边坡稳定计算书

路基边坡稳定性分析本设计计算内容为广西梧州绕城高速公路东段k15+400〜k16+800路段中出现的最大填方路段。该路堤边坡高22m，路基宽26m,需要进行边坡稳定性验算。确定本设计计算的基本参数本段路段路堤边坡的土为粘性土，根据《公路路基设计规范》，取土的容重Y=18.5kN/m3，粘聚力C=20kpa,内摩擦角C=24°,填土的内摩擦系数户tan24°=0.445。行车荷载当量高度换算高度为:2x2x5505.5x12.8x18.5=0.8446(m)h0—行车荷载换算高度;L—前后轮最大轴距，按《公路工程技术标准》(JTGB01-2003)规定对于标准车辆荷载为12.8m；Q——辆车的重力(标准车辆荷载为550kN)；N—并列车辆数，双车道N=2，单车道N=1；Y一路基填料的重度(kN/m3)；B一荷载横向分布宽度，表示如下：B=Nb+(N-1)m^d式中：b一后轮轮距，取1.8m；m一相邻两辆车后轮的中心间距，取1.3m；d一轮胎着地宽度，取0.6m。Bishop法求稳定系数k3.1计算步骤：按4.5H法确定滑动圆心辅助线。由表查得61=26°，p2=35°及荷载换算为土柱高度h0=0.8446(m),得G点。由坡脚A向下引竖线，在竖线上截取高度H=h+h0(h为边坡高度，h0为换算土层高)自仔点向右引水平线，在水平线上截取4.5H，得E点。根据两角分别自坡角和左点作直线相交于F点，EF的延长线即为滑动圆心辅助线。连接边坡坡脚A和顶点B,求得AB的斜度i=1/1.5，据此查《路基路面工程》表4-1得61,p2。kk图1（4.5H法确定圆心）（2） 在CAD上绘出五条不同的位置的滑动曲线（3） 将圆弧范围土体分成若干段。（4） 利用CAD功能读取滑动曲线每一分段中点与圆心竖曲线之间的偏角a|（圆心竖曲线左侧为负，右侧为正）以及每分段的面积S/口弧长L『（5） 计算稳定系数：首先假定两个条件：a,忽略土条间的竖向剪切力Xi及七作用；b,对滑动面上的切向力二的大小做了规定。根据土条i的竖向平衡条件可得：吧―Xj+X.+]—T.-N.cosa.=0即N.cosa.=吧—X.+X.+1—T.sina. ⑴若土坡的稳定安全系数为K，则土条i的滑动面上的抗剪强度T也只发挥了fi一部分，毕肖普假设T冥与滑动面上的切向力Ti相平衡，即：TT.=1（N.tan中.+cl）i= fi 1 1ii （2）将（1）代入式（2）得：W-X.+X-clsinaiI I I+1N；= z―小，K—cosa+tan中.sinRiK又已知土坡的稳定安全系数K为:

丫MK=__rMs£(Ntanp.+c")丫MK=__rMsw £W.sina(4)将式（3）i=1代入式（4）中得：(4)将式（3）i=1代入式（4）中得：£（W.-X.+X.+1）tanp.+cl.cosa.tanp.sina.K= cog广 K '_£Wsina. £Wsina.=1 .=1由于上式中Xj及Xi+1是未知的，故求解尚有困难。毕肖普假定土条间竖向剪切力均略去不计，则式（5）可简化为：£(N.tanp+cl)T=1 (5)^£W.tan中i+c.licosai(6)(7)Mai工W.Sina.(6)(7)iiM ；tangsina.其中 a.= .式（6）就是简化毕肖普法计算土坡稳定安全系数的公式。由于式（7）也包含K值，因此式（6）须用迭代法求解，即先假定一竹值，按式（7）求得的值，代入式（6）中求出K值。若值与假定值不符，则用此K值重新计算求得新的K值，如此反复迭代，直至假定K值于求得的K值相近为止。3.2具体计算过程及图表3.2.1以。为圆心过坡脚做一滑动面，R=41.723m。假设、=1.36，计算结果如表2-1所示：

土条编号或。）L(m)矿(kN)国如住侧）密的施侧）alicosaimai-^―(TF^tan皆+g£cos^x)11.094.1867.9060.20130.1638.660.7591.5221.043.58190.74164.49324.9436.250.79153.6330.993.21292.12244.22445.0935.230.82200.9340.922.94378.33301.00496.8435.620.87235.5050.882.75453.07349.19548.0635.040.89266.1060.822.6554.82405.65594.6035.480.92306.4470.752.48550.56375.28512.9036.290.95294.6380.692.38553.15352.10456.5336.710.98288.7890.612.3550.75315.50384.9337.701.01280.79100.562.23540.94287.34339.1437.791.02272.76110.482.18525.96242.88273.8238.671.04262.72120.422.14517.82211.14231.2439.081.05257.53130.362.1495.25174.46186.4139.311.05247.05140.32.07392.02115.85121.2639.551.05203.41150.242.04472.31112.27115.5839.631.05238.11160.172.03436.7973.9074.9840.011.04225.17170.132.01397.3851.5151.9539.861.03209.57180.0072354.462.482.4840.001.00197.29190.022307.846.166.1639.991.01175.8720-0.042281.39-11.25-11.2639.970.99167.5121-0.082.01252.53-20.18-20.2540.070.97157.0522-0.132.02196.47-25.47-25.6940.060.95134.3223-0.192.03135.42-25.58-26.0439.870.92108.8124-0.233.1880.48-18.35-18.8461.930.90108.71合计3744.795084.2计算可得£W.tan中.+c.l.cosa.M.ai=5084.20/3744.79=1.357£w.sina.=5084.20/3744.79=1.357iii=1计算出的匕与假设的K1相差很小，即K1=1.363.2.2以0为圆心过坡脚做一滑动面，R=46.968m。假设K2=1.32，计算结果如表2-2所示：土汆编号m（。）阻倒）JF顿网cicOSfiTtmat--^―(1朽tui0—C-iCOS-Ct)ms.10.93土EE54.3945.1781.0C30.SSD.更7S.EC2u.94?,23K.我125.94213.543S.1ju.ME124.710.89件F186.31切.01系■■31.7F0.72.£程.仕214.77321.773".氾0.菟1B0.43u.112.EE刘7.口214.17293.3238.?5LB3.660."手彳1「210.43279,89-MK0.68324.49204.D42&2.40M|〕，变Hl.心u.612.兖契土E1S5.57225.4C3S.691.?!L3j.关0,552.22322.09163.35197.4739.)51.无177.2ci).》工心336.331K,上183.74理,M1,MLSI.68二甘一_3u.452.IE342.148.95_65,4139.261.?5*U0.4N.H323.CE125.93136.723仇421.。匚174.210.H2.1130D.G31M,上If%33,足1.K164.4915u.32.032TJ.17£_.?234.3138.791.巧152.420.252.CC244.02GO.狞6S.31%.921.X141.12i).JJ.C<1匕1442.6043.46都，yj1.Jb129.32:713i'i.E2.r?202.39?4:Vi.河演.而1.初逐一110.112.Cl133.ir20.11212C39.)61.3C117.El0.){139.S69.HJ.yi都.担1.义C:2121n.0293.43■..可1.R739.99:.F41一「J-0.)22.C24：.El0.珀-),94时,79|).买73.14q：蹈,LJ计算可得£W.tan中i+cl.cosaiv MK= ai ''口’. =3079.18/2336.18=1.318乙W.sina.iii=1计算出的KH假设的K2相差很小，即K2=1.323.2.3以。为圆心过坡脚做一滑动面，R=52.645m。

土条编号阪°)/i(m)时kN)崎咿）[f加讷）Cilicosaimcti--^伽tan俱+裁cosai)10.933.3148.4738.8564.9939.580.8572.3120.883.07109.8984.70132.9339.120.88100.5630.832.89145.60107.44159.2039.010.90114.9640.782.74165.95116.71164.1638.960.93121.5250.732.62182.23121.52163.0839.050.95126.2960.682.52192.40120.98155.5939.190.97128.4270.632.43199.80117.71145.6839.270.99129.4630.592.36223.85124.54149.8839.221.00138.4490.552.3240.32125.61147.3439.221.01144.13100.52.25230.88110.69126.1339.491.03138.66110.452.2217.3894.55105.0039.621.03131.76120.412.16200.1779.7987.0039.621.04123.71130.362.13179.2763.1567.4839.871.04114.52140.322.1162.0650.9853.7039.871.05107.01150.282.07155.4042.9544.6939.791.05104.10160.242.05146.3434.7835.8139.821.04100.44170.192.04112.6721.2821.6740.071.0486.70180.162.0276.2212.1412.3039.881.0471.21190.122.7739.414.724.7555.001.0370.44合计1473.092124.65计算可得£W.tan中.+c.l.cosa.M.ai=2124.65/1473.09=1.442=1.44£w.sina.=2124.65/1473.09=1.442=1.44iii=1计算出的K3与假设的K3相差很小，即K33.2.4以。为圆心过坡脚做一滑动面，R=45.935m。

土条编号或°)li(m)照Ml施例CiliCOSatmai—+曷cos4X)113.6656.6147.6488.1639.550.8378.3720.963.29161.51132.30230.6937.740.85128.6830.913.02251.60198.64323.6537.070.88168.9840.862.82330.41250.40383.7936.800.91202.0450.82.67376.66270.20387.8237.200.94217.8060.752.55390.54266.20363.8237.320.96219.1670.682.45396.27249.17320.4538.100.99216.3580.632.36396.27233.46288.9338.141.01212.7490.572.29390.72210.85250.4438.561.03207.21100.512.23380.55185.77212.8638.921.04200.53110.462.18378.14167.87187.3539.071.05198.06120.42.14394.98153.81166.9939.421.05204.29130.362.11373.33131.51140.5239.491.06194.80140.32.08346.51102.40107.1939.741.06183.70150.252.05316.1778.2280.7339.731.05171.35160.22.03281.7655.9857.1139.791.05157.67170.152.02245.3136.6637.0739.951.04143.44180.12.01220.8922.0522.1640.001.03134.41190.052208.8710.4410.4539.951.02130.84200.012163.171.631.6340.001.00112.2321-0.042112.85-4.51-4.5239.970.9991.5022-0.083.1267.53-5.40-5.4162.200.9795.14合计2795.303669.29计算可彳K=一计算出1得£七i=1的匕与假tan中i£W.ii=1设的K相+cLckf.aisina.i1差很小)osa.i,即K4==3669.21.31,9/2795.30=1.3123.2.5以。为圆心过坡脚做一滑动面，R=43.022m。土条编号li(m)朋血块侧）您tan邮）CiliCOSatmcti-gtrn^+&oosix)11.053.9563.0954.72109.9839.310.7787.1420.013.45178.341.781.7869.001.00147.9030.953.13275.10223.77384.6936.410.84189.0340.912.89358.35282.91460.9635.470.86225.4450.842.71430.87320.84480.6936.180.90252.0760.782.57472.86332.55467.7836.540.93264.2970.722.46482.67318.26423.3336.990.96261.8880.662.37483.41296.38375.1737.450.98256.4490.62.29478.41270.13327.3037.801.00249.49100.532.23468.05236.61274.2438.481.02241.10110.472.18453.07205.19230.1438.871.04232.24120.412.14455.66181.63198.0439.251.04231.86130.362.1456.58160.84171.8639.311.05231.41140.32.07431.42127.49133.4539.551.05220.66150.242.05400.1695.1297.9239.821.05208.13160.192.03364.8268.9070.1639.871.04194.06170.132.01326.1642.2842.6439.861.03179.13180.082.01283.7922.6822.7540.071.02162.74190.032260.117.807.8139.981.01154.3320-0.022235.51-4.71-4.7139.990.99145.7521-0.072190.37-13.31-13.3539.900.98127.7722-0.122.01129.32-15.48-15.5939.910.95102.0723-0.163.2780.66-12.85-13.0264.560.94107.26合计3203.544472.19计算可得£W.tan中i+cl.cosaiMK= a Yu,. =4472.19/3203.54=1.396乙W.sina.iii=1计算出的％与假设的K相差很小，即%=1.403.3.2Ki=1计算出的％与假设的K相差很小，即%=1.40min因为%=1.36，K2=1.32，K3=1.44，K4=1.31，K5=1.40,所以、「=K4=1.31>1.30，满足要求挡土墙设计1.设计资料1.1墙身构造本设计路段在K15+400~K15+560段左侧地面横坡较长。为了收缩边坡，减少填方工程量，保证边坡的稳定性，避免因过高而造成边坡的可能滑动。特设置路堤府斜式挡土墙120m，沿墙长每10m设置伸缩缝，缝宽2cm，缝内沿墙内、外、顶三边添塞沥青板。采用最不利荷载，即挡墙最高的断面进行设计计算验算，取K15+540横断面左侧挡土墙进行分析。墙身拟采用7.5号浆砌片石结构，墙高10m，墙顶填土高a=12.62m宽b=20.94（m），顶宽1.3m，面坡背坡倾斜坡度分别为1：0.25,1：0.2,基底倾斜，倾斜角。。二11。为增加抗滑性能在墙底采用0.7X0，7（m）墙趾坡度与面坡一致。（如下 图 ）1'J[■挡土墙示意图（单位：m）1.2地质情况填方部分，假设都为粘土，内摩擦角二24。，土的容重Y=18.5KN/M3，粘聚力C=20Kpa，等效内摩擦角二27.96。，墙背与填土间的摩擦角0=11.31。，。墙底与地基摩擦力系数f=3.5，地基承载应力标准值［6。:|二350Kpa。1.3墙身材料墙体采用浆砌片石结构，7.5号砂浆，25号片石，墙体容重「24KN/m3。按规范：容许压应力为［。a］=720Kpa，容许剪应力［t］=147Kpa。2车辆荷载根据《路基设计规范》(JTG2004)，车辆荷载为计算的方便，可简化换算为路基填土的均布土层，并采用全断面布载。7NQ2x550换算土层厚:h= = =0.8446(m)换算土层厚:0BL5.5x12.8x18.53土压力计算对于墙前土的被动土压力，在挡墙基础一般埋深的情况，考虑各种自然力和人畜活动的影响，偏于安全。一般不计被动土压力，只要考虑主动土压力，用库伦理论进行计算，3.1破裂面计算假设破裂面交于荷载中部，则A0=1/2(a+H+2ho)(a+H)=1/2(12.62+10+2X0.84)X(12.62+12)=598.27B0=1/2ab+(b+d) ho-1/2H(H+2a+2h0)tana=1/2X12.62X20.94+(20.94+0)X0.84-1/2X10X(10+2X12.62+2X0.84)X0.19=115.235I Bqtan9--tanp土J(cot©+ +tany)…c: ~~115.23*口…ctan9=tan50.27。(cot27.96。+tan50.27。)x(59827+tan50.27。)=0.96其中：甲*+6+a=27.96°+11.31°+11°=50.27°则。=arctg0=44.94°3.2破裂面验算堤顶破裂面距墙踵距离(H+a)tg9=(10+12.62)X0.96=21.27m荷载内边缘距墙踵距离 b+d-Htga=20.94+0-10X0.20=19.64m荷载外边缘距墙踵b+d+l0-Htga=20.94+0+26-10X0.20=44.94m由以上数据可得：19.64V21.27V44.94,破裂面交于路基面外边坡，故假设与实际相符合。3.3计算主动土压力3.3.1系数的计算cos。+甲)sin(0+w)3.3.2土压力的计算h=d/(tan0+tana)=0.7/(tan44.940+tanllo)=0.587(m)h=(b-atan0)/(tan0+tana)=(22.94-12.94xtan44.94。)/(tan44.94。+tanll。)二8.41(m)h3=H-h1-h2=10-0.59-8.41=1 (m)叱,2a h 2hh 12x12.94,1 8.41 2x0.59x8.41K1=1+H(1-由+才 x(1-和)+ 102二2.603.3.3主动土压力Ea的计算E=2yh2KK1=0.5X18.5X102X0.35X2.60=841.75kNEx=Ecos(a+5)=841.75xcos(11°+11.31°)=778.74kNEy=Esin(x+5)=841.75xsin(11°+11.31°)=319.54kN3.4土压力作用点位置rHa(H—h)2+hh(3h—2H)气=T+ 23H很'—11012.94x(10—8.41)2+0.59x1x(3x1-2x10)=—+ 3 3x102x2.60=3.37Zx=B]—Zytga=3.46-3.37Xtan11°=2.80其中：B=3.4/cos11°=3.4614设计验算墙身面积：S=40.39(m2)墙重：G=747.23kN力臂Zg的计算：Zg=4.10(m)4.1墙身稳定性验算4.1.1抗滑稳定性验算抗滑稳定性验算在计算中我们不考虑墙前被动土压力的影响，所以E，=0\1.1G+丫Q(E+Etan以)]日