Foundation Design [Compatibility Mode]

1

Foundation designFoundation design

�� Present by Mr. Sieng PEOUPresent by Mr. Sieng PEOU�� Master science of geotechnical Master science of geotechnical

engineeringengineering

�� TelTel--011 874 974011 874 974�� email: [email protected]: [email protected]

Type of foundationType of foundation

�� Shallow foundationShallow foundation11--Spread footing : support the load from Spread footing : support the load from building by columnbuilding by column22--Strip footing : support the load from Strip footing : support the load from building by wallsbuilding by walls33--Mat foundation: combined all footingMat foundation: combined all footing

2

Type of foundationType of foundation�� Deep foundationDeep foundation

11-- End bearing pile : pile stand on End bearing pile : pile stand on rocks or very dense soils, so we rocks or very dense soils, so we have only end bearing capacityhave only end bearing capacity22-- Combined bearing pile : pile stand Combined bearing pile : pile stand on normal soils, so we have end on normal soils, so we have end bearing capacity and skin frictionbearing capacity and skin friction33-- Floating pile : pile stand on very Floating pile : pile stand on very loose or very soft soil, so we have loose or very soft soil, so we have only skin frictiononly skin friction

Spread footingSpread footing

Q

B

3

Strip footingStrip footing

q

B

Mat foundationMat foundation

B

4

End bearing pileEnd bearing pile

Rock layer

Soft soil layerPile

Combined bearing pileCombined bearing pile

Stiff soil layer

Soft soil layerPile

5

Floating pileFloating pile

Soft soil layerPile

Bearing capacity for Shallow Bearing capacity for Shallow foundationfoundation

�� Type of failureType of failure11--General shear failure for dense soil,we General shear failure for dense soil,we

can use can use C & C & φφφφφφφφ for design soils bearing for design soils bearing capacitycapacity

22--Local shear failure for loose soil, we can Local shear failure for loose soil, we can use use C’=2/3 C & C’=2/3 C & φφφφφφφφ’’ ========arctg(2/3tgarctg(2/3tgφ)φ)φ)φ)φ)φ)φ)φ) for for design soils bearing capacitydesign soils bearing capacity

33--Punching shear failure for very loose Punching shear failure for very loose soil,not recommendedsoil,not recommended

6

General shear failureGeneral shear failure

Q

D

Shear line

Local shear failureLocal shear failure

Q

D

Shear line

7

Punching shear failurePunching shear failure

Q

D

Failure mechanisms and derivation of Failure mechanisms and derivation of equationsequations

8

Failure mechanisms and derivation of Failure mechanisms and derivation of equationsequations

�� A relatively undeformed wedge of soil below the foundation A relatively undeformed wedge of soil below the foundation forms an active Rankine zone with angles (45º + forms an active Rankine zone with angles (45º + φφ'/2). '/2).

�� The wedge pushes soil outwards, causing passive Rankine The wedge pushes soil outwards, causing passive Rankine zones to form with angles (45º zones to form with angles (45º -- φφ'/2). '/2).

�� The transition zones take the form of log spiral fans. The transition zones take the form of log spiral fans. �� For purely cohesive soils (For purely cohesive soils (φφ = 0) the transition zones become = 0) the transition zones become

circular for which Prandtl had shown in 1920 that the solution circular for which Prandtl had shown in 1920 that the solution is is qq ff = (2 + = (2 + ππππππππ) Cu = 5.14 Cu) Cu = 5.14 Cu

�� This equation is based on a weightless soil. Therefore if the This equation is based on a weightless soil. Therefore if the soil is nonsoil is non--cohesive (c=0) the bearing capacity depends on cohesive (c=0) the bearing capacity depends on the surcharge qthe surcharge qoo. For a footing founded at depth D below the . For a footing founded at depth D below the surface, the surcharge surface, the surcharge qqoo = = γγγγγγγγDD. Normally for a shallow . Normally for a shallow foundation (D<B), the shear strength of the soil between the foundation (D<B), the shear strength of the soil between the surface and the founding depth D is neglected. surface and the founding depth D is neglected.

SemiSemi--circular slip mechanismcircular slip mechanism

�� Moment causing rotationMoment causing rotation= load x lever arm = load x lever arm = [(q = [(q -- qqoo) x B] x [½B] ) x B] x [½B]

�� Moment resisting rotationMoment resisting rotation= shear strength x length of arc x lever arm = shear strength x length of arc x lever arm = [s] x [= [s] x [ππ.B] x [B] .B] x [B]

�� At failure these are equal: At failure these are equal: (q (q -- qqoo ) x B x ½B = s x ) x B x ½B = s x ππ.B x B .B x B

�� Net pressure (q Net pressure (q -- qqoo ) at failure) at failure= = 2 2 ππππππππ x shear strength of the soilx shear strength of the soil�� This is an upperThis is an upper--bound solution. bound solution.

9

Circular arc slip mechanismCircular arc slip mechanism�� Moment causing rotationMoment causing rotation

= load x lever arm = load x lever arm = [ (q = [ (q -- qqoo) x B ] x [B/2] ) x B ] x [B/2]

�� Moment resisting rotationMoment resisting rotation= shear strength x length of arc x lever arm = shear strength x length of arc x lever arm = [s] x [2= [s] x [2αα R] x [R] R] x [R]

�� At failure these are equal: At failure these are equal: (q (q -- qqoo) x B x B/2 = Cu x 2 ) x B x B/2 = Cu x 2 αα R x R R x R

�� Since R = B / sin Since R = B / sin αα : : (q (q -- qqoo ) = Cu x 4) = Cu x 4αα /(sin /(sin αα)² )²

�� The worst case is when The worst case is when �� tantanαα=2=2αα at at αα = 1.1656 rad = 66.8 deg = 1.1656 rad = 66.8 deg

�� The net pressure (q The net pressure (q -- qqoo) at failure ) at failure �� == 5.52 x shear strength of soil5.52 x shear strength of soil

Ultimate bearing capacityUltimate bearing capacity

Settlement

σqu

10

Bearing capacity for strip footingBearing capacity for strip footinggeneral equation general equation

After Terzaghi (1943)After Terzaghi (1943)qd = CNc + γγγγs DNq +0.5 γγγγBNγγγγ

Nc = (Nq – 1 ) . Cotgφφφφ Prandtl 1921

Nγγγγ = 2(Nq + 1)tgφφφφ Caquot and Kerisel 1953 Vessic 1973

1924Re)2

45(tan tan2 issnereNqφπφ+=

Bearing capacity for footingBearing capacity for footingFrom TSA equationFrom TSA equationAfter Vessic (1973)After Vessic (1973)

qd = 5.14 Cu(1+0.2B/L) + γγγγs D

Cu:Undrained cohesion

B: Width of footing

L: Length of footing

11

Bearing capacity for footingBearing capacity for footingFrom TSA equationFrom TSA equation

After Skemton (1951)After Skemton (1951)

qd = 5 Cu(1+0.2B/L)(1+0.2D/B) + γγγγs D


B: Width of footing


D: Depth of footing

D/B<2.5

Bearing capacity for footingBearing capacity for footingFrom TSA equationFrom TSA equation

After Meyerhof (1951 to 1963)After Meyerhof (1951 to 1963)

qd = 5.14 Cu(1+0.2B/L)(1+0.2D/B) + γγγγs D


B: Width of footing


D/B<2.5

12

Bearing capacity for footingBearing capacity for footingFrom ESA equationFrom ESA equationAfter Vessic (1973)After Vessic (1973)

qd = γγγγs D Nq(1+B/L.tgφφφφ)+0.5γγγγBNγγγγ(1-0.4B/L)

φφφφ: Internal friction angle

B: Width of footing


Bearing capacity for footingBearing capacity for footingFrom ESA equationFrom ESA equation

After Meyerhof (1951 to 1963)After Meyerhof (1951 to 1963)qd = γγγγs D Nq.Sq.dq+0.5γγγγBNγγγγSγγγγdγγγγ


B: Width of footing


Sq=Sγγγγ=1+0.1Κ=1+0.1Κ=1+0.1Κ=1+0.1ΚPB/L ; K p= tg2(45+φ/φ/φ/φ/2)

dq=dγγγγ=1+0.1Kp0.5D/B

Nq the same Nq Terzaghi ; Nγγγγ=(Nq-1)tg(1.4φ)φ)φ)φ)

13

Bearing capacity for footingBearing capacity for footingFrom general equationFrom general equationAfter Meyerhof (1963)After Meyerhof (1963)

qd = C.Nc.Fcs.Fcd.Fci+γγγγs D Nq. Fqs.Fqd.Fqi +0.5γγγγBNγ γ γ γ Fγγγγs.Fγγγγd.Fγγγγi


B: Width of footing


Nq by Reissner1924 ; Nc by Prandtl1921 ; Nγγγγ by Caquot and Kerisel 1953 and by Vessic 1973

qnet =C.Nc.Fcs.Fcd.Fci+γγγγs D (Nq-1). Fqs.Fqd.Fqi +0.5γγγγBNγ γ γ γ Fγγγγs.Fγγγγd.Fγγγγi

Bearing factorBearing factor�� Shape factor by De Beer 1970Shape factor by De Beer 1970

FFcscs=1+B/L.N=1+B/L.Nqq/N/Ncc

FFqsqs=1+B/L.tg=1+B/L.tgφφFFγγss==11--0.4.B/L0.4.B/L

�� Depth factor by Hansen 1970Depth factor by Hansen 1970Condition D/B<1Condition D/B<1

�� FFcdcd=1+0.4D/B=1+0.4D/B�� FFqdqd=1+2.tg=1+2.tgφφ(1(1--sinsinφφ))22D/BD/B�� FFγγdd==11

14

Bearing factorBearing factor�� Depth factor by Hansen 1970Depth factor by Hansen 1970

Condition D/B>1Condition D/B>1FFcdcd=1+0.4.arctg(D/B)=1+0.4.arctg(D/B)FFqdqd=1+2.tg=1+2.tgφφ(1(1--sinsinφφ))22.arctg(D/B).arctg(D/B)FFγγdd==11

�� Inclined factor by Meyerhof 1963 Meyerhof and Inclined factor by Meyerhof 1963 Meyerhof and Hanna 1981Hanna 1981

FFcici=F=Fqiqi=(1=(1--α/α/90)90)22

FFγγii=(1=(1--α/φα/φ))22

Bearing capacity of mat foundationBearing capacity of mat foundation�� The gross ultimate bearing capacity of a mat The gross ultimate bearing capacity of a mat

foundation can be determined by the same foundation can be determined by the same equation used for shallow foundation.equation used for shallow foundation.

�� A suitable factor of safety should be used to A suitable factor of safety should be used to calculate the net allowable bearing capacity.For calculate the net allowable bearing capacity.For rafts on clay, the factor of safety should not be rafts on clay, the factor of safety should not be less than 3 under dead load and maximum live less than 3 under dead load and maximum live load.However, under the most extreme load.However, under the most extreme conditions,the factor of safety should be at least conditions,the factor of safety should be at least 1.75 to 2. For rafts constructed over sand,a 1.75 to 2. For rafts constructed over sand,a factor of safety of 3 should normally be used.factor of safety of 3 should normally be used.

15

Ultimate bearing capacity equation Ultimate bearing capacity equation for mat foundation on saturated clayfor mat foundation on saturated clay

)4.01)(195.0

1(14.5)( B

D

L

BCq f

uunet ++=

Allowable bearing capacityAllowable bearing capacity

�� Net ultimate bearing capacity Net ultimate bearing capacity qqnetnet=q=qdd--γγγγγγγγss.D.D

�� Net allowable bearing capacityNet allowable bearing capacityqqnetnet

allall =q=qnetnet/FS/FS�� Gross allowable bearing capacityGross allowable bearing capacity

qqallall =q=qnetnetallall ++γγγγγγγγss.D.D

16

Verify the stable of footingVerify the stable of footing

B&L

Q

QsQf

Qtotal=Q+Qf+Qs

Q- load apply by column

Qf –load of footing

Qs –load of soil above footing

Verify stable of footingVerify stable of footing

BL

Qq all

net = We find value of B

BL

Qq total

all >

And verify the stable of footing from equation

17

When effect water tableWhen effect water table

Water level case I

Water level case II

D

d

D1

D2

B

When effect water tableWhen effect water table

11--In case I if the water table is located so thatIn case I if the water table is located so that0<D0<D11<D, so we will change the factor<D, so we will change the factor

γγss.D .D γ.γ.DD11+D+D22((γγsatsat--γγww))Also value Also value γγ in the last term of the equation has in the last term of the equation has

to be replaced by to be replaced by γγ’= (’= (γγsatsat--γγww))22--In case II for a water table located so 0<d<BIn case II for a water table located so 0<d<Bvalue value γγ in the last term of the equation has to be in the last term of the equation has to be

replaced by replaced by γγcalcal= = γγ’+d/B.(’+d/B.(γγ−−γγ’)’)

18

Stable of footing when effect Stable of footing when effect inclined loadinclined load

B

Q

D

α

V

H

QT

qall>V/(BL)

Tall>H

V=Q.Cos αααα

Η=Η=Η=Η=Q.Sin αααα

T=V.tg(2/3φφφφ)+2/3.C.B.L.

Tall=T/1.5

When effect 0ne way bending When effect 0ne way bending momentmomentQ

MB

B

We change B to B’ for calculate bearing capacity

B’=B-2eB

eB=MB/Q

19

Verify stable of footing when Verify stable of footing when effect one way bending momenteffect one way bending moment

maxqqall >

)6

1(max B

e

BL

Qq B+=)

61(min B

e

BL

Qq B−=

MB

Q

When eB<B/6

Verify stable of footing when Verify stable of footing when effect one way bending momenteffect one way bending moment

maxqqall >

)2(3

4max

BeBL

Qq

−=

MB

Q

When eB>B/6

Not recommended

20

Foundation with two way Foundation with two way EccentricityEccentricity

For calculate bearing capacity we have to change:

B to B’=B-2eB

L to L’=L-2e L

A’=B’*L’

eB=MB/Q

eL=ML/Q

ML

MB

B

L

Q

Verify stable of footing when Verify stable of footing when effect two way bending momenteffect two way bending moment

Qult= qu’.A’

Case eL/L>1/6

eB/B>1/6

B1=B(1.5-3eB/B)

L 1=L(1.5-3eL/L)

B’=A’/L

21

Verify stable of footing when effect Verify stable of footing when effect two way bending momenttwo way bending moment

Qult= qu’.A’

Case 1/6<eL/L<0.5

0<eB/B<1/6

A’=0.5(L 1+L2)B

B’=A’/L 1


Qult= qu’.A’

Case eL/L< 1/6

1/6<eB/B< 0.5

A’=0.5(B1+B2)L

B’=A’/L

22


Qult= qu’.A’

Case eL/L< 1/6

eB/B< 1/6

A’= L 2B+0.5(B+B2)(L-L 2)

B’=A’/L

Footing on two layerFooting on two layer

D γs

B d1c1 γ1 φ1

c2 γ2 φ2

23

Bearing capacity of footing Bearing capacity of footing on two layeron two layer

11-- Determine influenced thicknessDetermine influenced thicknessH=0.5Btg(45+H=0.5Btg(45+φφ11/2)/2)

If H<dIf H<d11 : our footing not effect on second layer, : our footing not effect on second layer, so we calculate the soils bearing capacity by so we calculate the soils bearing capacity by using values Cusing values C11,,γγ11,φ,φ11

If H>dIf H>d11 : our footing effect on second layer, so : our footing effect on second layer, so we calculate the soils bearing capacity by using we calculate the soils bearing capacity by using condition as follows:condition as follows:

Bearing capacity of footing on Bearing capacity of footing on two layertwo layer

From TSA conditionFrom TSA condition11-- Design CDesign CRR=C=CU2U2/C/CU1U1

If CIf CRR<1 : < 5.14 for strip footing <1 : < 5.14 for strip footing

< 6.05 for spread footing < 6.05 for spread footing

so we calculate the soils bearing capacity by using so we calculate the soils bearing capacity by using equationequation

qqnetnet=C =C u1u1NNCCIf CIf CRR>0.7 the value of N>0.7 the value of NCC is decrease 10%is decrease 10%

RCB

dNc 14,5

5,1 1 +=

RCB

dNc 05,6

3 1 +=

24


If CIf CRR>1 : >1 : for strip footing for strip footing

for spread footing for spread footing

so we calculate the soils bearing capacity by so we calculate the soils bearing capacity by using equationusing equation

qqnetnet=C =C u1u1NNCC

14,45,0

11

+=d

BN 14,4

1,12

1

+=d

BN

05,533,0

11

+=d

BN 05,5

66,02

1

+=d

BN

221

21 ×+×=

NN

NNNc


From general equation 1From general equation 1

11-- Determine the average values of soils Determine the average values of soils parameterparameter

22-- Determine the soils bearing capacity by Determine the soils bearing capacity by using valuesusing values

C’ and C’ and φφφφφφφφ’’

H

dHd 2111 )('

φφφ −+=

H

cdHcdc 2111 )('

−+=

25



11-- Determine the bearing capacity for first layerDetermine the bearing capacity for first layer

22-- Determine the soils bearing capacity for Determine the soils bearing capacity for second layersecond layer

qnet1 = C1.Nc.Fcs.Fcd.Fci+γγγγs D (Nq-1). Fqs.Fqd.Fqi +0.5γγγγ1111BNγ γ γ γ Fγγγγs.Fγγγγd.Fγγγγi

qnet2= C2.Nc.Fcs.Fcd.Fci+(γγγγs D+γγγγ1d1) (Nq-1). Fqs.Fqd.Fqi +0.5.γγγγ2222....BNγ γ γ γ Fγγγγs.Fγγγγd.Fγγγγi



33-- Determine the bearing capacity Determine the bearing capacity < q< qnet1net1

P = 2(B+L)P = 2(B+L)

Pv = 0.5 Pv = 0.5 γγ11 dd1122+ + γγssD dD d11

KK s s =1=1--sinsinφφ11

Af =BLAf =BL

Af

CPd

Af

KPPqq sV

netnet111

2

tan +×+= φ

26

Bearing capacity from in situ Bearing capacity from in situ testtest

�� From static cone penetration testFrom static cone penetration test11-- for B<1.22mfor B<1.22m

22-- for B>1.22mfor B>1.22m

�� From dynamic cone penetration testFrom dynamic cone penetration test

qallowable =

15cq

qallowable = ( )2

28,3128,3

25 B

Bqc +

qallowable =

20

Rd

Bearing capacity from in situ testBearing capacity from in situ test�� From standard penetration test SPTFrom standard penetration test SPT11-- for B<1.22m and settlement 25mmfor B<1.22m and settlement 25mm

but after Bowles 1977but after Bowles 1977

22-- for B>1.22m and settlement 25mmfor B>1.22m and settlement 25mm

but after Bowles 1977but after Bowles 1977

corall NKPaq 98,11)( =

2

28,3128,3

99,7)(

+=B

BNKPaq corall

+=25

33,0116,19S

B

DNq corall

+

+=25

33,0128,3

128,398,11

2S

B

DN

B

Bq corall

33,133,01 <

+B

D

27

Combined footingCombined footingRectangular combined footingRectangular combined footing

Q1Q2

L3 L2L1X

q

B

L

Section

Plan

Q1+Q2

Design dimension of rectangular Design dimension of rectangular combined footingcombined footing

�� Determine the area of the footingDetermine the area of the footing

�� Determine the location of the resultant of the Determine the location of the resultant of the column loadscolumn loads

�� For uniform distribution of soil pressure under For uniform distribution of soil pressure under the foundation, the resultant of the column the foundation, the resultant of the column loads should pass through the centroid of the loads should pass through the centroid of the foundation.Thus,foundation.Thus,

)(

21

netallq

QQA

+=

21

32 .

QQ

LQX

+=

)(2 1 XLL +=

28

Design dimension of rectangular Design dimension of rectangular combined footingcombined footing

�� Once the length L is determined,obtain the Once the length L is determined,obtain the value of Lvalue of L11

�� Note that the magnitude of LNote that the magnitude of L22 will be known and will be known and depends on the location of the property linedepends on the location of the property line

�� The width of the foundation then is The width of the foundation then is

321 LLLL −−=

L

AB =

Combined footing Combined footing Trapezoidal combined footingTrapezoidal combined footing

L2 L3L1

XQ1 Q2

Q1+Q2

B1 B2L

Section

Plan

29

Design dimension of trapezoidal Design dimension of trapezoidal combined footingcombined footing

�� Determine the area of the footingDetermine the area of the footing

�� And we have relationAnd we have relation

�� Determine the location of the resultant of the Determine the location of the resultant of the column loadscolumn loads

)(

21

netallq

QQA

+=

21

32 .

QQ

LQX

+=

LBB

A2

21 +=

Design dimension of trapezoidal Design dimension of trapezoidal combined footingcombined footing

�� From the property of a trapezoid, we haveFrom the property of a trapezoid, we have

�� With Known values of With Known values of A,L,XA,L,X and and LL22 we can find we can find values of values of BB11 and and BB22, Note that for a trapezoid,, Note that for a trapezoid,

3

2

21

212

L

BB

BBLX

++=+

23 2

LLX

L <+<

30

Combined footing Combined footing Cantilever footingCantilever footing

Q1 Q2

R1

R2

S

e

L1 B2

Section

Plan

Design dimension of Design dimension of Cantilever footingCantilever footing

�� Design arm moment for soils reaction Design arm moment for soils reaction strength Rstrength R11

S’=SS’=S--e (value of e (value of ee is proposed by designer)is proposed by designer)�� Design soils reaction strengthDesign soils reaction strength

'11 S

SQR =

'

.122 S

eQQR −= 1212 RQQR −+=

31

Design dimension of Design dimension of Cantilever footingCantilever footing

�� Design the dimension of first footingDesign the dimension of first footing

�� C is length of columnC is length of column�� Design the dimension of second footingDesign the dimension of second footing

allnetq

RA 1

1 =

+=2

21

CeL

1

11 L

AB =

allnetq

RA 2

2 =2

22 L

AB =

Rock qualityRock quality�� Rock quality designation(RQD) is an index or Rock quality designation(RQD) is an index or

measure of the quality of a rock mass(Stagg and measure of the quality of a rock mass(Stagg and Zienkiewicz 1968) used by many engineers.RQD Zienkiewicz 1968) used by many engineers.RQD is computed from recovered core samples asis computed from recovered core samples as

�� A core advance of 1500mm produced a sample A core advance of 1500mm produced a sample length of 1310mm consisting of dust,gravel,and length of 1310mm consisting of dust,gravel,and intact pieces of rock.The sum of length of pieces intact pieces of rock.The sum of length of pieces 100mm or larger in length is 890mm.The recovery 100mm or larger in length is 890mm.The recovery ratio Lratio Lrr=1310/1500=0.87 and RQD=890/1500=0.59=1310/1500=0.87 and RQD=890/1500=0.59

advance core ofLength

100mmcore of piecesintact oflength >∑=RQD

32

Allowable Bearing capacity of Allowable Bearing capacity of rockrock

�� The allowable bearing capacity is The allowable bearing capacity is depending on geology,rock type,and depending on geology,rock type,and quality(as RQD).quality(as RQD).

�� If RQD>0.8 would not require as high an If RQD>0.8 would not require as high an FS as for RQD=0.4.FS as for RQD=0.4.

�� We take FS from 6 to 10 for RQD less We take FS from 6 to 10 for RQD less than about 0.75than about 0.75

Bearing capacity for sound Bearing capacity for sound rockrock

1

)2

45(tan5

)2

45(tan

4

6

+=

+=

+=

q

c

q

NN

N

N

γ

φ

φ

Φ=45 degree for most rock except

limestone or shale where values

between 38 to 45 degree.

Similarly we could in most cases

estimate Cu=5MPa as a conservative

value.

And finally we may reduce the ultimate

bearing capacity base on RQD as:

qult=qult(RQD)2

For calculate bearing capacity we use equation Terzaghi

33

Rang of properties for selected rock Rang of properties for selected rock groups;data from several sourcesgroups;data from several sources

Type of Type of rockrock

Unit wt.(KN/mUnit wt.(KN/m33)) E(MPa.10E(MPa.103)3) µµ qqu(u((Mpa)(Mpa)

BasaltBasalt 2828 1717--103103 0.270.27--0.320.32 170170--415415

GraniteGranite 26.426.4 1414--8383 0.260.26--0.300.30 7070--276276

SchistSchist 2626 77--8383 0.180.18--0.220.22 3535--105105

LimestoneLimestone 2626 2121--103103 0.240.24--0.450.45 3535--170170

Porous Porous limestonelimestone

-- 33--8383 0.350.35--0.450.45 77--3535

SandstoneSandstone 22.822.8--23.623.6 33--4242 0.200.20--0.450.45 2828--138138

ShaleShale 15.715.7--2.22.2 33--2121 0.250.25--0.450.45 77--4040

concreteconcrete 15.715.7--23.623.6 variablevariable 0.150.15 1515--4040

Settlement of shallow Settlement of shallow foundationfoundation

�� There are two types of settlementThere are two types of settlement11--Immediate settlement or elastic settlement SImmediate settlement or elastic settlement See

for sandy soilsfor sandy soils22--Consolidation settlement SConsolidation settlement Scc for fine grained for fine grained

soilssoils22--11--Primary consolidation settlement for soils Primary consolidation settlement for soils normalnormal22--22--Secondary consolidation settlement for Secondary consolidation settlement for organics soilsorganics soils

34

Immediate settlement on sandy soilsImmediate settlement on sandy soils

�� Foundation could be considered fully flexible or Foundation could be considered fully flexible or fully rigidfully rigid11--A uniformly loaded, perfectly flexible A uniformly loaded, perfectly flexible foundation resting on an elastic material such foundation resting on an elastic material such as saturated clay will have a sagging profile as as saturated clay will have a sagging profile as shown in shown in figure 1figure 1 ,because of elastic ,because of elastic settlement.settlement.22--If the foundation is rigid and is resting on an If the foundation is rigid and is resting on an elastic material such as clay,it will undergo elastic material such as clay,it will undergo uniform settlement and the contact pressure will uniform settlement and the contact pressure will be redistributed as shown in be redistributed as shown in figure 2figure 2 ..

Type of foundation settlementType of foundation settlement

Figure 1Settlement profile

Settlement profileFigure 2

35

Calculate immediate Calculate immediate settlementsettlement

Q

D

H

Soil

Rock

q0

µ−Poisson’s ratio

E-Modulus of elasticity

Calculate immediate Calculate immediate settlementsettlement

�� At corner of the flexible foundationAt corner of the flexible foundation

�� At center of the flexible foundationAt center of the flexible foundation

�� Average settlement for flexible foundationAverage settlement for flexible foundation

�� Settlement for rigid foundationSettlement for rigid foundation

2)1( 20 αµ−=

E

BqSe

αµ )1( 20 −=E

BqSe

−++++

−+++=

11

11ln

1

1ln

12

2

2

2

m

mm

mm

mm

πα

B

Lm =

ave E

BqS αµ )1( 20 −=

re E

BqS αµ )1( 20 −=

36

Value of Value of αα

Shape of Shape of foundationfoundation

Flexible foundationFlexible foundation Rigid Rigid foundationfoundationCenterCenter CornerCorner AverageAverage

CircularCircular 11 0.640.64 0.850.85 0.790.79

SquareSquare 1.121.12 0.560.56 0.950.95 0.820.82

RectangularRectangular

L/B=1.5L/B=1.5 1.361.36 0.680.68 1.151.15 1.061.06

L/B=5.0L/B=5.0 2.12.1 1.051.05 1.831.83 1.71.7

L/B=10L/B=10 2.542.54 1.271.27 2.252.25 2.12.1

Immediate settlement of Immediate settlement of foundation on saturated clayfoundation on saturated clay

�� Janbu et al.(1956)proposed an equation Janbu et al.(1956)proposed an equation for evaluating the average settlement of for evaluating the average settlement of flexible foundations on saturated clay soils flexible foundations on saturated clay soils (Poisson’s ratio (Poisson’s ratio µ=0.5)µ=0.5)

E

BqAASe

021.=

37

Variation of AVariation of A11 With H/B by Christian and With H/B by Christian and Carrier(1978)Carrier(1978)

H/BH/B AA 11

CircleCircle L/BL/B11 22 33 44 55

11 0.360.36 0.360.36 0.360.36 0.360.36 0.360.36 0.360.3622 0.470.47 0.530.53 0.630.63 0.640.64 0.640.64 0.640.6444 0.580.58 0.630.63 0.820.82 0.940.94 0.940.94 0.940.9466 0.610.61 0.670.67 0.880.88 1.081.08 1.141.14 1.161.1688 0.620.62 0.680.68 0.900.90 1.131.13 1.221.22 1.261.261010 0.630.63 0.700.70 0.920.92 1.181.18 1.301.30 1.421.422020 0.640.64 0.710.71 0.930.93 1.261.26 1.471.47 1.741.743030 0.660.66 0.730.73 0.950.95 1.291.29 1.541.54 1.841.84

Variation of AVariation of A22With D/B by Christian With D/B by Christian and Carrier(1978)and Carrier(1978)

D/BD/B AA 22

00 11

22 0.90.9

44 0.880.88

66 0.8750.875

88 0.870.87

1010 0.8650.865

1212 0.8630.863

1414 0.860.86

1616 0.8560.856

1818 0.8540.854

2020 0.850.85

38

Consolidation settlementConsolidation settlement�� For normally consolidated clay For normally consolidated clay σσ’’0 0 ≥ ≥ σσ’’pp

o

o

oe

HCcS

σσσ ∆+

+×= .log.

1

)4(6

1bmt σσσσ ++=∆

s

LWCc ∫

= .100

(%)2343.0

uPP CI 48.022' −=σ83.004.7' uP C=σ

689.0

00 ''.193.0'

=

σσσ N

P

By Mayne & Mitchell

By Mitchell(1988)

By Mayne & Kemper(1988)

Consolidation settlementConsolidation settlement

�� For over consolidated clay For over consolidated clay σσ’’00<<σσ’’pp11--

22--

σ'O + ∆∆∆∆σσσσ ≤≤≤≤ σ'P

o

o

oe

HCsS

σσσ ∆+

+= log

1

. s

LWCs ∫

= .100

(%)0463,0

σ'O + ∆∆∆∆σσσσ > σ'P

∆++

++

=P

o

oo

P

o e

HCc

e

HCsS

'

'log.

1

.'log.

1

.

σσσ

σσ

39

Tolerable Settlement of Tolerable Settlement of buildingbuilding

�� Settlement analysis is an important part of Settlement analysis is an important part of the design and construction of foundationthe design and construction of foundation

�� Large settlement of various component of Large settlement of various component of structure may lead to considerable structure may lead to considerable damage or may interfere with the proper damage or may interfere with the proper functioning of the structure. functioning of the structure.

Settlement of foundationSettlement of foundation

δi-total displacement at

point i

δij-different settlement

between point i and j

∆− relative deflection

ηij= angular

distortion

∆/L=deflection ratio

ωδ

−ij

ij

l

40

Limiting angular distortion as recommended Limiting angular distortion as recommended by Bjerrum(Compiled from Wahls,1981)by Bjerrum(Compiled from Wahls,1981)

damage Category of potential damage Category of potential ηηηηηηηηDanger to machinery sensitive to settlementDanger to machinery sensitive to settlement 1/7501/750

Danger to frames with diagonalsDanger to frames with diagonals 1/6001/600

Safe limit for no cracking of buildingSafe limit for no cracking of building 1/5001/500

First cracking of panel wallsFirst cracking of panel walls 1/3001/300

Difficulties with overhead cranesDifficulties with overhead cranes 1/3001/300

Tilting of high rigid building becomes visibleTilting of high rigid building becomes visible 1/2501/250

Considerable cracking of panel and brick wallsConsiderable cracking of panel and brick walls 1/1501/150

Danger of structure damage to general buildingDanger of structure damage to general building 1/1501/150

Safe limit for flexible brick walls L/H>4Safe limit for flexible brick walls L/H>4 1/1501/150

Safe limit include a factor of safetySafe limit include a factor of safety

Allowable settlement criteria:1955 U.S.S.R Allowable settlement criteria:1955 U.S.S.R Building code(compiled from walhls,1981)Building code(compiled from walhls,1981)

Type of structureType of structure Sand and hard claySand and hard clay Plastic clayPlastic clay

ηηηηηηηηCivil and industrial building column foundationCivil and industrial building column foundation

For steel and reinforced concrete structureFor steel and reinforced concrete structure 0.0020.002 0.0020.002

For end rows of columns with brick claddingFor end rows of columns with brick cladding 0.0070.007 0.0010.001

For structure where auxiliary strain does not arise duringFor structure where auxiliary strain does not arise during

Nonuniform settlement of foundationNonuniform settlement of foundation 0.0050.005 0.0050.005

Tilt of smokestacks,tower,silos,and so onTilt of smokestacks,tower,silos,and so on 0.0040.004 0.0040.004

Crane ways Crane ways 0.0030.003 0.0030.003

∆∆∆∆∆∆∆∆/L/L

Plain brick wallsPlain brick walls

For multistory dwelling and civil buildingFor multistory dwelling and civil building

At L/H<3At L/H<3 0.00030.0003 0.00040.0004

At L/H>5At L/H>5 0.00050.0005 0.00070.0007

For oneFor one--story millsstory mills 0.00100.0010 0.00100.0010

41

Allowable average settlement for different building Allowable average settlement for different building type(compiled from Wahls,1981)type(compiled from Wahls,1981)

Type of buildingType of building Allowable average Allowable average settlement(mm)settlement(mm)

Building with plain brick wallsBuilding with plain brick walls

L/H>2.5L/H>2.5 8080

L/H<1.5L/H<1.5 100100

Building with brick walls,reinforced with Building with brick walls,reinforced with reinforced concrete or reinforced brickreinforced concrete or reinforced brick

150150

Framed buildingFramed building 100100

Solid reinforced concrete foundation of Solid reinforced concrete foundation of smokestacks,silos,towers,and so onsmokestacks,silos,towers,and so on

300300

Deep foundationDeep foundation�� Need for pile foundationNeed for pile foundation

11--When the upper soils layers are highly compressible When the upper soils layers are highly compressible and too weak to support the load transmitted by the and too weak to support the load transmitted by the superstructure, piles are used to transmit the load to superstructure, piles are used to transmit the load to underlying bedrock or stronger soil layer.underlying bedrock or stronger soil layer.22--When subjected to horizontal force, pile foundations When subjected to horizontal force, pile foundations resist by bending while still supporting the vertical load resist by bending while still supporting the vertical load transmitted by superstructure.This situation is transmitted by superstructure.This situation is generally encountered in the design and construction generally encountered in the design and construction of earthof earth--retaining structures and foundations of tall retaining structures and foundations of tall structures that are subjected to strong wind and/or structures that are subjected to strong wind and/or earthquake forces. earthquake forces.

42

Deep foundationDeep foundation33--The expansive and collapsible soils may extend to a The expansive and collapsible soils may extend to a great depth below the ground surface.These soils great depth below the ground surface.These soils swell and shrink as the water content increase and swell and shrink as the water content increase and decrease.If shallow foundations are used, the decrease.If shallow foundations are used, the structure may suffer considerable damage.The pile structure may suffer considerable damage.The pile have to extend into stable soil layer beyond the zone have to extend into stable soil layer beyond the zone of possible moisture change.of possible moisture change.44--The foundation of some structures, such as The foundation of some structures, such as transmission towers,offshore platforms, and basement transmission towers,offshore platforms, and basement mats below the water table, are subjected to uplifting mats below the water table, are subjected to uplifting forces.Pile are sometime used for these foundations to forces.Pile are sometime used for these foundations to resist the uplifting force.resist the uplifting force.

Deep foundationDeep foundation55--Bridge abutments and piers are usually constructed Bridge abutments and piers are usually constructed over pile foundations to avoid the possible loss of over pile foundations to avoid the possible loss of bearing capacity that a shallow foundations might bearing capacity that a shallow foundations might suffer because of soil erosion at the ground surface.suffer because of soil erosion at the ground surface.Although numerous investigations, both theoretical Although numerous investigations, both theoretical and experimental, have been conducted to predict the and experimental, have been conducted to predict the behavior and the loadbehavior and the load--bearing capacity of piles in bearing capacity of piles in granular and cohesive soils,the mechanisms are not granular and cohesive soils,the mechanisms are not yet entirely understood and never be clear.The design yet entirely understood and never be clear.The design of pile foundations may be considered somewhat of of pile foundations may be considered somewhat of an”art”as a result of the uncertainties involved in an”art”as a result of the uncertainties involved in working with some subsoil condition.working with some subsoil condition.

43

Types of pilesTypes of piles

�� Different types of piles are used in construction Different types of piles are used in construction work,depending on the type of load to be work,depending on the type of load to be carried, the subsoil conditions,and the water carried, the subsoil conditions,and the water table.Pile can be divided into these categories:table.Pile can be divided into these categories:--Steel pilesSteel piles--Concrete pilesConcrete piles--Wooden(timber)pilesWooden(timber)piles--Composite pilesComposite piles

44

Comparisons of piles made of different materialsComparisons of piles made of different materialsPile typePile type Usual Usual

length of length of pile(m)pile(m)

Maximum Maximum length of length of pile(m)pile(m)

Usual load Usual load (KN)(KN)

Approximate Approximate maximum maximum load(KN) load(KN)

SteelSteel 1515--6060 Practically Practically unlimitedunlimited

300300--12001200 --

Advantages: Advantages: aa--Easy to handle with respect to cutoff and extension to the Easy to handle with respect to cutoff and extension to the desired lengthdesired length

bb--Can stand high driving stressesCan stand high driving stresses

cc--Can penetrate hard layer such as dense gravel,soft rockCan penetrate hard layer such as dense gravel,soft rock

dd--High loadHigh load--carrying capacitycarrying capacity

disadvantages: disadvantages: aa--Relatively costly materialRelatively costly material

bb--High level of noise during pile drivingHigh level of noise during pile driving

cc--Subject to corrosionSubject to corrosion

dd--HH--piles may be damaged or deflected from the vertical piles may be damaged or deflected from the vertical during driving through hard layers or past major obstructions during driving through hard layers or past major obstructions

Comparisons of piles made of different materialsComparisons of piles made of different materials

Pile typePile type Usual length Usual length of pile(m)of pile(m)


Usual Usual load load (KN)(KN)


Precast Precast concreteconcrete

precast::10precast::10--1515Prestressed:Prestressed:

1010--3535

precast::30precast::30Prestressed:Prestressed:

6060

300300--30003000

precast::800precast::800--900900

Prestressed:Prestressed:75007500--85008500

Advantages: Advantages: aa--Can be subjected to hard drivingCan be subjected to hard driving

bb--Corrosion resistantCorrosion resistant

cc--Can be easy combined with concrete superstructureCan be easy combined with concrete superstructure

disadvantages: disadvantages: aa--Difficult to achieve proper cutoffDifficult to achieve proper cutoff

bb--Difficult to transportDifficult to transport

45


Pile typePile type Usual Usual length of length of pile(m)pile(m)




Cased castCased cast--in place in place concreteconcrete

55--1515 1515--4040 200200--500500 800800

Advantages: Advantages: aa--Relatively cheapRelatively cheap

bb--Possibility of inspection before pouring concretePossibility of inspection before pouring concrete

cc--Easy to extendEasy to extend

disadvantages: disadvantages: aa--Difficult to splice after concretingDifficult to splice after concreting

bb--Think casings may be damages during drivingThink casings may be damages during driving






uncased uncased castcast--in place in place

concreteconcrete

55--1515 3030--4040 300300--500500 700700

Advantages: Advantages: aa--Initially economicalInitially economical

bb--Can be finished at any elevationCan be finished at any elevation

disadvantages: disadvantages: aa--Voids may be created if concrete is placed rapidly Voids may be created if concrete is placed rapidly

bb--In soft soils,the sides of the hole may cave in thus In soft soils,the sides of the hole may cave in thus Squeezing the concreteSqueezing the concrete

46






WoodWood 1010--1515 3030 100100--200200 270270

Advantages: Advantages: aa--EconomicalEconomical

bb--Permanently submerged piles are fairly resistant to decayPermanently submerged piles are fairly resistant to decay

cc--Easy to handleEasy to handle

disadvantages: disadvantages: aa-- Decay above water tableDecay above water table

bb--Can be damaged in hard driving Can be damaged in hard driving

cc--Low loadLow load--bearing capacitybearing capacity

dd--Low resistance to tensile load when splicesLow resistance to tensile load when splices

Typical concrete pileTypical concrete pile

Pile Shape*

D (mm)

Area of cross section (cm²)

Perimeter (mm)

Number of strands 12.7-mm 11.1-mm diameter diameter

Minimum effective prestress force (kN)

Section modulus

(m³ x 10-3)

Design bearing capacity (kN)

Concrete strength (MN/m²)

34.5 41.4

S O S O S O S O S O S O S O S O

254 254 305 305 356 356 406 406 457 457 508 508 559 559 610 610

645 536 929 768 1265 1045 1652 1368 2090 1729 2581 2136 3123 2587 3658 3078

1016 838 1219 1016 1422 1168 1626 1346 1829 1524 2032 1677 2235 1854 2438 2032

4 4 5 4 6 5 8 7 10 8 12 10 15 12 18 15

4 4 6 5 8 7 11 9 13 11 16 14 20 16 23 19

312 258 449 369 610 503 796 658 1010 836 1245 1032 1508 1280 1793 1486

2.737 1.786 4.719 3.097 7.489 4.916 11.192 7.341 15.928 10.455 21.844 14.355 29.087 19.107 37.756 34.794

556 462 801 662 1091 901 1425 1180 1803 1491 2226 1842 2694 2231 3155 2655

778 555 962 795 1310 1082 1710 1416 2163 1790 2672 2239 3232 2678 3786 3186

47

Practical list of typical air and steam hammersPractical list of typical air and steam hammers

Maker of hammer*

Model no.

Type of hammer

Rated energy (kN-m)

Blows per minute

Ram weight (kN)

V V V

MKT V V R

MKT R V R

MKT V V

MKT MKT MKT

V

3100 540 060

OS-60 040

400C 8/0 S-20 5/0

200-C 150-C S-14 140C 08 S-8

11B3 C-5 30-C

Single acting Single acting Single acting Single acting Single acting Differential Single acting Single acting Single acting Differential Differential Single acting Differential Single acting Single acting Double acting Double acting Double acting

407 271 244 244 163 154 110 82 77 68 66 51 49 35 35 26 22 10

58 48 62 55 60 100 35 60 44 98

95-105 60 103 50 55 95 110 133

449 182 267 267 178 178 111 89 78 89 67 62 62 36 36 22 22 13

Practical list of typical diesel hammersPractical list of typical diesel hammers

Maker of hammer*

Model no.

Rated energy (kN-m)

Blows per minute Piston weight (kN)

K M K K M K MKT K V L M V L MKT MKT L

K150 MB70 K-60 K-45 M-43 K-35 DE70B K-25 N-46 520 M-14S N-33 440 DE20 DE-10 180

379.7 191.2-86

143.2 123.5

113.9-51.3 96

85.4-57 68.8 44.1 35.7

35.3-16.1 33.4 24.7

24.4-16.3 11.9 11.0

45-60 38-60 42-60 39-60 40-60 39-60 40-50 39-60 50-60 80-84 42-60 50-60 86-90 40-50 40-50 90-95

147 71 59 44 42 34 31 25 18 23 13 13 18 9 5 8

48

Pile driven formulasPile driven formulas�� To develop the desired loadTo develop the desired load--carrying capacity,a point bearing carrying capacity,a point bearing

pile must penetrate the dense soil layer sufficiently or have pile must penetrate the dense soil layer sufficiently or have sufficient contact with a layer of rock.This requirement cannot sufficient contact with a layer of rock.This requirement cannot always be satisfied by driving a pile to a predetermined depth always be satisfied by driving a pile to a predetermined depth because soil profile vary.For that reason, several equations because soil profile vary.For that reason, several equations have been developed to calculate the ultimate capacity of a pile have been developed to calculate the ultimate capacity of a pile during driving.These dynamic equations are widely used in the during driving.These dynamic equations are widely used in the field to determine whether the pile has reached a satisfactory field to determine whether the pile has reached a satisfactory bearing value at the predetermined depth.One of the earliest of bearing value at the predetermined depth.One of the earliest of these dynamic equationsthese dynamic equations--commonly referred to as the commonly referred to as the Engineering News Record (ENR) formulaEngineering News Record (ENR) formula--is derived from the is derived from the workwork--energy theory;that is : Energy imparted by the hammer energy theory;that is : Energy imparted by the hammer per blow =(pile resistance)(penetration per hammer blow)per blow =(pile resistance)(penetration per hammer blow)

ENR equationsENR equations

�� Where WWhere WRR--Weight of the ramWeight of the ramhh--height of fall of ram(Cm)height of fall of ram(Cm)SS--penetration of the pile per penetration of the pile per

hammer blow(Cm)hammer blow(Cm)CC--a constanta constant

C=2.54 Cm for drop hammerC=2.54 Cm for drop hammerC=0.254Cm for steam hammerC=0.254Cm for steam hammer

Factor of safety FS=6Factor of safety FS=6

CS

hWQ R

u +=

49

ENR equations for single and double acting ENR equations for single and double acting hammerhammer

�� Where EWhere E--hammer efficiencyhammer efficiencyHHEE--rated energy of the hammerrated energy of the hammerSS--penetration of the pile per hammer penetration of the pile per hammer

blow(Cm)blow(Cm)CC--a constanta constantC=0.254 Cm C=0.254 Cm

Factor of safety FS=4 to 6Factor of safety FS=4 to 6

CS

HEQ E

u += .

Modified ENR equations Modified ENR equations

�� Where EWhere E--hammer efficiencyhammer efficiencyhh--height of fall of the ram(Cm)height of fall of the ram(Cm)SS--penetration of the pile per hammer penetration of the pile per hammer blow(Cm)blow(Cm)WWPP--weight of the pileweight of the pilenn--coefficient of restitution between coefficient of restitution between

the ram and the pile capthe ram and the pile capC=0.254 Cm C=0.254 Cm

Factor of safety FS=4 to 6Factor of safety FS=4 to 6

PR

PRRu WW

WnW

CS

hEWQ

++

+=

2

50

Michigan state highway commission equations Michigan state highway commission equations

�� After testing on 88 pile(1965)After testing on 88 pile(1965)�� Where WWhere WRR--weight of the ramweight of the ram

WWPP--weight of the pileweight of the pileHHEE--rated energy of the hammerrated energy of the hammerSS--penetration of the pile per hammer penetration of the pile per hammer

blow(M)blow(M)CC--a constanta constantC=2.54.10C=2.54.10––33MM

Factor of safety FS= 6Factor of safety FS= 6

PR

PREu WW

WnW

CS

HQ

++

+=

225,1

Danish equations Danish equations

�� Where EWhere E--hammer efficiencyhammer efficiencyEEPP--modulus of elasticity of the pilemodulus of elasticity of the pileHHEE--rated energy of the hammerrated energy of the hammerSS--penetration of the pile per hammer penetration of the pile per hammer

blow(M)blow(M)LL--length of the pilelength of the pile

AAPP--area of the pile cross sectionarea of the pile cross sectionFactor of safety FS= 6Factor of safety FS= 6

PP

E

Eu

EA

LEHS

EHQ

2+

=

51

Pacific Coast Uniform Building Code equations Pacific Coast Uniform Building Code equations

After International Conference of building After International Conference of building officials,1982officials,1982

�� Where EWhere E--hammer efficiencyhammer efficiencyHHEE--rated energy of the hammerrated energy of the hammerSS--penetration of the pile per hammer penetration of the pile per hammer blow(M)blow(M)LL--length of the pilelength of the pile

EEPP--modulus of elasticity of pilemodulus of elasticity of pilen=0.25 for steel piles and n=0.1 for another n=0.25 for steel piles and n=0.1 for another

pilespilesFactor of safety FS= 4 to 5Factor of safety FS= 4 to 5

PP

u

PR

PRE

u

EA

LQS

WW

nWWEH

Q+

++

=)(

Value of E & nValue of E & nHammer typeHammer type Efficiency,EEfficiency,E

Single and double acting hammersSingle and double acting hammers 0.70.7--0.850.85

Diesel hammersDiesel hammers 0.80.8--0.90.9

Drop hammersDrop hammers 0.70.7--0.90.9

Pile materialPile material Coefficient of restitutionCoefficient of restitutionnn

Cast iron hammer and concrete pile Cast iron hammer and concrete pile without capwithout cap

0.40.4--0.50.5

Wood cushion on steel pileWood cushion on steel pile 0.30.3--0.40.4

Wooden pileWooden pile 0.250.25--0.30.3

52

Equation for estimation of pile Equation for estimation of pile capacitycapacity

�� QQUU=Q=QPP+Q+Qss

Where QWhere QUU is ultimate load carrying capacity is ultimate load carrying capacity of pileof pile

QQPP is load carrying capacity of the pile is load carrying capacity of the pile pointpoint

QQSS is frictional resistance is frictional resistance

Pile foundationPile foundation

L Weak soil

Rock

Qp

Qu= Qp

L

Lb

Weak soil

Qp

Qs

Strong soil layer

Qu= Qp+Qs

Strong soil layer

L Weak soil

Qp

Qu= Qs

Qs

53

Minimum pile embedment depth Minimum pile embedment depth into founding soil stratainto founding soil strata

�� From civil engineering association forum the From civil engineering association forum the minimum pile embedment depth into bearing minimum pile embedment depth into bearing stratum is 3 times diameter of pile.stratum is 3 times diameter of pile.

�� Replace the pile with one having a different helix Replace the pile with one having a different helix configuration. The replacement pile must not configuration. The replacement pile must not exceed any applicable maximum embedment exceed any applicable maximum embedment length and either (A) meet the minimum effective length and either (A) meet the minimum effective torsion resistance criterion and all applicable torsion resistance criterion and all applicable embedment criteria shown in Table for the embedment criteria shown in Table for the design load type (s), or (B) pass proof testing.design load type (s), or (B) pass proof testing.

Replacement pile embedment Replacement pile embedment criteria criteria

Design Load typeDesign Load type Replacement Pile Embedment CriterionReplacement Pile Embedment Criterion

Tension Tension The last helix must be embedded atThe last helix must be embedded atleast three times its own diameterleast three times its own diameterbeyond the position of the first helixbeyond the position of the first helixof the replaced pile.of the replaced pile.

Compression Compression The last helix must be embeddedThe last helix must be embeddedbeyond the position of the first helixbeyond the position of the first helixof the replaced pile.of the replaced pile.

Shear/Overturning Shear/Overturning Embedment must satisfy the specifiedEmbedment must satisfy the specifiedminimum.minimum.

54

LoadLoad--carrying Capacity of the pile point,Qcarrying Capacity of the pile point,QPPfrom Terzaghi’s equationfrom Terzaghi’s equation

�� QQPP=A=APP.q.qPP=A=APP(CN (CN **cc+q’N +q’N **

qq))Where AWhere APP--area of pile tiparea of pile tip

CC--cohesion of the soil supporting the pile tipcohesion of the soil supporting the pile tipqqPP--unit point resistanceunit point resistanceq’q’--effective vertical stress at the level of the pile effective vertical stress at the level of the pile

tiptipNN**

CC,N,N**qq--bearing capacity factor after Caquot & bearing capacity factor after Caquot &

KeriselKerisel

φtgeNq

7* = φcot)1( ** −= qC NN

LoadLoad--carrying Capacity of the pile point,Qcarrying Capacity of the pile point,QPPfrom Eric Gervreau in Euro code 2000from Eric Gervreau in Euro code 2000

�� QQPP=A=APP.q.qPP=A=APP(1.3CN (1.3CN **cc+50N +50N **qq))

Where AWhere APP--area of pile tiparea of pile tipCC--cohesion of the soil supporting the pile cohesion of the soil supporting the pile

tiptipqqPP--unit point resistanceunit point resistance

NN**CC,N,N**

qq--bearing capacity factor after bearing capacity factor after Caquot & KeriselCaquot & Kerisel

φtgeNq

7* = φcot)1( ** −= qC NN

55

Critical depthCritical depth

�� In the case of calculation of In the case of calculation of q’q’ , the normal , the normal practice is to assume that practice is to assume that q’q’ increases increases linearly with depth from zero at ground linearly with depth from zero at ground level to a maximum value level to a maximum value q’q’ (max)(max) at the tip at the tip of pile.of pile.

�� However, extensive research carried out However, extensive research carried out by Vessic(1967) has indicated that by Vessic(1967) has indicated that q’q’varies linearly from the ground surface up varies linearly from the ground surface up to a limited depth only beyond which to a limited depth only beyond which q’q’ ,,remains constant irrespective of the depth remains constant irrespective of the depth of embedment of pile.of embedment of pile.

Critical depthCritical depth�� This phenomenon was attributed to arching of This phenomenon was attributed to arching of

SANDSAND..�� This depth within which This depth within which q’q’ varies linearly with varies linearly with

depth may be called as the depth may be called as the critical depthcritical depth DDcc..�� From the curves given by From the curves given by PoulosPoulos (1980), we (1980), we

may writemay write

�� For 28<For 28<φφ<36.5 we have D<36.5 we have Dcc/B=5+0.24(/B=5+0.24(φφ--28)28)

�� For 36.5<For 36.5<φφ<42 we have D<42 we have Dcc/B=7+2.35(/B=7+2.35(φφ--36.5)36.5)

56

Critical depthCritical depth�� From Caquot & Kerisel DFrom Caquot & Kerisel Dcc=B/4.N*=B/4.N*qq

(2/3)(2/3)

�� In Bearing Capacity Technical Guidance by Career In Bearing Capacity Technical Guidance by Career Development and Resources for Geotechnical Development and Resources for Geotechnical Engineers Engineers --Dc = 10B, for Dc = 10B, for looseloose silts and sandssilts and sands--Dc = 15B, for Dc = 15B, for mediummedium dense silts and sandsdense silts and sands--Dc = 20B, for Dc = 20B, for densedense silts and sands silts and sands

--loose when loose when N<10 or N<10 or φφφφφφφφ<30<30

--medium dense when medium dense when 10<N<30 or 30<10<N<30 or 30<φφφφφφφφ<36<36

--dense when dense when 30<N or 36<30<N or 36<φφφφφφφφ

Critical depthCritical depth

�� This critical concept implies that This critical concept implies that ff ss for cohesionless for cohesionless soil for a driven pile varies linearly with depth up to soil for a driven pile varies linearly with depth up to depth depth DDcc only and beyond this depth only and beyond this depth ff ss remains remains constant.constant.

�� Note that the application concept Note that the application concept DDcc in case the soil is in case the soil is homogeneous for the whole depth of embedment homogeneous for the whole depth of embedment DD..

�� Since no information is available on the layered Since no information is available on the layered system of soil, this approach has to be used with system of soil, this approach has to be used with caution. Tomlinson(1986) Bowles(1988) has not use caution. Tomlinson(1986) Bowles(1988) has not use of this concept .of this concept .This indicates that this method has This indicates that this method has not yet found favor with the designer.not yet found favor with the designer.

57

LoadLoad--carrying Capacity of the pile point in sand carrying Capacity of the pile point in sand from ESA condition after Meyerhof (1976)from ESA condition after Meyerhof (1976)

QQPP=A=APP.q.qPP=A=APPq’N q’N **qq

Where AWhere APP--area of pile tiparea of pile tipqqPP--unit point resistanceunit point resistanceq’q’--effective vertical stress at the level of effective vertical stress at the level of

the pile tipthe pile tipNN**

qq--bearing capacity factorbearing capacity factorQQPP=A=Appq’Nq’N **

qq<A<Appqq ii

qq ii=50N=50N**qqtgtg φφφφφφφφ(KN/M(KN/M22))

As per Tomlinson, the maximum base resistance As per Tomlinson, the maximum base resistance qqpp is normally limited 11000KPa.is normally limited 11000KPa.

φtgeNq

7* =

LoadLoad--carrying Capacity of the pile point in carrying Capacity of the pile point in sand from ESA condition after Meyerhof sand from ESA condition after Meyerhof

(1976)(1976)

�� The angle The angle φφ to be use for determination to be use for determination NNqq

** areare

�� For driven pile For driven pile φ φ = = φφ11

�� For bored pile For bored pile φ φ = = φφ11--3 3

Where Where φφ11 is angle of internal friction prior to is angle of internal friction prior to installation of pile.installation of pile.

58

LoadLoad--carrying Capacity of the pile point in carrying Capacity of the pile point in saturated clay from TSA conditionsaturated clay from TSA condition

�� QQPP=A=APP.q.qPP=A=AppCCUU N N **cc= 9C= 9CUUAAPP

Where AWhere APP--area of pile tiparea of pile tipqqPP--unit point resistanceunit point resistance

NN**cc--bearing capacity factor for bearing capacity factor for φφ=0 N=0 N**

CC=9=9

Carrying capacity of piles in layered soilCarrying capacity of piles in layered soilfrom meyerhof equationfrom meyerhof equation

59

Carrying capacity of piles in layered soilCarrying capacity of piles in layered soil

�� If the pile toe terminates in a layer of dense sand or If the pile toe terminates in a layer of dense sand or stiff clay overlying a layer of soft clay or loose sand stiff clay overlying a layer of soft clay or loose sand there is a danger of it punching through to the weaker there is a danger of it punching through to the weaker layer.layer.

�� To account for this, Meyerhof's equation is used. To account for this, Meyerhof's equation is used. �� The base resistance at the pile toe is The base resistance at the pile toe is

qqp p = q= q22 + (q+ (q11 --qq22)H / 10B but < q)H / 10B but < q 11

�� where where --B is the diameter of the pile B is the diameter of the pile --H is the thickness between the base of the pile and H is the thickness between the base of the pile and the top of the weaker layerthe top of the weaker layer--qq22 is the ultimate base resistance in the weak layeris the ultimate base resistance in the weak layer--qq11 is the ultimate base resistance in the strong layer. is the ultimate base resistance in the strong layer.

Relation between ultimate point resistance of pile Relation between ultimate point resistance of pile and depth in sand stratum beneath weak soil layer and depth in sand stratum beneath weak soil layer

from Terzaghi 1982from Terzaghi 1982

60

Relation between ultimate point resistance of pile Relation between ultimate point resistance of pile and depth in sand stratum beneath weak soil layer and depth in sand stratum beneath weak soil layer

from Terzaghi 1982from Terzaghi 1982

Frictional Resistance QFrictional Resistance QSS

Where PWhere P--perimeter of the pile sectionperimeter of the pile section∆∆LL--incremental pile length over incremental pile length over

which P and f are taken constantwhich P and f are taken constantff--unit friction resistance at any unit friction resistance at any

depth Zdepth Z

fLPQs ..∆∑=

61

Skin friction from Skin friction from ββ MethodMethod

f=βσ’0

σ’0-effective vertical stress at center of layer

As Tomlinson, the maximum frictional resistance is

limited 110KPa

From Meyehof 1976

φ<28 we have β=0.44

28<φ<35 we have β=0.75

35<φ<37 we have β=1.20


62


�� The angle The angle φφ to be use for determination to be use for determination ββ areare

�� For driven pile For driven pile φ φ = 0.75= 0.75φφ11+10+10

�� For bored pile For bored pile φ φ = = φφ11--3 3

Where Where φφ11 is angle of internal friction prior to is angle of internal friction prior to installation of pile.installation of pile.

Skin friction from Skin friction from αα MethodMethod

Skin friction for clayey soil for driven pilef=αxCu α=1 for Cu=<25KPa

α=0.5 for Cu=>70KPaα=1-(Cu-25)/90 for 25KPa<Cu<70KPa API(1984)

α=1 for Cu<=35KPaα=0.5 for Cu=>80KPaα=1-(Cu-35)/90 for 35KPa<Cu<80KPa Semple and Rigden(1984)

Skin friction for clayey soil for Bored pile or drilled shaftsf=αxCu α=0.45 for London clay Skempton(1959)

α=0.7 time value for driven diplacement pile Flaming et al(1985)α=0 for Z<1.5 Reese and Oneill(1985)

63

Tomlinson Tomlinson αα methodmethod

�� Case 1:pile driven through sands or sandy Case 1:pile driven through sands or sandy gravels into stiff clay strata.gravels into stiff clay strata.

�� Case 2:pile driven through soft clay into Case 2:pile driven through soft clay into stiff clay strata.stiff clay strata.

�� Case 3:pile driven into a firm to stiff clay Case 3:pile driven into a firm to stiff clay without any overlying strata.without any overlying strata.

�� The value of The value of αααααααα vary with Cvary with Cuu and L/B ratioand L/B ratio

Tomlinson Tomlinson αα methodmethod

64

Negative skin frictionNegative skin friction�� Negative skin friction is a downward drag force exerted Negative skin friction is a downward drag force exerted

on the pile by the soil surrounding it.This action can on the pile by the soil surrounding it.This action can occur under conditions such as the following:occur under conditions such as the following:11--if a fill of clay soil is placed over a granular soil layer if a fill of clay soil is placed over a granular soil layer into witch a pile is driven, the fill will gradually into witch a pile is driven, the fill will gradually consolidate. This consolidation process will exert a consolidate. This consolidation process will exert a downward drag force on the pile during the period of downward drag force on the pile during the period of consolidation.consolidation.22--if a fill of granular soil is placed over a layer of soft if a fill of granular soil is placed over a layer of soft clay,it will induce the process of consolidation in the clay clay,it will induce the process of consolidation in the clay layer and thus exert a downward drag on the pilelayer and thus exert a downward drag on the pile33--lowering of the water table will increase the vertical lowering of the water table will increase the vertical effective stress on the soil at any depth,which will induce effective stress on the soil at any depth,which will induce consolidation settlement in clay.If a pile is located in the consolidation settlement in clay.If a pile is located in the clay layer,it will be subjected to a downward drag force. clay layer,it will be subjected to a downward drag force.

Clay fill over granular soil

Hf

Sand

Clay

fillL

Granular soil fill over clay

Hf

L

Sand

fill

Clay

Neutral

plane

L1

65

Clay fill over granular soilClay fill over granular soil

�� Where:Where:K’=earth pressure coefficient =Ko=1K’=earth pressure coefficient =Ko=1--sinsinφφσσ’’oo=vertical effective stress at any depth Z =vertical effective stress at any depth Z

= = γγ’’ff.Z..Z.γγ’’f f =effective unit weight of fill Clay=effective unit weight of fill Clayδδ=soil=soil--pile friction angle = 0.5pile friction angle = 0.5φφ to 0.7to 0.7φφ

δσ tan'' 0Kfn =

2

tan'')tan''(

2

0

δγδγ

HPKZdPKQ f

z

H

fn == ∫

Granular soil fill over clayGranular soil fill over clay�� In this case, the evidence indicates that the In this case, the evidence indicates that the

negative skin stress on the pile may exist from negative skin stress on the pile may exist from Z=0 to Z=LZ=0 to Z=L11,which is referred to as the neutral ,which is referred to as the neutral depth.The neutral depth may be given as depth.The neutral depth may be given as (Bowles 1982)(Bowles 1982)

�� Hence,the total drag force isHence,the total drag force is

'

'2

'

'

211 γ

γγ

γ ffffff HHHL

L

HLL −

+

−−=

∫ +=+=1

0

21

1 2

tan''tan''tan)''('

L

ffZffn

PKLHLPKdZHPKQ

δγδγδγγ

66

Determine End bearing capacity of Determine End bearing capacity of pile foundation from SPT testpile foundation from SPT test

Driven MethodC

Sand qp=CN(Mpa) 0.45 N=average SPT value in By Martin et al(1987)

qp=CN(Mpa) 0.4 local failure zone By Decourt(1982)

qp=CN(Mpa) 0.04 Ls/D Ls=Length of pile in sand Mayerhof(1976)D=width of pile C<=0.4

Silt, sandy silts qp=CN(Mpa) 0.35 N=average SPT value in Matin et al.(1987)

Glacial Coarse to fine siltqp=CN(Mpa) 0.25 local failure zone Thorburn and Mac Vicar(1987)

Residual sandy silt qp=CN(Mpa) 0.25 Decourt(1982)

Residual Clayey silt qp=CN(Mpa) 0.2 Decourt(1982)

Clay qp=CN(Mpa) 0.2 Matin et al.(1987)

Clay qp=CN(Mpa) 0.12 Decourt(1982)

All soil qp=CN(Mpa) 0.3 ForL/D>=5 Shioi and Fukui(1982)If L/D<5,C=0.1+0.04L/D

for closed end pileand C=0.06L/D

for open end pile

Determine End bearing capacity of Determine End bearing capacity of pile foundation from SPT testpile foundation from SPT test

Cast in place method

Coarse grained soil qp=CN(Mpa) 0.15 qp<3.0MPa Shioi and Fukui(1982)

qp=CN(Mpa) 0.15 qp<7.5MPa Yamashita et al(1987)

Fine grained soil qp=CN(Mpa) 0.15 qp=0.09(1+0.16Lt) Yamashita et al(1987)Lt=pile length

Bored pileSand qp=CN(Mpa) 0.1 Shioi and Fukui(1982)

Clay qp=CN(Mpa) 0.15 Shioi and Fukui(1982)

67

Determine skin friction from SPT Determine skin friction from SPT testtest

Driven Methode A BCoarse grained soil qf=A+BN(Kpa) 0 2 N=average SPT Mayerhof(1956)

along Shaft Shioi and Fukui(1982)Coarse grained &fine soilqf=A+BN(Kpa) 10 3.3 3<N<50 Decourt(1982)

Fine grained soil qf=A+BN(Kpa) 0 10 Shioi and Fukui(1982)Cast in place methodeCoarse grained soil qf=A+BN(Kpa) 30 2 qf<200Kpa Yamashita et al(1987)

qf=A+BN(Kpa) 0 5 Shioi and Fukui(1982)

Fine grained soil qf=A+BN(Kpa) 0 5 qf<150Kpa Yamashita et al(1987)

qf=A+BN(Kpa) 0 10 Shioi and Fukui(1982)Bored pileCoarse grained soil qf=A+BN(Kpa) 0 1 Findlay(1984)&Shioi & Fukui(1982)

qf=A+BN(Kpa) 0 3.3 Wright &Reese(1979)

Fine graned soil qf=A+BN(Kpa) 10 3.3 qf<170Kpa Decourt(1982)

LoadLoad--Carrying capacity of pile point resting on Carrying capacity of pile point resting on rockrock

�� The ultimate unit point resistance in The ultimate unit point resistance in rock(Goodman,1980) is approximatelyrock(Goodman,1980) is approximately

qqpp=q=quu--RR(N(Nφφφφφφφφ+1)+1)Where NWhere Nφφ=tg=tg22(45+(45+φφ/2)/2)

qquu--RR=unconfined compression strength of rock=unconfined compression strength of rockφφ=drained angle of friction of rock=drained angle of friction of rock

The allowable loadThe allowable load--carrying capacity of the pilecarrying capacity of the pilepoint.thuspoint.thus

[ ]FS

ANqQ PRu

allp

)1()(

+= − φ FS=3

68

Typical unconfined compressive strength of rockTypical unconfined compressive strength of rock

Rock typeRock type qq uu--RR(Mpa)(Mpa)

Sandstone Sandstone 7070--140140

LimestoneLimestone 105105--210210

ShaleShale 3535--7070

GraniteGranite 140140--210210

MarbleMarble 6060--7070

5)(

)(labRu

designRu

qq −

− =

Drilled Shafts Extending into Drilled Shafts Extending into RockRock

�� Based on the procedure developed by Reese and Based on the procedure developed by Reese and O’Neill(1988O’Neill(1988--1989),we can estimate the bearing load 1989),we can estimate the bearing load capacity of drilled shafts extending into Rock as capacity of drilled shafts extending into Rock as follows:follows:

11--Calculate the ultimate unit side resistance as:Calculate the ultimate unit side resistance as:f=6.564qf=6.564quu

0.50.5≤0.15q≤0.15quu

Where qWhere quu=unconfined compression strength or Rock =unconfined compression strength or Rock corecore

22--Calculate the ultimate capacity based on side Calculate the ultimate capacity based on side resistance only:resistance only:

Qu=Qu=ππDDssLfLf

69

�� Calculate the settlement Se of the shaft at the top of the Rock Calculate the settlement Se of the shaft at the top of the Rock socked:socked:

Se=Se(s)+Se(b)Se=Se(s)+Se(b)Where Se(s)=elastic compression of the drilled shaft within the Where Se(s)=elastic compression of the drilled shaft within the

socket, assuming on side resistancesocket, assuming on side resistanceSe(b)=settlement of the baseSe(b)=settlement of the base

However Se(s)=However Se(s)=

And Se(b)=And Se(b)=

CC EA

LUQ

massS

f

ED

IUQ

70

Where QWhere Quu=Ultimate friction load=Ultimate friction loadAAcc=Cross=Cross--section area of the drilled shaft section area of the drilled shaft

in the sockedin the sockedDDss=Diameter of the drilled shaft=Diameter of the drilled shaftEEcc=Young’s modulus of the concrete=Young’s modulus of the concreteEEmassmass=Young’s modulus of the rock mass=Young’s modulus of the rock mass

IIff=Elastic influence coefficient (read on =Elastic influence coefficient (read on chart)chart)

L=Depth of embedment in rock L=Depth of embedment in rock If Se is less than 10mm, then the ultimate loadIf Se is less than 10mm, then the ultimate load--

carrying capacity from this way is correct.carrying capacity from this way is correct.

If Se≥ 10mm, there way be rapid, progressive side If Se≥ 10mm, there way be rapid, progressive side shear failure in the rock socket ,resulting in a shear failure in the rock socket ,resulting in a complete loss of side resistance. In that case the complete loss of side resistance. In that case the ultimate capacity is equal to the point resistance :ultimate capacity is equal to the point resistance :

+

+= 5.0

300110

3

3

S

S

S

cU

C

D

C

AqQuδ

71

Where CWhere Css=Spacing of discontinuities=Spacing of discontinuitiesδδ=Thickness of individual discontinuity=Thickness of individual discontinuityqquu=unconfined compression strength of =unconfined compression strength of

the rock beneath the base of the socket or the rock beneath the base of the socket or drilled shaft concrete, whichever is smaller.drilled shaft concrete, whichever is smaller.

Note that applies for horizontally stratified Note that applies for horizontally stratified discontinuities with Cdiscontinuities with Css>305 mm and >305 mm and δδ<5mm<5mm

Typical values of angle of friction of rocksTypical values of angle of friction of rocks

Rock typeRock type Angle of friction Angle of friction φφφφφφφφ(deg)(deg)

Sandstone Sandstone 2727--4545

LimestoneLimestone 3030--4040

ShaleShale 1010--2020

GraniteGranite 4040--5050

MarbleMarble 2525--3030

72

Group pile Group pile Pile cap

L

d d

Bg

Lg

d

d

d d

Group pile efficiencyGroup pile efficiency�� Determination of the load bearing capacity of group Determination of the load bearing capacity of group

piles is extremely complicated and has not yet been piles is extremely complicated and has not yet been fully resolved.When the piles are placed close to each fully resolved.When the piles are placed close to each other,a reasonable assumption is that the stress other,a reasonable assumption is that the stress transmitted by the piles to the soil will overlap,thus transmitted by the piles to the soil will overlap,thus reducing the load bearing capacity of the reducing the load bearing capacity of the pile.Ideally,the piles in a group should be spaced so pile.Ideally,the piles in a group should be spaced so that the load bearing capacity of the group should be that the load bearing capacity of the group should be no less than the sum of the bearing capacity of the no less than the sum of the bearing capacity of the individual piles.In practice,the minimum center to individual piles.In practice,the minimum center to center pile spacing center pile spacing d d is is 2.5D2.5D and in ordinary situations and in ordinary situations is actually about is actually about 3D3D to to 3.5D3.5D..

73

Efficiency factorEfficiency factor�� Many structural engineers used a simplified Many structural engineers used a simplified

analysis to obtained the group efficiency for analysis to obtained the group efficiency for friction piles (ratio between friction piles (ratio between QQss & Q& Quu is over is over 80%80%),particularly in sand.The piles may act in one ),particularly in sand.The piles may act in one of two way:of two way:11--as a block with dimension as a block with dimension LLgg*B*Bgg*L*L22--as individual pilesas individual piles

If the piles act as the block, the frictional capacity isIf the piles act as the block, the frictional capacity isQQg(u)g(u)=f=favavPPggL note PL note Pgg=2(n=2(n11+ n+ n22--2)d+4D2)d+4DFor each pile acting individuallyFor each pile acting individuallyQQ(u)(u)=f=favavLPLP

Efficiency factorEfficiency factor

�� Where Where ηη=group efficiency =group efficiency QQg(u)g(u)=ultimate load bearing capacity of =ultimate load bearing capacity of

group pilegroup pileQQ(u)(u)=ultimate load bearing capacity of =ultimate load bearing capacity of

each pileeach pile

)(

)(

u

ug

Q

Q

∑=η

21

21 4)2(2

nPn

Ddnn +−+=η

74

Converse Labarre equationConverse Labarre equation

θη

−+−−=21

1221

90

)1()1(1

nn

nnnn

)/((deg) dDarctg=θ

Pile in sandPile in sand�� Model test results on group piles in sand have shown Model test results on group piles in sand have shown

that group efficiency can be that group efficiency can be greater than 1greater than 1 because because soil compaction zones are created around the piles soil compaction zones are created around the piles during driving.Based on the experimental observations during driving.Based on the experimental observations of the behavior of group piles in sand to date,two of the behavior of group piles in sand to date,two general conclusions may be drawn:general conclusions may be drawn:11--for driven group piles in sand with for driven group piles in sand with d>3D, Qd>3D, Qg(u)g(u)==ΣΣΣΣΣΣΣΣQQ(u)(u)

22--for bored group piles in sand at conventional for bored group piles in sand at conventional spacingspacingd=3D,Qd=3D,Qg(u)g(u) may be taken may be taken 2/3 to 3/4 time 2/3 to 3/4 time ΣΣΣΣΣΣΣΣQQ(u)(u)

75

Pile in clayPile in clay

�� The ultimate load bearing capacity of group piles in clay The ultimate load bearing capacity of group piles in clay may be estimated with the following procedure:may be estimated with the following procedure:

11--Determine Determine ΣΣΣΣΣΣΣΣQQuu=n=n11nn22(Q(QPP+Q+Qss) ;) ;ΣΣΣΣΣΣΣΣQQuu=n=n11nn22[9C[9CuuAApp++ΣαΣαΣαΣαΣαΣαΣαΣαPCPCuuL]L]22--determine the ultimate capacity by assuming that the determine the ultimate capacity by assuming that the piles in the group act as a block with dimension piles in the group act as a block with dimension Lg*BLg*Bgg*L.The skin resistance of the block is:*L.The skin resistance of the block is:

QQs(g)s(g)==Σ2αΣ2αΣ2αΣ2αΣ2αΣ2αΣ2αΣ2αCuCu((((((((LLgg+B+Bgg)L)LCalculate the point bearing capacity fromCalculate the point bearing capacity from

QQP(g)P(g)=N=N**ccCCuuLLggBBgg , N, N**

CC=5.14(1+0.2B=5.14(1+0.2Bgg/L/Lgg)(1+0.2L/B)(1+0.2L/Bgg)<9)<9

ΣΣΣΣΣΣΣΣQQ(u)(u)=Q=Qs(g)s(g)+Q+QP(g)P(g)

33--Compare the 2 results,The lower of the two valu e isCompare the 2 results,The lower of the two value isQQg(u)g(u)

Piles in rockPiles in rock

�� For point bearing piles resting on For point bearing piles resting on rock,most building codes specify that rock,most building codes specify that QQg(u)g(u)==ΣΣΣΣΣΣΣΣQQ(u)(u),provided that the minimum ,provided that the minimum center to center spacing of pile is center to center spacing of pile is D+300mmD+300mm.For H.For H--piles and piles with piles and piles with square cross sections,the magnitude of D square cross sections,the magnitude of D is equal to the diagonal dimension of the is equal to the diagonal dimension of the pile cross sectionpile cross section

76

Settlement of piles and groups in Settlement of piles and groups in sands and Gravelssands and Gravels

�� The present Knowledge is not sufficient to The present Knowledge is not sufficient to evaluate of pile and pile groups. For most evaluate of pile and pile groups. For most engineering structures, the loads to be applied engineering structures, the loads to be applied to a pile group will be governed by consideration to a pile group will be governed by consideration of consolidation settlement rather than by of consolidation settlement rather than by bearing capacity of the groups divided by an bearing capacity of the groups divided by an arbitrary factor of safety of 2 or 3. It has been arbitrary factor of safety of 2 or 3. It has been found from field observation that the settlement found from field observation that the settlement of a pile groups is many times the settlement of of a pile groups is many times the settlement of a single pile at the corresponding working load.a single pile at the corresponding working load.

Settlement of piles and groups in Settlement of piles and groups in sands and Gravelssands and Gravels

�� The settlement of a group is affected by the The settlement of a group is affected by the shape and size of group, length of pile, method shape and size of group, length of pile, method of installation of pile and possibly many other of installation of pile and possibly many other factors.factors.

�� There are no equations that would There are no equations that would satisfactorily predict the settlement of pile in satisfactorily predict the settlement of pile in SAND. It is better to rely on load tests for SAND. It is better to rely on load tests for piles in SAND.piles in SAND.

�� In this chapter we try to show some equations In this chapter we try to show some equations for estimation the settlement of pile in SAND.for estimation the settlement of pile in SAND.

77

Settlement of pile shaftSettlement of pile shaft

�� Where : LWhere : L--pile lengthpile lengthEEPP--elastic modulus of pile elastic modulus of pile

material,for concrete pile Ematerial,for concrete pile EPP=21000MPa=21000MPaζζ=0.5=0.5AAPP--area of pile tiparea of pile tip

pp

allf

allp

EA

LQQSe

)(1

ξ+=

Settlement of pile cause by load at Settlement of pile cause by load at the pile pointthe pile point

�� Where : BWhere : B--Width of pileWidth of pileEE--elastic modulus of soil elastic modulus of soil

µµ--Poisson ratioPoisson ratio

)1(85.0 22 µ−=

E

BqSe

allp

78

Settlement of pile cause by the load Settlement of pile cause by the load transmitted along the pile shafttransmitted along the pile shaft

�� Where : BWhere : B--Width of pileWidth of pileEE--elastic modulus of soil elastic modulus of soil µµ--Poisson ratioPoisson ratioLL--pile lengthpile lengthPP--perimeter of the pile sectionperimeter of the pile section

f

allf I

E

B

PL

QSi )1( 2

3 µ−=B

LI f 35.02+=

Consolidation settlement of group pilesConsolidation settlement of group piles

�� The settlement of pile group in clay can be estimated The settlement of pile group in clay can be estimated by assuming that the total load is carried by an by assuming that the total load is carried by an equivalent raft located at depth of equivalent raft located at depth of 2L/3 where L2L/3 where L is the is the length of the piles.It may be assumed,that the load is length of the piles.It may be assumed,that the load is spread from the perimeter of the group at a slope of spread from the perimeter of the group at a slope of 1 1 horizontal to 4horizontal to 4 vertical vertical to allow for that part of the to allow for that part of the load transferred to the soil by skin friction.The vertical load transferred to the soil by skin friction.The vertical stress increment at any depth below the equivalent stress increment at any depth below the equivalent raft may be estimated by assuming in turn that the raft may be estimated by assuming in turn that the total load is spread to the underlying soil at slope of total load is spread to the underlying soil at slope of 1 1 horizontal to 2 verticalhorizontal to 2 vertical.The consolidation settlement .The consolidation settlement is than calculated as the shallow foundation.is than calculated as the shallow foundation.

79

Equivalent raft conceptEquivalent raft concept

2L/3L

1:4

1:2

Q

B’&L’

''LB

Qq =

d d

L’=D+2d+L/3

Bg

Lg

dd

d

B’=D+d+L/3

q

Thank you for your attentionThank you for your attention

��Mr. Sieng Mr. Sieng PEOUPEOU

��Master Master science of science of geotechnical geotechnical engineeringengineering

Foundation Design [Compatibility Mode]

Documents

Transcript of Foundation Design [Compatibility Mode]