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1、清华大学版土力学课后答案第一章 1-1：已知：V=72cm3 m=129.1gms =121.5gG s =2.70 则：129.1 121.56.3%121.5ssm mwm- -= = =3333129.1 *1017.9 /72121.5452.772 45 271.0*27 121.5*10 20.6 /72sssV ssat w V ssat satmg g KN mvmV cmV V V cmm V mg g g KN mV Vg rrrg r= = = = = = - = - =+ += = = = = 3320.6 10 10.6 /121.5*10 16.9 /72sat w
2、sdsat dKN mmg KN mVg g ggg g g g =- = - = = = 则 1-2：已知：G s =2.72 设 V s =1cm3则33332.72 /2.722.72*10 16 /1.72.72 0.7*1 *1020.1 /1.720.1 10 10.1 /75%1.0*0.7*75% 0.5250.52519.3%2.720.525 2.721.sssd ds V wwrw w V rwsw sg cmm gmg g KN mVm Vg g KN mVKN mm V S gmwmm mg gVrg rrg rg g grg r= = = =+ += = = = =
3、- = - = = = = =+ += = =当S 时，3*10 19.1 /7KN m =1-3：3 4 77 77 7331.70*10 *8*10 13.6*1013.6*10 *20% 2.72*1013.6*10 2.72*10850001.92*10s dw ss wm V kgm m w kgm mV mrr= = = = =+ += =挖1-4：甲：333340 25 151* 2.72.7*30% 0.81100%0.812.7 0.811.94 /1 0.8119.4 /2.71.48 /1.8114.8 /0.81p L Pss s swrwVws ws wsds wd
4、dvsI w wVm V gm gSmVm mg cmV Vg KN mmg cmV Vg KN mVeVrrrg rrg r= - = - = = = = =+ += = =+ += = = =+= = =设 则又因为乙：3333381 2.682.68*22% 0.47960.47962.68 0.47962.14 /1 0.47962.14*10 21.4 /2.681.84 /1.47961.84*10 18.4 /0.4796p L ps s s sw sVs ws Vsds wd dVsI w wV m V gm m w gV cmm mg cmV Vg KN mmg cmV Vg
5、 KN mVeVrrg rrg r= - = = = = =+ += = =+ += = = = =+= = = =设 则则 g g 乙 甲d dg g 乙 甲p pI I 乙 甲则（1）、（4）正确1-5：1s wdGerr =+则 2.7*11 1 0.591.7022%*2.71 85%0.59s wdsrGewGSerr= - = - = = = 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：min maxmax min( )( )d d drd d dDr r rr r r-=- 式中 D r =0.7 3max1.96 /dg cm r = 3min1.46 /dg
6、 cm r =则可得：31.78 /dg cm r = 1-7：设S=1， 则sV Sh h = =则压缩后：2.7s s sm V G h = = 2.7 *28%w sm m w h = =则 2.7 *28%wwwmV hr= =2.7 *28% 1.95s wV V h h + = + = 则 1.11 h cm =2.0 1.11 0.89Vh cm = - =0.890.81.11V VsV heV h= = = = 1-8：甲：45 251.3340 25pLL pw wIw w-= = =- - 流塑状态 乙：20 250.3340 25pLL pw wIw w-= = = -
7、 -坚硬（半固态）15p L pI w w = - = 属于粉质粘土（中液限粘质土）乙土较适合作自然地基1-9：0.00253 360.31 0.7555pIAP-= = = 乙乙乙属活性粘土 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其次章 2-1 解：依据渗流连续原理，流经三种土样的渗透速度 v 应相等，即A B Cv v v = =依据达西定律，得：C A BA B CA B Ch h hR R RL L LD D D= =: : 1:2:4A B Ch h h D D D =又 35A B Ch h h cm D +D +D =5 , 10 , 20A B Ch
8、cm h cm h cm D = D = D =31*10 /AAAhV k cm sL-D= =3* *t=0.1cm V V A =加水 2-2 解：1 2.70 11.0761 1 0.58scrGie- -= = =+ + 2-3 解：（1）土样单位体积所受的渗透力201* 1*9.8* 6.5330whj r NLD= = =（2）1 2.72 11.0551 1 0.63scrGie- -= = =+ + 200.66730hiLD= = =cri i 时，会发生流土破坏，crhiL 即 时* 30*1.055 31.65crh L i cm = =水头差值为 32cm 时就可使土
9、样发生流土破坏 2-4 解：（1）6 , 7.5 , 6.752A CA C Bh hh m h m h m+= = = =3* 3.675 /wwr hj r i kN mlD= = =（2）若要保持水深 1m， 0.625hiLD= =而8 6 320*10*1.5*10 *0.625 1.875*10 / Q Akv m s- -= = =故单位时间内抽水量为6 31.875*10 / m s- 2-5：解：1 1s ssat wG e G ee er r+ += =+ +，而11scrGie-=+ (1 )1 11 1s scr satG e e G eie er+ - + + = =
10、 - = -+ + 又sat satr r 砂层 粘土，故只考虑satr粘土 就可以 31 2.04 1 1.04 /cr sati g cm r = - = - =粘土 又7.5 ( 3) 4.53 3crh h hiLD - + - = =则 1.38 h故开挖深度为 6m 时，基坑中水深至少 1.38m 才能防止发生流土现象 2-6：解：（1）地基中渗透流速最大的不为在等势线最密集处，故在其次根流线上 (5 1)0.2671 16 1H H mh mN nD D -D = = = =- - 0.2670.40.667hiLD= = =3 41*10 *0.4 4*10 / v ki cm
11、 s = = =（2）0.2670.10682.5hiLD= = =均均 1 2 1 1cr sati r = - = - =则cri i 均 故地基土处于稳定状态 （3）5 5 25*1*10 *0.267 1.335*10 / q M q Mk h m s- -= S = D = =2-7：解：（1）3.6 H m D = ，3.60.25714 14Hh mDD = = =4 4 3 2 36*1.8*10 *0.257 2.776*10 / 1.666*10 / min q M q Mk h m s m- - -= D = D = = =（2）18.51 1 0.8889.8satcr
12、w wr rir r= = - = - =0.2570.5140.5hiLD= = = ，故cri i ，则附加应力0 0p p 乙 甲 ，故可能有sz 0 zpzK s s = =乙 乙 乙 甲方案二：使2 1B B = ，则s sz zK K =乙 甲 ，即增加 H1 或减小 H2 方案三：增大 B2，使1 2 12 B B B =改用式：2 23 1 3tan (45 /2) 260*tan (45 15 ) 86.67 90f mkpa s s f s = - = - = =∴该点未剪破 （3）当τ值增加至 60KN/m2 时 21 32 23 1 3tan (4
13、5 / 2) 236.7 271tan (45 / 2) 271*tan (45 15 ) 90.33ff m mkpa s s fs s f s= + = （3 3 1175 96, 79 , 271 ) kpa kpa s s s = = = 则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：17.5 , 16cu cuC kpa f = = （2）计算有效应力 1 13145 31 11460 31 29u kpakpas ss =- = - = =- = 1 3228 55 173 , 100 55 45
14、kpa kpa s s = - = = - =1 3310 92 218 , 150 92 58 kpa kpa s s = - = = - =1 3401 120 281 , 200 120 80 kpa kpa s s = - = = - =绘有效应力圆图如下 由图可量得：7.5 , 32 c kpa f = = （3）破坏主应力线如上图中的虚线表示：可得 7 , 27.4 a kpa a = = ∴1 1sin (tan ) sin (tan27.4 ) 31.2 f a- -= = = 78.18cos cos31.2ac kpaf= = =5-5 解：（1）砾砂粘聚力
15、c=0 2 2 2 23350 150( ) 2 100 ( 100) 250 100 22 2 2z x z xs s s ss+ - + = + = + - = 1 3250 100 2 391.4 , 250 100 2 108.6 kpa kpa s s = + = = - =M 点处于极限平衡状态，则 1 11 31 3sin sin 0.5656 34.4s sfs s- -= = = + （2）求大主应力方向：100*2tan2 1350 1502z xtas s= = =- 2 45 , 22.5 a a = = 由于裂开面与最大主应力面成 45°+Φ/2 的夹
16、角，故：45 /2 22.5 45 34.4 /2 84.7 b a f = + + = + + = 滑裂面通过 M 点的方向如图： 5-6 解：3 1 3( ) u A s s s = D + -12350 0.2*85 67100 0.2*83 116.5150 0.2*87 167.4u kpau kpau kpa= + = + = + = 试件：3 3 1 1 1 133 , 118 u kpa u kpa s s s s = - = = - =试件：3 3 2 1 1 233.4 , 116.4 u kpa u kpa s s s s = - = = - =试件：3 3 3 1 1
17、332.6 , 119.6 u kpa u kpa s s s s = - = = - = 5-7 解：由图可知 1 3 1 3sin ( ) sin * cos2 2uC cctg cs s s sf f f f + + = + = +1 32uCs s -=即1 32uC s s = +3( )sin cosu uC C c s f f = + +3sin cos1 sinucCfs ff += - 5-10 解：σ 3 等于常量，增大σ 1 直至试件剪切破坏 当起先固结1 3 3 1 332, 02 2 2P qs s s s ss+ -= = = = =当起先剪
18、切时，σ 3 等于常量 3 1 1 1 32 1 11 3 1 1 32 1 112 2 212 2 2P P Pq q qs s s s sss s s s ss+ +D +D = - = - =- +D -D = - = - = p-q 坐标上的三轴试验应力路径为：σ 1 等于常量，减小σ 3 直至试件剪切破坏 ，固结同剪切过程，σ 1 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：01 sin 0.357 K f = - =主动土压力系数：2tan (45 /2) 0.217aK f = - =被动土压力系数：2tan (4
19、5 /2) 4.6pK f = + =静止土压力：20 0180.33 /2E H K kN m g = =主动土压力：2148.8 /2a aE H K kN m g = =被动土压力：211035 /2p pE H K kN m g = =20 d = 时：主动土压力系数为：0.199aK =主动土压力：2144.775 /2a aE H K kN m g = =6-2：解：（1）2tan (45 /2) 0.455aK f = - =2 20*0.455 18*0.455 2*16 0.455 8.19 12.5a a a ap qK zK c K z z g = + - = + - =
20、 -z 0 1 1.53 2 3 4 5 6 pa 0 0 0 3.88 12.07 20.26 28.45 36.64 （2）01( ) 0.5*36.64*(6 1.53) 81.92a zE p H z = - = - =作用点在 z=4.51m 处 （3）01.53 z m =6-4：解：查表得：0.236aK =水位以上土压力：a ap zK g =水位以下土压力：1( 1.5)a a ap z K H K g g = - +结果如下：z 0 1 1.5 2 3 4 5 pa 0 4.248 6.372 7.67 10.266 12.862 15.45804.2486.3727.67
21、10.26612.86215.458-6-5-4-3-2-100 5 10 15 2005152535-6-5-4-3-2-100 10 20 30 40 主动土压力分布图水压力分布图 水压力：1( )w wp z H g = - 结果如下：z 1.5 2 3 4 5 pw 0 5 15 25 356-5：解：22 2cos ( )0.755sin( )*sin( )cos *cos( )*1 cos( )*cos( )aKf af d f ba a da d a b-= =+ -+ + - 21278.1422a a aE H K qHK g = + =方向与水平向呈 64 度角，指向墙背；
22、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于地面故θ=0 tan 0.364 i a = =土坡的稳定平安系数 / cos sin( )tan0.755sin cos( )wswV iVFV iVg a g a q fg a g a q- -= =+ - 7-2：解：从无限长坡中截取单宽土柱进行稳定分析，单宽土柱的平安系数与全坡相同 土柱重量：W H g =沿基面滑动力：sin T W a =沿基面抗滑力：cos tan R W a f =粘性土的粘聚力：* /coscF c l cb a = =cos tan /cossincsR F W cbF
23、T Wa f aa+ += =1.0, 7.224sF H m = = 又 则7-3：解：3(1 )* 19.31 /1ssatG wg kN meg+= =+ tansin 0.342/cosh bis baaaD= = = =D 9.8*4*0.342 13.41wJ Aj H i kN g = = = =平安系数：cos tan1.097sincsH FFH Jg a fg a += = +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Φ的增加，N r ，N q ，N
24、 c 增加很大，承载力也增大许多。（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 （3）由极限承载力公式uv192q c rP N cN BN g = + + 可知，基础宽度的增加会引起承载力的增加。8-2：解：均布荷载219*2 38 / q D kN m g = = =查表可得 1.51, 4.65, 10.0r q cN N N = = =极限承载力21 1*19*3*1.51 38*4.65 10*10 299.4 /2 2vh r q cP BN qN cN kN m g = + + = + +
25、 = 8-3：解：（1）地基产生整体剪切破坏时， 12u r q cP BN qN cN g = + +查表得：5.0, 7.5, 18.0r q cN N N = = =21(19.2 9.8)*2.4*5.0 18.4*2*7.5 8*18.0 476.4 /2uP kN m = - + + =（2）局部剪切破坏时：22 3u r q cBP N qN cNg = + +查图 8-18 可得：0.5, 4.0, 12.0r q cN N N = = =21*(19.2 9.8)*2.4*0.5 18.4*2*4.0 8*12.0 248.84 /2uP kN m = - + + = 8-5
26、：解:（1）自然容重：301.79*9.8 17.54 / kN m g = =浮容重：3(1.96 1)*9.8 9.41 / kN m g =- =（2）求公式 8-57 中承载力系数3.14* 256.672525 3.14/2 *22 360cctg ctgNctg ctgp fpf f p= = =- + - + 1 1 4.112q cN N tgctgpfpf f= + = + =- + 1( )411.5622rNctgppf f= =- + 1( )322.0732rNctgppf f= =- + （3）求1 14 3, ,crP P P2022.5*1.5*4.11 15*
27、6.67 208.18 /cr q cP DN cN kN m g = + = + =211 1( )4 41*9.41*3*1.56 208.18 230.20 /2 2crrBP N P kN mg= + = + =211 1( )3 31*9.41*3*2.07 208.18 237.40 /2 2crrBP N P kN mg= + = + =按普朗德尔理论：2 2 2525(45 ) (45 ) 10.662 2tg tgqN tg e tg ep f pf= + = + =( 1) 25 20.7c qN N tg = - =2022.54*1.5*10.66 15*20.7 590.96 /u q c q cP qN cN DN cN kN m g = + = + = + =按太沙基理论：2u r c qBP N cN qNg = + +查图 8-18 有：4.5, 26.0, 13.0r c qN N N = = =20.5*9.41*3*4.5 15*26.0 17.54*1.5*13.0 795.55 /uP kN m = + + =（4）12u r r r q q q c c cP BN S d qN S d cN S d g = + +13.0*31 1 1.187526.0*8qccN BSN L= + = + =