TCBE 3104 - Design of Geotechnical Structures - Notes 2024_edited.pdf

Geotechnical structural design

1. 1
Bearing Capacity -
Foundations

2. Knappett,J.A.andCraig,R.F.,
“Craig’sSoilMechanics”8
th
edition,SponPress,2012.
2
RecommendedText
Coduto,D.P.,Kitch,W.A.,and
Yeung,M.R.“FoundationDesign,
PrinciplesandPractices”,3
rd
edition,
PrenticeHall,NewJersey,2016.

3. Whatarefoundations?
Whydoweneedfoundations?
Whatarethetypesoffoundations?
Whatareshallow foundations?
Whataredeepfoundations?
3
OverviewofShallow andDeepFoundations
Introduction

4. Introduction
Allengineeringstructuresneedsomeformoffoundation.In
thesimplestcases,foundationsarebuiltontopofexisting
topsoilorrockmaterialonsite.Inmanycases,thesoilmay
provetobetooweaktosupportthefoundationdirectly.
Thisisespeciallythecaseintheconstructionanderection
oflargeandheavybuildingsorstructures.
4

5. Introduction
Thesuperstructurebringsloadstothesoilinterfaceusing
column-typemembers.The loadcarryingcolumnsare
usuallyofsmallcross-sectionalareawithallowabledesign
compressivestressesontheorderof140MPa(Steel)to
10MPa(concrete).
Thesupportingcapacityofsoilisseldomabove1000kPa
andaremoreoftenontheorderof 200-250kPa .
Therefore,thefoundationisinterfacingbetweentwo
materialswherethestrengthratioisontheorderof
severalhundred.
Thefoundationthereforemustacttospreadtheloadsuch
thatthelimitingstrengthofthesoilisnotexceeded,and
theresultingdeformationsaretolerable.
5

6. Whatarefoundations?
Afoundationispartof
astructurewhich
transmitloadsdirectly
totheunderlyingsoil,
aprocessknownas
soil-structure
interaction.
6

7. Whydoweneedfoundations
Foundations transmitbuilding loadstothesoilsafely.
Structuralelementsinbuildingsaretypicallymadeof
steelorconcrete.Thesematerialshavehighstrength,
about100or1000timesthatofsoils.
7
Grade30Concrete–30MPa
Steel~400MPa
P = 200 kN
A = 0.5 m
2
= P/A = 200/0.5 = 400 kPa

8. VisualIdentificationUndrainedshear
strength(kPa)
Consistency
Thumbcanpenetratemorethan25mm<12Verysoft
Thumbcanpenetrateabout25mm12-25Soft
Penetratedwiththumbwithmoderateeffort25-50Medium
Thumbwillindentsoilabout8mm50-100Stiff
Thumbwillnot indentsoil,butreadilyindented
bythumbnail
100-200Verystiff
Indentedbythumbnailwilldifficultyorcannot
indentwiththumbnail
>200Hard
8
Relationshipbetweenconsistencyofcohesive soils
andundrainedshearstrength(fromCoduto2001)

9. Whydoweneedfoundations
9
Foundationstransmitbuildingloadstothesoilsafely.
Toperforminasatisfactoryway,thefoundationmustmeettwo
principalperformancerequirements(knownaslimitstates),
namely:
Strength–Thecapacityorresistancemustbesufficientto
supporttheloads(actions)applied(i.e.sothatitdoesn’t
collapse)
Serviceability -Excessivedeformationmust be avoided
undertheseloads, whichmightdamage thesupported
structureorlead toalossoffunction.
Otherperformancerequirements:costandconstructibility

10. Whatarethetypesoffoundations?
10
Foundations
Shallow
D/B < 1
Deep
1<D/B<3
1.Spread
2.Sloped
3.Stepped
etc.....
Raft / Mat
Foundations
1.Piers
2.Caissons
Piles >3
1.Piles
2.Piers
3.Caissons
Isolated
Footings
Combined
Footings
1. Rectangular
2. Trapezoidal
3. Strap

11. Figure10.1Foundationsystems:(a)pads/strips,(b)raft,(c)piled,(d)piledraft
11

12. 66-stories
30-Stories
Podium
Fill
Marineclay
Residualsoil
BoulderyclayCaissons
Boredpiles
Diaphragm
wall
Raft
12

13. Whatareshallow foundations?
Ifasoilstratumnearthesurfaceiscapableof
adequatelysupportingthestructuralloadsitis
possibletouseshallowfoundations.
Arbitrarydefinition: Shallow foundations referto
those whered/B<1orwhend<3m (accordingto
CP4).
B
13
d

14. Whatareshallow foundations?
Examples of shallow foundations:
■
Spreadfootings
Afooting
supportinga
singlecolumn
isreferred as
anindividual
footingor
pad.
14
Astripfooting
supportsaload
bearingwall.
Shallow foundations transfer the structural load to the soil
by spreading the load laterally and therefore the term
“spread footing”. A spread footing supports a single column.

15. Whatareshallow foundations?
15
Examplesofshallowfoundations:
■
Combinedfootings
Acombinedfooting
supportmorethanone
loadbearingstructural
elements.Theyareused
wherethereisspace
constraintorneara
propertyline.

16. Whatareshallow foundations?
Examplesofshallowfoundations:
■
Raft/matfoundations
Amatfoundationisalarge
spreadfootingusedto
supportmorethanoneload
bearingstructuralelements
inseverallines.Usually,it
encompassestheentire
footprintofthestructure.
16

17. Whatareshallow foundations?
■
Raft/matfoundations
17
A raft is a special footing that supports several randomly
spaced columns or several rows of parallel columns. They
usually occupy the footprint of the entire building.

18. Whataredeepfoundations?
18
Ifthesoilnearthesurfaceisincapableof
supportingthestructuralloads,piles,orother
formsofdeepfoundationssuchas piersor
caissons,areusedtotransmittheappliedloads
tosuitablesoil(orrock)atgreaterdepthwhere
theeffectivestresses(andhenceshearstrength)
arelarger.

19. Whataredeepfoundations?
19
Ifthesoilnearthesurfaceisincapableof
supportingthestructuralloads,piles,orother
formsofdeepfoundationssuchas piersor
caissons,areusedtotransmittheappliedloadsto
suitablesoil(orrock)atgreaterdepthwherethe
effectivestresses(andhenceshearstrength)are
larger.
Whyarethestressesofthesesoilsatdeeper
depthslarger?

20. SampleA
SampleB
Homogeneousclay
20
Whichsamplehasahighershearstrength?
Why?

21. Whataredeepfoundations?
Examples of deep foundations:
(a)PrecastRCpile
(b)SteelH-pile
(c)Steeltubularpile
21
(d)Shellpile(e)CFApile (d)Under-reamedboredpile

22. ShallowvsDeepFoundations
(Coduto2001)
22
Inshallow
foundations,load
transferisby
lateralspreading
i.e.endbearing
throughpassive
resistanceofsoil
Indeepfoundations,side
resistancebecomesthe
dominantloadtransfer
mechanismcomparedto
endbearing.
Deep foundationstransfer most of the applied structural loads to
deeper strata

23. Foundation Selection and Design Criteria
The type of foundation most suitable for a given
structure depends upon several factors:
1.Function of structure and the loads it must carry
2.Subsurface conditions
3.Cost of the foundation in comparison with the
cost of the superstructure
23

24. FoundationDesignProcess
24
1.
Establishrequirementsforstructural
conditionsandsitecharacterization
2.
PreliminaryStudy:Obtaingeneralsitegeology,
Collectfoundationexperiencefromthearea
3.
Planandexecutesubsurfaceinvestigation
4.
Evaluateinformationandselectfoundationsystem:
DeepFoundationorShallowFoundation(see
performancerequirements)

25. The foundation design must satisfy the following
criteria:
25
1.Factor of safety/Resistance w.r.t.collapse of building
or failure of the foundation is adequate both during the
construction and the life of the structure
2.Settlements at working load are sufficiently small that
they do not affect the function of the building
3.Adverse effects on adjacent structures and on the
environment are as small as possible during
construction as well as during the life of the structure

26. Site Investigation
26
Asiteinvestigationinoneformoranotherisalwaysrequiredforany
engineeringstructure.Theinvestigationmayrangeinscopefroma
simpleexaminationofthesurfacesoilswithorwithoutafewshallow
trialpits,toadetailedstudyofthesoilandgroundwaterconditions
toaconsiderabledepthbelowthesurfacebymeansofboreholes
andin-situandlaboratorytestsonthematerialsencountered.
1.Todeterminethesequence,thicknessandlateralextentof
thesoilstrata,whereappropriatethelevelofbedrock
2.Toobtainrepresentativesamplesofsoils(androck)for
identificationand,ifnecessary,foruseinlabteststo
determinerelevantsoilparameters
3.Toidentifygroundwaterconditions

27. Site Investigation –Eurocode 7 Pt 1
27
3.2.1
(1)PGeotechnicalinvestigationsshallprovidesufficientdata
concerningthegroundandthegroundwaterconditions atand
aroundthesiteforaproperdescriptionoftheessentialground
propertiesandareliableassessmentofthe characteristic
valuesofthegroundparameterstobeusedindesign
calculations.
(2)PTheinvestigationandtheamountofthegeotechnical
investigationsshallbeadjustedtotheparticularinvestigation
phaseandgeotechnicalcategories.

28. Site Investigation –Eurocode 7 Pt 2
28
6.1–Clause2:
PTheGroundInvestigationReportshallconsistofthe
following:
-apresentationofallappropriategeotechnicalinformation
includinggeologicalfeaturesandrelevantdata;
-ageotechnicalevaluation,statingtheassumptionsmade
intheinterpretationofthetestresults.3

29. Numberof boreholes
29
EN 1997-2:2007 –Appendix B recommends for:
1.High-rise and industrial structures: grid pts at 15m
–40m spacing
2.Large area structures: grid pts at ≤ 60m spacing
3.Linear structures (e.g.roads, railway, tunnels etc.):
20m –200m
4.Special structures (e.g.bridges, stacks, machine
foundations): 2 –6 pts per foundation
5.Dams and weirs: 25m –75m along relevant section

30. ShallowFoundations
30

31. Soilsgenerallyfailinshear:
31

32. ModulusofElasticityE,ShearModulusG
x
32
z z
E

1

''

E
1

x


'
x
'
z

xzzxxz
 G

33. 33
Typical Values of Poisson’s Ratio for Soils and Rocks

34. BearingCapacity
34
Ultimate bearing capacity, q
ult
, is the least pressure that would
cause shear failure of the soil immediately below or adjacent to a
foundation.
Allowable bearing capacity, q
a
, is the maximum pressure that may
be applied to the soil such that:
1.settlement is limited to a tolerable amount, and
2.ultimate bearing capacity is not exceeded
Bearing capacity (q
f
) is defined as the pressure which would cause
shear failure of the supporting soil immediately below and adjacent
to a foundation.

35. Shear Failure modes
Threedistinctmodes:
35
General shear
Local shear
Punching shear

36. 36
Example of shear failure:

37. BearingCapacity
GeneralShear:
q
f
Continuousfailuresurfacesdevelopbetweentheedges
ofthefootingandthegroundsurface.Asthe pressureis
increasedtowardsthevalueq
f
astateofplastic
equilibriumisreachedinitiallyinthesoilaroundthe
edges ofthefooting,whichsubsequentlyspreads
downwards andoutwards.Ultimately,thestateofplastic
equilibriumis fullydevelopedthroughoutthesoilabove
thefailure surfaces.Heaveofthegroundsurfaceoccurs
onboth sidesofthefooting,althoughinmanycasesthe
finalslip movementoccursonlyononeside,
accompaniedby tiltingofthefooting,asthefootingwill
notbeperfectly leveledandhencewillbebiasedtofail
towardsoneside.
37
Typicalofsoilsoflow
compressibilityi.e.
densecoarse-grained
orstifffine-grained
soils.

38. BearingCapacity
LocalShear:
Thereissignificantcompressionofthesoilbelow
thefooting,andonlypartialdevelopmentofthestate
ofplasticequilibrium.Thefailuresurfaces,therefore,
donotreachthegroundsurfaceandonlyslight
heavingoccurs.Tiltingofthefoundationwouldnot
beexpected.
Localshearfailureisassociatedwithsoilsofhigh
compressibilityandischaracterizedbythe
occurrenceofrelativelylargesettlements(which
wouldbeunacceptableinpractice)andthefactthat
q
f
isnotclearlydefined.
Acceptable
settlement
38
q
f
= ?

39. BearingCapacity
Punchingshear:
Punchingshearfailureoccurswhenthereis
relativelyhighcompressionofthesoilunderthe
footing,accompaniedbyshearinginthevertical
directionaroundtheedgesofthefooting.Thereis
noheavingofthegroundsurfaceawayfromthe
edges,andnotiltingoftheofthefooting.Relatively
largesettlementsarealsocharacteristicofthis
modeandagainq
f
isnotwelldefined.
Punchingshearfailurewillalsooccurinasoiloflow
compressibilityifthefoundationislocatedat
considerabledepth.
39
q
f
= ?

40. Commoncauseoffoundationproblems
Lackor inadequate
site investigation
40
In case of the leaning tower of Pisa where
inadequate soil tests were performed,
even though they might have saved
money in the process, it jeopardized the
stability of the structure and therefore led
to the ultimate loss of levelling.
Such practice can seriously lead to a
complete collapse of a structure.

41. UltimateBearingPressure/Capacity,q
f
41
-Maximumbearing pressure that the soil can
sustainbeforebearingcapacityfailure.
The resistance for the general shear bearing
capacity failure can be derived

42. PlasticityTheory
Thebearing capacityproblemcanbeconsideredinterms
ofplasticitytheorybyassuming thatthestressstrain
behaviorofthesoilisrepresentedbytherigid-perfectly
plasticidealisation inwhichbothyieldingandshearfailure
occursatthesamestateofstress:unrestrictedplasticflow
takesplaceatthisstresslevel.
42

43. PlasticityTheory
43
Plasticcollapse occursafterthestateofplasticequilibrium
hasbeenreachedinpartofasoilmass,resultinginthe
formationofanunstable mechanism: thatpartofthesoil
massslipsrelative totherestofthemass.
Theappliedloadsystem,includingbodyforces,forthis
conditionisreferredtoasthecollapseload.

44. ShallowvsDeepFoundations
(Coduto2001)
44
Inshallow
foundations,load
transferisby
lateralspreading
i.e.endbearing
throughpassive
resistanceofsoil
Indeepfoundations,side
resistancebecomesthe
dominantloadtransfer
mechanismcomparedto
endbearing.

45. UpperBound(UB)Theorem
45
Ifakinematicallyadmissiblemechanismofplasticcollapse is
postulatedandif,inanincrementifdisplacement,the work
done byasystemofexternal loadsisequaltothe dissipation
ofenergybytheinternalstresses,then collapsemustoccur;
theexternalloadsystemthus constitutesanupper boundto
thetruecollapse load.
LowerBound(LB) Theorem
Ifastateofstresscanbefoundwhichatnopointexceeds
thefailurecriterionforthesoilandisinequilibriumwitha
systemofexternalloads,thencollapsecannotoccur;the
externalloadsystemthusconstitutesalowerboundtothe
truecollapseload.

46. Rankine Method is alowerboundsolution
–It overestimatesactivepressureand
underestimates passive pressure.
46
CoulombMethodisanupperboundsolution
–Itunderestimatesactivepressureand
overestimatespassivepressure.

47. Shearstrength–undrained,drained
47
Undrained:c
u
,ors
uDrained:c’, ’

48. BearingCapacityinUndrainedMaterials
48
Undrainedmaterial–considersonly
undrainedshear strengthofsoili.e.c
u
ors
u
.
Analysisusingtheupperboundtheorem
Forundrainedconditionsthefailuremechanism
withinthe soilmassshouldconsistofsliplines
whichareeither straightlinesorcirculararcs
(oracombination).

49. Upperboundapproach,mechanismUB1
ProposedmechanismUB1
(Planestrain)
Slipvelocities
Dimensions
A
B
C
49

50. BearingCapacityFailure
50
Bearingpressure,q
Overburdenpressureorsurcharge,
q

51. Upperboundapproach,mechanismUB1
ProposedmechanismUB1
(Planestrain)
51
Slipvelocities
Dimensions

52. Upperboundapproach,mechanismUB1
Energydissipatedduetoshearingatrelative
velocityv
i
alongsliplineoflengthL
i
:
E
i

f
L
i
v
i
WorkdoneW
i
byapressureq
i
actingoveran
areaperunitlengthB
i
movingatvelocityv
i
:
W
i
q
i
B
i
v
i
52

53. Upperboundapproach,mechanismUB1
Hodograph
53
v 2 v 2
v 2
v 2

54. Upperboundapproach,mechanismUB1
E
i
dissipatedEnergyv
i
RelativevelocityL
i
Length
fi
StresslineSlip
c
u
Bvv2B/2c
u
OA
2c
u
Bv2vBc
u
OB
c
u
Bvv2B/2c
u
OC
c
u
Bvv2B/2c
u
AB
c
u
Bvv2B/2c
u
BC
TotalEnergyDissipated,E
i
=6c
u
Bv
-v
54
E
i

f
L
i
v
i
Energydissipated:

55. Upperboundapproach,mechanismUB1
Energydissipated:
Workdone:
E
i
dissipatedEnergyv
i
RelativevelocityL
i
Length
fi
StresslineSlip
c
u
Bv2vB/2c
u
OA
2c
u
Bv2vBc
u
OB
c
u
Bv2vB/2c
u
OC
c
u
Bv2vB/2c
u
AB
c
u
Bv2vB/2c
u
BC
TotalEnergyDissipated,E
i
=
55
6c
u
Bv
W
i
doneWorkv
i
velocityRelativeB
i
Areap
i
PressureComponent
q
f
BvvBq
f
Footing
pressure
-
q
Bv-vB
q
Surcharge
q
f
Bv-
q
BvTotalWorkDone,W
i
=
W
i
q
i
B
i
v
i

56. Upperboundapproach,mechanismUB1
56

q
f

q

Bv6c
u
Bv
q
f
6c
u

q
W
i
E
i

57. Upperboundapproach,mechanismUB2
Amoreefficientmechanism:
57

58. Upperboundapproach,mechanismUB2
q
f


2

c
u

q
W
i
E
i

q
f

q

Bv

2

c
u
Bv
E
i
dissipatedEnergyv
i
RelativevelocityL
i
Length
fi
StresslineSlip
c
u
Bvv2B/2c
u
OA
c
u
Bvv
fan
= v2R=B/2c
u
Fanzone
c
u
Bvv2B/2c
u
OC
TotalEnergyDissipated,E
i
=
58
(2+)c
u
Bv

59. BearingCapacityinUndrainedMaterials
59
Analysisusingthelowerboundtheorem
Inthelowerboundapproach,theconditions ofequilibrium
andyieldaresatisfiedwithoutconsiderationofthemodeof
deformation.
2 2
4
f u q u
f u q
q z c z c
q c
    
  

60. Lowerboundapproach,stressstateLB2
60
c
u

1
= 
f
+ z

3
= 
q
+ z
  

2
f u q u u
f u q
q z c z c c
q c
      
   

61. ComparingUpperandLowerBoundsq
f
LowerBoundUpperBound
4c
u

q
6c
u

q
1

2

c
u

q

2

c
u

q
2
61

62. BearingCapacityFactors (Undrained)
Generalformofbearingcapacityofashallowfoundation
onanundrainedmaterial maybewrittenas:
q
f
s
c
N
c
c
u

q
Wheres
c
=shapefactor,forstripfooting=1.0
N
c
=bearingcapacityfactor=(2+)
d
62
Foundingplane

q
d

63. BearingCapacityFactors(Undrained)




B
 d
N
c


2

10.27
Salgadoetal.(2004):
Eqn(a)
L
63
Eurocode7:
c
s10.2
B
BL
Inpractice,N
c
islimitedatavalue
of9.0foradeeplyembedded
squareorcircularfoundation.
Eqn(b)
a and b
N
c
= 9.0

64. Terzaghi’sBearingCapacityFormula
for undrained material
64
q
f
s
c
c
u
N
c

q
s
c
Footingshape
1.0Continuous
1.3Square
1.3Circular
N
c
=5.7–differentassumptions

65. BearingCapacityFactors (Undrained)
For two layer soil,
Merifieldetal.(1999)
SolidlinesforUB,dashedlinesforLB
65
MerifieldandNguyen(2006)
forsquarefootings
Note:incalculatingq
f
,c
u
=c
u1

66. BearingCapacityFactors (Undrained)
Forstripfooting closeto aslope,
Georgiadis(2010)
Lowestupperboundsolution
Not the exact solution
66

67. BearingCapacityFactors (Undrained)
ForGibson’ssoil:c
u
(z)c
u0
Cz
DavisandBooker(1973)
67

f z
CB
q 2 F
4
u
c
 
  
 
 

68. BearingCapacityFactors (Undrained)
68
Bearingcapacityisaffectedby:
(a)Equation used
(b)Depth of embedment
(c)Layeredsoil
(d)Proximityofslope

69. Astripfoundation 2.0mwide islocatedatadepthof2.0 m
inastiffclayofsaturatedunitweightof21kN/m
3
.The
undrainedshearstrengthisuniformwithdepth,withc
u
=
120kPa.Determinetheundrainedbearingcapacityofthe
foundationunderthefollowingconditions:
69
(a)
Thefoundationisconstructedinlevelground;
(b)
Acuttingatagradientof1:2issubsequentlymade
adjacenttothefoundation,withthecrest1.5mfromthe
edgeofthefoundation.
Example 1

70. Solution (1a)
d=2.0m
Foundingplane

q
d

21kN/m
3


2.0m

42kN/m
2
B=2.0mc
u
=120kPa
70
B2.0
UsingSkempton'svalues,N
c
6.4
Forstripfooting,s
c
1.0
d

2.0
1.0
6.4
q
f
s
c
N
c
c
u

q


1.0x6.4x120

42
810kPa

71. Solution (1b)
2.9
2.0

1.5
0.75
o

120
B

21

2.0

26.6
1

2
tan
c
u
1
71
d=2.0m
Foundingplane

q
d

21kN/m
3


2.0m

42kN/m
2
B=2.0mc
u
=120kPa
B=1.5m
1
2
Interpolating,N
c
4.7
4.7
q
f
s
c
N
c
c
u

q


1.0x4.7x120

42
606kPa

72. BearingCapacityinDrainedMaterials
72
Drainedmaterial–considerseffectiveshearstrength
parametersofsoili.e.c’and’.
Analysisusingtheupperboundtheorem
Fordrainedconditions,theslipsurfaceswithinakinematically
admissiblefailuremechanismshouldconsistofsliplinesof
eitherstraightlinesorcurvesofaspecificformknownaslog
spirals(oracombination).
Logspiral,rr
0
e
tan

73. Upperboundapproach,mechanismUB-1
73

74. TheDirectShearTest
=dilationangle
Forthespecialcase,=’(normalityprinciple)
74
Associativeflowrule

75. Upperboundapproach,mechanismUB1
Workdone:
W
i
doneWorkv
i
velocityRelativeB
i
Areap
i
PressureComponent
q
f
BvVBq
f
Footing
pressure
Surcharge
75

q
Asaresultofthenormalityprinciple,E
i
0.
W
i
0.

'
42
tan

tan'
Be
2
'
42
tan

tan'
ve
2

tan
42
2
'
tan'
'
q
Bve
tan 2
f q q q
q = e tan N
4 2
       
   
  
  
tan 2
f q
q Bv - Bve tan = 0
4 2
     
  
 
 

76. Lowerboundapproach,stressstateLB1
q
f

1-sin'

2
f
'
q
e
'

q
eq
e
tan'




N
q
'
q



42
'
tan
tan'






1sin'


1sin'

tan'


1sin'

'
q
Sameq
f
asupperboundapproach!



2
(8.26)
76

77. q
f
s
c
N
c
c's
q
N
q
'
q
0.5Bs

N

77
GeneralBearingCapacityforDrainedMaterials
(Coduto2001)
d
’
q

78. q
f
s
c
d
c
N
c
c's
q
d
q
N
q
'
q
0.5Bs

d

N

GeneralBearingCapacityforDrained Materials
78


2

(Salgado2008)
(EC7)
N

2

N
q
1

tan'
N



N
q
1

tan

1.32'

 

tan
q
1+ sin
N e
1- sin
 



q
c
N 1
N
tan



As'0,Nc
q q
c
q
S N 1
S
N 1



q
B
S 1+ sin
L
 
B
S 1- 0.3
L



79. q
f
s
c
N
c
c's
q
N
q
'
q
0.5Bs

N

GeneralBearingCapacityforDrained Materials
79
d
c
,d
q
,d

–depthfactors.EC7doesnotrecommendtheuseofdepthfactorsi.e.
d
c
=d
q
=d

=1.0


2

(Salgado2008)
(EC7)
N

2

N
q
1

tan'
N



N
q
1

tan

1.32'

 

tan
q
1+ sin
N e
1- sin
 



q
c
N 1
N
tan



As'0,Nc
q q
c
q
S N 1
S
N 1



q
B
S 1+ sin
L
 
B
S 1- 0.3
L



80. Case1:
Case2:
Case3:
d
’
q
80
GeneralBearingCapacityforDrained Materials
'
w
Case1:d
w
d
Case2:d<d
w
<d+B
'Case3:d+Bd
w
1
w
w
d d
B
   
 
  
  
d
w
=depthofgroundwatertable
Unitweighttobeused 0.5Bs

N


81. Example2
Afooting 2.25mx2.25mislocatedatadepthof1.5min asandfor
whichc’=0and ’=38
o
.Determinethebearing resistancefor:
(a)ifthegroundwater tableiswellbelowthe foundation level,and
(b)ifthegroundwatertableisatthe groundsurface.Theunitweightof
thesandabovethe groundwater tableis18kN/m
3
;thesaturatedunit
weightis 20kN/m
3
.
(notusedasc'0)
For

'38
o
,
81
(notusedasc'0)
 

tan
q
1+ sin
N e 49
1- sin
 

 

q
c
N 1
N
tan



N

2

N
q
1

tan= 75
q q
c
q
S N 1
S
N 1



q
B
S 1+ sin
L
 
B
S 1- 0.3 = 0.7
L



82. Solution2(a)
(a)ifthegroundwatertableiswellbelowthefoundationlevel
q
f
s
q
N
q
'
q
0.5Bs

N

s
q
N
q
d0.5Bs

N



1.62x49x 18x 1.5



0.5x 18x2.25x0.7x75

3206kPa
Case 3
82
(b)ifthegroundwatertableisatthegroundsurface
s
q
N
q
'd0.5'Bs

N



1.62x49x

20-9.81

x 1.5



0.5x

209.81

x2.25x 0.70x75

1815kPa
q
f
s
q
N
q
'
q
0.5Bs

N

Case 1

83. Terzaghi’s Bearing CapacityFormulas
q
ult
cN
c

zD
N
q
0.5BN

83

84. Terzaghi’s Bearing CapacityFormulas
q
ult
s
c
cN
c

zD
N
q
0.5s

BN

s

s
c
Footingshape
1.01.0Continuous
0.81.3Square
0.61.3Circular
84
Terzaghi’sBearingCapacityFormula for
undrained material
q
f
s
c
c
u
N
c

q

85. Terzaghi’s Bearing CapacityFormulas

Approximate formulaby Coduto
2
N
q
2
q
c
a
2
q
10.4sin4'
2

N1

tan'





1
cos'
tan'
for'0
tan'
N1
N
for'0N
c
5.7




tan'
360
o




 '
whereaexp0.75

2
2 cos 45

2
 '
o

N


K
p
N






85

86. 86

87. Example 1 –A square footing is to be constructed as shown inFigure
below. The groundwater table is at a depth of 50 ft below the ground
surface. Compute the ultimate bearing capacity and the column load
required to produce a bearing capacity failure.
19kN/m
3
0.6m
1.0m
7kPaD

19kN/m
3


0.6m

11.4kPa
For ' 30
o
:N37.2,N22.5,N20.1
zD
q
ult
1.3cN
c

zD
N
q
0.4BN

1.3

7kPa

37.2



11.4kPa

22.5

0.4(19kN/m
3
)

1m

20.1

338.5256.5152.8
748kPa
c q 
87

88. Example 1(cont’d)
A
0.6m

ult
3
f

P W
f
q q

23.6kN/m

14.2kNW

1m

1m

748
P 14.2kN
(1m)(1m)
P 733.8kN
“Nearly half of this capacity comes from the first term in the bearing
capacity formula and is therefore dependent on the cohesion of thesoil.
……it is prudent to use conservative values of c’ in bearing capacity
analyses. In contrast, the frictional strength is more reliable and does not
need to be interpreted asconservatively.”
88

89. Example 2 –The proposed continuous footing shown in the Figure belowwill
support the exterior wall of a new industrial building. The underlying soil is an
undrained clay, and the groundwater table is below the bottom of the footing.
Compute the ultimate bearing capacity,and compute the wall load required to
cause a bearing capacityfailure.
D

18kN/m
3


0.4m

7.2kPa
For 0
o
:N5.7,N1,N0
zD
q
ult
s
u
N
c

zD
N
q
0.4BN



120kPa

5.7



7.2kPa

1

0.5(18kN/m
3
)

0.7m

0

6847.20
691kPa
c q D=
89

90. Example 2(cont’d)
The zone above the bottom of the footing is partly concrete
andpartlysoil.The weight of this zone is small compared
to the wall load, so compute it using (18+23.6)/2 21
kN/m
3
as the averaged.
q
2b
ult
33
P
11kN/m
3
691kPa 
b
0.7m
P 473kN
P

W
f
q
bb
B

21kN/m

11kN/m

 

 W
f
0.4m1.1m


0.7m

90

91. Vesic’s Bearing CapacityFormulas
q
ult
cN
c
s
c
d
c
i
c
b
c
g
c

zD
N
q
s
q
d
q
i
q
b
q
g
q
0.5BN

s

d

i

b

g

91
Notation for Vesic’sload inclination, base inclination, and ground inclination factors. All angles are expressed
in degrees

92. Vesic’s Bearing CapacityFormulas
92

93. Vesic’s Bearing CapacityFormulas
BearingCapacity
Factors
GroundInclination
Factors
BaseInclination
Factors
N
N
q
1
for'0
c
tan'
N
c
5.14for '0
Ne
tan'
tan
2

45
o

'
q
2

N

2

N
q
1

tan'
g1

c
147
o
gg

1tan

2
q 
b1

c
147
o
tan'
2
b
q
b

1
57
o


93

94. GroundwaterEffects

ApparentCohesion

Soils may be unsaturated, additional shear
strength contribution due tosuction.

Sampling disturbance, storage and handling
conditions may alter water content and change
the shearstrength.

Pore WaterPressure
94

95. GroundwaterEffects
q'
ult
c'N
c
s
c
d
c
i
c
b
c
g
c
'
zD
N
q
s
q
d
q
i
q
b
q
g
q
0.5'BN

s

d

i

b

g

95
Three groundwater cases for bearing capacity analyses

96. GroundwaterEffects
'
w
Case 1: D
w
D
Case 2: D < D
w
<D+B
'
Case 3: D+B D
w
q'
ult
c'N
c
s
c
d
c
i
c
b
c
g
c
'
zD
N
q
s
q
d
q
i
q
b
q
g
q
0.5'BN

s

d

i

b

g

q
ult
q'
ult
u
D
For totalstressparameters, 
T
and c
T
,
q
ult
c
T
N
c
s
c
d
c
i
c
b
c
g
c

zD
N
q
s
q
d
q
i
q
b
q
g
q
0.5BN

s

d

i

b

g

96
w
w
D D
1
B
   
 
  
  

97. GroundwaterEffects
B


w
B

D
w
D


B

B

D
w
D




w

+
97
w
w
D D
1
B
   
 
  
  

98. q

50



L


s10.4
B
10.4
30
0.76


L
 
50


s1
B
tan'1
30
tan30
o
1.35
'
zD
Du
D


18.5

10

0185kPa
For ' 30
o
:N18.5,N22.4
9.4kN/m
3
Example 3 –A 30-m by 50-m mat foundation is to be built as shown
in the Figure below. Compute the ultimate bearingcapacity.
Case2:DD
w
DB

q 
q
d1
d12ktan'

1sin'

2
12

0.33

tan30
o

1sin30
o

2
1.10

q
ult
'
zD
N
q
s
q
d
q
0.5' BN

s

d

u
D
185

18.4

1.35

1.10

0.5

9.4

30

22.4

0.76

1.0

0
7455kPa
(Check q
ult
using undrained strengthparameters)
Using Vesic'smethod:
98
w
w
D D
1
B
   
 
  
  
12 10
9.81 1
30
   
 
  
  

99. Allowable BearingCapacity
F
q
a_net

q
ult _ net
99
Factors affecting the deign factor of safety, and typical values of F

100. Example 4 –A column has the following design vertical loads: P
D
=1335 kN,
P
L
=623 kN will be supported on a spread footing located 0.9 m below the
ground surface. The underlying soil has an undrained shear strength of 96 kPa
and a unit weight of 17 kN/m
3
. The groundwater table is at a depth of 1.2m.
Determine the required footing width to maintain a factor of safety of 3 against
a bearing capacityfailure.
D
P
D
P
L
13356231958kN
UsingTerzaghi'smethod(Square footing):
For0,N
c
5.7,N
q
1.0

zD
D

17

0.9

15.3kPa
q
ult
1.3s
u
N
c

zD
N
q
1.3

96

5.7



15.3

1.0

727kPa
P1335kN
q
0.9m
3F
q
B
2
a
22 3
f
a
B 3m
A
195821.2B
2
242

P W
f
W

B

23.6 kN/m

21.2B

727
242kPa
Determinedesignworkingloadasworstcaseof:

q
ult
100

101. Bearing Capacity on LayeredSoils

Evaluate bearing capacity based on lowest
values of c’, ’ and in the zone B below the
footing –mostconservative

Use weighted values c’, ’ and based on
layer thickness in the zone B below the
footing –may or may not beconservative

Use a series of trial failure surfaces similar to
slope stability analysis. The trial failure
surface that gives the lowest q
ult
is the critical
failure surface –complexanalysis
101

102. Example 5 –Using the second method, compute the factor of safety
against a bearing capacity failure in the square footing shown below.
102

103. 
1.8
 
1.8

 

1.1

18.2kN/m
3


0.7

20.1kN/m
3

18.9kN/m
3

1.8

1.8


'
1.1

32
o


0.7

38
o

34
o


1.8

1.8


c'
1.1

5kPa


0.7

0kPa

3kPa
Example 5(cont’d)
Find weighted c',' and :
A
1.8m

0.4m

1.5m


1.8m

2
zD D
3232
f
w
27kPa
q 
P W
f

800kN 116kN
283kPa

17.5kN/m




23.6kN/m

116kNW

1.8m

'18.99.89.1kN/m
3
Groundwatercase1(D
w
D):
'

Hu

17.5kN/m
3


1.2m



18.2kN/m
3


0.7m



9.8kN/m
3


0.7m

103

104. Example 5(cont’d)
UseTerzaghi'sFormula(Squarefooting):
q
zD
28333.7

zD

145733.7
5.7F 
q
ult
For'34
o
,N52.6,N36.5,N39.6
c q 
q
ult
1.3c'N
c
'
zD
N
q
0.4'BN

u
D
1.3

3

52.6



27

36.5

0.4

9.1

1.8

39.6

6.9
2059862596.91457kPa
Note: F=5.7 is much greater than typical values of 2.5 to 3.5.
Footing size is overdesigned for bearing capacity. However, need
to consider settlement before reducing size of footing. Usually,
settlement governs size of shallow foundations.
104
(17.5x1.2+0.7x18.2) = 33.74
ult
q1457
Coduto uses gross pressures: F = 5.15
q 283
 
 
 
 

105. Publication DateTitleEurocodePart
Oct2010Eurocode 7. Geotechnical
design. Generalrules
SS EN1997-1
Oct2010Eurocode 7. Geotechnical
design. Ground investigation
andtesting
SS EN1997-2
CP4: 2003 FoundationDesign
(BS 8004:1986 Code of practice forfoundations)
105

106. 106

107. EC 7 –Limit statedesign
Characteristicvalue
Partial factors for limit statesDesign Approach 1 for footings &piles
Combination1 Combination2 Combination2 -piles
R4M2M1orA2R1M2A2R1M1A1
1.01.01.35UnfavourablePermanentActions
1.01.01.0Favourable
1.31.31.5UnfavourableVariable
1.251.01.251.0
tan

'
Soil
1.251.01.251.0Effective cohesion,c'
1.41.01.41.0Undrained strengthc
u
1.41.01.41.0Unconfined strengthq
u
1.01.01.01.0
Weight

1.01.0BearingSpreadfooting
1.01.0Sliding
1.7/1.51.0BaseDrivenpiles
1.5/1.31.0Shaft(compression)
1.7/1.51.0Total/combined
2.0/1.71.0Shaft intension
2.0/1.81.0BaseBoredpiles
1.6/1.41.0Shaft(compression)
2.0/1.71.0Total/combined
2.0/1.81.0Shaft intension
107
Design Approach 1 (DA1)
Combination 1: A1+M1+R1
Combination 2: A2+M2+R1
Design Approach 2 (DA2)
Combination: A1+M1+R2
Combination: (A1 or A2)+M2+R3
Design Approach 3 (DA3)

108. EC 7 –Limit statedesign
The characteristic value of a geotechnical
parameter shall be selected as a cautiousestimate
of the value affecting the occurrence of the limit
state.
108

109. Example 6.4 –A column has the following design vertical loads: P
D
=1335 kN,
P
L
=623 kN will be supported on a spread footing located 0.9 m below the
ground surface. The underlying soil has an undrained shear strength of 96 kPa
and a unit weight of 17 kN/m
3
. The groundwater table is at a depth of 1.2m.
Determine the required footing width according to EC 7 Approach 1
Combination1.
Factoredload:
P
D

D
P
L

L
1335

1.35

623

1.5

2734 kN
Vertical effectof actions: E
d
27341.35W
f
kN
Design strength of soil 
s
u

96
96kPa
γ
m
1
EC7partialfactorsassumeHansenbearingfactorsasbasis
forbearingcapacitycalculations(Square footing):
For0,N
c
5.14,N
q
1.0,s
c
1.3
q
ult
forlimitstates
u
N
c
s
c
q


96

5.14

1.3

0.9(17) 657 kPa
Bearing resistance R
d
657

B

B

657BkN
2
A
ult
657B
2
273428.7B
2
B 2.1m
qforlimitstate
27341.35W
f
109

110. Foundations with Eccentric or Moment Loads
These are called eccentric loads, and they produce a non-
uniform bearing pressure distribution. Assuming a linear
distribution of the bearing pressure, the eccentricity, e, of the
bearing pressure is given by
L
f
Pe
e
P W


where
e = eccentricity of bearing pressure distribution
P = applied vertical load (or the same per unit length for continuous footings)
e
L
= eccentricity of applied vertical load
W
f
= weight of footing (or the same per unit length for continuous footings)
M = applied moment load (or the same per unit length for continuous footings)
or
f
M
e
P W


110

111. One-Way Eccentric or Moment Loading
If the eccentric or moment loads occur only in the B direction,
then the bearing pressure distribution
Distribution of bearing pressure beneath
footings with various eccentricities:
(a) e < B/6;
(b) e = B/6; and
(c) e > B/6.
111

112. One-Way Eccentric or Moment Loading
If e B/6 the bearing pressure distribution is trapezoidal, as shown in Figure a,
the minimum and maximum bearing pressures on square or rectangular
foundations are:
where;
q
min
= minimum bearing pressure
q
max
= maximum bearing pressure
P = column load
A = base area of foundation
u
D
= pore water pressure along base of foundation
e = eccentricity of bearing pressure distribution
B = width of foundation
min
6
1
f
D
P W e
q u
A B
   
     
  
max
6
1
f
D
P W e
q u
A B
   
     
  
If e = B/6 (i.e., the resultant force acts at the third-point of the foundation), then
q
min
= 0 and the bearing pressure distribution is triangular (Fig.b).
If e > B/6, the resultant of the bearing pressure acts outside the third-point and the
pressure distribution is as shown in Figure c
112

113. Two-Way Eccentric or Moment Loading
If the resultant load acting on the base is eccentric in both the B and L directions,
it must fall within the diamond-shaped kern shown in Figure b for the contact
pressure to be compressive along the entire base of the foundation.
Pressure distribution beneath spread
footing with vertical load
that is eccentric in both the B and L
directions;
where
e
B
= eccentricity in the B direction
e
L
= eccentricity in the L direction
6 6
1
f
D
P W e e
q u
A B L
   
      
  
113

114. A 3 m square footing supports a vertical column load of 200 kNand two moment
loads with axes parallel to the sides of the footing, each having a magnitude of 124
kN-m. The embedment of the footing is 1 m, and the groundwater table is at a great
depth.
Determine whether eccentric loading requirements will be met. If these
requirements are not met, determine the minimum footing width, B, needed to
satisfy these requirements.
Example 1
114

115. Therefore, a 3.6 m x 3.6 m square footing would be required to keep the
resultant within the kern.
115

116. Equivalent Uniformly Loaded Footing
When we have a footing with an eccentric or moment loading, it is
convenient, for computational purpose, to determine the dimensions of an
equivalent uniformly loaded footing.
The equivalent footing is a footing that carries the same applied load
as the original footing but has different dimensions such that the resulting
bearing pressure is uniform (Meyerhof, 1963; BrinchHansen, 1970).
This is accomplished by decreasing the
footing width, B, to an equivalent width, B,
and the length, L, to an equivalent length, L.
The equivalent dimensions Band Lare
selected such that the net resultant load acts
through the centroid of the equivalent
footing.
116

117. Equivalent dimensions Band L
First determine the eccentricities e
B
and e
L
from the applied load and moments
(Slide 1).
Next, check to ensure that the resultant footing load acts within the kern. If not,
increase B and L such that this criterion is satisfied. Then compute the
dimensions Band Lusing
B= B -2e
B
L= L -2e
L
The equivalent bearing pressure for this footing, q
eq
, will be greater than the
average bearing of the true footing and computed as:
f
eq
P W
q
BL
 


117

118. ShallowFoundations
Settlement
118

119. PerformanceRequirements
119

Strength

Geotechnical (For shallow foundations, this is
referred as bearingcapacity)

Structural

Serviceability

Constructibility

Economic

120. Introduction

Settlement analysis is important for shallow
foundations especially when B islarge.

Besides structural loading, settlement may
also be causedby:

Weight of recently placedfill

Lowering of groundwatertable

Underground mining ortunneling

Formation ofsinkholes

Secondarycompression

Lateral movements due to adjacentexcavation
120

121. Design Requirements forSettlement


a


D

Da
The values of allowable total settlement,
a
and 
Da
depends on the type of structure supported by the
foundation.
and 
D
arecomputed using the unfactored
downward loads (Actualloads).
121

122. Design Requirements forSettlement
122
Typical Allowable Total Settlements For Foundation Design
If the predicted settlement, , is greater than 
a
, we could consider any or all ofthe
following measures:
1.Adjust the foundation design
2.Use a more elaborate foundation -Piles
3.Improve the properties of the soil –removal of compressible soil and replacing
it with engineering soils
4.Redesign the structure so it is more tolerant of settlements —For example,
flexible joints could be installed on pipes

123. Design Requirements forSettlement
123
Engineersnormallydesignthefoundationsforastructuresuch
thatallofthemhavethesamecomputedtotalsettlement.Thus,in
theory,thestructurewillsettleuniformly.Unfortunately,the
actualperformanceofthefoundationswillusuallynotbeexactly
aspredicted,withsomeofthemsettlingmorethanexpectedand
othersless.
Thedifferentialsettlement,
D
,isthedifferenceintotal
settlementbetweentwofoundationsorbetweentwopointsona
singlefoundation.Differentialsettlementsaregenerallymore
troublesomethantotalsettlementsbecausetheydistortthe
structure,
DifferentialSettlement
Therefore, we define a maximum allowable differential settlement, 
Da
, and design
the foundations so that: 
D

Da

124. SettlementAnalysis
124

Based on plate loadtest

Based on laboratory or in-situtests

125. Plate LoadTest
125
Theplateloadtestisasemi-directmethodtoestimate
theallowablebearingpressureofsoiltoinduceagiven
amountofsettlement.Plates,roundorsquare,varying
insize,from30to60cmandthicknessofabout2.5cm
areemployedforthetest.Theloadontheplateis
appliedbymakinguseofahydraulicjack.Thereaction
ofthejackloadistakenbyacrossbeamorasteeltruss
anchoredsuitablyatboththeends.
Thesettlementoftheplateismeasuredbyasetofthree
dialgaugesofsensitivity0.02mmplaced120°apart.
Thedialgaugesarefixedtoindependentsupportswhich
remainundisturbedduringthetest.

126. Plate LoadTest
126

127. Plate LoadTest
127
Plate load test arrangement

128. Plate LoadTestProcedure
128
1.Excavateapitofsizenotlessthan4to5timesthe
sizeoftheplate.Thebottomofthepitshould
coincidewiththelevelofthefoundation
2.Ifthewatertableisabovethelevelofthefoundation,
pumpoutthewatercarefullyandkeepitatthelevel
ofthefoundation.
3.Asuitablesizeofplateisselectedforthetest.
Normallyaplateofsize30cmisusedinsandysoils
andalargersizeinclaysoils.Thegroundshouldbe
levelled,andtheplateshouldbeseatedoverthe
ground.

129. Plate LoadTestProcedure Cont….
129
Aseatingloadofabout70gm/cm
2
isfirstappliedandreleased
aftersometime.Ahigherloadisnextplacedontheplateand
settlementsarerecordedbymeansofthedialgauges.Observations
oneveryloadincrementshallbetakenuntiltherateofsettlement
islessthan0.25mmperhour.
Loadincrementsshallbeapproximatelyone-fifthoftheestimated
safebearingcapacityofthesoil.
Theaverageofthesettlementsrecordedby2or3dialgaugesshall
betakenasthesettlementoftheplateforeachoftheload
increments.
The test should continue until a total settlement of 2.5 cm or the
settlement at which the soil fails, whichever is earlier, is obtained.
After the load is released, the elastic rebound of the soil should be
recorded.

130. Pressure
Settlement
25mm
130
q
a
Uses a plate size of approximately 0.3
m to 1m square or diameter. The
allowable bearing pressure for design
is taken as that corresponding to a
settlement of 1 inch (typically2.5cm).

131. 131
If the sand were to behave like an elastic material, then the settlement could be calculated
from
where
p
p
is the plate settlement,
q
ap
is the applied stress,
B
p
is the width or diameter of the plate,
vis Poisson’s ratio,
Eis the elastic modulus, and
I
p
is an influence factor (0.82 for a rigid plate).
The settlement of the real footing () of width B is related to the plate settlement by

132. 132
Load-settlement curve of a plate-load test

133. 133
For a plate of 1 ft square, the above equation may be expressed as

134. 134
ThepermissiblesettlementS
f
,foraprototypefoundation
shouldbeknown.Normallyasettlementof2.5cmis
recommended.Inaboveequations,thevaluesofS
f
,andbp
areknown.TheunknownsareS
p
andB.ThevalueofS
p
for
anyassumedsizeBmaybefoundfromtheequation.
Usingtheplateloadsettlementcurvethevalueofthe
bearingpressurecorrespondingtothecomputedvalueofS
P
isfound.
Thisbearingpressureisthesafebearingpressureforagiven
permissiblesettlementS
f
.Thevalueofimmediatesettlement
only.Iftheunderlyingsoilissandyinnatureimmediate
settlementmaybetakenasthetotalsettlement.Ifthesoilisa
clayeytype,theimmediatesettlementisonlyafractionofthe
totalsettlement.

135. 135
1.Thetestisreliableonlyifthesandlayeristhickandhomogeneous.
2.Thedepthofsandthatisstressedbelowtheplateissignificantlylower
thantherealfooting.Aweaksoillayerbelowtheplatemaynotaffect
thetestresultsbecauseitmaybeatadepthatwhichthestresses
imposedontheweaklayerbytheplateloadsmaybeinsignificant.
3.Localconditionssuchasapocketofweaksoilnearthesurfaceofthe
platecanaffectthetestresults,butthesemayhavenosignificanteffect
ontherealfooting.
4.Thecorrelationbetweenplateloadtestresultsandtherealfootingis
generallyproblematic.
5.Performanceofthetestisdifficult.Onexcavationofsandtomakeapit,
thesoilbelowtheplateinvariablybecomeslooser,andthishas
considerableinfluenceonthetestresults.Goodcontactmustbe
achievedbetweentheplateandthesandsurface,butthisisoften
difficult.
Problems associated with the Plate LoadTest

136. Settlement Analysis(cont’d)
136

137. Settlement Analyses by Lab or In-situ Test
137
Laboratory tests will involve some determination
of compressibility interms of amodulus
SPT and CPT has been commonlyused.
Especially suitable for sandy deposits where
sampling isdifficult.

138. Induced Stresses BeneathShallow
Foundations
Boussinesq’sMethod

z
I


q 
zD

138

139. Induced Stresses Beneath Shallow
Foundations

z
I


q 
zD

Newmark (1935)solution
139

140. SimplifiedMethod
Note:
(q -’
zD
)
shouldbe
(q -
zD
)
140

141. Divide1.2mx1.2mfootinginto4equalquadrants(i.e.0.6mx0.6m).
141
z
2
f
B
2
L
2
f
222
1.5
2

0.6

2

0.6

2
z
2
f
B
2
L
2
f
B
2
BLz
0.0576
L
2
z
2
0.6
2
0.6
2
1.5
2
2.97
I

0.0602
Forthe4equalquadrants,

z
4I


q
zD

4

0.0602

1816

42kPa
A l.2m x 1.2m square footing supports a column load of 250 kN.The
bottom of this footing is 0.3m below the ground surface, the groundwater
table is at a great depth, and the unit weight of the soil is 19.0 kN/m3.
Compute the induced vertical stress, 
z
, at a point 1.5 m below the
centreof this footing.

142. Example 2 –Using SimplifiedMethod
Forsquarefooting:
142

143. Approximate Method(2:1)
(from Geotechnical Engineering, Coduto1999)
143
Averagestress beneathfoundation

144. Example 2 –Using ApproximateMethod
144
Forsquarefooting:

145. SettlementAnalyses
145

Immediate settlement (independent oftime)

Consolidation settlement(time-dependent)

Secondary compression(time-dependent)
Total settlement = Immediate + consolidation +secondary
Immediatesettlement,S
e
,isthatpartofthetotalsettlement,S,
whichissupposedtotakeplaceduringtheapplicationofloading.
Theconsolidationsettlement,S
c
isthatpartwhichisduetothe
expulsionofporewaterfromthevoidsandistime-dependent
settlement.
Secondarysettlement,S
s
normallystartswiththecompletionof
theconsolidation.Itmeans,duringthestageofthissettlement,the
porewaterpressureiszeroandthesettlementisonlyduetothe
distortionofthesoilskeleton.

146. ImmediateSettlement
Skempton and BjerrumMethod:
Distortion settlement, 
d
, (also called immediate
settlement, initial settlement, or undrainedsettlement)
146
Distortion settlement beneath a spread
footing

147. ImmediateSettlement
(Assumes0.5)
147
d 0 1
qB
E
I I 
Influence factors for computing settlement (after Christian and Carrier, 1978).
where
q = average bearing stress (Total service load)
B = footing width
E = average soil modulus over depth of compressible layer
I
0
= influence factor accounting for footing depth
I
1
= influence factor accounting for footing shape and thickness
of compressible layer

148. ImmediateSettlement
148
100 E1500
(Duncan andBuchignani,1976)
Typically, E300s
u
d 0 1
qB
E
II 
Chart for estimating the undrained modulus of clays based on
overconsolidationratio, plasticity, and undrained shear strength
Undrained
E

s
u

Drained
E= 0.67 1 E
where = 0.1 to 0.4
 


149. Immediate settlement –Cohesivesoils
For a uniformly loaded circular or rectangular areas
near the surface of a relatively deep stratum, the
vertical settlementis:
149
C
s
Shapeandrigidityfactor
quniformload
Bwidth offoundation

s
Poisson' s ratio ofsoil
E
s
Young' s modulusof soil
2
s
d s
s
1
CqB
E
 
   
 

150. C
s
for infinitedepth
150
AverageEdge / Middleof
longside
CornerCentreShape &Rigidity
0.850.641.0Circle(flexible)
0.790.790.79Circle(rigid)
0.950.760.561.12Square(flexible)
0.820.820.820.82Square(rigid)
Rectangle(flexible)
1.301.120.761.53L/B =2
1.821.681.052.10L/B =5
2.242.101.282.56L/B =10
Rectangle(rigid)
1.121.121.121.12L/B =2
1.601.601.601.60L/B =5
2.02.02.02.0L/B =10

151. C
s
for
limited
depth
over
Rigid
Base
151

Corner of Flexible RectangularAreaCentre of Circular
Area
(diameter=B)
H/B
L/B=∞L/B=10L/B=5L/B=2L/B=1
=0.50
0.00.00.00.00.00.00
0.040.040.040.040.050.140.5
0.100.100.100.120.150.351.0
0.180.180.180.220.230.481.5
0.260.260.270.290.290.542.0
0.270.380.390.400.360.623.0
0.520.540.550.520.440.695.0
0.730.770.760.640.480.7410.0
=0.33
0.00.00.00.00.00.00
0.080.080.080.080.090.200.5
0.160.160.160.180.190.401.0
0.250.250.250.280.270.511.5
0.340.340.340.340.320.572.0
0.450.450.460.440.380.643.0
0.610.610.600.560.460.705.0
0.810.820.800.660.490.7410.0

152. 43200
B1.8m

1.8m

1.8m

A
1
z
h
B
0
u u
zD
f


152-11

1.8


0.98

0.7

4mm
 I0.7
L
1,
B
I0.98
D

0.6m
0.3

43200kPaE300s300144kPa


0.6m


18kN/m
3

11kPa
q 
P W
f

445kN 46kN
152 kPa
W(1.8m)(1.8m)(0.6m)(23.6kN/m
3
) 46kN
152
Example Find the allowable settlement for the proposed square footing in
Figure
d 0 1
qB
E
I I 
Proposed square footing

153. One-Dimensional Consolidation Settlement –
ClassicalAnalysis
153
where

c
= ultimate consolidation settlement
r = rigidity factor
C
c
= compression index
C
r
= recompression index
e
0
= initial void ratio
H = thickness of soil layer

z0
= initial vertical effective stress at midpoint of soil layer

zf
= final vertical effective stress at midpoint of soil layer

c
= preconsolidation stress at midpoint of soil layer
'
zf
'
z0

z

154. Definition of OC-I andOC-II
CaseI CaseII
154

155. One-Dimensional Consolidation
Settlement –ClassicalAnalysis
155
r -VALUES FOR COMPUTATION OF TOTAL SETTLEMENT AT THE CENTER OF A SHALLOW FOUNDATION, AND METHODOLOGY
FOR COMPUTING DIFFERENTIAL SETTLEMENT
Entirefootingsettlesuniformly,solongasbearing
pressureisuniform.Computedifferential
settlementbetweenfootingsoralonglengthof
continuousfootingusingthetotalsettlement
analysisorthe
D
/ratiomethod

156. One-Dimensional Consolidation
Settlement –ClassicalAnalysis
156
APPROXIMATE THICKNESSES OF SOIL LAYERS FOR MANUAL
COMPUTATION OF CONSOLIDATION SETTLEMENT OF SPREAD FOOTINGS
Theclassicalmethoddividesthesoilbeneaththe
footingintolayers.Thebestprecisionisobtained
whentheuppermostlayeristhin,andtheybecome
progressivelythickerwithdepth

157. One-Dimensional Consolidation Settlement –
ClassicalAnalysis
157
Influence factors for induced vertical
stress under a semi-infinite strip load
Influence factors for induced vertical
stress under a square loaded area

158. Example –The allowable settlement for the proposed continuous
footing in the Figureis 25 mm. Using the classical method, compute its
settlement and determine if it satisfies thiscriterion.
158
zD
B1.2
W
f


18kN/m
3


0.5m

9kPa
P

W
f
q
bb

65 14
66kPa


1.2m

0.5m


23.6kN/m
3

14kN/m
P 65kN/m
Proposed footing

2.60
z zD 2
f
1
1 q -
B
1
2z
 
 
 
 
   
    
 
  
 
  
 
    

159. Example (cont’d)
23.40.040.13OCI3186951180.501.01
7.10.040.13OCI3365418361.501.02
2.90.040.13OCI61576512.751.53
1.60.040.13OCI75683654.502.04
0.70.050.15OCI487881877.003.05
35.7=
For continuous footing, 
c
= r x 35.7mm = 0.85 x 35.7mm 31mm

c
= 31 mm>
a
Settlement criterion is notsatisfied
159
'
zf
'
z0

z

160. Skempton and BjerrumMethod

d

c
160

161. Settlement Analyses based on In-situTests
161

Standard Penetration Test(SPT)

Cone Penetration Test(CPT)

Dilatometer Test(DMT)

Pressuremeter Test(PMT)
Suitable method for all soil types,
especially suitable for sandy soils as they
are difficult tosample.

162. Settlement Analyses based on In-situ Tests
(cont’d)
162

Schmertmann’s Method (1970,1978)

Meyerhof’s Method(1965)

Burland and Burbidge’s Method(1985)

163. Schmertmann’sMethod
163

Perform appropriate in-situ tests to define
subsurfaceconditions.

SPT

CPT

DMT

PMT

164. Schmertmann’s Method(cont’d)

Consider the soil from the base of the foundation
depth of influence below the base. Divide this zone into
several soil layers.

Compute the peak strain influencefactor,
164
zD
p
zp

I = 0.5 + 0.1
q



where
I
p
= peak strain influence factor
q = bearing stress

zD
= initial vertical effective stress at a depth D below the ground surface

zp
= initial vertical effective stress at depth of peak strain influence factor
(for square and circular footings (L/B = 1), compute 
zp
at a depth of
D + B/2 below the ground surface; for continuous footings (L/B 10),
compute it at a depth of D + B)

165. Induced Stresses Beneath Shallow
Foundations
Boussinesq’sMethod

z
I


q
zD

I



p

165

166. Schmertmann’s Method(cont’d)

Compute the strain
influence factor, I
є
, in the
middle of each soillayer.
Soil layers are determined from
stiffnessprofile
166
Distribution of strain influence factor with
depth under square and continuous footings
(Adapted from Schmertmann1978)

167. Schmertmann’s Method(cont’d)

Compute the correction factors, C1, C2 andC3:
167
Footingshape,
Secondarycreep,
Depthfactor,
zD
1
zD
C = 1 - 0.5
q -
 
 

 
2
t
C = 1 + 0.2log
0.1
 
 
 

Compute thesettlement:
s
zD123
E


I

H
CCC

q 
for t 0.1 year
3
L
1.03 0.03
C = max B
0.73
 

 
 
 

168. Schmertmann’s Method(cont’d)
168
where
= settlement of footing
C
1
= depth factor
C
2
= secondary creep factor
C
3
= shape factor
q = bearing stress

zD
= effective vertical stress at a depth D below the ground surface
I
e
= influence factor at midpoint of soil layer
H = thickness of soil layer
E
s
= equivalent modulus of elasticity of soil layer
t = time since application of load (year) (t 0.1 year)
B = footing width
L = footing length
These formulas may be used with any consistent set of units, except that t must be
expressed in years. Typicallywe use t = 50 years (C
2
= 1.54).

169. E
s
from CPTResults
169
ESTIMATING EQUIVALENT SOIL MODULUS, E
S
, VALUES FROM CPT
RESULTS [Adapted from Schmertmannet al. (1978), Robertson and
Campanella (1989), and other sources.]

170. E
s
from SPTResults
170
0.6
N
60

E
m
C
B
C
S
C
R
N
E
s

0
OCR 
1
N
60

171. Example –The results of a CPT sounding performed at McDonald’s Farm near Vancouver,
British Columbia, are shown in Figure shown. The soils at this site consist of young, normally
consolidated sands with interbedded silts. The groundwater tables is at a depth of 2.0 m
below the ground surface.
171
A 375 kN/m load is to be
supported on a 2.5 m x 30 m
footing to be founded at a depthof
2.0 m in this soil. Use
Schmertmann’s method to
compute settlement of this footing
soon after construction and the
settlement 50 yearsafter
construction.
=17kN/m
3
=20kN/m
3
CPT results

172. Solution
1kPa0.0102kg/cm
2
172
From the Table, use E
s
= 2.5q
c
Depth of influence = D + 4B = 2.0 + 4(2.5) = 12.0 m

173. Example (cont’d)
W
f


2.5m

2.0m


23.6kN/m
3

118kN/m
59kPa
173
zp
zD
0.50.1
197kPa 34kPa
0.666
'
q 
zD
I0.50.1
'
D

17kN/m
3


2m

34kPa
p

375kN/m 118kN/m –0
197kPa
2.5m
'
zp

atzDB



Hu


17kN/m
3


2m



20kN/m
3


2.5m



9.8kN/m
3


2.5m

59kPa
f
D
p w
q u
B

 

174. Example(cont’d)
174

175. Example 7.6(cont’d)
s
175
zD123
E


I

H
CCC

q 

176. E
s
zD123
At t50yr,
At t = 0.1 yearC
2
1.0
C C C

q 


I

H
Example (cont’d)


0.896

1.0

0.73

19734


35.73x10
5

38mm
176
zD
1
zD
34
C = 1 - 0.5 = 1 - 0.5 = 0.896
q - 197 - 34
   
   
   
3
3
L
C = 1.03 0.03 0.73
B
30
= 1.03 - 0.03x = 0.67 Use C = 0.73
2.5
 
 
 
 


0.896

1.54

0.73

19734


35.73x10
5

59mm
2
t 50
C = 1 + 0.2log = 1 + 0.2log = 1.54
0.1 0.1
   
   
   

177. Accuracy of SettlementPredictions
177

Uncertainties in defining soilprofile

Disturbances of soilsamples

Errors in in-situtests

Errors in laboratorytests

Uncertainties in defining serviceloads

Constructiontolerances

Errors in determining degree ofconsolidation

Inaccuracies in the analysismethodologies

Neglecting soil-structure interactioneffects

178. Accuracy of Settlement Predictions
(cont’d)
•
Settlement predictions are
more oftenconservative
•
Settlement predictions based
on Schmertmann’s method
with CPT data are more
reliable than with those based
onSPT.
•
Settlement predictions in
clays, especially clays that
are overconsolidated, are
usually more reliablethan
those insands.
178
Comparison between computed and measured settlements of spread footings. Each bar
represents the 90 percent confidence interval (i.e., 90 percent of the settlement
predictions will be within this range). The line in the middle of each bar represents the
average prediction, and the number to the right indicates the number of data points used
to evaluate each method (based on data from Burlandand Burbridge, 1985; Butler, 1975;
Schmertmann, 1970; and Wahls, 1985).