Principles Of Foundation Engineering, 8th Edition Solution Manual

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Principles Of Foundation Engineering, 8th Edition Solution Manual - Page 1 preview image

An’s Solutions Manual to AccompanyPRINCIPLES OF FOUNDATIONENGINEERING, 8THEDITION, SI EDITIONBRAJA M. DAS

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ContentsChapterPage2.................................................................................................................................13...............................................................................................................................134...............................................................................................................................255...............................................................................................................................376...............................................................................................................................497...............................................................................................................................558...............................................................................................................................699...............................................................................................................................7510..............................................................................................................................9511............................................................................................................................10712............................................................................................................................11313............................................................................................................................12114............................................................................................................................13315............................................................................................................................14716............................................................................................................................157Note that there are no end of chapter questions for Chapter 1.

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1Chapter 22.1d.3kN/m17.17)05.0)(1000()81.9)(5.87(c.3kN/m14.9315.0117.171wda.Eq. (2.12):0.76eeeGws;1)81.9)(68.2(14.93.1b.Eq. (2.6):0.4376.0176.01eene.Eq. (2.14):53%)100(76.0)68.2)(15.0(eGwVVSsvw2.2a.From Eqs. (2.11) and (2.12), it can be seen that,3kN/m16.4822.011.201wdb.eGeGsws1)81.9(1kN/m16.483Eq. (2.14):).)(22.0(ssGwGeSo,2.67sssGGG;22.0181.948.162.3a.0.55eeewGws;1)12.01)(81.9)(65.2(18.79.1)1(

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2b.0.35555.0155.0nc.57.8%10055.0)65.2)(12.0(eGwSsd.3kN/m16.7812.0179.181wd2.4a.weGs.0.97eeeewewd;1)81.9(36.018.43.1)(b.0.4997.0197.01eenc.2.6936.097.0weGsd.3kN/m18.2397.01)81.9)(97.069.2(1)(sateeGws2.5From Eqs. (2.11) and (2.12):3kN/m98.1608.0134.18dEq. (2.12):53.0;1)81.9)(65.2(16.98;1eeeGwsdEq. (2.23):0.94maxmaxmaxminmaxmax;44.053.082.0eeeeeeeDr3kN/m13.494.01)81.9)(65.2(1max(min)eGwsd

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32.6Refer to Table 2.7 for classification.Soil A:A-7-6(9)(Note: PI is greater than LL30.)GI=(F200–35)[0.2 + 0.005(LL–40)] + 0.01(F200–15)(PI–10)==(65–35)[0.2 + 0.005(42–40)] + 0.01(65–15)(16–10)9.39Soil B:A-6(5)GI==(55–35)[0.2 + 0.005(38–40)] + 0.01(55–15)(13–10)5.45Soil C:A-3-(0)Soil D:A-4(5)GI==(64–35)[0.2 + 0.005(35–40)] + 0.01(64–15)(9–10)4.585 ≈5Soil E:A-2-6(1)GI=0.01(F200–15)(PI–10) = 0.01(33–15)(13–10) = 0.54≈1Soil F:A-7-6(19)(PI is greater than LL30.)GI==(76–35)[0.2 + 0.005(52–40)] + 0.01(76–15)(24–10)19.2≈192.7Soil A:Table 2.8: 65% passing No. 200 sieve.Fine grained soil;LL = 42; PI = 16Figure 2.5:MLFigure 2.7: Plus No. 200 > 30%; Plus No. 4 = 0% sand > % gravel–sandy siltSoil B:Table 2.8: 55% passing No. 200 sieve.Fine grained soil;LL = 38; PI = 13Figure 2.5: Plots below A-line–MLFigure 2.7: Plus No. 200 > 30%% sand > % gravel–sandy silt

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4Soil C:Table 2.8: 8% passing No. 200 sieve.% sand > % gravel–sandy soil–SPFigure 2.6: % gravel = 100–95 = 5% < 15%–poorly graded sandSoil D:Table 2.8: 64% passing No. 200 sieveFine grained soil;LL = 35, PI = 9Figure 2.5–MLFigure 2.7:% sand (31%) > % gravel (5%)–sandy siltSoil E:Table 2.8: 33% passing No. 200 sieve; 100% passing No. 4 sieve.Sandy soil;LL = 38; PI = 13Figure 2.5: Plots below A-line–SMFigure 2.6: % gravel (0%) < 15%–silty sandSoil F:Table 2.8: 76% passing No. 200 sieve;LL = 52; PI = 24Figure 2.5:CHFigure 2.7: Plus No. 200 is 100–76 = 24%% gravel > % gravel–fat clay with sand2.843.0139.18)81.9)(68.2(1e;1dwswsdGeGEq. (2.37):43.0143.063.0163.022.0;1133223213121keeeekk;k2=0.08 cm/s2.9From Eq. (2.41):nnneeeekk)6316.0(1667.0;9.12.12.29.21091.0102.0or;1166211221898.31996.0778.0)6316.0log()1667.0log(n

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5cm/s100.0756-)10216.0(9.19.0110216.02.1)2.2)(102.0()1(6998.336998.36111eeCkeekCnn2.10The flow net is shown./m/sm1017.0636-8)4)(25.5(10105.6So,m.25.575.17cm/s;105.62421m ax4qHHhk2.11a.7825.0327825.032106.016.0)2.0(4622.2)1(4622.2eeDkcm/s0.041b.32.26.0332.2106.03)2.0(2.04.06.016.0)35()(135DCeekucm/s0.171

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62.123dry(sand)kN/m16.8455.01)81.9)(66.2(1eGws3sat(sand)kN/m81.0248.01)48.066.2(81.91eeGwws3sat(clay )kN/m18.55)74.2)(3478.0(1)3478.01)(81.9)(74.2(1)1(swswGwGAtA:σ =0;u=0; σ=0AtB:σ = (16.84)(3) =50.52 kN/m2u=0σ=50.52 kN/m2AtC:σ = σB+ (20.81)(1.5) = 50.52 + 31.22 =81.74 kN/m2u= (9.81)(1.5) =14.72 kN/m2σ= 81.74–14.72 =67.02 kN/m2AtD:σ=σC+ (18.55)(5) = 81.74 + 92.75 =174.49 kN/m2u= (9.81)(6.5) =63.77 kN/m2σ= 174.49–63.77 =110.72 kN/m22.13Eq. (2.54):Cc= 0.009(LL–10) = 0.009(42–10) = 0.288Eq. (2.65):mm87.2110155log82.01mm)10007.3)(288.0(log1oooccceHCS2.14Eq. (2.69):mm56.69128155log82.01)3700)(288.0(110128log75.01mm)3700(5288.0log1log1cooccococsceHCeHCS

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72.15a.Eq. (2.53):0.392150300log792.091.0log1221eeCcFrom Eq. (2.65):oooccceHCSlog1Using the results of Problem 2.12,2kN/m87.88)81.955.18(25)81.981.20(5.1)84.16)(3(o953.0)74.2)(3478.0(sowGemm194.5487.885087.88log953.01mm)5000)(392.0(cSb.Eq. (2.73):2HtCTvv.ForU= 50%,Tv= 0.197 (Table 2.11). So,days609sec105262;cm)500(1036.9197.0424tt2.16a.Eq. (2.53):0.377120360log64.082.0log1221eeCcb.0.736221221;120200log82.00.377;logeeeeCc2.17Eq. (2.73):2HtCTvv. For 60% consolidation,Tv= 0.286 (Table 2.11).Lab time:min649min861t

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8/mincm5057.0;)cm8.3(649286.022vvCCField:U= 50%;Tv= 0.197days6.29min9060;cm210005.3)5057.0(197.02tt2.185.06030UttHtCTvv6221)1()1(102210002))(2(ttHtCTvv6222)2()2(108210001))(2(So,Tv(1)= 0.25Tv(2). The following table can be prepared for trial and errorprocedure.Tv(1)Tv(2)U1U2UHHHUHU212211(Figure 2.22)0.050.100.1250.11250.20.40.50.450.260.360.400.3850.510.700.760.730.340.4730.520.50So,Tv(1)= 0.1125 = 210-6t;t= 56,250 min =39.06 days2.19Eq. (2.84):.2HtCTcvctc= 60 days = 60246060 sec;22Hm = 1000 mm.

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90415.0)1000()60602460)(108(23cTAfter 30 days:0207.0)1000()60602430)(108(232HtCTvvFrom Figure 2.24 forTv= 0.0207 andTc= 0.0415,U= 5%. SoSc= (0.05)(120) =6 mmAfter 100 days:069.0)1000()606024100)(108(232HtCTvvFrom Figure 2.24 forTv= 0.069 andTc= 0.0415,U23%. SoSc= (0.23)(120) =27.6 mm2.20NS1tanNormal force,N(N)Shear force,S(N)NS1tan(deg)222.4489.3667.2193.5424.8587.141.0240.9641.35From the graph,412.21Normally consolidated clay;c= 0.245tan231;38;245tan2074.66220722.22245tan231;30;245tan1382761382

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102.23c= 0. Eq. (2.91):2kN/m387.822845tan140245tan22312.24Eq. (2.91):245tan224531c245tan2245tan1403682c(a)245tan2245tan2807012c(b)Solving Eqs. (a) and (b),=24;c=12 kN/m22.25257.898.2207.898.220sinsin13131131311sin2323kN/m75.5195.377.89;kN/m85.18295.378.2203475.5185.18275.5185.182sin1Normally consolidated clay;c=0andc=02.26245tan231.221kN/m305.922045tan1502kN/m61.9uuu;22845tan1509.305;245tan2231

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112.27a.)log(6.14.0102650DCDur30.7)]13.0)[log(6.1()1.2)(4.0()53.0)(10(26b.bae1327.209.021.0097.0101.2097.0101.21585DDa081.0)327.2)(398.0(845.0398.0845.0ab33.67081.0)68.0)(327.2(1

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13Chapter33.1Eq. (3.3):8.96%10073732.76100(%)222222iioRDDDA3.2Depthfromgroundsurface(m)N60cu(kN/m2)[Eq. (3.8)]o(MN/m2)OCR[Eq. 3.9)]3.0592.403854.0)]81.919)(5.1()5.16)(5.1[(100015.514.58129.60490.0)81.98.16)(5.1(1000103854.06.466.0892.40595.0)81.98.16)(5.1(100010490.05.657.59141.107.0)81.98.16)(5.1(100010595.05.489.010152.20805.0)81.98.16)(5.1(1000107.05.35Averagecu=121.5 kN/m2Average OCR =5.69

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