土力学课后习题答案清华大学出版社

1、第一章1-1:已知：V=72cm3 m=129.1g ms=121.5g Gs=2.70则:m msw =ms129.1-121.5121.5= 6.3%m129 13=注g = g =q-*10 =17.9KN /m3v72mqVss空=45M2.7Vv =V -Vs3= 72-45 = 27cmsat = :satg sat satmsat;wVvms一 g =gVV1.0*27121.572_ 一3*10 =20.6KN /m3铲= wsat - w =20.6 -10 = 10.6KN /m3d =m!g =121*10 =16.9KN /m3V 721-2:已知:则 sat . dG

2、s=2.72设 Vs=1cm3s = 2.72g /cm3ms = 2.72 gd =注dg tg =272*10 =16KN /m3d " V 1.7、g =mwg 二27!”。“0.1KN /m3V1.7则 =- w =20.1 -10 =10.1KN /m3当 S=75% 时，mw =wVVSr =1.0*0.7*75% = 0.525gww =ms0 -19.3%2.72wsg0.525 2.72*10 =19.1kn /m31.71-3:ms = ,：dV =1.70*103*8*10 4 =13.6*107kgmw =msw =13.6*107*20% =2.72*107

3、 kg典 m, 13.6*107 2.72*1071.92*10385000m31-4:甲：I p = Wl - Wp = 40 - 25 = 15设vs =侦Ums = :s*Vs =2.7gmw =2.7*30% = 0.81g又因为 Sr =100%.VV =业=0.81、ms mw2.7 0.81-=1.94g / cmVs Vw1 0.81=E =19.4KN /m3:、=m = 7 = 1.48 g / cm3Vs Vw1.81d =注dg =14.8KN /m3e =四=0.81Vs乙:Ip =Wl * =8设 Vs =侦U ms =sVs =2.68gmw =msw =2.68

4、*22% -0.4796g则 VV =0.4796cm3一 ms mw2.68 0.47963-=2.14g /cmVs VV1 0.4796=-g =2.14*10 =21.4 KN /m3一ms2.6833 = = = 1.84 g / cmVs Vw 1.4796d =，dg =1.84*10 =18.4 KN / m3e* =0.4796Vsms则（1）、（4）正确1-5:伤=。则1 eGs 七 12.7*1丁一 - 1.70一1 =0.59cwGsSr Se22%*2.70.59=1 85%所以该料场的土料不适合筑坝，建议翻晒，使其含水率降低1-6:(| d - :d min ) :

5、 d maxur ()孔dmax d min d式中Dr=0.7岩 max = 1. 96g/cm3;?dm i 1 . 4gp of rri则可得：=1.78g/cm31-7:设 S=1,则 Vs=Sh=h则压缩后：ms =VsGs =2.7hmw =mw=2 . h * 2 8% WJ Vw =告=2.7h*28%.wVs+Vv =2.7h*28% +h =1.95贝U h=1.11cm.尾=2.0-1.11 =0.89cmVl 4 =四Vsh 1.11= 0.81-8:甲：流塑状态w - Wp45-25Il = = =1.33WL-Wp 40-25W Wc 2025乙：Il= =四4 =

6、 -0.33坚硬(半固态)WL-Wp 40-25Ip =WlWp =15届丁粉质粘土（中液限粘质土）乙土较适合作天然地基1-9:气=-匕=臣笠=0.31 <0.75届非活性粘土P0.002 甲552=上匚J3=13>125届活性粘土P0.002乙27乙土活动性高，可能为伊利石，及少量的高岭石，工程性质乙土的可能较第二章2-1 解:根据渗流连续原理，流经三种土样的渗透速度v应相等，即=Vb = Vc根据达西定律，得:hAhB尼RA- B - C LALbLC. ：hA : :hB : . ：hC =1: 2:4又. ：hA、hB、hC =35cm.: hA =5cm, . ：hB =

7、10cm, . ：hC = 20cm.：hA-3V =kAA =1*10 cm/sV加水=V* A*t=0.1cm 32-2 解:"玄厦EE1 e 1 0.582-3 解：(1) 土样单位体积所受的渗透力=1* .竺=1*9 8* Z0L306.53NG -1 2.72-1=1.0551 e 1 0.6320=0.66730则土体处于稳定状态，不会发生流土现象h(3)当i A”时，会发生流土破坏，即icr时h L* icr =30*1.055 = 31.65cm ci水头差值为32cm时就可使土样发生流土破坏2-4 解:(1) hA =6m, ho =7.5m, hB = "

8、;" = 6.75m2C* h3j =rw*i3.675kN/m3lh(2) 右要保持水深1m, i = =0.625L而 Q =Akv =20*10*1.5*10*0.625 = 1.875*10 *m3 / s故单位时间内抽水量为 1.875*10_63m / s2-5: sati cr解：Gs e - _ Gs e1 w =，1 e 1 eGs e-(1 e) Gs e而icr ci1 = :' sat 11 e sat又f Psat粘土 < Rat砂层，故只考虑PsatB土就可以3icr =sat粘土 -1 =2.04-1 =1.04g/cm7.5-(h 3)

9、4.5-h又 7 icr 一；则 h _1.38故开挖深度为2-6:解：(1)地基中渗透流速最大的不为在等势线最密集处，故在第二根流线上H H (5-1)m 八=h = = 0.267mNn -116-1：h 0.267 i L6m时，基坑中水深至少 1.38m才能防止发生流土现象0.4 0.667v = ki= 1*103*0.4 =4*10 4cm/s(2)i crn -1 =2 -1 =1 sat< icr故地基土处于稳定状态(3)q =M Eq =Mk ：h =5*1*10 '*0.267 =1.335*10 'm.巾 0.267 /s2-7:解:(1)AH =3

10、.6m, Ah =组= 36 = 0.257m1414q = M q = Mk ：h =6*1.8*10 "4*0.257 =2.776*10 ' m3 / s = 1.666*10 m3/ min(2)icrrMt18.5一=熟 -1 =-1=0.888rwrw9.8hi = 一LicrFs =一i0.257 =0.514,故i <&,不可能发生流土破坏0.50.888 ,1.730.514第三章土体中的应力计算3-1:解:41.0m： ；s1 = 1H1 =1.70*10*3 =51kpa40.0m：二2=s12H2=51(1.90 -1.0)*10*1=6

11、0kpa38.0m：os3-；.s23H3=60(1.85-1.0)*10*2=77kpa35.0m：；二4=；：s34H4=77(2.0 -1.0)*10*3=107kpa水位降低到35.0m41.0m： ；：s1 =51kpa40.0m：二s2 =cs12H2 =51 1.90*10*1 =70kpa38.0m： ；s3 = ；s2 3H3 =70 1.85*10*1 =88.5kpa35.0m：二s4 -； s34H4 =88.5 1.82*10*3 =143.1kpa3-2:解：偏心受压:e = 0.2m6epmaxp_(1 )BB7006*0.2(1 1010)=78.4kNpmin

12、 =61.6kNz(m)n=z/B均布荷载p=61.6三角形荷载p16.8水平附加应力总附加 应力b(kPa)K(TK(T0.10.010.99961.53840.58.4069.938410.10.99761.41520.4988.3664069.781620.20.97860.24480.4988.3664068.611240.40.88154.26960.4417.4088061.678460.60.75646.56960.3786.3504052.9280.80.64239.5472 10.3215.3928044.941010.54933.81840.2754.62038.43841

13、21.20.47829.44480.2394.0152033.46141.40.4225.872 10.213.528029.42020.30618.84960.1532.5704021.42由丁是中点，故Fs 二H cos-：、tanFccH sin 工"J= 1.0973-3:解：(1)可将矩形分为上下两部分，则为 2者叠加m=,n=，查表得 K, crzo = 2K *。B B(2)可将该题视为求解条形基础中线下附加应力分布，上部荷载为50kN/m2的均布荷 载与100 kN/m2的三角形荷载叠加而成。3-4:解：只考虑B的影响：用角点法可分为4部分，m1 =上=1.5,n1

14、=三=0.5，得 K1 = 0.2373B1B1L2八 z, 风.m2 = = 3,n2 = = 1 ,碍 K2 = 0.2034m3 =2, n3=1 ,得 K3 = 0.1999m4 =务=1,n4 =春=1 '得 K4 = 0.1752L =(K1 -K2 -K3 K4)c- =2.76kN/m2只考虑A:为三角形荷载与均布荷载叠加 _ 2m=1,n=1, Ki =0.1752, Li = Ki=0.1752*100 =17.52kN/mK2 =0.0666, r FL =0.066*100 =6.6kN/m2；z =匚1；z2 =24.12kN / m22贝 Uc盘=2.76

15、24.12 =26.88kN/m3-6:解：（1）不考虑毛兰田管升局:深度z (nj)b (kN/m2)u (kN/m2)b , (kN/m2)0.516.8*0.5=8.408.4216.8*2=33.6033.6433.6+19.4*2=72.42*9.8=19.652.88（上）72.4+20.4*4=1546*9.8=58.895.28 （下）72.4+20.4*4=15410*9.8=985612154+19.4*4=231.614*9.8=137.294.4（2）毛细管升高1.5m深度z (nj)b (kN/m2)u (kN/m2)b , ( kN/m2)0.516.8*0.5=8

16、.49.8* (-1.5 ) = (-14.7 )23.128.4+19.4*1.5=37.5037.5437.5+19.4*2=76.3.2*9.8=19.656.78（上）76.3+20.4*4=157.96*9.8=58.899.18 （下）76.3+20.4*4=157.910*9.8=9859.912157.9+19.4*4=235.514*9.8=137.298.33-7:解:点号b (kN/m2)u (kN/m2)b , ( kN/m2)A2*9.8=19.62*9.8=19.60B19.6+2*20=59.65.5*9.8=53.95.7C59.6+2*20=99.67.5*9

17、.8=73.526.13-8:解:试件饱和，则B=1一Ua可得 A =1一 =0.5.'，二一上:二13_ .'. _ .3匚 uA2 = A(1 - : ；3) = 75kN / m则水平向总应力 Q3=100kN/m3 有效应力。 =。3 - AuA2 = 25kN/m3竖直向总应力 B1 = 叫2 =l50kN/m3 有效应力 L=b1 AuA2 =75kN/m33-10：解：(1)粉质粘土饱和，Gs = 2.7,w = 26%s = 2.7 g / cm3m =msmw = "M'Mw =我(1 w)m”，：?cV =Vs Vw =Vs；=Vs(1?w

18、).w. w如w)3=2g/cm由图可知，未加载前M点总应力为：竖直向：可=药砂+。粉粘=丫砂 H + Y 粉粘 H2 =1.8*10*2 +2*10*3 =96kN/m ；x3 =K°L =0.6*116 -69.6kN /m ux3 =99.6kN/m2 (4) K0 =0.6，即。；=0.6。；2孔隙水压力为：u1 = wH2 =1.0*10*3 =30kN/m» 、, 2有效应力：二-1 =二- u = 66kN / m水平向：Cx1 =K0'= 0.6*66 =39.6kN/m222Ux1 = U1 =30kN / m , 。x1 = ux1 0x1= 6

19、9.6kN / ms2(2)加荷后，M点的竖宜向附加应力为：bz = Ksp =0.5*100 =50kN/m22 水平向附加应力为：二-x =0.3弓=0.3*50 =15kN/m在加荷瞬间，上部荷载主要有孔隙水压力承担，贝U：一 .，_ _ _2竖宜向：二 2 二二-1 二z =96 50 =146kN/m_ _ 2u2 = u；弓=30 50 = 80kN / m2L =二2 -也=146 -80 = 66kN /m水平向：ux2 =虫=80kN/m2；x2 = K0o2 =0.6*66 =39.6kN/m2二2 = J ux2 =119.6kN /m2(3) 土层完全固结后，上部荷载主

20、要由有效应力部分承担竖直向：二 3 =二-1 L =96 50 =146kN/m2 2u3 = u1 = 30kN / m二3 =二1 七=66 50 =116kN/m22水平向：ux3 =u3 =30kN/m上P =（'z Lx） =0.8 上。z1上q = （，z -»x） =0.2上。z4-1 :解:试验结束时,w 二 0.278*2.7 e =Sr 1第四章= 0.7506此过程中，土样变化 s = 2.0 -1.98 = 0.02cm初始孔隙比e =e s/H。1 -s/ H00.7506 0.02/2.0 =0.7681-0.02/2.0s孔隙比e = e0 -(

21、1 e0)H°= 0.768 -(1 0.768)*= 0.768 0.884s当 © =200kPa 时,& =2.0 -1.990 = 0.01cm,e = 0.768 - 0.884*0.01 = 0.7592当 s2 =300kPa 时,§ =2.0 -1.970 = 0.03cm ,e1 =0.768 -0.884*0.03 =0.7415a? j34-4:(1)(2)Act解：e 0.7592-0.741510.18MPa 100两基础中心点沉降量不相同 通过调整两基础的H和B,可使两基础的沉降量相近调整方案有：方案一：增大B2使B2 a2B，

22、则附加应力P0乙< p0甲ss而K亿A K z甲，故可目匕有方案二：使B2 = B1，则Ksz乙=Ksz甲，即增加H1或减小H2方案三：增大B2 ,使B1 < 3 <2B1，同时，减小H2或增大H1(3)方案三较好，省原料，方 便施工4 5:解：(1) t=0 , t=4个月，t=无穷大时土层中超静水压力沿深度分布如图所示:(2)由图可知4个月时Ut财45.3%L = U2t = *0.4532 =0.161 v 4 t 4Cv =上 = 39.22m2/ 年 t当 U t =90% 时，Tv =-0.933lg(1 Ut)-0.085 =0.933lg 0.10.085 =

23、 0.848T0H2Cv0.848*9239.22=1.7弭4-6:解:(1) S/ AH =0”5MPa *240 kN/m2*10m = 0.33m- 1 e1 0.8k2=2.0cm/年 *1.8 3=i4.69m2 / 年a w0.25MPa*9.8kN/m3Cv 14.69m2/ 年Tv=Tt2*1 年=0.1469H 100mUt =1 -争七“=0.4359则St= Ut*S 一一 =0.142m(2)当 § = 0.2m 时,宜=匹=0.6061UtS： 0.33查表有：Tv =0.293=1.9欧财2年t TvH20.392*100 m2一 Cv 一 14.69m2

24、/ 年 故加荷历史2年地面沉降量可达 20cm第五章 土的抗剪强度5-2解：由剪破面与大主应力面交角 60°600 =a =45° +/2得：中=30°由试样达到破坏状态的应力条件：2_-；i = ；：.-3tan (45/2) 2ctan(45 /2)二3 = "an2(45 -、/2)-2ctan(45 - /2)已知：2。1 =500KN /m2。3 =100KN /m2则 c =100/ .3 =57.7KN /m2法向应力：二- =(。1 匚)/2 (r -j)*cos 2- /2 =300 200*( -0.5) = 200kpa剪应力：.=

25、(二1 - 二3)*sin2 ： / 2 = 200*、3 / 2 = 173kpa5-3解：(1)求该点主应力值；x J z . . x 22；3 = z 2 x _, ( z 2 x).=175 _85；了3 =90kpa(2) 该点破坏可能性判断c=0= &mtan2(45/2) =90*tan 2(45 15 ) =270=260kpa改用式：c * = ；1mtan2(45 - /2) =260*tan2(45 -15 ) =86.67 ：匚=90kpa该点未剪破(3) 当c值增加至60KN/ nf时；了仃=；： 3tan2(45/ 2) = 236.7 ： 271kpa=c

26、1mtan2(45 - /2) =271* tan2(45 -15 )= 90.33 ；3m(<V=175±96,则。3 =79kpa,Q1 =271kpa)即实际的小主应力低于维持极限平衡状态所要求的小主应力，故土体破坏5-4解：(1)绘总应力圆图如下由图可量得，总应力强度指标：Ccu =17.5kpa"cu =16(2) 计算有效应力 。1 =c1 -u =145-31 =114kpa二3 =60-31 =29kpa ；：1 = 228 -55 = 173kpa,二 3 = 100 - 55 = 45kpa c1 =310 -92 =218kpa, ；3 =150

27、-92 =58kpa ；：=401 -120 =281kpa,；3 =200-120 = 80kpa绘有效应力圆图如下由图可量得：c'=7.5kpa, * =32(3) 破坏主应力线如上图中的虚线表示：可得 a = 7kpa, : = 27.4- ' =sin(tan_：、)= sin(tan 27.4 ) =31.2- = 7 =8.18kpacos cos31.25-5解：(1)砾砂粘聚力c=0七 c、 二 7 一二、22 350 15022-二3 = 一( x)2 22 = 1002 (-100)2 =250 一100、22,22。1 =250 100.2 =391.4k

28、pa,j =250-100'、2=108.6kpa.M点处于极限平衡状态，贝U"-1 一、- 3= sin=sin 0.5656 =34.4二1匚3100*2(2)求大主应力方向:tan2 ： = = = 1七-七350-15022. =45 , .=22.5由于破裂面与最大主应力面成45 ° +/2的夹角，故：二=45/2 =22.54534.4 /2=84.7滑裂面通过M点的方向如图:5-6解：u -：；3 A（二1 - ；3）ui = 50 0.2*85 = 67kpaU2 =100 0.2*83 =116.5kpau3 =150 0.2*87 =167.4k

29、pa试件：二一3 - -3 7 =33kpa,o - .-U| = 118kpa试件：c-3-3-u2 =33.4kpa,厂-u2 =116.4kpa试件：二 3 -；-3 -u3 =32.6kpa,-如 - u =119.6kpa5-7解:由图可知FFFFCusin （ t 2 "3 cctg ） = sin，* '-1 ? 3 c cosCu二 1 - ；了32Q = （Q ；了3 ）sin L cosc sin 。3 c cosC =3u 1-sin5-10解：b 3等于常量，增大 b 1直至试件剪切破坏当开始固结P二1 ,二32七当开始剪切时,O'1 J%,q

30、= =0b 3等于常量二1 .二3可-cro-cro,qW" 3131=一 J2p-q坐标上的三轴试验应力路b 1等于常量，减小径为：b3直至试件剪切破坏 ，固结同剪切过程，b1为常量第六章挡土结构物上的土压力6-1:解：静止土压力系数：K0 =1-sin * =0.357主动土压力系数：Ka =tan中=P2 -P(45 - /2) =0.217被动土压力系数：Kp =tan2(45 、/2) =4.61 O 静止土压力：E0H 2K0 = 80.33kN / m2 1王动土压力：EaH2Ka =48.8kN/m2被动土压力： E =1 H2Kq =1035kN/m p 2 p&a

31、mp; =20心时：主动土压力系数为：Ka =0.199 12王动土压力：Ea H2Ka =44.775kN/m 26-2:解：(1) K a = tan2(45*-妇2) =0.455pa =qKazKa-2c K? =20*0.455 18*0.455 z - 2*16、. 0.455 = 8.19z -12.5z011.5323456pa0003.8812.0720.2628.4536.64(2)1Ea =1pz(H y)-0.5*36.64*(6-1.53)=81.92作用点在z=4.51m处(3)z0 = 1.53m6-4:解：查表得：Ka =0.236水位以上土压力：pa=zKa水

32、位以下土压力：pa =(z-1.5)KaH1Ka结果如下：z011.52345pa04.2486.3727.6710.26612.86215.458主动土压力分布图水压力：Pw = K(z-Hi)结果如下：水压力分布图z1.523pw051525356-5:解:Kacos2（二二）cos2 ： *cos（ qW）*1si” ； ）*sin（ - ：）2 cosG " ）*cos（：-）= 0.755_12Ea = H2Ka qHK =278.142 aaa方向与水平向呈64度角，指向墙背；作用点为梯形面积重心第七章 土坡稳定分析7-1:解：渗流出段内水流的方向平行丁地面故 9 =0i

33、 =tan： =0.364土坡的稳定安全系数Fs=0.755/V cose-侦iV sin(: - Qtan /V sin .二:u' ;'wiV cos()7-2:解：从无限长坡中截取单宽土柱进行稳定分析，单宽土柱的安全系数与全坡相同土柱重量：W =斗沿基面滑动力：T =Wsin :沿基面抗滑力：R = W cos ： tan粘性土的粘聚力：Fc = c* l = cb /cos:l R Fc Wcos： tan cb/cos：Fs =cTWsin：又 Fs =1.0,贝U H =7.224m7-3:解：sat =Gs(1 w)*g =19.31kN/m3 1 e5 btan

34、:i =sin ： = 0.342s b/cos；J =Aj =H wi =9.8*4*0.342 =13.41kN安全系数：巳=Hcos：tan Fc =1.097 s H sin ： J第八章8-1:解：(1) 基础宽度、基础埋深和粘聚力同时增加1倍时，地基的承载力也增加1 倍，地基的承载力随基础宽度、基础埋深和粘聚力成倍增长，随着内摩擦 角的增加，Nr, Nq, Nc增加很大，承载力也增大很多。(2) 对砂土地基，其c=0,这时基础的埋深对极限承载力起重要作用，若此时 基础埋深太浅(D<0.5B),地基的极限承载力会显著下降1(3) 由极限承载力公式 氐=9Nq +cNc+ YBNr可知，基础宽度的增加会引起承载q 2力的增加。8-2:解：均布荷载 q = D =19*2 =38kN/m2查表可得 Nr =1.51,Nq =4.65, Nc =10.0极限承载力1 12% =日 BNr qNq cNc*19*3*1.5138*4.65 10*10