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《土力学》作业答案第一章1—1根据下列颗粒分析试验结果,作出级配曲线,算出Cu及Cv值,并判断其级配情况是否良好。粒径/mm粒组含量/%粒径/mm粒组含量/%20~1010.25~0.101910~530.10~0.0585~250.05~0.0142~170.01~0.00531~0.5200.005~0.00220.5~0.2528解:粒径(mm)20105210.50.250.10.050.010.005小于该粒径的含量百分数(%)10099969184643617952级配曲线见附图。1001010.10.011E-30102030405060708090100小于某直径之土重百分数%土粒直径以毫米计习题1-1颗粒大小级配曲线由级配曲线查得:d60=0.45,d10=0.055,d30=0.2;18.8055.045.01060ddCu62.1055.045.02.026010230dddCcCu5,1Cc3;故,为级配良好的土。1—2有A、B、C三种土,各取500g,颗粒大小分析结果如表:粒径/mm质量/g粒径/mm质量/gABCABC10~5001450.1~0.05112.517.5105~2025750.05~0.0297.51002~117.56000.02~0.0135001~0.542.5210300.01~0.00520000.5~0.2557.51401000.005~0.00220000.25~0.197.537.5140要求:(1)绘出级配曲线;(2)确定不均匀系数Cu及曲率系数Cv,并由Cu、Cv判断级配情况。解:粒径(mm)105210.50.250.10.050.020.010.0050.002小于该粒径的含量百分数(%)10096.58876.55734.515840A100958341135.520B1007156565.3.20C级配曲线见附图。1001010.10.011E-30102030405060708090100土粒直径以毫米计小于某直径之土重百分数%习题1-2颗粒大小级配曲线由级配曲线查得d10、d30、d60,并计算Cu、Cc:d10d30d60CuCc级配情况A0.01250.0450.129.61.35良好B0.190.410.693.631.28不良C0.1350.252.9021.480.16不良1—3某土样孔隙体积等于颗粒体积,求孔隙比e为若干?若Gs=2.66,求d=?若孔隙为水所充满求其密度和含水量W。解:111svVVe;3/33.1266.2cmgVMsd;3/83.12166.2cmgVMMws;%6.3766.21swMM。1—4在某一层土中,用容积为72cm3的环刀取样,经测定,土样质量129.1g,烘干后质量121.5g,土粒比重为2.70,问该土样的含水量、密度、饱和密度、浮密度、干密度各是多少?解:3457.25.121cmGMVsss;3274572cmVVVsV;%26.60626.05.1215.1211.129swMM;3/79.1721.129cmgVM;3/06.2722715.121cmgVVMvwssat;3/06.1724515.121'cmgVVMsws;[或3/06.1106.2'cmgwsat];3/69.1725.121cmgVMsd。1—5某饱和土样,含水量w=40%,密度g/cm3,求它的孔隙比e和土粒比重Gs。解:365.04.083.14.1sV;74.2365.01wsssVMG;10.1365.04.0svVVe。1—6某科研试验,需配制含水量等于62%的饱和软土1m3,现有含水量为15%、比重为2.70的湿土,问需湿土多少公斤?加水多少公斤?解:1m3饱和软土中含土粒:tMs01.17.2162.01;折合%15的湿土:kgtMMMMsws116016.1)15.01(01.1)1(;需要加水:kgtMMsw475475.0)15.062.0(01.1)(12。1—7已知土粒比重为2.72,饱和度为37%,孔隙比为0.95,问孔隙比不变的条件下,饱和度提高到90%时,每立方米的土应加多少水?解:1m3土原有水:tMw18.095.1137.095.01;Sr提高到90%后:tMw438.095.119.095.02;1m3土需加水:tMw258.018.0438.0。1—8某港回淤的淤泥sat=1.50t/m3,w=84%,Gs=2.70。现拟用挖泥船清淤,挖除时需用水将淤泥混成10%浓度的泥浆(土粒占泥浆质量的10%)才可以输送。问欲清除1*106m3淤泥共需输送泥浆多少立方米?解:31100000mV;7.2197.2174.021VV;解得:32844000mV。1—9饱和土体孔隙比为0.7,比重为2.72,用三相图计算干重度、饱和重度sat和浮重度,并求饱和度Sr为75%时的重度和含水量w。(分别设Vs=1、V=1和M=1进行计算,比较哪种方法更简单些?)解:3/6.17.0172.2cmgVMsd;3/0.27.0117.072.2cmgVVMwvssat;3/91.17.01175.07.072.2cmgVM;%3.1972.2175.07.0swMM。1—10证明下列等式:(1)dGs-1Gs;(2)rGses。解:(1))(1)1(11'aegGeggGVVWswswss;)(1begGVWssd;(a)/(b):ssdGG1';故,dssGG1'。(2))1(wwsvwreGVVS;故,eGSsr。1—11经勘探某土料场埋藏土料250000m3,其天然孔隙比e1=1.20,问这些土料可填筑成孔隙比e2=0.70的土堤多少立方米?解:212111eeVV;故311221931812500002.1170.0111mVeeV。1—12从甲、乙两地粘性土层中各取出土样进行界限含水量试验,两土样液、塑限相同:L=40%,P=25%。但甲地天然含水量w=45%,而乙地天然含水量w=20%,问两地的液性指数I1各为多少?各处何种状态?作为地基哪一处较好?解:甲地:33.125402545PLPLI流动;乙地:33.025402520PLPLI坚硬;故,乙地好!1—13已知饱和软粘土的塑性指数Ip=25,液限L=55%,液性指数IL=1.5,土粒比重Gs=2.72,求孔隙比。解:由PLPI,P5525;得到:%30P;由PLPLI,3055305.1;得到:%5.67;由三相图:836.172.21675.0svVVe。1—14测得某饱和土样质量为40g,体积为21.5cm3,放在烘箱内干燥一段时间后,测得质量为33g,体积为15.7cm3,饱和度Sr=75%,试求土样天然状态下的含水量、孔隙比、干密度。解:水量损失:gMw73340;相应体积:37cmVw;总体积减小量:38.57.155.21cmV;烘干后空气体积:32.18.57cmVa;由vavvwrVVVVVS,即vvVV2.175.0;得到:38.4cmVv;烘干后含水:gMw6.31)2.18.4(2;土粒质量:gMs4.296.333;天然状态含水:gMw6.104.29401;天然状态含水量:%364.296.10;天然状态孔隙比:972.06.105.216.10e;天然状态干密度:3/37.15.214.29cmgd。1—15某饱和砂层天然密度=2.01cm3,比重Gs=2.67,试验测得该砂最松状态时装满1000cm3容器需干砂1550g,最紧状态需干砂1700g,求其相对密实度Dr,并判断其松密状态。解:VVMwvssat,即ee167.201.2;得到:653.0e;由试验:3min/55.110001550cmgd;3max/70.110001700cmgd;由三相图:eVMsd167.2;得到:dde67.2;故,723.055.155.167.267.2minminmaxdde;571.070.170.167.267.2maxmaxmindde;相对密度:461.0571.0723.0653.0723.0minmaxmaxeeeeDr。1-16某粘性土土样的击实试验结果如表:含水量/%14.716.518.421.823.7干密度/g.cm-31.591.631.661.651.62该土土粒比重Gs=2.70,试绘出该土的击实曲线及饱和曲线,确定其最优含水量op与最大干密度dmax,并求出相应于击实曲线峰点的饱和度与孔隙比e各为多少?解:由试验结果绘出击实曲线如下图:饱和曲线干密度g/cm31.81.71.61.512152025ωopρdmaxω(%)又sdG1求出各所对应的ρd。绘出饱和曲线ρd(g/cm2)1.821.751.691.64(%)18202224由图知:%5.19op3max/665.1cmgdgMMsw5265.07.2195.062.1665.17.2maxdsMV%9.84162.15265.01vwVVS62.01162.1svVVe1-17某粘土在含水量为20%时被击实至密度为1.84g/cm3,土粒比重为2.74,试确定此时土中空气含量百分数(Va/V)及干密度。解:gws288.374.2548.0gMMsw5265.07.2195.05265.0wV1sVggGWs7.2548.074.22.0swMM0548wV1sV74.2GMs3787.184.1288.3cmV3787.01787.1cmVVVsv3239.0548.0787.0cmVVVwva%4.13787.1239.0VVa3/533.1787.174.2cmgVMsd1-18推证饱和曲线的方程为wd1Gs证:GGVGVMvsd111—19有A、B、C、D四种土,颗粒大小分析试验及液限、塑限试验结果如表1-7,试按《建筑地基基础设计规范》予以分类。表1-7习题1-19表ABCD颗粒大小分析试验粒径d/mm〈d的含量百分数/%20901082562.51002.02797.51.013/0.58841000.25470960.1047831000.053367.597.50.022247880.011533780.005102368.50.00241358GGVGVMvsd11ddGGGd1证毕液限L/%无塑性333578塑限P272631注
本文标题:土力学习题答案(完整版)
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