209672999 slope stability

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SLOPE STABILITY CONTENTS: 1.0 Introduction 1.1 Types of slips 1.2 Procedure for estimating stability 2.0 Total stress analysis 2.1.0 Rotational analysis 2.1.1 Effects of tension crack 2.1.2 Partly submerged slopes 2.1.3 Rapid draw-down 2.2 Estimation of stability (Taylor) 2.2.1 Undrained cohesive soil 2.2.2 Frictional resistance 2.2.3 Layered soil 2.2.4 Location of the critical circle 3.0 Effective stress analysis 3.1 Soil strength considerations 3.2 Effective strength parameters 3.3 Total stress conditions 3.4 Method of slices 3.4.1 Swedish method 3.4.2 Bishops method 3.5 Summary of methods of slices 4.0 Choice of factor of safety 5.0 Cuttings in over-consolidated clays REFERENCES

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Slope stability


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1.0 Introduction

1.1 Types of slips

1.2 Procedure for estimating stability

2.0 Total stress analysis 2.1.0 Rotational analysis

2.1.1 Effects of tension crack

2.1.2 Partly submerged slopes

2.1.3 Rapid draw-down

2.2 Estimation of stability (Taylor)

2.2.1 Undrained cohesive soil 2.2.2 Frictional resistance

2.2.3 Layered soil

2.2.4 Location of the critical circle

3.0 Effective stress analysis

3.1 Soil strength considerations 3.2 Effective strength parameters

3.3 Total stress conditions

3.4 Method of slices

3.4.1 Swedish method

3.4.2 Bishop‟s method 3.5 Summary of methods of slices

4.0 Choice of factor of safety

5.0 Cuttings in over-consolidated clays


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The degradation of natural or artificial slopes (in deep cut excavations or high embankments) is the result of mass movements which occur chiefly owing to the

action of gravitational forces, sometimes supplemented by earthquake forces.

The downward movement of rock or soil masses occurs when the equilibrium is disturbed along a certain plane (within the slope) and the shear stresses along it exceed the available shearing resistance; this can occur as the result of either an

increase in the shear stresses or a reduction or deterioration of the shear strength. A distribution of landslides (nearly 7000) in the UK is shown in Fig.1A.

The manner in which a slope fails is chiefly controlled by the following factors;


hydrological topographical

climatic conditions extent of weathering of rocks / soils

This results in a great variety of types of mass movements see Fig.1 below.

1.1 Types of slips

Figure 1 Basic types of mass movement

Falls are usually associated with short-term failure of steep slopes (in artificial excavations or river banks) usually in rock with vertical joints, Fig.1(a). As the lateral support is removed, bulging occurs at the slope foot and tension cracks

open behind its crest, usually along the pre-existing fissures. This leads to progressive increase of stresses in the root of the separating mass and to the

eventual collapse; the process is frequently accelerated by water entering the tension cracks.

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Figure 1A Landslide distribution in the UK

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Rotational slides occur characteristically in slopes of fairly uniform clay. The failure surface is curved and usually deep seated. The slipping mass slumps,

sinking at the rear and heaving at the toe, Fig.1(b). Approximately circular rotational slips are associated with cut slopes of uniform clays whereas non-

circular rotational slips with natural slopes of over-consolidated clays in which weathering has produced a softened upper layer.

Translational slides generally result from the presence of a heterogeneity, in the form of a weak soil layer, located at shallow depth beneath the slope. The

failure surface is approximately planar and parallel to the ground surface, Fig.1(c). If the plane of weakness is at a moderate depth beneath the slope, a compound slide of partly rotational and partly translational character may occur.

Flows are mass movements in which there are no well-defined failure planes,

Fig.1(d). Skempton and Hutchinson (1969) differentiate between earth flows and mud flows; the latter are glacier-like in form whereas the former are considered to be transitional in character between the slides and mud flows.

From the slope stability point of view the most dangerous conditions are

encountered in areas where in the past the soils have been deformed and folded by tectonic activities or ice advances or where previous mass movements have

taken place. In these circumstances pre-existing slip planes (slickensides) are usually present along which the shear strength is only a fraction of the strength of the intact soil.

The typical features of landslide areas are the following:

a) The presence of depressions and bulges on natural slopes. b) The presence of deformed trees with trunks bent in random

directions. c) The existence of springs on slopes and outcrops of water-bearing

strata. d) The existence of slickensides and deformed layers of clays (these

can be best observed in trial pits or by breaking down undisturbed

tube samples).

Above information may be found in Z Wiln and K Starzewski 1972, Vol 2 Chapter 3, section 3.1 pp73-74 [Book out of print].

In summary, a slip can occur due to

(A) An increase in shear stress (e.g. adding a surcharge) OR

(B) A deterioration of soil strength over a number

of years (20 100 years) OR

(C) A combination of the above These notes cover the analysis of rotational circular slips.

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1.2 Procedure for estimating stability

There are 3 steps in process of estimating stability:

1. Estimate disturbing forces The components are;

gravity acting on body of soil

super-imposed loads (if any) seepage force due to water flow (if any) earthquake forces (not dealt with in these notes)

2. Shearing resistance of soil

- Determine the number, thickness and average strength parameters of each soil layer.

- Soil strength equation for;

total stress, f = cu + n tan u

effective stress, ‛f = c‛ + ( n - u) tan ‛


f = shearing resistance at failure.

- Include a factor of safety, F, to limit the maximum mobilised shearing resistance on a failure plane;

= f

F REF: DW Taylor, 1948 Fundamentals of soil Mechanics; section 16.6 pp 414-

417 describes in detail how different F values for Fc and F may be combined

3. Select appropriate analysis

There are 2 approaches;

(a) Limit state equilibrium:

- Determines the overall stability of the sliding mass.

- Method is to analyse various potential failure surfaces to determine which has the lowest F.

- This method of analysis is generally not sensitive to the chosen shape of failure surface.

- A circular arc is chosen because it is the simplest to analyse and

is usually sufficiently accurate. [These notes consider limit state analysis in detail]

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NOTE: The computed critical failure arc may not coincide with the actual failure surface, however, their factor of safety (F) values will be similar.

(b) Stress analysis: [Not covered in these notes]

- Uses the principles of elasticity to evaluate stress and strain

throughout a slope.

- Finite element analysis is a powerful analytical tool, but its potential accuracy is limited by the highly variable and non-

linear characteristics of most soils.


(Stability of cohesive soils)

- Undrained conditions apply (i.e. during or at end of construction).

- Soil strength parameters are cu and u (N.B. if Sr=1 then u =0o).

- The excess pore water pressure due to loading has not had time to dissipate to any extent.

2.1.0 Rotational Analysis

A series of trial slip surfaces with various centres of rotation, O and radii, R (see Fig.2) are analysed to determine the most critical combination of O and R values, i.e. to find which potential failure circle has the lowest F value.

Figure 2 Rotational analysis

Where; G = centre of gravity

Length of arc, A-B = R x rad

= R x o π / 180

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Taking moments about O;

W d = cu R rad R = cu R2 rad

[NB analysis for a 1m wide strip of arc AB]

F = Restraining moment

= cu R

2 rad

Disturbing moment W d

2.1.1 Effect of Tension Crack A tension crack, see CD in Fig.3, always develops behind the crest of a slope as

failure start to occur in cohesive soil. Therefore, regular inspections of the ground behind the crest is advisable when a cutting is being excavated.

Figure 3 Rotational analysis with tension crack

Cohesive soil can support a vertical face, C-D in Fig.3, to a maximum height, zc;

zc = 2cu

γ It is likely that the tension crack will fill water - this causes an extra thrust

adding to the disturbing moment (lever arm, yc);

= 0.5 γw zc2


NOTE: The tension crack reduces the weight of the arc to Wt and its lever arm

to dt and the sector angle to c radians;

F = cu R

2 c (π / 180)

Wtdt + 0.5 γw zc2 yc

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2.1.2 Partly submerged slopes

Figure 4 Partly submerged slope In the above diagram;

W = total weight of soil and water in area B

W‛ = submerged weight of soil particles in area A

The pressure of water provides and additional restoring moment (i.e. resisting movement about O). This moment exactly balances the moment (about O) of a

mass of water filling the space below the external water level and above the rupture surface.

Therefore, use submerged unit weight, sub to allow for external water level;

sub = ( sat - w)


w = 9.81 kN/m2

F = cu R

2 (π/180)

(w x d) + (w‛ x d‛) 2.1.3 Rapid Drawdown

If the retained water is quickly lowered there is no water pressure on the face of

the slope. Also water levels (in the slope) immediately after draw down remain

constant. Thus sat is now used for area A moment and since the moment of

cohesion is not altered the F value will be lowered - possibly to < 1 and failure.

F = cu R

2 (π/180)

(w x d) + (wsat x d‛)

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Class example 1

A cutting, with side slopes of 1.0 vertical : 1.5 horizontal, is excavated in saturated clay to a vertical height of 10.0m as shown below;

Soil properties are as follows;

saturated unit weight, γ = 18.50 kN/m3

undrained cohesion, cu = 40.0 kN/m2

Using the trial slip circle shown in the diagram, determine the factor of safety against short term failure for each of the following conditions:

a) Ignoring the tension crack, given;

sector angle, θ = 84.06o

area of slipped mass, A = 102.1m2

lever arm, d = 6.54m

b) Allowing for a dry tension crack, given;

sector angle, θc = 67.44o

area of slipped mass, A = 71.64m2

lever arm, d = 5.86m

c) Allowing for a water filled tension crack.


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2.2 Rapid estimation of stability Taylor (1948)

2.2.1 For undrained cohesive soil ( u = 0o)

resisting force = cu x arc length (l)

which is proportional to H (H = vertical height of slope)

disturbing force = x area of sector which is proportional to H2

Stability number, Ns = cu H

= cu

H2 H

The above equation is for a F of S = 1 (i.e. only just stable)

For a “safe” design, we must include a factor of safety, F;

Stability number, Ns = cu


Note: Ns is the same for slopes of similar geometry in DIFFERENT soils.

Taylor devised a chart, Fig.5 that relates Ns to angle of slope, β, for u=0 soils

(saturated clay, Sr=1) and u>0 (partly saturated soils, Sr<1);

Figure 5 Taylor’s chart

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The inset defines parameters H, and DH where D = depth factor Note the following features;

- Gives the Ns value for the most critical failure circle

- Toe failures occur > 54 in u = O soil

- At < 54 the failure circle will pass below the level of the toe - A firm stratum (hard clay, rock) may limit the depth of penetration of

failure circle, thus lowering Ns value (i.e. makes slope more stable) see

dashed lines Fig. 5 Taylor extended his analysis to enable the influence of a hard stratum at depth

DH to be found more accurately, see Fig.6.

Figure 6 Extended Taylor’s chart

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- also shows where the failure plane will emerge in from of toe,

see inset (a) [If this is of any use!]


1. Taylor‟s method is an approximate analysis intended for feasibility studies on temporary works. It provides an indication of F and if low

(i.e. only just above 1) whether a more rigorous analysis is required. 2. It does NOT provide information on the location of „O‟ or length of

radius „R.‟

Class example 2

A cutting in a saturated clay has a vertical height of 10.0m. Hard bedrock is located at a depth of 6.0m below the floor of the cutting. The clay has the following properties;

Undrained cohesion, cu = 34.0kN/m2

Undrained angle of friction, u = 0o

Unit weight, = 19.0kN/m3

Determine the maximum safe slope that will provide a factor of safety of 1.25 against short term failure.


2.2.2 Frictional resistance in undrained soil ( u>0 )

For example, partly saturated soils (e.g. clay fill used in embankments).

Frictional resistance must be taken into account, but the value of frictional

resistance is not constant along slip surface. A factor of safety must now apply to the frictional component also;

= cu

+ n tan u

Fc F

A working value of F can be found by successive approximations, as shown in the solution to Class example 3

Class example 3

An earth embankment with side slopes of 1 vertical to 2 horizontal and vertical height of 31.0m, is constructed of compacted clay soil and rests on a strong foundation soil. The embankment soil has the following properties;

Undrained cohesion, cu = 25.0 kN/m2

Undrained angle of friction, u = 20.0o

Unit weight, = 16.0 kN/m3

Estimate the factor of safety against short term failure. [1.35]

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2.2.3 Layered soil

Figure 7 Non-homogeneous soil

Note: B < sat

The mirror images in the lower part of slope counterbalance each other so the

net disturbing force is provided by areas A and B.

For av - use weighted B and sat for the two areas;

av = (A x B) + (B x sat)

A + B


Ns = cu

F av H

2.2.4 Location of the critical circle

The most critical circle (as defined by the location of centre of rotation, O and radius, R) is the one for which the calculated factor of safety has the lowest

value. Taylor‟s curves, however, give no indication of where the centre of rotation is

located, although toe failures are likely in a slope where >54 and u = 0

Fig. 8 shows a chart (taken from Whitlow, 2001, p371) from which an indication of the most critical circle can be obtained as a starting point for detailed


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Figure 8 Approximate location of critical circle

The above chart applies to saturated (undrained conditions) uniform clay soil.

For a given slope angle β, values of Xc/H and Yc/H are read off the chart and

converted to Xc and Yc coordinates measured from the toe

NOTE: A hard stratum at or just below toe level will limit the depth of the slip

circle, so that a result from Fig.8 would not apply.

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3.1 Soil Strength Considerations

Any change in total stress ( ) in a saturated cohesive soil immediately alters pore water pressure (p.w.p.) by an amount du (termed the excess p.w.p.).

But du takes a long time to dissipate because of;

a) low permeability (k) of clay and b) large volume (hence long drainage paths) of soil in a slope.

For an excavated slope there are 3 stability conditions which need to be evaluated;

1) Short term (total stress analysis) During or at end of construction du at maximum.

Applied load carried initially by increased porewater pressure

The du that is developed due to a change in total stress is given by Skempton‟s (1954) equation.

du = B[ 3 + A( 1 - 3)]

A and B are pore pressure parameters

3, 1 are changes in total principal stress

See Note set “THE PORE PRESSURE PARAMETERS A and B” which outlines;

(1) Principles of effective stress (2) Derivation of pore pressure parameters

(3) Soil tests for A and B. 2) Long term (effective stress analysis)

- As du dissipates more of shear force is carried by soil structure. - After many years (5-10-20 or longer) du = 0 and all load is carried

by the soil particle as effective stress, ‟. - When du returns to 0 clay soil is at its “strongest”

The strength parameters, c‛ and ‛, with respect to effective stress, are obtained

from either:

(i) Fully drained triaxial tests.

or (ii) Consolidated undrained tests with pore pressure


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3) Long term (heavily over-consolidated soils)

Weathering from the surface downwards causes the strength of heavily over-consolidated soils to deteriorate over a long period of time, 20 - 100

years +, geologically „old‟ cays become weaker (e.g. London, Gault and Kimmeridge Clays).

- Soil strength reduces from a “peak” value (Sf)to a constant

“residual” value (Sr).

The concept of residual strength is discussed in R Whitlow, 2001, section

7.13, p 254-56 (describes tests to measure residual strength). See also Fig. 7.3, p.213.

3.2 Effective Strength Parameters

c‛ and ‛ are obtained from either;

(i) Fully drained triaxial tests. or (ii) Consolidated undrained tests with p.w.p. measurement.

c‛ and ‛, parameters depend on;

d t - change in total stress

and du – a consequent change in p.w.p.

It is essential that the du due to construction is predicted;

During construction (controlling du) Immediately after construction Long term - when du = 0

Possible future loading conditions e.g. construction / placing fill / removal of toe.

Pore water pressure results from;

i) Static water level with/without

(ii) Conditions of steady seepage and/or (iii) Changes in total stress.

Granular Soils

- Coarse soils (clean or <5%, fines) have a high permeability, k, value.

- du full dissipates AS is applied (e.g. during construction.)

- thus a ‛ analysis is used at all times.

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3.3 Total stress conditions

- Applies during or immediately after construction. - There is a lack of p.w.p. dissipation.

- Conditions are similar to the undrained triaxial test. - Applies to saturated clay soils, see Fig.9

Figure 9 Failure envelope of fully saturated clay

- If a clay soil is partially saturated, e.g. compacted clay, then u >0o

(see Fig.10) and stability may be analysed to include a friction component in addition to cohesion.

Figure 10 Failure envelope of partly saturated clay

NOTE: u is not constant as air readily compresses as total stress,

t , increases, until the air dissolves when u becomes zero.

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3.4 Method of Slices

- The soil mass above the potential arc A-B in Fig.11, is divided into an arbitrary number of vertical slices (It simplifies analysis if they are

mostly the same width).

- In terms of effective stress, the factor of safety on soil strength is

defined by the expression;

f = c‛

+ ( n – u) tan ‛


- The frictional resistance at the base of each slice varies and is given


N‛ tan ‛ = (N - ul) tan ‛ Where;

N = total normal stress N‛ = effective normal stress

l = chord length of slice (NOTE: this is not a one “1”)

- N‛ depends on total weight of slice and pore pressure (u) at base.

- A complicating factor is that the slices are not physically separated.

- Forces (vertical, z and lateral, x) act between slices - In practice it is difficult to evaluate x and z.

NOTE: (zn+1 - zn) affects the vertical weight of a slice.

(xn+1 - xn) affects shear force T on the base of a slice.

- Various approximate solutions (“methods”) take x and z account to an increasing extent and therefore become more accurate.

Symbols used in Figs. 11 and 12;

O = centre of rotation R = radius of failure arc b = width of slice

h = average height of slice l = chord length

xn = horizontal lateral force between slices (total stress)

zn = vertical shear force between slices

(total stress) W = weight of slice (total stress)

N = normal stress vector (total stress)

N‛ = normal stress vector (effective stress)

u = water pressure at centre of base of slice T = shear stress

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Figure 11 Principle of slices

Figure 12 Polygon of forces acting on a slice

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Two “methods” of analysis (Swedish and Bishop‟s) are detailed below:

3.4.1 The Swedish method (Also known as the Fellenius Method, 1936)

Assumes that Ln and Ln+1 (see below) the resultants of zn and xn etc, are

equal and opposite and act parallel to tangent of failure arc. Thus the resultant of inter-slice forces is zero.

- Also assumes base of slice is the chord.

- Clearly the lateral force assumption for the centre slice in Fig.11 is reasonable.

- However it is not approximately true for slices near A and B. - The solution UNDERESTIMATES F by 5 -> 20% depending on slope

geometry (compared to more accurate methods).


Taking moments about centre of rotation, “0”;

disturbing moment = R T

restraining moment = R(c‛R rad + N‛tan ‛)

Hence F = R(c‛R rad + N‛tan ‛)


N.B. top “R” outside bracket cancels out with lower “R”

R rad = length of arc A-B

- BUT N‛ and T are not convenient quantities and are replaced by;

N‛ = W cos - ul


T = W sin



Centre line

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F = C‛R rad + (Wcos - ul) tan ‛


N.B. Total stress analysis can be applied to this method and if u = 0

F = cuR rad


NOTE: The objective is to find the most critical centre and

radius for a given slope.

It is common practice to select a grid of centers, (based on experience) and find critical radius. Contours of F can then be drawn above the slope, see Fig.13;

Figure 13 Location of critical circle

REF: G. N. Smith, 6th Ed - p161/162 “Location of the most critical circle”

and Fig. 5.10, for general comments.

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Class example 4

Determine by means of the Swedish method the long term factor of safety for the trial slip circle shown in the diagram below;

Soil properties are; Strength with respect to effective stress;

c‟ = 10.0 kN/m2

‛ = 28o

Unit weight, = 18.0 kN/m3

Note that slices No. 1 -7 have a width, b, of 3.0 m and No.8 has a width of 3.8 m.

Assume the unit weight of the soil has the same value above and below the water table.


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3.4.2 Bishop’s method [also known as Bishop‟s (1955) Conventional method]

Figure 14 Interslice forces

T= c‛ l

+ N‛tan ‛


Where l = length of base of slice in metres (see B-C above)

N = N‛ + ul

And the shear strength mobilised;

= c‛ + ( n – u) tan ‛


n = stress (total) normal to a failure surface

n = N/l on base of slice (NB - 1m thick)

= 1

[c‛ + (N/l – u) tan ‛] F

Shear resistance of whole of base of a slice = x l

ASSUMES: The resultant of the forces acting on the sides of any slice is horizontal, i.e. there is NO net force in the vertical direction

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At equilibrium;

disturbing moment = restoring moment

Wx = TR

Wx = l R

Wx = 1

[c‛l + (N – ul)tan ‛] F

F = 1

[c‛l + (N – ul)tan ‛] Wx

Given that; N = Wcos and x = Rsin

F = 1

[c‛l + W(cos – ul)tan ‛] Wsin

(Note: This is an equivalent expression to the Swedish method)

Pore pressure ratio, ru, can be used to represent u;

ru = pore water pressure

= U

total stress z Where;

= unit weight

z = depth to an element of soil Typically saturated soil has a unit weight of 20 kN/m3 and water 10 kN/m3

(approx), thus for an element of soil at a depth of, say 5m;

ru = (10 x 5)/(20 x 5) = 0.5

Given that ru is a measure of the average p.w.p in a slope, 0.5 is a reasonable “guestimate”.

The water pressure at the base of a slice is given by;

u = ru z = ru W


Now b = l cos

u = ru W

l cos

Given that; sec = 1/cos

u = ru W

sec l

Thus, the final conventional Bishop equation becomes

F = 1

[c‛l + W(cos – rusec )tan ‛] Wsin

The Bishop conventional solution under-estimates F by 5% to 30% (deep seated failure) compared to more accurate methods.

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Class example 5

Determine, using Bishop‟s conventional method, the long term factor of safety for the trial slip circle shown in Class Example 4.


3.5 Summary of methods of analysis

Swedish Circle Method

- Interslice forces are ignored. - Should not use for submerged slopes or where earthquakes forces are

expected. - Note the assumed force distribution does not satisfy conditions of vertical and

horizontal equilibrium.

- Thus for > 0o soils F of S values are on low side compared to methods below.

- Errors, whilst on safe side, can be large (up to 60%).

- NB: For = 0o soils the accuracy is the same as Bishop‟s method.

Bishop‟s Conventional Method

- Interslice forces are considered to be horizontal. - Satisfies conditions of overall vertical equilibrium (but not horizontal). - The resultant errors are generally small and on safe side.

Bishop‟s Rigorous Method (Not covered in these notes)

- Satisfies conditions of vertical and horizontal and moment equilibrium for

whole slipped mass.

- Assumes inter-slice forces are parallel but not necessarily horizontal. - This method is preferred for steep slopes.


Factors to be considered in making a choice of F value fall into two broad categories;

i) Consequences of failure: A higher factor would be chosen where there is

a risk to life and/or adjacent structures. A lower factor when instabilities are localised requiring simple remedial measures (pipes, drains or road surfaces) or where the period of exposure is short (e.g. temporary works).

ii) Confidence in the information available: As a result of complexity of the

ground conditions, in-adequacy of the S.I. information, uncertainty about the design parameters, e.g. pore pressure. Potential future changes to water table levels must also be considered (see below);

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*BS 6031 (1981) Code of practice for earthworks


i) The strength of freshly exposed over-consolidated (o.c.) clays gradually decreases with time due to weathering. Table 1 shows the magnitude and costs due to minor instability M1, M10 and M45 from opening in 1959 to


Hertfordshire Bedfordshire Northamptonshire

Length of M-way in county (km) 27 27 45

Amount of cutting/embankment,

% 80 50 90

Number of failures 7 45 55

Total expenditure (1966 prices) £10,200 £36,500 £40,000

These are shallow failures due to surface weathering

Deeper major failures can occur after longer period (50-100 yrs)

ii) Time dependent softening is not measured in a standard test.

iii) Fig.15 shows the peak (Sf) and residual (Sr) strengths from a ring shear

test. This test is designed to measure residual strength - BUT at strains

much larger than occur during initiation of failure.

Figure 15 Stress – strain curve for an o.c. clay

End of construction (embankments & cuttings) 1.30* Long term steady state seepage condition 1.25 – 1.5* After sudden drawdown 1.20*

Slides where pre-existing surfaces exist 1.20* Natural slope of long standing 1.10 – 1.20

Spoil tip 1.5 Problems involving buildings 2.0

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NOTE: Also Sr is not a “weathered” residual strength.

iv) Fig.16 shows the loss of cohesion (c‛) residual state. ‛ is also reduced by

a few degrees.

Figure 16 Strength envelopes for an o.c. clay

The data below illustrates the reduction in strength in 40+ years;

c‛ from 15.3 to 3.8 kN/M2

‛ from 20 to 17

Residual strength (for c‛ and ‛) are the only available parameters to use

for long term stability.

Morgenstern has looked at natural slopes in London Clay (heavily o.c.) see

Fig.17, - majority of slopes are below 10 !

Figure 17 Histograms of angles of natural slopes in the London Clay

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Knowing the age of cuttings of various slope angles which have failed has enabled Morgenstern to determine an idea of the RATE of strength

reduction for London Clay see Fig.18.

Figure 18 Rate of progressive reduction in shear strength in the London Clay


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