边坡稳定经典英文教材及翻译之原文.docx

CHAPTER 9Stability of Slopes9.1 IntroductionGravitational 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.9.1.In rotational slips the shape of the failure surface in section may be a circular arc or a non-circular curveIn general，circular slips are associated with homogeneous soil conditions and non-circular slips with non-homogeneous conditionsTranslational and compound slips occur where the form of the failure surface is influenced by the presence of an adjacent stratum of significantlydifferent strengthTranslational 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 sectionsIn 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 surfaceThe 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 surfaceThe problem is considered in two dimensions，conditions of plane strain being assumedIt has been shown that a two-dimensional analysis gives a conservative result for a failure on a three-dimensional(dish-shaped) surface9.2 Analysis for the Case of u =0This analysis, in terms of total stress，covers the case of a fully saturated clay under undrained conditions, i.e. For the condition immediately after constructionOnly moment equilibrium is considered in the analysisIn section, the potential failure surface is assumed to be a circular arc. A trial failure surface(centre O，radius r and length La)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 surfaceFor equilibrium the shear strength which must be mobilized along the failure surface is expressed aswhere F is the factor of safety with respect to shear strengthEquating moments about O： Therefore (9.1) The moments of any additional forces must be taken into accountIn the event of a tension crack developing ，as shown in Fig.9.2，the arc length La is shortened and a hydrostatic force will act normal to the crack if the crack fills with waterIt is necessary to analyze the slope for a number of trial failure surfaces in order that the minimum factor of safety can be determined Based on the principle of geometric similarity，Taylor9.9published stability coefficients for the analysis of homogeneous slopes in terms of total stressFor a slope of height H the stability coefficient (Ns) for the failure surface along which the factor of safety is a minimum is (9.2)For the case ofu =0，values of Ns can be obtained from Fig.9.3.The coefficient Ns depends on the slope angleand the depth factor D，where DH is the depth to a firm stratumGibson and Morgenstern 9.3 published stability coefficients for slopes in normally consolidated clays in which the undrained strength cu(u =0) varies linearly with depthExample 9.1A 45slope is excavated to a depth of 8 m in a deep layer of saturated clay of unit weight 19 kNm3:the relevant shear strength parameters are cu =65 kNm2 andu =0Determine the factor of safety for the trial failure surface specified in Fig.9.4.In Fig.9.4, the cross-sectional area ABCD is 70 m2.Weight of soil mass=7019=1330kNmThe centroid of ABCD is 4.5 m from OThe angle AOC is 89.5and radius OC is 12.1 mThe arc length ABC is calculated as 18.9mThe 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 safetyThe minimum factor of safety can be estimated by using Equation 9.2.From Fig.9.3，=45and assuming that D is large，the value of Ns is 0.18.Then9.3 The Method of SlicesIn this method the potential failure surface，in section，is again assumed to be a circular arc with centre O and radius rThe 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.9.5.The base of each slice is assumed to be a straight lineFor any slice the inclination of the base to the horizontal isand the height, measured on the centre-1ine,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. 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 slicesThe forces (per unit dimension normal to the section) acting on a slice are：1.The total weight of the slice，W=b h (sat where appropriate)2.The total normal force on the base，N (equal to l)In general thisforce has two components，the effective normal force N(equal tol ) and the boundary water force U(equal to ul )，where u is the pore water pressure at the centre of the base and l is the length of the base3.The shear force on the base，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 interslice 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 ABCDFor any slice the lever arm of W is rsin,thereforeTr=Wr sinNow, For an analysis in terms of effective stress，Or (9.3)where La is the arc length ACEquation 9.3 is exact but approximations are introduced in determining the forces NFor a given failure arc the value of F will depend on the way in which the forces N are estimated The Fellenius SolutionIn this solution it is assumed that for each slice the resultant of the interslice forces is zeroThe solution involves resolving the forces on each slice normal to the base，i.e.N=WCOS-ulHence the factor of safety in terms of effective stress (Equation 9.3) is given by (9.4)The components WCOSand Wsincan be determined graphically for each sliceAlternatively，the value of can be measured or calculatedAgain，a series of trial failure surfaces must be chosen in order to obtain the minimum factor of safetyThis 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 Cu andu are used and the value of u in Equation 9.4 is zeroIf u=0 ,the factor of safety is given by (9.5)As N does not appear in Equation 9.5 an exact value of F is obtainedThe Bishop Simplified SolutionIn this solution it is assumed that the resultant forces on the sides of theslices are horizontal，i.e.Xl-X2=0For equilibrium the shear force on the base of any slice is Resolving forces in the vertical direction: (9.6)It is convenient to substitute l=b secFrom Equation 9.3，after some rearrangement， (9.7) The pore water pressure can be related to the total fill pressure at anypoint by means of the dimensionless pore pressure ratio，defined as (9.8)(sat where appropriate)For any slice, Hence Equation 9.7 can be written: (9.9) As the factor of safety occurs on both sides of Equation 9.9,a process of 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 computerMore 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 designAgain，the factor of safety determined by this method is an underestimate but the error is unlikely to exceed 7and in most cases is less than 2 Spencer 9.8 proposed a method of analysis in which the resultant Interslice forces are parallel and in which both force and moment equilibrium are satisfiedSpencer 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 interslice forces Dimensionless stability coefficients for homogeneous slopes，based on Equation 9.9，have been published by Bishop and Morgenstern 9.2.It can be shown that for a given slope angle and given soil properties the factor of safety varies linearly with u and can thus be expressed asF=m-nu (9.10)where，m and n are the stability coefficientsThe coefficients，m and n arefunctions of,，the dimensionless number c/and the depth factor D.Example 9.2Using the Fellenius method of slices，determine the factor of safety，in terms of effective stress，of the slope shown in Fig.9.6 for the given failure surfaceThe unit weight of the soil，both above and below the water table，is 20 kNm 3 and the relevant shear strength parameters are c=10 kN/m2 and=29.The factor of safety is given by Equation 9.4.The soil mass is divided into slices l.5 m wide. The weight (W) of each slice is given by W=bh=201.5h=30h kNmThe height h for each slice is set off below the centre of the base and thenormal and tangential components hcosand hsinrespectively are determined graphically，as shown in Fig.9.6.ThenWcos=30h cosW sin=30h sinThe pore water pressure at the centre of the base of each slice is taken to bewzw，where zw is the vertical distance of the centre point below the water table (as shown in figure)This procedure slightly overestimates the pore water pressure which strictly should be) wze，where ze is the vertical distance below the point of intersection of the water table and the equipotential through the centre of the slice baseThe error involved is on the safe sideThe arc length (La) is calculated as 14.35 mmThe results are given inTable 9.1Wcos=3017.50=525kNmW sin=308.45=254kNm(wcos -ul)=525132=393kNm9.4 Analysis of a Plane Translational SlipIt 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 ignoredThe slope is inclined at angle to the horizontal and the depth of the failure plane is zas shown in section in Fig.9.7.The water table is taken to be parallel to the slope at a height of mz (0m1)above the failure planeSteady seepage is assumed to be taking place in a direction parallel to the slopeThe 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 planeIn terms of effective stress，the shear strength of the soil along the failure plane is and the factor of safety isThe expressions for,andare:The following special cases are of interestIf c=0 and m=0 (i.e. the soilbetween the surface and the failure plane is not fully saturated),then (9.11)If c=0 and m=1(i.e. the water table coincides with the surface of the slope)，then： (9.12)It should be noted that when c=0 the factor of safety is independent ofthe depth zIf c is greater than zero，the factor of safety is a function of z, and may exceed provided z is less than a critical valueFor a total stress analysis the shear strength parameters cu andu are used with a zero value of u.Example 9.3A long natural slope in a fissured overconsolidated clay is inclined at 12to the horizontalThe water table is at the surface and seepage is roughly parallel to the slopeA slip has developed on a plane parallel to the surface at a depth of 5 mThe saturated unit weight of the clay is 20 kNm3The peak strength parameters are c=10 kN/m2 and=26；the residual strength parameters are cr=0 andr=18.Determine the factor of safety along the slip plane(a)in terms of the peak strength parameters (b)in terms of the residual strength parametersWith the water table at the surface(m=1)，at any point on the slip plane， Using the peak strength parameters， Then the factor of safety is given by Using the residual strength parameters，the factor of safety can beobtained from Equation 9.12: 9.5 General Methods of AnalysisMorgenstern and Price9.4developed a general analysis in which all boundary and equilibrium conditions are satisfied and in which the failure surface may be any shape，circular，non-circular or compoundThe soil mass above the failure plane is divided into sections by a number of vertical planes and the problem is rendered statically determinate by assuming a relationship between the forces E and X on the vertical boundaries between each sectionThis assumption is of the formX=f(x)E (9.13)where f(x)is an arbitrary function describing the pattern in which the ratio X/E varies across the soil mass andis a scale factorThe value ofis obtained as part of the solution along with the factor of safety FThe values of the forces E and X and the point of application of E can be determined at each vertical boundaryFor any assumed function f(x) it is necessary to examine the solution in detail to ensure that it is physically reasonable (i.e. no shear failure or tension must be implied within the soil mass above the failure surface). The choice of the function f(x) does not appear to influence the computed value of F by more tha