used and indeed has a slightly lower Fs. If time permitted, a further iteration would be done, this time between 38o and 40o. However, the error in using the value at 38o will be very small. Tension crack depth =

2. c 2 x 15   1. 9 m  16


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ENV-2E1Y: Fluvial Geomorphology 2004– 2005
failure surface length area of wedge weight of cohesion x wedge L Normal Force Shear Force

Section 5
Factor of Safety

[1] [2] [3] [4] [5] [6] [7] [8] 38 13.1 539 197 146 332 1.033 so the failure surface length (at 38o) is reduced by 1.9 / sin 38 = 3.1 m to 13.1 m and the weight lost beyond The slope is still stable (see Table 5.3), but is now the crack will be 0.5 x1.9 x 1.9.tan38 x unit weight = becoming somewhat critical as Fs is only just over 22.6. 1.000. Table 5.3 Revised Factor of Safety Calculations incorporating DRY tension crack. It is now necessary to out the effects of the crack being filled with water after heavy rain. This is done in the manner shown in section 5.8.7. i.e. the pore pressure in the crack is
2 U  0. 5  w hc

= 0.5 x 10 x 1.92 = 18 kPa

where hc is the depth of the tension crack. The normal force is now W cos  tan  - U sin  tan  and the Shear Force is ( W sin  + U cos ) and the final, and most critical value of the factor of safety is now recalculated. Note that in this case it is only columns [5] to [8] that have to be recalculated as the values in the other columns remain the same. failure surface angle [1] ---failure surface length [2] as area of wedge [3] above weight of cohesion wedge xL [4] ---[5] 197 Normal Force [6] 142 Shear Force [7] 346 Factor of Safety [8] 0.979

Table 5.4 Final calculation of Factor of Safety at critical angle – water now fills the crack. The factor of safety is now less than unity and failure is likely.


General Statement on Factors of Safety.

As a guidance, the factor of safety required in Hong Kong is:1.4 if the slope threatens residential buildings 1.2 if it threatens lines of communication 1.1 if it threatens other man-made resources 1.0 ( or less) for other regions Remember, that all approximations tend to under estimate the factor of safety and that is why there are several slopes which have factors of safety around 0.9 and yet appear stable. What we can say is that their stability is critical and in adverse weather conditions failure is likely. A further reason why the factor of safety is underestimated is that when we conduct shear strength tests on the soil we will always draw an envelope to the weakest strengths measured rather than take a mean. In this way we always work on the safe side, but this has the effect of underestimating the factor of safety further. 71

While all analyses compute a factor of safety and this forms the basis of whether a slope is stable, a further discussion is important. A factor of safety of unity indicates that a slope is just stable, while one which is above unity should be stable and one which is less than unity is critical. The choice of what factor of safety to use in a risk analysis is a matter of judgement, but there is a growing awareness that this should take account of the consequences of a potential slope failure. Normally, we estimate the factor of safety under the worst conceivable conditions (e.g. a 1000 year storm). We would estimate the likely ground water level at this stage and carry out the analysis by one of the methods outline previously. We would then test alternative modes of failure until we found the most critical one. If this was above a given threshold then we should be satisfied regarding its stability.

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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

5.11 Predicting the form and position of potential failures.
In the analysis we will always strive to find the failure mechanism which gives the minimum factor of safety. More often than not, we will find a band of possible slip surfaces which have approximately the same stability. Defining this critical band will indicate the position of the slip surface zone should failure occur.

In early October 1996 the largest glacier in Europe in Iceland became the centre of attention when a volcano erupted beneath the glacier causing a large amount of the ice to melt with the threat of extensive and perhaps catastrophic flooding. This section has been written at a time when the outcome is not known but we can consider possible modes of failure that would lead to flooding. Fig. 5.24 shows a simplified possible cross section of the toe of the glacier some 150m long and 100 m high. Behind this is a crevasse in which water from the melted glacier behind is accumulating and starting to rise in the crevice. The first question to address is what possible modes of flooding are likely. Four situations could lead to extensive flooding

5.12 Summary of Slope Stability 
Methods for analysing the stability of slopes all require the identification of a potential failure mechanism. Several such mechanisms should be tested to determine the most critical one. Analysis can be done on current water level conditions, to determine the current stability, or more usually, a particularly severe rainfall event will be used as the basis for analysis (e.g. 1000 year storm). Many approximations can be made to simplify the analysis and in all cases, the approximations reduce the factor of safety, and hence are all SAFE ASSUMPTIONS. The choice of what factor of safety to use should be determined on the basis of the consequences of failure following a model such as that used in Hong Kong. Rarely is this done.




The water level in the crevasse may reach such a height that the flow of water beneath the toe glacier become critical causing a quicksand at the down steam end and excessive erosion and undermining of the toe of the glacier. As the erosion takes place, the resistance to flow will decrease causing an increasing flow of water until the level of the lake eventually dissipates. The level of water continues to rise as more ice melts (up valley of the toe] causing the toe glacier to be over-topped and flow of water over the surface. This is unlikely to be catastrophic as the flow will eventually equal the rate of ice melting. Erosion of the surface of the toe glacier is likely causing further release of water, but this should be controllable.



5.13 Some comments about the stability of the Glacier in Iceland.

Fig. 5.25 Schematic representation of toe of glacier with build of water in crevasse behind.


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ENV-2E1Y: Fluvial Geomorphology: 2002 - 2003

Section 5


Iceland is volcanic, and if the underlying part of the glacier is volcanic ash rather than rock, it is possible that the glacier will have carved out deep valleys. The same would be true in rock (e.g. Ushaped valleys). If as a result these valleys are steep, then exposed to a water level rise could trigger landslides which could suddenly displace a large quantity of water causing a massive overtopping of the toe glacier and catastrophic flooding (as happened to the dam at Vaiont in Italy in 1962). Note, however, that since the slope is solely submerged, then that stability is the same as in airi.e. it is not a seepage situation. If the underlying material is rock, then rock slides could also occur, depending on the dip of the strata - fissures may have opened up while the ice was present, but with water filling the gaps, the effective normal force would be reduced.

distribution as was should in section 2 in the lecture notes for flow beneath a weir. To investigate whether the toe glacier is stable we need to work out the weight of the glacier. This we can do by dividing it into say 5 slices (as in the general slope stability method), or by counting squares. Let us assume that we have worked this out (remember the unit weight of ice is 9.00 kN m-3 (see data sheets) Let this value be W The total resisting horizontal force provided by the surface of the sand/silt would be

( W  U ). tan ( )
and the factor of safety against sliding would be:-


Uplift on the base of the toe glacier because of seepage pressures could reduce the effective normal stress and hence the ability of the toe glacier to resist sliding following the build up of horizontal pressure on the side of the crevasse. This would be the most catastrophic failure mode.

Fs 

( W  U ). tan (  ) H

The horizontal force (H) from the hydrostatic pressure is proportional to depth below the surface of the water, so the mean force will be


1  h2 2

As previously, the situation is critical if Fs < 1. We can easily investigate how stability would be subsequently affected by changing the height of the water (h) and reworking U and hence Fs. As in many previous examples we would plot Fs against h. However, this time the value of Fs would continue to decrease as h increased. What we would be looking for is the value of h where Fs becomes 1.0. The total area of the toe glacier (which was determined by slicing the section) = 9650 m2, and since the unit weight of ice = 9.00 kN m-2 - from data tables, the total weight of the toe ice is 86850 kN per metre width of the glacier. Remember we have to work this out only once in the calculations (unlike the case of the slope stability as the failure will occur at the base of the glacier, if any where, and the weight above remains constant. As a first guess try a value of h = 50m. This gives a high factor of safety. Then try say 60, 70 m until Fs fall below 1,0. Finally perhaps, and if time permits add one intermediate point. Now plot the factor of safety against height of water and read of value when Fs = 1.0. In this case it is 62.9 m.

Since the length of the toe glacier is large compared with the depth of the sediment, and since the base is nearly horizontal in this case, the hydrostatic uplift on the base of the toe glacier will decline linearly from the above value at the crevasse to zero at the toe. Thus the total uplift (U) on the base of the glacier will be


1  hd 2

where d is the length of the toe glacier (150 m in this case) If the base of the toe glacier was not straight or the depth of the sandy/silt was large compared to the length d, then it would be more accurate to plot the seepage flownet beneath the glacier to work out the pressure Height of water (h) (m) 50 60 70 65 Horizontal force H (kN m-1) 12500 18000 24500 21125 Uplift force U (kN m-1) 37500 45000 52500 48750


( W  U ) tan (  ) H

Fs =

49350 41850 34350 38100

1.974 1.163 0.902 0.902

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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5

Table 5.5 Calculation of Factor of Safety against failure

Fig. 5.26 Factor of safety against height of water

We can thus predict that failure of the ice block will occur when the water in the crevass reaches 62.9m. space for additional notes.


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ENV-2E1Y: Fluvial Geomorphology 2004– 2005

Section 5


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