ce222 sm 12 stresses in soil
DESCRIPTION
Soil MechanicsBasicsTRANSCRIPT
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SoilMechanicsICE222CE 222
StressesinSoil
1
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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.
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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.
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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
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Boussinesqformulaforpointload 5
( ) ( )
++= 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
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Boussinesqformulaforpointload 6
( ) 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
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Boussinesqformulaforpointload 7
( )[ ] Bz IzPzrzP 225 22 1123 =
+=
P P
( )[ ]zr 1 +
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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+
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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.
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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
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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
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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
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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
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Verticalstresscausedbyastripload 14
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
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Verticalstresscausedbyastripload 15
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
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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
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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
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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
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Embankmentloading example 20
Example4:Anembankmentisshownbelow.DeterminethestressincreaseundertheembankmentatpointsA1 andA2.
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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
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22
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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
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24Stressesunderuniformlyloadedcircularfooting
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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.
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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
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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
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28
Stressescausedbyrectangularloadedarea
LnBm == ,z
nz
m ,
Logscale
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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
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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
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Rectangularloadedarea example 31
Example6A20x30ftrectangularfootingcarryingauniform loadof6000lb/ft2isappliedtothegroundsurface.
RequiredTheverticalstressincrementduetothisuniformloadatadepthof20ftbelowthecenterofloadedarea
A BE
G20ft
F
D C
30ft
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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.
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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
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Newmarks chart example 34
Example7Araftfoundationofthesizegivenbelowcarriesauniformlydistributedloadof300kN/m2.Estimatetheverticalpressureatadepthof9mbelowpointOmarkedinthefigure.
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Newmarks chart example 35
( )qMINz =IN =influencevalueofchart=0.005q =pressureonloadedarea=300kPaM =numberofelementsofchartenclosed by plan of loaded area 62enclosedbyplanofloadedarea=62
z =0.005x300x62=93kN/m2.
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Pressureisobars 36
Anisobarisalinewhichconnectsallpointsofequalstressbelowthegroundsurface.Inotherwords,anisobarisastresscontour.
Eachisobarrepresentsafractionoftheloadappliedatthesurface.
Sinceisobarsformclosedfiguresresemblingtheformofabulb,theyaretermedaspressurebulb.
Isobarscanbedrawnforvertical,horizontal&shearstresses.
Verticalpressureisobarsareimportant as these are used inimportantastheseareusedincalculationoffootingsettlement.
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Pressureisobars 37
PressureisobarsbasedonBoussinesqequationforuniformlyloadedcircularfootings
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Pressureisobars square&cont.footing 38
UsingWestergaardequation UsingBoussinesqequation
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Pressureisobars 39
ThedepthofstressedzoneresponsibleforsettlementistermedassignificantdepthDs.Forallpracticalpurposesonecantakeastresscontourwhichrepresents20%offoundationcontactpressureq or0.2q (Terzaghi).
Significantdepthofstressedzoneforsinglefooting Effectofcloselyplacedfootings
Pressurebulbgivestheideaabout the depth of soil affectedaboutthedepthofsoilaffectedbyfoundation.
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Pressureisobars 40
Boreholesinasiteinvestigationshouldbetakendowntoadepthatleast1.5to2timesthewidthoftheproposedfoundationoruntilrock is encountered whichever is the lesserrockisencountered,whicheveristhelesser.
Smallfoundationswillacttogetherasonelarge
c/cdistance
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Pressurebulb Importance 41
Resultsofplateloadtestcanbemisleadingiftheproposedfoundationismuchlarger.
Thesoftlayerofsoilinthefollowingdiagramisunaffectedbytheplateloadingtestbutwouldbeconsiderablystressedbyfoundation.