Introduction Gravitational and seepage forces tend to cause instability in natural slopes, in slopes formed by excavation and in the slopes of embankments and earth dams. The most important types of slope failure are illustrated in Fig.1. In rotational slips the shape of the failure surface in section may be a circular arc or a noncircular curve. In general, circular slips are associated with homogeneous soil conditions and noncircular slips with nonhomogeneous conditions. Translational and compound slips occur where the form of the failure surface is influenced by the presence of an adjacent stratum of significantly different 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 sections.
In practice, limiting equilibrium methods are used 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 twodimensional analysis gives a conservative result for a failure on a threedimensional (dishshaped) surface. Analysis for the Case of This analysis, in terms of total stress, covers the case of a fully saturated clay under undrained conditions, i.e. for the condition immediately after construction. Only moment equilibrium is considered in the analysis. In section, the potential failure surface is assumed to be a circular arc. A trial failure surface (centre O, radius r and length is shown in Fig.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 expressed as
where F is the factor of safety with respect to shear strength. Equating moments about O:
therefore
The moments of any additional forces must be taken into account. In the event of a tension crack developing, as shown in Fig.2, the arc length 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.
For the case of , values of can be obtained from Fig 3. The coefficient depends on the slope angle and the depth factor D, where DH is the depth to a firm stratum. Gibson and Morgenstern published stability coefficients for slopes in normally consolidated clays in which the undrained strength varies linearly with depth.
Example A 45° slope is excavated to a depth of 8 m in a deep layer of saturated clay of unit weight 19 kN/m3: the relevant shear strength parameters are cu = 65.0 kN/m2 and . Determine the factor of safety for the trial failure surface specified in Fig. 4.
crosssectional area ABCD is 70 m^{2}. Weight of soil mass = 70 x 19 = 1330 kN/mThe centroid of ABCD is 4.5 m from O. The angle AOC is 89.5° and radius OC is 12.1 m. The arc length ABC is calculated as 18.9 m. The factor of safety is given by:
This is the factor of safety for the trial failure surface selected and is not necessarily the minimum factor of safety. The minimum factor of safety can be estimated by using Equation (2). From Fig. 3, and assuming that D is large, the value of is 0.18. Then
The Method of Slices In this method the potential failure surface, in section, 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 series of slices of width b, as shown in Fig.5. The base of each slice is assumed to be a straight line. For any slice the inclination of the base to the horizontal is a and the height, measured on the centreline, is h. The factor of safety is defined as the ratio of the available shear strength to the shear strength
which must be mobilized to maintain a condition of limiting equilibrium, i.e.
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 section) acting on a slice are:
Any external forces must also be included in the analysis. The problem is statically indeterminate and in order to obtain a solution assumptions must be made regarding the interstice forces E and X: the resulting solution for factor of safety is not exact. 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 , therefore
where La is the arc length AC. Equation is 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 base, i.e. N' = Wcosa — ul Hence the factor of safety in terms of effective stress(Equation 3) is given by:
The components W cos a and W sin a can be determined graphically for each slice. Alternatively, the value of a can be measured or calculated. Again, a series of trial failure surfaces must be chosen 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 520%.
As N' does not appear in Equation (5) an exact value of F is obtained. The Bishop Simplified Solution In this solution it is assumed that the resultant forces on the sides of the slices are horizontal, i.e. X1X2=0 For equilibrium the shear force on the base of any slice is
Resolving forces in the vertical direction:
It is convenient to substitute l = b sec a From Equation(3), after some rearrangement,
The pore water pressure can be related to the total 'fill pressure' at any  point by means of the dimensionless pore pressure ratio, defined as
As the factor of safety occurs on both sides of Equation(9), a process of I successive approximation must be used to obtain a solution but convergence is rapid. Due to the repetitive nature of the calculations and the need to select an adequate number of trial failure surfaces, the method of slices is particularly suitable for solution by computer. More complex slope geometry and different soil strata can be introduced. In most problems the value of the pore pressure ratio ru is not constant over the whole failure surface but, unless there are isolated regions of high pore pressure, an average value (weighted on an area basis) is normally used in design. Again, the factor of safety determined by this method is an underestimate but the error is unlikely to exceed 7% and in most cases is less than 2%. Spencer proposed a method of analysis in which the resultant interstice forces are parallel and in which both force and moment equilibrium are satisfied. Spencer showed that the accuracy of the Bishop simplified method, in which only moment equilibrium is satisfied, is due to the insensitivity of the moment equation to the slope of the interstice forces. Dimensionless stability coefficients for homogeneous slopes, based on Equation (9), have been published by Bishop and Morgenstern. It can be shown that for a given slope angle and given soil properties the factor of safety varies linearly with ru and can thus be expressed as F = mnru (10) where m and n are the stability coefficients. The coefficients m and n are functions of the dimensionless number c'/gH and the depth factor D.
Example Using the Fellenius method of slices, determine the factor of safety, in terms . of effective stress, of the slope shown in Fig.(6) for the given failure surface. The unit weight of the soil, both above and below the water table, is 20kN/m3 and the relevant shear strength parameters are c' = 10kN/m2 and. The factor of safety is given by Equation (4). The soil mass is divided into slices 1.5 m wide. The weight (W) of each slice is given by W= gbh =20 x 1.5 x h = 30h kN/m
Table (1)
The height h for each slice is set off below the centre of the base and the normal and tangential components hcosa and hsina respectively are determined graphically, as shown in Fig (6). Then
The pore water pressure at the centre of the base of each slice is taken to be gw,zw, where zw is the vertical distance of the centre point below the water  table (as shown in the figure). This procedure slightly overestimates the pore "water pressure which strictly should be gwze, where ze is the vertical distance below the point of intersection of the water table and the equipotential through the center of the slice base. The error involved is on the safe side. The arc length (La) is calculated as 14.35 mm.The results are given Table 1.
Analysis of a Plane Translational Slip It is assumed that the potential failure surface is parallel to the surface of the slope and is at a depth that is small compared with the length of the slope. The slope can then be considered as being of infinite length, with end effects being ignored. The slope is inclined at angle b to the horizontal and the depth of the failure plane is z, as shown in section in Fig (7). The water table is taken to be parallel to the slope at a height of mz (0 < m < 1) above the failure plane. Steady seepage is assumed to be taking place in a direction parallel to the slope. The forces on the sides of any vertical slice are equal and opposite and the stress conditions are the same at every point on the failure plane.
In terms of effective stress, the shear strength of the soil along the failure plane is
The following special cases are of interest. If c' = 0 and m = 0 (i.e. the soilil between the surface and the failure plane is not fully saturated), then
If c' = 0 and m = 1 (i.e. the water table coincides with the surface of the J slope), then:
Example
