CE222 SM 12 Stresses in Soil

SoilMechanicsICE222CE 222

StressesinSoil

1

Stressesinsoilfromsurface&interiorloads 2

Itisimportanttoknowhowthesurfacestressesaredistributedwithinthesoil&theresultingdisplacementsto estimate:toestimate:

whetherthesoilundergeosystemwouldfail,or

the resulting displacements are excessive or theresultingdisplacementsareexcessive,or

whethernearbystructureswouldbenegativelyaffected.

Soil is considered a semi infinite homogeneous linear Soilisconsideredasemiinfinite,homogeneous,linear,isotropic,elasticmaterial.

Semiinfinite mass is bounded on one side and extendsSemi infinitemassisboundedononesideandextendsinfinitelyinallotherdirections.Thisiscalledelastichalfspace.

Forsoils,horizontalsurfaceistheboundingside.

Stressesinsoilfromsurface&interiorloads 3

Sinceweareassumingsoilisanelasticmaterial,wecanusetheprincipleofsuperpositiontodeterminep p p pstressdistributionforcomplexloadingsby

decomposing the complex loading into simple loads (e.g.decomposingthecomplexloadingintosimpleloads(e.g.rectangularorcircular)and

addingthesolutionofeachofthesesimpleloads.g p

Thestressincreasefromsurfaceloadsaretotalstressincreasesincreases.

Ifsoilsisdryorverypermeable(e.g.cleancoarsei d il ) th th t th tgrainedsoils),thenwecanassumethatthestress

increaseareeffectivestressincreases.

Boussinesqformulaforpointload 4

Boussinesq(1883)solvedtheproblemofstressproducedbyanypointloadonfollowingassumptions.

The soil mass is elastic, isotropic, homogeneous and semiinfinite.Thesoilmassiselastic,isotropic,homogeneousandsemi infinite. Thesoilmassisweightless. Theloadisapointloadactingonthesurface.

xyr

P P

y xy

zL z Lz

z yx


( ) ( )

++= 23

2

2

22

5

2

2132 rL

zyzLLr

yxL

zxPx ( ) 2522

3

5

3

23

23

zrzP

LPz

z +== h( ) ( )

++= 23

2

2

22

5

2

2132 rL

zxzLLr

xyL

zyPy

22222

22 zxr +=where

P

22222 zrzyxL +=++=Vertical normal stress z is independent of Poissons ratio.

x

yx

yrP

= poissonsratio

yzL z

z yx


( ) 25223

5

3

23

23

zrzP

LPz

z +== ( )The above relationship for z can be re-written as

( )[ ] Bz IzPzrzP 225 22 1123 =

+= P

where ( )[ ] 25 2 1123 +=IB ( )[ ]12 +zr


( )[ ] Bz IzPzrzP 225 22 1123 =

+=

P P

( )[ ]zr 1 +

Westergaards formulaforpointloads 8

Westergaard,aBritishScientist,proposed(1938)aformulaforthecomputationofverticalstressz byapointload,P,atthesurfaceas

( ) ( )( ) ( ) ( )[ ] Wz IzPzrzP 223 22 221 22212 =+ = ( ) ( ) ( )[ ]zr221 +

Ifpoissons ratio,,istakenaszero,theaboveequationsimplifiesto

( )[ ] Wz IzPzrzP 223 22 21 1 =+= where ( )[ ] 23 221 11 zrIW += ( )[ ]21 zr+

Westergaardvs Boussinesqcoefficient 9

( )[ ] 23 221 11 zrIW += [ ] 25123=IB

( )[ ]( )[ ] 252 12 +zrB

Th l f I t / 0 iThevalueofIW atr/z =0is0.32whichislessthanthatofIB by33%.

Geotechnicalengineerspreferto use Boussinesqs solutiontouseBoussinesq ssolutionasthisgivesconservativeresults.

Pointload examples 10

Example1:Aconcentratedloadof1000kN isappliedatthegroundsurface.Computetheverticalstress(i)atadepthof4mbelo the load (ii) at a distance of 3 m at the same depth Usebelowtheload,(ii)atadistanceof3matthesamedepth.UseBoussinesqsequation.

PP=1000KNZ=4matx,y=0I1 0 4775I1=0.4775verticalstress=????

Z=4m,r=3m

Verticalstresscausedbyalineload 11

d fBoussinesqequationisusedforcomputingz atanypointPinanelasticsemiinfinitemass.Usingthistheory,thestressesatanypointinthesoilmassduetolineloadofinfiniteextentactingatp gsurfacecanbeobtained.

ThestrainatanypointPintheYdirectionparalleltolineloadis

q/unit length

assumedequaltozero.Thisistermedasplanestraincondition.

3 x

zq/unit length

y( )22232zx

zqz +=

z

A

( )

zx

Lineload example 12

Example2:Followingfigureshowstwolineloadsandapointloadactingatthegroundsurface.DeterminetheincreaseinverticalstressatpointA,whichislocatedatadepthof1.5m.

P = 30 kN q2 = 10 kN/m q1 = 15 kN/m

2 m 2 m

3 m

zA

1.5 m

Overburdenpressureisnotincooperated

Verticalstresscausedbyastripload 13

( )22232zx

zqz += For line load => ( ) 32 d

Substitute qdr for q and (x r) for x =>( )

( )[ ]22232

zrx

zqdrd z +=

B=2bq = Load per unit area

Assume line loadqdr (force/length)

x

zdr

r x r zz

A

r xr

z


Applying the principal of superposition, the total stress by strip load is

B 2

( )[ ] drzrx qzdB

Bzz +==

2

2222

32

( )[ ]zrxB +2

( ) ( )

+

11

2tan

2tan

Bxz

Bxz

q ( )[ ]( )[ ]

=

222222

222

44BzxBz

qz

( )[ ] ++ 222222 4 zBBzx


In non-dimensional form

( )( )

( )( )

+

11

122tan

122tan

BBz

BBz

( ) ( )( ) ( ) ( )[ ]

+

= 22 122212121

BzBxBz

BxBx

qz

( ) ( ) ( )[ ]

( ) ( )[ ] ( ) ++

2222 2212221 BzBzBx

q 2

16

00 0.2 0.4 0.6 0.8 1

z/q

0

Verticalstresscausedbyastripload

1

Graphical representation 2

2

z

/

B

p pof equation

32

( )( )

( )( )

+

11

122tan

122tan

1 BxBz

BxBz

4

( ) ( ) ( )[ ]( ) ( )[ ] ( )

++

=

2222

22

2212221

12221

BzBzBx

BzBxBzqz

5

Stripload example 17

Example3Considerthefollowingfigure.Givenq =200kPa,B=6m,z =3m.Determinetheverticalstressincreaseatx =9,6,3,and0m.Plotthegraph.

B=2bq = Load per unit area

g p

x

zz

z

Ax

Verticalstressduetoembankmentloading 18

( ) ( )

+

+=

BB

BBBqo

z 22

121

2

21

=

+= zB

zB

zBBradians 1121

12111 tan ,tantan)(

B2 B1A simplified form of above equation is

H qo = Habove equation is

embankoz Iq=

z21 2

19B2/z log scale

Verticalstressduetoembankmentloading

embankoz Iq=I

e

m

b

a

n

k

B2 B1 Osterbergs chartf dH qo = H

B /z

fordeterminationofverticalstress

duetoembankment

z21 B1/z loading

Embankmentloading example 20

Example4:Anembankmentisshownbelow.DeterminethestressincreaseundertheembankmentatpointsA1 andA2.

21

qo=H =(17.5)(7)=122.5kN/m2Considerleftsideoffig.B 2 5 B 14 > B /z 2 5/5 0 5 B /z 14/5 2 8B1 =2.5,B2 =14=>B1/z=2.5/5=0.5,B2/z=14/5=2.8FromfigureIembank =0.445z =2(qoIembank)=2(122.5x0.445)=109.03kN/m2

23StressesunderuniformlyloadedcircularfootingConsiderelementaryareadA.LetdQ bethepointloadactingonthisareawhichisequaltoqdA. ( )( )[ ]drrdqqdAdQ == ( )( )[ ]drrdqqdAdQ Theverticalstressd atdepthz duetopointloaddQmaybeexpressedas

3 dd( ) 25223

23

zrdrrdzqd z +=

The integral form of the equation for entireTheintegralformoftheequationforentirecircularareamaybewrittenas

=== =

==

232 3 oo RrRr drrdqzd ( ) = == = +== 0 0 25220 0 2 rr zz zrdOnintegrationwehave 1

( )[ ]

+= 232 111

zRq

oz

24Stressesunderuniformlyloadedcircularfooting

Circularloadedarea example 25

Example5Awatertankisrequiredtobeconstructedwithacircularfoundationhavingadiameterof16mfoundedatadepthof2mbelowthegroundsurface.Theestimateddistributedloadonthefoundation is 325 kPa. Assuming that the subsoil extends to afoundationis325kPa.Assumingthatthesubsoilextendstoagreatdepthandisisotropicandhomogeneous.Determinethestressz atpoints(i)z =8m,r =0(ii)z =8m,r =8,(iii)z =16m,r0 and (iv) z 16 m r 8 where r is the radial distance from the=0,and(iv)z =16m,r =8,wherer istheradialdistancefromthe

centralaxis.Neglecttheeffectofthedepthofthefoundationonthestresses.

Stressescausedbyrectangularloadedarea 26

x

Considerasmallareadxdy.thepressureactingonthisareacanbereplacedbyaconcentratedloaddQ actingatitscenter.Hence

qdxdyqdAdQ ==q

y dxdy

qdxdyqdAdQTheincreaseinstressdz duetodQcanbewrittenas

z( ) 25223

23

zrqdxdyzd z +=

A

TheincreaseinstressatpointA duetoentireloadedrectangularareacanbedeterminedby integrating above eq.byintegratingaboveeq.

( ) recB L

zz qIdxdyqzd === 2522

33z( ) recy xzz qyzr + = =0 0 25222

Stressescausedbyrectangularloadedarea 27

( ) recB L

zz qIdxdyqzd === 2533 ( ) recy xzz qyzr + = =0 0 25222

nmnm

nmnmnmmn

++++

+++++

12

112

122

22

2222

22

nmnmnmmn

Irec

+++++

=

112tan

41

2222

221

LnBm

nmnm

== ++

,

1

zz

28

Stressescausedbyrectangularloadedarea

LnBm == ,z

nz

m ,

Logscale

Rectangularloadedarea differentcases 29

G

A BEA B

GF

D CD CCaseI CaseII

LoadonABCD=4 x Load on EBFG

A BE

4xLoadonEBFG

IFH

CaseIIILoadonABCD=LoadonEBFI+IFCG+IGDH+AEIH

D CG

Rectangularloadedarea differentcases 30

A B A BE

EF

D C D CF

A B E

CaseIVLoadonABCD=2xLoadonABEF

CaseVLoadonABCD=2xLoadonEBCF

A B E

CaseVIL d ABCD L d

DC

F

LoadonABCD=LoadonAEGI BEGH DFGI+CFGH

G

C

HI

Rectangularloadedarea example 31

Example6A20x30ftrectangularfootingcarryingauniform loadof6000lb/ft2isappliedtothegroundsurface.

RequiredTheverticalstressincrementduetothisuniformloadatadepthof20ftbelowthecenterofloadedarea

A BE

G20ft

F

D C

30ft

Newmarks influencechart 32

Stressunderuniformlyloadedcircularareais

( )[ ]

= 232 1

11R

qz ( )[ ] +2 1zRRearrangingtheaboveequation,weget

32 ABInfluencevalue=0.005

11

=

qzR z

UsingvaluesofR/zforvariouspressureratios,Newmark (1942) presentedNewmark(1942)presentedaninfluencechartthatcanbeusedtodeterminevertical stress at any pointverticalstressatanypointbelowauniformlyloadedflexibleareaofanyshape.

Newmarks influencechart 33

z/q=0.9, R = 1.9084z/q=0.8, R = 1.3871 /q=0 8 R = 1 3871z/q 0.8, R 1.3871N element in the chart

I fl l 1/NInfluence value = 1/N = 1/200 = 0.005

( )qMIN=A

( )qMINzIN = influence value of chartq = pressure on loaded areaM number of elements of

AB = z (depth below loaded area at which stress is required)

BInfluencevalue=0.005

M = number of elements of chart enclosed by plan of loaded area

Newmarks chart example 34

Example7Araftfoundationofthesizegivenbelowcarriesauniformlydistributedloadof300kN/m2.Estimatetheverticalpressureatadepthof9mbelowpointOmarkedinthefigure.

Newmarks chart example 35

( )qMINz =IN =influencevalueofchart=0.005q =pressureonloadedarea=300kPaM =numberofelementsofchartenclosed by plan of loaded area 62enclosedbyplanofloadedarea=62

z =0.005x300x62=93kN/m2.

Pressureisobars 36

Anisobarisalinewhichconnectsallpointsofequalstressbelowthegroundsurface.Inotherwords,anisobarisastresscontour.

Eachisobarrepresentsafractionoftheloadappliedatthesurface.

Sinceisobarsformclosedfiguresresemblingtheformofabulb,theyaretermedaspressurebulb.

Isobarscanbedrawnforvertical,horizontal&shearstresses.

Verticalpressureisobarsareimportant as these are used inimportantastheseareusedincalculationoffootingsettlement.

Pressureisobars 37

PressureisobarsbasedonBoussinesqequationforuniformlyloadedcircularfootings

Pressureisobars square&cont.footing 38

UsingWestergaardequation UsingBoussinesqequation

Pressureisobars 39

ThedepthofstressedzoneresponsibleforsettlementistermedassignificantdepthDs.Forallpracticalpurposesonecantakeastresscontourwhichrepresents20%offoundationcontactpressureq or0.2q (Terzaghi).

Significantdepthofstressedzoneforsinglefooting Effectofcloselyplacedfootings

Pressurebulbgivestheideaabout the depth of soil affectedaboutthedepthofsoilaffectedbyfoundation.

Pressureisobars 40

Boreholesinasiteinvestigationshouldbetakendowntoadepthatleast1.5to2timesthewidthoftheproposedfoundationoruntilrock is encountered whichever is the lesserrockisencountered,whicheveristhelesser.

Smallfoundationswillacttogetherasonelarge

c/cdistance

Pressurebulb Importance 41

Resultsofplateloadtestcanbemisleadingiftheproposedfoundationismuchlarger.

Thesoftlayerofsoilinthefollowingdiagramisunaffectedbytheplateloadingtestbutwouldbeconsiderablystressedbyfoundation.

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CE222 SM 12 Stresses in Soil