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-rockinterface given H = 2.4m.H, will give a factor of safety, F, of 2 against sliding along (2) What height, s the soil-rock interface?．??25?1kk1HSoilRockSolution⑴Equation is?na C t?F?, s2???natna?r?H?cost?? Given ,,,r,HC We have 24?F1.s(2) Equation isC, ?H?nat2??n??cotsa?r?(F)s?nta??,,F,C,r Given s We have m11?1.H32??. 2. A cut is to be made in a soil that has,, and mkN/16.5?m?29kN/c?15?The side of the cut slope will make an angle of 45°with the horizontal. WhatFS, of 3?depth of the cut slope will have a factor of safety,S2?.If, andthenSolution We are given 3FS?mkN/c?29??15C FSFS andshould both be equal to 3. We have?C c?FS c c d Orcc292mkN/??c??9.67d FSFS3SC Similarly,?tan?FS??tan d??tan15tantan???tan?d3FSFS?s Ortan15???1?tan5.1?????d3???into equation givesand Substituting the preceding valuesof c dd??????cos4c sin45cos5.19.67sin?4dd m?H?7.1????????????5.1??1cos1?16.5cos45?????d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，3.。

，计算作用在抗滑结构物上的滑坡推力P取安全系数为1.053抗滑滑传下滑倾系KN/KN/0.7312 0005 500453?1 17 000 ②19 000 17? 2 400③2 70017?R??TP?PFF解：余推力，其中1.05为安全系数is?1ii?1iis RT?P?F则1s11=1.05*1200-5500=7100N?RT??FPP?221s12 =7100*0.733+1.05*1700-1900=4054.3N?RTP??FP?333s22 =4054.3*1+1.05*2400-2700=3874.3N3874.3N 则滑坡推力为?30?某岩性边坡为平面破坏形式，已知滑面AB上的C=20kPa，，当滑面4.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：?C2?? Z?tg45???O?2??2?20 ?2.tg6077mZ?O代入得：255. 岩质边坡坡角35°，重度，岩层为顺坡，倾角与坡角相同，3?m/.3kN?25厚度t=0.63m，弹性模量E=350MPa，内摩擦角，则根据欧拉定理计算 ?30?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??12???EI3L???????tg cos?tg0.49t??13tI?取12：得22?Et?L3?? ????tg cos?tg6代入上述数值得：L=93m为极限长度，???m53??HL sin?则，岩坡极限高度：6．已探明某岩石边坡的滑面为AB,坡顶裂缝DC深，裂缝内水深m?15z??，岩石角容角重，滑坡倾，，坡高坡m10Z???28?H?45m60?w3??，计算此边坡的稳定内摩檫角，滑面粘结力，m/?25kN?26?kPa?80c系数。

22?gZ?0.5V=490kN109.8×解：①作用于BC上的静水压力=0.5×1×wwH?Z?ww gZ?0.5U为上的静水压力U1××9.8×10×②作用于AB=0.5ww?sin40?10=3133kN ?sin28______③=（H-Z）÷sinβ=（45-15）÷sin28°=63.9m AB______?=（45+15）×cos28°×0.5×0.5G=(H+Z)××cosβ××25=331kN AB________AB???CtgV?U?sin?)(G cos jj?边坡稳定性系数为???cos g sin V?63.9?80000tg28?）26??490000sin3133000?(331100cos28??=?490000cos?9.8sin28?28=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，kN?700W1，，。

外荷载，分kN?1500WkN?kN900??W2400kN1800kNPP?500W31242?，块上，其作用线通过相应块的重心。

滑面角别作用在第一块﹑第二??401???。

滑面上内摩擦角均为，粘聚力c为。

，，??5???10?20?kPa015?5.342滑块长度，，，。

试计算滑坡推力并判断其稳m?9lm14ll?15m?l?15m3421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：?)sin P?T(W?；下滑力iiii???c)cos ltgR?(W?P；抗滑力iiiiiii?????cos a sin（)??（?a)tg传递系数；iiiii?1?i1将已知值分别代入上式，可得：第一滑块：T=(700+900)×sin40°=1600×0.64=1024kN 1 R=(700+900)×cos40°tg15°+5×15=403kN1ψ=cos(-40°) - tg15°sin(-40°)=0.94-0.09=0.93810.34=992kN=2900sin20=(2400+500)T第二滑块：×°×2R15=805kN×+5°tg15°×cos20×=(2400+500)2ψ=cos(40°-20°) - tg15°sin(40°-20°)=0.94-0.09=0.852第三滑块：T=1500×sin（-5°）=-131kN 3 R=1500×cos（-5°）×tg15°+5×9=445kN3ψ=cos(20°+5°)-tg15°sin(20°+5°)= 0.793 3第四滑块：T=1800×sin10°=312kN 4 R=1800×cos10°×tg15°+5×14=545kN4（2）计算滑坡推力 ?F??RF?TK。

滑坡推力1iiii?i1? Fs=1.05，由上式计算可得：当F=1024×1.05-403=672kN1F=992×1.05-805+0.938×672=867kN2F=-131×1.05-445+0.85×867=154kN 3 F=312×1.05-545+0.793×154= -95kN4F<0，边坡稳定。

4时，计算可得：当Fs=1.5 F=1024×1.5 -403=1133kN1F=992×1.5-805+0.938×1133=1746kN 2F=-131×1.5-445+0.85×1746=1236kN 3 F=312×1.5-545+0.973×1236=1126kN4F>0，安全系数不能达到1.5。

48. Use Limit Equilibrium Equations to analyse the stability of a slope subject to aplanar instability. The design slope (in rock 2.7 g/cc) is 30 m high and dips due southat 75?.Base case:?? = 30?? c = 150 kPa? slip 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 constant).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 caseparameters constant).3) Assume water pressure exists along the slip plane –with a triangular pressure distribution. Provide a plot of FS versus peak hydraulic head for pressure headrange from 0 to 10 m.4) Assume you can add a single row of high capacity cable bolts at mid-height of theslope. Each cable has a working load of 2000 kN and is spaced 2 m apart (intopage). The cables are installed perpendicular to the slope strike. Assumeworst-case water conditions in the tension crack. Provide a plot of FS versuscable plunge; include both upholes and downholes (keep other parametersconstant).5) You have just completed a simple sensitivity study. Comment on the findings –what did you learn from your plots, what are the controlling 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 probabilityof failure. What would be the advantages and disadvantages of the analysisyou performed using your excel spreadsheet compared to a analysis using aMonte Carlo method?30mCalculate the factor of safety against A block of rock lies on a slope as shown. 9.If the slope and rock become completely submerged by water in sliding for this block.For both cases, assume the sheara reservoir, recalculate the factor of safety.plus a cohesion ?strength at the base of the block is governed by a friction angle of 32The width of the block into the page is 3 m and the density of the rock is of 100 kPa.2400 kg/m3.3 m2 4????，32?40?，，，C=100KPaSolution Length=3m，Height=2m，Width=3m3density=2400kg/m3m?18?V olume?33?2kg4230018?Weight?2400?8.?942300KN36?423.?Gravity1000??tan?L?W?cos?C?Fs?sin?W??tan32.36?cos40?100?3?423??423.36?sin4085.?1：When the block is completely submerged by water112??Z???L??cos?（Z40?）UV?；；22wwww2?m，L?3KN?2?9.81/Zm，m其中：ww1KN.51347?11?9.81?2.V? 21KN43?29.?9.81??U2?32?????tan?V?C?L?W?cossin?U?Fs??cos?sin V?W???tan3240??sin1129cos?40??.43?.51?36??1003423.? ?cos?40511140?.42336sin??.17.1?。