January 1986
T
H
Delft
Prof.dr.ir. A. Verruijt
Laboratori"LllII v o o r G e o t e c h n i e k
OFFSHORE SOIL MECHANICS
A. V e r r u . i j t
D e l f t , 1 9 8 5
Acknowledaements :
wordstar, copyriaht by MicroPro Int., San Rafael, California Prlntstar, copyrlaht by Mlcrostar Inc., Emeryville, California SANYO MBC-550
CONTENTS 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. ll. Introduction ... . Subgrade modulus of soils Linear flexibility of piles 3.1 Axially loaded pile 3.2 Laterally loaded pile Dynamic response of piles 4.1 Axially loaded pile 4.2 Laterally loaded pile Elasto-plastic axial response 5.1 Description of the model Elasto-plastic lateral response
6.1 Active and passive soil pressure
6,2 API
6.3 Generalized elasto-plastic model A numerical model for axially loaded piles 7.1 Basic equations ....
7.2 Computer program 7.3 Example
A numerical model for laterally loaded piles 8.1 Basic equations 8.2 Computer program 8, 3 Example ...•••.•..•. 8.4 Approximate solution Damping . . . , ... . 9.1 Viscoelastic damping 9.2 Numerical model
9.3 Damping in lateral loading Pile driveability Gravity foundations 11.1 Bearing capacity 11.2 Cyclic loading References ... . Appendices
API Reconunended practice, Foundation design DNV Rules, Appendix F, Foundations
J. Brinch Hansen, Bearing capacity
F.P. Smits, Geotechnical design of gravity structures
1 2 4 4 6 8 8 9 10 10 13 13 14 16 19 19 23 26 28 28 32 36 38 41 41 45 47 52 53 53 54 58
1, INTRODUCTION
The recent development of offshore engineering, in particular the
construc-tion of platforms for drilling and producconstruc-tion of oil and gas, leads to a
growing demand for realistic predictions of the behaviour of the foundation, and of the sea bottom. Even though the soils usually encountered on the sea bottom are of the same nature as the soils on land (mainly sand and clay) there are various factors that lead to a difference in the approach and in the type of problems to be considered. The main factors are:
- Investigation of the soil in situ is much more difficult offshore than it is on land,
- One of the main loading conditions to be considered is of a cyclic nature, due to wave action.
- The bearing capacity of piles consists mainly of friction.
In this report these factors will be discussed, for three types of
struc-tures: pile foundations, gravity foundations, and pipelines. Special atten-tion will be paid to the determinaatten-tion of parameters describing the soil behaviour suitable for the analysis of the structures as a whole.
2. SUBGRADE MODULUS OF SOILS
Many problems of foundation engineering can be formulated in a rather simple form when using a so-called subgrade modulus to characterize the soil
res-ponse. In this section this subgrade modulus is correlated with other
elastic properties of the soil.
Consider a square footing, with dimensions b*b. The pressure at the base of
the footing is denoted by p. The stresses in the soil under the footing
will decrease with depth, because of the spatial spreading. It can be
expected that at a depth z the stresses will be distributed over an area b+az, where the dimensionless constant a is of the order of magnitude
of
1. Thus the stresses at depth z are(J' = p
(l+az/b)2
Assuming a linear relationship between stresses and strains, with a modulus of elasticity E, the strains are
Because t=dw/dz, where w is the vertical displacement, one now obtains,
after integration from z=O to z=oo, the following expression for the displa-cement of the footing,
w = ~
aE
If this is compared with the usual expression in terms of a subgrade modu-lus c,
w = p/c
it follows that the parameters are related by
c = aE/b
Because the value of a is about 1 this means that the subgrade modulus can be considered to be approximately equal to the modulus of elasticity, di-vided by the dimension b.
In the case of a bemn on elastic foundation the subgrade constant k is usu-ally defined as
k = cb
This means that k and E are related as follows.
k.= aE
It should of course be noted that this correlation is only justified as a
the actual value·on the basis of a site investigation. The formulas given above can serve as an estimate, however, if no further information is avai-lable.
The modulus of elasticity of a soil is sometimes correlated with the com-pression constant C in Terzaghi's logarithmic formula
t = _cl log( L )
<To
where <ro is the original stress and <r the actual stress. small stress increment one may write <r=<ro+dcr,, and then
In the case of a
t = _cl log(l + d<r) ~ d<r
<To C<ro
If this is compared with the elastic relation t=d<r/E it follows that E = C<ro
..
Thus the modulus of elasticity of a soil can be constant C and the stress level <ro are known. C=200 for sand, and C=lO to C=50 for clay,
estimated if the compression
;L___iINEAR FLEXIBILITY OF PILES
Although the response of a foundation pile to a load applied at its top is
in general strongly non-linear, it is of some interest to consider the
linear case first. The purpose of these considerations is to serve as a
reference for later non-linear calculations, and to provide some .simple approximate formulas.
3.1 Axially loaded pile
Consider a pile of constant cross section, consisting of a homogeneous
linear elastic material, with modulus of elasticity Ep, The cross-sectional
area is denoted by A, and the circumference by
o.
The normal force Nin the pile can be related to the friction .r by the equation of equilibriumdN + n> = O
dz
where. is the shear stress acting on the pile. Assuming a linear
relation-ship between this shear stress and the displacement w of the pile one may write
T = -cw
where the constant c has the character (and dimension) of a subgrade
modu-lus. The minus sign has been introduced to express that the shear stress
acting upon the pile is opposed to the direction of the displacement.
The normal force Nin the pile can be expressed into the displacement w by Hooke's law,
dw N = EA
---dz
where EA is the extension stiffness of the pile, defined as
Combination of the three equations given above leads to the following diffe-rential equation
EA dZw - cOw = 0 dz2
The solution of the differential equation, for the case of a pile of infi-nite length, loaded at its top by a force P is
Ph
w = EA exp(-z/h)
where his a parameter defined by h = v(EA/cO)
"The normal force in the pile is N = -P exp(-z/h)
and the shear stress distribution is p
r = Oh exp(-z/h)
The maximum shear stress occurs at the top of the pile, and its magnitude is P/Oh.
The value of the parameter h determines whether a pile can be considered as infinitely long. For a tubular pile the cross sectional area A is about Od,
where d is the wall thickness of the pile, If furthermore the subgrade
modulus c is expressed as c=Es/D, where Es is the modulus of elasticity of the soil, and Dis the pile diameter, one obtains
h = v(DdEp/Es)
In general the pile will be much stiffer than the soil, say by a factor 2000. For a pile having a diameter of 1 meter and a wall thickness of 50 nun
the value of his about 10 meter. Foundation pile of offshore structures
are usually much longer, so that they can indeed be considered, as far as their axial flexibility is concerned, as infinitely long.
The flexibility of the pile as a whole can be expressed by a single spring constant K relating the applied force P and the displacement w,
w = P/K where
K = EA/h = v(EAcO)
For a tubular pile this can also be written as K = nv(EpEsDd)
where again the subgrade modulus c has been replaced by Es/D,
circumference O of the pile has been expressed as ~D, and
nDd,
and where the the area as
3,2 Laterally loaded pile
For a laterally loaded pile the basic differential equation is the well known equation for a beam on elastic foundation,
EI d4 !! + ku = f dz4
where EI is the bending stiffness, k the subgrade constant (k=cD), and f the lateral load.
The solution for a beam of infinite length, loaded by a lateral force Q at
its top is
where}. is a characteristic length, defined by >.4 = 4EI/k
For a tubular pile one may write I=¼nD3d, and therefore, if k is replaced by Es (see chapter 1),
If Ep/Es=2000 and d/D=0.05 this gives >./D=3.5. From the theory of beams on elastic foundation· it is known that the "wave length" of the deflection curve is 2n>.. In this case this is about 22D. Thus a pile can be consi-dered as infinitely long, as far as lateral flexibility is concerned, if it is longer than about 20 pile diameters. This will often be the case.
The response of the pile top can be expressed by a spring constant K such that
u = Q/K where
For the more general case of loading by a lateral load Q and a moment Tone
may write
where now
u = Q/K-T/R du/dz = -Q/R+T/L
It is interesting to note that the lateral stiffness factor K is mainly
determined by the subgrade modulus, whereas the rotational stiffness factor
L is mainly determined by the bending stiffness EI. This can be seen by
noting that K=kA/2, where A is defined by (4EI/k)114 , which means that K is
proportional to k314 , and to (EI)114 , This means that the dependence on k
is 3 times stronger than that on EI. On the other hand the value of L is proportional to (EI)3 1 4 and k1 1 4 , which means that the rotational
flexibili-ty depends mainly on the bending stiffness EI of the beam. This difference· in behaviour can be explained on an intuitive basis by notating that for rotation the main deformation characteristic is the curvature of the beam, whereas for the lateral displacement of the pile top the main phenomenon is
the generation 9f soil resistance.
It will be seen later, see paragraph 8, that these conclusions are not valid
if the soil is considered as an elasto-plastic material. Then the soil
4. DYNAMIC RESPONSE OF PILES
The external loads on an offshore structure consist for a large part of
dynamic forces, due to wave action. On the North Sea, and in many other
locations in the world, these dynamic loads have a typical period of the
order of magnitude of 10 seconds. In this chapter the importance of certain inertia effects for a pile foundation are investigated.
4.1 Axially loaded pile
In the case of an axially loaded pile, supported by friction of the soil, the basic differential equation for the axial displacement w of the pile is, when ,inertia of the pile is taken into account
EA::: -
cOw = pA : : :where p is the density of the pile material. It is asswned that the load is periodic, with angular frequency w, so that the response of the pile is also periodic, with the same frequency, Then one may write
w = W(z)exp(iwt)
The differential equation for W(z), the complex amplitude of the
displace-ment is
EA d2_W - cO[l- pAcOw2 )W = 0
dz2
The relative importance of inertia of the pile can be investigated by
esti-mating the value of the dimensionless factor pAw2/cO. If this is a small
quantity inertia effects can be disregarded. If it is greater than 1 the
basic behaviour changes from exponentially damped to sinusoidal fluctua-tions. If this factor is equal to 1 resonance can be considered to occur, As mentioned before the usual type of an offshore pile is a steel tubular pile. For such a pile the area divided by the circumference is about equal
to the wall thickness (A/O=d). Furthermore the subgrade modulus c can be
estimated as c=Es/D, where Es is the modulus of elasticity of the soil, and
D the pile diameter, The modulus of.elasticity of the soil can be
expres-sed, approximately, in terms of Terzaghi's compression factor C by the
correlation Es = C<T
where <T is the stress level in the soil. This can be estimated to be
<T = ps gz
where ps is the (effective) soil density, g is the gravity constant, and z is the local depth, One now obtains
pAw2 =
Q!. dDw2 cO p Cgz
Thus the resonance frequency wo is wo = v((ps/pp)Cgz/(dD)]
This seems to be a large quantity, In order to obtain a low estimate one may assume that ps/pp=0.5, that C=50, z=D, anq that d=O,l m. Then one
ob-tains wo=50/s. The major frequency to be considered is the one
correspon-ding to a wave having a period of about 10 seconds, The angular frequency w then is about 0.6/s, This is much smaller than the critical frequency fJo, The conclusion can be that loading conditions with a period greater than about 0.5 seconds can be considered as quasi-static, i.e. inertia effects of the pile can be disregarded.
4,2 Laterally loaded pile
A similar consideration for a laterally loaded pile leads to a formula for the critical (resonance) frequency fJo of the following form
wo = v[k/(ppA)
Using the same estimations as used in the previous paragraph this can also be written as
wo = v[(ps/pp)Cgz/(2ndD)J
With C=50, z=D and d=O.l m the critical frequency is now found to be about 28/s, which means that for lateral loading inertia effects of the pile can also be disregarded.
The general conclusion from this paragraph can be that for the analysis of the response of a pile foundation of an offshore structure to wave action, inertia effects of the pile can usually be disregarded. The analysis can be static or quasi-static.
5. ELASTO-PLASTIC AXIAL RESPONSE
For a realistic analysis of the load transfer from the top of a pile to the soil it is necessary to include the main non-linear effect, which is the possibility of local slip of the pile relative to the soil. The skin fric-tion between pile and soil in general is limited in both direcfric-tions, and these limiting values are usually denoted as positive and negative skin
friction. The positive friction can be determined in a relatively easy way
by the cone penetrometer test (CPT). In this chapter a simple model
incor-porating elasto-plastic deformations for an axially loaded pile is presen-ted.
!h_L_pescription of th~ model
The simplest elasto-plastic model assumes a linear branch between two limi-ting values, as shown in figure 5.1.
tp ... , , ~
-DisplaceMent
- - - ·
· · · ta
Figure 5,1. Elasto-plastic model.
The response is characterized by three parameters the maximum positive
friction (tp), the maximum negative friction (ta), and a spring constant. This spring constant can conveniently be described by the length of the
elastic branch. This represents the displacement necessary to generate the
maximum (positive or negative) friction. This value, often denoted as the
quake, can often be estimated more easily from a soil investigation than the
spring constant itself, The use of this quantity has the additional
advan-tage that the rigidity of the soil will automatically increase with depth if the quake is constant, and the maximum friction is increasing with depth, which is often observed in practice. It may be expected that the length of the linear branch varies within narrower bounds than the spring constant, and therefore spring constants will be defined through parameters such as
the quake in this report,
The values of the maximum positive and negative friction are usually almost
equal. It is sometimes observed, however, that the negative friction is
somewhat smaller than the positive friction, and this may be attributed to
the fact that negative friction is usually associated with tension in the pile, during which the pile contracts, whereas positive friction in the pile
occurs ·during compression, with the pile dilating laterally. In order to
take this possibility into account the soil will be characterized by inde-pendent parameters for positive and negative friction, see figure 5.2.
Shear stress
tp ... ·
-wa
DisplaceMent
WP
ta
Figure 5.2. Asymmetric elasto-plastic model.
In figure 5.2 the positive and negative quake (wp and wa) are considered to
be completely independent. A mathematical formulation of this
elasto-plastic behaviour is as follows (in PASCAL). if (w>O) and (w~wp) then t:=-tp*w/wp; if (w>=wp) then t:=-tp;
if (w<O) and (w>-wa) then t:=ta*w/wa; if (w<=-wa) then t:=ta;
The sign convention used here is that the displacement w is positive in downward direction (the positive z-direction), and that the shear stress t acting on the pile is also positive when acting in downward direction,
A complete formulation of the model should also be applicable during unloa-ding and subsequent reloaunloa-ding, and therefore this behaviour should also be
specified, The formulation given above, when incorporated in a computer
program, merely describes the behaviour in monotonic loading, and does not
correctly describe the possibility of permanent plastic displacements. The
general notion of plastic deformations is that the response in unloading
(and in subsequent reloading up to the previous stress level) follows a
branch parallel to the elastic branch. This can most conveniently be
formu-lated by introducing a possible shift of the origin, which occurs during any
plastic deformation. In order to take into account the different behaviour
in the positive and negative branches one may consider the a summation of two independent elasto-plastic springs, one tp and an initial value wo, and the other described by wa, complete fonnulation now is as follows.
if (w-wo>O) and (w-wo<wp) then t:=-tp*(w-wo)/wp; if (w-wo>=wp) then
begin t:=-tp; wo:=w-wp end;
if (w-wo<O) and (w-wo>-wa) then t:=ta*(w-wo)/wa; if (w-wo<=-wa) then begin t: =ta; wo:=w+wa end; soil response as described by wp, ta and wo. A
As can be seen from this algorithm the value of wo (which is initially zero) is changed whenever plastic slip occurs, so that the length of the elastic branch remains constant during the deformation history,
6 ELASTO-PLASTIC LATERAL RESPONSE
In this paragraph the response of the soil to a laterally loaded pile is
considered. Two possibilities of describing this response will be
presen-ted, and these will be correlapresen-ted, The mathematical description of the
models will be given in PASCAL.
6.1 Activ~ and passive soil pressures
An elementary way to describe the soil response is by using quantities such as active, passive and neutral soil pressure, which are widely used in
classical soil mechanics, In this model it is postulated that a
displace-ment of a pile eledisplace-ment, say towards the right, will generate a local respon-se on both sides of the pile. On the right side a force will act towards the left, and this force cannot be smaller than the active pressure (fa),
and not greater than the passive pressure (fp). Similarly a force acting
towards the right will be acting on the left side of the pile, and this force cannot be smaller than fa, and not greater than fp, Furthermore it is assumed that the soil pressures in the case of zero displacement are fn
(neutral soil pressure) on both sides. For intermediate values of the
dis-placement the relationship between force and disdis-placement is assumed to be
linear, The force acting on the right side (fr) can now be described as
follows.
if (u>-ua) and (u<up) then fr:=-fn-(fp-fa)*u/du; if (u>=up) then fr:=-fp;
if (u<=-ua) then fr:=-fa; in which O<fa<fn<fp, du>O, and
ua:=du*(fn-fa)/(fp-fa); up:=du*(fp-fn)/(fp-fa);
The fundamental parameters are fa,fn,fp and du, where du is the total length of the linear branch, to be denoted as the stroke, This is the displacement
necessary to pass from active to passive pressure. The minus sign in the
expressions for fr indicate that this force is acting in negative x-direc-tion.
In a similar way the force on the left side is described by the following algorithm.
if (u<ua) and (u>-up) then fl:=fn-(fp-fa)*u/du; if (u<=-up) then fl:=fp;
if (u>ua) then fl:=fa;
This force acts in positive x-direction.
In order to complete the formulation it is again necessary to also specify
during unloading the non-linear behaviour is such that the soil reacts elastic (stiff), This is described by introducing two additional parameters ur and ul, which express the pennanent displacements of the right and left side springs, respectively. The complete formulation is now as follows.
if (u-ur>-ua) and (u-ur<up) then fr:=-fn-·(fp-fa)*(u-ur)/du; if (u-ur>=up) then begin fr: =-fp; ur:=u-up end; if (u-ur<=-ua) then begin fl: =-fa; ur:=u+ua end;
if (u-ul<ua) and (u-ul>-up) then fl:=fn-(fp-fa)*(u-ul)/du; if (u-ul<=-up) then begin fl: =fp; ul:=u+up end; if (u-ul>=ua) then begin fl: =fa; ul:=u-ua end; f: =fr+fl;
The physical meaning of the parameters ul and ur is the value of the displa-cement for which the soil pressure acting on the left or right side of the
pile is equal to the neutral value. These values constantly change during
plastic deformation. In a complete numerical model, in which the response of a pile to a varying lateral load is calculated, the values of ul and ur must be updated after each loading step.
§..2 API
A description which more closely follows the Recommendations of the American Petroleum Institute (API) is to assume that the total response of the soil can be described by ·a force-displacement relation consisting of two linear branches and a constant maximum value. Such a relation can be characterized by the location of two points in the force displacement diagram, see figure
6. l. The bi-linear elasto-plastic relation ·is characterized by the
coor-dinates of the points 1 and 2: ul and fl, respectively u2 and f2. It
should be noted that the directions of displacement and force in general are
opposed; hence in a formal mathematical formulation such as used in the
previous paragraph, a minus sign will appear.
If it is now assumed that the total response is generated by a physical model as described in the previous paragraph, it follows that
ul:=ua; u2:=up;
fl:=-fa+fn+(fp-fa)*ua/du; f2:=fp-fa;
These relations can be established by noting that the discontinuity in point 1, assuming a displacement to the right, must be due to the fact that on the
NorMal stHss
Displacei.ent
Figure 6.1. API-model.
left side the minimum (active) soil pressure is reached. The second
discon-tinuity, in point 2, corresponds to reaching the maximum (passive) soil
pressure on the right side. Thus the API model seems to be a
straightfor-ward representation of the physical model used in developing the algorithms in the previous paragraph.
Relations between parameters
The two models seem to be fully equivalent, both characterized by four parameters: fa, fn, fp and du for the first model, and fl, f2, ul and u2 for the second model. This is not the case, however, as can be seen from a more detailed inspection of the properties of the two modeis.
It follows from the description of the first model that (fn-fa)=(fp-fa)*ua/du,
from which it can be derived that fl=2*(fp-fa)*ul/du,
f2=fp-fa.
2*(f2-fl)/(u2-ul)=fl/ul.
This means that the slope of the first branch of the force-displacement
relation is twice the slope of the second branch. Actually this is n9t so
surprising, taking into account the physical model underlying the basic
formulas, which consists of two elasto-plastic springs. In the first branch
both springs are elastic, and in the second branch only one elastic spring
remains.
It is now assumed that the basic (physical) model is valid, so that the
relationship obtained above must indeed be satisfied. This implies a
res-triction in the API model such that the first slope is exactly twice the
second slope, which entails that only three parameters remain, say ul, fl
and f2. The value of the displacement u2 now follows from the condition u2:=ul+2*ul*(f2-fl)/fl;
It remains to explain the apparent occurrence of four parameters in the original first model. This can be done by first noting that it follows from the description of the two models that
ua=ul, up=u2, du=up-ua.
Furthermore it can be seen that fl/ul=2*(fp-fa)/du,
which means that the value of (fp-fa) can be determined from fl and ul, (fp-fa)=0.5*fl*du/ul.
The value of (fn-fa) can also be determined, because (fn-fa)=(fp-fa)*ua/du.
One of the three values fa, fn or fp can be chosen arbitrarily (e.g. fa=O)
without affecting the total force on the pile. Actually this is physically
fully acceptable: the first model indeed contains only three parameters,
because the total soil response is composed of a superposition of the forces left and right, which means that a constant value may be added to the force
on both sides without affecting the resultant force. It can be concluded
that only the differences (fn-fa) and (fp-fa) are physically meaningful.
6.3 Generalized elasto-plastic model
The elasto-plastic model introduced in section 6,1 has the advantage that
its parameters have a physical meaning which closely follows classical soil
mechanics, using parameters such as active and passive soil pressure. This
model can be generalized so that it covers all the possibilities of the API model in the following way.
A model with an elastic branch which is limited on both sides is in general a schematization of a curved relation between displacement and soil
pres-sure. A possible refinement in the schematization is to assume that the
slope of the linear branch is different in the active and passive sections, see figure 6.2.
The model illustrated in this figure contains four basic parameters, namely the values of ua, up, fp-fn, and fn-fa. As stated before the actual value of the parameter fn can be considered to be irrelevant, because in the neutral state this stress acts on both sides of the pile.
NofMal
stress
'tp·''' ''' '''
ta
DisplaceMent
M ~
Figure 6.2, Generalized elasto-plastic model.
The model illustrated in figure 6.2 can be described mathematically as
follows.
if (u-ur>=O) and (u-ur<up) then fr:=-fn-(fp-fn)*(u-ur)/up; if (u-ur>=up) then
begin fr:=-fp; ur:=u-up end;
if (u-ur<=O) and (u-ur>-ua) then fr:=-fn-(fn-fa)*(u-ur)/ua; if (u-ur<=-ua) then
begin fr:=-fa; ur:=u+ua end;
if (u-ul<=O) and (u-ul>-up) then fl:=fn-(fp-fn)*(u-ul)/up; if (u-ul)<=-up) then
begin fl:=fp; ul:=u+up end;
if (u-ul>=O) and (u-ul<ua) then fl:=fn-(fn-fa)*(u-ul)/ua; if (u-ul>=ua) then
fl:=fa; ul:=u-ua end;
f:=fl+fr;
The relationship with the API model can be established by noting (see also the previous section) that
ul=ua, u2=up,
fl=(fn-fa)+(fp-fn)*ua/up, f2=fp-fa.
The value given for fl can be derived from the description
given above, with u-ur=ua and u-ul=ua. In that case fl=fa
fn)*ua/up. By adding these values (and changing the sign)
for fl is obtained.
The inverse relations are ua=ul, up=u2, (fp-fn)=(f2-fl)*up/(up-ua), (fn-fa)=(up*fl-ua*f2)/(up-ua). of the model and fr=-fn-(fp-the expression
All these quantities can be shown to be positive (as required), provided
that the general shape of the API curve is as shown in paragraph 5.2, with the slope of the seconq branch being smaller than the slope of the first branch, which is almost self-evident.
The generalized model presented in this section has been found to be fully
equivalent to the API model. Because its parameters have been defined in
terms of familiar soil properties such as active, passive and n~utral soil
properties, it has been found to be possible to formulate the behaviour of
the model for a general type of deformation, including the possibilities of
7. A NUMERICAL MODEL FOR AXIALLY LOADED PILES
In this section complete numerical model for the transfer of axial loads to the soil is developed. Load transfer takes place through friction along the
shaft of the pile, and at the point by point resistance. The
characteris-tics of both forms of stress transfer are in general non-linear. An elemen-tary computer program will be presented.
7.1 Basic equations
Consider a pile of length L, loaded at its top by a force P, see figure.7.1. The pile is subdivided into N small elements, each having a length D(I), and axial stiffness EA(I). The elements are numbered from 1 to N, and the nodal points are numbered from Oto N, so that element I is located between points I-1 and I. The basic equations can be developed in the following way.
1-1
f(l-1) ... (
-( R-(I)
- - ) f(I)
Figure 7.1. Pile with axial load.
The forces acting on element I are the normal forces F(I-1) and F(I) at the
two.ends, and a friction force R(I), acting along the shaft. Equilibrium
requires that
F(I)-F(l-1)-R(I)=O, (I=l. .. N).
The strain in this element is determined by the average normal force N.( I),
N(I)=(F(I-l)+F(I))/2.
With Hooke's law the increment of the length of this element is DW(I)=N(I)*D(I)/EA(I),
where D(I) is the length of the element, and EA(!) its axial stiffness. Because DW(I)=W(I)-W(I-1), where W(I) is the displacement of point I, one may write
(EA(I)/D(I))*(W(I)-W(I-l))=N(I)=(F(I)+F(I-1))/2,
(EA(I+l)/D(I+l))*(W(I+l)-W(I))=N(I+l)=(F(I+l)+F(I))/2,
Subtraction of these two equations gives, after multiplication by -2,
where
A(I+l)*W(I+l)+A(I)*W(I)+A(I-l)*W(I-l)=-F(I+l)+F(I-1),
A(I+l)=-2*EA(I+l)/D(I+l),
A(I)=2*EA(I+l)/D(I+l)+2*EA(I)/D(I), A(I-1)=-2*EA(I)/D(I).
The forces F(I+l) and F(I~l) can be related to the friction forces R(l+l)
and R(I) by the equations of equilibrium F(I+l)-F(I)=R(I+l),
F(I)-F(I-l)=R(I),
It follows from these two equations that F(I+l)-F(I-l)=R(I)+R(I+l),
The basic equation can now be written as follows.
A(I+l)*W(I+l)+A(I)*W(I)+A(I-l)*W(I-1)=-R(I)-R(I+l).
This will lead to a linear system of equations, if the forces R(I) are
given. In general these friction forces depend upon the local displacement,
as described in section 4. It is assumed that the friction force is deter-mined by the average displacement V(I), defined as
V(I)=(W(I-l)+W(I))/2.
Furthermore it is assumed that in the linear branch the friction force is
proportional to the displacement difference V(I)-WM(I), where WM(I) is a
memory function, initially zero, which represents the permanent
displace-ments that have occurred during plastic deformations, The formulation of
the model is as follows.
if (V(I)-WM(I)>-B(I)) and (V(I)-WM(I)<B(I)) then begin
S(l):=C(l)/B(I); T(I):=O
if (V(I)-WM(I)<=-B(I)) then begin S(I):=O; T(I):=-C(I) end; if (V(I)-WM(I)>=B(I)) then begin S(I):=O; T(I):=C(I) end; R(I):=S(I)*(V(I)-WM(I))+T(I);
Here C(I) is the maximum value of the friction (here assumed to be
inde-pendent of the direction of the displacement), B(I) is the quake, i.e. the
displacement necessary to generate the maximum friction force. For reasons
of simplicity this parameter has also been assumed to be independent of the direction of the displacement. The parameter S(I) is a spring constant, and T(I) is a possible given force.
After completion of each lpading step the memory function WM(!) should be
updated as follows.
if (V(I)-WM(I)<=-B(I)) then WM(I):=V(I)+B(I); if (V(I)-WM(I)>=B(I)) then WM(I):=V(I)-B(I);
In this way it is ensured that during unloading after plastic deformation the response is again elastic, and that in further reloading the response is elasto-plastic.
The general expression for the shear force R(I) is R(I)=S(I)*(V(I)-WM(I))+T(I), or where R(I)=S(I)*V(I)+TT(I), V(I)=(W(I-l)+W(I))/2, TT(I)=T(I)-S(I)*WM(I). The system of equation now becomes
where AA(I+l)*W(I+l)+AA(I)*W(I)+AA(I-l)*W(I-1)=-TT(I)-TT(I+l), (I=l •.• N-1), AA(I+l)=A(I+l)+S(I+l)/2, AA(I)=A(I)+S(I)/2+S(I+l)/2, AA(I-l)=A(I-l)+S(I)/2. . I
The system of equations is now completely defined for all interior points. The boundary conditions will next be specified,
It is assumed that at the pile top the external force is given. With
Hooke's law for the first element one may write (EA(l)/D(l))*(W(l)-W(O))=(F(O)+F(l))/2.
The normal force F(l) can be eliminated from this equation by using the equilibrium pondition for element 1,
F(l)-F(O) =R(l).
One now obtains, with F(O)=-P, and with R(l)=S(l)*V(l)+TT(l), (2*EA(l)/D(l)+S(l)/2)*W(0)+(-2*EA(l)/D(l)+S(l)/2)*W(l)=
=2*P-TT(l).
This equation can be added to the system of equations. It can be considered as equation number 0,
Pile point
At the point of the pile a resistance is assumed to be generated by any
downward displacement. This resisting force is written as F(N)=-CP*W(N)-FP,
where CP=FM/WM is a spring constant, and where FP is either zero (in the
elastic branch, for small displacements) or FM (a given constant) if the
displacement is greater than \'JM, the quake. The mathematical formulation of this elasto-plastic response is as follows.
if(W(N)-WP>O) and (W(N)-WP<\'JM) then begin CP:=FM/\'JM; FP:=O end; if (W(N)-WP<=O) then begin CP:=O; FP:=O end; if (W(N)-WP)=\'JM) then begin CP:=O; FP: =FM end;
Here it has been assumed that no tension can be transmitted at the pile point.
After each loading step the memory function WP should be updated, to account for the permanent plastic deformations,
if (W(N)-WP<=O) then WP:=W(N); if (W(N)-WP)=\'tM) then WP:=W(N)-\'tM;
The boundary condition can be introduced by starting from a consideration of the deformation of the last element,
(EA(N)/D(N))*(W(N)-W(N-l))=(F(N-l)+F(N))/2.
The normal force F(N-1) can be eliminated by using the equation of equili-brium of the last element,
F(N)-F(N-l)=R(N). This gives
(2*EA(N)/D(N))*(W(N)-W(N-1))=2*F(N)-R(N).
Now by using the relation F(N)=-CP*W(N)-FP and expressing R(N) into W(N-1) and W(N) the following result is obtained
(2*EA(N)/D(N)+2*CP+S(N)/2)*W(N)+
+(-2*EA(N)/D(N)+S(N)/2)*W(N-l)=-2*FP-TT(N).
This is the final formulation of the boundary condition at the pile point. Again an equation in terms of the basic variables W(I) has been obtained,
with given forces on the right hand side. This equation can be considered
to be equation number N.
The system of equations, for I=O ... N, can be solved numerically if all the
coefficients are known. Because of the non-linear spring behaviour this
must be done iteratively, by first assuming elastic or plastic behaviour in each element, and subsequent correction, if necessary.
7.2 Computer program
An elementary computer program (ALP-1,0) that performs the calculations des-cribed above, is given in,this section, The program has been kept as simple as possible, by a restriction to a homogeneous pile in a homogeneous soil. The handling of input and output is also very simple, without any possibili-ties of correction of typing errors, or possible output on a printer. In the program all input is performed interactively, with the user responding to
question prompts from the program. As these questions are self-explanatory
no further manual is needed. Output is given on the screen of the computer, in the form of a list of values for each node, After presentation of output
data the program will ask for a new value of the axial load. The program
can only be stopped by an external interrupt, for instance by BREAK or AC.
The program has been written in Microsoft BASIC, and has been tested ·and
used successfully on various computers. No responsibility for any errors
Program ALP-1.0
1000 REM===== ALP-1.0
---1010 GOSUB 2000:DEFINT I-N:B$=" Impossible":Z=4
1020 DIM X(lOO),W(lOO),F(lOO),D(lOO),S(lOO),T(lOO),WM(lOO) 1030 DIM EA(lOO),SS(lOO),TA(lOO),TB(lOO),K%(100,4),P(l00,4) 1040 DIM DA(lOO') ,DB(lOO) ,NP(lOO)
1050 INPUT"Length of the pile (m) . . . . • "; WL 1060 IF WL<=O THEN PRINT B$:GOT0 1050
1070 INPUT"Axial stiffness EA (kN) . , .• ";EA 1080 IF EA<=O THEN PRINT B$:GOT0 1070
1090 INPUT"Circumference (m) •••••••••• ";00 1100 IF OO<=O THEN PRINT B$:GOT0 1090
1110 INPUT"Maximum friction (kN/m"2) , . "; FR 1120 IF FR<O THEN PRINT B$:GOT0 1110
1130 INPUT" Quake (m) .. .. • .. .. .. .. .. "; QR 1140 IF QR<=O THEN PRINT B$:GOT0 1130
1150 INPUT"Point resistance (kN) ,, .... ";FM 1160 IF FM<O THEN PRINT B$:.GOT0 1150
1170 INPUT" Quake (m) ... , "; UM 1180 IF UM<=O THEN PRINT B$:GOT0 1170
1190 INPUT"Nwnber of elements ...•... ";N 1200 IF N<lO THEN N=lO ELSE IF N>lOO THEN N=lOO 1210 DA=WL/N:TA=DA*OO*FR:SA=TA/QR:CP=FM/UM:X(O)=O 1220 FOR I=l TO N:D(I)=DA:EA(I)=EA:SS(l)=SA:T(I)=TA
1230 TA(I)=TA:TB(I)=TA:DA(I)=QR:DB(I)=QR:X(I)=X(I-l)+DA:NEXT I 1240 GOSUB 2000:INPUT"Force (kN) •••.••••.•... ";FA
1250 KP=O:NQ=O:GOSUB 2000
1260 IF IQ()O AND ABS(FA)>=FB THEN 1290 1270 MP=O:CP=FM/UM:FP=O
1280 FOR I=l TO N:S(I)=SS(I):NP(I)=O:T(I)=-SS(I)*WM(I):NEXT I 1290 PRH~T"Force (kN) •... , ... : "; FA: FB=ABS(FA)
1300 FOR I=O TO N:FOR J=l TO Z:K%(I,J)=O:P(I,J)=O:NEXT J,I 1310 IQ=IQ+l:FOR I=l TO N-1
1320 K%(I,l)=I:K%(I,2)=I-l:K%(I,3)=I+l:K%(I,Z)=3:NEXT I
1330 K%(0,l)=O:K%(0,2)=1:K%(0,Z)=2:K%(N,l)=N:K%(N,2)=N-l:K%(N,Z)=2 1340 PRINT:PRINT"Generation of matrix ••• : +";
1350 FOR I=l TO N-l:Al=2*EA(I)/D(I):A2=2*EA(I+l)/D(I+l) 1360 Bl=S(I)/2:B2=S(I+l)/2:PRINT 11
+11 ;
1370 P(I,l)=Al+A2+Bl+B2:P(I,2)=-Al+Bl:P(I,3)=-A2+B2 1380 P(I,Z)=-T(I)-T(I+l) :NEXT !:PRINT "+"
1390 Al=2*EA(1)/D(l):B1=S(l)/2
1400 P(O,l)=Al+B1:P(0,2)=-Al+Bl:P(O,Z)=2*FA-T(l) 1410 Al=2*EA(N)/D(N):B1=S(N)/2:Cl=2*CP
1420 P(N,l)=Al+Bl+Cl:P(N,2)=-Al+B1:P(N,Z)=-2*FP-T(N) 1430 PRINT: PRINT"Elimination •. , •..••. , .. : ";
1440 FOR I=O TO N:PRINT "+";:KC=K%(I,Z):IF KC=l .THEN 1570 1450 FOR J=2 TO KC:C=P(I,J)/P(I,l):JJ=K%(I,J)
1460 P(JJ,Z)=P(JJ,Z)-C*P(I,Z):L=K%(JJ,Z) 1470 FOR JK=2 TO L:IF K%(JJ,JK)=I THEN 1490 1480 NEXT JK
1490 K%(JJ,JK)=K%(JJ,L):K%(JJ,L)=O
1500 P(JJ,JK)=P(JJ,L):P(JJ,L)=O:L=L-l:K%(JJ,Z)=L
1520 NEXT IJ:L=L+l:IJ=L 1530 K%(JJ,Z)=L:K%(JJ,IJ)=K%(I1II) 1540 P(JJ,IJ)=P(JJ,IJ)-C*P(I,II) 1550 NEXT II 1560 NEXT J 1570 C=l/P(I,l):FOR J=l TO KC:P(I,J)=C*P(I,J):NEXT J 1580 P(I,Z)=C*P(I,Z):NEXT I 1590 PRINT:PRINT:PRINT"Back substitution ....•. : ";
1600 FOR I=O TO N:J=N-I:PRINT "+";:L=K%(J,Z):IF L=l THEN 1620 1610 FOR K=2 TO L:JJ=K%(J,K):P(J,Z)=P(J,Z)-P(J,K)*P(JJ,Z):NEXT K 1620 NEXT I:PRINT:F(O)=-FA:FOR I=O TO N:W(I)=P(I,Z):NEXT I
1630 FOR I=l TO N:F(I)=2*EA(I)*(W(I)-W(I-l))/D(I)-F(I-l):NEXT I· 1640 PRINT:PRINT"Check springs";:LL=O
1650 FOR I=l TO N:A=(W(I)+W(I-1))/2:AA=A-WM(I):IF NP(I)=O THEN 1690 1660 IF NP(I)=l THEN 1720
1670 IF AA<=-DB(I) THEN 1740
1680 S(I)=SS(I):T(I)=-SS(I)*WM(I):NP(I)=O:LL=LL+l:GOTO 1740 1690 IF AA>=-DB(I) AND AA<=DA(I) THEN 1740
1700 LL=LL+l:IF AA>=DA(I) THEN NP(I)=l:S(I)=O:T(I)=TA(I):GOTO 1740 1710 NP(I)=-1:S(I)=O:T(I)=-TB(I):GOTO 1740
1720 IF AA>=DA(I) THEN 1740
1730 S(I)=SS(I):T(I)=-SS(I)*WM(I):NP(I)=O:LL=LL+l 1740 NEXT I:AA=W(N)-WP:IF MP=O THEN 1790
1750 IF MP=l THEN 1780 1760 IF W(N)<=WP THEN 1820
1770 MP=O:CP=FM/UM:FP=-CP*WP:LL=LL+l:GOTO 1820 1780 IF AA>=UM THEN GOTO 1820 ELSE GOTO 1770 1790 IF AA<=UM AND W(N)>=WP THEN 1820
1800 LL=LL+l:IF AA>=UM THEN MP=l:CP=O:FP=FM:GOTO 1820 1810 MP=-1:CP=O:FP=O
1820 I=O:IF LL=O THEN GOTO 1870 ELSR IF NQ=lOO THEN GOTO 1860 1830 NQ=NQ+l:LP=O:FOR I=O TO N:IF NP(I)<>O THEN LP=LP+l
1840 NEXT I:GOSUB 2000:PRINT"Plastic springs •.••..•• :";LP 1850 PRINT:GOTO 1290
1860 GOSUB 2000:PRINT"No convergence":GOSUB 1980:CLS:END
1870 GOSUB 2000: PRINT" I Z W N"
1880 A$="### ####,### ####,#### ######,##":PRINT
1890 FOR I=O TO N:PRINT USING A$;I,X(I),W(I),F(I):NEXT !:PRINT 1900 A$=11####,####11:PRINT"Top: F = ";:PRINT USING A$;FA
1910 PRINT" W = 11
; : PRINT USING A$; W(O): PRINT
1920 GOSUB 1980:FOR I=l TO N:A=(W(I)+W(I-1))/2:AA=A-WM(I) 1930 IF AA>=DA(I) THEN WM(I)=A-DA(I)
1940 IF AA<=-DB(I) THEN WM(I)=A+DB(I) 1950 NEXT I:IF W(N)>=WP+UM THEN WP=W(N)-UM 1960 IF W(N)<=WP THEN WP=W(N)
1970 GOTO 1240
1980 LOCATE 25,25,0:COLOR 0,7:PRINT" Touch any key to continue"; 1990 COLOR 7,0:A$=INPUT$(1):RETURN
2000 CLS:LOCATE 1,31,0:COLOR 0,7
7.3 Example
As an example the following case will be considered. A pile of 50 m length
has been installed in a homogeneous soil. The soil properties are a maximum
friction of 50 kN/m2 , with a corresponding quake of 0.01 m. Point resistance
is neglected. The pile diameter is 1,00 m, and its wall thickness is
0.05 m. This means that the orea of the pile section is 0.1492 m2 , and that
the circumference of the pile is 3.1416 m. The load-displacement curve for
this pile, as calculated by a computer program of the type presented above, is shown in figure 7.2. I ·FMax Forice I
DisplaceMent
, ·
Figure 7.2. Example,It is interesting to note that plastic deformations start to occur only for
large forces, close to failure. Actually this could have been expected by
considering the load transfer just before failure. The maximum force is
50 kN/m2*50 m*3.1416 m = 7854 kN. Failure will occur if in the last spring,
at the bottom of the pile, the displacement reaches the quake, i.e. 0.01 m.
Then the average force in the pile is 3927 kN, and the elastic deformation
of the pile corresponding to this force is 0.00627 m. This means that the
displacement just before failure is 0,01627 m, whereas the displacement
necessary to cause the first spring to fail is 0.01 m, which is reached only
for a force of 6000 kN. Apparently this pile is relatively stiff. In
engineering practice it is important to compare the.deformation of the pile
itself (at failure) with the soil deformations at failure. If the pile
deformations are relatively small the pile behaves almost as an infinitely
stiff body, and in a homogeneous soil failure may occur rather suddenly. In
cases like this, with a relatively stiff pile, the analysis of the stress
distribution can directly be done, without using a computer program. The
displacement of the pile will be such that the total soil reaction,
gene-rated all along the pile, equals the total load.
Another interesting observation from the example is that the behaviour in
unloading and subsequent re-loading is linear. In the terminology of
plas-ticity theory this can be called shakedown. The permanent deformations are
occur, provided that the maximum load does not exceed the previous maximum value. It should be realized that this is a feature of the present model,
in nature this phenomenon is usually not observed, and therefore it is a
shortcoming of the model,
It should be noted that the program presented above is restricted to a pile in a homogeneous soil. In a more advanced version of the program (ALP-1.3) the soil may consist of a number of layers with different properties. Input is stored in a datafile on diskette, which enables to inspect the data, and
if necessary, to correct or modify them. Input data can be created and
modified interactively. Output can be given either on the screen of the
terminal, or on a line printer. This program is written, for the IBM
Personal computer or a similar computer using MS-DOS, in compiled Basic,
using Microsoft's BASCOM compiler. The program is distributed by the
8. A NUMERICAL MODEL FOR 1ATERALLY L9ADED PILES
In this section a complete numerical model for the transfer of lateral loads to the soil is developed. Load transfer takes place along the shaft of the pile to the soil on either side. The characteristics of this form of stress
transfer will be considered non-linear, as described in paragraph 5.3. An
elementary computer program will also be presented,
8.1 Basic equations
Just as for the case of axial loading the pile 'is subdivided into N small elements, see figure 8.1.
Pm
l
~<l+ll
Q( I) ... , ... · ,1:, :· I\ ~·.,\':/ ,:,i., ,,.._~,~·:· ~lo:---+-i..~~;,++---.lio~
'•i· .. .: .. ,,. ,· Q(I 1)
.
z
,::,,:.,,<H'f I+
:c.~ ...
·.i,·?.'t(l+l)
Figure 8.1. Element of beam.
The element located between points Z(I) and Z(I+l) is loaded by a
distri-buted load, the resulting force of which is denoted by F(I+l). It is
assumed that the elements are small enough that this force can be considered
to be acting in the center of the element. Another possible load is a
concentrated force P(I), acting immediately to the right of point I. The
soil reaction is a distributed load, which is assumed to be acting in the center of the element, and the magnitude of which depends upon the average displacement of the element.
Equilibrium of forces in the direction perpendicular to the axis of the pile now requires, for element I+l,
Q(I+l)-Q(I)+F(I+l)+P(I)=R(I+l), where R(I+l) represents the soil reaction. equilibrium is
Q(I)-Q(I-l)+F(l)+P(I-l)=R(I).
Equilibrium of moments for element I+l requires that M(I+l)-M(I)=(Q(I+l)+Q(I)-P(I))*D(I+l)/2. Similarly for element I
M(I)-M(I-l)=(Q(I)+Q(I-1)-P(I-l))*D(I)/2. It follows from these last two equations that
Q(I+l)-Q(I-l)=P(I)-P(I-1)+2*(M(I+l)-M(I))/D(I+l)-2*(M(I)-M(I-1))/D(I). Similarly from the first two equations one obtains, by addition of the two equations
Q(I+l)-Q(I-l)=R(I+l)+R(I)-F(I+l)-F(I)-P(I)-P(I-1).
The shear forces can now be eliminated from the equations, which leads to the following equation in terms of bending moments only
A(I+l)*M(I+l)+A(I)*M(I)+A(I~l)*M(I-l)+R(I+l)+R(I)=F(I+l)+F(I)+2*P(I), (8,1) where
A(I+l)=-2/D(I+l), A(I)=2/D(I+l)+2/(D(I), A(I-1)=-2/D(I).
Equation (8.1) is the numerical equivalent of the equation of equilibrium for the moments in the analysis of bending of a beam,
d2M/dZ2=-f,
If all intervals are of the same length the standard molecule
finite differences (-1,2,-1) is obtained, because then
A(I)=4/D, and A(I+l)=-2/D,
for central A(I-1)=-2/D,
In eq. (8.1) the soil resistance R will be dependent upon the lateral
displacement U. In general this will be a non-linear relation, which will
be specified later. At this point it suffices to note that in each node
there are two basic variables, U(I) and M(I), and therefore a second
equa-tion is needed to complete the model. This second equation can be obtained
from a consideration of the deformation of the beam itself. If the beam
material is linear elastic this leads to the classical bending equation
d2U/dZ2=-M/EI, (8.2)
where EI is the bending stiffness of the beam: This equation will next be
approximated, taking special care to include the possibility of irregular
intervals, because these intervals may be determined by the natural varia-tions in the soil properties.
In order to derive a numerical expression for the bending equation a part of the beam between nodes I-1 and I+l is considered, see figure 8.2.
u
2
1-1
1+1
Figure 8.2. Bending of beam.
For simplicity it is assumed that the origin Z=O is located in node I. The deflection of the beam is now assumed to be as follows.
Z(I-l)<Z<Z(I) U=U(I)+A*Z+B*Z*Z, Z(I)<Z<Z(I+l) U=U(I)+A*Z+c*Z*Z,
where A, Band Care constants. It can be seen that this approximation
ensures that the displacement U and its first derivative are continuous in Z=O. For Z=Z(I-1) and Z=Z(I+l) one now obtains
U(I-l)=U(I)-A*D(I)+B*D(I)*D(I),
U(I+l)=U(I)+A*D(I+l)+C*D(I+l)*D(I+l), and hence, after elimination of A,
(U(I+l)-U(I))/D(I+l)-(U(I)-U(I-1))/D(I)=B*D(I)+c*D(I+l),
On the other hand the second derivative in the interval between I-1 and I is 2*8, and in the interval between I and I+l is 2*C, and these can be related to the average bending moment in these intervals as follows,
Z(I-l)<Z<Z(I) d2U/dZ2=2*B=-(M(I-l)+M(I))/(2*EI(I))
Z(I)<Z<Z(I+l) d2U/dZ2=2*C=-(M(I)+M(I+l))/(2*EI(I+l))
Using these expressions for Band C the final equation is
where
B(I-l)*U(I-l)+B(I)*U(I)+B(I+l)*U(I+l)=
B(I-1)=-l/D(I), B(I)=l/D(I)+l/D(I+l), B(I+l)=-1/D(I+l), C(I-l)=D(I)/(4*El(l)), C(I)=D(I)/(4*EI(I))+D(I+l)/(4*EI(I+l)), C(I+l)=D(I+l)/(4*EI(I+l)).
Equation (8.3) is the numerical equivalent of the bending equation (8,2). A
simplified version (used in earlier editions of this report) is to use only the value of M(I) in the right hand side, with a coefficient 2*C(I). This seems to be somewhat less accurate, especially for non-homogeneous piles. Boundary conditions
The boundary conditions deserve some special attention, especially because it is preferable that in each point an equation can be formulated for each of the two variables: the lateral displacement U and the bending moment M. It is then avoided that in the matrix describing the system of equations a zero coefficient appears on the main diagonal, which would be numerically
inconvenient. In the numerical model it is assumed that the upper end of
the beam (the top of the pile) is the nodal point I=O, and that the lower end of the beam (the point of the pile) is the nodal point I=N.
The simplest boundary condition would be a hinged support, because then both the bending moment and the lateral displacement are directly given, which
can inunediately be incorporated in the numerical model, Unfortunately this
is not the most realistic boundary condition. The most realistic boundary
condition is a free end, with a given shear force and a given moment. The given moment (zero or a prescribed value) can inunediately be incorporated, but the given shear fore~ should be transformed into a condition for the
lateral displacement. This can be done, for example for the point I=O, by
starting from the equation of equilibrium of moments in the first element, M(l)-M(O)=(Q(l)+Q(O)-P(O))*D(I)/2,
The shear force Q(O) in this equation can be assumed to be zero, with the
external load being introduced as P(O). The shear force Q(l) can be
elimi-nated by using the equation of horizontal equilibrium for element 1, Q(l)-Q(O)+F(l)+P(O)=R(l),
With Q(O)=O it now follows that
R(l)+(2/D(l))*(M(O)-M(l))=F(l)+2*P(O),
Because the soil reaction R(l) will depend upon the lateral displacements W(O) and W(l) this can be considered as an equation formulated in terms of W(O).
Soil response
The response of the soil has been introduced above as a force R(I). In
accordance with the considerations of chapter 5 this should be related to
the local lateral displacement, taking into account that the soil pressure
cannot be greater than in the passive state, and not smaller than in the
active state. The precise formulation will not be repeated here. The
reader may also inspect the computer program presented on the next pages.
An elementary computer program (LLP-1.0) is reproduced below. Again the
program has been kept as simple as possible, with interactive input, and output as a list on the screen. For simplicity the pile has been assumed to be homogeneous (EI constant), and the soil properties have also been assumed as constant. This does not mean that the elastic and plastic soil reactions
are constant, however, because the parameters describing the soil reaction
are the vertical stress (which increases with depth), active and passive
coefficients, and a value for the total stroke (i.e. the displacement
diffe-rence between the generation of active and passive soil pressures). In this
program no difference is made between the stiffnesses in the active and passive branches, as suggested in paragraph 6,3.
Program LLP-1.0
1000 REM===== LLP-1.0 ---1010 GOSUB 2210:DEFINT I-N:B$=11
Impossible":Z=4 1020 DIM X(lOO),R(lOO),D(lOO),F(lOO),U(lOO),Q(lOO) 1030 DIM NP(lOO),NQ(lOO),BM(lOO),FF(lOO)
1040 DIM K%(100,3),A(l00,3,l,l),UM(l00),UN(l00),G(l,1) 1050 INPUT"Length of the pile (m) •.•...•.. ";UL
1060 IF UL<=O THEN PRINT 8$:GOTO 1050
1070. INPUT"Width of the pile (m) , •. , • , , .• , "; BB 1080 IF BB<=O THEN PRINT 8$:GOTO 1070
1090 INPUT"Unit weight of soil (kN/m"3) ,,, ";DG
1100 IF DG<=O THEN PRINT 8$:GOTO 1090
1110 INPUT"Active soil pressure,,,,,,,,,,, ";CA 1120 IF CA<=O OR CA>l THEN PRINT B$:GOT0 1110 1130 INPUT"Passive soil pressure ...•. , .. ,, ";CP 1140 IF CP<l THEN PRINT B$:GOT0 1130
1150 INPUT"Neutral soil pressure , . , , ..•.• , "; CK 1160 IF CK<=CA OR CK)=CP THEN PRINT B$:GOT0 1150 1170 INPUT" Total stroke (m) • , , , .. , . , .• , "; UU 1180 IF UU<=O THEN PRINT B$:GOT0 1170
1190 INPUT"Bending stiffness EI (kN*m"2) .. ";EI 1200 IF EI<=O THEN PRINT B$:GOT0 1190
1210 INPUT"Number of elements (10,,.100) ,. ";N 1220 IF N<lO THEN N=lO ELSE IF N>lOO THEN N=lOO 1230 IF CA=l AND CP=l THEN UC=O:UP=O:GOTO 1250 1240 UC=(CK-CA)*UU/(CP-CA):UP=(CP-CK)*UU/(CP-CA) 1250 DG=DG*BB:ZZ=(CP-CA)/UU:Z=ZZ*DG*UL/2:FT=O 1260 A=UL/N:FOR K=l TO N:X(K)=X(K-l)+A:NEXT K 1270 GOSUB 2210:INPUT"Lateral force (kN) ,,,,, ";FL
1280 IF ABS(FL)(FT THEN FOR K=l TO N:NP(K)=O:NQ(K)=O:NEXT K 1290 FOR K=l TO N:GOSUB 1880:NEXT K
1300 NP(O)=O:NQ(O)=O:LI=O:GOSUB 2210:FT=ABS(FL) 1310 FOR I=O TO N:FOR J=O TO 3:K%(I,J)=O
1320 FOR K=O TO l:FOR L=O TO l:A(I,J,K,L)=O:NEXT L,K,J,I 1330 PRINT"Force (kN) •.•. , . , .. , , , . : "; FL: PRINT
1340 PRINT"Generet ion of matrix • ·• , : "; : FOR I= 1 TO N-1 1350 K%(I,O)=I:K%(I,l)=I-l:K%(I,2)=I+l:K%(I,3)=2:NEXT I 1360 K%(0,1)=1:K%(0,3)=1:K%(N,O)=N:K%(N,l)=N-l:K%(N,3)=1 1370 I=O:PRINT "+";:FOR I=l TO N-1:PRINT "+";
1380 AA=l/(D(I)*D(I+l)):Al=AA*D(I):A2=AA*D(I+l) 1390 A(I,O,O,O)=-Al-A2:A(I,l,O,O)=A2:A(I,2,0,0)=Al 1400 A(I,O,O,l)=-(R(I)+R(I+l))/4:A(I,l,O,l)=-R(I)/4 1410 A(I,2,0,l)=-R(I+l)/4 1420 A(I,3,0,0)=-(F(I)+F(I+l))/2:A(I,0,1,l)=-Al-A2:A(I,l,1,l)=A2 1430 A(I,2,l,l)=Al:A(I,0,1,0)=(D(I)+D(I+l))/(4*EI) 1440 A(I,l,l,O)=D(I)/(4*EI):A(I,2,1,0)=D(I+l)/(4*EI) 1450 NEXT I: PRINT "+" 1460 A(O,O,O,O)=l:A(0,3,0,0)=0:A(O,O,l,l)=R(l)/4 1470 A(O,l,l,l)=R(l)/4:A(0,0,1,0)=l/D(l):A(O,l,1,0)=-l/D(l) 1480 A(0,3,l,l)=F(l)/2+FL:A(N,O,O,O)=l 1490 A(N,3,0,0)=0:A(N,O,l,l)=R(N)/4:PRINT 1500 A(N,l,1,l)=R(N)/4:A(N,O,l,O)=l/D(N):A(N,l,l,O)=-l/D(N) 1510 A(N,3,l,l)=F(N)/2:PRINT"Elimination . .. .. .. .. .. . ";
..
1520 H=3:FOR I=N TOO STEP -1
1530 KC=K%(I.,H):PRINT "+";:FOR KV=O TO l:CC=l/A(I,O,KV,KV)
1540 FOR II=O TO KC:FOR LV=O TO l:A(I,II,KV,LV)=CC*A(I,II,KV,LV) 1550 NEXT LV,II
1560 A(I,H,KV,KV)=CC*A(I,H,KV,KV):FOR LV=O TO l:IF LV=KV THEN 1600 1570 CC=A(I,O,LV,KV):FOR II=O TO KC:FOR IJ=O TO 1
1580 A(I,II,LV,IJ)=A(I,II,LV,IJ)-CC*A(I,II,KV,IJ):NEXT IJ,II 1590 A(I,H,LV,LV)=A(I,H,LV,LV)-CC*A(I,H,KV,KV)
1600 NEXT LV,KV:IF KC=O THEN 1750
1610 FOR J=l TO KC:JJ=K%(I,J):L=K%(JJ,H) 1620 FOR JK=l TO L:IF K%(JJ,JK)=I THEN 1640 1630 NEXT JK
1640 FOR KV=O TO l:FOR LV=O TO l:G(KV,LV)=A(JJ,JK,KV,LV) 1650 NEXT LV,KV:K%(JJ,JK)=K%(JJ,L):K%(JJ,L)=O
1660 FOR KV=O TO 1:FOR LV=O TO l:A(JJ,JK,KV,LV)=A(JJ,L,KV,LV) 1670 A(JJ,L,KV,LV)=O
1680 A(JJ,H,LV,LV)=A(JJ,H,LV,LV)-G(LV,KV)*A(I,H,KV,KV) 1690 NEXT LV,KV:L=L-l:K%(JJ,H)=L:FOR II=l TO KC
1700 FOR IJ=O TO L:IF K%(JJ,IJ)=K%(I,1I) THEN 1720 1710 NEXT IJ:L=L+l:IJ=L:K%(JJ,H)=L:K%(JJ,IJ)=K%(I,I1) 1720 FOR KV=O TO l:FOR LV=O TO l:FOR JV=O TO 1
1730 A(JJ,IJ,KV,LV)=A(JJ,IJ,KV,LV)-G(KV,JV)*A(I,II,JV,LV) 1740 NEXT JV,LV,KV,II,J
1750 NEXT I
1760 PRINT:PRINT:PRINT"Back substitution ... : "; 1770 FOR J=O TO N:PRINT "+";:L=K%(J,H):IF L=O THEN 1810 1780 FOR K=l TO L:JJ=K%(J,K):FOH KV=O TO l:FOR LV=O TO 1 1790 A(J,H,KV,KV)=A(J,H,KV,KV)-A(J,K,KV,LV)*A(JJ,H,LV,LV) 1800 NEXT LV,KV,K
1810 NEXT J:FOR I=O TO N:U(I):::A(I,H,1,1):BM(I)=A(l,H,O,O):NEXT I 1820 Q(O)=-FL:FF(O)=O:FOR I=l TO N:AA=(BM(I)-BM(I-1))/D(I)
1830 Q(I)=-Q(I-1)+2*AA:FF(I)=(Q(I-l)-Q(I))/D(I):NEXT I 1840 LI=LI+l:GOSUB 2010:IF LI<lOO THEN 1860
1850 CLS:PRINT"After 100 iterations no convergence":PRINT:END 1860 IF LL>O THEN 1310
1870 GOSUB 2060:GOSUB 2190:GOTO 1270
1880 REM Calculation of stresses and springs
1890 A=X(K)-X(K-l):X=X(K)-A/2:D(K)=A:U=(U(K)+U(K-1))/2 1900 VM=UM(K):VN=UN(K)
1910 SZ=DG*X:IF U-VM>UC THEN SX=CA*SZ:R(K)=O:NR=l:LP=LP+l:GOTO 1940 1920 IF U-VM<-UP THEN SX=CP*SZ:R(K)=O:NR=-1:LP=LP+l:GOTO 1940
1930 R(K)=ZZ*SZ:SX=CK*SZ+R(K)*VM:NR=O
1940 F(K)=SX*A:R(K)=R(K)*A:IF NP(K)<>NR THEN NP(K)=NR:LL=LL+l 1950 IF U-VN<-UC THEN SX=CA*SZ:RR=O:NR=l:LP=LP+l:GOTO 1980 1960 IF U-VN>UP THEN SX=CP*SZ:RR=O:NR=-1:LP=LP+l:GOTO 1980 1970 RR=ZZ*SZ:SX=CK*SZ-RR*VN:NR=O
1980 F(K)=F(K)-SX*A:R(K)=R(K)+RR*A 1990 IF NQ(K)<>NR THEN NQ(K)=NR:LL=LL+l 2000 RETURN
2010 REM Ver~fication of springs 2020 PRINT: PRINT: PRINT"Check springs"
2030 LL=O:LP=O:FOR K=l TO N:GOSUB 1880:NEXT K:PRINT:GOSUB 2210 2040 PRINT"Plastic springs ... :";LP:PRINT
2050 RETURN 2060 REM Output
2070 PRINT C$:FOR J=l TO N:U=(U(J)+U(J-1))/2 2080 IF U-UM(J)>=UC THEN UM(J)=U-UC
2090 IF U-UM(J)<=-UP THEN UM(J)=U+UP 2100 IF U-UN(J)<=-UC THEN UN(J)=U+UC 2110 IF U-UN(J)>=UP THEN UN(J)=U-UP 2120 NEXT J
2130 PRINT" Z (m) U (m) M (kNm) Q (kN) F(kN/m)"
2140 A$="###,### ###,### #####,### #####,### #####,###":PRINT
2150 FOR J=O TO N:PRINT USING A$;X(J);U(J);BM(J);Q(J);FF(J):NEXT J 2160 A$="####.######":PRINT:PRINT"Top: F = ";
2170 PRINT USING A$;FL:PRINT" U = ";:PRINT USING A$;U(O):PRINT
2180 RETURN
2190 LOCATE 24,25,0:COLOR 0,7:PRINT" Touch any key to ~ontinue "; 2200 COLOR 7,0:A$=INPUT$(1):RETURN
22]0 CLS:LOCATE 1,31,0:COLOR 0,7:PRINT" LLP-1.0 11
In the computer program presented above the elasticity of the soil has been characterized by a value for the displacement difference between the
genera-tion of active and passive lateral stress, denoted as the stroke. By
defi-ning the elastic properties of the soil in this way the spring constant is
automatically increasing with depth, which is not unrealistic. As mentioned
in paragraph 2 the spring constant in a soil is often correlated to the
modulus of elasticity (Vesic [1961]) by writing
c = aE/D,
where a is a coefficient which describes the spreading of stresses, of the
order of magnitude of 1, Eis the modulus of elasticity of the soil, and D
is the pile diameter. The modulus of elasticity of a soil can be expressed
in terms of the compressibility C in Terzaghi's [1941) logarithmic formula,
by assuming a small stress increment. This gives
E = Ccr,
where cr is the stress level. Thus one obtains
c = aCcr/D
In the present analysis the spring constant in the elastic branch is defined as
c = Kcr/Au
where K is a pressure coefficient, for example K=3. By equating these two
expressions one obtains
Au/D = 1/aKC
The possibility to correlate the stroke Au to such a familiar soil parameter
as the compressibility may serve to further illustrate the relevance of the
description of the elastic branch in this way. Conunon values for the
com-pressibility of sandy soils are 50-200. This means that the stroke Du is of
the order of 1 % of the pile diameter.
8,3 Example
As an example a steel pile of 50 meter length, and l meter diameter will be
considered. The wall thickness is supposed to be 0,05 m, so that the bending
stiffness is EI=4.12 GNm2 • The soil is homogeneous, with an effective
volu-metric weight of 10 kN/1113, and with a response characterized by a minimum
stress equal to 0.3333 times the vertical (effective) stress, a maximum stress equal to 3 times that vertical stress, and a displacement difference
between these two extremes of 0.01 m. The displacement caused by a
varia-ble lateral load, up to 2000 kN, is shown in figure 8.3. It should be noted
that, in contrast with the case of an axial load (see the previous
para-graph), plastic deformations start to occur for small loads. This must be
due to the fact that near the soil surface the effective stresses, and therefore the soil strength, are very small. Another important factor may be
that the pile, although of considerable dimensions, is relatively flexible in bending, Another interesting feature of the behaviour shown in figure 8.3
r - - - = = = : : : : = - - - : . - - Q
11
Figure 8.3. Example.
is that plastic deformations continue to occur during unloading and
subse-quent reloading, thus leading to a continuous loss of energy. The
deforma-tion corresponding to the maximum force also tends to increase in each
loading cycle, All these seemingly realistic characteristics of the
load-displacement diagram have been obtained using the simple local
elasto-plastic stress-strain relation of paragraph 6, and have not been introduced
as such.
By varying the elastic range of the soil deformations (the stroke 6u) it is
found that this parameter has only limited influence, especially for large
lateral forces, see figure 8.4. The three curves shown were calculated using
