Solutions for Introduction To Geotechnical Engineering 3rd Us Edition by Holtz

INSTRUCTOR’S SOLUTIONS MANUAL

AN IN T R O D C T I O N TO GEOTECHNICAL

THIRD

EDITION

Robert D. Holtz, Ph.D., P.E., D.GE

University of Washington

William D. Kovacs, Ph.D., P.E., D.GE

University of Rhode Island

Thomas C. Sheahan, Sc.D., P.E.

Northeastern University

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ISBN-13: 978-0-13-561930-8

ISBN-10: 0-13-561930-0

CHAPTER2:INDEXANDCLASSIFICATIONPROPERTIESOFSOILS

2.1.Prepareaspreadsheetplotofdrydensityinkg/m3 astheordinateversuswatercontentin percentastheabscissa.Assume ρs =2750kg/m3 andvarythedegreeofsaturation, S,from100% to40%in10%increments.Amaximumof50%watercontentshouldbeadequate(exceptforvery softclays,tobedefinedshortly).[Note:Suchagraphisveryusefultocheckproblemsthatyoudo inthischapter,asthereisauniquerelationshipbetweendrydensity,watercontent,andthedegree ofsaturationforanygivendensityofsolids.Voidratio,wetdensity,andsoonarethenreadily computed.Similargraphsfordensityofothersolidsmaybeeasilycomputedandplotted.]Show allyourequationsascommentsonyoursheetforfuturereference.

Solution:SolveEq.2.12andEq.2.15for ρd = f (ρs,ω,S,Gs),oruseEq.5.1.

Finally,

ρw (kg/m3)1000

ρs (kg/m3)2750

Gs 2.75

S(%)100908070605040302010

w(%)

ρd (kg/m3)

02750275027502750275027502750275027502750 52418238623472299223721572047188616301158 10215721062047197418861774163014351158733 1519471886181417301630150713541158898537 201774170716301540143513101158971733423 251630155914791387128211581011835620349 30150714351354126211581038898733537297 3514011329124811581056940807653473259 401310123811581069971859733589423229 45122911581080994898791672537383206 50115810881011928835733620493349186

2.2.PrepareagraphlikethatinProblem2.1,onlyusedrydensityunitsofkN/m3 andpounds percubicfeet.

Solution:FromEq.2.12andEq.2.15

P.2.1

2.2.a.IntheSIunitsystem:

γw (kN/m3)9.8

γs (kN/m3)26.95 Gs 2.75 S(%)100908070605040302010

w(%) γd (kN/m3)

026.9526.9526.9526.9526.9526.9526.9526.9526.9526.95 523.6923.3823.0022.5321.9321.1420.0618.4815.9711.35 1021.1420.6420.0619.3518.4817.3915.9714.0611.357.19 1519.0818.4817.7816.9615.9714.7713.2711.358.805.26 2017.3916.7315.9715.0914.0612.8311.359.517.194.15 2515.9715.2814.4913.6012.5611.359.918.196.073.42 3014.7714.0613.2712.3711.3510.178.807.195.262.91 3513.7313.0212.2311.3510.359.217.916.404.642.54 4012.8312.1311.3510.489.518.427.195.784.152.25 4512.0411.3510.589.748.807.766.585.263.752.01 5011.3510.669.919.098.197.196.074.833.421.83

IntheEnglishunitsystem:

ρw (pcf)62.4

ρs (pcf)171.6 Gs 2.75

S(%)100908070605040302010

w(%)

ρd (pcf)

0171.56171.56171.56171.56171.56171.56171.56171.56171.56171.56 5150.82148.82146.40143.39139.57134.56127.67117.64101.6772.24 10134.56131.41127.67123.17117.64110.68101.6789.5172.2445.75 15121.46117.64113.19107.95101.6794.0184.4672.2456.0233.48 20110.68106.49101.6796.0789.5181.7072.2460.5545.7526.39 25101.6797.2692.2786.5579.9572.2463.1052.1238.6621.79 3094.0189.5184.4678.7572.2464.7456.0245.7533.4818.55 3587.4282.9077.8772.2465.8858.6550.3740.7729.5216.15 4081.7077.2072.2466.7260.5553.6145.7536.7626.3914.30 4576.6872.2467.3661.9856.0249.3741.9133.4823.8712.83 5072.2467.8763.1057.8852.1245.7538.6630.7321.7911.63

2.3.PrepareagraphlikethatinProblem2.1,onlyforS=100%,andvarythedensityofsolids from2600to2800kg/m3.Youdecidethesizeoftheincrementsyouneedto“satisfactorily”evaluatetherelationshipas ρs varies.Prepareaconcludingstatementofyourobservations.

Solution:Therelationshipbetween ρd and w isnotoverlysensitiveto ρs.

P.2.2.b

ρw (kg/m3)1000 S(%)100

ρs (kg/m3)26002650270027502800 w(%) ρd (kg/m3) 026002650270027502800 523012340237924182456 1020632095212621572188 1518711896192219471972 2017111732175317741795 2515761594161216301647 3014611476149215071522 3513611375138814011414 4012751286129813101321 4511981209121912291239 5011301140114911581167

2.4.Thedrydensityofacompactedsandis19.5kN/m3 andthedensityofthesolidsis27.1 kN/m3.WhatisthewatercontentofthematerialwhenS=50%?Phasediagram!

FromEq.5.1:

Solvingforw: w = Sγw 1 γd 1 γs =(0 5)(9 8) 1 19 5 1 27 1 =0 07=7%

Forphasediagram,assume Vt =1m3

γd = Ws Vt ⇒ Ws = γsVt =19 5kN w = Ww Ws × 100 ⇒ Ww = wWs 100 = (7)(19 5) 100 =1.37kN Wt = Ww + Ws =1.37+19.5=20.87kN

ρw =9 8kN/m 3 = Ww Vw ⇒ Vw = 1 37 9 8 =0 14m3

Va = Vt Vw Vs =0 14m3

Vv = Va + Vw =0 28m3

TBEXAM.COM

Checkingthat S =50 S = Vw Vw = 0 14 0.28 =0 5=50%(OK) P.2.4.Phasediagram

2.5 Asoilthatiscompletelysaturatedhasatotaldensityof2050kg/m3 andawatercontentof 20%.Whatisthedensityofthesolids?Whatisthedrydensityofthesoil?Phasediagram!

Solution:

a.)Solveusingequationsorphasediagrams:

ρs = ρwρd ρd + ρw ρsat = (1000)(1708 3) (1708 3+1000 2045) =2594.9kg/m 3

b)Solveusingphasediagramrelationships:Assume Vt =1.0m3

Mt =2050kg

Mw Ms

ρd = Ms Vt = 1708 3 1 =1708.3kg/m 3

ρs = Ms Vs = 1708.3 1 0.34 =2594 9kg/m 3

2.6 Whatisthewatercontentofafullysaturatedsoilwithadrydensityof1750kg/m3?Assume ρs =2820kg/m3

Solution:

FromEq.2-12andEq.2-15:

Solvingfor w

2.7 Adryquartzsandhasadensityof1675kg/m3.Determineitsdensitywhenthedegreeof saturationis65%.Thedensityofsolidsforquartzis2680kg/m3

Solution:

Recognizethat ρd(initial)= ρd (final); S =75%

FromEq.2-12andEq.2-15:

Solvingfor w:

Then,usingEq.2.14:

P.2.5.Phasediagram

ρt = ρd(1+ w)=1675kg/m 3(1+0 168)=1956 3kg/m 3

2.8 Thedryunitweightofasoilis15.7kN/m3 andthesolidshaveaunitweightof26kN/m3 Assumingthesoilissaturated,find(a)watercontent,(b)voidratio,and(c)totalunitweight. a.)Watercontent.FromEq.5.1:

Solvingforw:

= Sγw 1

b.)Voidratio.FromEq.2.15:

c.)Totaldensitywhenthesoilissaturated.UsingEq.2.14:

)=15

TBEXAM.COM

2.9 Anaturaldepositofsoilisfoundtohaveawatercontentof20%andtobe90%saturated. Whatisthevoidratioofthissoil?

Solution: w =20%, S =90%;assume Gs =2 70FromEq.2.15: γwSe = wγ

2.10 Achunkofsoilhasawetweightof62lbandavolumeof0.56ft3.Whendriedinanoven, thesoilweighs50lb.IfthespecificgravityofsolidsGs=2.64,determinethewatercontent,wet unitweight,dryunitweight,andvoidratioofthesoil.

Solution: Solveusingphasediagramrelationships.

a.)Watercontent:

w = Ww Ws × 100= 12(100) 50 =24.0%

b.)Wetunitweight:

γt = Wt Vt = 62 0 56 =110 7pcf

c.)Dryunitweight:

γd = Ws Vt = 50 0.56 =89 29pcf

d.)Voidratio:

Vw = Ww γw = 12 62 4 =0.1923ft3

Vs = Ws Gsγw = 50 (2 64)(62 4) =0.3035ft3

Vv = Vt Vv =0 56 0303=0 256

e = Vv Vs = 0 256 0 303 =0.84

TBEXAM.COM

2.11 Inthelab,acontainerofsaturatedsoilhadamassof115.5gbeforeitwasplacedintheoven and102.4gafterthesoilhaddried.Thecontaineralonehadamassof49.31g.Thespecificgravity ofsolidsis2.70.Determinethevoidratioandwatercontentoftheoriginalsoilsample.

Solution:

Solveusingphasediagramrelationships.

Ms =102.4 49.31=53.09g

Mw =115 5 102 4=13 10g

w = Mw Ms × 100%=24.67%

FromEq.2.15: e = Gsw S = (2 70)(24 67) 100 =0 66

2.12 Thenaturalwatercontentofasampletakenfromasoildepositwasfoundtobe20.0%.It hasbeencalculatedthatthemaximumdensityforthesoilwillbeobtainedwhenthewatercontent reaches25.0%.Computehowmanygramsofwatermustberemovedtoeach1000gofsoil(inits naturalstate)inordertodecreasethewatercontentto25.0%.

Solution:

Naturalstate:

w =0 20= Mw Ms ⇒ Mw =0 20Ms

Mt = Ms + Mw = Ms +0.20Ms =1.20Ms =1000g

Ms =1000/1 20=833 33g

Mw =(0.20)(833.3)=166.67g

Targetstate:

(Note: Ms doesnotchangebetweenthenaturalstateandthetargetstate)

Mw = wMs =0 25(833 33)=208 33g

additionalwaternecessary=208.33-166.67=41.67g

2.13 Acubicmeterofdryquartzsand(Gs =2.65)porosityof60%isimmersedinanoilbath havingadensityof0.92g/cm3 .Ifthesandcontains0.27m3 ofentrappedair,howmuchforceis requiredtopreventitfromsinking?Assumethataweightless,permeablemembranesurroundsthe specimen.(Prof.C.C.Ladd.)

Solution:

Vt =1m3 =1, 000, 000cm3

TBEXAM.COM

n = Vv Vt ⇒ Vv = nVt =0.6(1.0)=0.60m3

Vs = Vt Vv =1.0 0.6=0.40m3

Ms = GsρwVs =2 65(1000kg/m 3)(0 40m3)=1060kg= Mt (dryquartzsand)

ρt = Mt Vt = 1060 1 0 =1060kg/m 3

ρbouy = ρt ρoil =1060 920=140kg/m 3

γbouy = ρbouy × g =140(9 81)=1373 4N/m 3

Finallytheforce(F)iscalculatedas:

F = γbouy × (entrappedair)=1373 4N/m 3 × 0 27m3 =370 8N

2.14 Asoilsampletakenfromaborrowpithasanaturalvoidratioof0.89.Thesoilwillbe usedforahighwayprojectwhereatotalof100,000m3 ofsoilisneededinitscompactedstate;its compactedvoidratiois0.69.Howmuchvolumehastobeexcavatedfromtheborrowpittomeet thejobrequirements?

Solution:

Vs = Vv e = Vt Vs e ⇒ Vs = Vt e +1

Embankment:

Vt =100, 000m3

Vs(emb) = 100, 000 0.69+1 =59171 6m3

Borrowpit:

Vs(borr) = Vs(emb) =59171.6m3

Vs(emb) = 100, 000 0.89+1 = Vt e +1

Vt(emb) = Vs(eborr +1)=(59171 6)(0 89+1)=111, 834 3m3

2.15 Asampleofmoistsoilwasfoundtohavethefollowingcharacteristics:

Totalvolume:0.0138m3

Totalmass:28.31kg

Massafterovendrying:23.40kg

Specificgravityofsolids:2.71

TBEXAM.COM

Findthedensity,unitweight,voidratio,porosity,anddegreeofsaturationforthemoistsoil.?

Solution:

a.)Density:

ρt = Mt Vt = 28 31 0.0138 =2051 45kg/m 3

b.)unitweight:

γt = ρt × g =(2051 45)(9 81)=20124 72N/m 3 =20 12kN/m 3

c.)voidratio:

Vs = Ms Gsρw = 23 40 2.71(1000) =0 00863m3

Vv = Vv Vs =0 0138 0 00863=0 00516m3

e = Vv Vs = 0.00516 0.00863 =0 60

d.)porosity:

n = e 1+ e × 100= 0 60 1+0 60 × 100=37.5%

e.)degreeofsaturation:

Mw =28 31 23 40=4 91g

Vw = Mw ρw = 4 91 1000 =0.00491

S = Vw Vv × 100= 0 00491 0 00516 =95.1%

2.16 Agraysiltyclay(CL)issampledfromadepthof12.5feet.The“moist”soilwasextruded froma6-inch-highbrasslinerwithaninsidediameterof7.18cmandweighed823grams.

(a)Calculatethewetunitweightinpoundspercubicfeet.

(b)Asmallchunkoftheoriginalsamplehadawetweightof154.3gramsandweighed74.3grams afterdrying.Computethewatercontent,usingthecorrectnumberofsignificantfigures.

(c)Computethedrydensityinkg/m3

Solution:

a.)Wetunitweightinpcf: Vt = πd2 4 ×L = π 7 18cm(1in/2 54cm) 2 4 ×6in=37

γt = Mt Vt = 823g(1lb/453 6g) 0.0218ft3 =83 26lb/ft3

b.)watercontent

Mw = Mt Ms =140 9 85 2=55 7 g

w = Mw Ms × 100= 55 7g 85 2g × 100=654%

c.)drydensity

2.17 Acylindricalsoilspecimenistestedinthelaboratory.Thefollowingpropertieswereobtained:

Samplediameter:3inches

Samplelength:6inches

Wt.beforedryinginoven:2.95lb

Et.afterdryinginoven:2.54lb

Oventemperature:110oC

Dryingtime:24hours

Specificgravityofsolids:2.65

Whatisthedegreeofsaturationofthissample?

Solution:

Vt = π(3)2 4 × 6=42 4115in3

Vs = Ws Gsγw = 2 54lb 2 65(62 4pcf) =0 01536ft3 =26 542in3

Vv = Vt Vs =42 4115 26 542=15 869in3

Ww = Wt Ws =2.95 2.54=0.41lb

Vw = Ww γw = 0 41lb 62 4pcf =0 00657ft3 =11 354in3

S = Vw Vv × 100%= 11 354 15 869 × 100=71 5%

2.18 Asampleofsaturatedsiltis10cmindiameterand2.5cmthick.Itsvoidratiointhisstate is1.42,andthespecificgravityofsolidsis2.68.Thesampleiscompressedtoa2-cmthickness withoutachangeindiameter.

(a)Findthedensityofthesiltsample,ing/cm3,priortobeingcompressed.

(b)Findthevoidratioaftercompressionandthechangeinwatercontentthatoccurredfrominitial tofinalstate.

Solution: a.)density Vt = π(10)2 4 × 2 5=196 35cm3

S =100%= Vw Vv ⇒ Vw = Vv

e = Vv Vs ⇒ Vv = eVs =1 42Vs

Vt = Vv + Vs =1 42Vs + Vs =2 42Vs =196 35cm3 ⇒ Vs =81 14cm3

Vv =1 42Vs =1 42(81 14)=115 21cm3

Mw = ρwVw =(1g/cm3)(115.21)=115.21g

Ms = GsρwVs =2 68(1g/cm3)(81 14cm3)=217 44g

Mt = Mw + Ms =115.21+217.44=332.66g

ρt = Mt Vt = 332.66 196.35 =1 694g/cm3 =1694kg/m 3

winitial = Mw Ms × 100= 115 21 217 44 × 100=53 0%

b.)voidratioandchangeinwatercontent

finalconditions:

Vt,2 = π(10)2 4 × 2.0=157.08cm3

Vs =81 14cm3 (nochange)

Vv = Vt,2 Vs =157.08 81.14=75.94cm3

efinal = Vv Vs = 75.94 81.14 =0 94

Vw = Vv =75 94cm3 (Saturated)

Mw = ρwVw =(1g/cm3)(75 94cm3)=75 94g

wfinal = 75 94 217 44 × 100=34.9%

∆w = winitial wfinal =53.0 34.9=18.1%

2.19 Asampleofsandhasthefollowingproperties:totalmass Mt =182g;totalvolume Vt =90 cm3;watercontent w =14%;specificgravityofsolids Gs =2.71.Howmuchwouldthesample volumehavetochangetoget90%saturation,assumingthesamplemass Mt stayedthesame?

Solution:

initialconditions:

Mw = wMs =0 14Ms

Mt = Ms +0.14Ms =182g ⇒ Ms = 182 1 14 =159.65g

Mw =0 14(159 65)=22 35g

Vw = γwMw =(1g/cm3)(22 35)=22 35cm3

Vs = Ms Gsρw = 159.65 2.71(1) =58 91cm3

Vv = Vt Vs =90 58 91=31 09cm3

S = Vw Vv × 100= 22.35 31.09 × 100=71 9%

desiredcondition:S=90%:

Vt changes,but Vs, Vw, Ms and Mw remainthesame

S = Vw Vv ⇒ Vv = Vw S = 22 35 0 90 =24 834cm3

Vt = Vs + Vv =58.91+24.834=83.74cm3

∆V =90 83 74=6 26cm3

TBEXAM.COM

2.20 Asampleofsoilpluscontainerweighs413.7gwhentheinitialwatercontentis9.5%.The containerweighs258.7g.Howmuchwaterneedstoberemovedtheoriginalspecimenifthewater contentistobedecreasedby4.2%?(AfterU.S.Dept.ofInterior,1990.)

Solution:

initialconditions:

Ms+c =413 7g

Mc =258.7g

Ms = Ms+c Mc =155g

w = Mw Ms ⇒ Mw = Msw =155(0 095)=14 72g

Desiredconditions:

wf = wi 4.2=9.5 4.2=5.3%

Mw,f = Ms(w)=155(0 053)=8 22%

w = Mw Mw,f =14 725 8 22=6 51g

2.21 Asoilsampleisdriedinamicrowaveoventodetermineitswatercontent.Fromthedata below,evaluatethewatercontentanddrawconclusions.Theoven-driedwatercontentis19.4%. Themassofthedishis131.2grams.(AfterU.S.Dept.ofInterior,1990.)

TimeinOven,minTotalOvenTime,minMassofSoil+Dish(g)

Solution:

a Thistotaltimeof4minutesisfrom3minutesbeforeandonemore minute,givingamassofsoil+dish=216.22grams,andsoon.

Mt =231 62 131 2=100 42g

Mw =(0.194)Ms

Mt =0 194Ms + Ms

Ms = Mt 1.194 =84 10g(thisvalueisconstantthroughoutdryingperiod)

w =column5= Mw Ms × 100= 231 62 column3 84 10 × 100

Conclusion:Thelossofadditionalwaterinthesoilsamplebecomesnegligibleafter8to10minutes inthemicrowaveovenusedintheexperiment.

TimeinOven,minTotalOvenTime,minMassofSoil+Dish(g)Massofwater(g) wn (%) 00231.62

a Thistotaltimeof4minutesisfrom3minutesbeforeandonemoreminute,givingamassofsoil+ dish=216.22grams,andsoon.

2.22 Thevolumeofwaterinasampleofsoilwith S =85%is0.32m3.Thevolumeofsolids Vs is 0.34m3.Giventhatthedensityofsoilsolids ρs is2690kg/m3,find(a)thewatercontent,(b)the voidratio,(c)theporosity,(d)thetotalorwetdensityand(e)thedrydensity.Giveyouranswers inkg/m3,andlbf/ft3

Solution:

a.)watercontent

Ms = ρsVs =2690(0 34)=914 6kg

Mw = ρwVw =1000(0 32)=320kg

w = Mw Ms × 100= 320 914.6 × 100=35 0%

b.)voidratio

P.2.21

Gs = ρs ρw = 2690 1000 =2 69

e = Gsw S = 2 69(35) 85 =1.11

c.)theporosity

n = e 1+ e × 100= 1 107 1+1 107 × 100=52 5%

d.)totalorwetdensity

Vt = Vs + Vv = Vs + eVs = Vs(1+ e)=0 34(1+1 107)=0 716m3

ρt = Mt Vt = 1234 6 0 716 =1723.17kg/m 3 =107.57pcf

ρd = Ms Vt = 914 94 0 716 =1276 54kg/m 3 =79 69pcf

2.23 A592-cm3 volumeofmoistsandweighs1090g.Itsdryweightis920gandthedensityof solidsis2680kg/m3.Computethevoidratio,porosity,watercontent,degreeofsaturation,and totaldensityinkg/m3

Solution:

ρs =2680kg/m 3 =2 68g/cm3

Vs = Vs ρs = 920 2 68 =343.284cm3

Vv = Vt Vs =592 343 284=248 716cm3

a.)voidratio

e = Vv Vs = 248 716 343 284 =0 724=0 72

c.)theporosity

n = e 1+ e × 100= 0 7245 1+0 7245 × 100=42 0%

d.)watercontent

Mw =1090 920=170g

w = Mw Ms × 100= 170 920 × 100=18 5%

d.)saturation

Vw = Mw ρw =170cm3

S = Vw Vv × 100= 170 248.716 × 100=68 4%

e.)totalorwetdensity

ρt = Mt Vt = 1090 592 =1.841g/cm3 =1841kg/m 3

2.24 Thesaturatedunitweight γsat ofasoilis128lbf/ft3.Findthebuoyantunitweightofthis soilinlbf/ft3 andbuoyantdensityinkg/m3 .

Solution:

γ = γsat γw =128 62.4=65.6lbf/ft3

ρ =(65.6lbf/ft3) 16 018kg/m3 1lbf/ft3 =1050.8kg/m 3

2.25 Asandiscomposedofsolidconstituentshavingaunitweightof168.5lb/ft3.Thevoidratio is0.62.Computetheunitweightofthesandwhendryandwhensaturatedandcompareitwith theunitweightwhensubmerged.

Solution:

Assume Vs =1ft3

Vv = eVs =0.62ft3

Vt = Vs + Vv =0 62+1=1 62ft3

Ws = ρsVs =168 5(1)=168 5lb

γd = Ws Vt = 168.5 1.62 =168 5lb/ft3

γd = Ws Vt = 168 5 1 62 =168 5lb/ft3

WhensoilissaturatedS=100%,then Vw = Vv =0 62lb/ft3

Ww = γwVw =62 4(0 62)=38 69lb

γsat = Ws + Ww Vt = 168 5+38 69 1 62 =127.9lb/ft3

γ = ρsat ρw =127.9 62.4=65.49lb/ft3

2.26 Asampleofnaturalglacialtillwastakenfrombelowthegroundwatertable.Thewatercontentwasfoundtobe43%.Estimatethewetdensity,drydensity,buoyantdensity,porosity,and voidratio.Clearlystateanynecessaryassumption.

Solution:

Assume Vs =1m3 and Gs =2.70

a.)totalorwetdensity

Ms = GsVsρs =2 7(1)(1000)=2700kg/m 3

Mw = wMs =0.43(2700)=1161kg/m 3

Mt = Ms + Mw =2700+1161=3861kg/m 3

S = Vw Vv =1 ⇒ Vw = Vv = Mw ρw =1.161m3

TBEXAM.COM

Vt = Vv + Vs =1 161+1=2 161m3

ρt = Mt Vt = 3861 2 161 =1786.7kg/m 3

b.)drydensity

ρd = Ms Vt = 2700 2 161 =1249.4kg/m 3

c.)buoyantdensity

ρ = ρsat ρw =1786.67 1000=786.7=786.7kg/m 3

c.)porosity

n = Vv Vt = 1.161 2 161 =53 7%

d.)voidratio

e = Vv Vs = 1 161 1 =1 2

2.27 Thevoidratioofclaysoilis0.58andthedegreeofsaturationis80%.Assumingthespecific gravityofsolidsis2.69,compute(a)thewatercontentand(b)dryandwetdensitiesinbothSI andBritishimperialunits.

Solution:

Assume Vs =1 0m3

Vv = eVs =0 58(1 0)=0 58m3

Vw = SVv =0 80(0 58)=0 464m3

Vt =0.58+1=1.58m3

Mw = ρwVw =(1000)(0.464)=464kg

Ms = ρsVs =(2690)(1)=2690kg

Mt = Ms + Mw =2690+464=3154kg

a.)watercontent:

w = Mw Ms × 100= 464 2690 =17 2%

b.)drydensity

ρd = Ms Vt = 2690 1 58 =1702kg/

c.)wetdensity

ρt = Mt Vt = 2690 1 58 =1996kg/m 3 =124 6lb/ft3

2.28 Thevaluesofminimumnandmaximumnforapuresilicasand(Gs =2.7)werefoundtobe 35%and42%,respectively.Whatisthecorrespondingrangeinthesaturatedunitweightinlb/ft3?

Solution:

maximum: ρsat =

minimum: ρsat =

168 5+62 4(0 538) 1+0 538 =131.35pcf

168 5+62 4(0 724) 1+0 724 =123 926pcf

2.29 Calculatethemaximumpossibleporosityandvoidratioforacollectionof(a)basketballs (9in.indiameter),(b)tennisballs(2.7in.indiameter)and(c)tinyballbearings0.01in.in diameter.Writeasentenceaboutwhatcausesthedifferencesacrossthesesize“particles.”

Solution:

Three-dimensionalparticlearrangementofequalsphereshasbeenstudiedindepthbymathematicians,statisticians,andmaterialsscientistssincethe1600s.Aquickinternetsearchonpacking ofequalsphereswillrevealnumerousmathematicaltheoriesandapproachesforestimatingthe densestandloosestpossiblepacking.Ingeneral,theloosestarrangementofequalspheresyields avoidfractionofabout0.48,regardlessofspheresize.Asanaside,thedensestpossiblepacking ofequal-sizespheresyieldsasolidsvolumeofabout: Vs = π/√18=0.7405(Thesevaluesare approximate–thereisnotaunifiedconsensusintheliterature.)

Loosestpacking

For Vt =1.0,Vv =0.48, and Vs =1 0.48=0.52;thus:

nmax = Vv Vt × 100= 0.48 1 0 × 100=48%

emax = Vv Vs = 0 48 0.52 =0 92

TBEXAM.COM

Densestpacking (notrequiredinproblemstatement)

Vs =0 7405,Vv =1 0 7405=0 2595, thus:

nmax = Vv Vt × 100= 0 2595 1 0 × 100=26%

emax = Vv Vs = 0.595 0 7405 =0 35

2.30 Aplastic-limittesthasthefollowingresults:

Wetweight+container=24 75g

Dryweight+container=19 37g

Containerweight=1 46g 23 Co pyrig ht © 2 0 2 3 Pe a rson

ComputethePLofthesoil.

Solution:

Mw =24 75 19 37=5 38g

Ms =19.37 1.46=17.91g

PL = Mw Ms × 100= 5 38 17 91 × 100=30%

2.31 Duringaliquid-limittest,thefollowingdatawasobtainedforoneofthesamples::

Wetweight+container=24 29g

Dryweight+container=17 62g

Containerweight=1 50g

AndthegrooveintheCasagrandecupclose1/2”at22blows.WhatistheLLofthsoil.

Solution:

Usetheone-pointmethod.

Mw =24.29 17.62=6.67g

Ms =17 62 1 5=16 12g

wn = Mw Ms × 100= 6 67 16 12 × 100=41.4%

UseEq.(2.38)

2.32 The“chunkdensity”methodisoftenusedtodeterminetheunitweight(andothernecessary information)ofaspecimenofirregularshape,especiallyoffriablesamples.Thespecimenatits naturalwatercontentis(1)weighed,(2)paintedwithathincoatowaxorparaffin(toprevent waterfrommovingintooroutofthepores),(3)weighedagain(Wt + Wmax),and(4)weighed inwater(togetthevolumeofthesample+waxcoating-rememberArchimedes?).Finally,the naturalwatercontentofthespecimenisdetermined.Aspecimenofcementedsiltysandistreated inthiswaytoobtainthe“chunkdensity.”Fromtheinformationgivenbelow,determinethe(a) wetdensity,(b)drydensity,(c)voidratio,and(d)degreeofsaturationofthesample.

Given:

Weightofspecimenatnaturalwatercontentinair=181 8g

Weightofspecimen+waxcoatinginair=215 9g

Weightofspecimen+waxinwater=58.9g

Naturalwatercontent=2.5%

Soilsoliddensity, ρs =2650kg/m 3

Waxsoliddensity, ρwax =940kg/m 3

Phasediagram!!

Solution: w = Mt Ms Ms ⇒ Ms = Mt 1+ w = 181 8 1+0 025 =177.366g

Mw = Mt Ms =181.8 177.4=4.43g

Vs = Ms ρs = 177.366g 2 650g/cm3 =66 93cm3

Vw = Mw ρw = 4.43g 1 0g/cm3 =4 43cm3

Mwax = Mt+wax Mwaterdisplaced ⇒ Mwaterdisplaced =215.9 58.9=157.0g

Vt+wax = Mwaterdisplaced ρw = 157 1.0 =157 0cm3

TBEXAM.COM

Vair = Vt+wax Vwax Vw Vs =157 0 36 28 4 43 66 93=49 36cm3

Vv = Vair + Vw =49 36+4 43=53 79cm3

Vt = Vt+wax Vwax =157.0 36.28=120.72cm3

a.)totaldensity

ρt = Mt Vt = 181 8 120 72 =1.50g/cm3 =1506kg/m 3

b.)drydensity

ρd = Ms Vt = 177 366 120.72 =1 47g/cm3 =1469kg/m 3

c.)voidratio

e = Vv Vs = 53.79 66.93 =0 80

c.)saturation

S = Vw Vv × 100= 4 43 53 79 =8 2%

P.2.32.Phasediagram

2.33 Asensitivevolcanicclaysoilwastestedinthelaboratoryandfoundtohavethefollowing properties:

(a) ρ =1280kg/m3 (b)e=9.0 (c)S=95%

(d) ρs =2750kg/m3 (e)w=311%

Inrecheckingtheabovevalues,onewasfoundtobeinconsistentwiththerest.Findtheinconsistent valueandreportitcorrectly.Showallyourcomputationsandphasediagrams.

Solution:

Assume Vs =1m3

Ms = ρsVs =2750kg

Mw = wMs =3 11(2750)=8552 5kg

Mt = Ms + Mw =2750+8552.5=11302.5kg

Vt = Mt ρt = 11302.5 1280 =8 83m3

Vv = eVs =9(1)=9m3

Vt = Vv + Vs =9+1=10m3 =8 83m3

Check: Gsw = Se

2.65(3.11)=(0.95)(9.0)

8 55=8 55Thesevaluesareincorrectproportion

Consequently,theerrormustbein ρt value.

Re-calculate ρt = Mt Vt = 11302.5 10 =1130 2kg/m 3

Solution ρt =1130 2kg/m 3

P.2.33.Phasediagram

2.34 A15cmhighcylindercontains847cm3 ofloosedrysandwhichweighs843g.Undera staticloadof200kPathevolumeisreduced3%,andthenbyvibrationitisreduced12%fromthe originalvolume.Assumethesoliddensityofthesandgrainsis2710kg/m3.Computethevoid ratio,porosity,drydensity,andtotaldensitycorrespondingtoeachofthefollowingcases:

(a)Loosesand

(b)Sandunderstaticload

(c)Vibratedandloadedsand

Solution:

Vs = Ms ρs = 843 2 71 =311 1cm3 ; Mt =843g

drysand: Mw = Vw =0

a.)loosesand-Initialcondition

Vt =847cm3

Vv = Vt Vs =847 311 1=535 93cm3

e = Vv Vs = 535 93 311 1 =1.72

n = Vv Vt × 100= 535 93 847 × 100=63.3%

ρd = ρt = Ms Vt = 843 847 =0.995g/cm3 =995kg/m 3

b.)Sandunderstaticload

Vt =847 847(0 03)=821 59cm3

Vv = Vt Vs =821 59 311 1=510 52cm3

e = Vv Vs = 510 52 311 1 =1.64

n = Vv Vt × 100= 510 52 821.59 × 100=62 14%

ρd = ρt = Ms Vt = 843 821 59 =1 026g/cm3 =1026kg/m 3

c.)Vibratedandloadedsand

Vt =847 847(0.12)=723cm3

Vv = Vt Vs =723 311 1=411 93cm3

e = Vv Vs = 411.93 311.1 =1 32

n = Vv Vt × 100= 411 93 723 × 100=57 0%

TBEXAM.COM

ρd = ρt = Ms Vt = 843 723 =1.166g/cm3 =1166kg/m 3

2.35 Onfive-cyclesemilogarithmicpaper,plotthegrain-sizedistributioncurvesfromthefollowing mechanicalanalysisdataonsixsoils,AthroughF.Foreachsoildeterminetheeffectivesizeas wellastheuniformitycoefficientandthecoefficientofcurvature.Determinealsothepercentages ofgravel,sandsilt,andclayaccordingto(a)ASTM,(b)AASHTO,(c)USCS,and(d)theBritish Standard.

U.S.StandardSieveNo.PercentPassingbyWeight orParticleSizeSoilASoilBSoilCSoilDSoilESoilF 75mm(3in.)100.00100 38(1-1/2)70— 19(3/4)4910091

9.5(3/8)36—87 No.4278881100 No.1020827010089 No.20—80—99— No.40878499163 No.60—74—37— No.1005——9— No.140—6535460 No.20045532—57100

40 µm331274199

20 µm219223592 10 µm113182082 5 µm <11014871 2 µm——11—52 1 µm—210—39

Note:Missingdataisindicatedbyadashinthecolumn.

Solution:

Effectivesize,uniformitycoefficientandcoefficientofcurvature:

Effectivesize

A0.662846.72.1 B0.0050.040.0918.03.6 C0.0010.0611000.03.6 D0.160.220.31.91.0

E0.0060.0150.116.70.4 FN/DN/D0.003N/DN/D

P.2.35.Granulometriccurve

P.2.35.Soilclassificationchartafter(BardetandJean-Pierre,1997)

(a)Table.PercentagesaccordingtoASTM

Percentage(%)SoilASoilBSoilCSoilDSoilESoilF

Gravel731219000

Sand233349100430

Silt4451804929

Clay010140871

Fines(silt+clay)45532057100

Check:100100100100100100

(b)Table.PercentagesaccordingtoAASHTO

Percentage(%)SoilASoilBSoilCSoilDSoilESoilF

Gravel8018300110

Sand162738100320

Silt4451804929

Clay010140871

Fines(silt+clay)45532057100

Check:100100100100100100

(c)Table.PercentagesaccordingtoUSCS

TBEXAM.COM

Percentage(%)SoilASoilBSoilCSoilDSoilESoilF

Gravel731219000

Sand233349100430

Silt------

Clay------

Fines(silt+clay)45532057100

Check:100100100100100100

(d)Table.PercentagesaccordingtotheBritishStandard

Percentage(%)SoilASoilBSoilCSoilDSoilESoilF

Gravel8018300110

Sand162738100320

Silt4452105748

Clay010110052

Fines(silt+clay)45532057100

Check:100100100100100100

2.37 ThesoilsinProblem2.35havethefollowingAtterberglimitsandnaturalwatercontents. DeterminethePIandLIforeachsoilandcommentontheirgeneralactivity.

Solution:

Fromequations:

SoilA:verysensitive,highlyactive

SoilB:mostlikelyaclayabovethewatertablethathasexperiencedadecreaseinmoisture

SoilC:mostlikelyaclayabovethewatertablethathasexperiencedadecreaseinmoisture

SoilD:mostlikelyafinesand

SoilE:mostlikelyasilt

SoilF:slightlysensitiveandactive

2.38 CommentonthevalidityoftheresultsofAtterberglimitsonsoilsGandH.

Solution:

BasedonAtterberg’sdefinitions:LL >> PL > SL.SoilGviolatesthedefinitionsbecausetheSL > PL(25 > 20).SoilHviolatesthedefinitionsbecausethePL > LL(42 > 38).

2.39 Thefollowingdatawereobtainedfromaliquid-limittestonasiltyclay.

Twoplastic-limitdeterminationshadwatercontentsof24.2%and23.5%.DeterminetheLL,PI,the flowindex,andthetoughnessindex.Theflowindexistheslopeofthewatercontentversuslogof numberofblowsintheliquid-limittest,andthetoughnessindexisthePIdividedbytheflowindex.

Solution:

Theflowindexistheslopeofthetrendlineequalto15.7%

Thetoughnessindexisdefinedas:=

2.40 ClassifythefollowingsoilsaccordingtotheUSCS:

(a)Asampleofwell-gradedgravelwithsandhas73%finetocoarsesubangulargravel,25%fineto coarsesubangularsand,and2%fines.Themaximumsizeoftheparticlesis75mm.Thecoefficient ofcurvatureis2.7,whiletheuniformitycoefficientis12.4.

Solution: GW–Well-gradedgravelwithsand.

(b)Adarkbrown,wet,organic-odorsoilhas100%passingtheNo.200sieve.Theliquidlimitis 32%(notdried,andis21%whenovendried!)andtheplasticindexis21%(notdried).

Solution: OL–OrganicclayorOrganicsilt.

(c)Thissandhas61%predominatelyfinesand,23%siltyfines,and16%finesubroundedgravel size.Themaximumsizeis20mm.Theliquidlimitis33%andtheplasticlimitis27%.

Solution: SM–Siltysandwithgravel

(d)Thismaterialhas74%finetocoarsesubangularreddishsandand26%organicandsiltydark brownfines.Theliquidlimit(notdried)is37%whileitis26%whenovendried.Theplasticindex (notdried)is6.

Solution: SM–Siltysandwithorganicfines

(e)Althoughthissoilhasonly6%nonplasticsiltyfines,ithaseverythingelse!Ithasgravelcontentof78%finetocoarsesubroundedtosubangulargravel,and16%finetocoarsesubrounded tosubangularsand.Themaximumsizeofthesubroundedbouldersis500mm.Theuniformity

SievePercentFinerbyWeight 1/2–83 No.452 No.1040 No.2032 No.4017 No.6010 No.1008

coefficientis40,whilethecoefficientofcurvatureisonly0.8.(AfterU.S.Dept.oftheInterior,1990.)

Solution: GP-GM–Poorlygradedgravelwithsilt,sand,cobbles,andboulders(orGP-GC)

2.41 Youknowwhatiscoming.Classifythefivesoilsintheprecedingquestionaccordingtothe AASHTOmethodofsoilclassification.Youcanfindproceduresfordoingthisinthereferences giveninSection2.9orontheWeb

Solution:

(a)A-1-a (b)A-8

(c)A-2-4

(d)A-2-4orA-8

(e)A-1-a

2.42 Theresultsofasievetestbelowgivethepercentagepassingthroughthesieve.

(a)Usingaspreadsheet,plottheparticle-sizedistribution.

(b)Calculatetheuniformitycoefficient.

(c)Calculatethecoefficientofcurvature.

Solution:

(a)particle-sizedistribution

(b)Uniformitycoefficient:

(c)coefficientofcurvature:

P.2.42.Particle-sizedistribution

2.43 Forthedatagivenbelow,classifythesoilsaccordingtotheUSCS.Foreachsoil,giveboth thelettersymbolandthenarrativedescription.

(a)65%materialretainedonNo.4sieve,32%retainedonNo.200sieve.Cu=3,Cc=1.

Solution: GP–Poorlygradedgravelwithsand (b)100%materialpassedNo.4sieve,90%passedNo.200sieve.LL=23,PL=17.

Solution: CL-ML–Siltyclay

(c)70%materialretainedonNo.4sieve,27%retainedonNo.200sieve.Cu=5,Cc=1.5.

Solution: GW–Well-gradegravelwithsand

2.44 Asampleofsoilwastestedinthelaboratoryandthefollowinggrain-sizeanalysisresultswere obtained.LL=34,PL=21,28%iscoarserthanthe1/2-inchsieve.Classifythissoilaccordingto theUSCS,providingthegroupsymbolforit.

Solution: PI = LL PL =34 21=13% Cu = D60 D10 = 2 8 0 065 =43.1(2.24)

SieveNo.SieveOpening(mm)PercentCoarserbyWeightPercentfinerbyweight 44.752872

P.2.44.Particle-sizedistribution

SC-CL–Clayeysandwithgravel

2.45 AminusNo.40materialhadaliquidityindexof0.67,anaturalwatercontentof38.7%,and aplasticityindexof18.9.ClassifythissoilaccordingtotheUSCS,providingthegroupsymbolfor it.Youdonotneedtographthisdata;uselinearinterpolationifyouneedspecificvaluesnotgiven.

2.46 Asampleofsoilwastestedinthelaboratoryandthefollowinggrain-sizeanalysisresultswere obtained.LL=60,PL=26.ClassifythissoilaccordingtotheUSCS,providingthegroupsymbol forit.

Solution:

P.2.46.Particle-sizedistribution PI = LL PL =60 26=34%

u = D60 D10 = 4 75 0.075 =63(2.24)

SieveNo.SieveOpening(mm)PercentFinerbyWeight

2.47 Asampleofsoilwastestedinthelaboratoryandthefollowinggrain-sizeanalysisresultswere obtained.AtterberglimitsonminusNo.40materialwere:LL=38,PL=19.

Solution:

PI = LL PL =38 19=19%

2.48 Laboratorytestingwasperformedontwosoilsamples(AandB)andthedataissummarized inthetable.DeterminetheUSCSclassificationforsamplesAandB.Useloginterpolationas

SW-SC–Well-gradedsandwithclay

P.2.47.Particle-sizedistribution

SC-CL–Clayeysand

necessary.

Solution:

SieveNo.SampleA-%Passing(mm)SampleB-%Passing 3in.(76.2mm)1001.5in.(38.1mm)980.75in.(19.1mm)964(4.75mm)77100 10(2.00mm)Notused96 20(0.85mm)5594 40(0.425mm)Notused73

100(0.150mm)30Notused 200(0.075mm)1855

Liquidlimit3252 Plasticlimit2532

TBEXAM.COM

(a)ForSampleA: PI = LL PL =32 25=7% Cu = D60 D10 = 1 3 0 044 =29 5(2.24)

P.2.48

P.2.48.Particle-sizedistribution

SM—Siltysandwithgravel.

(b)ForSampleB:

PI = LL PL =52 32=20%

MHorOH—Elasticsandysiltororganicsandysilt/clay.

2.49 Asampleofsoilwastestedinthelaboratoryandthefollowinggrain-sizeanalysisresultswere obtained.AtterberglimitsonminusNo.40material:LL=56,PL=25.DeterminetheUSCS lettersymbol(e.g.,GP)forthissoil.

PI = LL PL =56 25=31%

Cu = D60 D10 = 0 31 0 055 =5.6(2.24) Cc = D2 30 D60 × D10 = 0.152 0.31(0.055) =1 3 ⇒

SC-CH—Clayeysand

P.2.49.Particle-sizedistribution

2.50 AsampleofabrownsandyclaywasobtainedtodetermineitsAtterberglimitsandthen classifyitssoiltypeaccordingtotheUnifiedSoilClassificationSystem.ForoneofthePLdeterminations,thewetweight+dish=11.53gandthedryweight+dish=10.49g.Thedishonly weighed4.15g.Computetheplasticlimit.Anotherplasticlimitwas16.9%.Threedeterminations oftheliquidlimitweremade.For17blows,thewatercontentwas49.8%;for26blows,thewater contentwas47.5%;andfor36blows,thewatercontentwas46.3%.Evaluatethesoiltype,indicate theinformationonaplasticitychart,andgivetheUnifiedSoilClassificationsymbol.

Solution: Mw =11 53 10 49=1 04g

Ms =10.49 4.15=6.34g

wPL = PL = Mw Ms × 100= 1 04 6.34 × 100=16 4

PLavg = 16.4+16.9 2 =16 65;Fromplotbelow:LL=48

PI =48 16 7=31

BasedonCasagrande’splasticitychart,thesoilfinesclassifyasCL—Sandyleanclay.

CHAPTER3:GEOLOGY,LANDFORMS,ANDTHEORIGINOFGEOMATERIALS

CHAPTER3:PROPERTIESOFROCKMASSES,SOIL PARTICLES

3.15 Inonecorerunof1500mmselectedfromcoresobtainedduringdrillingforabridgefoundation inhardlimestone,thefollowingcorerecoveryinformationwasobtained: Determine(a)thepercentcorerecovery,and(b)theRQD.BasedonthisRQD,whatistherock quality?Why?

P.3.15 CoreRecoveryLengthofCorePieces (mm) > 100mm

Solution:

(a)Corerecovery:

CR = Totallengthofrockrecovered Totalcorerunlength × 100 CR =

(b)RQD:

RQD = Lengthofsoundpieces

× 100

RQD = 150+100+125+100+125 1500 = 850 1500 × 100=56.7%

BasedontheRockMassClassificationsshowninTable3.6,thelimestonerockqualitywouldbe deemedfair.

3.16 Calculatethespecificsurfaceofacube(a)25mm,(b)2.5mm,(c)2.5 µm,and(d)2.5nm onaside.Calculatethespecificsurfaceintermsofbothareasandm2/kg.Assumeforthelatter casethat γs =2710kg/m3

Solution:

Solveusingthefollowingequation:

specificsurface= surfacearea unitvolume

(a)specificsurface= 6(625mm2) 253 mm3 = 3750mm2

intermsofmass:specificsurface= 0 24/ mm × (1000mm/1m) 2710kg/m3 =0.089m2/kg

(b)specificsurface= 6(6 25mm2) 2.53 mm3 = 37 5mm2 15.625mm3 =2.4/mm

TBEXAM.COM

intermsofmass:specificsurface= 2 4/ mm × (1000mm/1m) 2710kg/m3 =0.89m2/kg

(c)specificsurface= 6(6.25 µm2) 2.53 µm3 = 37.5 µm2 15.625 µm3 =2 4/µm=2400/ mm

intermsofmass:specificsurface= 2400/ mm × (1000mm/1m) 2710kg/m3 =885 6m2/kg

(d)specificsurface= 6(6 25nm2) 2 53 nm3 = 37 5nm2 15 625nm3 =2.4/nm=2, 400, 000/ mm

intermsofmass:specificsurface= 2, 400, 000/ mm × (1000mm/1m) 2710kg/m3 =885, 608.8m2/kg

3.18 Aspeciallyprocessedclayhasparticlesthatare600nmthickand12,000nm × 12,000nm wide.Thespecificgravityofsolidsis2.72.Theparticleslieperfectlyparallelwithanedge-to-edge spacingof375nm(i.e.,theylooklikethinbricksstackedperfectlyparallel).(a)Initially,thecation valenceinthedoublelayeris+1,resultinginaface-to-facespacingof1500nm.Howmanyparticlespercm3 willtherebeatthisspacing?Whatarethevoidratioandwatercontent,assuming

thatthesoilisat100%saturation?(b)Anothersampleoftheclayismixedsuchthatthecation valenceis+2.Whatarethenewvoidratioandwatercontentundertheseconditions?Assume theedge-to-edgespacingremains375nmandthatS=100%.(AfterC.C.Ladd.)

Solution:

Assumesymmetricaledge-to-edgespacingof375nmandonlyincludewholeparticles.

(a)thehorizontalspacingbetweentwoparticleisequalto:12,000+375=12,375nm.Then,the numberofparticlesinthisplaneisequalto:

no.ofparticlesinplan= 1, 000, 000 12, 000 12, 375 +1=808 1=808particles

Totalno.ofparticlesinplan=808 ∗ 808=652, 864particles

Theverticalspacefortwoparticlesisequalto600+1500=2100nm.Then,thenumberofparticles intheverticalis:

no.ofparticlesinvertical= 1, 000, 000 600 2100 +1=4762 6=4762particles

Totalno.ofparticlesin1cm3 =652, 864 × 4762=3108 × 106 particles

Thevolumeofofeachparticleis: V =600 × 1200 × 1200nm3 =8.64 × 1010 nm 3 =8.

× 10 11 cm3 thevolumeofthesolidsis: Vs =8 6 × 10 11 cm3 × 3108 × 106 particles=0 269cm3

ForS=100% ⇒ Vw = Vv =1cm3 0 269cm3 =0 731cm3

Thevoidratiocanbecalculateas:

e = Vv Vs = 0 731 0 269 =2 72

Forthewatercontent: Ms = VsGsρw =(0.269)(2.72)(1g/cm3)=0.73

w = Mw Ms = 0 73 0 73 × 100=100%

(b)theonlychangerespecttoparta.istheverticalspacing,whichisequalto=600+1500*2= 3600nm.

no.ofparticlesinvertical= 1, 000, 000 600 3600 +1=2778 6=2778particles

Totalno.ofparticlesin1cm3 =652, 864 × 2778=1840 × 106 particles

thevolumeofthesolidsis

Vs =8 64 × 10 11 cm3 × 1840 × 106 particles=0 157cm3

ForS=100% ⇒ Vw = Vv =1cm3 0 157cm3 =0 84cm3

Thevoidratiocanbecalculateas:

e = Vv Vs = 0.84 0.157 =5 38

Forthewatercontent:

Ms = VsGsρw =(0 157)(2 72)(1g/cm3)=0 426

w = Mw Ms = 0 84 0 426 × 100=197 9%

3.21 Canasoilhavealowvoidratioandahighdensityatthesametime?Explain.

Solution:

Thespecifygravityofthesolidshavetobeverylarge.

TBEXAM.COM

CHAPTER4:COMPACTIONANDSTABILIZATIONOF SOILS

4.1 ForthedatainFig.4.2:

(a)Estimatethemaximumunitweightandoptimumwatercontentforboththestandardcurve andthemodifiedProctorcurve.

(b)Whatisthewatercontentrangetoachieve90%relativecompactionforthemodifiedProctor curveand95%relativecompactionforthestandardProctorcurve?

(c)IfahighercompactiveeffortthanthemodifiedProctorwasusedandachievedamaximumunit weightof120lb/ft3,whatisyourpredictionofthewatercontentatwhichthatwasachieved?

Solution:

(a)Forthestandardproctor: ρd max =1730kg/m 3 =108lb/ft3 ; wopt =16%

Forthemodifiedproctor: ρd max =1875kg/m 3 =117lb/ft3 ; wopt =12%

(b)95%ofStandardProctorcurve:0.95×ρd max =(0 95)(1730)=1643kg/m3

watercontentrangeforStandardProctor: w =11to20%

90%ofModifiedProctorcurve:0.90×ρd max =(0.90)(1875)=1687kg/m3

watercontentrangeforStandardProctor: w =5to18%

(c)thewatercontentforthemaximununitweightof120lb/ft3 is10.3%

4.2 Thenaturalwatercontentofaborrowmaterialisknowntobe12%.Assuming4.75kgofwet soilisusedforlaboratorycompactiontestpoints,computethevolumeofwateristobeaddedto other4.75samplestobringtheirwatercontentsupto14%,17%,20%,23%,and26%.

Solution: w =0 12= Mw Ms ⇒ Mw =0 12Ms; Mt =4 75kg

Ms remainsconstant.Determine Ms forthenaturalmoisturecondition.

Mt = Mw + Ms =4 75kg; 4.75=0.12Ms + Ms ⇒ Ms =4.241kg; Mw fornaturalcondition:Mw =(0 12)(4 241)=0 509kg

(a) w =14%

Mw = wMs =(0.14)(4.241)=0.509kg

additionalwaternecessaryforw=14%:0 594 0 509=0 085kg

(b) w =17%

Mw = wMs =(0.17)(4.241)=0.721kg

additionalwaternecessaryforw=17%:0 721 0 509=0 212kg

(c) w =20%

Mw = wMs =(0.20)(4.241)=0.848kg

additionalwaternecessaryforw=20%:0 848 0 509=0 339kg

(d) w =23%

Mw = wMs =(0.23)(4.241)=0.975kg

additionalwaternecessaryforw=23%:0 975 0 509=0 467kg

(e) w =26%

Mw = wMs =(0.26)(4.241)=1.103kg

additionalwaternecessaryforw=26%:1 103 0 509=0 594kg

4.3 ForthesoilshowninFig.4.2,afielddensitytestprovidedthefollowinginformation:

Watercontent=13%

Wetunitweight=115lb/ft3

ComputethepercentrelativecompactionbasedonthemodifiedProctorandthestandardProctor curves.

Solution:

Dryunitweightinfield:

FromFig.4.2:StandardProctor γdmax =108.0pcf;ModifiedProctor γdmax =117.1pcf Therelativecompactionisequalto: RC= γdfield γ

4.4 Forthedatagivenbelow ρs =2710kg/m3:

(a)Plotthecompactioncurves.

(b)Establishthemaximumdrydensityandoptimumwatercontentforeachtest.

(c)ComputethedegreeofsaturationattheoptimumpointfordatainColumnA.

(d)Plotthe100%saturation(zeroairvoids)curve.Alsoplotthe70%,80%,and90%saturation curves.Plotthelineofoptimums.

A(modified)B(modified)C(lowenergy) ρd (kg/m3) w (%) ρd (kg/m3) w (%) ρd (kg/m3) w (%) 18739.316919.3162710.9 191012.8171511.8163912.3 180315.5175514.3174016.3 169918.7174717.6170720.1 164121.1168520.8164722.4 161923

Solution:

(a)SeeFig.P4.4.

(b)

Test wopt (%) ρmax

Modified12.01918

Standard15.51761

Lowenergy17.31747

(c)FromEq.4.1:

(d)FromEq.4.1:

P.4.4

P.4.4