外文翻译
1.1 Introduction
Stability of Slopes
Gravitational and seepage forces tend to cause instability in natural slopes, in slopes of embankments and earth dams. The most important types of slope failure arc illustrated in Fig.1.1. In rotational slips the shape of the failure surface in section may be a circular are 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 failure surface is influenced by the presence of an adjacent stratum is at a relatively shallow depth bellow 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.
Figure 1.1 Type of slope failure
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 the limiting equilibrium is compared with the available shear strength of the soil, giving the average factor safety along the failure surface. The problem is considered in two dimensions, conditions of plane strain being assumed. It has been shown that two-dimensional analysis gives a conservative result for a failure on a three-dimensional (dish-shaped) surface. Figure 1. 2 The ?u=0 analysis
1.2 Analysis for the Case of ?u=0
The analysis, in term 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 circle arc. A trial failure surface (centre O, radius and length La) is shown in Fig 1.2.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: ?m=
?fCu= FFwhere F is the factor of safety with respect to shear strength. Equation moment
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about O:
CuLar FCuLar F= (1.1)
Wd Wd=
The moments of any additional forces must be taken into account. In the event of a tension crack developing, as shown in Fig1.2,the arc length La 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.
Example 1.1
A 45°slope is excavated to a depth of 8m in a deep layer of unit weight 19kN/m3: the relevant shear strength parameters are cu=65kN/m3 and ?u=0.Determine the factor of safety for the trial surface specified in Fig1.3. In Fig1.3, the cross-sectional area ABCD is 70m2. The weight of the soil mass=70×19=1330m2. The cent roid 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:
CuLar Wd65?18.9?12.165?18.9?12.1 =
1330?4.51330?4.5 F=
=2.48
This is the factor of safety for the trial failure surface selected and is not
necessarily the minimum factor of safety.
Figure 1.3 example 1.1
1.3 The ?-Circle Method
The analysis is in terms of total stress. A trial failure surface , a circular arc (centre o, radius r) is selected as shown in Fig 1.4.If the shear strength parameters are cu and ?u ,the shear strength which must be mobilized for equilibrium is: ?m=
?f=?l F =cm+?tan?m
Figure 1.4 The ?-circle method
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Where F is the factor of safety with respect to shear strength .For convenience the following notation is introduced: cm=
cu (1.2) Fc tan?m=
tan?u (1.3) F?it being a requirement that: FC=F?=F
An element ab, of length l, of the failure surface is considered, the element being short enough to be approximated to a straight line. The forces acting on ab (per unit dimension normal to the section) are as follows: (1) the total normal force ?1;
(2) the component of shearing resistance cml;
(3) the component of shearing resistance ?1tan?m.
If each force cml along the failure surface is split into components perpendicular and parallel to the chord AB, the perpendicular components sum to zero and the sum of the parallel components is given by:
C=cmLc (1.4) where Lc is the chord length AB. The force C is thus the resultant, acting parallel to the chord cml. The line of application of the resultant force C can be determined by taking moments about the centre O, then: C rc=r?cml i.e.
cmLcrc=rcmLa where La=?l is the arc length AB. Thus,
rc=
Lar (1.5) LCThe resultant of the forces ?1 and ?1tan?m on the element ab acts at angle ?m to the normal and is the force tangential to a circle, centre O, of radius r sin?m: this circle is referred to as the ?-circle. The same technique was used in Chapter 5.The overallresult (R) for the arc AB is assumed to be tangential to the ?-circle. Strictly, the resultant R is tangential to a circle of radius slightly greater than r sin?m but the error involved in the above assumption is generally insignificant.
The soil mass above the trial failure surface is in equilibrium under its total
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weight (W) and the shear resultants C and R. The force W is known in magnitude and direction; the direction only of the resultant C is known. Initially a trial value of F? is selected and the corresponding value of ?m is calculated from equation 1.3.For equilibrium the line of application of the resultant R must be tangential to the ?-circle and pass though the point of intersection of the forces W and C. The force diagram can then be drawn, from which the value of C can be obtained .Then:
cm =and
FC=
Cu CmC LCIt is necessary to repeat the analysis at least three times, starting with different values of F?.If the calculated values of FC are plotted against the corresponding values of F?,the factor of safety corresponding to the requirement FC=F? can be determined .The whole procedure must be repeated for a series of trial failure surfaces in order that the minimum factor of safety is obtained.
For an effective stress analysis the total weight W is combined with the resultant boundary water force on the failure mass and the effective stress parameters c′and ?′used.
Based on the principle of geometric similarity, Taylor（1.13）published stability coefficients for the analysis of homogeneous slopes in terms of total stress. For a slope of height H the stability coefficients for the analysis of homogeneous slopes in terms of total stress. For a slope of height H the stability coefficient (Ns) for the failure surface along which the factor of safety is a minimum is:
Ns=
Cu (1.6) F?HValues of Ns , which is a function of the slope angle ? and the shear strength parameter ?u,can be obtained from Fig 1.5.For ?u=0,the value of Ns also depends on the depth factor D, where DH is the depth to a firm stratum. Firm stratum
Figure 1.5 Taylor′s coefficients.
In example 1.1, ?=45°, ?u=0,and assuming D is large, the value of Ns is 0.18.Then from equation 1.6:
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F= =
Cu Ns?H65
0.18?19?8 =2.37
Gibson and Morgenstern〔1.4〕published stability coefficients for slopes in normally-consolidated clays in which the untrained strength cu(?u=0) varies linearly with depth.
Figure 1.6 Example 1.2
Example 1.2
An embankment slope is detailed in Figure 1.6.Fir the given failure surface. Determine the factor of safety in terms of total stress using the ?-circle method. The appropriate shear strength parameters are cu=15kN/m2 and ?u=15°: the unit weight of soil is 20 kN/m2.
The area ABCD is 68 m2 and the centroid (G) is 0.60m from the vertical through D. The radius of the failure arc is 11.10m.The arc length AC is 19.15 m and the chord length AC is 16.85m. The weight of the soil mass is:
W=68×20=1360 k N/m The position of the resultant C is given by:
Lar LC19.15
= ×11.10
16.85
r c=
Now:
?m=tan-1(
tan15) F?Trial value of F? are chosen, the corresponding values of r sin ?m are calculated and the ?-circle drawn shown in Fig.1.6.The resultant C(for any value of F?) acts in a directions parallel to the chord AC and at distance rc from O. The resultant C(for any value F?) acts in a direction parallel to the chord AC and at distance rc from O. The forces C and W intersect at point E. The resultant R, corresponding to each value of F?, passes through E is tangential to the appropriate ?-circle. The force diagrams are drawn and the values of C determined.
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