土力学地基基础课后答案清华大学出版社陈希哲第四版

1、习 题答案土木工程专业（建筑工程方向）3.1（p134）0cokpahca0 .275 . 1*0 .1811kpahcacb84.606 . 3*)104 .19(0 .2722 kpahcbcc48.788 . 1*)108 .19(84.6033 上kpahcccc48.132) 8 . 16 . 3(*1048.78上下当第四层为强风化岩时：kpahcbcc48.788 . 1*)108 .19(84.6033 天然地面1.5m3.6m坚硬整体岩石层1.8m粉土=19.8kn/m=18.0kn/m地下水素填土=18.0kn/m中砂土sc(kpa)zoabc27.060.8478.481

2、32.48天然地面1.5m3.6m1.8m=18.0kn/m素填土中砂土粉土=18.0kn/m=19.8kn/m地下水强风化岩石sc(kpa)zoabc27.060.8478.483.2(p135)解：kpahniiic48.59) 1 . 18 . 4(*)101 .20(1 . 1*1 .2013.3(p135)解：kpapo100)50150(21)(21minmax0.1)0,0(),(fbzbxf96.0)25.0,0(),(fbzbxfkpapoz0.96100*96.0bz25. 00zkpapoz0.100100*0.1552.0)0 .1 ,0(),(fbzbxf82.0)5

3、0.0,0(),(fbzbxfbz50. 0kpapoz0.82100*82.0bz0 . 1kpapoz2.55100*552.0bz0 . 2306.0)0 .2,0(),(fbzbxfkpapoz6.30100*306.0208.0)0 .3 ,0(),(fbzbxfkpapoz8.20100*208.0bz0 . 3kpa50minkpa150max2/b2/bz100)(kpazbz25. 0bz50. 0bz0 . 1bz0 . 2bz0 . 3bz0 . 40 .960 .822 .556 .308 .200 .163.4(p135)b/2b/2l5b解：取一半l/b10，按均布

4、条形荷载边点下考虑z/b=0, x/b=0.5, =0.5, z=1.0*0.5*100=50.0kpaz/b=0.50, x/b=0.5, =0.481, z= 1.0* 0.496*100=48.1kpaz/b=1.0, x/b=0.5, =0.410, z= 1.0* 0.410*100=41.0kpaz/b=2.0, x/b=0.5, =0.275, z= 1.0* 0.275*100=27.5kpaz/b=4.0, x/b=0.5, =0.153, z= 1.0* 0.153*100=15.3kpaz/b=6.0, x/b=0.5, =0.104, z= 1.0* 0.104*100

5、=10.4kpakpao100z50)(kpazbz25. 0bz5 . 0bz0 . 1bz0 . 6bz0 . 21 .480 .415 .274 .103 .153.5(p135)ep曲线0.920.9250.930.9350.940.9450.950.9550.960.9650.97050100150200250300ep1112212116. 000016. 0100936. 0952. 0mpakpappeeampaaees2 .1216. 0952. 0112112112115 . 01 . 0mpaampa中压缩性土3.6(p135)14.0m5.0m5.0m6.0m基础中心点

6、下1034.0,0.25/10/,4.15/7/bzbloozopp1034. 0*44am点下：1350. 0, 25/10/, 0 . 45/20/1bzbl0947. 0, 25/10/, 2 . 15/6/1bzbloozapp*)0947. 01350. 0(*0 . 2*)(*0 . 221%487.191034. 0*4/ )0947. 01350. 0(*0 . 2/zoza3.7(p135)6.0m0.25mp=2400kn25. 00 . 16/66/eb0 .666*1622hbw100400625. 0*240062400maxminwmagfpo3.0m9.0m300

7、500100500300kpapo=300kpa, b=9.0m, z=9.0m的均布荷载x/b=0.5, z/b=1.0, 1=0.410, z1=0.41*300=123.0kpapo=400kpa, b=9.0m, z=9.0m的三角形荷载x/b=0.5, z/b=1.0, 2=0.25, z2=0.25*300=75.0kpa6.0m0.25m3.0m9.0m500100po=500kpa, b=3.0m, z=9.0m的均布荷载x/b=0.5, z/b=3.0, 3=0.198, z3=0.198*500=99.0kpapo=100kpa, b=3.0m, z=9.0m的三角形荷载x

8、/b=0.5, z/b=3.0, 4=0.10, z4=0.10*100=10.0kpaz= z1+ z2 z3-z4= 123.0+75.0-99-10.0=89.0kpa6.0m0.25mp=2400kn3.0m9.0m200500100po=200kpa, b=9.0m, z=9.0m的均布荷载x/b=0.5, z/b=1.0, 1=0.410, z1=0.41*200=82.0kpapo=300kpa, b=9.0m, z=9.0m的三角形荷载x/b=-0.5, z/b=1.0, 2=0.16, z2=0.16*300=48.0kpapo=200kpa, b=3.0m, z=9.0m的

9、均布荷载x/b=0.5, z/b=3.0, 3=0.198, z3=0.198*200=39.6kpapo=100kpa, b=3.0m, z=9.0m的三角形荷载x/b=-0.5, z/b=3.0, 4=0.10, z4=0.10*100=10.0kpaz= z1+ z2 z3-z4= 82.0+48.0-39.6-10.0=80.4kpa6.0m0.25mp=2400kn3.0m9.0m200500100po=300kpa, b=6.0m, z=9.0m的均布荷载x/b=6.0/6.0=1.0, z/b=9.0/6.0=1.5, 1=0.211, z1=0.211*300=63.3kpap

10、o=200kpa, b=6.0m, z=9.0m的三角形荷载x/b= 6.0/6.0=1.0, z/b=9.0/6.0=1.5, 2=0.0.13, z2=0.13*200=26.0kpa 直接按条形荷载计算（计算点位于大边下）：z= z1+ z2 =63.3+26.0=89.30kpa（计算点位于小边下）：x/b=- 6.0/6.0=-1.0, z/b=9.0/6.0=1.5, 2=0.09, z2=0.09*200=18.0kpaz= z1+ z2 =63.3+18.0=81.30kpa3.8(p135)d1n1b1b1粉土31/20mkn12125. 0mpaa6b12n12b132/1

11、9mkn12150. 0mpaa粘土问两基础的沉降量是否相同？何故？通过整d和b，能否使两基础沉降量接近？说明有几种方案，并给出评介。111120dbnpo11112222202220oopdbndbnposbpezs1)0(1111*在第1层土内中心点下的沉降差osbpezs1)0(2121*807.0),0(),0(111)0(11bbfbzfb940.0)2,0()2,0(111)0(21bbfbzfbososbpbepbes1111)0(111807.01ososbpbepbes1111)0(212940.0111111133.0)807.0940.0(sosoepbepb在第2层土内

12、中心点下的沉降差osbbpebbs1)0(11)70(1121117)7,0(),0(111)70(11bbfbzfb条形平均附加应力系数查不到，借用矩形面积上均布荷载角点下的平均附加应力系数2768. 00692. 0*4)5 . 07,10(4)5 . 0,10(4111)70(11bbfbzfb111121*1306.1807.02768.0*7soosepbpbesosbbpebbs2)0(21)70(2122117444.0)27,0()2,0(111)70(21bbfbzfb211222*168.294.0444.0*7soosepbpbes设212ssee111122336.45

13、.0*168.2sosoepbepbs111122054. 3)1306. 1336. 4(sosoepbepb2、1基础间的沉降差1111*3384.3)2054.3133.0(sosoepbepb最有效的方法：调整d，加大基础2的埋深。调整基底宽度b，加大基础1的b，亦可使0202220111121dbndbnppoo3.9(p135)4*102*2010*1010*4*10*1036008000dagfpmokpa400s3.10(p136)2.0mf4.0m4.0人工填土31/5 .17mkn1.6m32/16mkn160. 0mpaa粘土0.11e卵石dagfpmo2*5 .174*

14、6 . 5) 2*4*6 . 5*20(6600kpa)64.299(64.259mzi6 . 5mzi0 . 414.128.2bl8 . 226 . 5bzi1574. 0i求粘土层沉降量。0 . 220 . 41bzi1875.01i)(11iiiisozzeps粘土粘土)1875. 0*0 . 41574. 0*6 . 5(1000*333. 3)64.299(64.259*4mpaaees333.360.00.1111粘土mmm)2 .47( 1 .41)0472. 0(0410. 0用中心点下的系数：4.146.5bl4 . 146 . 5bzi629. 0i0 . 1441bzi

15、749.01i)(11iiiisozzeps粘土粘土)749. 0*0 . 4629. 0*6 . 5(1000*333. 3)64.299(64.259mmm)3 .47( 1 .41)0473. 0(0410. 03.11(p136)2.0mn=4720kn4.0m4.0细砂31/5 .17mkn3.0m32/16mknmpaes33. 32粉质粘土碎石dagfpmo2*5 .174*42*4*4*204720kpa0 .300用分层总和法计算粉质粘土层沉降量。mpaes0 . 814.5mmpaes0 .223分层，取层厚1.0m，各层面处的附加应力如下：序号zz/bz=4po (kpa

16、)14.0m2.00.0840100.8025.0m2.50.060472.4836.0m3.00.044753.6447.0m3.50.034441.28l/b=2.0/2.0=1.0附加应力计算表序号 zi (kpa) zi-1(kpa) z (kpa) si=zh/es172.48100.886.6426.02mm253.6472.4863.0618.94mm341.2853.6447.4614.25mms=si =59.21mm沉降量计算表 h=1.0m3.12(p136)1.0mn=900kn2.0m3/0 .16mkn14 . 0mpaa粉质粘土dagfpmo1*0 .162*6

17、. 31*2*6 . 3*20900kpa0 .129用规范法计算基础中心点的最终沉降量。0 . 11e8 . 12/6 . 3/bl取mzi2 . 56 . 222 . 5bzi442. 0impaaees0 .540.00 .1111)(111iiiisozzeps)(05929. 0442. 0*2 . 5*1000*0 . 50 .129m返算厚度：mbfz3 . 0)(取：mzi5 . 575. 225 . 5bzi424. 0imzi2 . 51)(11iiiisonzzeps)(000867. 0)442. 0*2 . 5424. 0*5 . 5(*1000*0 . 50 .12

18、9m025. 0014. 0000867. 005929. 0000867. 0inssmsi06016.0000867.005929.0)5 .500006016.0424.0*5 .5*0 .129(kpaeaaesiiismpaes0 . 5kpafak200设9 . 0s4.0 7.01.0 0.7mmmssis5405414.006016.0*9 .03.13(p136)1.5mn=576kn2.0m4.4m杂填土32/0 .18mkn6.5m33/16mkn卵石31/0 .17mkn粉土mpaes0 . 32mpaes0 .203用规范法计算基础中心点的最终沉降量。天然地面dagf

19、pmo5 . 1*0 .170 . 2*0 . 25 . 1*0 . 2*0 . 2*20576kpa5 .1480 . 10 . 2/0 . 2/blmz4 . 412 . 224 . 41bz414. 0i)(111iiiisozzeps)(0902.0414.0*4 .4*1000*0 .35 .148m返算厚度：mbfz3 . 0)(取：mzi7 . 435. 227 . 4bzi3935. 0imzi4 . 41)(11iiiisonzzeps)(00021. 0)414. 0*4 . 43935. 0*7 . 4(*1000*0 .205 .148m025. 00022. 0000

20、21. 00902. 000021. 0insskpaeaaesiiis75.303700021.00902.03935.0*7 .4*5 .148kpafak200设06. 1s2.5 4.01.10 1.0mpaes038.3mmmssis9609583. 0)00021. 00902. 0( *06. 13.14(p136)1.5mn=706kn2.0m2.5m粉质粘土32/0 .17mkn6.6m33/0 .20mkn卵石31/0 .18mkn粘土0 . 1empaes0 . 53用规范法计算基础中心点的最终沉降量。天然地面6 . 0limpaes0 . 32粉土5.8mmpaes0

21、.254dagfpmo5 . 1*0 .180 . 2*4 . 25 . 1*0 . 2*4 . 2*20706kpa08.1502 . 10 . 2/4 . 2/blmz5 . 2125. 125 . 21bz648. 0i1111zepsso)(0810. 0648. 0*5 . 2*1000*0 . 308.150mmz0 . 62mz5 . 210 . 320 . 62bz346. 02)(112222zzepsso)(00137. 0)648. 0*5 . 2346. 0*0 . 6(*1000*0 . 508.150m返算厚度：mbfz3 . 0)(取：mzn3 . 615. 32

22、3 . 6bzn3325. 0nmzn0 . 61)(11iiiisonzzeps)(00056. 0)346. 0*0 . 63325. 0*3 . 6(*1000*0 . 508.150m025. 00068. 000056. 000137. 00810. 000056. 0insskpaeaaesiiis9 .379008293.03325.0*3 .6*08.150msi08293. 000056. 000137. 00810. 0kpafak200设014. 1s2.5 4.01.10 1.0mpaes79.3mmmssis8408409. 008293. 0*014. 13.15(

23、p136)4.0m8.0m粘土mpaes5 . 7scmk/10*6 . 08kpapo214砂层kpaz160砂层z计算地基沉降与时间的关系：需补充：a、e及t后才能计算。基底尺寸42.5m*13.5m4.1(p185)67119162216050100150200050100150200250300350400450ts67119162216050100150200050100150200250300350400450tskpac205 .0502045tano6.26kpakpacf1055 .132205 . 0*225tan未坏。4.2(p185)050100150200250050

24、100150200250300350400450500550600650tskpa2502100600231maxo45(1)(2)(3)ooo6030902cos223131kpao22560*2cos210060021006002sin231kpao5 .21660*2sin21006004.3(p185)(1)tanfof69.33300200tantan11(2) 2/45(tan231o)2/69.3345(tan23oo31491.3off845.64*2sin2491.32sin23331o845.64*2sin2455. 12003oooof845.612/69.33452/4

25、5kpao99.192845.64*2sin*2455. 12003kpa73.673491.331ooofo155.28845.6190904.4(p186)300180031kpa2303kpa1870112224.5(p186)170kpao37kpao77.13537cos170kpao31.10237sin170kpakpaof31.10239.17810030tan77.135不会破坏。4.6(p186)剪切位移1/100(mm)040100140180240320剪力/n06.156.0110.3169.5233.0 125.002040608010005010015020025

26、0300350te)(kpa)1001(mmkpa93f1tan62. 0tan)10*25(37510*93tan1431o80.314.7(p186)iichkpa0 .800 . 6*0 . 80 . 2*0 .16kpac0 .2080*25. 025. 0nknp6006.010*30*0.2041np1202np0.18010*30*80*75.013np0.24010*30*80*0.1144.8(p186)14521831040160100150200315792126k线表示法：kpa/1kpa/3kpa/42.55980100.50255075100050100150200

27、250300ts42.55980100.50255075100050100150200250300ts总应力42.55980100.542.55980100.50255075100050100150200250300ts总应力42.55980100.542.55980100.50255075100050100150200250300ts总应力有效应力总应力： a=13kpa由图知：=125， =49o07.16288. 0tan1251349tan11cos casintanoo73.16)07.16(tansin)(tansin11kpaaco58.1373.16cos13cos有效应力：a

28、=3kpa, 由图知： =100， =59o25.2956.0tan100359tan11oo06.34)25.29(tansin)(tansin11kpaaco62. 306.34cos3cos4.9(p186)22)2(213xzxz8517540)2100250(21002502213)2/45(tan231okpakpaoo260270)2/3045(tan*9012未坏；当kpa60时，05.9617560)2100250(2100250221326090kpa05.27195.78kpa)2/45(tan231okpakpaoo26085.236)2/3045(tan*95.781

29、2破坏；4.10(p186)dctgdctgcpmmcr2/)(2.1*0.18180/152/15)2.1*0.181525(oooctgctgkpa46.1704.11(p187)dctgbdctgcpmm2/)4/(410.2*0.11180/302/30)4/0.3*110.2*0.110(ooctgkpa59.160mdmb0 .2,0 .3(1)(2)dctgbdctgcpmm2/)4/(410.2*0.11180/302/30)4/0.6*110.2*0.110(ooctgkpa4 .198mdmb0 .2,0 .6(3)dctgbdctgcpmm2/)4/(410.4*0.11

30、180/302/30)4/0.3*110.4*0.110(ooctgkpa40.283mdmb0 .4,0 .3kpa59.160b=3.0,d=2.0 b=6.0,d=2.0b=3.0,d=4.041pkpa4 .198kpa40.283地基的临界荷载随基底宽度与埋深的增加而增加，相比之下，随埋深增大，临界荷载增加的更显著。4.12(p187)dctgbdctgcpmm2/)4/(412 .1*0 .20180/252/25)4/0 .2*0 .202 .1*0 .2025*10(oooctgctgkpa27.196mdmb2 .1,0 .2kpap2502500不满足按太沙基极限承载力公式

31、计算根据o25查p173图4.30得23cn11qn5rncqmruncndnbp2123*0 .1011*2 .1*0 .205*0 .2*0 .20*21kpa0 .59438.22500 .594k按汉森极限承载力公式计算根据o25查表4.5得79.20cn72.10qn21. 8rn79.20*0 .1072.10*2 .1*0 .2021.8*0 .2*0 .20*21upkpa38.62952. 225038.629k4.13(p187)knkapu1200按太沙基极限承载力公式计算根据查p173图4.30得35cn18qn20rno30cqmruncndnbp21qmrundnl

32、ap2118*0 . 1*1920*11*21lapu2*12001200aka设2ba 260018*0.1*1920*11*21bb045.511.323bb075. 130911. 3392221bbp22. 345. 511. 3*272*31*272332131bbbbq032213bxbxbx0350. 1)075. 1 ()222. 3()2(3232pqd772. 2350. 1222. 323dqu405. 1u448. 0350. 1222. 323dqv765. 0v)(1333. 13/11. 3765. 0405. 13/1mbvub4.14(p187)按汉森极限承载

33、力公式计算根据查p179表4.5得35. 8cn47.2qn47. 0rno10cqmruncndnbp2135.8*0 .1047.2*0 .1*0 .1947.0*0 .3*0 .19*21kpa8 .143（1）（2） 当地下水上升至基础底面时35.8*0 .1047.2*0 .1*0 .1947.0*0 .3*0 .9*21upkpa8.1364.15(1)极限荷载按斯肯普顿极限承载力公式计算dbdlbcpmu)2.01()2.01(52 .1*0 .18)4 .22 .1*2 .01 ()0 .34 .2*2 .01 (*0 .16*5kpa68.123（2）按浅基础地基的临界荷载公

34、式41p查p165表面4.4得：0 . 1dn0.3cn041nbncndnpcmd4141kpa6.6900.16*0.32.1*0.18*0.14.16(p187)按汉森极限承载力公式计算根据查p179表4.5得：62.11cn33. 4qn72. 1rno16ccccqqqqmrrrudisncdisndisnbp21233. 1*68. 0*15. 1*33. 4*0 . 2*6 .18462. 0*7 . 0*72. 1*0 . 3*6 .18*21kpa68.2477 . 043*4 . 014 . 01lbsr15. 143*2 . 012 . 01lbssqc233. 10 .

35、 30 . 2*35. 0135. 01bdddqc2 . 01998. 03 .11tan8111tantanooo462.0ri583.0ci680.0qi233. 1*583. 0*15. 1*62.11*0 . 8干砂3/18mkno36m0 .45.1(p236)填土面（1）求静止土压力值412.036sin1sin1ookmknkhpoo/33.59412. 0*0 . 4*0 .18*212122（2）求产生主动土压 力所需的位移（3）求主动土压力值（1）密实砂土，产生主动土压力所需的位移约为墙高的0.5%，即：mmmh0 .2002. 0%5 . 0*0 . 4%5 . 0*（

36、2）(3)2596. 0)2/3645(tan)2/45(tan22oooakmknkhpaa/38.372596. 0*0 . 4*0 .18*2121225.2(p237)砂土3/18mkno36m0 .2填土面m0 .2 地下水位面3/0 .21mknsat（1）求静止土压力值（2）求主动土压力值 （3）求墙后水压力值（1）412.036sin1sin1ookookhe根据得：12301oekpakheoo832.14412.0*0.2*1812kpakhheowo896.23412.0*)2*0 .110 .2*18()(213m0 .2填土面m0 .2123832.14896.23m

37、knpo/56.530 . 2*)896.23832.14(*210 . 2*832.14*21(2)ookhe根据得：2596.0)2/3645(tan)2/45(tan22oooak01aekpakheaa346.92596.0*0.2*1812kpakhheawa057.152596. 0*)2*0 .110 . 2*18()(213m0 .2填土面m0 .2123346.9057.15mknpa/75.330 . 2*)057.15346. 9(*210 . 2*346. 9*21(3)mknhpwww/0 .200 . 2*10*2121225.3(p237)根据0o36o24查表5

38、.1得：25. 0akmknkheaa/288.36252.0*0.4*0.18*2121225.4(p237)根据0o30o20查表得：3548. 0ako12mknkheaa/40.753548.0*0 .5*0 .17*212122mkneoax/85.7020cos*4 .75mkneoay/79.2520sin*4 .755.5(p237)1.5m2.5mo125.0mmkneoax/85.7020cos*4 .75mkneoay/79.2520sin*4 .755/3axeaye3 . 139. 185.704 . 0*79.255 . 0*5*) 5 . 25 . 1 (*22)

39、(axayseegk可以3/5*85.70) 2/5 . 10 . 1 (*) 2/5*5 . 1*22(3/ 1*) 2/5*0 . 1*22(3/5axaytebeagk6 . 138. 1不可以5.6(p237)3.0mkpaq20中砂31/5 .18mkno301粗砂32/0 .19mkno35132/0 .20mknsat3.0m地下水4.0m求主动土压力和水压力123kpakqpaa666.6333.0*0 .200111112*)*(aakhqp上333. 0) 2/3045(tan) 2/45(tan2121oooak271. 0) 2/3545(tan) 2/45(tan2222oooak4kpa17.25333.0*)0.3*5.1820(222113*)*(aakhhqp23222114*)*(aakhhhqp2112*)*(aakhqp下kpa46.20271.0*)0.3*5.1820(kpa91.35271.0*)0 .3*0 .190 .3*5 .1820(kpa75.46271. 0*)0 . 4