土力学地基基础 第四版 (陈希哲 著) 清华大学出版社 3—5章 课后答案 39页

'课后答案网您最真诚的朋友www.hackshp.cn网团队竭诚为学生服务，免费提供各门课后答案，不用积分，甚至不用注册，旨在为广大学生提供自主学习的平台！课后答案网：www.hackshp.cn视频教程网：www.efanjy.comPPT课件网：www.ppthouse.com课后答案网www.hackshp.cn 习题土木工程专业（建筑工程方向）khdaw.com3.1（课后答案网P134）σ=0coσ=γ⋅h=180.5.1*=270.kPacawww.hackshp.cn11σcb=σca+γ2′⋅h2=27.0+(194.−10*)3.6=60.84kPaσcc上=σcb+γ3′⋅h3=60.84+(198.−10*)1.8=78.48kPaσcc下=σcc上+γω′⋅hω=7848.+103(*.6+1.)8=132.48kPa当第四层为强风化岩时：σcc=σcb+γ3′⋅h3=6084.+(198.−10*)1.8=78.48kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 天然地面s(kPa)co1.5m地下水素填土=18.0kN/m27.0a3.6m粉土=18.0kN/m60.84b1.8m中砂土=19.8kN/m78.48khdaw.com132.48c坚硬整体岩石层Z课后答案网天然地面sc(kPa)o1.5m地下水素填土=18.0kN/m27.0www.hackshp.cna3.6m粉土=18.0kN/m60.84b1.8m中砂土=19.8kN/m78.48c强风化岩石Zkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.2(P135)n解：σc=∑γi⋅hi=20.1*1.1+(20.1−104(*).8−1.)1=59.48kPai=13.3(P135)11解：p=(σ+σ)=(150+50)=100kPaomaxmin22xzz=0α=f(,)=f(0,0)=1.0bbσ=α⋅p=1.0*100=100.0kPakhdaw.comzoz=.025bxzα=f(,)=f(0,0.25)=0.96bbσ=α⋅p=0.96*100=96.0kPazo课后答案网xzz=0.50bα=f(,)=f.0,0(50)=.082bbσz=α⋅po=0.82*100=82.0kPaz=www.hackshp.cn1.0bxzα=f(,)=f0(,1.)0=0.552bbσ=α⋅p=0.552*100=55.2kPazoz=2.0bα=f(x,z)=f0(,2.)0=0.306bbσ=α⋅p=0.306*100=30.6kPazoz=0.3bxzα=f(,)=f0(,3.)0=0.208bbσ=α⋅p=0.208*100=20.8kPazokhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com b/2b/2σ=150kPaσ=50kPamaxminσ(kPa)z100z=.025b96.0z=.050b82.0z=0.1b55.2khdaw.comz=0.2b30.6z=0.3b20.8z=4.0b16.0z3.4(P135)课后答案网L>5bb/2b/2www.hackshp.cn解：取一半L/b>10，按均布条形荷载边点下考虑z/b=0,x/b=0.5,α=0.5,σz=1.0*0.5*100=50.0kPaz/b=0.50,x/b=0.5,α=0.481,σz=1.0*0.496*100=48.1kPaz/b=1.0,x/b=0.5,α=0.410,σz=1.0*0.410*100=41.0kPaz/b=2.0,x/b=0.5,α=0.275,σz=1.0*0.275*100=27.5kPaz/b=4.0,x/b=0.5,α=0.153,σz=1.0*0.153*100=15.3kPaz/b=6.0,x/b=0.5,α=0.104,σz=1.0*0.104*100=10.4kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com σ=100kPaoσ(kPa)z50z=.025b48.1z=5.0b41.0khdaw.comz=0.1b27.5z=0.2b15.3z=6.0b10.4z3.5(P135)课后答案网e～p曲线e0.970.9650.960.955中压www.hackshp.cn0.95缩性0.9450.94土0.9350.930.9250.92p050100150200250300e1−e2.0952−.0936−1−1a===.000016kPa=.016MPa1−2p−p100211+e1+0.952E=1==12.2MPas1−2a0.161−2khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.6(P135)6.0m14.0m5.0mA5.0m基础中心点下L/b=7/5=1.4,z/b=10/5=2.0,α=0.1034σ=4⋅α⋅p=4*.01034pzoookhdaw.comM点下：L/b=205/=,0.4z/b=105/=,2α=.013501L/b=5/6=,2.1z/b=105/=,2α=.009471σ=0.2*(α−α)*p=0.2*.0(1350−.00947)*pzA12ooσ/σ=0.2*.0(1350−.009474/)*.01034=19.487%zAzo6.0mP=2400kN3.7(P135)课后答案网0.25m3.0m100b6/=6/6=0.1>e=.02522b⋅h1*6W===6.03005006www.hackshp.cn69.0mF+GM24002400*0.25Pmax=±=±=400±100ominAW66=500kPa300po=300kPa,b=9.0m,z=9.0m的均布荷载x/b=0.5,z/b=1.0,α1=0.410,σz1=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,α2=0.25,σz2=0.25*300=75.0kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 6.0m0.25m3.0m100500po=500kPa,b=3.0m,z=9.0m的均布荷载9.0mx/b=0.5,z/b=3.0,α3=0.198,σz3=0.198*500=99.0kPakhdaw.compo=100kPa,b=3.0m,z=9.0m的三角形荷载x/b=0.5,z/b=3.0,α4=0.10,σz4=0.10*100=10.0kPaσz=σz1+σz2–σz3-σz4=123.0+75.0-99-10.0=89.0kPa课后答案网3.0m6.0mP=2400kN0.25m100500200www.hackshp.cn9.0mpo=200kPa,b=9.0m,z=9.0m的均布荷载x/b=0.5,z/b=1.0,α1=0.410,σz1=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,α2=0.16,σz2=0.16*300=48.0kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.0m6.0mP=2400kN0.25m1005002009.0mkhdaw.compo=200kPa,b=3.0m,z=9.0m的均布荷载x/b=0.5,z/b=3.0,α3=0.198,σz3=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,α4=0.10,σz4=0.10*100=10.0kPaσz=σz1+σz2–σz3-σz4=82.0+48.0-39.6-10.0=80.4kPa直接按条形荷载计算课后答案网（计算点位于大边下）：po=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,αwww.hackshp.cn1=0.211,σz1=0.211*300=63.3kPapo=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,α2=0.0.13,σz2=0.13*200=26.0kPaσz=σz1+σz2=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,α2=0.09,σz2=0.09*200=18.0kPaσz=σz1+σz2=63.3+18.0=81.30kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.8(P135)2N1N1d1b12b1b13γ=20kN/m粉土1−1a=0.25MPa1−23γ=19kN/mkhdaw.com2粘6b1−1土a=0.50MPa1−2问两基础的沉降量是否相同？何故？通过整课后答案网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。Np=1−20⋅do11b1N2⋅N21po2www.hackshp.cn=−20⋅d2=−20⋅d1=po1b2⋅b21在第1层土内中心点下的沉降差z*αz*α0(2−b1)10(−b)s=1⋅ps=⋅p12o11oEEs1s1zb1α=f0(,)=f0(,)=0.807(10−b1)bb11zb1α=f0(,)=f0(,)=0.9400(2−b1)2⋅b2⋅b11khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com α1(0−b).0807s=1⋅b⋅p=⋅b⋅p111o1oEEs1s1α2(0−b1).0940s=⋅b⋅p=⋅b⋅p121o1oEEs1s1b⋅pb⋅p∆=.0(940−.0807)⋅1o=.0133⋅1o1EEkhdaw.coms1s1在第2层土内中心点下的沉降差课后答案网71bα1(0−7b1)−b1α1(0−b1)s=⋅p21oEs1z7bα=f(0,)=f(0,1)1(0−7b)www.hackshp.cn1bb11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z7b1α=4f(10,)=4f(10,)=4*0.0692=0.27681(0−7b)10.5b0.5⋅b117*.02768−.0807b⋅ps=⋅b⋅p=.11306*1o211oEEs1s1khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 7bα−bα12(07−b)12(0−b)s=11⋅p22oEs2z7bα=f(0,)=f(0,1)=0.4442(0−7b)12b2b117*.0444−.094b⋅ps=⋅b⋅p=.2168*1o221oEEs2s2设E≈2Ekhdaw.coms1s2b⋅pb⋅p1o1os=2.168*=4.336225.0EEs1s1b⋅pb⋅p∆=4(.336−1.1306)⋅1o=3.2054⋅1o2EEs1s12、课后答案网1基础间的沉降差b1⋅pob1⋅po∆=0(.133+3.2054)⋅=3.3384*EEwww.hackshp.cns1s1最有效的方法：调整d，加大基础2的埋深。N2⋅N亦可使p=p=1−20⋅d=1−20⋅d=0o1o2b2⋅b11调整基底宽度b，加大基础1的b，3.9(P135)F+G8000+3600−10*10*4*10p=−γ⋅d=−20*2−10*4omA10*10=−4kPa≈0s≈0khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.10(P136)FF+Gpo=−γm⋅d2.0mA6600+(20*56.*4*)24.0m=−17.5*24.05.6*43人工填土γ1=17.5kN/m=259.64(299.64)kPa粘土γ=16kN/m3z=6.5mz=0.4m2ii−1khdaw.com1.6ma=0.60MPa−1e=1.01l2.8==1.4b2卵石zi5.6==2.8b2求粘土层沉降量。α=.01574izi−课后答案网14.0==2.0α=.01875i−1b21+e1+1.0E=1==3.333MPas粘土www.hackshp.cna0.60po∆S=⋅(α⋅z−α⋅z)粘土iii−1i−1Es粘土4*259.64(299.64)=⋅5(.6*0.1574−4.0*0.1875)3333.*1000=.00410.0(0472)m=41(1.47)2.mm用中心点下的系数：l5.6zi=5.6=1.4α=.0629==1.4ib4b4khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com zi−14==1.0α=.0749i−1b4po∆S=⋅(α⋅z−α⋅z)粘土iii−1i−1Es粘土259.64(299.64)=⋅6.5(*.0629−0.4*.0749).3333*1000=.00410.0(0473)m=41(1.47)3.mmkhdaw.com3.11(P136)课后答案网N=4720kNF+Gpo=−γm⋅d2.0mwww.hackshp.cnA4720+20*4*4*24.0m=−17.5*24*44.03γ=17.5kN/m细砂1=3000.kPaE=8.0MPas13粉质γ2=16kN/m3.0m分层，取层厚粘土E=3.33MPas21.0m，各层面处的附加碎石E=22.0MPa4.5ms3应力如下：用分层总和法计算粉质粘土层沉降量。khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 附加应力计算表L/b=2.0/2.0=1.0序号zz/bασz=4αpo(kPa)14.0m2.00.0840100.8025.0m2.50.060472.4836.0m3.00.044753.6447.0m3.50.034441.28沉降量计算表h=1.0mkhdaw.com序号σzi(kPa)σzi-1(kPa)σz(kPa)ΔSi=σzh/Es172.48100.886.6426.02mm253.6472.4863.0618.94mm341.2853.6447.4614.25mmS=ΣSi=59.21mm3.12(P136)课后答案网N=900kNF+Gpo=−γm⋅d1.0mwww.hackshp.cnA900+20*6.3*21*粉2.0m=−160.1*质*6.323粘γ=16.0kN/m=1290.kPae=0.1土11+e1+1.01−1E===5.0MPaa=0.4MPasa0.40用规范法计算基础中L/b=2/6.3=8.1心点的最终沉降量。取z=2.5miz5.2iα=.0442==2.6ib2khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com po∆S=⋅(α⋅z−α⋅z)1iii−1i−1Es1290.=*2.5*.0442=.005929(m)0.5*1000返算厚度：∆z=f(b)=0.3m取：z=5.5mz=2.5mii−1zi5.5==2.75α=.0424ikhdaw.comb2po∆S=⋅(α⋅z−α⋅z)niii−1i−1Es1290.=5.5(**.0424−2.5*.0442)=.0000867(m)0.5*1000课后答案网∆s.0000867n==.0014<.0025∑∆si.005929+.0000867∑∆www.hackshp.cnsi=.005929+.0000867=.006016m4.07.0E=0.5MPa设f=200kPaψs=0.9sak1.00.7∑Ai129.0*5.5*0.424(E===5000.5kPa)sAi0.06016∑Esis=ψs⋅∑∆si=9.0.0*06016=.005414m=54mmkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.13(P136)N=576kN天然地面31.5m杂填土γ1=17.0kN/m2.0m粉土4.4m3E=3.0MPaγ=18.0kN/ms22卵3E=20.0MPaγ=16kN/mkhdaw.com6.5m3s3石用规范法计算基础中心点的最终沉降量。课后答案网F+Gpo=−γm⋅dA576+20*2.0*2.0*1.5=−17.0*1.5www.hackshp.cn2.0*2.0=1485.kPaL/b=0.2/0.2=0.1z14.4z=4.4m==2.2α=.04141ib2po∆S=⋅(α⋅z−α⋅z)1iii−1i−1Es1485.=*4.4*.0414=.00902(m)0.3*1000khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 返算厚度：∆z=f(b)=0.3m取：z=7.4mz=4.4mii−1zi4.7==2.35α=.03935ib2po∆S=⋅(α⋅z−α⋅z)niii−1i−1Es148.5=4(*.7*0.3935−4.4*0.414)=000021.(m)khdaw.com20.0*1000∆s.000021n==.00022<.0025∑∆si.00902+.000021∑Ai148.5*4.7*0.3935E===3037.75kPasAi0.0902+0.00021∑Esi课后答案网2.54.0E=3.038MPa设f=200kPaψs=1.06sak1.101.0s=ψs⋅∑∆si=.106.0(*0902+.000021)=.009583m=96mmwww.hackshp.cnkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.14(P136)N=706kN天然地面31.5m粉质粘土γ1=18.0kN/m2.0m粘土e=1.0I=6.0L2.5m3γ=17.0kN/m2E=3.0MPas2粉土3E=5.0MPaγ=20.0kN/mkhdaw.com6.6m3s35.8m卵E=25.0MPas4石用规范法计算基础中心点的最终沉降量。课后答案网F+Gpo=−γm⋅dA706+20*2.4*2.0*1.5=−18.0*1.5www.hackshp.cn2.4*2.0=150.08kPaL/b=0.2/4.2=2.1z12.5z=5.2m==1.25α=.06481ib2po150.08∆S1=⋅α1⋅z1=*2.5*0.648=0.0810(m)Es13.0*1000z26.0z=0.6mz=5.2m==3.0α=.0346212b2khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com po∆S=⋅(α⋅z−α⋅z)22211Es2150.08=6(*.0*0.346−2.5*0.648)=000137.(m)5.0*1000返算厚度：∆z=f(b)=3.0m取：z=3.6mzn−1=0.6mnzn=6.3=3.15α=.03325nkhdaw.comb2po∆S=⋅(α⋅z−α⋅z)niii−1i−1Es150.08=6(*.3*0.3325−6.0*0.346)=000056.(m)5.0*1000∆sn课后答案网.000056==.00068<.0025∑∆si.00810+.000137+.000056∑∆si=0.0810+0.00137+0.00056=0.08293mwww.hackshp.cn∑Ai150.08*6.3*0.3325E===3790.9kPasAi0.08293∑Esi2.54.0E=3.79MPa设f=200kPaψ=1.014saks1.101.0s=ψs⋅∑∆si=.1014.0*08293=.008409m=84mmkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 3.15(P136)砂层4.0mp=214kPao粘土基底尺寸8.0mE=7.5MPa42.5m*13.5msσz−8k=0.*610cm/skhdaw.comσz=160kPa砂层计算地基沉降与时间的关系：需补充：a、e及t后才能计算。4.1(P185)课后答案网t216tc=20kPa20021620045−20162tanϕ==0.5150www.hackshp.cn16250150119119oϕ=26.6100100676750500ss05050100100150150200200250250300300350350400400450450τ=σ⋅tanϕ+c=225*05.+20=132.5kPa>τ=105kPaf未坏。khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 4.2(P185)t250200150100500s050100150200250300350400450500550600650(1)σ1−σ3600−100(2)α=45oτ===250kPamaxkhdaw.com22(3)α=90o−30o=60oσ+σσ−σσ=13+13⋅cos2αα22600+100600−100o=+⋅cos2*60=225kPa22τ课后答案网=σ1−σ3⋅sin2αα2600−100o=⋅sin2*60=216.5kPa24.3(P185)www.hackshp.cnτ200−1f−1o(1)τ=σ⋅tanϕϕ=tan=tan=33.69fσ300(2)tan2(45o)2/2ooσ=σ⋅+ϕ=σ⋅tan(45+33.69)2/133σ=3.491⋅σ<1>13ooooαf=45+ϕ2/=45+33.692/=61.845σ1−σ3.3491⋅σ3−σ3oταf=⋅sin2αf=⋅sin2*64.84522o200=.12455⋅σ⋅sin2*64.8453khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 200σ==192.99kPa3o1.2455*sin2*64.845σ=3.491⋅σ=673.73kPa13ooooα=90−α=90−61.845=28.155f4.4(P186)σ=230kPa3σ=1870kPa1khdaw.com2β2σ2β11σ33001800170kPa4.5(P186)课后答案网oθ=37www.hackshp.cnoσ=170⋅cos37=135.77kPaαoτ=170⋅sin37=102.31kPaαoτ=135.77⋅tan30+100=178.39kPa>τ=102.31kPafαα不会破坏。khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 4.6(P186)剪切位移0401001401802403201/100(mm)剪力/N06.156.0110.3169.5233.0125.0t(kPa)10093kPa8060khdaw.com402010e(mm)0501001502002503003501003−1τf=tan−193*10=−1=31.80oϕ=tan−4tan.062σ375÷(25*10)4.7(P186)课后答案网σc=∑γi⋅hi=160.*0.2+0.8*0.6=800.kPa0.25⋅σ=0.25*80=20.0kPawww.hackshp.cnc−4P=20.0*30*10=0.06kN=60N1P=120N2−1P=0.75*80*30*10=180.0N3−1P=1.0*80*30*10=240.0N4khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 4.8(P186)σ/kPa1452183104011σ3/kPa60100150200µ/kPa315792126K线表示法：ttt100100100100100.5100.5100.5100.5100.5100.5有效应力808080808080khdaw.com7575757559595959总应力5050总应力总应力42.542.542.542.52525250sss050100100150150150200200200250250250300300300总应力：课后答案网a=13kPa由图知：σ=125，τ=49−149−13−1oα=tan=tan0.288=16.07125tanwww.hackshp.cnα=sinϕ−1−1ooϕ=sin(tanα)=sin(tan16.07)=16.73a=c⋅cosϕc=a=13=13.58kPaocosϕcos16.73有效应力：a"=3kPa,由图知：σ"=100，τ"=59−159−3−1oα′=tan=tan0.56=29.25100−1−1ooϕ′=sin(tanα′)=sin(tan29.25)=34.06a′3c′===3.62kPaocosϕ′cos34.06khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 4.9(P186)σz+σxσz−σx22σ1=±()+τ322260250+100250−10022kPaσ1=±()+40=175±85=322902oσ1=σ3tan⋅(45+ϕ)2/2oo=90*tan(45+30)2/=270kPa>σ=260kPa1未坏；当τ=60kPa时khdaw.com250+100，250−10022271.05σ1=±()+60=175±96.05=kPa32278.952oσ=σ⋅tan(45+ϕ)2/132oo=78.95*tan(45+30)2/=236.85kPa<σ=260kPa1破坏；4.10(P186)课后答案网π⋅(c⋅ctgϕ+γ⋅d)mpcr=+γm⋅dctgϕ−π/2+ϕowww.hackshp.cnπ⋅(25⋅ctg15+180.*)2.1=+180.*2.1ooctg15−π/2+15⋅π/180=170.46kPa4.11(P187)(1)b=0.3m,d=0.2mπ⋅(c⋅ctgϕ+γ⋅d+γ⋅b/)4mp1=+γm⋅d4ctgϕ−π/2+ϕπ⋅0(+11.0*2.0+11*3.0/4)=+11.0*2.0ooctg30−π/2+30⋅π/180=160.59kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com (2)b=6.0m,d=2.0mπ⋅(c⋅ctgϕ+γ⋅d+γ⋅b/)4mp1=+γm⋅d4ctgϕ−π/2+ϕπ⋅0(+110.*0.2+11*0.6/)4=+110.*0.2ooctg30−π/2+30⋅π/180=1984.kPa(3)b=0.3m,d=0.4mπ⋅(c⋅ctgϕ+γ⋅d+γ⋅b/)4mp1=+γ⋅dkhdaw.comm4ctgϕ−π/2+ϕπ⋅0(+110.*0.4+11*0.3/)4=+110.*0.4ooctg30−π/2+30⋅π/180=283.40kPa课后答案网b=3.0,d=2.0b=6.0,d=2.0b=3.0,d=4.0p14160.59kPa1984.kPa283.40kPa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。www.hackshp.cn4.12(P187)500b=2.0m,d=1.2mp==250kPa2π⋅(c⋅ctgϕ+γ⋅d+γ⋅b/)4mp1=+γm⋅d4ctgϕ−π/2+ϕoπ⋅(10*ctg25+200.*2.1+200.*)4/0.2=+200.*2.1ooctg25−π2/+25⋅π/180=196.27kPa不满足khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 按太沙基极限承载力公式计算根据ϕ=25o查P173图4.30得Nc=23Nq=11Nr=51p=⋅γ⋅b⋅N+γ⋅d⋅N+c⋅Nurmqc21=*20.0*20.*5+20.0*12.*11+10.0*232=594.0kPa594.0K==2.38khdaw.com250按汉森极限承载力公式计算oN=20.79Nq=10.72Nr=.821根据ϕ=25查表4.5得c1p=*20.0*2.0*8.21+20.0*1.2*10.72+10.0*20.79u2629.38=629.38kPaK==2.522504.13(P187)课后答案网按太沙基极限承载力公式计算oN=35Nq=18Nr=20根据www.hackshp.cnϕ=30查P173图4.30得cp⋅A⋅K=1200kNu1p=⋅γ⋅b⋅N+γ⋅d⋅N+c⋅Nurmqc21Ap=⋅γ⋅⋅N+γ⋅d⋅Nurmq2l1A12001200p=*11**20+19*1.0*18==u2lA⋅KA*22设A=b1600*11*b*20+19*1.0*18=22bkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 32b+3.11b−5.45=032x+b⋅x+b⋅x+b=012322bb3.11012p=−+=−+=−1.075939323123q=*b1−*b1*b2+b3=*3.11−5.45=−3.2227327q23−3.2223D=()+p=()−(1.075)=1.350>0223q3.22khdaw.comu=−+D=+1.350=2.772u=.1405223q3.22v=−−D=−1.350=0.448v=.076522b=u+v−b3/=.1405+.0765−.3113/=.11333(m)1课后答案网4.14(P187)按汉森极限承载力公式计算根据o查P179表4.5得Nc=.835Nq=.247N=.047ϕ=10r1（1）pwww.hackshp.cn=⋅γ⋅b⋅N+γ⋅d⋅N+c⋅Nurmqc21=*19.0*3.0*0.47+19.0*1.0*2.47+10.0*8.352=143.8kPa（2）当地下水上升至基础底面时1p=*9.0*3.0*0.47+19.0*1.0*2.47+10.0*8.35u2=136.8kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 4.15(1)极限荷载按斯肯普顿极限承载力公式计算bdp=5⋅c⋅1(+0.⋅2)⋅1(+0.2⋅)+γ⋅dumlb2.41.2=5*16.01(*+0.2*)⋅1(+0.2*)+18.0*1.23.02.4=123.68kPakhdaw.com（2）按浅基础地基的临界荷载公式p14Nc=0.3N1=0查P165表面4.4得：=1.04Ndp1=Nd⋅γm⋅d+Nc⋅c+N1⋅γ⋅b44=1.0*18.0*1.2+3.0*160.+0=69.6kPa4.16(P187)课后答案网按汉森极限承载力公式计算oN=.172根据ϕ=16查P179表4.5得：Nc=11.62q=.433Nrb3b3S=1−04.=1−0.4*=0.7S=S=1+02.=1+0.2*=1.15rcqwww.hackshp.cnl4l4d2.0d=d=1+0.35=1+0.35*=1.233cqb3.0ootanδ=tan1181′=tan113.=.01998=2.0oi=.0462i=.0583iq=.0680rc1p⋅=⋅γ⋅bN⋅S⋅i⋅+γ⋅d⋅NS⋅i⋅d⋅+c⋅NS⋅i⋅durrrmqqqqcccc21=*18.6*3.01*.72*0.7*0.462+18.6*2.0*4.331*.15*0.681*.2332+8.0*11.62*1.15*0.583*1.233=247.68kPakhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 填土面5.1(P236)（1）求静止土压力值干砂3γ=18kN/m（2）求产生主动土压o0.4mϕ=36力所需的位移（3）求主动土压力值（1）oK≈1−sinϕ′≈1−sin36=.0412okhdaw.com1212P=⋅⋅γH⋅K=*18.0*4.0*0.412=59.33kN/moo22（2）密实砂土，产生主动土压力所需的位移约为墙高的0.5%，即：∆≈H*5.0%=0.4*5.0%=.002m=200.mm(3)2o2ooKtana=(45−ϕ)2/=tan(45−36)2/=0.2596课后答案网1212P=⋅⋅γH⋅K=*18.0*4.0*0.2596=37.38kN/maa225.2(P237)www.hackshp.cn填土面13γ=18kN/m0.2m地下水位面砂oϕ=362土（1）求静止土压力值3γsat=21.0kN/m（2）求主动土压力值2.0m（3）求墙后水压力值3oK≈−ϕ′（1）o1sin≈1−sin36=.0412根据e=γ⋅h⋅K得：ookhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com e=0o1e=γ⋅h⋅K=18*0.2*.0412=14.832kPao21oeo3=(γ⋅h1+γw⋅h2)⋅Ko=(18*2.0+11.0*2)*0412.=23.896kPa填土面12.0m14.8322khdaw.com2.0m323.89611P=*14.832*2.0+(*14.832+23.896*)2.0=53.56kN/mo222o2oo(2)课后答案网Ka=tan(45−ϕ/2)=tan(45−36/2)=.02596根据e=γ⋅h⋅K得：ooea1=0ea2=www.hackshp.cnγ⋅h1⋅Ka=18*2.0*0.2596=9.346kPae=(γ⋅h+γ⋅h)⋅K=(18*0.2+110.**)2.02596=15.057kPaa31w2a填土面10.2m9.34620.2m315.057khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 11P=*9.346*2.0+9(*.346+15.057*)2.0=33.75kN/ma221212(3)Pw=⋅γw⋅hw=*10*20.=20.0kN/m225.3(P237)oo根据ε=β=0ϕ=36δ=24查表5.1得：Ka=0.251212E=⋅γ⋅H⋅K=*18.0*4.0*0252.=36.288kN/maa22khdaw.com5.4(P237)ooo根据ε=0ϕ=30δ=20β=12查表得：Ka=0.35481212E=⋅⋅γH⋅K=*17.0*5.0*0.3548=75.40kN/maa22ooEax=754.*cos20=70.85kN/mEay=754.*sin20=25.79kN/m5.5(P237)课后答案网1.5moE=754.*cos20=70.85kN/maxo12oE=75.4*sin20=25.79kN/mEayaywww.hackshp.cn5.0mEax5/32.5m(G+Eay)⋅µ[221(*.5+2.*)55*0.5+25.79]*0.4K===1.39>1.3sE70.85ax可以G⋅a+Eay⋅b(221*.0*3/1*)2/5+(221*.51(*)2/5*.0+1.)2/5K==tE⋅3/570.85*3/5ax=1.38<1.6不可以khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com q=20kPa5.6(P237)求主动土压1力和水压力3中γ=185.kN/m13.0m砂oϕ=30123γ=190.kN/m2地下水粗3.0moϕ=353砂13γ=200.kN/m2sat4.0m42o2ooK=tan(45−ϕ)2/=tan(45−30)2/=0.333a11khdaw.comK=tan2(45o−ϕ)2/=tan2(45o−35o)2/=0.271a22p=0+q⋅K=20.0*0.333=6.666kPaa1a1p=(q+γ*h)*Ka2上11a1=(20+18.5*3.0)*0.333=25.17kPap=(q+γ*h*)Ka2下课后答案网11a2=(20+18.5*3.0)*0.271=20.46kPap=(q+γ*h+γ⋅h)*Ka31122a2www.hackshp.cn=(20+18.5*3.0+19.0*3.)0*0.271=35.91kPap=(q+γ*h+γ⋅h+γ′⋅h*)Ka4112223a2=(20+18.5*3.0+19.0*3.0+10*4.*)00.271=46.75kPa6.66125.173.0m220.463.0m335.914.0m446.75khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 1E=*(6.666+25.17)*3.0=47.75kN/ma126.666*3.0*1.5+(25.17−6.666)*3.0*1.0z==1.791mp147.751E=(*20.46+35.91*)3.0=84.55kN/ma2220.46*3.01*.5+(35.91−20.46*)3.0*1.0z==1.637mp284.551E=*(35.91+46.75)*4.0=165.32kN/ma3khdaw.com235.91*4.0*2.0+(46.75−35.91*)4.0*4.3/0z==2.087mp3165.32E=E+E+E=47.15+84.55+165.32=297.02kN/maa1a2a31212E=⋅γ⋅h=*10*4.0=80kN/mwww22课后答案网1E=47.75kN/m3.0ma1.1791m2E=84.55kN/mwww.hackshp.cna23.0m.16373E=165.32kN/ma34.2087m4.0mγ⋅h=400.kPaww∑zai⋅Eai(.1791+0.7)*47.75+(.1637+0.4)*84.55+.2087*165.32z==a∑Eai297.02=.418mkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 5.7(P237)1.0mo6Eay20o3中γ1=17.0kN/m砂7.0mEaxoϕ=301µ=0.41.5m3细γ2=180.kN/mkhdaw.com砂ϕ=20o5.0m2（1）求作用在墙背上的土压力；（2）作用在墙前趾上的土压力，（3）验算挡土墙抗滑稳定性。（课后答案网1）作用在墙背上的土压力；oooo根据ε=20ϕ=30δ=15β=6查表得：Ka=0.3811212E=⋅⋅γH⋅K=*17.0*7.0*0.381=158.69kN/maawww.hackshp.cn22oE=158.69*sin(δ+ε)=158.69sin35=91.02kN/mayoE=158.69*cos(δ+ε)=158.69cos35=129.99kN/max（2）作用在墙前趾上的土压力−1ooε=tan[(5.0−1.0−7.0tan207/).0]=11.72εo根据ε=11.72oδ=10oβ=0ϕ=201.5m按式5.16计算得：Kp=2.228khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 1212E=⋅⋅γH⋅K=*18.0*1.5*2.228=45.117kN/mpp22oE=45.117*sin(ε−δ)=45.117sin.172=.135kN/mpyoE=45.117*cos(ε−δ)=45.117cos.172=45.10kN/mpx（3）验算挡土墙抗滑稳定性。εoδ=10khdaw.comε(G+Eay+Epy)⋅µ[24*(5.0+1.0)*7.0/2+91.02+1.35]*0.4Ks===2.81Eax−Epx129.99−45.10>3.1安全5.8(P237)课后答案网oo根据θ=30ϕ=20查图5.48得：N=0.026cc5N=H===12mwww.hackshp.cnγ⋅Hγ⋅N160.*.00265.9(P237)c0.7N===.00389γ⋅H180.*100.o根据ϕ=20N=0.0389查图5.48得：oθ=35khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com 5.10(P237)i=1:13γ=18kN/m3.0moϕ=20c=4.5kPa3γ=19kN/m3.0moϕ=16c=10kPa7.0mkhdaw.com课后答案网www.hackshp.cn6842khdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com khdaw.com642课后答案网第1个小土条,b=1.0mwww.hackshp.cnkhdaw.com若侵犯了您的版权利益，敬请来信通知我们！℡www.khdaw.com'