Circular Failure-Hoek&Bray.pptx

Circular Failure - Hoek & Bray
Prof. K. G. Sharma
Department of Civil Engineering
Indian Institute of Technology Delhi, New Delhi, India

Conditions for Circular Failure
i. Soil consisting of sand, silt and smaller particle sizes will
exhibit circular slide surfaces, even in slopes only a few
meters in height.
ii. In the case of a closely fractured or highly weathered rock, the
slide surface is free to find the line of least resistance through
the slope. This slide surface generally takes the form of a
circle.
iii. Highly altered and weathered rocks, as well as rock with
closely spaced, randomly oriented discontinuities such as
some rapidly cooled basalts, will also tend to have circular
slide surface. Similarly overburden material.
iv. Crushed rock in a large waste dump will tend to behave like
soil and failure will occur in a circular mode.
v. Finely ground waste material, disposed of after completion of
milling and metal recovery process will exhibit circular failure
surface.

Circular Failure Analysis
The assumptions made in deriving the Stability Charts:
a. The material forming the slope is homogeneous, with uniform
shear strength properties along the slide surface.
b. The shear strength τ of the material is characterized by
cohesion, c and a friction angle, , that are related by the
equation τ = c + σ tan .
c. Failure occurs on a circular slide surface, which passes through
the toe of the slope.
d. A vertical tension crack occurs in the upper surface or in the
face of the slope.
e. The locations of the tension crack and of the slide surface are
such that the factor of safety of the slope is a minimum for the
slope geometry and ground water conditions considered.
f. A range of ground water conditions, varying from a dry
(drained) slope to a fully saturated slope under heavy recharge,
are considered in the analysis.

Circular Failure Analysis
Circular failure charts are optimized for a rock mass density of
18.9 kN/m3. Densities higher than this give high factors of safety,
densities lower than this give low factors of safety. Detailed
circular analysis may be required for slopes in which the material
density is significantly different from 18.9 kN/m3.

Shape of Sliding Surface
The actual shape of the “circular” slide surface is influenced by
the geological conditions in the slope.
In a homogenous weak or weathered rock mass, or a rock fill,
the failure is likely to form as a shallow, large radius surface
extending from a tension crack close behind the crest to the toe of
the slope (Figure (a)).
Failures in high cohesion, low friction materials such as clays,
the surface may be deeper with a smaller radius that may exit
beyond the toe of the slope.
The circular surface in the upper, weathered rock may be
truncated by the shallow dipping, stronger rock near the base
(Figure (b)).

For each combination of slope parameters there will be a slide
surface for which the factor of safety is a minimum—this is
usually termed the critical surface.
The procedure to find the critical surface is to run a large
number of analyses in which the center co-ordinates and the
radius of the circle are varied until the surface with the lowest
factor of safety is found. This is an essential part of circular
slope stability analysis.
Shape of Sliding Surface

The stability analysis of circular failure is carried out using
the limit equilibrium procedure.
This procedure involves comparing the available shear
strength along the sliding surface with the force required to
maintain the slope in equilibrium.
The application of this procedure to circular failures involves
division of the slope into a series of slices.
Since the shear strength available to resist sliding is
dependent upon the distribution of the normal stress  along
this surface, and since this normal stress distribution is
unknown, the problem is statically indeterminate.
Various researchers made assumptions for this distribution.
Stability Analysis Procedure

In order to calculate the forces due to water pressures acting on
the slide surface and in the tension crack, it is necessary to
assume a set of ground water flow patterns that coincide as
closely as possible with conditions that are believed to exist in
the field.
In the case of slopes in soil or waste rock, the permeability of
the mass of material is generally several orders of magnitude
higher than that of intact rock and, hence, a general flow pattern
will develop in the material behind the slope.
Within the rock mass, the equipotentials are approximately
perpendicular to the phreatic surface. Consequently, the flow
lines will be approximately parallel to the phreatic surface for
the condition of steady-state drawdown.
Groundwater Flow Assumptions

The phreatic surface is assumed to coincide with the ground
surface at a distance x, measured in multiples of the slope
height, behind the toe of the slope. This may correspond to the
position of a surface water source, or be the point where the
phreatic surface is judged to intersect the ground surface.
The phreatic surface itself has been obtained, for the range of
the slope angles and values of x considered, by solution of the
equations proposed by Casagrande (1934).
For the case of a saturated slope subjected to heavy surface
recharge, the equipotentials and the associated flow lines used
in the stability analysis are based upon the work of Han (1972).
Charts 1 to 5

Ground Water Flow Pattern
Ground water flow pattern under steady state drawdown
conditions where the phreatic surface coincides with the
ground surface at a distance x behind the toe of the slope.

Ground Water Flow Pattern
Ground water flow pattern in a saturated slope subjected
to surface recharge by heavy rain.

Ground water flow
models used with
Circular failure
Analysis charts

The circular failure charts presented in this chapter were
produced by running a search routine to find the most critical
combination of slide surface and tension crack for each of a
wide range of slope geometries and ground water conditions.
Provision was made for the tension crack to be located in either
the upper surface, or in the face of the slope. Detailed checks
were carried out in the region surrounding the toe of the slope
where the curvature of the equipotentials results in local flow
which differs from that illustrated in Figure (a) – Slide 12.
Charts 1 to 5

1. Decide upon the ground water conditions which are
believed to exist in the slope and choose the chart which is
closest to these conditions, using the figure indicating
Charts 1 to 5. Select rock strength parameters applicable to
the material forming the slope.
2. Calculate the value of the dimensionless ratio c/(γ H tan )
and find this value on the outer circular scale of the chart.
3. Follow the radial line from the value found in step 2 to its
intersection with the curve which corresponds to the slope
angle.
4. Find the corresponding value of tan /FS or c/(γ H FS),
depending upon which is more convenient, and calculate
the factor of safety.
Steps in using the Charts

Steps in Using Circular Failure Charts

Chart 1 – GW Condition 1
Fully Drained Slope

Chart 5 – GW Condition 5
Fully Saturated Slope

A 15.2 m high cut with a face angle of 40 is to be excavated in
overburden soil with a density γ = 15.7 kN/m3, a cohesion of 38
kPa and a friction angle of 30. Find the factor of safety of
the slope, assuming that there is a surface water source 61 m
behind the toe of the slope.
The ground water conditions indicate the use of Chart number 3
(61/15.2 ∼ 4).
The value of c/(γ H tan ) = 0.28.
The corresponding value of tan  /FS, for a 40 slope, is 0.32.
Hence, the factor of safety of the slope of 1.80.
Charts quick and simple to use and Sensitivity analysis of FS of
a slope to a very wide range of conditions can be carried out.
For example, if the cohesion were to be halved to 20 kPa and
the ground water pressure increased to that represented by
Chart number 2, the factor of safety drops to 1.28.
Example

During the production of the circular failure charts presented in
this chapter, the locations of both the critical slide surface and the
critical tension crack for limiting equilibrium (FS = 1) were
determined for each slope analyzed. These locations are
presented, in the form of charts.
It was found that, once ground water is present in the slope, the
locations of the critical circle and the tension crack are not
particularly sensitive to the position of the phreatic surface and
hence only one case, that for chart number 3, has been plotted.
It will be noted that the location of the critical circle centre with
water differs significantly from that for the drained slope.
These charts are useful for the construction of drawings of
potential slides and for estimating the friction angle when back-
analyzing existing circular slides. They also provide a start in
locating the critical slide surface when carrying out more
sophisticated circular failure analysis.
Location of Critical Sliding Surface

Drained Slope

Location of Critical Tension Crack
Drained Slope

Ground Water Chart 3

Location of Critical Tension Crack
Ground Water Chart 3

Example for Location of Critical Sliding Surface
Drained Slope with Slope Face Angle 30 and
Friction Angle 20
The critical slide circle center is located at X = 0.2H
and Y = 1.85H
The critical tension crack is at a distance b = 0.1H
behind the crest of the slope

Example-1: China Clay Pit Slope
Ley (1972) investigated the stability of a China clay pit slope
which was considered to be potentially unstable, and that a
circular failure was the likely type of instability.
The slope profile is illustrated in Figure.
The material, a heavily kaolinized granite, was tested in direct
shear to determine the friction angle,  and cohesion, c.
The input data used for the analysis is
H = 76.8 m, c = 6.9 kPa,  = 37 and  = 21.5 kN/m3.
Chart number 2 corresponds most closely to the position of
phreatic surfaces estimated from two piezometers in the slope
and a known water source some distance behind the slope.
c/(γ H tan ) = 0.0056
tan  /FS = 0.76
FS = 1.01 From Janbu’s Method FS = 1.03
Stability inadequate.

Example-2
A 22 m high rock cut with a face angle of 60 has been
excavated in a massive, very weak volcanic tuff. A tension
crack has opened behind the crest and it is likely that the slope
is on the point of failure, that is, the factor of safety is
approximately 1.0. The friction angle of the material is
estimated to be 30, its density is 25 kN/m3, and the position
of the water table is shown on the sketch of the slope. The
rock contains no continuous joints dipping out of the face, and
the most likely type of failure mode is circular
failure.

Slope Geometry with Ground Water Table

Example-2
a) The ground water level corresponds to ground water
condition 3, so circular failure Chart 3. H = 22 m.
When  = 30 and FS = 1.0, tan/FS = 0.58.
For a slope angle of 60, c/(H FS) = 0.086
 c = 0.086x25x22x1 = 47.3 kPa
b) If the slope were completely drained, circular failure Chart
1 could be used for analysis. With c = 47.3 kPa,
c/(H tan) = 47.3/(25x22xtan30) = 0.15
tan /FS = 0.52 from Chart 1 giving FS = 1.11.
This factor of safety is less than that usually accepted for a
temporary slope, that is, FS = 1.2, so draining the slope
would not be an effective means of stabilization.
c) When FS = 1.3 and  = 30, then tan /FS = 0.44.
Using Chart 3 and For a slope angle of 60,
c/(H FS) = 0.11
 H = 47.3/(25x1.3x0.11) = 13.2 m

Example-2
This shows that the slope height must be reduced by 8.8m
to increase the factor of safety from 1.0 to 1.3.
d) The critical circle and critical tension crack for a slope
with ground water present are located using the Chart. For
a slope angle of 60 and a friction angle of 30, the
coordinates of the center of the circle are:
X = −0.35H = −7.7 m, that is, 7.7 m horizontally beyond
the toe.
Y = H = 22 m, that is 22 m above the toe.
The location of the tension crack behind the crest is:
b/H = 0.13
or b = 2.9 m.

Position of Critical Sliding Surface & Tension Crack

Example-3
A highway plan called for a cut at an angle of 42. The total
height of the cut would be 61 m and it was required to check
whether the cut would be stable. The slope was in weathered
and altered material, and failure, if it occurred, would be a
circular type. Insufficient time was available for ground water
levels to be accurately established or for shear tests to be
carried out. The stability analysis was carried out as follows:
For the condition of limiting equilibrium, FS = 1 and tan /FS
= tan . By reversing the procedure outlined earlier (i.e. the
steps), a range of friction angles were used to find the values
of the ratio c/(γ H tan ) for a face angle of 42. The value of
the cohesion c which is mobilized at failure, for a given
friction angle, can then be calculated. This analysis was
carried out for dry slopes using chart number 1 (line B in
Figure), and for saturated slopes using chart number 5 (line
A). Figure shows the range of friction angles and cohesions
that would be mobilized at failure.

Comparison between
shear strength
mobilized and shear
strength available
Example-3

Example-3
The shaded circle (D) included in Figure indicated the range
of shear strengths that were considered probable for the
material under consideration. This figure shows that the
available shear strength may not be adequate to maintain
stability in this cut, particularly when the cut is saturated.
Consequently, the face angle could be reduced, or ground
water conditions investigated to establish actual ground water
pressures and the feasibility of drainage.
The effect of reducing the slope angle can be checked very
quickly by finding the value of the ratio c/(γ H tan ) for a
flatter slope of 30, in the same way as it was found for the
42 slope. The dashed line (C) in Figure indicates the shear
strength, which is mobilized in a dry slope with a face angle of
30. Since the mobilized shear strength C is less than the
available shear strength D, the dry slope is likely to be stable.

Example-4
A catastrophic slide in colliery waste material at Aberfan in
Wales (UK) on October 22, 1966.
Since then, attention focussed on potential danger associated
with large waste dumps from mining operations.
The shear strength vs. normal stress relation is usually
nonlinear for colliery waste material.
Till now linear strength criterion. To use nonlinear strength
criterion, we require instantaneous friction angles (i) and
cohesive strengths (ci) for different effective normal stress
levels. This is done by drawing a series of tangents to the
strength envelope at different normal stress.
Strength envelope for colliery waste with three tangents-
Figure

Shear Strength of Typical Colliery Waste Material

Example-4
Strength envelope for colliery waste with three tangents-
Figure.
Tangent Number Cohesion, ci (kPa) Friction Angle, i
1 0 38
2 20 26
3 40 22
Relationship between Slope Height and Slope Angle for
Limiting Equilibrium (FS=1) investigated for
i. Dry dump (using Chart 1)
ii. Dump with some groundwater flow (using Chart 3)

Example-4
Tangent Number 1
Cohesion ci = 0 giving c/(Htan) = 0.
For tan/F = tan38/1 = 0.78, the slope is:
Chart 1: 38
Chart 3: 25
For zero cohesion, the dump face angle would be independent
of slope height.
It is normally assumed that the angle of repose of a waste
dump is independent of the height of the dump and is equal to
the friction angle of the material.
But this assumption is only correct for a dry slope of limited
height.
Any build up of water pressure within the dump causes a
serious reduction in the stable face angle and, once the normal
stress across the potential surface becomes high enough for the
next tangent to become operative, the high initial friction angle
no longer applies and the dump face assumes a flatter angle.

Relationship between Slope Height & Slope Angle for a
Typical Colliery Waste Dump

Example-4
Tangent Number 2
For c = 20 kPa,  = 26 and  = 18 kN/m3
and tan/F = tan26/1 = 0.49, Hence
H
H
H
c 28
.
2
26
tan
18
20
tan






H (m) c/(Htan) Slope Angle (degrees)
Chart 1 Chart 3
20 0.114 50 39
40 0.057 39 25
60 0.038 34 20
80 0.029 32 18
100 0.023 31 17
Plotting these values gives Curve No. 2 for the drained and the
wet dumps.

Example-4
Tangent Number 3
For c = 40 kPa,  = 22 and  = 18 kN/m3
and tan/F = tan22/1 = 0.4, Hence
H
H
H
c 5
.
5
22
tan
18
40
tan






H (m) c/(Htan) Slope Angle (degrees)
Chart 1 Chart 3
20 0.275 61 56
40 0.138 43 31
60 0..092 34 22
80 0.069 31 18
100 0.055 30 16
Plotting these values gives Curve No. 3 for the drained and the
wet dumps.

Example-4
Envelopes to the curves (1, 2 and 3) drawn for both drained
and wet dumps, which give the relationship between dump
face angle and dump height.
It is clear from the envelopes that continuing to increase the
dump height on the assumption that it will remain stable at an
angle of repose equal to friction angle is not correct.
The danger associated with poor dump drainage is also
evident.
The slope height versus slope angle relationship is extremely
sensitive to the shape of the shear strength envelope.

Circular Failure-Hoek&Bray.pptx

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