边坡稳定分析与计算例题.doc

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1、【精品文档】如有侵权，请联系网站删除，仅供学习与交流边坡稳定分析与计算例题.精品文档.边坡工程计算例题1. Consider the infinite slope shown in figure.(1) Determine the factor of safety against sliding along the soil-rock interface given H = 2.4m.(2) What height, H, will give a factor of safety, F, of 2 against sliding along the soil-rock interface?HR

2、ockSoilSolution Equation is Given We have (2) Equation isGiven We have 2. A cut is to be made in a soil that has, and. The side of the cut slope will make an angle of 45with the horizontal. What depth of the cut slope will have a factor of safety, of 3?Solution We are given and .If, then andshould b

3、oth be equal to 3. We haveOrSimilarly,OrSubstituting the preceding values ofandinto equation gives3.某滑坡的滑面为折线，其断面和力学参数如图和表所示，拟设计抗滑结构物，取安全系数为1.05，计算作用在抗滑结构物上的滑坡推力P。下滑力（KN/m）抗滑力（KN/m）滑面倾角传递系数12 0005 500450.73317 00019 0001712 4002 70017P3解：余推力，其中为安全系数1.05 则 =1.05*1200-5500=7100N =7100*0.733+1.05*1700-

4、1900 =4054.3N =4054.3*1+1.05*2400-2700 =3874.3N则滑坡推力为3874.3N 4. 某岩性边坡为平面破坏形式，已知滑面AB上的C=20kPa， ，当滑面上岩体滑动时，滑动体后部张裂缝CE的深度为多少米？解：单一滑动面滑动时，后部张裂缝深度的理论公式为：代入得： 5. 岩质边坡坡角35，重度 ，岩层为顺坡，倾角与坡角相同，厚度t=0.63m，弹性模量E=350MPa，内摩擦角 ，则根据欧拉定理计算此岩坡的极限高度为多少米？解：根据欧拉定理，边坡顺向岩层不发生曲折破坏的极限长度计算式为取得： 代入上述数值得：L=93m为极限长度， 则，岩坡极限高度： 6

5、已探明某岩石边坡的滑面为AB,坡顶裂缝DC深，裂缝内水深，坡高，坡角，滑坡倾角，岩石容重，滑面粘结力， 内摩檫角，计算此边坡的稳定系数。解：作用于BC上的静水压力=0.519.8102=490kN作用于AB上的静水压力U为=0.519.810=3133kN=（H-Z）sin=（45-15）sin28=63.9mG=(H+Z) cos0.5=（45+15）cos280.525=331kN边坡稳定性系数为 =2.4527. 某一滑坡下卧稳定基岩，断面如图所示。滑块各块重量分别为，。外荷载，分别作用在第一块第二 块上，其作用线通过相应块的重心。滑面角，。滑面上内摩擦角均为，粘聚力c为。滑块长度，。试

6、计算滑坡推力并判断其稳定性（安全系数Fs取1.05）能否达到1.5。解：（1）计算各滑块抗滑力、下滑力和传递系数：下滑力 ；抗滑力 ；传递系数 ；将已知值分别代入上式，可得：第一滑块：T1=(700+900)sin40=16000.64=1024kNR1=(700+900)cos40tg15+515=403kN1=cos(-40) - tg15sin(-40)=0.94-0.09=0.938第二滑块：T2=(2400+500)sin20=29000.34=992kNR2=(2400+500)cos20tg15+515=805kN2=cos(40-20) - tg15sin(40-20)=0.9

7、4-0.09=0.85第三滑块：T3=1500sin（-5）=-131kNR3=1500cos（-5）tg15+59=445kN3=cos(20+5)-tg15sin(20+5)= 0.793第四滑块：T4=1800sin10=312kNR4=1800cos10tg15+514=545kN（2）计算滑坡推力滑坡推力 。当Fs=1.05，由上式计算可得：F1=10241.05-403=672kNF2=9921.05-805+0.938672=867kNF3=-1311.05-445+0.85867=154kNF4=3121.05-545+0.793154= -95kNF40，安全系数不能达到1.

8、5。8. Use Limit Equilibrium Equations to analyse the stability of a slope subject to a planar instability. The design slope (in rock 2.7 g/cc) is 30 m high and dips due south at 75.Base case: f = 30 c = 150 kPa slip plane dips 40 due south and daylights above the toe of the slope1) Provide a plot of

9、FS versus slip plane dip (keep all other base case parameters constant).2) Provide two plots of FS versus slip plane friction angle (f = 10 to 40) on the same graph, one with c = 0 and the other with c = 150kPa (keep other base case parameters constant). 3) Assume water pressure exists along the sli

10、p plane with a triangular pressure distribution. Provide a plot of FS versus peak hydraulic head for pressure head range from 0 to 10 m.4) Assume you can add a single row of high capacity cable bolts at mid-height of the slope. Each cable has a working load of 2000 kN and is spaced 2 m apart (into p

11、age). The cables are installed perpendicular to the slope strike. Assume worst-case water conditions in the tension crack. Provide a plot of FS versus cable plunge; include both upholes and downholes (keep other parameters constant).5) You have just completed a simple sensitivity study. Comment on t

12、he findings what did you learn from your plots, what are the controlling parameter(s)? Put some intelligent words on paper, neatly! Avoid stating the obvious (e.g. steeper slip planes have lower factors of safety) as your main conclusion.6) Discuss how you would do a Monte Carlo simulation to determ

13、ine the probability of failure. What would be the advantages and disadvantages of the analysis you performed using your excel spreadsheet compared to a analysis using a Monte Carlo method?9. A block of rock lies on a slope as shown. Calculate the factor of safety against sliding for this block. If t

14、he slope and rock become completely submerged by water in a reservoir, recalculate the factor of safety. For both cases, assume the shear strength at the base of the block is governed by a friction angle of 32 plus a cohesion of 100 kPa. The width of the block into the page is 3 m and the density of the rock is 2400 kg/m3. Solution Length=3m，Height=2m，Width=3m， ，C=100KPa，density=2400kg/m3When the block is completely submerged by water：其中：