清华版土力学与地基基础(陈希哲)课后习题答案ppt课件

1、习 题土木工程专业建筑工程方向3.1P1340cokPahca0 .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)Zoabc27.060.8478.48132.48天

2、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(P)解：kPahniiic48.59) 1 . 18 . 4(*)101 .20(1 . 1*1 .2013.3(P)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)50.0,0(),(f

3、bzbxfbz50. 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(P)b/2b/2L5b解：取一半L/b10，按均布条形荷载边点下思索z/b=

4、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=10.4kPakPao1

5、00z50)(kPazbz25. 0bz5 . 0bz0 . 1bz0 . 6bz0 . 21 .480 .415 .274 .103 .153.5(P)ep曲线0.920.9250.930.9350.940.9450.950.9550.960.9650.97050100150200250300ep1112212116. 000016. 0100936. 0952. 0MPakPappeeaMPaaeEs2 .1216. 0952. 0112112112115 . 01 . 0MPaaMPa中紧缩性土3.6(P)14.0m5.0m5.0m6.0m根底中心点下1034.0,0.25/10/,4.

6、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(P)6.0m0.25mP=2400kN25. 00 . 16/66/eb0.666*1622hbW100400625. 0*240062400maxminWMAGFPo3.0m9.0m300500100500300kPapo=300kP

7、a, 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/b=0.5, z/b=3.0, 4=0.10

8、, 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的均布荷载x/b=0.5, z/b=3.0, 3

9、=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=63.3kPapo=200kPa, b=6.0m, z=9.0

10、m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(P)d1N1b1b1粉土31/20mkN12125. 0MPaa6b12N12b132/19mkN12150. 0MPaa粘土问两根底的沉降量能否一样

11、？何故？经过整d和b，能否使两根底沉降量接近？阐明有几种方案，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层土内中心点下的沉降差osbbpEbbs1)0(11)70(112

12、1117)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212ssEE111122336.45.0*168.2sosoEpbEpbs111122054.

13、3)1306. 1336. 4(sosoEpbEpb2、1根底间的沉降差1111*3384.3)2054.3133.0(sosoEpbEpb最有效的方法：调整d，加大根底2的埋深。调整基底宽度b，加大根底1的b，亦可使0202220111121dbNdbNppoo3.9(P)4*102*2010*1010*4*10*1036008000dAGFpmokPa400s3.10(P)2.0mF4.0m4.0人工填土31/5 .17mkN1.6m32/16mkN160. 0MPaa粘土0.11e卵石dAGFpmo2*5 .174*6 . 5) 2*4*6 . 5*20(6600kPa)64.299(6

14、4.259mzi6 . 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 . 1441bzi749.01i)(11iiiisozzEpS粘土粘土)749. 0*0

15、. 4629. 0*6 . 5(1000*333. 3)64.299(64.259mmm)3 .47( 1 .41)0473. 0(0410. 03.11(P)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)14.0m2.00.0840100.8025.0m2.50.060472.4

16、836.0m3.00.044753.6447.0m3.50.034441.28L/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(P)1.0mN=900kN2.0m3/0 .16mkN14 . 0MPaa粉质粘土dAGFpmo1*0 .162*6 . 31*2*6 . 3*20900kPa0 .129用规范法计算根底中心点的最终沉降量

17、。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 .129m025. 0014. 0000867. 005929. 0000867. 0inss

18、msi06016.0000867.005929.0)5 .500006016.0424.0*5 .5*0 .129(kPaEAAEsiiisMPaEs0 . 5kPafak200设9 . 0s4.0 7.01.0 0.7mmmssis5405414.006016.0*9 .03.13(P)1.5mN=576kN2.0m4.4m杂填土32/0 .18mkN6.5m33/16mkN卵石31/0 .17mkN粉土MPaEs0 . 32MPaEs0 .203用规范法计算根底中心点的最终沉降量。天然地面dAGFpmo5 . 1*0 .170 . 2*0 . 25 . 1*0 . 2*0 . 2*2057

19、6kPa5 .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. 000021. 00902. 000021. 0insskPaEAAEsiiis75.30370002

20、1.00902.03935.0*7 .4*5 .148kPafak200设06. 1s2.5 4.01.10 1.0MPaEs038.3mmmssis9609583. 0)00021. 00902. 0( *06. 13.14(P)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用规范法计算根底中心点的最终沉降量。天然地面6 . 0LIMPaEs0 . 32粉土5.8mMPaEs0 .254dAGFpmo5 . 1*0 .180 . 2*4 . 25 . 1*0 . 2*4 . 2

21、*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. 323 . 6bzn3325. 0nmzn0 . 61)(11iiiisonzzEpS)(00056.

22、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(P)4.0m8.0m粘土MPaEs5 . 7scmk/10*6 . 08kPapo214砂层kPaz

23、160砂层z计算地基沉降与时间的关系：需补充：a、e及t后才干计算。基底尺寸42.5m*13.5m4.1(P185)67119162216050100150200050100150200250300350400450ts67119162216050100150200050100150200250300350400450tskPac205 .0502045tano6.26kPakPacf1055 .132205 . 0*225tan未坏。4.2(P185)050100150200250050100150200250300350400450500550600650tskPa250210060023

24、1maxo45(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. 12003oooof845.612/69.33452/45kPao99.192845.64*2sin*2455. 12003kPa73.673491.331ooo

25、fo155.28845.6190904.4(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/N06.156.0110.3169.5233.0 125.0020406080100050100150200250300350te)(kPa)1001(mmkPa93f1tan62. 0tan)10*25(37510*

26、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/42.55980100.50255075100050100150200250300ts42.55980100.50255075100050100150200250300ts总应

27、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=3kPa, 由图知： =100， =59o25.2956.0tan100359tan11oo06.34)

28、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*95.7812破坏；4.10(P186)dctgdctgcpmmcr2/)(2.1*0.18180/152/15)2.

29、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*0.11180/302/30)4/0.3*110.4*0.110(ooctgkPa40.283mdmb0 .4,0

30、 .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添加而添加，相比之下，随埋深增大，临界荷载添加的更显著。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计算根据o25查P173图4.30得23cN11qN5rNcqmruNcNdNbp2123*0 .1011

31、*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35cN18qN20rNo30cqmruNcNdNbp21qmruNdNlAp2118*0 . 1*1920*11*21lApu2*12001200AKA设2bA 260018*0

32、.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)按汉森极限承载力公式计算根据查P179表4.5得35. 8cN47.2qN47. 0rNo10cqmruNcNdNbp2

33、135.8*0 .1047.2*0 .1*0 .1947.0*0 .3*0 .19*21kPa8 .14312 当地下水上升至根底底面时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*5kPa68.1232按浅根底地基的临界荷载公式41p查P165外表4.4得：0 . 1dN0.3cN041NbNcNdNpcmd4141kPa6.6900.16*

34、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 . 30 . 2*35. 0135. 01bdddqc2 . 01998. 03 .11tan8111tantanooo

35、462.0ri583.0ci680.0qi233. 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*2121222求产生自动土压 力所需的位移3求自动土压力值1密实砂土，产生自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*2(3)2596. 0)2/3645(tan)2/45(tan22oooaKmkNKHPaa/38.372596.0*0 .4*0 .1

36、8*2121225.2(P237)砂土3/18mkNo36m0 .2填土面m0 .2 地下水位面3/0 .21mkNsat1求静止土压力值2求自动土压力值 3求墙后水压力值1412.036sin1sin1ooKooKhe根据得：12301oekPaKheoo832.14412.0*0.2*1812kPaKhheowo896.23412.0*)2*0 .110 .2*18()(213m0 .2填土面m0 .2123832.14896.23mkNPo/56.530 . 2*)896.23832.14(*210 . 2*832.14*21(2)ooKhe根据得：2596. 0)2/3645(tan)

37、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5.1得：25. 0aKmkNKHEaa/288.36252.0*0.4*0.18*2121225.4(P237)根据0o30o20查表得：3548. 0aK

38、o12mkNKHEaa/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)(axaysEEGK可以3/5*85.70) 2/5 . 10 . 1 (*) 2/5*5 . 1*22(3/ 1*) 2/5*0 . 1*22(3/5ax

39、aytEbEaGK6 . 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求自动土压力和水压力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