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CHAPTER 9
Stability of Slopes
9.1Introduction
Gravitational and epage forces tend to cau instability in natural slopes,in slopes fo rmed by excavation and in the slopes of embankments and earth dams.The most important types of slope failure are illustrated in Fig.9.1.In rotational slips the shape of the failure surface in ction may be a circular arc or a non-circular curve ．In general ，circular slips are associated with homogeneous soil conditions and non-circular slips with non-homogeneous conditions ．Translational and compound slips occur where the form of the failure surface is influenced by the prence of an adjacent stratum of significantlydifferent strength ．
Translational slips tend to occur where the adjacent stratum is at a relatively shallow depth below the surface of the slope:the failure surface tends to be plane and roughly parallel to the slope.Compound slips usually occur where the adjacent stratum is at greater depth ，the failure surface consisting of curved and plane ctions
．
In practice,limiting equilibrium methods are ud in the analysis of slope stability.It is considered that failure is on the point of occurring along an assumed or a known failure surface ．The shear strength required to maintain a condition of limiting equilibrium is compared with the available shear strength of the soil ，giving the average factor of safety along the failure surface ．The problem is considered in two dimensions ，conditions of plane strain being assumed ．It has been shown that a two-dimensional analysis gives a conrvative result for a failure on a three-dimensional(dish-shaped)surface ．
9.2Analysis for the Ca of φu =0
This analysis,in terms of total stress ，covers the ca of a fully saturated clay under undrained For the condition immediately after construction ．Only moment equilibrium is considered in the analysis ．In ction,the potential
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failure surface is assumed to be a circular arc.A trial failure surface(centre O ，radius r and length L a )is shown in Fig.9.2.Potential instability is due to the total weight of the soil mass(W per unit Length)above the failure surface ．For equilibrium the shear strength which must be mobilized along the failure surface is expresd
as
where F is the factor of safety with respect to shear strength ．Equating moments about O
：
Therefore
(9.1)
The moments of any additional forces must be taken into account ．In the event of a tension crack developing ，as shown in Fig.9.2，the arc length L a is shortened and a hydrostatic force will act normal to the crack if the crack fills with water ．It is necessary to analyze the slope for a number of trial failure surfaces in order that the minimum factor of safety can be determined ．
Bad on the principle of geometric similarity ，Taylor[9.9]published stability coefficients for the analysis of homogeneous slopes in terms of total stress ．For a slope of height H the stability coefficient (N s )for the failure surface along which the factor of safety is a minimum
is
(9.2)
For the ca of φu =0，values of N s can be obtained from Fig.9.3.The coefficient N s depends on the slope angle βand the depth factor D ，where DH is the depth to a firm stratum ．
Gibson and Morgenstern [9.3]published stability coefficients for slopes in normally consolidated clays in which the undrained strength c u (φu =0)varies linearly with depth ．
Example 9.1
A 45°slope is excavated to a depth of 8m in a deep layer of saturated clay of
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unit weight 19kN ／m 3:the relevant shear strength parameters are c u =65kN ／m 2and φu =0．Determine the factor of safety for the trial failure surface specified in Fig.9.4.
In Fig.9.4,the cross-ctional area ABCD is 70m 2.
Weight of soil mass=70×19=1330kN ／m
The centroid of ABCD is 4.5m from O ．The angle AOC is 89.5°and radius OC is 12.1m ．The arc length ABC is calculated as 18.9m ．The factor of safety is given by
：
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This is the factor of safety for the trial failure surface lected and is not necessarily the minimum factor of safety ．
The minimum factor of safety can be estimated by using Equation 9.2.From Fig.9.3，β=45°and assuming that D is large ，the value of N s is
0.18.Then
9.3The Method of Slices
In this method the potential failure surface ，in ction ，is again assumed to be a circular arc with centre O and radius r ．The soil mass (ABCD)above a trial failure surface (AC)is divided by vertical planes into a ries of slices of width b,as shown in Fig.9.5.The ba of each slice is assumed to be a straight line ．For any slice the inclination of the ba to the horizontal is αand the height,measured on the centre-1ine,is h.The factor of safety is defined as the ratio of the available shear strength(τf )to the shear strength(τm )which must be mobilized to maintain a condition of limiting equilibrium,
<
The factor of safety is taken to be the same for each slice ，implying that there must be mutual support between slices ，i.e.forces must act between the slices ．The forces (per unit dimension normal to the ction)acting on a slice are ：
1.The total weight of the slice ，W=γb h (γsat where appropriate)．
2.The total normal force on the ba ，N (equal to σl)．In general this
force has two components ，the effective normal force N '(equal to σ'l )and the boundary water force U(equal to ul )，where u is the pore water pressure at the centre of the ba and l is the length of the ba ．
3.The shear force on the ba ，T=τm l .
4.The total normal forces on the sides,E 1and E 2.
5.The shear forces on the sides ，X 1and X 2.
Any external forces must also be included in the analysis ．
The problem is statically indeterminate and in order to obtain a solution
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Considering moments about O ，the sum of the moments of the shear forces T on the failure arc AC must equal the moment of the weight of the soil mass ABCD ．For any slice the lever arm of W is rsin α,
therefore
∑Tr=∑Wr sin α
Now,
For an analysis in terms of effective stress
，
Or
(9.3)
where L a is the arc length AC ．Equation 9.3is exact but approximations are introduced in determining the forces N '．For a given failure arc the value of F will depend on the way in which the forces N 'are estimated ．
The Fellenius Solution
In this solution it is assumed that for each slice the resultant of the interslice forces is zero ．The solution involves resolving the forces on each slice normal to the ba ，i.e.
N '=WCOS α-ul
Hence the factor of safety in terms of effective stress (Equation 9.3)is given
by
(9.4)
The components WCOS αand Wsin αcan be determined graphically for each slice ．Alternatively ，the value of αcan be measured or calculated ．Again ，a ries of trial failure surfaces must be chon in order to obtain the minimum factor of safety ．This solution underestimates the factor of safety ：the error ，compared with more accurate methods of analysis ，is usually within the range 5-2%.
For an analysis in terms of total stress the parameters C u and φu are ud and the value of u in Equation 9.4is zero ．If φu =0,the factor of safety is given
by
(9.5)
