边坡稳定分析与计算例题

1、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 s, of 2 against sliding alongSolution Equation isFsCr H co2s tant an ,t an ,Given C,r,H , ,We hav

2、e Fs 1.24(2) Equation isr (Fstant an) c o 2sGiven C,r,Fs, ,We have H 1.11m322. A cut is to be made in a soil that has16.5kN /m3,c 29kN /m2, and 15 .The side of the cut slope will make an angle of 45with the horizontal. Whatdepth of the cut slope will have a factor of safety, FSS, of 3?Solution We ar

3、e given 15 and c 29kN /m2 .If FSC 3 , thenFSC andFS should both be equal to 3. We haveFScOrc c 29 9.67kN /m2FSC FSS 3Similarly,FStan tan dOrtan tantan dFSFSstan153tan 1tan315 5.1Substituting the preceding values ofcd and d into equation givessincoscos4 9.67 sin 45cos5.17.1m16.5 1 cos 45 5.13. 某滑坡的滑面

4、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 P3下滑力 T （KN/m）抗滑力R（ KN/m）滑面 倾角传递 系数12 0005 500450.73317 00019 0001712 4002 70017解：余推力 Pi Pi 1 i 1 TiFs Ri，其中 Fs为安全系数 1.05 则 P1 FsT1 R1=1.05*1200-5500=7100NP2 P1 1 FsT2 R2=7100*0.733+1.05*1700-1900=4054.3NP3 P2 2 FsT3 R3=4054.3*1+1.05*2400-270

5、0=3874.3N则滑坡推力为 3874.3N4. 某岩性边坡为平面破坏形式，已知滑面 AB 上的 C=20kPa， 30 ，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：2CZ O tg 45 O22 20ZOtg60 2.77mO 255. 岩质边坡坡角 35，重度25.3kN /m3 ，岩层为顺坡，倾角与坡角相同，厚度 t=0.63m ，弹性模量 E=350MP，a 内摩擦角30 ，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1L2EI30.49t co

6、s tg tg2EI取得：I 112t3222Et2L36 cos tg tg代入上述数值得： L=93m为极限长度，则，岩坡极限高度：H L sin 53 m6已探明某岩石边坡的滑面为AB,坡顶裂缝 DC深 z 15m ，裂缝内水深Zw 10m，坡高 H 45m，坡角 60 ，滑坡倾角 28 ，岩石容重25kN /m3 ，滑面粘结力 c 80kPa， 内摩檫角 26 ，计算此边坡的稳定 系数 。解：作用于 BC上的静水压力 V 0.5 wgZw2 =0.519.8102=490kNHZ作用于 AB上的静水压力 U为U 0.5 wgZw Hw Zw =0.519.8 10 sin40 10si

7、n28=3133kN AB =( H-Z) sin =(45-15 ) sin28 =63.9mG=(H+Z) AB cos0.5 =(45+15) cos28 0.5 25=331kN边坡稳定性系数为(Gcos U Vsin )tg j Cj ABgsin Vcos(331100cos28 3133000 490000sin 28 )tg26 80000 63.99.8sin 28 490000cos28=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W1 700kN ，W2 2400kN ，W3 1500kN ，W4 1800kN 。外荷载 P2 500kN ，P1

8、 900kN 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 1 40 ，2 20 ， 3 5 ， 4 10 。滑面上内摩擦角均为 15 ，粘聚力 c 为 5.0kPa。 滑块长度 l1 15m， l2 15m ，l3 9m， l4 14m 。试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 Ti (Wi Pi )sin i； 抗滑力 Ri (Wi Pi )cos itg i cili ； 传递系数 i cos( i 1 ai ) tg i sin( i 1 ai ) ；将已知值分别代入上式

9、，可得：第一滑块：T1=(700+900)sin40=16000.64=1024kNR1=(700+900)cos40tg15+515=403kN1=cos(-40) - tg15 sin(-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)=0.94-0.09=0.85第三滑块：T3=1500sin(-5)=-131kNR3=1500cos(-5) tg15+59=445kN3=cos(20+5)-tg15s

10、in(20+5)= 0.793第四滑块： T4=1800 sin10=312kNR4=1800cos10tg15+514=545kN(2)计算滑坡推力滑坡推力 Fi TiK Rii 1Fi 1。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 1.5。8. Use Limit Equilibrium Equations to analyse the stability

11、 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:= 30c = 150 kPaslip plane dips 40 due south and daylights above the toe of the slope1) Provide a plot of FS versus slip plane dip (keep all other base case parameters con

12、stant).2) Provide two plots of FS versus slip plane friction angle (= 10 to 40 ) on thesame graph, one with c = 0 and the other with c = 150kPa (keep other base case parameters constant).3) Assume water pressure exists along the slip plane with a triangular pressuredistribution. Provide a plot of FS

13、 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 page). The cables are installed perpendicular to the slope strike.Assu

14、meworst-case water conditions in the tension crack.Provide a plot of FS versuscable plunge; include both upholes and downholes (keep other parameters constant).5) You have just completed a simple sensitivity study. Comment on the findings what did you learn from your plots, what are the controlling

15、parameter(s)?Putsome intelligent words on paper, neatly!Avoid stating the obvious (e.g. steeperslip planes have lower factors of safety) as your main conclusion.6) Discuss how you would do a Monte Carlo simulation to determine the probability of failure. What would be the advantages and disadvantage

16、s 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 againstsliding for this block.If the slope and rock become completely submerged by water ina reservoir, recalculat

17、e the factor of safety.For both cases, assume the shearstrength at the base of the block is governed by a friction angle of 32plus a cohesionof 100 kPa. The width of the block into the page is 3 m and the density of the rock is 2400 kg/m3.40Solution Length=3m ， Height=2m ， Width=3m ， 40 ， 32 ， C=100KPa ，density=2400kg/m3Volume 3 3 2 18m3Weight 2400 18 42300kgGravity42300 9.81000423.36KNC L W cos tanFsW sin100 3 423.36 cos40 tan32423