Foundation design Present by Mr. Sieng PEOU Master science of geotechnical engineering Tel-011...

Foundation designFoundation design

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

engineeringengineering

Tel-011 874 974Tel-011 874 974 email: [email protected]: [email protected]

Type of foundationType of foundation

Shallow foundationShallow foundation

1-Spread footing : support the load from 1-Spread footing : support the load from building by columnbuilding by column

2-Strip footing : support the load from 2-Strip footing : support the load from building by wallsbuilding by walls

3-Mat foundation: combined all footing3-Mat foundation: combined all footing

Type of foundationType of foundation Deep foundationDeep foundation

1- End bearing pile : pile stand on 1- End bearing pile : pile stand on rocks or very dense soils, so we have rocks or very dense soils, so we have only end bearing capacityonly end bearing capacity2- Combined bearing pile : pile stand 2- 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 friction3- Floating pile : pile stand on very 3- 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

Strip footingStrip footing

q

B

Mat foundationMat foundation

B

End bearing pileEnd bearing pile

Rock layer

Soft soil layerPile

Combined bearing pileCombined bearing pile

Stiff soil layer

Soft soil layerPile

Floating pileFloating pile

Soft soil layerPile

Bearing capacity for Shallow Bearing capacity for Shallow foundationfoundation

Type of failureType of failure1-General shear failure for dense soil,we 1-General shear failure for dense soil,we

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

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

3-Punching shear failure for very loose 3-Punching shear failure for very loose soil,not recommendedsoil,not recommended

General shear failureGeneral shear failure

Q

D

Shear line

Local shear failureLocal shear failure

Q

D

Shear line

Punching shear failurePunching shear failure

Q

D

Failure mechanisms and derivation of Failure mechanisms and derivation of equationsequations

Ultimate bearing capacityUltimate bearing capacity

Settlement

qu

qu

S

S

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 + '/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 - '/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 ( = 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 qqff = (2 + = (2 + ) 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 non-cohesive (c=0) the bearing capacity depends on soil is non-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 = = DD. 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.

Semi-circular slip mechanismSemi-circular slip mechanism

Moment causing rotationMoment causing rotation = load x lever arm = load x lever arm = [(q - q= [(q - qoo) 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 = [Cu] x [= [Cu] x [.B] x [B] .B] x [B]

At failure these are equal: At failure these are equal: (q - q(q - qoo ) x B x ½B = Cu x ) x B x ½B = Cu x .B x B .B x B

Net pressure (q - qNet pressure (q - qoo ) at failure ) at failure = = 2 2 x Cu x Cu

This is an upper-bound solution. This is an upper-bound solution.

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

= load x lever arm = load x lever arm = [ (q - q= [ (q - qoo) 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 = [Cu] x [2= [Cu] x [2 R] x [R] R] x [R]

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

Since R = B / sin Since R = B / sin : : (q - q(q - qoo ) = Cu x 4 ) = Cu x 4 /(sin /(sin )² )²

The worst case is when The worst case is when tantan=2=2 at at = 1.1656 rad = 66.8 deg = 1.1656 rad = 66.8 deg

The net pressure (q - qThe net pressure (q - qoo) at failure ) at failure == 5.52 x Cu5.52 x Cu

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

After Terzaghi (1943)After Terzaghi (1943)qd = CNc + s DNq +0.5 BN

Nc = (Nq – 1 ) . Cotg Prandtl 1921

N = 2(Nq + 1)tg Caquot and Kerisel 1953 Vessic 1973

1924Re)2

45(tan tan2 issnereNq

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

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

Cu:Undrained cohesion

B: Width of footing

L: Length of footing

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) + s 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) + s D


B: Width of footing


D/B<2.5

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

qd = s D Nq(1+B/L.tg)+0.5BN(1-0.4B/L)

: 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 = s D Nq.Sq.dq+0.5BNSd


B: Width of footing


Sq=SPB/L ; Kp= tg2(45+2)

dq=d=1+0.1Kp0.5D/B

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

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

qd = C.Nc.Fcs.Fcd.Fci+s D Nq. Fqs.Fqd.Fqi +0.5BNFs.Fd.Fi


B: Width of footing


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

qnet =C.Nc.Fcs.Fcd.Fci+s D (Nq-1). Fqs.Fqd.Fqi +0.5BNFs.Fd.Fi

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.tg

FFss==1-0.4.B/L1-0.4.B/L

Depth factor by Hansen 1970Depth factor by Hansen 1970

Condition D/B<1Condition D/B<1

FFcdcd=1+0.4D/B=1+0.4D/B

FFqdqd=1+2.tg=1+2.tg(1-sin(1-sin))22D/BD/B

FFdd==11

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

Condition D/B>1Condition D/B>1

FFcdcd=1+0.4.arctg(D/B)=1+0.4.arctg(D/B)

FFqdqd=1+2.tg=1+2.tg(1-sin(1-sin))22.arctg(D/B).arctg(D/B)

FFdd==11

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

FFcici=F=Fqiqi=(1-=(1-90)90)22

FFii=(1-=(1-))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.

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

Structural Structural Design of Design of

Mat Mat FoundationFoundation

Conventional Rigid MethodConventional Rigid Method

Step1: Calculate the total column loadStep1: Calculate the total column load iQQ

Step2: Determine the pressure on the soil (q) Determine the pressure on the soil (q) below the mat at point A, B, C…by using the below the mat at point A, B, C…by using the equationequation

x

x

y

y

I

YM

I

XM

A

Qq

Where Where

A=BLA=BL

IIxx=(1/12)BL=(1/12)BL33 : moment of inertia about the X : moment of inertia about the X

axisaxis

IIYY=(1/12)LB=(1/12)LB33 : moment of inertia about the Y : moment of inertia about the Y

axisaxis

MMxx : moment of the column load about the X : moment of the column load about the X

axis = Q.eaxis = Q.eYY

MMYY : moment of the column load about the Y : moment of the column load about the Y

axis = Q.eaxis = Q.exx

Step 3: Compare the values of the soil pressures Step 3: Compare the values of the soil pressures determine in step 2 with the net allowable soil determine in step 2 with the net allowable soil pressure to check if q<qpressure to check if q<qall(net)all(net)

Step 4: Divide the mat into several strips in X and Step 4: Divide the mat into several strips in X and Y direction. Let the width of any strip be BY direction. Let the width of any strip be B11..

Step 5: Draw the shear and moment diagrams for Step 5: Draw the shear and moment diagrams for each individual strip in X and Y direction. For each individual strip in X and Y direction. For example, take bottom strip in the X direction its example, take bottom strip in the X direction its average soil pressure can be given as:average soil pressure can be given as:

qqavav=1/2(q=1/2(qll+q+qFF))

Where qWhere qll and q and qFF soil pressures at point I and F soil pressures at point I and F

The total soil reaction is equal to qThe total soil reaction is equal to qavavBBBB11 because the because the

shear between the adjacent strips has not been taken shear between the adjacent strips has not been taken into account. for this reason, the soil reaction and the into account. for this reason, the soil reaction and the column load need to be adjustedcolumn load need to be adjusted

1)(mod BBq

Qqq

av

avavifiedav

21

iavav

QBBqQ

eeX X and eand eYY are the load eccentricities in the direction are the load eccentricities in the direction

of the X and Yof the X and Y

Q

XQX ii

''

Q

YQY ii

''

2'B

XeX 2

'L

Yey

Also, the column load modification factor isAlso, the column load modification factor is

i

av

Q

QF

So the modified column load are FQSo the modified column load are FQ ii

Now the shear and moment diagram for this strip can Now the shear and moment diagram for this strip can be drawn. this procedure can be repeated for all be drawn. this procedure can be repeated for all strips in the X and Y direction.strips in the X and Y direction.

Step 6: determine depth of the mat d. This can be done Step 6: determine depth of the mat d. This can be done by checking for diagonal tension shear near various by checking for diagonal tension shear near various column. According to ACI Code 318-95(section column. According to ACI Code 318-95(section 11.122.1c). For critical section11.122.1c). For critical section

cfdbU ')34.0(0

Where:Where:

-U : Factored column load (MN)=F-U : Factored column load (MN)=F

-- Reduction factor =0.85 Reduction factor =0.85

-f’-f’cc : Compressive strength of concrete 28 days : Compressive strength of concrete 28 days

(MN/M (MN/M22))

The unit of bThe unit of b00 and d in the preceding equation and d in the preceding equation

are in meters. The expression of bare in meters. The expression of b0 0 in term of in term of

d, which depends on the location of the d, which depends on the location of the column with respect to the plan of the mat, column with respect to the plan of the mat, can be obtained from Figure 4.8c.can be obtained from Figure 4.8c.

Step 7: from the moment diagrams of all strips in Step 7: from the moment diagrams of all strips in a given direction (that is X or Y), obtain the a given direction (that is X or Y), obtain the maximum positive and negative moments per maximum positive and negative moments per unit width M’=M/Bunit width M’=M/B11

Step8: Determine the areas of steel per unit width for Step8: Determine the areas of steel per unit width for positive and negative reinforcement in X and Y positive and negative reinforcement in X and Y directions from the following equations.directions from the following equations.

)2

().('a

dfAfactorloadMM ySU bf

fAa

c

yS

'85.0

Where:Where:

-A-Ass: area of steel per unit width: area of steel per unit width

-f-fYY :Yield stress of reinforcement in tension :Yield stress of reinforcement in tension

-M-Muu :Factored moment :Factored moment

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

Allowable bearing capacityAllowable bearing capacity

Net ultimate bearing capacity Net ultimate bearing capacity

qqnetnet=q=qdd--ss.D.D

Net allowable bearing capacityNet allowable bearing capacity

qqnetnetallall=q=qnetnet/FS/FS

FS-Safety factor =3FS-Safety factor =3 Gross allowable bearing capacityGross allowable bearing capacity

qqallall=q=qnetnetallall++ss.D.D

Verify stable of footingVerify stable of footing

BL

Qqallnet We find value of B

BL

Qq totalall

And verify the stable of footing from equation

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

1-In case I if the water table is located so that1-In case I if the water table is located so that

0<D0<D11<D, so we will change the factor<D, so we will change the factor

ss.D .D DD11+D+D22((sat-sat-ww))

Also value Also value in the last term of the equation has in the last term of the equation has to be replaced by to be replaced by ’= (’= (sat-sat-ww))

2-In case II for a water table located so 0<d<B2-In case II for a water table located so 0<d<Bvalue value in the last term of the equation has to in the last term of the equation has to

be replaced by be replaced by calcal= = ’+d/B.(’+d/B.(’)’)

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

B

Q

D

V

H

QT

qall>V/(BL)

Tall>H

V=Q.Cos

Q.Sin

T=V.tg(2/3)+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

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

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-2eL

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)

L1=L(1.5-3eL/L)

B’=A’/L

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(L1+L2)B

B’=A’/L1

Qult= qu’.A’

Case eL/L< 1/6

1/6<eB/B< 0.5

A’=0.5(B1+B2)L

B’=A’/L

Qult= qu’.A’

Case eL/L< 1/6

eB/B< 1/6

A’= L2B+0.5(B+B2)(L-L2)

B’=A’/L

Footing on two layerFooting on two layer

D s

B d1c1

c2

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

1- Determine influenced thickness1- Determine influenced thickness

H=0.5Btg(45+H=0.5Btg(45+

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,,

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 condition1- Design C1- Design 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 u1u1NNCC

If 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


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

1- Determine the average values of soils parameter1- Determine the average values of soils parameter

2- Determine the soils bearing capacity by using 2- Determine the soils bearing capacity by using valuesvalues

C’ and C’ and ’’

H

dHd 2111 )('

H

cdHcdc 2111 )('



1- Determine the bearing capacity for first layer1- Determine the bearing capacity for first layer

2- Determine the soils bearing capacity for 2- Determine the soils bearing capacity for second layersecond layer

qnet1 = C1.Nc.Fcs.Fcd.Fci+s D (Nq-1). Fqs.Fqd.Fqi +0.5BNFs.Fd.Fi

qnet2= C2.Nc.Fcs.Fcd.Fci+(s D+1d1) (Nq-1). Fqs.Fqd.Fqi +0.5.BNFs.Fd.Fi



3- Determine the bearing capacity 3- Determine the bearing capacity

< q< qnet1net1

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

Pv = 0.5 Pv = 0.5 11 d d1122+ + ss D d D d

11

KKs s =1-sin=1-sin11

Af =BLAf =BL

Af

CPd

Af

KPPqq sVnetnet

1112

tan

Bearing capacity from in situ Bearing capacity from in situ testtest

From static cone penetration testFrom static cone penetration test

1- for B<1.22m1- for B<1.22m

2- for B>1.22m2- for B>1.22m

From dynamic cone penetration testFrom dynamic cone penetration test

qallowable =

15cq

qallowable = 2

28,3

128,3

25 B

Bqc

qallowable =

20

Rd

Bearing capacity from in situ testBearing capacity from in situ test

From standard penetration test SPT by Meyerhof(1965)From standard penetration test SPT by Meyerhof(1965)

1- for B<1.22m1- for B<1.22m

2- for B>1.22m2- for B>1.22m

25

)(33.01

05.0)/( 602 mmS

B

DNmKNq e

netall

25

)(33.01

3.0

08.0)/(

2

602 mmS

B

D

B

BNmKNq e

netall

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

)(

21

netallq

QQA

21

32.

QQ

LQX

)(2 1 XLL

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

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

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’=S-e (value of S’=S-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

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

221

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

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 rockBearing capacity for sound rock

1

)2

45(tan5

)2

45(tan

4

6

q

c

q

NN

N

N

=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(cal)=qult(RQD)2

For calculate bearing capacity we use equation Terzaghi

245tan'2

cquc

Table 4.2 Range of the Unconfined Compression Strength of Various Types of Rocks

quc Phi

Rock type MN/m2 kip/m2 (deg)Granite 65–250 9.5–36 45–55Limestone 30–150 4.0–22 35–45Sandstone 25–130 3.5–19 30–45Shale 5–40 0.75–6 15–30

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) qqu(u((Mpa)(Mpa)

BasaltBasalt 2828 17-10317-103 0.27-0.320.27-0.32 170-415170-415

GraniteGranite 26.426.4 14-8314-83 0.26-0.300.26-0.30 70-27670-276

SchistSchist 2626 7-837-83 0.18-0.220.18-0.22 35-10535-105

LimestoneLimestone 2626 21-10321-103 0.24-0.450.24-0.45 35-17035-170

Porous Porous limestonelimestone

-- 3-833-83 0.35-0.450.35-0.45 7-357-35

SandstoneSandstone 22.8-23.622.8-23.6 3-423-42 0.20-0.450.20-0.45 28-13828-138

ShaleShale 15.7-2.215.7-2.2 3-213-21 0.25-0.450.25-0.45 7-407-40

concreteconcrete 15.7-23.615.7-23.6 variablevariable 0.150.15 15-4015-40

Settlement of shallow Settlement of shallow foundationfoundation

There are two types of settlementThere are two types of settlement1-Immediate settlement or elastic settlement S1-Immediate settlement or elastic settlement See

for sandy soilsfor sandy soils

2-Consolidation settlement S2-Consolidation settlement Scc for fine grained for fine grained

soilssoils

2-1-Primary consolidation settlement for 2-1-Primary consolidation settlement for inorganic soils inorganic soils

2-2-Secondary consolidation settlement for 2-2-Secondary consolidation settlement for organics soilsorganics soils

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 rigid

1-A uniformly loaded, perfectly flexible foundation 1-A uniformly loaded, perfectly flexible foundation resting on an elastic material such as saturated clay resting on an elastic material such as saturated clay will have a sagging profile as shown in will have a sagging profile as shown in figure figure 11,because of elastic settlement.,because of elastic settlement.

2-If the foundation is rigid and is resting on an 2-If the foundation is rigid and is resting on an elastic material such as clay,it will undergo uniform elastic material such as clay,it will undergo uniform settlement and the contact pressure will be settlement and the contact pressure will be redistributed as shown in redistributed as shown in figure 2figure 2..

Type of foundation settlementType of foundation settlement

Figure 1Settlement profile

Settlement profileFigure 2

Calculate immediate Calculate immediate settlementsettlement

Q

D

H

Soil

Rock

q0

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

E

BqSe

)1( 20 E

BqSe

11

11ln

1

1ln

12

2

2

2

m

mm

mm

mm

B

Lm

ave E

BqS )1( 20

re E

BqS )1( 20

Value of Value of

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

E

BqAASe

021.

Variation of AVariation of A11 With H/B by Christian and With H/B by Christian and

Carrier(1978)Carrier(1978)H/BH/B AA11

CircleCircle L/BL/B

11 22 33 44 55

11 0.360.36 0.360.36 0.360.36 0.360.36 0.360.36 0.360.36

22 0.470.47 0.530.53 0.630.63 0.640.64 0.640.64 0.640.64

44 0.580.58 0.630.63 0.820.82 0.940.94 0.940.94 0.940.94

66 0.610.61 0.670.67 0.880.88 1.081.08 1.141.14 1.161.16

88 0.620.62 0.680.68 0.900.90 1.131.13 1.221.22 1.261.26

1010 0.630.63 0.700.70 0.920.92 1.181.18 1.301.30 1.421.42

2020 0.640.64 0.710.71 0.930.93 1.261.26 1.471.47 1.741.74

3030 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 AA22

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

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

o

o

oe

HCcS

.log.1

)4(6

1bmt

s

LWCc

.

100

(%)2343.0

uPP CI 48.022' 83.004.7' uP C

689.0

00 ''.193.0'

N

P

By Mayne & Mitchell

By Mitchell(1988)

By Mayne & Kemper(1988)

Consolidation settlementConsolidation settlement

For over consolidated clay For over consolidated clay ’’00<<’’pp

1-1-

2-2-

'O + 'P

o

o

oe

HCsS

log

1

.

sLW

Cs

.

100

(%)0463,0

'O + > 'P

P

o

oo

P

o e

HCc

e

HCsS

'

'log.

1

.'log.

1

.

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

i-total displacement at point i

ij-different settlement between point i and j

relative deflection

ij= angular

distortion

L=deflection ratio

ij

ij

l

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

Danger 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

Civil 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 one-story millsFor one-story mills 0.00100.0010 0.00100.0010

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

1-When the upper soils layers are highly compressible 1-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.

2-When subjected to horizontal force, pile foundations 2-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 earth-retaining structures and foundations of tall of earth-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.

Deep foundationDeep foundation

3-The expansive and collapsible soils may extend to a 3-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.

4-The foundation of some structures, such as 4-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 foundation5-Bridge abutments and piers are usually constructed 5-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 load-bearing capacity of piles in behavior and the load-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.

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 piles-Steel piles

-Concrete piles-Concrete piles

-Wooden(timber)piles-Wooden(timber)piles

-Composite piles-Composite piles

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 15-6015-60 Practically Practically unlimitedunlimited

300-1200300-1200 --

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

b-Can stand high driving stressesb-Can stand high driving stresses

c-Can penetrate hard layer such as dense gravel,soft rockc-Can penetrate hard layer such as dense gravel,soft rock

d-High load-carrying capacityd-High load-carrying capacity

disadvantages: disadvantages: a-Relatively costly materiala-Relatively costly material

b-High level of noise during pile drivingb-High level of noise during pile driving

c-Subject to corrosionc-Subject to corrosion

d-H-piles may be damaged or deflected from the vertical d-H-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::10-15precast::10-15

Prestressed:Prestressed:

10-3510-35

precast::30precast::30

Prestressed:Prestressed:6060

300-300-30003000

precast::800-precast::800-900900

Prestressed:Prestressed:

7500-85007500-8500

Advantages: Advantages: a-Can be subjected to hard drivinga-Can be subjected to hard driving

b-Corrosion resistantb-Corrosion resistant

c-Can be easy combined with concrete superstructurec-Can be easy combined with concrete superstructure

disadvantages: disadvantages: a-Difficult to achieve proper cutoffa-Difficult to achieve proper cutoff

b-Difficult to transportb-Difficult to transport


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



(KN)(KN)


Cased cast-Cased cast-in place in place concreteconcrete

5-155-15 15-4015-40 200-500200-500 800800

Advantages: Advantages: a-Relatively cheapa-Relatively cheap

b-Possibility of inspection before pouring concreteb-Possibility of inspection before pouring concrete

c-Easy to extendc-Easy to extend

disadvantages: disadvantages: a-Difficult to splice after concretinga-Difficult to splice after concreting

b-Think casings may be damages during drivingb-Think casings may be damages during driving





(KN)(KN)


uncased uncased cast-in place cast-in place

concreteconcrete

5-155-15 30-4030-40 300-500300-500 700700

Advantages: Advantages: a-Initially economicala-Initially economical

b-Can be finished at any elevationb-Can be finished at any elevation

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

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





(KN)(KN)


WoodWood 10-1510-15 3030 100-200100-200 270270

Advantages: Advantages: a-Economicala-Economical

b-Permanently submerged piles are fairly resistant to decayb-Permanently submerged piles are fairly resistant to decay

c-Easy to handlec-Easy to handle

disadvantages: disadvantages: a- Decay above water tablea- Decay above water table

b-Can be damaged in hard driving b-Can be damaged in hard driving

c-Low load-bearing capacityc-Low load-bearing capacity

d-Low resistance to tensile load when splicesd-Low 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

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

Pile driven formulasPile driven formulas To develop the desired load-carrying capacity,a point bearing To develop the desired load-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.bearing value at the predetermined depth.One of the earliest of One of the earliest of these dynamic equations-commonly referred to as the these dynamic equations-commonly referred to as the Engineering News Record (ENR) formula-is derived from the Engineering News Record (ENR) formula-is derived from the work-energy theory;that is : Energy imparted by the hammer work-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 ram-Weight of the ram

h-height of fall of ram(Cm)h-height of fall of ram(Cm)

S-penetration of the pile per S-penetration of the pile per hammer blow(Cm)hammer blow(Cm)

C-a constantC-a constant

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

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

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

CS

hWQ R

u

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

Where E-hammer efficiencyWhere E-hammer efficiency

HHEE-rated energy of the hammer-rated energy of the hammer

S-penetration of the pile per hammer S-penetration of the pile per hammer blow(Cm) blow(Cm)


C=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 E-hammer efficiencyWhere E-hammer efficiency h-height of fall of the ram(Cm)h-height of fall of the ram(Cm)

S-penetration of the pile per hammer S-penetration of the pile per hammer blow(Cm) blow(Cm)

WWPP-weight of the pile-weight of the pile

n-coefficient of restitution between n-coefficient of restitution between the ram and the pile capthe ram and the pile cap

C=0.254 Cm C=0.254 Cm Factor of safety FS=4 to 6Factor of safety FS=4 to 6

))((2

PR

PRRu WW

WnW

CS

hEWQ

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 ram-weight of the ram

WWPP-weight of the pile-weight of the pile


S-penetration of the pile per hammer S-penetration of the pile per hammer blow(M) blow(M)


C=2.54.10C=2.54.10–3–3MM

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

PR

PREu WW

WnW

CS

EHQ

225,1

Danish equations Danish equations


EEPP-modulus of elasticity of the pile-modulus of elasticity of the pile



L-length of the pileL-length of the pile

AAPP-area of the pile cross section-area of the pile cross section

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

PP

E

Eu

EA

LEHS

EHQ

2

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




L-length of the pileL-length of the pile

EEPP-modulus of elasticity of pile-modulus of elasticity of pile

n=0.25 for steel piles and n=0.1 for another n=0.25 for steel piles and n=0.1 for another pilespiles

Factor 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 0.7-0.850.7-0.85

Diesel hammersDiesel hammers 0.8-0.90.8-0.9

Drop hammersDrop hammers 0.7-0.90.7-0.9

Pile materialPile material Coefficient of restitutionCoefficient of restitution

nn

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

0.4-0.50.4-0.5

Wood cushion on steel pileWood cushion on steel pile 0.3-0.40.3-0.4

Wooden pileWooden pile 0.25-0.30.25-0.3

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

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 at

least three times its own diameterleast three times its own diameter

beyond the position of the first helixbeyond the position of the first helix

of the replaced pile.of the replaced pile.

Compression Compression The last helix must be embeddedThe last helix must be embedded

beyond the position of the first helixbeyond the position of the first helix

of the replaced pile.of the replaced pile.

Shear/Overturning Shear/Overturning Embedment must satisfy the specifiedEmbedment must satisfy the specified

minimum.minimum.

Load-carrying Capacity of the pile point,QLoad-carrying Capacity of the pile point,QPP

from 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 tip-area of pile tip

C-cohesion of the soil supporting the pile tipC-cohesion of the soil supporting the pile tip

qqPP-unit point resistance-unit point resistance

q’-effective vertical stress at the level of the pile q’-effective vertical stress at the level of the pile tiptip

NN**CC,N,N**

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

KeriselKerisel

tgeNq

7* cot)1( ** qC NN

Load-carrying Capacity of the pile point,QLoad-carrying Capacity of the pile point,QPP

from 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))


C-cohesion of the soil supporting the pile C-cohesion of the soil supporting the pile tiptip


NN**CC,N,N**

qq-bearing capacity factor after -bearing capacity factor after

Caquot & KeriselCaquot & Kerisel

tgeNq

7* cot)1( ** qC NN

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

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

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 sands silts and sands -Dc = 15B, for -Dc = 15B, for mediummedium dense silts and sands dense 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 <30<30

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

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

Critical depthCritical depth

This critical concept implies that This critical concept implies that ffss 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 ffss 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.

Load-carrying Capacity of the pile point in sand Load-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



q’-effective vertical stress at the level of the q’-effective vertical stress at the level of the pile tippile tip

NN**qq-bearing capacity factor-bearing capacity factor

QQPP=A=Appq’Nq’N**qq<A<Appqqii

qqii=50N=50N**qqtgtg(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.

Load-carrying Capacity of the pile point in Load-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 to be use for determination to be use for determination NNqq

** are are

For driven pile For driven pile = = 11

For bored pile For bored pile = = 11-3 -3

Where Where 11 is angle of internal friction prior to is angle of internal friction prior to

installation of pile.installation of pile.

Load-carrying Capacity of the pile point in Load-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



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

CC=9=9

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

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 - q - q22)H / (10B) but < q)H / (10B) but < q11 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

-q-q22 is the ultimate base resistance in the weak layer is the ultimate base resistance in the weak layer

-q-q11 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 and Relation between ultimate point resistance of pile and depth in sand stratum beneath weak soil layer from depth in sand stratum beneath weak soil layer from

Terzaghi lectures, 1974-1982Terzaghi lectures, 1974-1982

Skin friction from Skin friction from Method Method

f=’0

’0-effective vertical stress at center of layer

As Tomlinson, the maximum frictional resistance is limited 110KPa

From Meyehof 1976

<28 we have =0.44

28<<35 we have0.44< <0.75

35<<37 we have 0.75<<1.20

Skin frictionSkin friction


The angle The angle to be use for determination to be use for determination are are

For driven pile For driven pile = 0.75= 0.7511+10+10

For bored pile For bored pile = = 11-3 -3

Where Where 11 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 Method MethodSkin friction for clayey soil for driven pilef=xCu =1 for Cu=<25KPa

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

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

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

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

for driven Pile

Tomlinson Tomlinson method method

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 vary with C vary with Cuu and L/B ratio and L/B ratio

Tomlinson Tomlinson method method

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:1-if a fill of clay soil is placed over a granular soil layer 1-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.2-if a fill of granular soil is placed over a layer of soft 2-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 pile3-lowering of the water table will increase the vertical 3-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

Clay fill over granular soilClay fill over granular soil

Where:Where:

K’=earth pressure coefficient =Ko=1-sinK’=earth pressure coefficient =Ko=1-sin’’oo=vertical effective stress at any depth Z =vertical effective stress at any depth Z

= = ’’ff.Z..Z.

’’f f =effective unit weight of fill Clay=effective unit weight of fill Clay

=soil-pile friction angle = 0.5=soil-pile friction angle = 0.5 to 0.7 to 0.7

tan'' 0Kfn

2

tan'')tan''(

2

0

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

ffffff HHHL

L

HLL

1

0

21

1 2

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

L

ffZffn

PKLHLPKdZHPKQ

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<=0.4Silt, 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/Dfor closed end pile

and C=0.06L/Dfor 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)

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)

Load-Carrying capacity of pile point resting on Load-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=qu-Ru-R(N(N+1)+1)

Where NWhere N=tg=tg22(45+(45+/2)/2)

qqu-Ru-R=unconfined compression strength of rock=unconfined compression strength of rock

=drained angle of friction of rock=drained angle of friction of rock

The allowable load-carrying capacity of the pileThe allowable load-carrying capacity of the pile point.thuspoint.thus

FS

ANqQ PRu

allp

)1()(

FS=3

Typical unconfined compressive strength of rockTypical unconfined compressive strength of rock

Rock typeRock type qqu-Ru-R(Mpa)(Mpa)

Sandstone Sandstone 70-14070-140

LimestoneLimestone 105-210105-210

ShaleShale 35-7035-70

GraniteGranite 140-210140-210

MarbleMarble 60-7060-70

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(1988-1989),we can estimate the bearing load O’Neill(1988-1989),we can estimate the bearing load capacity of drilled shafts extending into Rock as capacity of drilled shafts extending into Rock as follows:follows:

1-Calculate the ultimate unit side resistance as:1-Calculate the ultimate unit side resistance as:

f=6.564qf=6.564quu0.50.5≤0.15q≤0.15quu

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

corecore

2-Calculate the ultimate capacity based on side 2-Calculate the ultimate capacity based on side resistance only:resistance only:

Qu=Qu=ππDDssLfLf

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 resistance

Se(b)=settlement of the baseSe(b)=settlement of the base

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

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

CCEA

LUQ

massS

f

ED

IUQ

Where QWhere Quu=Ultimate friction load=Ultimate friction load

AAcc=Cross-section area of the drilled shaft =Cross-section area of the drilled shaft

in the socketin the socket

DDss=Diameter of the drilled shaft=Diameter of the drilled shaft

EEcc=Young’s modulus of the concrete=Young’s modulus of the concrete

EEmassmass=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 load-If 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 may be rapid, progressive side If Se≥ 10mm, there may 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

DC

AqQu

Where CWhere Css=Spacing of discontinuities=Spacing of discontinuities

δδ=Thickness of individual discontinuity=Thickness of individual discontinuity

qquu=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 (deg)(deg)

Sandstone Sandstone 27-4527-45

LimestoneLimestone 30-4030-40

ShaleShale 10-2010-20

GraniteGranite 40-5040-50

MarbleMarble 25-3025-30

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..

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:

1-as a block with dimension 1-as a block with dimension LLgg*B*Bgg*L*L

2-as individual piles2-as individual piles

If the piles act as the block, the frictional capacity isIf the piles act as the block, the frictional capacity is

QQg(u)g(u)=f=favavPPggL note PL note Pgg=2(n=2(n11+ n+ n22-2)d+4D-2)d+4D

For each pile acting individuallyFor each pile acting individually

QQ(u)(u)=f=favavLPLP

Efficiency factorEfficiency factor

Where Where =group efficiency =group efficiency

QQg(u)g(u)=ultimate load bearing capacity of =ultimate load bearing capacity of

group pilegroup pile

QQ(u)(u)=ultimate load bearing capacity of =ultimate load bearing capacity of

each pileeach pile

)(

)(

u

ug

Q

Q

21

21 4)2(2

nPn

Ddnn

Converse Labarre equationConverse Labarre equation

21

1221

90

)1()1(1

nn

nnnn

)/((deg) dDarctg

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:

1-for driven group piles in sand with 1-for driven group piles in sand with d>3D, Qd>3D, Qg(u)g(u)==QQ(u)(u)

2-for bored group piles in sand at conventional 2-for bored group piles in sand at conventional spacingspacing

d=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 QQ(u)(u)

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:

1-Determine 1-Determine QQuu=n=n11nn22(Q(QPP+Q+Qss) ;) ; QQuu=n=n11nn22[9C[9CuuAApp++PCPCuuL]L]

2-determine the ultimate capacity by assuming that the 2-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)==CuCuLLgg+B+Bgg)L)L

Calculate 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

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

3-Compare the 2 results,The lower of the two value is3-Compare the 2 results,The lower of the two value is QQg(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)==QQ(u)(u),,provided that the minimum provided that the minimum

center to center spacing of pile iscenter to center spacing of pile is D+300mmD+300mm..For H-piles and piles with For H-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

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.

Settlement of pile shaftSettlement of pile shaft

Where : L-pile lengthWhere : L-pile length

EEPP-elastic modulus of pile -elastic modulus of pile

material,for concrete pile Ematerial,for concrete pile EPP=21000MPa=21000MPa

=0.5=0.5

AAPP-area of pile tip-area of pile tip

pp

allf

allp

EA

LQQSe

)(1

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

Where : B-Width of pileWhere : B-Width of pile

E-elastic modulus of soil E-elastic modulus of soil -Poisson ratio-Poisson ratio

)1(85.0 22

E

BqSe

allp

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

Where : B-Width of pileWhere : B-Width of pile

E-elastic modulus of soil E-elastic modulus of soil -Poisson ratio-Poisson ratio

L-pile lengthL-pile length

P-perimeter of the pile sectionP-perimeter of the pile section

B

LI f 35.02

f

allf I

E

B

PL

QSe )1( 2

3

For group piles in sand and gravel, for elastic settlement, Meyerhof (1976) suggested the empirical

relation

60)(

96.0)(

N

IBqmmS

g

eg

q = Qg/ (Lg.Bg) (in kN/m2)

I = influence factor = 1 – L/8.Bg > 0.5

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.

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 Present by Mr. Sieng PEOU Master science of geotechnical engineering Tel-011...

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