土力学第九章习题.doc

习题9-1（1）由题意可得=0，所以，将c=20kPa，W=346kN,d=5m代入公式得 （2）平行四边形的重心在图形的形心处，所以当移去土体后剩下土体的重心作用点距转动中心线的距离为计算习题9-3安全系数由下式计算：计算过程为：将上式右端各项（除含Fs项m）求出，然后假设Fs值（一般初值可取1.0），代入公式右端，看求出的Fs是否与假设的相等，如果不相等，再以计算出的Fs值代入公式右端，仪此方法迭代，直到Fs的初值与计算值相等为止。计算过程见下表：土条号（1）（2）（3）（4）（5）（6）（7）（8）（9）（10）（11）b(m)()W(kN)sincosWsinu(kPa)ub(3)-(8)tgcb(9)+(10)13.457208.170.83870.5446174.586121.5873.37277.82576.66484.48972435452.240.57360.8192259.394221.5886.32211.26407.84219.10403416.5440.470.28400.9588125.100215.762.8218.04797.84225.887942.82.5212.190.04360.99909.25563.9210.976116.17105.488121.659053.2-10153.82-0.17360.9848-26.71060088.80806.27295.080064-25.5109.87-0.43050.9026-47.30030063.43357.8471.273572-4017.66-0.64280.7660-11.35160010.19603.9214.1160482.9737设Fs1设Fs1.8设Fs1.82（12）（13）（12）（13）（12）（13）m(11)/(12)m(11)/(12)m(11)/(12)1.0288 82.1208 0.8136 103.8412 0.8107 104.2199 1.1503 190.4745 1.0031 218.4211 1.0011 218.8622 1.1228 201.1834 1.0499 215.1482 1.0489 215.3535 1.0242 118.7807 1.0130 120.0930 1.0129 120.1113 0.8846 107.4895 0.9291 102.3345 0.9297 102.2671 0.6540 108.9759 0.7645 93.2290 0.7660 93.0443 0.3949 35.7430 0.5599 25.2130 0.5621 25.1114 844.7677878.28003878.96966Fs=(13)/(6)=1.7491 Fs=1.8184844Fs=1.8199 习题9-4安全系数由下式计算：计算过程见下表：土条号(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)bh1(m)h2(m)()sincosW1(kN)W2(kN)W(kN)WsinWcos885.715.0853.50.80390.5948685.220.32705.52567.137419.659387116.478.245.70.71570.69841423.41795.93219.2822304.022248.39586117.9712.14380.61570.78801753.42658.84412.18142716.413476.84645119.4813.8310.51500.85722085.63022.35107.9382630.784378.357541110.9813.6924.50.41470.91002415.62998.25413.84692245.094926.39131112.4911.5618.40.31560.94892747.82531.85279.55561666.495009.643721115.19.2412.40.21470.976733222023.75345.65241147.95220.9505111175.186.60.11490.993437401134.54874.4718560.2584842.167501118.912.7300.00001.00004160.2597.94758.097304758.0973-11114.620-4.8-0.08370.99653216.403216.4-269.143205.1196-21110.460-10.2-0.17710.98422301.202301.2-407.512264.8309-3104.540-15.8-0.27230.962237.5037.5-10.21136.08317513151.2土条号(12)(13)(14)=(12)(13)(15)(16)(17) =(15)(16)(18)=(14)(15)(19)(20)(21)(11)-(18)(20)(22)(17)+(21)Bh(kPa)u (kPa)lc (kPa)clul()tan(Wcos-ul)tancl+(Wcos-ul)tan80.25215.342853.835713.449449.5665.7445724.0570280.5317-161.8511503.8934270.25292.66273.165515.749949.5779.62211152.3523280.5317582.776631362.398760.25401.1074100.27713.959249.5690.98041399.7846280.53171104.39331795.373750.25464.358116.0912.833049.5635.23191489.7728280.53171535.88772171.119640.25492.1679123.04212.088449.5598.37711487.3839280.53171828.55252426.929630.25479.9596119.9911.592749.5573.83681391.0025280.53171924.06562497.902420.25485.9684121.49211.262749.5557.50531368.3332280.53172048.47292605.978310.25443.1338110.78311.073449.5548.13261226.7479280.53171922.35272470.485300.25432.5543108.13911.000024.75272.25001189.524331.50.61282186.82432459.0743-10292.4011.0387000350.70022244.24892244.2489-20209.2011.1766000350.70021585.85171585.8517-3090.8010.3927000350.700225.2657125.2657122148.522习题8-2由，代入P176公式（8-21），得（或查表可得）沿A的铅直线墙高为习题8-3由查表8-2可得：Ka=0.343则地下水位处墙底处墙底处水压力为习题8-55mq=20kPa37.1kPa9.806kPa由，查P177表8-1，得水位处的土压力为：墙角处的土压力为：墙角处的水压力为：习题8-7kPa0.7936mkPakPa习题8-8荷载q在墙背上产生侧压力的起点距墙顶的距离为由于荷载较宽，其右端按角度向墙背投