driven piles oe4624

7/27/2019 Driven Piles OE4624

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OE 4624 - Offshore Soil Mechanics Axially and laterally loaded piles

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6 Axially and laterally loaded piles

6.1 Introduction

In this chapter the theory of axially and laterally loaded piles will be discussed. After an

introduction of different offshore piles, the axially loaded pile theory will be discussed. This will be

followed by the theory of pile plugging. Finally the laterally loaded pile theory will be discussed.

The theories will be completed with several examples from current design methods.

6.1.1 Working area

Before designing an offshore foundation, an extensive investigation of the working area isnecessary. This is already discussed in chapter Offshore site investigation . What aspects do we

have to take into account? Is the working area shallow or deep water? This is important for the

determination of the type of foundation. In shallow water usually a pile foundation is chosen,

floating constructions are usually used in deep water. What conditions (vertical/horizontal loads,

waves, weather, geotechnical hazards) do we have to take into account? Examples of the possible

hazards are illustrated in figure 6-1.

figure 6-1: hazards;


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Case Foundation type (text van Baaijens, gebruiken?)

The OSM Field is an extensive one under a relatively sloping sea bed where the water depths

vary from about 90 meters at one edge of the field located about 150 km from shore to 800

meters at the other extreme some 100 km further from the coast. Seismic work indicates that

there are two relatively high points in the reservoir: One near the shallow water end - in about

115 meters water depth and the other at the deeper end in roughly 750 meters of water. No

drilling has been done, yet.

There is no existing offshore infrastructure in the vicinity of this field, but sufficient onshore

facilities - for fabrication of equipment or for the storage and export or utilization of the produced

oil or gas - are available along the coast.

The wave climate is usually relatively calm with occasional severe typhoons.

Exploration drilling will have to be carried out to check the seismic data. This will involve at

least two wells - one at each of the reservoir peaks. This drilling will be done with a floating

drilling rig using a spread mooring and drag-embedded anchors.

The field is expected to be large enough to justify two production platforms - one near each of

the reservoir high points. The shallow end of the field will be developed using a fixed, pile-

supported steel tower structure. A floating production system with a spread mooring and suction

anchors is one option for the deeper area. An alternative to be considered here is a tension leg

platform.

A jack-up rig will be brought in to provide tender-assisted drilling from the main production

platform at the shallower location as well as to drill a few satellite water injection wells in the

area.

The deeper production centre will have to be developed with floating equipment - both for

drilling and for production.

Field export pipelines will carry the oil and gas from the production facilities in the deeper part

of the field to the platform in the shallower part for booster pumping to shore. These - as well as

any satellite well flow lines - will be laid using a dynamically positioned lay vessel. The two

pipelines from the shallower platform to shore will be laid using a conventional lay-barge using

anchor handling tugs and a spread of up to 12 drag-embedded anchors. This latter contractor is

also responsible for the shore approach of these pipelines.

6.1.2 Pile foundations

Pile foundations are used for offshore platforms, but they are as well used as anchors for

floating structures. Examples are a tension leg platform and floating production vessels (FPSO).

The mean focus of this course can be found in foundation in the shallow part of the sea. In this


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part often pile foundations are used. Why do we use deep pile foundations instead of shallow

foundations on the bottom? Pile foundations are used in situations if there is an existence of soft

soils, when high horizontal loads are present and if there is a danger of scouring. Offshore piles

are mostly made of steel. This because steel pipes are easy to handle and they can handle high

axial loads. The offshore piles are open piles. Its not possible to use closed piles in the offshore

field. It is not possible to drive a closed pile into the ground, because the pile will break. Another

advantage of open piles is the fact that the piles can be transported in pieces and connected in

place. Open steel piles can handle a high axial capacity (>50 MN 5000 cars). The diameter

varies from 30 to 155 (= 0.76 to 4.0 m). Typical wall diameters are between 1/30 - 1/60 of the

diameter. The installation depth could be up to 120 m depth. The number of piles per platform

leg depends on the size of the structure. Do we have to put one pile at each corner for small

platforms or do we have to put more piles for big platforms? This will be solved using the

available experience with the use of open steel piles onshore and offshore. The Dutch Standard

Rules the NEN 6743 deals with open steel piles.

The piles can be installed:

x Driven (by far most common);

x Drilled and Grouted (Calcareous sediments or rock);

x Grouted Driven (new technology).

Trends in offshore industry are the use of larger diameter piles and underwater hydraulic

hammers replacing above water steam hammers.

6.1.3 Driven Piles

The installation consists of stabbing the pile through guides; followed by the driving of the pile

to the design depth, see figure 6-2. Problems could be the pile drivability (if unknown rock/hard

layers will be penetrated), premature plugging (the pile is open and it has to behave as a open

pile pending the driving in, more resistance) and potential low friction in calcareous sands.

Offshore piles are relatively long (friction piles) and depend on the friction part.

6.1.4 Drilled and Grouted Piles

The installation consists of drilling of a hole with a larger diameter than the pile and inserting

the pile, see figure 6-2. The installation of the pile goes simultaneously with the grouting process.

The grout is injected to get more friction. A rougher surface is created, resulting in more capacity

of the pile.

Problems could concern the stability of the hole. The drilled hole could collapse before

installing the pile. This technique is expensive, because it is time consuming. End bearing is

usually ignored. The pressure of the grout is important for the friction. Grouting pressure may


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limit the friction; less pressure gives less friction. The problems with low friction in calcareous

sands can be solved by the use of drilled and grouted piles. Uncontrolled grout flow is also a

problem.

6.1.5 Grouted Driven Piles

This will be the technique of the future. The installation consists of the driving a pile and

subsequently grouting the pile under pressure through holes, see figure 6-2.

Problems could be an unknown extent of grout coverage. The effects of the pressure of the

grout are unknown. More experience and research is needed before this technique can be used.

figure 6-2: pile installations;

6.1.6 Soil parameters

In offshore cases the shaft friction dominates the axial bearing capacity of a pile. What factors

influence the shaft friction? Off course the initial ground conditions. Correlations are available with

certain tests to get the design strengths/parameters. The soil in the surrounding pile area gets

disturbed during construction (driving and drainage), resulting in stress level changes. An

important fact plays the pile surface roughness (skin friction). Is the surface rough or smooth,

steel or teflon? The time effect: the time between loading and installation. Another important fact

is anisotropy. The soil could have stronger directions.

The strength of clay is determined by the shear strength ( uc , u s ) and by CPTs (Cone

Penetration Test), an example is given in figure 6-3. The relation between the undrained shear

strength uc and the cone resistance q c is about 1/15 to 1/20.


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figure 6-3: undrained shear strength profile against the depth;

The strength of sand is determined by the internal angle of friction ( M ), the weight of overburden and by CPTs, an example CPT-plot is given in figure 6-4.

figure 6-4: CPT plot, cone resistance, friction and pore pressure;

The key parameters in pile designing are the friction and the end bearing ( W and bq ). The

design parameters should easily follow from the site investigation tests. Relations are needed

between design parameters and the undrained shear strength ( uc ), the vertical effective stress

( 'vV ) and with CPTs ( cq ). The design parameters in terms of in-situ soil parameters must be

assessed. The maximum resistance value is called the quake value. The displacements continueusing the same force, this is failure of the soil. For clay the maximum resistance is related to the

measured soil strength (shear strength) multiplied with factor D. In figure 6-5 two examples are

illustrated of strength determination for clay and sand.


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V 1, V 3

V 2 cu

V 1, V 3

V 2 cu

V1, V3

V2 Strength

M

V1, V3

V2 Strength

M

figure 6-5: soil parameters for clay and sand;

6.2 Axial pile capacity

The response of a foundation pile to an axial load is applied at its top is in general strongly

non-linear, and may involve large irreversible deformations. Throughout this chapter the

behaviour of the soil surrounding the pile will be characterized by the response of a non-linear

spring, see figure 6-6. The response of this spring is completely determined by the local

displacement of the pile. This means that all vertical stress transfer in the soil is disregarded.

Models to take this effect into account have been developed (Poulos, 1986) for elastic soils.

figure 6-6: load transfer analysis simulated by means of springs;

6.2.1 Pile in linear material

Consider a pile of constant cross sectional area A , consisting of a linear elastic material, with

modulus of elasticity E . The circumference of the pile is denoted by O . The normal force N in

the pile can be related to the friction along the circumference by the equation of equilibrium

0dN

Odz

W . (6.2)


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N

N N '

W W

figure 6-7: element of axially loaded pile;

The normal force N is related to the stress by

N AV ,and the stress is related to the strain by Hookes law for the pile material

E V H .Finally, the strain is related to the vertical displacement w by

dwdz

H .

Thus the normal force N is related to the vertical displacement w by

dw N EA

dz . (6.3)

Substitution of equation (6.3) into equation (6.2) gives

2

2 0d w

EA Odz

W . (6.4)

The further analysis depends upon the relation between the shear stress W and thedisplacement w . In this section it will be assumed that the shear stress W is proportional to thevertical displacement w

cwW , (6.5)where c has the character of a sub-grade modulus.

Substitution of equation (6.5) in (6.4) gives

2

2 0d w

EA cOwdz

. (6.6)

This is the basic differential equation for an axially loaded pile supported by continuous linear

springs.

When all the physical and geometrical parameters are constant the general solution of the

differential equation (6.6) is

1 2exp exp z z

w C C h h

, (6.7)

where h is a characteristic length, defined by


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E Ah

O c, (6.8)

and where 1C and 2C are integration constants, to be determined from the boundary

conditions. For a pile of length l , loaded at its top these boundary conditions are

0 z : N P , (6.9)and

z l : 0 N , (6.10)

when it is assumed that there is no point resistance.

Using the boundary conditions the constants 1C and 2C can be determined. The final

expression for the vertical displacement then is

cosh[( ) / ]sinh( / )

Ph l z hw

EA l h. (6.11)

The corresponding formula for the normal force in the pile issinh[( ) / ]

sinh( / )l z h

N P l h

. (6.12)

It can easily be seen that this solution satisfies the two boundary conditions. The solution is

shown graphically in figure 6-8.

P

N

z

P

figure 6-8: normal force in pile;

The depth of influence of the load at the top is determined by the value of the parameter h .

This parameter also determines how long a pile must be to be considered infinitely long.Therefore it is interesting to estimate the value of h . This will be done for an important particular

case, namely a tubular steel pile. Let the diameter of the pile be D and the wall thickness d .

Then O DS and A Dd S . Furthermore an expression of the type / sc E D relates the

sub-grade constant to the elasticity of the soil ( s E ). Thus equation (6.8) becomes

/ sh EDd E . (6.13)


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The ratio of the elastic moduli of steel and soil is of the order of magnitude 1000 (or perhaps

even more), the radius of the pile may be 1 m, and the wall thickness may be 0.050 m. In that

case one obtains 7h | m, indicating that the order of magnitude of this parameter is 10 m, orperhaps more. Thus, if the mechanism of load transfer considered here is applicable (and it will

appear that this is certainly not always the case), a pile longer than say 10 m or 20 m can be

considered as infinitely long.

If l o f the solution (6.11) reduces to

exp( / ) Ph

w z h EA

, (6.14)

and the formula for the normal force in the pile becomes

exp( / ) N P z h . (6.15)

It may be noted that in all these cases the shear stress is proportional to the displacement,

see equation (6.5). Thus the maximum shear stress occurs at the top of the pile, and the shear

stress decreases exponentially with depth. Although this seems perfectly reasonable under theassumptions made, it may be in conflict with the shear strength properties of the soil, as will be

elaborated in the next sections.

6.2.2 Pile in cohesive material

For a pile in cohesive material, such a clay, the shear stress along the shaft of the pile can not

be larger than some limiting value, see figure 6-9. In the plastic branch the shear stress remains

constant, even if the displacement increases.

w

W

0W

0w

figure 6-9: cohesive material;

The maximum shear stress is denoted by 0W . Its magnitude will be close to the undrained

shear strength u s of the clay, but is not necessarily the same, because it may be influenced by

the roughness of the pile wall, and by the behaviour of the clay (e.g. dilatant properties). It has


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also been observed that the maximum shear stress is larger when the pile is in compression than

when it is in tension. This may be due to the lateral deformation of the pile.

The elastic branch can be characterized by the sub-grade modulus, see equation (6.5).

Another way to characterize the elastic branch is through the maximum shear stress 0W and the

quake 0w , which is the displacement necessary to generate the maximum shear stress. Actually,this parameter is often considered more convenient, and easier to estimate. The relation with the

sub-grade modulus is

0

0

cwW

. (6.16)

The quake may be estimated by noting that the elastic shear deformation of the soil is limited

to say about 1 % or 2 %, and that the deformation will occur around the pile, in a zone which can

be expected to be about as large as the radius of the pile. This means that the order of

magnitude of the quake will be about 1 % of the pile diameter,

0 0.01w D| . (6.17)

It may be noted that this means that the sub-grade modulus is about 100 /uc s D .

Comparing this to the correlation / sc E D shows that these correlations are in agreement if

the modulus of elasticity of a clayey soil is about 100 times the undrained shearing strength. This

is a relation that is indeed often found in engineering practice.

It may also be noted that in kick deposits of natural clay the shear strength often increases

with depth. Keeping the quake constant in such a material automatically ensures that the stiffness

in the elastic branch also increases with depth, which is in agreement with experimental evidence.

All these considerations support the preference for the parameter 0w , the quake, to characterize

the elastic branch of the response.

6.2.3 Solution for an infinitely long pile

For an axially loaded pile in a homogeneous cohesive material the distribution of shear

stresses along the pile shaft is shown in figure 6-10, provided that the load P is large enough to

produce sliding of the pile along the soil at the top part. If it is assumed that the depth over

which plastic deformations occur is d , the problem consists of two parts, one ( d z l ) inwhich the shear stress is proportional to the displacement, and another one ( 0 z d ) in which

the shear stress is constant. If it is assumed that the pile is long enough to be considered as

infinitely long the solution in the elastic part is similar to the solution (6.14) for a pile in a

completely elastic material, except that the top is now at the level z d ,

0 exp[ ( ) / ]w w z d h . (6.18)


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The normal force in this part is

0 exp[ ( ) / ] EAw

N z d hh

, (6.19)

which means that the normal force at the top of the elastic part is

0d

EAw N h . (6.20)

P

d

z

W

figure 6-10: shear stress along pile in clay;

In the plastic part the differential equation is , with (6.4),

2

02

d w EA O

dz W . (6.21)

The solution of this problem will not be given in detail here. It is a relatively simple differential

equation, and the boundary conditions follow immediately from the given load at the top, and the

known force and displacement at the bottom of the plastic region.

The final solution of the problem in the plastic part is

200 (1 ) ( )2

Od z w w d z

h EAW

. (6.22)

The maximum displacement t w occurs at the top of the pile, of course. Its value is

20

0 (1 ) 2t Od d

w wh EA

W . (6.23)

The normal force in the plastic region is found to be0

0 ( ) EAw

N O d z h

W . (6.24)

The normal force at the top is

00

EAw P Od

hW . (6.25)

By noting that


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0 0/ /h EA cO EAw OW , (6.26)

the relation between the force at the top ( P ) and the displacement at the top ( t w ) can be

written as

2

0 2(1 )2t d d

w w h h , (6.27)

0 (1 ) Ph d

w EA h

. (6.28)

This is a parameter representation of the force-displacement relation of the pile. After

elimination of the parameter /d h one obtains

2

0 0

1[1 ( ) ]

2t w Ph

w EAw. (6.29)

This solution applies only if 0d ! , otherwise the relation between t w and P is linear,

passing through the origin and the point 0w w , 0 / P EAw h . The relation between P and

t w is shown in figure 6-11.

P

w

figure 6-11: load-displacement curve for pile in clay;

After a first linear branch the displacement increases more rapidly with the load, as an ever-

increasing part at the top of the pile becomes plastic. There is no limit load, as the pile is infinitely

long.

It should be noted that the analysis presented in this section assumes that the shear stress

along the pile shaft is zero when the displacement is zero. This will be the case if the pile is

loaded for the first time, and if there are no initial normal forces in the pile. After several times of

loading to a level at which plastic deformations are produced, an equilibrium system of stresses

may remain in the pile after unloading. The analysis then is much more complicated, and is best

performed numerically.


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6.2.4 Solution for a pile of finite length

For a pile of finite length in a homogeneous cohesive material, loaded by an axial force, it can

be assumed that the plastic deformations will start in a zone of length d near the top of the pile.

The analysis is restricted to monotonic loading of an initially unstressed pile. It should be noted

that after once loading the pile beyond the initiation of plastic deformations and subsequentunloading, permanent displacements of the pile top are generated and a system of stresses

remains in the pile after unloading. The actual stresses and deformations then depend upon the

loading history of the pile. This is excluded here by assuming that the pile is free of stress before

the loading.

In the elastic zone, for d z l , the differential equation is (6.6),2

2 0d w

EA cOwdz

, (6.30)

with the solution

1 2exp( / ) exp( / )w C z h C z h , (6.31)

where /h EA cO . Using the condition that the normal force is zero at the bottom end of

the pile, i.e. for z l , it follows that the solution can be written as

12 exp( / ) cosh[( ) / ]w C l h l z h . (6.32)

At the depth z d this displacement must be equal to the value of the quake, 0w , i.e. the

displacement necessary to reach plasticity. This determines the constant 1C , and the solution can

be written as

d z l : 0 cosh[( ) / ]cosh[( ) / ]l z hw wl d h

. (6.33)

The normal stress in the elastic part of the pile is

d z l : 0sinh[( ) / ]cosh[( ) / ]

wdw l z h N EA EA

dz h l d h. (6.34)

It can easily be verified that the solution in the elastic part of the pile satisfies the conditions

that for z l the normal force is zero, and that for z d the displacement is equal to 0w .

In the plastic part of the pile the differential equation is equation (6.21), which can also be

written as

0 z d :2

02 2

wd wdz h

. (6.35)

The solution of this linear differential equation is

0 z d :2

0 1 22

12

z w w D z D

h, (6.36)


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where 1 D and 2 D are constants, to be determined from the boundary conditions. These

constants can best be determined from the continuity conditions at the level z d , where the

displacement w and the normal force N must be continuous. This gives

0 01 2 tanh[( ) / ]

w d w D l d h

h h, (6.37)

2

2 0 0 02

1tanh[( ) / ]

2d d

D w w w l d hh h

. (6.38)

The displacement t w of the top of the pile is found to be

2

20

11 tanh[( ) / ]

2t w d d l d h

w h h. (6.39)

If the normal force at the top of the pile is denoted by P , the minus sign denoting that P is

a compressive force, it is found that

0

/ tanh[( ) / ]tanh( / )

P d h l d h P l h

, (6.40)

where 0 P is the normal force at the moment when the displacement of the top of the pile

reaches the value of the quake, i.e.

00 tanh( / )

w P EA l h

h. (6.41)

Equations (6.39) and (6.40) form a parameter representation of the relation between the force

P at the top of the pile and the displacement t w of the top of the pile. They apply only if the

force P is sufficiently large for plastic deformations to develop. In the elastic range the responseis a linear relation, through the point 0w w for 0 P P .

The maximum possible value of the length of the plastic zone is d l . Then equation (6.40)

gives

max

0

/tanh( / )

P l h P l h

. (6.42)

With (6.41) thiss gives

0max 2

w P EAl

h. (6.43)

Using the definition of h and the relation 0 0w cW it finally follows that

max 0 P Ol W , (6.44)

which simply states that the maximum possible force is the maximum shear stress multiplied

by the area of the pile. That is a trivial result, but it is encouraging that the general solution

indeed leads to this maximum value.


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As an example figure 6-12 shows the load displacement curve for the case that / 4.0l h .

The figure has been constructed by letting /d l vary from 0.00 to 0.99 and then by calculating

the force and displacement according to equations (6.40) and (6.39) respectively.

It can easily be verified that the solution of this section, as expressed by equations (6.39) and

(6.40), reduces to the solution given in the previous section for a pile of infinite legth, see

equations (6.27) and (6.28), when the pile length l tends towards infinity.

P

w

figure 6-12 load-displacement curve;

6.2.5 Pile in frictional material

For an axially loaded pile in a homogeneous frictional material such as sand, it can be

expected that the shear stress can not be larger than the maximum value as given by Coulombs

relation' tan f hW V G , (6.45)

where 'hV is the effective horizontal normal stress, and G is the friction angle between pile

and soil. It has been assumed, for reasons of simplicity, that the soil is cohesionless. The

expected distribution of shear stresses along the pile shaft is shown schematically in figure 6-13.

This figure is based upon the notion that the normal stresses in a soil increase linearly with depth.

Using Coulombs relation (6.45) it then follows that this is also the case for the maximum shear

stresses.


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P

d

W

z

figure 6-13: Shear stress along pile in sand;

In the simplest case, of a homogeneous soil under water, one may write for the vertical

effective stresses in the soil

' 'v z V J , (6.46)

where 'J is the volumetric weight of the soil under water, i.e.,

' wJ J J , (6.47)

where J is the volumetric weight of the soil as a whole, and wJ is the volumetric weight of

water.

The horizontal effective stresses are usually considered to be a certain fraction of the vertical

effective stresses,

0 0' ' 'h v K K z V V J , (6.48)

where 0 K , the coefficient of lateral earth pressure at rest, is supposed to be given constant.

Usually 0 K is somewhat smaller than 1, say 0 0.5 K or 0 0.8 K , depending upon the soil

properties and upon the geological and geotechnical history.

With (6.48) the maximum shear stress, see (6.45), is found to be

0 0' tan ' tan f v K K z W V G J G . (6.49)

The order of magnitude of the product coefficient 0 tan K G will be in the range from 0.2 to

0.4 .

Again the transition from elastic to the plastic range is characterized by the quake 0w . This

means that the sub-grade coefficient c now is

0

0 0

' tan f K z cw w

W J G . (6.50)

In the elastic region the basic differential equation now is

2

2 3- 0d w zwdz b

, (6.51)


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where b is the characteristic length, defined by

3 0

0 ' tan EAw

b K OJ G

. (6.52)

The general solution of the differential equation (6.51) is

1 2 z z w C Ai C Bib b

, (6.53)

where the functions ( ) Ai z and ( ) Bi z are Airy functions , see appendix Airy functions.


For an infinitely long pile the solution vanishing at infinity is

1 ( / )w C Ai z b . (6.54)

The boundary condition at the upper boundary of the elastic region is

z d : 0w w . (6.55)

The constants 1C can now easily be determined, and the final expression for the displacement

in the elastic range is

0( / )( / )

Ai z bw w

Ai d b. (6.56)

In the plastic region, for 0 z d , the differential equation is2

0 02 3

tan ' K O w z d w z

dz EA bGJ

. (6.57)

This differential equation can easily be integrated,

30

3 436w z

w C z C b

. (6.58)

Using the boundary condition at the top of the pile ( N P ), and the continuity conditions

at the interface z d , the integration constants can be determined, and a relation to determine

the load P for a given plastic depth d is obtained. It is left to the reader to verify that this

relation between P and d is2

02

'( / )[ ]2 ( / )

w P d Ai d b EA b b Ai d b , (6.59)

and that the displacement at the top of the pile is given by

3

30

'( / )1

( / )t w d d Ai d b

w b b Ai d b. (6.60)

Equations (6.59) and (6.60) are a parameter representation of the load-displacement curve,

see figure 6-14. As the pile is infinitely long, there is no finite limit load. As the load increases the


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plastic region will increase, but there always remains an elastic part, although this may be located

infinitely deep.

P

w

figure 6-14: load-displacement curve;

6.2.7 Bearing capacity

The analytic solutions presented in the previous sections may give some insight into the

significance of the various physical parameters. They are also useful to validate more general

numerical solution techniques. The solutions given are not very useful for civil engineering

practice, however, because they are mainly restricted to piles of infinite length. Both for clay and

for sand no finite limit load was found, for instance, and such a limit load is of great importance

for the design of pile foundations. Therefore in this section this limit load will be considered in

some detail. It will be appear that no information is obtained about the deformations prior to

failure. The principle of load transfer by a pile is sketched in f igure 6-15.

P

figure 6-15: load transfer by a pile;


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Part of the load is transferred to the soil by friction along the shaft of the pile, and another

part is transferred to the soil below the pile by point bearing. We will write

c f q P P P , (6.61)

where f P is the contribution to the limit load c P of the friction, and q P is the contribution of

the point resistance. There are several approaches possible for the determination of the maximum

load that a pile can carry (the bearing capacity of the pile).

6.2.8 Theoretical approach

A theoretical approach to the bearing capacity problem is to determine both components from

theoretical considerations.

For the shear stress along the shaft of the pile one may write

max 0' tan ' tanh vc c K W V G V G . (6.62)

The total contribution of the friction to the bearing capacity of the pile is now found by

integration along the pile shaft,

00( ' tan )

l

f v P c K Odz V G . (6.63)Thus the value of f P can be determined if the values of the parameters c , 0 K , G and 'vV

and O are known, or can be estimated. Unfortunately, this is not a trivial matter, and may

require careful soil investigation.

The point bearing capacity can be estimated from a theoretical analysis following the theory

developed by Prandtl, Terzaghi and Brinch Hansen. The formula to be used is'c c c q q vq s N c s N V . (6.64)

The shape factors c s and q s for a square or a circular region are often taken as 1.2c s and

1 sinq s I . The values of the bearing capacity coefficients c N and q N strongly depend upon

the value of the friction angle I . For 30I D they are 30c N and 18q N . In order to obtain

the total force the limit stress cq is multiplied by the area A of the pile foot, thus

q c P q A . (6.65)

Although this theoretical method requires knowledge about a great number of soil parameters,

these may well estimated, on the basis of experience and global information about the site. Thus,

an experienced engineer may perform a first estimation on the basis of data from a geological

map or survey, and some information about the type of soil.


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6.2.9 Cone penetration test

Very valuable information about the soil may be obtained from a cone penetration test (CPT).

Actually, this test was developed in the Netherlands with the explicit purpose to serve as a model

test for a pile. By measuring the point resistance of a small rod being pushed into the ground, the

value of cq is measured directly. It may be noted that the bearing capacity formulas providetheoretical support for the CPT-analysis, as the maximum stress cq is independent of the area of

the pile or rod.

The friction can also be measured, by measuring the stress in the cone at two locations. Thus

the maximum shear stress is also measured directly, and this can immediately be integrated along

the shaft of the pile.

max0

l

f P Odz W . (6.66)Thus the CPT data lead directly to a prediction of the bearing capacity.

6.2.10 Static pile loading test

In some countries it is required to perform at least ine large scale pile loading test for a pile

foundation. Of course this is an excellent method to determine the bearing capacity, but it

requires heavy equipment, and a test pile, so that it is a very expensive method.

6.2.11 Dynamic pile loading test

A modern technique to predict the static bearing capacity of a pile is to use the information

obtained from a dynamic test on a pile. The dynamic test may be the driving of the pile, or it may

be a special test performed after installation of the pile. The principle of the method is to load the

pile by a dynamic pulse (dropping a weight on its top), and then to measure the response of the

pile top, for instance by measuring the displacement, the velocity or the acceleration. The

interpretation of the test is usually done by comparing the response of the pile with the

theoretical response of a model in which the soil-pile-interaction is taken into account.


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figure 6-16: load transfer by a plug;

The driving formulas that can be found in some early books on foundation engineering, and

that are sometimes still used by experienced engineers, can be considered as an early version of

dynamic pile testing. It should be noted that there is a large variety of driving formulas, which

indicates that the correlation is not so simple as these formulas suggest. The modern way of

analyzing dynamic pile testing is more sophisticated, and at least has a theoretical basis.

6.2.12 Plug in tubular piles

In the case of hollow tubular steel piles friction may develop not only with the soil surrounding

the pile, but also with the soil in the pile. When the force is very small the pile will not slip along

the soil, and the soil in the pile will act as an internal plug, going down with the pile. At the tip of

the pile resistance will act on the plug, causing shear stress transfer to the pile. As the point

resistance develops these shear stresses will increase, and may become so large that the soil slips

along the inner pile wall. This will start at the bottom of the plug, and may gradually extend in

upward direction. There are two limiting situations. The first is that the soil below the pile reaches

its limiting bearing capacity. From that moment one the force at the pile tip can no longer

increase. Part of the soil inside the pile (near the pile top) will never reach its maximum shear

stress. The other limiting condition is when the point resistance is so large that the entire plug

slips along the inner pile wall. The point resistance at the bottom of the plug will never reach its

maximum.

The first limiting force is

1 c p P q A , (6.67)

where p A is the area of the plug and the pile wall, i.e. the total area of the pile and the soil

inside it. For a circular tube this is 214 p

A DS , where D is the outer diameter of the pile.

The second limiting force is

2 max0

l

i c w P O dz q AW , (6.68)


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where iO is the circumference of the plug. For a circular pile this is i iO DS , where i D is

the inner diameter of the pile. The point resistance of the steel wall has been added to the

integral of the friction, as this can always develop.

The smallest of the two forces 1 P or 2 P may be added to the force representing the maximum

friction of the outer wall to give the total bearing capacity of the pile. A positive effect in the case of hollow piles is that the compressive stresses in the soil plug will

result in an increase of the horizontal stresses inside the pile, and thus in an increase of the

maximum possible shear stress. Because of the difficulty to evaluate this effect quantitatively it is

usually disregarded. It gives some additional sfaty, however.

6.2.13 Dynamic effects

One may wonder under what conditions of dynamic loading of a pile inertia effects have to be

taken into account. This problem is considered in this section, concentrating on the dynamic loadsof structures such as an offshore platform loaded by wave action. The basic differential equation

is the extension of equation (6.6) to the dynamic case,

2

2 2

w w EA cOw A

z t U w w

w w, (6.69)

where U is the density of the pile material (steel or concrete). It is now assumed that theload is periodic, with an angular frequency Z , so that the response of the pile can also beexpected to be periodic, with the same frequency. Thus one may write

exp( )w w i t Z , (6.70)

where Z is assumed to be a function of z only. Substitution into equation (6.69) now gives2

2 (1 ) 0dw A

EA cO wdz cO

U Z . (6.71)

The relative importance of inertia effects of the pile is determined by the value of the

parameter 2 / A cO U Z .

For a steel turbular pile the ratio / A O d , where d is the wall thickness of the pile.

Furthermore the subgrade coefficient can be estimated to be

max

0

0.2 '0.01

v

c w DW V

| . (6.72)

If the vertical effective stress is evaluated at a depth of one half of the pile length, one obtains

10 ' 10 'l gl c

D DJ U | | , (6.73)

where ' U is the density of the soil under water, and g is the gravity constant. The inertiaparameter now becomes


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2 2 2

10 ' A dD dDcO gl gl

U Z U Z Z U

| | . (6.74)

For a typical pile one may estimate 1 D m, 0.05d m, 10 g m/s 2, 20l m. Then

2 2

2

4000

A

cO s

U Z Z | . (6.75)

For normal loading conditions of a structure, with dynamic loads due to wave loading of an

offshore platform, for instance, ths parameter is always very small. Typical wave periods are

about 10 seconds, and in that case 10.6 sZ | . Thus inertia effects may usually be disregarded.It may be noted that this conclusion is not true for the dynamic loads occurring during pile

driving. These effects are not considered here, however.

6.2.14 Friction and end-bearing

Dit hoofdstuk moet nog aangevuld worden!

Note: total capacity = friction + end bearing

S e t t l e m e n t

Force

FrictionIncl end bearing

S e t t l e m e n t

Force

FrictionIncl end bearing

Mobilization of the resistance

Skin friction mobilizes fast ( z W curves), friction develops fastTip resistance mobilizes slow ( q z curves)

Resistance o

S e t t l em

en

t o

Total skin friction ( 0 paaltip t dz)

Tip resistanceTotal

Remark:

Reduction of pile length nottaken intoaccount!

Resistance o

S e t t l em

en

t o

Total skin friction ( 0 paaltip t dz)

Tip resistanceTotal

Remark:

Reduction of pile length nottaken intoaccount!


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figure 6-17: mobilization of the resistance;

Friction will mobilize 10 times faster than end-bearing. The displacement needed for the total

development of the end-bearing is in the order of 10 times higher than fr iction.

For shallow foundations: slip circles are used to model the resistance.

For deeper foundations, the slip circles will go all the way down, extra compensation of the soil

next to the pile. This is also the fact with deep foundations, as can be seen in figure 6-18; also

the soil next to the pile influences the end-bearing.

Limit Value!

Comparable with CPT results

Limit Value!

Comparable with CPT results

figure 6-18: slip circles;

These results are comparerable with the results from CPT tests:

For clay, a relation between the end-bearing and the undrained shear strength ( uc ) has been

derived; the end-bearing has a value of approximately 9 times the undrained shear strength.

Note, this is very conservative. Also the cone resistance of a CPT ( cq ) has a relation with the

end-bearing; the end-bearing has approximately the same value as cone resistance.

For sand, relations have been derived with formulas of Brinch Hansen, Vesic and Meyerhof.

And also the cone resistance of a CPT ( cq ) has a relation with the end-bearing; the end-bearing

has approximately the same value as cone resistance.

An API recommendation is given as an example. In figure 6-19 the relation between the pile

tip load and the displacement is illustrated.


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figure 6-19: relation between the pile tip load and the displacement, (Q-z);

6.3 Pile plugging

The axial Capacity ( Q ) consists of the shaft capacity ( sQ ) and the base capacity ( bQ ). The

shaft capacity depends on the skin friction and the shaft area, the pile dimensions length and

diameter ( L , D ), see figure 6-20. The base capacity depends on the base area or Annulus and if

the behaviour is plugged or unplugged.

Pile load

Skin friction

End-BearingPressure

Pile load

Skin frictionSkin friction

End-BearingPressure

figure 6-20: pile reactions;

In a hollow tubular foundation pile a soil plug may develop at the bottom of the pile. This plug

consists of soil under a high compressive stress, resulting in a high stiffness and strength. During

pile driving the formation of such a plug may result in a very high driving resistance that may

hinder the driving process. In static conditions the presence of a plug may give a considerablecontribution to the bearing capacity of the pile.

The open pile is schematised as in figure 6-21. Important parts are the outer friction, the inner

friction and the wall resistance. If the behaviour is plugged the inner friction is smaller than the

plug resistance. If the behaviour is unplugged the plug resistance is smaller than the inner

friction.


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Tip resistance

Skin friction

Pile load

Tip resistance

Skin friction

Pile load

Skin friction

Pile load

Skin friction

Pile load

Tip resistance

Skin friction

Pile load

Tip resistance

Skin friction

Pile load

figure 6-21: unplugged and plugged behavior;

Important fact is the difference between static and dynamic plugging. In case of static

plugging more resistance could be build up. The silo-theory is valid, large stresses/forces can

occur, see figure 6-22. An assumption has been made: at penetrations deeper than 2 D to 8 D ,

an open ended pile acts like closed pile. This assumption is not taken into account in currentdesign methods.

figure 6-22: plugging;

6.3.1 Linear elastic model

For the purpose of reference a simple linear elastic model will be developed first. The soil

column is supposed to have a constant cross sectional area A , consisting of a constant modulus

of elasticity E , and an effective volumetric weight J . The circumference of the pile is denoted by

O . Because the soil column is circular we may write 2 / 4 A DS and O DS , where D is thediameter of the column. The normal stress V in the soil can be related to the friction along thecircumference by the equation of equilibrium

0d

A O Adz V W J . (6.76)


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V

V V '

W W J

figure 6-23: element of axially loaded column of soil;

It can be assumed that the stress due to the weight of the soil column are already acting in

the soil before the pile is installed. This means that the deformation of the soil will be determined

only by the incremental stresses. These are defined as

' z V V J . (6.77)The equation of equilibrium for the incremental stress now is

' 0d A Odz V W

. (6.78)

The incremental stress 'V is related to the strain by Hookes law for the soil material

'dw

E E dz

V H . (6.79)

Here the minus sign is necessary because the stresses have been assumed to be positive for

compression. Substitution of equation (6.79) into equation (6.78) gives

2

2 0d w

EA Odz

W . (6.80)

The further analysis depends upon the relation between the shear stress W and thedisplacement w .

In this section it will be assumed that the shear stress W is proportional to the relativedisplacement of the pile and the soil, v w , where v is the (constant) vertical displacement of

the pile. Hence

( )k v wW , (6.81)

where k has the character of a sub-grade modulus. Substitution of equation (6.81) in (6.80)

gives

2

2 ( ) 0d w EA kO w vdz . (6.82)

This is the basic differential equation for an elastic plug, with stress transfer from the pile

through linear springs. The general solution of the differential equation (6.82) is

exp exp1 2

z z w v C C

h h

, (6.83)

where h is a characteristic length, defined by


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EAh

kO, (6.84)

and where 1C and 2C are integration constants, to be determined from the boundary

conditions. For a plug of length L on a rigid base the boundary conditions are

0 z : ' 0V , (6.85)and

z L : 0w . (6.86)

The first boundary condition expresses that the top of the plug is free of stress, and the

second one expresses that the bottom of the plug rests on a rigid base. Using the boundary

conditions the constants 1C and 2C can be determined. The final expression for the vertical

displacement then is

cosh( / )cosh( / )

z hw v v

L h. (6.87)

The corresponding formula for the incremental stress in the pile is

sinh( / )'

cosh( / ) Ev z hh L h

V , (6.88)

and the shear stress is found to be

cosh( / )cosh( / )

z hkv

L hW . (6.89)

It can easily be seen that this solution satisfies the two boundary conditions. The actual stress

in the pile is obtained by adding the initial stress (due to the weight of the soil) to the incremental

stress. The shear stress distribution is shown graphically in figure 6-24, for the case that 2 L h .

W

z

figure 6-24: shear stress, elastic springs;


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The zone of influence of the force at the bottom of the pile is determined by the value of the

parameter h . Therefore it is interesting to estimate the value of h , see equation (6.84). The

ratio of the modulus of elasticity E and the spring constant k can be estimated to be of the

order of magnitude of the diameter D . This means that the value of h will be of the order of

magnitude of / 2 D . It may be concluded that for normal piles (for which the ratio of length todiameter is large), the value of / L h will be large. In that case the formula for the shear stress

distribution (6.89) can be approximated by

( )exp

L z kv

hW |

. (6.90)

The shear stresses (and thus the plug) will be concentrated in a zone near the bottom of the

pile, with a length of about 2 times the diameter.

6.3.2 Fully plastic model What is the maximum force on soil plug? How can we calculate the plugging resistance of the

pile?

Two assumptions for friction are made:

1. Linear elastic friction (related to clay)

2. Friction related to effective stress (related to sand)

Linear elastic friction (neglect own weight of the soil)

Difference between the displacement of the soil and the displacement of the pile:

(Randolph, 1988) has considered the limiting situation, in which the shear stress along the pile

shaft has reached its maximum value everywhere. The basic equation is again the equation of

equilibrium (6.76),

0d

A O Adz V W J . (6.91)

The shear stress W is now supposed to be determined by a friction factor times the horizontalstress, which in its turn is related to the vertical normal stress V . The two factors can becombined into one, by writing

W EV , (6.92)

Where E can be considered to be of the order of magnitude of the product of the coefficient

of neutral earth pressure 0 K and the friction coefficient tan G ,

0 tan K E G . (6.93)


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It should be noted that in this case the total stress V , which includes the initial stress due tothe weight of the soil, should be used in the analysis, and not the incremental stress. With (6.92)

the basic equation (6.91) becomes

4d dz DV E V J . (6.94)

The solution of this differential equation satisfying the condition that 0V at the top of theplug ( 0 z ) is

4exp 1

4 D z

DJ E V E

-

. (6.95)

The stress at the bottom of the plug is obtained by taking z L in equation (6.95). When the

factor 4 / L D E is large compared to 1 the force at the bottom of the plug will be very large.The shear stress distribution is, with (6.92) and (6.95),

4exp 14

D z D

J E W

- . (6.96)

For the case that 0.25 E and 2 L D the shear stress distribution is shown in figure 6-25.

W

/ z L

1

0

figure 6-25: shear stress, plastic springs, L/D=2;

In most practical cases the value of the parameter / L D will be much larger than 2. Then the

shear stresses are concentrated near the bottom of the plug, again indicating that the plug

develops near the bottom end of the pile. This behaviour is shown in figure 6-26 for the case that

0.25 E and .


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W

/ z L

0

1

figure 6-26: shear stress, plastic springs, L/D=20;

6.4 Design methods


Existing design methods for pipe piles are among others (see literature/library for details):

x NEN 6743 (Nederlands Normalisatie Instituut);

x API (American Petroleum Institute, mostly used);

x MTD (IC) (Marine Technology Directorate).

Different (offshore) design methods take into account the calculation of the failure pile

capacity instead of ultimate pile capacity. This is based on theory and experiments and is

approved by insurance companies worldwide (= conservative). Changes are made or new

methods are added regularly (e.g. Kolk&van der Velde for clay, and the CUR2001 only for sand,

open steel piles)

An important fact in the design methods is the skin friction, are the piles friction piles? Is the

behaviour plugged or not? Due to pile length the pile nearly always behaves plugged. Other

facts are scour around pile (relative reduction in 'vV ), negative adhesion (in Dutch negatieve

kleef) and the assumption of uniform soil (no reduction factor [ , due to bad/little investigation).

The pile resistance ( F ) in formulas:

outer friction 0

l

z F f Odz , (6.97)tip z F q A , (6.98)

pile total outer friction tip F F F . (6.99)


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The API formulas for clay:

Friction:

z u f cD , (6.100)

where

,u

v

c\ V

and 1.0D d ,

1.0\ d : 0.50.5D \ , (6.101)

1.0\ ! : 0.250.5D \ . (6.102)For end-bearing:

z c uq N c , (6.103)

where 9c N .

The API formulas for sand:

Friction:,

0 limtan( ) z v f K f V G , (6.104)

where 0 0.8 K is the earth pressure coefficient and 5G M D

is the friction angle

between pile and soil. Thus lim f depends on M .For end bearing:

,lim z q vq N qV . (6.105)

Thus q N and limq depend on M .

Limit values (the design parameters API for sand), for the friction angle and friction, see table

6-1.

M [] 20 25 30 35 40

lim f [kPa] 47.8 67.0 81.3 95.7 114.8

table 6-1: limit values, API for sand;

Limit values (the design parameters API for sand), for the q N and the end-bearing, see table

6-2.

q N [-] 8 12 20 40 50

limq [MPa] 1.9 2.9 4.8 9.6 12.0


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table 6-2: limit values, API for sand;

The typical skin friction profile is in reality even more extreme. API gives a complete different

path than what is measured, but the total area, the surface of two graphs is equal (same total

skin friction). The effect of length is not taken into account. The longer the pile, the higher the

total friction.

API

NEN

Measured

API

NEN

Measured

figure 6-27: API, NEN and measured values;

A typical result, the ultimate resistance versus depth for certain pile diameter, is given in figure

6-28.

Resistance [MN]

D e

p t h [ m ]

Resistance [MN]

D e

p t h [ m ]

D e

p t h [ m ]

figure 6-28: typical result;

6.4.1 Pile in layered elastic material



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In the models presented in the previous paragraphs the soil reaction has been represented by

a system of non-linear springs. This means that the soil response at a certain depth is assumed to

be determined only by the local displacement. Thus, the actual stress transfer in the soil has no

influence on the interaction of the soil and the pile. In more refined models the soil is represented

by an elastic continuum. Examples of this are the models developed by (Poulos, 1971), (Banerjee,

Davis, 1978), (Randolph, 1981), (Verruijt, Kooijman, 1989). This latter model, which applies to a

laterally loaded pile only, will be presented in this section. Compared to other models it has the

advantage that it can be generalized to include plastic deformations of the soil and to the

interaction of a pile group wit the soil (Kooijman, 1989).

Zone of influence of tip resistance:

Tip resistance [MN]

Clay

Sand

Clay

50%2D

3D

Tip resistance [MN]

Clay

Sand

Clay

Tip resistance [MN]

Clay

Sand

Clay

50%2D

3D

50%2D

3D

50%2D

3D

figure 6-29: layered soil;

MTD Method or IC (Imperial College) method, formulas

CUR 2001-8 Bearing Capacity of steel pipe piles

Unit friction for typical offshore pipe piles (note, relation cone resistance and friction):0.13 0.38,

*0.029 tan( )v

f c a

f hab

q p RV G

| , (6.106)

with * 8h

Rt and * 2 20 i R R R , where i R is the inner diameter and o R is the outer

diameter. And 1.0ab for compression and 0.72ab for tension.

Unit end-bearing for pipe piles (API not possible):

Non-plugging: 0.02( 30)r D D!

w cq q on wall only

Plugging: 0.02( 30)r D D


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0.5 0.25loge cCPT

Dq q

D

CUR 2001-8 formulas for tension piles:0.15 0.85,

*0.045v

c a

f h

q p R

V |

, (6.107)

where * 4h

Rt .

CUR 2001-8 formulas for compression piles:

* 4h

Rt :

0.05 0.9

*0.08c v

c a

f hq p R

V

, (6.108)

* 4h

R:

0.05 0.9

* *0.08 4c v

c a

f h hq p R R

V

, (6.109)

and

0.5

0.25,8.5 c avea a

qq DR

p p

, (6.110)

where

21 /i o DR R R . (6.111)

The average cone resistance ,c avq runs from 1.5 D above the pile tip to 1.5 D below the pile

tip.

6.4.2 The effect of length


The relation length of the pile and friction:

A longer pile has at the same point less friction than a shorter pile. This is illustrated in figure

6-30 where different pile lengths and their friction are shown. This phenomenon is called friction

degradation.


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hhh

figure 6-30: friction degradation;

Database used for CUR 2001-8 method

To derive the equations different tests have done, all done in sand. Results of the tests are

given in table 6-3.

Piles Tension Tests Compression Tests Total

Open Ended 16 (+6) 12 (+2) 28 (+8)

Closed Ended 2 7 9

Total 18 (+6) 19 (+2) 37 (+8)

table 6-3: tests results;

The tests have been plotted, see figure 6-31 and figure 6-32:

0

0.5

1

1.5

2

2.5

3

3.5

0 20 40 60 80 100 120 140

L'/D

Q_

A P I /

Q_

M E A S

[ - ]

open pipe piles

open pipe piles with DQF 4 or 5

closed pipe piles

best fit through all data points

Best fit though data point with DQF 4 or 5

figure 6-31: open and closed piles in tension-API RP2A;


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0

0.5

1

1.5

2

2.5

3

0 20 40 60 80 100 120 140L'/D

Q_

C U R / Q

_ M E A S [ - ]

open pipe piles

open pipe piles with DQF 4 or 5

closed pipe piles

best fit through all data points

Best fit though data point with DQF 4 or 5

figure 6-32: open and closed piles in tension-new;

The measured values are different from the calculated values. The main conclusions after

comparing design methods with measurements:

For API:

x Less conservative with increasing / L D ;

x More conservative with increasing relative density.

For MTD:

x Slightly unconservative for tension piles/pile friction;

x Conservative for end bearing.

6.4.3 Load-settlement diagram


Settlements of constructions in the offshore field are not as important as for onshore

constructions. But still it is important to understand this phenomenon. The pile is settling down

when a load is put on top. The three components influencing a load-settlement diagram are:

length reduction, development of skin friction and the tip resistance.

The length reduction can be calculated using the following equation

average LF

l EA' . (6.112)

The development of skin friction can be determined by using z W curves, the skin friction W and settlement z . Formulas can be found in literature.

It may be noted, that the skin friction is mobilized much sooner than the end-bearing

resistance.


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For API:

clay maxt when 0.01 z D!

Attention!

Reduction of t after 0.01 D possiblesand maxt when 0.1 z D! Inch

Development of tip resistance:

q z curves (formulas see literature)

where q is the tip resistance and z is the settlement

For API:

claymax

q when 0.1 z D!

sand maxq when 0.1 z D!

The tip resistance is mobilized after a much larger settlement than skin friction.

figure 6-33: API z W curve and q z curve;

For NEN 6743, use graphs for displacement of piles

Quake value of 10 mm, see figure 6-34.


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- 190 -

S e t t l em en

t ( %D

) o

End Bearing o

10%

S e t t l em en

t ( %D

) o

End Bearing o

10%

S e t t l em en

t ( mm

) o

Friction load o

10 mm

S e t t l em en

t ( mm

) o

Friction load o

10 mm

figure 6-34: NEN 6743 curves;

Remark:

Reduction of pile length is not taken into account.

Progressive failure can cause, the first and the second spring fail.

Due to compressibility of slender / long piles and strain softening

6.5 Lateral Pile Capacity

Although foundation piles usually have as their main purpose to

carry axial loads, they may also be subject to lateral loads. This is especially the case for offshore

platforms, which my be loaded by large lateral forces due to wind and water waves. The response

of a pile to a lateral load is studied in this chapter. Some analytic solutions for simplified problems

will be presented first.

Throughout this chapter the response of the soil next to the pile will be assumed to be

determined only by the lateral displacement of the pile at that point. This means that all load

transfer in the soil in vertical direction is disregarded. It also means that effects of plasticity can

only taken into account in as functions of the local displacement of the pile. Elastic solutions, for

piles in a linear elastic medium, have been developed by (Poulos, 1986). A refined model, in

which the soil is modelled as a system of elastic-plastic layers, has been developed by (Kooijman,

1989). This model also applies to the analysis of pile groups, even when the plastic zones of the

individual piles in the pile group overlap. The model also has the feature that a gap between pile

and soil may be develop at the upward side of the pile.


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6.5.1 Theory lateral pile capacity


Offshore constructions are subjected to axial loads, significant lateral loads and moments. The

lateral loads are caused by forces due to wind and waves. Laterally pile capacity is especiallyimportant for Monopods. This is a platform that is founded on one single pile. The main

difference between axial and lateral loading is that only the lateral loads only affect the top of the

pile 10 D . The most important design parameters are the lateral stiffness ( EI ) and the

maximum bending moments of the piles ( M ).

Lateral loading will cause soil pressures around the pile. These pressures give resistance to the

movement of the pile. The soil resistance is expressed as a force per unit length (kN/m), rather

than an average pressure. In order to avoid problems with scales, different pile sizes are used.

Failure mechanisms of short piles occur by rotation of the whole pile around a rotation at adepth, z below the soil surface, see figure 6-35.

H

Centre of rotation

z

H

Centre of rotation

z

figure 6-35: failure mechanism short pile;

Failure mechanisms for long piles occur by the formation of a plastic hinge at a depth z below

the surface, see figure 6-36. If the pile is uniform this will be the point of maximum bending

moment.


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H

z

plastic hinge

H

z

plastic hinge

figure 6-36: failure mechanism long pile;

Failure can occur in three different ways, see figure 6-37:

x Lateral displacement of complete pile;

x Plastic hinge at pile cap, rotation of whole pile;

x 2 plastic hinges, rotation of upper part of pile.

H H HHH HH HHHH

figure 6-37: failure mechanisms fixed head piles;

The failure mechanisms in cohesive soil can occur due to the following causes, namely the

formation of a failure wedge near the ground surface, a gap between the ground and the pile

behind the pile and the flow of the ground around the pile at deeper depths see figure 6-38.


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H

gap

H

gap

figure 6-38: failure mechanism cohesive soil;

The limiting soil resistance for uniform u s , according to (Broms, 1964):

x 2u u P Ds near surface;

x 9u u P Ds depths greater than 3 D .

Recent attention to this topic has been given by (Murff, Hamilton, 1993), (Randolph, Houlsby,

1984). They recommend a design value of 10.5u u P Ds . However, the conservative approach of

Broms 9u u P Ds is still mostly adopted.

figure 6-39: flow mechanism;

Limiting Soil Resistance for non-cohesive Soils:

Near the surface expected to be close to

'u p P K zDJ , (6.113)

with p K as the passive earth pressure coefficient.


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Research from (Broms, 1964) 3 p N K (low value near surface ignored)

Research from (Barton, 1964) 2 p N K (low value near surface ignored)

(Brinch Hanssen, 1961) and (Meyerhof, 1995) incorporated shallow wedge failure in design

charts (increasing N values with depths)

Literature / papers used in API design rules:

x soft clay: correlations for design of laterally loaded piles in soft clay, (Matlock, 1970);

x stiff clay: field testing and analysis of laterally loaded piles in stiff clay, (Reese et al,

1975);

x sand: analysis of laterally loaded piles in sand, (Reese et al, 1974).

6.5.2 Pile in linear material

Consider a pile consisting of a linear elastic material, with modulus of elasticity E . The second

order moment of the pile cross-section in the direction of bending of the pile is I , so that the

bending stiffness is EI . The lateral load f on the pile, illustrated in figure 6-40, is related to the

shear force Q by the equation of equilibrium in lateral direction

dQ f

dz . (6.114)

Equilibrium of moments requires that

dM Q

dz . (6.115)

From these two equations it follows that

2

2

d M f

dz . (6.116)

This is the well-known relation between the bending moment in a beam and the load upon it.

Q Q'

Q

f

x

z

figure 6-40: element of laterally loaded pile;


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The second basic equation describes the deformation of the pile. Using Bernoullis assumption

that plane normal sections remain plane this equation is found to be

2

2

d u EI M

dz , (6.117)

where u is the lateral displacement if the pile, in -direction. The bending moment M can

be eliminated from equation (6.116) and (6.117), to give

4

4

d u EI f

dz . (6.118)

This is the well known basic equation of the theory of bending of beams.

In the case of a pile with lateral support, the load is generated by the lateral displacement u .

In the simplest case the load is proportional to the displacement,

f ku , (6.119)

where k is the subgrade modulus (in kN/m 3), and the minus sign is needed because when the

displacement is to the right, the soil reaction will be to the left, see figure 6-40. Assuming that

this is the only load on the pile, substitution of (6.119) into (6.118) gives

4

4 0d u

EI kudz

, (6.120)

or

4

4 4

40

d u udz O

, (6.121)

where O is a parameter of dimension length, defined as

4 4 EI k O . (6.122)


1 2

3 4

exp( / ) cos( / ) exp( / ) sin( / )

exp( / ) cos( / ) exp( / ) sin( / )

u C z z C z z

C z z C z z

O O O O O O O O

. (6.123)

The constants 1C , 2C , 3C and 4C must be determined from the boundary conditions. In

general both boundaries will give rise to two conditions, so that the four constants can indeed be

determined.


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Another definition of lateral load f is the lateral resistance of soil. For clay, the maximum

resistance is related to the measured soil strength, the shear strength. For sand, the maximum

resistance is related to measured friction angle and stress.

Stroke

Displacement [m]

R e s

i s t a

n c e

[ k P a

]

Stroke

Displacement [m]

R e s

i s t a

n c e

[ k P a

]

Stroke

Displacement [m]

R e s

i s t a

n c e

[ k P a

]


In case of an infinitely long pile one of the boundaries is at z f . The terms with exp( / ) z O

then must be eliminated by taking 1 0C and 2 0C . If it is assumed that the boundary

conditions at the top af the pile are

0 z :

2

2 0d u

M EI dz , (6.124)

and

0 z :3

3

d uQ EI P

dz , (6.125)

the final solution becomes


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3

exp( / ) cos( / )2 P

u z z EI O O O . (6.126)

This solution is shown graphically in figure 6-41.

u

z

figure 6-41: laterally loaded pile in elastic material;

The displacement at the top of the pile will be denoted by t u . Its value is

3

2t P

u EI O

. (6.127)

The meaning of the parameter O can be seen from figure 6-41. The displacement is zero if / 2 z SO , and after the next zero at 3 / 2 z SO the displacement is practically zero. Thus the

influence of the load at the top can be felt to a depth of about 5O . This also means that a pilelonger than 5O can be considered as infinitely long.

It is not surprising to note that the lateral displacement of the top of the pile t u is proportional

to the lateral load P . This is an immediate consequence of the linearity of the model used. It can

also be seen from the solution (6.127) that the displacement is inversely proportional to the

parameters 3/ 4k and 1/ 4 EI . This suggests that the pile stiffness EI is relatively unimportant,

and that the soil stiffness k is very important for the soil response. It will appear later, when

considering more realistic soil models, that these conclusions are not generally valid.

A variant of the problem solved in this section is to consider that the subgrade modulus

increases linearly with depth. This may be a realistic assumption, as often the stiffness of the soil

increases with the initial stress, which increases with depth. The analytical solution of this

problem will not be considered here, however.


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6.5.4 Pile in perfectly plastic material

In this section an approximate analytic solution will be presented for the problem of a pile in a

homogeneous material, taking into account the plastic deformations of the soil.

A fairly realistic assumption for the response of the soil to a lateral displacement is to consider

this to be elasto-plastic, see figure 6-42.

'hV

u'

' 2 p v p K c K V

u

' 2a v a K c K V

'hV

u'

' 2 p v p K c K V

u

' 2a v a K c K V

figure 6-42: elasto-plastic soil response;

In this figure the response is assumed to be elastic if the displacement is small. It reaches its

maximum value (the passive lateral soil pressure),

max' ' 2h p v p K c K V V , (6.128)

when the displacement reaches a certain positive value, and it reaches its minimum value (theactive lateral soil pressure),

min' ' 2h a v a K c K V V , (6.129)

when the displacement reaches a certain negative value. If the value obtained from equation

(6.129) is found to be negative the horizontal effective stress is zero, because tensile stresses

cannot be transferred in a granular material. In the equations given above c is the cohesion of

the soil, and a K and p K are the active and passive pressure coefficients,

1 sin

1 sina K

I

I , (6.130)

1 sin1 sin p

K I I

, (6.131)

where I is the friction angle of the soil. The length of the range of elastic displacements ( u'

in figure 6-42) is called the stroke . This is similar quantity as the quake, used in the analysis of


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axially loaded piles. It represents the displacement difference between generating active and

passive soil pressures, respectively.

The response shown in figure 6-42 will be used in the next section to develop a numerical

model. In the present section a simplified analysis will first be presented, due to (Blum, 1931). In

this analysis it is assumed that the elastic range is extremely small as illustrated in figure 6-43.

This is called a perfectly plastic response.

'hV

u

' 2v p K c K V

' 2a v a K c K V

'hV

u

' 2v p K c K V

' 2a v a K c K V

figure 6-43: perfectly plastic soil response;

As soon as there is a displacement the response is either at its maximum or its minimum. If

the load acts towards the right the force on the right side of the pile is the passive earth pressure,

and the force on the left side of the pile is the active soil pressure.

Thus the total force is

( ) ' 2 ( ) p a v p a f K K D cD K K V , (6.132)

or, assuming that the vertical effective stress increases linearly with depth,

( ) ' 2 ( ) p a p a f K K D z cD K K J , (6.133)

where 'J is the submerged unit weight of the soil.It is furthermore assumed that the distribution of the soil reaction is as shown in figure 6-44.


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P

z

P

z

figure 6-44: soil reaction according to Blum;

The basic idea is that the pile is displaced towards the right by the applied force, except at the

lower end, where a displacement towards the left occurs. The soil reaction on this deeper part of

the pile, is replaced by a concentrated force (Blums Ersatzkraft ), further assuming that the pile is

clamped at that depth.

The differential equation now is

4

4 ( ) ' 2 ( ) p a p ad u

EI f K K D z cD K K dz

J , (6.134)

and the boundary

0 z : Q P , (6.135)

0 z : 0 M , (6.136) z h : 0u , (6.137)

z h : 0dudz

, (6.138)

z h : 0 M . (6.139)

The value of the parameter h , which is the level at which the pile is considered to be

clamped, is unknown. A possible reasoning leading to the last three conditions is that the pile is

very well clamped at great depths, so that the displacement u and its first two derivatives are

zero there. The weak point in this argument, at least from the mathematical side, is that at the

depth z h the third derivative cannot be zero, because the concentrated force is acting there.

It may also be noted that the assumed soil reaction cannot be in equilibrium with the applied load

without the concentrated force, because then equilibrium of moments is impossible. The analysis

using the assumptions for the pressure distribution as shown in figure 6-44 is an excellent

example of the art of engineering, especially since it will appear later that the results are in

excellent agreement with those of more refined models.


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453 2

1 2 3 4

2 ( )( ) '

120 24 p a p a cD K K z K K Dz u C z C z C z C

EI EI

J . (6.140)

Using the boundary conditions the integration constants can be determined. Substitution of

these values in the solution (6.140) gives the final solution in the form

2 23

3

( ) ' (8 9 3 )( )

360

( ) ( )( )

12

p a

p a

K K D h hz z u h z

EI

K K cD h z h z

EI

J

, (6.141)

where the value of h must be determined from the relation

21 ( ) ' ( )6 p a p a

P K K Dh K K cDhJ . (6.142)

If the force P is given, and the properties of soil and pile are known, the value of h can be

determined from equation (6.142). Then the displacement can be obtained from equation (6.141)

.

When comparing this solution for a perfectly plastic material with the elastic solution of the

previous section it may be noted that here the most important parameters for the response

characteristic are the pile stiffness EI and the soil strength, as expressed by the parameters

( p a K K ) and ( ) p ac K K . The soil stiffness no longer appears in the solution because it

has been eliminated from the model by assuming a perfectly plastic response. At this stage it is a

matter of conjecture, or of engineering intuition, to prefer one of the two models in favour of the

other one.

6.5.5 API (P-y curves)


The pile bends in horizontal direction when a load is put on top of the pile. The three

components influencing a load-deformation diagram are: the pile stiffness, the lateral resistance

and the sliding of tip surface.

To determine the lateral resistance, p y curves (formulas can be found literature), p(passive - active resistance p/m) and the displacement y must be defined.

API

For clay : max 9 u p cd

For sand: max p depends on M


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ATTENTION: The cyclic resistance is weaker, see next paragraph.

Sliding of tip surface: tip p y curves

APIFor clay: tip u p c

For sand: ' tan( )tip v p V M

6.6 Cyclic lateral loading

Damping of energy in laterally loaded piles is an important physical phenomenon, which is for

instance used to great advantage in guardrails along motorways. During a crash large amounts of

energy are absorbed in the plastic deformation of the soil. As the soil retains its friction, thebarrier can easily be restored to its original form and function.

This hysteretic damping of energy is also of great importance for the behaviour in cyclic

loading of large structures founded on piles. It is essential for this type of damping that it is

generated by dry friction, and is independent of the velocity. This property is in sharp contrast

with viscous damping, as produced by the behaviour of a fluid, for instance in a dashpot, which is

proportional to the velocity. The effect of a viscous damper is very large when the frequency of

the vibrations is very high, and its influence practically vanishes when the load is applied very

slowly. This is not the case for a hysteretic, or plastic, damper. Here the effect is mainly

determined by the magnitude of the applied load, and not by the loading rate. It will appear later,however, that it may be possible to compare the two types of behaviour, and to determine an

equivalent viscosity for the pile-soil system, at least for certain forms of loading.

In order to further illustrate the effect of hysteretic damping occurring in a laterally loaded

pile, a pile is considered loaded by a cyclic load, varying between +1000 kN and -1000 kN, see

figure 6-45.


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figure 6-45: damping in laterally loaded pile;

All properties of the soil and pile are the same as those for the example in the previous

section. The pile length is 50 m, its width is 1 m, and its wall thickness is 0.05 m. The submerged

unit weight of the soil is 10 kN/m 3, the active pressure coefficient is 0.333, the passive pressure

coefficient is 3.0, the neutral pressure coefficient is 0.5, the cohesion is zero, and the stroke is

0.01 m. The response of the pile-soil system has been calculated using the numerical program

LLP, see figure 6-45, which shows the response in three full cycles of loading.

It appears from figure 6-45 that there is very little difference between the behaviour in the

various cycles, except in the first one, during which some additional deformations occur. It seems

that in the first loading step an initial state of stress is created, and that thereafter the pile

behaviour is independent of the cycle number. This type of behaviour is k

driven piles oe4624

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