The results are tabulated below.
If Fc is plotted against F?(Fig.1.6) it is apparent that: F=FC=F?=1.43
1.4 The Method of Slices
In this method the potential failure surface, in section, is against assumed to be a circle arc with centre O radius r. The soil mass (ABCD) above a trial failure surface(AC) is divided by vertical planes into series of slices of width b, as shown in Fig.1.7. The base of each slice is assumed to be a straight line. For any slice the inclination of the base to the horizontal is ? and the height, measured on the centerline, is h. The factor of safety is defined as the ratio of the available shear strength (?f) to the shear strength (?m) which must be mobilized to maintain a condition of limiting equilibrium. i.e. F=
?f ?mThe factor of safety is taken to be the same for each slice, implying that there must be mutual support between the slice. i.e. forces must act between the slices.
The forces (per unit dimension normal to the section) acting on a slice are listed below.
(1) The total weight of slice, W=?bh (?sat where appropriate)
(2) The total normal force on the base, N. In general this force has two components. The effective normal force N′ (equal to ?′l) and the boundary water force ul, where u is the pore water pressure at the center of the base and l is the length of the base.
(3) The shear force on the sides, T=?ml.
(4) The total normal forces on the sides,E1 and E2. (5) The shear forces on the sides, X1 and X2.
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 inter-slice 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 r sin ?, therefore:
?Tr=?Wsin?
r6
Figure 1.7 The method of slices.
Now,
T=?ml= ∵
?f?fFl
?Fl=?Wsin?
∴ F=
??l ?Wsin?fFor an analysis in terms of effective stress: F= or,
F=
c?La?tan???N??(c????tan??)l ?Wsin? （1.7）
?Wsin?where La is the arc length AC. Equation 1.7 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 the solution it is assumed that for each slice the resultant of the inter-slice forces is zero. The solution involves resolving the forces on each slice normal to the base, i.e.:
N??Wcos??ul
Hence the factor of safety in terms of effective stress ( equation 1.7 ) is given by: F=
c?La?tan???(Wcos??ul)?Wsin? (1.8)
The components Wcos? and Wsin? can be determined graphically for each slice. Alternatively, the value of ? 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 of 5-20﹪.
For an analysis in terms of total stress the parameters cu and ?u are used and the value of u in equation 1.8 is zero. If ?u=0 the factor of safety is given by:
F=
cuLa (1.9)
?Wsin?7
As N′does not appear in equation 1.9 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.
X1-X2=0
For equilibrium the shear force on the base of any slice is: T=
1(c?l?N?tan??) FResolving forces in the vertical direction:
c?lN?sin??tan??sin? FFc?ltan??sin?) (1.10) ∴ N??(W?sin??ulcos?)/(cos??FF W?N?cos??ulcos??It is convenient to substitute:
l= bsec? From equation 1.7, after some rearrangement: F?1sec???[{cb?(W?ub)tan?}] (1.11) ??tan?tan??Wsin?1?FThe pore water press can be related to the total‘fill pressure’ at any point by means of the dimensionless pore press ratio, defined as: ru?u (1.12) ?h(?sat where appropriate ) For any slice,
ru?u W/bHence equation 1.11 can be written: F?1sec???[{cb?W(1?ru)tan?}] (1.13) ??tan?tan?Wsin??1?FAs the factor of safety occurs on both sides of equation 1.13 a process of successive approximation must be used to obtain a solution but convergence is rapid. The method is very suitable for solution on the computer. In the computer program the slope geometry can be made more come complex, with soil strata having different properties and pore pressure conditions being introduced.
In most problems the value of the pore pressure ratio ru is not constant over
8
the whole failure surface but, unless there are isolates 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 [1.12] proposed a method of analysis in which the resultant inter-slice 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 inter-slice forces.
Dimensionless stability coefficients for homogeneous slopes, based on equation 1.13, have been published by Bishop and Morgenstern[1.3]. 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?m?nru (1.14)
where m and n are the stability coefficients m and n are functions of ?,??, the dimensionless number c?/?h and the depth factor D.
Example 1.3
Using the Fellenius method of slices, determined the factor of safety in terms of effective stress of the slope shown in Fig.1.8 for the given failure surface. The distribution of pore water pressure along the failure surface is given in the figure. The unit weight of the soil is 20 kN/m3 and the relevant shear strength parameters are c?=10kN/m2 and ??=29°.
The factor of safety is given by equation 9.8. The soil mass is divided into slices 1.5m wide. The weight(W) of each slice is given by: W??bh?20?1.5?h?30hkN/m
The height h for each slice is set off bellow the centre of the base and the normal and tangential components hcos? and hsin? respectively are determined graphically, as shown in Fig.1.8. Then:
Wcos??30hcos? and
Wsin??30hsin?
Figure 1.8 Example 1.3.
The arc length (La) is calculated as 14.35m. The results are tabulated below:
9
?Wcos??30?17.50?525kN/m ?Wsin??30?8.45?254kN/m
?(Wcos??ul)?525?132.8?392.2kN/m
c?L?tan???(Wcos??ul) F?
Wsin??
a ?(10?14.35)?(0.554?393.2)
2541.5 Analysis of a Plane Translational Slip
It is assumed that 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 ? to the horizontal and the depth of the failure plane z, as shown in section in Fig.1.9. 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. Figure 1.9 Plane translational slip.
In terms of effective stress, the shear strength of the soil along the failure plane is:
?f?c??(??u)tan?? and the factor of safety is: F??f ?The expressions for ?, ? and u are as follows: ??{(1?m)??m?sat}zcos2? ??{(1?m)??m?sat}zsin?cos? u?mz?wcos2?
The following special cases are of interest. If c?=0 and m=0(i.e. the soil between the surface and the surface plane is not fully saturated), then:
F?tan?? （1.15） tan?If c?=0 and m=1(i.e. the water table conditions with the surface of the slope),then: F?
??tan??? ?sattan?10