重慶大學土力學課后習題部分答案

1、第一章土的物理性質(zhì)及工程分類1.1解：(1)A曲線：卵石或碎石(>20mm)占100-77=23%礫粒(20-2mm)占77-50=27%砂粒(2-0.075mm)占50T0=40%粉粒(V0.075mm)占10%( 2) A曲線較平緩，說明A土土粒粒度分布范圍廣，顆粒不均勻，故級配良好；而曲線較A土曲線陡，說明其粒度分布范圍窄，土粒均勻，故級配不良。( 3) A土d100.08d300.6d603.6Cud60/d103.6/0.08455Cc(d30)2/(d10d60)(0.6)2/(3.60.08)1.25在1-3之間，故A土級配好B土d100.15d300.35d600.74C

2、ud60/d100.74/0.154.95Cc(d30)2/(d10d60)(0.35)2/(0.740.15)1.10只滿足一個條件，故級配不良6.72%3.2 解：ms119gmwmms127-1198gwmw/ms8/119eGspw(1w)/p12.71(10.0672)/1.76-10.6373pm/V127/721.76g/cm3ftat(Gse)w/(1e)(2.70.637)/(10.637)2.04g/cm33Psatw1.04g/cmams/V119/721.65g/cm3ne/(1e)0.637/(10.637)38.9%SrwGs/e0.06722.7/0.63728.

3、5%比較密度：3.3 解：VvVs,eVv/Vs1dGsw/(1e)2.6810/(11)13.4kN/m3VvVw_3sat(GsVv)w/(1e)(dse)w/(1e)(2.681)10/(11)18.4kN/mwWw/WsVww/VsGsw1v/Gs1/2.6837.3%.、一33.4 解：設V1mWsdV11kN一_3VsWs/(dsw)11/(2.710)0.407mVvVVs10.4070.593m3wWw/WsWwwWs0.5115.5kNVwWw/w5.5/100.55m3VaVvVw0.5930.550.043m3av3.5 解：飽和土ewGssat(Gse)w/(1e)即e

4、0.4Gs18(Gse)/(1e)以上兩式聯(lián)立求解解得e1.05Gs2.653.6 解：wWw/Ws(WWs)/Ws可求得w16%時，W21.64N時土粒的質(zhì)量WsW/(1w)21.64/(10.16)18.655N其中水量為WwwWS0.1618.6552.985N.-'.-.當w'25%時，土中水量為Www(Ws0.2518.6554.664N10因此增加的水量Ww4.6642.9851.679N3.7 解：設飽和土為V1m3VVsVw1飽和土wWw/WsVw/VsGs0.6由上兩式聯(lián)立可得：Vs0.382Vw0.618WsVsGsw0.3822.71010.314kNWw

5、wWS0.6010.316.186kN當w15%時，需要土W(w1)WS1.1510.31411.86kN其中水量WwwWS0.1510.3141.547kN故增加水的重量為W6.1861.5474.64kN333.8 解：e1eVv/VsVvVSV1mVvVS0.5mSrVw/VvVwSrVvm3當水蒸發(fā)掉一半時VwVw/20.4/20.2m3Vv'Vw'/sr0.2/0.9V0.50.2/0.90.72272.2%3.9 證：/d1(W/V)/(Ws/V)1W/Ws1(WWs)/WsWw/Ws1.10 解： e Gs w(1 w)/1(2.6510(10.

6、073)/17.110.663d minWs/V29.85/20000.0149N/cm314.9kN/m3dminGS/w/(1emax)14.92.6510/(1emax)emax0.779d minWs/V32.5/20000.0163N/cm316.3kN/m3dminGS/w/(1emin)16.32.6510/(1emin)emin0.626Dr( emaxe) /( emaxemin )(0.770 0.663)/(0.779 0.626) 0.7581.11 解：在壓實以前n e/(1 e) 0.72/(1 0.72) 0.419SrGsw/e2.650.073/0.66329

7、.2%在 1m2 的地基面積中，土粒體積為VS (1 n)V (1 0.419) 5 2.91m3在壓實以后，土粒體積減少0.3m3n (V VS)/V (5 0.3 2.91) /(5 0.3) 0.381e n /(1 n) 0.381 /(1 0.381) 0.616此時 D r(emax e)/(emax emin)(0.75 0.616)/(0.75 0.52)0.5831.12 解： mw 23 15 8gw mw / ms 8/15 0.533 53.3%I wL wP 40 24 16 pI L (w wP) /(wL wP) (53 24)/16 1.81310 I P 17

8、 粉質(zhì)粘土IL 1流態(tài)1.13 解： ( 1 ) A 土 I P 30 > B 土 I P 11故 A 土粘粒含量多2)由Gs w(1 w) /(1 e) dsw(1 w) /(1 w Gs /Sr )Sr100%13A土2.6910(10.53)/(10.532.69)16.97kN/m3B土2.7110(10.26)/(10.262.71)20.03kN/m3B土的天然重度大(3)d/(1w)A土d16.97/(10.53)11.09kN/m33B土d20.03/(10.26)15.89kN/m3B土的干重度大(4)因為是飽和土ewGsA土e0.532.691.43B土e0.262.

9、710.70A土的孔隙比大而0.5mm配所示屬砂土，而2mm粒徑的顆粒只占總質(zhì)量16%25%,故該土樣不是礫砂,粒徑的顆粒占總質(zhì)量（1639）55%50%,故該土樣定名為粗砂。第三章土中的應力3.1解：細砂、粗砂分界線處：CZ16.4349.2kPa粗砂n上e2/31e1n19.08kN/m3Gsw(1w)2.6510(10.2)Te12/3水位線處CZ49.219.082.494.8kPa粗砂與粘土分界處sat(Gse)w(2.652/3)1019.9kN/m31e12/3CZ94.8(19.910)2.1115.6kPa粘土wmw-VJ0.6eVw0.62.731.638msGsVsVss

10、at(Gse)w73.姆1016.56kN/m31e11.638粘土與泥巖分界處CZ115.6(16.5610)3135.28kPa泥巖頂面CZ 135.28 10 5.1 186.28kPa3.2解：見圖結(jié)論：隨著地下水位的下降，土有效應力增大，使得變形會增加。3.3 解：（1）填土前總應力（kpa）孔隙水壓力u（kpa）有效應力（kpa）中砂3m19x3=5757中砂3m57+20X3=11710X3=30117-30=87粉土3m117+18X3=17130+10X3=60171-60=111粗3m171+20X3=23160+10X3=90231-90=141砂（2）剛填土后（假定一次

11、將士驟然填上）總應力（kpa）孔隙水壓力u（kpa）有效應力（kpa）中砂3m9090+57=14790-0=90147-0=147中砂3m90+117=20710X3=30207-30=177粉土3m171+90=26190+30=12090+60=150207-120=87261-150=111粗砂3m231+90=32110X6=6010X9=90261-60=201321-90=231（3）填土后土層完全固結(jié)總應力（kpa）孔隙水壓力u（kpa）有效應力'（kpa）中砂3m9014790147中砂3m20730177粉土3m261602011粗砂3m321902313.4 解：

12、對基地面積為3m6m,Po200kPa點開L/bz/bCz4cP00200.250200121/1.50.228182.4222/1.50.170136323/1.50.12096424/1.50.08668.8525/1.50.06350.46240.04838.4對基地面積為1m2m,F0200kPa點開L/bz/bCz4cP00200.250200121/0.50.12096222/0.50.04838.4323/0.50.02419.2424/0.50.01411.2525/0.50.0097.262120.0064.8要畫圖結(jié)論：隨著深度增加，土的附加應力減小；基礎面積越小，減小的速

13、度越快。2t4rri12m3.5 解：(a)I區(qū)：l/b=2,Z/b=4,：=0.048n區(qū)：l/b=3,Z/b=4,n=0.060出區(qū)：l/b=1.5,Z/b=2,0.1065z ( i 234)P 14.9kPaN區(qū):l/b=1,Z/b=2,=0.084(b)I區(qū)：l/b=2,Z/b=1.33,i=0.170n區(qū)：l/b=6,Z/b=4,口=0.073z(12)P4.85kPa(c) I 區(qū):l/b=1,Z/b=2.67,I =0.055n區(qū):l/b=2,Z/b=2.67,n =0.138山區(qū):l/b=1,Z/b=1.33,rn =0.138IV區(qū):l/b=3,Z/b=4,=0.060Z(

14、1234)P 10.95kPa3.6 解：甲基礎作用下：Z=6.5m ,基地附加應力n2rn )4mPa3002.520250kPaA處l/b=0.6/0.6=1,Z/b=6.5/0.6=10.83,c=0.042ZA4CPA40.00422504.2kPa同理8處ZB2(12)Pa2(0.020.015)2502.5kPaC處zc2(i2)Pa2(0.0130.0062)2503.4kPa乙基礎作用下：Z=7.5m,基地附加應力PA3001.520270kPa人處ZA2(0.0160.01125)2702.57kPa8處ZB40.002752702.97kPaC處zc2(0.010.0046

15、)2702.92kPa故za4.22.576.77kPaZB2.52.975.47kPaZC3.42.926.32kPa3.7 解：l/b10z/b8/42aci=0.1375l/b10z/b8/24ac口=0.0765l/b4/22z/b8/24aw=0.0474A:z=(aci+ac口acm)Xp0=(0.1375+0.07650.0474)=0.2614X200=33.32kPaB:z=(aci+acn+aw)Xp0=(0.1375+0.0765+0.0474)=0.2614X200=52.28kPa3.8 解：求U形基礎形心座標，以X軸為對稱軸矩形AiAi=3X6=18m2Xi=1.5

16、myi=0矩形An,An=Am=1X3=3m2Xn=Xm=4.5my口=2.5mym=-2.5mAiXi181.5234.554m3-一一一一2Ai182324m2xAXi/Ai54/242.25m則偏心品巨e2.2520.25m基礎左端邊緣距形心軸y'距離為g2.25m基礎右端邊緣距形心軸y'距離為C23.75m58.5m4截面慣性距I1/12633630.7522/121332132.252WiI/C158.5/2.2526m3W2I/c258.5/3.7515.6m3p1F/AM/W160/24600.25/263.07kPap2F/AM/W160/24600.25/15

17、.61.54kPa3.9解：堆載處:Z=4m , b1=10m., b2=4ml/b>10,z/b1=0.4,z/b2=1，查表3.3得60.244,C20.205z 2( C1C2)P02(0.2440.205)40031.2kPa條基：PoF(g0)d100kPaAl/b>10,z/b=3/1=3,c=0.099Z4CPA40.09910040kPaz31.24071.2kPa3.10 道理同前，Z20.4kPa3.11 解：x/b0z/b3/21.5查表2-6sz0.4sx0.01豎向有效應力=豎向有效自重應力+豎向附加應力zszPo0.415060kPaczy'z(

18、2010)330kPa豎向有效應力=60+30=90kPa水平向有效應力=水平向有效自重應力+水平向附加應力cxK0cz0.43012kPacxczzxsxp00.011501.5kPa水平向有效應力=12+1.5=13.5kPa3.12(1)解：三角形荷載abd在M點所產(chǎn)生的附加應力3 P0 t2MafP0cMb 2 p0 t2Mbf1/ba/4.5a 0.22z/b 4.5a/4.5at2Maf0.03751/ba/3a0.33z/b 4.5a/3a 1.5t2Mbf0.02901/b3a/ az/b 4.5a/ a 4.5cMb0.051z10.03753p00.051p0 0.029

19、2p00.0035P0(2)三角形荷載ghn在M點所產(chǎn)生的附加應力z22 P0 t1gHGP0cHhP0 t2nMH1/ba/3a 0.33z/b4.5a/3a 1.5t1gGH0.02151/b3a/a 3z/b4.5a/a 4.5cHh 0.05101/ba/1.5a 0.67z/b4.5a/1.5a 3t2nMH 0.0163z20.0215 2p00.051p00.0163 P0 0.0083P0z1z2(0.00350.0083)P00.012P0t2(174)3P02z2(245)2P0t2(367)P00.003753P020.0292P00.0163P00.0128P0第四章土

20、的壓縮性及地基沉降計算4.1 解：(1)e0Gsw(1w)/)1(2.710(10.2)/19.9)10.628e1e0(1eo)/h0s10.628(10.628)/200.70.571e2e(1e1)/hiS20.571(10.571)/(200.7)(0.950.7)0.551(2)a12(e1e2)/(p2P1)(0.5710.551)/(200100)10000.2MPa1Es12(1e1)/a12(10.571)/0.27.85MPa(3) a120.2MPa,屬中壓縮性。65/100 0.654.2 解：（1）由公式4.12和4.13得K0-1K。Ko10.394由公式4.15得

21、12K。120.3940.650.488EsE。/10/0.48820.5MPa4.3解：(1)由a(e1e2)/(P2P1)和Es(1e)/a列表計算壓力段（kPa）0-5050-100100-200200-4001a(MPa1)0.35Es(MPa)1.1332.4383.1835.286(2)s,(e。e1)/(1e。)h(1.040.91)/(11.04)201.275mms2(e。0)/(1e。)h(1.040.85)/(11.04)201.863mmss2sl0.588mm4.4解：附加應力：F0F(G0)d800/6(2019)1134.33kPaA分層厚度：h

22、i0.4b0.420.8m點開l/bz(m)z/bcz4cP。(kPa)01.5000.25134.330.237127.3421.51.01.00.193103.700.128569.050.084545.400.058331.330.04222.570.031316.790.023412.575.3m深處z12.57kPa0.2c0,280,3616.07kPa,故取Zn5.3mzc土層占八、號Z自重應力附力口應力分層厚平均自重應力cic(i1平均附加

23、應力ziz(i1)平均自重應力+平均附受壓前孔隙比受壓前孔隙比si(eiies(2is)cZ度22加應力=P2ie1ie2i1eiah)s"粘0019134.33土10.528.5127.3450023.75130.84154.590.79580.7415.542138103.7050033.25115.52148.770.790.741913.44粉31.74569.0570041.586.38127.880.72430.6817.98質(zhì)42.45245.470048.557.23105.730.72080.688113.3粘53.15931.3370055.538.3793.87

24、0.71670.69379.38土63.86622.5770062.526.9589.450.71250.69636.6274.57316.7970069.519.6889.180.70830.69654.84粉土85.380.3612.5780076.6814.6891.360.80.7894.2885.44.6解：(1)基底附加壓力poF/A(go)d900/7.2(2018)1127kPa地基為均質(zhì)粘土，故不用分層，初按式(3-33)確定znZnb(2.50.4lnb)2(2.50.4ln2)4.45m取Zn4.5m,z0.3m將基底面積為相同的小塊(ll.8m,b1m)采用角點法當Z0

25、0時，Z000Z14.5m140.122850.4914sp0/Es(Z11z00)127/5(4.50,49140)56.2mm計算z(取0.3m)層土的壓縮量sn當z4.2m40.12890.5156snp0/Es(z11z)127/5(4.50.49144.20.5156)1.2mmSn/s 1.2/56.2 0.021 0.025 滿足要求由Es5MPa查0.9(取P00.75fak)sss0.956.250.6mm(2)考慮相鄰基礎的影響初定z6mz0.3mZ00Z000Zi6m自身荷載作用下0.099140.3964相鄰基礎的影響(荷載面積(oacdoabe)2)對面積查oacd1

26、/b7/1.83.89z/b6/1.83.33查得0.1613對面積查oabe1/b5/1.82.78z/b6/1.83.33查得0.1594故(0.16310.1594)20.0074實際上0.39640.00740.4038sPo/Es(Z11zo0)127/5(60.40380)61.5mm計算z土層的厚度z5.7m自身荷載作用下0.103240.4128相鄰基礎的影響(荷載面積(oacd-oabe)x2)對面積oacd1/b7/1.83.89z/b5.7/1.83.27查得0.1645對面積oabe1/b5/1.82.78z/b5.7/1.83.27查得0.1609故(0.16450.

27、1609)20.0072實際上0.41280.00720.4200sn p0/Es(z1 1 z ) 127/5(6 0.4038 5.7 0.4200)0.7mm 0.025s 1.5mm所以滿足要求s0.9s0.961.555.4mm4.8 解：（1）因為土的性質(zhì)和排水條件相同由Cv1Cv2Tv1Tv2得22ti/(H1/2)2t2/(H2/2)222t1H12/H2一2._2t2300/811406.25小時=59天t1_ 2 (H"2)4 59 236天t°由-2H12一2t3H-2/(H1/2)2t14t14.9 解：粘土層平均附加應力z(20050)/2125k

28、Pa(1)最終沉降量sa/(1e)zH0.5/(11.1)1255148.38mm Cv k(1 e)/( wa)0.0036(11.1)/(100.5103)1.512m2/年_2_2_TvCvt/H21.5122/520.121v200/504查表得U48%2 年后stUs0.48148.872mm3 3)Ust/s100/148.80.67267.2%v4查得Tv0.27tTvH2/Cv0.2752/1.5124.5年4 4)若為雙面排水v1,查得Tv0.38,tTvH2/(4Cv)0.3825/(41.512)1.57年第五章土的抗剪強度與地基承載力1 .1解：(1)ftan，即200300tan得33.692 2)1/2(13)1/2(13)cos21/2(13)sin229033.69123.69代入上式得：3001/2(13)1/2(13)cos123.692001/2(13)sin123.69解得1673kPa3193kPa3 3）45/24533.