1、外文文獻翻譯一、原文Stability 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 rotational slips the shape of the failure surface in section may be a circular arc or a non-circular cu

2、rveIn general，circular slips are associated with homogeneous soil conditions and non-circular slips with non-homogeneouconditionsTranslational and compound slips occur where the form of the failure surface is influenced by the presence of an adjacent stratum of significantly different strength . Tra

3、nslational 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 consis

4、ting 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

5、 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 th

6、ree-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 underfund rained conditions, i.e. For the condition immediately after constructionOnly moment equilibrium is considered in the analysisIn section, th

7、e 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 mob

8、ilized 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

9、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 stabi

10、lity 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 depen

11、ds on the slope angle and 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

12、 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 centroi

13、d 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 e

14、stimated by using Equation 9.2.，=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 divid

15、ed 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 ava

16、ilable 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

17、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 )，wh

18、ere 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

19、 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 weigh

20、t 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 depe

21、nd 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 eff

22、ective stress (Equation 9.3) is given by (9.4)The components WCOS and Wsin can 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 underest

23、imates 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 appea

24、r 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

25、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

26、: (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 sli

27、ces is particularly suitable for solution by computerMore complex slope geometry and different soil strata can be introducedIn 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 v

28、alue(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

29、 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 h

30、omogeneous 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 coeffi

31、cients，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 th

32、e 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 centr

33、e of the base and thenormal and tangential components hcos and hsin resWcos=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 ov

34、erestimates 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 resu

35、lts 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 c

36、an 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 planeSt

37、eady 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

38、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

39、)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

40、 u.Example 9.3A long natural slope in a fissured over consolidated 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

41、 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 poin

42、t 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

43、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

44、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 th

45、e 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 impli

46、ed 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 than about 5% and f(x)=l is a common assumption The analysis involves a complex process of iteration for the values of and F，described by Morgenstern and P

47、rice9.5，and the use of a computer is essential. Bell 9.1 proposed a method of analysis in which all the conditions of equilibrium are satisfied and the assumed failure surface may be of any shapeThe soil mass is divided into a number of vertical slices and statical determinacy is obtained by means o

48、f an assumed distribution of normal stress along the failure surface Sarma 9.6 developed a method，based on the method of slices，in which the critical earthquake acceleration required to produce a condition of limiting equilibrium is determinedAn assumed distribution of vertical interslice forces is

49、used in the analysisAgain，all the conditions of equilibrium are satisfied and the assumed failure surface may be of any shapeThe static factor of safety is the factor by which the shear strength of the soil must be reduced such that the critical acceleration is zero The use of a computer is also ess

50、ential for the Bell and Sarma methods and all solutions must be checked to ensure that they are physically acceptableReferences9.1Bell,J,M.(1968):General Slope Stability Analysis, Journal ASCE,V01.94,No.SM6.9.2Bishop,A.W.and Morgenstern,N.R.(1960)：Stability Coefficients for Earth Slopes Geotechnique

51、，9.3Gibson,R.E.and Morgenstern,N.R.(1962):A Note on the Stability of Cuttings in Normally Consolidated Clays9.4Morgenstern,N.R.and Price,V.E.(1965):The Analysis of the Stability of General Slip Surfaces,Geotechnique，Vo1.1 5,No.1.9.5Morgenstern,N.R.and Price,V.E.(1967): A Numerical Method for Solving

52、 the Equations of Stability of General Slip SurfacesComputer Journal，Voi.9,P.3889.6Sarma,S.K. (1973):Stability Analysis of Embankments and Slopes,Geotechnique，Vo1.23，No.2.9.7Skempton,A.W.(1970):First-Time Slides in Overconsolidated Clays(Technical Note)，9.8Spencer，E(1 967)：A Method of Analysis of th

53、e Stability of Embankments Assuming Parallel Inter-Slice Forces，Geotechnique，9.9Taylor,D.W.(1937):Stability of Earth Slopes,Journal of the Boston Society of Civil Engineers，Vo1.24,No.3二、譯文Stability of Slopes邊坡穩定9.1 引言重力和滲透力易引起天然邊坡、開挖形成的邊坡、堤防邊坡和土壩的不穩定性。最重要的邊坡破壞的類型如圖9.1所示。在旋滑中，破壞面局部的形狀可能是圓弧或非圓弧線。總的來說，

54、勻質土為圓弧滑動破壞，而非勻質土為非圓弧滑動破壞。平面滑動和復合滑動發生在那些強度差異明顯的相鄰地層的交界面處。平面滑動易發生在相鄰地層處于邊坡破壞面以下相對較淺深度的地方：破壞面多為平面，且與邊坡大致平行。復合滑動通常發生在相鄰地層處于深處的地段，破壞面由圓弧面和平面組成。 在實踐中極限平衡法被用于邊坡穩定分析當中。它假定破壞面是發生在沿著一個假想或破壞面的點上的。土的有效抗剪強度與保持極限平衡狀態所要求的抗剪強度相比，就可以得到沿著破壞面上的平均平安系數。問題以二維考慮，即假想為平面應變的情況。二維分析為三維碟形面解答提供了保守的結果。9.2 u =0情況的分析在這種分析方法中，應用總應力法，適用于完全飽和粘土在不條件排水下的情況。如建造完工的瞬間情況。這種分析中只考慮力矩平衡。此間，假定潛在破壞面為圓弧面。圖9.2展示了一個試驗性破壞面圓心O，半徑r，長度La。潛在的不穩定性取決于破壞面以上土體的總重量(單位長度上的重量W。為了