邊坡穩(wěn)定分析與計(jì)算例題

1、邊坡工程計(jì)算例題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、為折線，其斷面和力學(xué)參數(shù)如圖和表所示，擬設(shè)計(jì)抗滑結(jié)構(gòu)物，取安全系數(shù)為 1.05 ，計(jì)算作用在抗滑結(jié)構(gòu)物上的滑坡推力 P3下滑力 T （KN/m）抗滑力R（ KN/m）滑面 傾角傳遞 系數(shù)12 0005 500450.73317 00019 0001712 4002 70017解：余推力 Pi Pi 1 i 1 TiFs Ri，其中 Fs為安全系數(shù) 1.05 則 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則滑坡推力為 3874.3N4. 某巖性邊坡為平面破壞形式，已知滑面 AB 上的 C=20kPa， 30 ，當(dāng)滑面 上巖體滑動(dòng)時(shí)，滑動(dòng)體后部張裂縫 CE的深度為多少米？解：單一滑動(dòng)面滑動(dòng)時(shí)，后部張裂縫深度的理論公式為：代入得：2CZ O tg 45 O22 20ZOtg60 2.77mO 255. 巖質(zhì)邊坡坡角 35，重度25.3kN /m3 ，巖層為順坡，傾角與坡角相同，厚度 t=0.63m ，彈性模量 E=350MP，a 內(nèi)摩擦角30 ，則根據(jù)歐拉定理計(jì)算此巖坡的極限高度為多少米？ 解：根據(jù)歐拉定理，邊坡順向巖層不發(fā)生曲折破壞的極限長度計(jì)算式為1L2EI30.49t co

6、s tg tg2EI取得：I 112t3222Et2L36 cos tg tg代入上述數(shù)值得： L=93m為極限長度，則，巖坡極限高度：H L sin 53 m6已探明某巖石邊坡的滑面為AB,坡頂裂縫 DC深 z 15m ，裂縫內(nèi)水深Zw 10m，坡高 H 45m，坡角 60 ，滑坡傾角 28 ，巖石容重25kN /m3 ，滑面粘結(jié)力 c 80kPa， 內(nèi)摩檫角 26 ，計(jì)算此邊坡的穩(wěn)定 系數(shù) 。解：作用于 BC上的靜水壓力 V 0.5 wgZw2 =0.519.8102=490kNHZ作用于 AB上的靜水壓力 U為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邊坡穩(wěn)定性系數(shù)為(Gcos U Vsin )tg j Cj ABgsin Vcos(331100cos28 3133000 490000sin 28 )tg26 80000 63.99.8sin 28 490000cos28=2.4527. 某一滑坡下臥穩(wěn)定基巖，斷面如圖所示。滑塊各塊重量分別為W1 700kN ，W2 2400kN ，W3 1500kN ，W4 1800kN 。外荷載 P2 500kN ，P1

8、 900kN 分 別作用在第一塊第二 塊上，其作用線通過相應(yīng)塊的重心?；娼?1 40 ，2 20 ， 3 5 ， 4 10 ?；嫔蟽?nèi)摩擦角均為 15 ，粘聚力 c 為 5.0kPa。 滑塊長度 l1 15m， l2 15m ，l3 9m， l4 14m 。試計(jì)算滑坡推力并判斷其穩(wěn) 定性(安全系數(shù) Fs取 1.05 )能否達(dá)到 1.5 。解：(1)計(jì)算各滑塊抗滑力、下滑力和傳遞系數(shù)： 下滑力 Ti (Wi Pi )sin i； 抗滑力 Ri (Wi Pi )cos itg i cili ； 傳遞系數(shù) i cos( i 1 ai ) tg i sin( i 1 ai ) ；將已知值分別代入上式

9、，可得：第一滑塊：T1=(700+900)sin40=16000.64=1024kNR1=(700+900)cos40tg15+515=403kN1=cos(-40) - tg15 sin(-40)=0.94-0.09=0.938第二滑塊：T2=(2400+500)sin20=29000.34=992kNR2=(2400+500)cos20tg15+515=805kN2=cos(40-20) - tg15sin(40-20)=0.94-0.09=0.85第三滑塊：T3=1500sin(-5)=-131kNR3=1500cos(-5) tg15+59=445kN3=cos(20+5)-tg15s

10、in(20+5)= 0.793第四滑塊： T4=1800 sin10=312kNR4=1800cos10tg15+514=545kN(2)計(jì)算滑坡推力滑坡推力 Fi TiK Rii 1Fi 1。當(dāng) Fs=1.05，由上式計(jì)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，安全系數(shù)不能達(dá)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