Principles Of Foundation Engineering, 8th Edition Test Bank

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

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AnInstructor’s Solutions Manualto AccompanyPrinciples of FoundationEngineering8th EditionSI 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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143.3Depth fromground surface(m)N60oa(kN/m2)OCR[Eq. 3.10)]c(kN/m2)[Eq. (3.11)]3.0538.547.522354.5849.09.473766.0859.58.03767.5970.07.464239.01080.57.2470aSee Problem 3.23.4Depth (m)o(kN/m2)1.53.04.56.07.59.0181.5 =183.0 =184.5=186.0 =108 + (1.5)(20.2–9.81) =123.6 + (1.5)(20.2–9.81) =275481108123.6139.2Eq. (3.13):5.01 aoNpCDepth (m)N60o(kN/m2)CN(N1)60a1.56271.92123.08541.36114.59811.11106.081080.9687.513123.60.9129.014139.20.8512aRounded off to nearest whole number

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153.5 aoNpC12Depth (m)N60oa(kN/m2)CN(N1)60b1.56271.5793.08541.3104.59811.1106.081080.9687.513123.60.89129.014139.20.8412aSee Problem 3.4bRounded off to nearest whole number3.6From Problem 3.4, the average value of1183.10)12128101112(61)(601NEq. (3.31):3583.3420)11)(20(20)(20601N3.7From Problem 3.4Depth (m)o(kN/m2)pa(kN/m2)N60(deg)[Eq. (3.30)]1.527100634.73.054100834.94.581100934.06.0108100831.47.5123.61001334.99.0139.21001434.9Average= 34.134

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163.8Eq. (3.29):26060)(00054.03.01.27NNDepth (m)N60(deg)[Eq. (3.29)]1.5628.883.0829.474.5929.766.0829.477.51330.99.01431.19Average= 29.93303.9Depth (m)o(kN/m2)pa(kN/m2)N60Dr(%) [Eq. (3.22)]1.53.04.56.07.59.0275481108123.6139.21001001001001001006898131450.651.849.743.252.852.7AverageDr= 50.13%50%3.10Eq. (3.13):606015.0)(;1NCNpCNaoNDepth(m)o(kN/m2)pa(kN/m2)N60CN601)(08.1N5.060160)(08.1NDr[Eq. (3.28)]1.53.04.56.07.59.0275481108123.6139.2100100100100100100689813141.921.361.110.960.90.8512.4411.7510.788.2912.6412.8545.342.342.437.245.946.3Average = 43.23%43%

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173.11Depth(m)o(kN/m2)op(kN/m2)7.15006.023.0DN60Dr(%)[Eq. (3.23)]1.53.04.56.07.59.026.452.879.2105.6132.0158.41001001001001001000.1330.1330.1330.1330.1330.1335111418162152.955.551.150.242.344.33.12Depth (m)(kN/m3)o(kN/m2)pa(kN/m2)N60(deg)a[Eq. (3.30)]3.04.56.07.59.010.512.016.6616.6616.6618.5518.5518.5518.5549.9874.9799.66127.49155.31183.14210.9710010010010010010010079111618202234343537363636arounded to nearest whole numberAverage364.35)36363637353434(713.13Between depths 6 m and 9 m, average15)181611(3160N.Eq. (3.32):2kN/m15,000)15)(10)(100(60NpEas3.14Eq. (3.21):5.026050779OCR)(711231122275.02.12(%) uaorCpND

Principles Of Foundation Engineering, 8th Edition Test Bank - Page 22 preview image

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18At a depth of 3.0 m:%3.46)8.2)(50(10055)779()2()711(2311)9(22275.02.12(%)5.02rDAt a depth of 4.5 m:%2.48)8.2)(50(10082)779()2()711(2311)11(22275.02.12(%)5.02rDAt a depth of 6.0 m:%9.48)8.2)(50(10098)779()2()711(2311)12(22275.02.12(%)5.02rDAverageDr=48%%7.48)9.482.483.46(313.15Eq. (3.34):KTcu000994.0)127.0)(6(45cos0635.045cos0635.012)0635.0(6coscos12:(3.38)Eq.22hididdKBT22)VST(kN/m3.51N/m308,51000994.051ucEqs. (3.39) and (3.40a):

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192kN/m48.48)3.51)](2146log(54071[)3.51)](PIlog(54.07.1[)VST()(corrected.-.ccuu3.16Eq. (3.42):o)field(OCRuco= 59.5 kN/m2(from Problem 3.2)Eq. (3.43):= 22(PI)0.48= (22)(25)0.48= 4.694.045.593.51)69.4(OCR3.17a.28.506.101dhFrom Eq. (3.36):333m00048.06m)0508.0)()(7(67dK2kN/m58.32)VST(N/m333,58300048.0280KTcub.From Eqs. (3.39) and (3.40a):2kN/m53.07)3.58)](2958log(54.07.1[)1129)](PIlog(54.07.1)corrected(uc3.18Eq. (3.45):18.4)92(0055.008.01)PI(0055.008.01Eq. (3.42):3.82.643.58)18.4(OCRo)field(uc

Principles Of Foundation Engineering, 8th Edition Test Bank - Page 24 preview image

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203.19Depth(m)o(MN/m2)qc(MN/m2)(deg)1.53.04.56.07.59.00.0240.0480.0720.0960.1200.1442.064.236.018.189.9712.4242.0642.1541.9842.0341.9342.07(average)423.20Note: OCR = 1;Qc= 1Depth (m)qc(kN/m2)o(kN/m2)Dr(%) [Eq. (3.51)]1.53.04.56.07.59.020604230601081809970124202448729612014437.144.748.252.354.658.33.21From Eq. (3.55):58.3)2.0(44.544.526.026.05060DNpqacPa100 kN/m2Depth (m)qc(MN/m2)N601.53.04.56.07.59.02.064.236.018.189.9712.425.75611.81216.781722.852327.852834.6935

Principles Of Foundation Engineering, 8th Edition Test Bank - Page 25 preview image

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213.22a.o= (2)(18) + (20)(4) = 116 kN/m2Eq. (3.56):2kN/m45.615116800kocuNqcb.o= (2)(18) + (20–9.81)(4) = 76.76 kN/m2Eq. (3.60):3.3701.101.176.76116800)37.0(37.0OCRoocq3.23Eq. (3.61):2kN/m4121.6461804.425.326218046535)5.01)(2())(1(2vpvVEmosp3.24a.43.295)38)(81.9(280oooDupKEq. (3.68):0.656.05.143.26.05.147.047.0DoKKb.Eq. (3.69): OCR = (0.5KD)1.6= (0.52.43)1.6=1.372kN/m2131)280350)(7.34)(35.01())(7.34)(1()1(:(3.72)Eq.c.2122osDssppEE3.252.3)81.98.19)(4()5.14)(2()81.9)(4(260oooDupK

Principles Of Foundation Engineering, 8th Edition Test Bank - Page 26 preview image

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22Eq. (3.74a):38.2)2.3)(066.0(236.02.331066.0236.031DDKK3.26Eq. (3.79):;)1)(21(1ssssgEvs= 0.32)32.1)(36.0(68.081.918105sE;Es=14.136 kN/m23.27A time-distance plot is shown.5.71024.1510ofSlope31valayer)(top24.15105.731m/s492v19.71010ofslope32m/s1390abv9.51020ofslope33m/s3390bcvxc= 7.5 m

Principles Of Foundation Engineering, 8th Edition Test Bank - Page 27 preview image

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23m2.6)5.7(49213904921390212112121cxvvvvZEq. (3.81):s1020;))((22132222323132123122iiTvvvvvvvvZTZ0105.0)492)(3390()492()3390()6.2)(2())((2221321231vvvvZ1524)1390()3390()1390)(3390(22222323vvvvSo,m7.24)1524)(0105.002.0(212Z

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25Chapter44.1a.Eq. (4.8):BNNqNcqqcu21= 28. Table 4.1:Nc= 31.61;Nq= 17.81;N= 13.72kN/m248.5)7.13)(91.0)(29.17(21)81.17)(91.0)(29.17()61.31)(17.19(41FSalluqqb.= 35. Table 4.1:Nc= 57.75;Nq= 41.44;N= 45.412kN/m372.8)41.45)(5.1)(8.17(21)44.41)(2.1)(8.17()75.57)(0(41FSalluqqc.Table 4.1:= 30;Nq= 22.46;N= 19.13Eq. (4.17), withc= 0:2kN/m280)]13.19)(3)(5.16)(4.0()46.22)(5.162[(41)4.0(FS1FSallBNNqqqqu4.2Eq. (4.17);c= 0:)4.0(FS1FSallBNNqqqqu22allall1805BBQqFrom Table 4.1 for= 34,Nq= 36.5;N= 38.04)]40.38)()(9.15)(4.0()5.36)(9.15)(5.1[(3118052BB

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26By trial and error,B =2 m4.3a.Eq. (4.26):idsqiqdqsqcicdcscuFFFBNFFFNqFFFNcq21Vertical load, soFci,Fqi, andFiare equal to 1.Continuous foundation, soFcs,Fqs, andFsare equal to 1.= 28. Table 4.2:Nc= 25.8;Nq= 14.72;N= 16.72.Table 4.3:1321.1)28)(tan8.25(299.11299.1tan1299.191.091.0)28sin1(28tan21)sin1(tan2122dcqdqdcdfqdFNFFFBDF2kN/m271.4)1)(72.16)(91.0)(29.17(21)299.1)(72.14)(91.029.17()321.1)(8.25)(17.19(4121FS1FSalldqdqcdcuFBNFNqFNcqqb.c= 0dqdqFNBFNqq21FS1all= 35. Table 4.2:Nq= 33.3;N= 48.03.Table 4.3:2.15.12.1)35sin1(35tan21)sin1(tan2122BDFfqd

Principles Of Foundation Engineering, 8th Edition Test Bank - Page 31 preview image

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271dF2kN/m373.7)1)(03.48)(5.1)(8.17(21)2.1)(3.33)(8.172.1(41allqc.sdqsqdquFFBNFFNqq21= 30. From Table 4.2:Nq= 18.4;N= 22.4.Table 4.3:Fs= 0.6;Fd= 1;Fqs= 1 + tan= 1.577193.132289.01)sin1(tan212BDFfqd2kN/m368.8)1)(6.0)(4.22)(3)(5.16(21)577.1)(193.1)(4.18)(5.16)(2(41allq4.420cos12112allidsqiqdqsqcicdcscFFFBNFFFNqFFFNcBFSQNote:Qall= inclined allowable load.= 25. Table 4.2:Nc= 20.72;Nq= 10.66;N= 10.88.6.0)1(4.014.01466.1)25(tan52.152.11tan1514.172.2066.1052.152.111LBFLBFNNLBFsqscqcs206.1)25)(tan72.20(184.11187.1tan1184.152.19.0311.01)sin1(tan212cqdqdcdfqdNFFFBDF

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