12_12. fourier series aided finite element method (fsafem)

12. Fourier series aided finite elementmethod (FSAFEM)

12.1 SynopsisAs described in Chapter 11, a conventional three dimensional (3D) finite elementanalysis of a typical nonlinear geotechnical problem is complex and requires alarge amount of computer resources. The Fourier series aided finite elementmethod (FSAFEM) is a means of increasing the computational efficiency of theconventional finite element method, for a special class of 3D problems which havea geometry that does not vary in one of the coordinate directions (out of planedirection), but whose material properties and/or boundary conditions do. Thisefficiency is gained by assuming that the displacements in the geometrical out ofplane direction can be represented using a Fourier series and exploiting itsorthogonal properties. Two types of FSAFEM exist, the continuous FSAFEM (i.e.CFSAFEM) and the discrete FSAFEM (i.e. DFSAFEM) and they are described inthis chapter.

12.2 IntroductionConventional 3D analyses of nonlinear geotecnical boundary value problemsrequire large amounts of computer resources (Brown and Shie (1990)). A largeproportion of these resources are involved in inverting the global stiffness matrix.Consequently, one way of economising on computer resources is to use an efficientmethod of inverting this matrix. In this respect the use of iterative solutiontechniques was described in Chapter 11, where it was shown that, for currentcomputer hardware technology, such techniques can result in economies for linearproblems, but are unlikely to result in savings for nonlinear problems. However,the situation may improve with future developments in computer hardware.

Another way of simplifying 3D analyses is to exploit any geometricalsymmetries that exist. One approach that capitalises on such symmetries is theFourier series aided finite element method (FSAFEM). The theory behind thismethod and its implementation is the subject of this chapter. Conventionally, theFSAFEM has been applied to problems with an axi-symmetric geometry (but nonaxi-symmetric loading and/or variation of material properties) and it is theapplication of the method to such problems that is considered here. However, themethod can also be implemented for problems expressed in terms of Cartesiangeometries.

Fourier series aided finite element method / 345

All existing formulations of the FSAFEM have been based on linear elasticmaterial behaviour. Nonlinear problems have been analysed with some success(Winnicki and Zienkiewicz (1979), Griffiths and Lane (1990)). In these analysesthe nonlinear behaviour was dealt with using a finite element algorithm in whichthe global stiffness matrix was based on the linear elastic material properties andthe nonlinearity was dealt with by modifying the right hand side of the governingfinite element equations. In addition, most of the past implementations of theFSAFEM have assumed that the system forces and displacements have a symmetryabout the 6 = 0° direction, where 6 is the angular coordinate. This assumptionresults in a large saving of computer resources required for an analysis and alsoconsiderably simplifies the formulation. However, these assumptions arerestrictive. Recently a new nonlinear formulation has been developed (Ganendra(1993), Ganendra and Potts (1995)), which allows the stiffness matrix to beupdated during an analysis using the nonlinear material properties. It also placesno symmetry constraints on the variation of system forces and displacements.However, options can be included to capitalise on any such symmetry if it exists.The chapter begins by presenting the basic theory behind the FSAFEM, and thendescribes its implementation.

12.3 The continuous Fourier series aided finite elementmethod

12.3.1 Formulation for linear behaviourThe axi-symmetric geometry of the problemdomain allows a cylindrical coordinatesystem to be defined (r-z-6, see Figure 12.1)such that the r-z plane can be discretisedusing a 2D finite element mesh. Thus thedistribution of variables in the r-z plane canbe described using nodal values andconventional 2D finite element shapefunctions. The distribution of variables in the6 direction can be described using a Fourierseries, e.g. the incremental radialdisplacement, Au, can be written as:

Au = Au°

Figure 12.1 Cylindricalcoordinate system

Aw2cos2#+ Aw2sin2<9+ ... +Aw/cos/<9+ Aulsinl0+ ...(12.1)

where Au°, Au1 and Au' are the 0th, Ith order cosine and /th order sine harmoniccoefficients of variable Au respectively.

Consider the displacements at a point, Au, with components Aw, Av, and Aw inthe r, z, and 6 directions respectively. Displacements Aw can be expressed in vector

346 / Finite element analysis in geotechnical engineering: Theory

form in terms of the shape functions of the appropriate element in the 2D finiteelement mesh, and the Fourier series of the nodal displacements of the element:

{Au} =

tsu

Av

Aw

N/=o

I (12.2)

where:Nj is the shape function of the /lh node defined in the element;

Uj, Vf and Wf are respectively the /* cosine harmonic coefficient of radial,vertical and circumferential incremental displacement at the /* node;

Uj, Vf and WJ are respectively the Ith sine harmonic coefficient of radial,vertical and circumferential incremental displacement at the fh node;n is the number of nodes in the element;L is the order of the harmonic series used to represent displacement and is equalto the highest order harmonic used.

The ensuing procedure for formulating the stiffness matrix for the CFSAFEMis undertaken in a manner similar to that for a full 3D analysis. However, as willbecome evident, the CFSAFEM uncouples the full 3D stiffness matrix, which hasa form shown in Equation (12.3), into a series of smaller independent stiffnessmatrices of the form shown in Equation (12.4). The variables that are solved forin the CFSAFEM are the harmonic coefficients of incremental displacement at

each node in the 2D mesh, i.e. u\, Vj , WJ , uj, V/ , w} .

Arf, A/?,

Kr. Arf,. Afl, (12.3)


00000000

0[Kl

0000000

00

000000

000

00000

0000

\Kl]0000

00000

000

000000

o [0

0000000

0

0000000

] o[KL]

Ad1

Ad1

AdLl

AdL

Ad°* , Ad1**

* , Ad1**

* , AdL~1*** , AdL**

AR1

AR1

ARL~l

ARL

AR°* , AR1**

* , AR1**

* , ARL~]*** , ARL**

(12.4}

The incremental strains at a point, assuming a compression positive signconvention, are defined in cylindrical coordinates as:

{Ae} =

Aer ^AszAeeAerz

Aere

As

d(Au)dr

d(Av)dz

Au 1 d(Aw)r ' r dO

d(Au) d(Av)dz dr

1 d(Au) d(Aw) tr dO ' dr

1 a(Av) d(Aw)r d6 ' dz

Aw

(12.5)

Thus in the problem domain defined above, the incremental strains at a point canbe expressed in terms of Fourier series harmonic coefficients of incremental nodaldisplacements and element shape functions:

dr

dz

^-U Ujcosie + UJsinW ^-\ v/cosie + v/sidz [ dr

—L\ u'cosW + UJsinI0 ^-\-W/sinl0+Wjlcosl0

IN; -U sinl0 + U cosl0

^±\ W/cosl0+w/sinl0

(12.6)


These incremental strains can be rewritten as the sum of the product of a seriesof strain matrices, [B], and vectors of harmonic coefficients of incrementaldisplacements. The resulting expressions can then be divided into two parts,parallel symmetry terms and orthogonal symmetry terms. Parallel symmetry termsconsist of cosine harmonic coefficients of radial and vertical incrementaldisplacements and sine harmonic coefficients of circumferential incrementaldisplacements. Conversely, orthogonal symmetry terms consist of sine harmoniccoefficients of radial and vertical incremental displacements and cosine harmoniccoefficients of circumferential incremental displacements. Rearranging Equation(12.6) in this manner gives:

AS,:Asz

Ase

Asrz

Aere

n, = — V

/ = 1

L

Z/=0

dr-cos/<9 0

C 0 S/6>

dNj_dz

dr

codeIN,

0

0

code

dNi Ni\ • mdr r )

IN, sin/<9dz

-sin/6>

(12.7)

n L

- Z Zi = l / = 0

8r-sinlO

dz

0

'-sinie

0 IN,

0

0

sin/<9

dz dr

illLcos/(9r

0 lNi in—-codOr

^ L - i ^ - cos/(9ar r y

-—LCOs/<9az

u

-w

Splitting the equation into two parts in this manner results in the [B] matricesfor each part having a similar form. The dashed lines in the matrices separate thesine and cosine terms and split the [B] matrix into a top and bottom section. Notethat the cosine coefficient of circumferential incremental displacement is expressedwith a sign change. Thus the strains can be written as:

(12.8)


where:

BN;

dzHi. o —i

dz dr

and

w!(12.9)

r

0 -

dr rIN-,

dz

Note:

andi

{Ad/*} and {A<//**} are the parallel and orthogonal 7th order harmoniccoefficients of incremental displacement for the fi node respectively, and[B1 /] and [B2/] are the top and bottom sections of the Ith order harmonic [B]matrix for the ith node respectively.

The incremental internal work done can now be written as:

AW = J |{A^}T{Acr} <\6 r&area- 7 1

= J ]{Au}T[Bf[D][B] {AM} d0 r darea

-\)t\i k .{Acini -[D]- (12-10>

7=1 '=0

[51^.]sin/6>"-[B2lj]cosl0

The [D] matrix for a material relates the incremental stress vector {Ac} to theincremental strain vector {As}, i.e:

{A<7} = [D] {As} (12.11)

It can be divided into four submatrices, [Du], [Dn], [^2i]5 an<i L^L ^ accordance

with the subdivision of the [B] matrix into two submatrices [B\] and [B2]. Thiscan be written in matrix form as:


A<7,.

Aa:

Acre

ACT,,

[Ail

As,.Asz

Aee

As,r

As,a

(12.12)

Thus the equation for incremental internal work can be rewritten as:

AW = | [zv rn ^ ^[Bl}]sh

[52f]cos^<9j [D22]\ h

rJ2J 2

[B2'j]sin!0

k=0 -[B2?]cosk0 [D2l][D22]\ to

r dOdarea

d'j }

{Ad'j

(12.13)

When integrating with respect to 6, a large number of terms in the stiffnessmatrix become zero due to the orthogonal properties of the Fourier series:

J sin£<9 cos/6> &6 = 0 for all k and / ;- 7 1

f s i n £ < 9 s i n / < 9 d < 9 = 0 i f k * /, = % i f k=I* 0 , = 0 ifA: = / = O ; ( 1 2 . 1 4 )- 7 1

JcosA:6> cos/6» &6 = 0 ifk*I, =nifk = l*0, =2nifk = l = 0 .

Thus the incremental internal work done can now be simplified to:


} r darea

-2{Arf/0*}T[JBl/

0]T[JD12][^.]{A^/0"}- 2{Arff*}T[JB2,0]T[/)21][51^]{Arff} +

} + {Adj } [52,] [Z>21][j?l ]{Atf } +

I } r darea

£

(12.15)The stiffness matrix has been uncoupled into L+l smaller stiffness matrices of

the form shown in Equation (12.4), with one independent set of equations for eachharmonic order. This form of uncoupling is called harmonic uncoupling. However,for each set of equations associated with a particular harmonic order the parallelsymmetry displacements, {Ad/*}, and orthogonal symmetry displacements,{Ad/**}, are coupled in the manner shown below:

'[KlY [Kl]po

[Kl]l]op

{Ad1"}{Ad1**}

(12.16)

The terms above the dashed line in Equation (12.15) give the independentdiagonal terms [Kl]p and [K'Y, and those below give the cross coupling terms [K!]op

and [K'Y°- It is noted that for materials with a [D] matrix which has zero offdiagonal submatrices, [Z>12] and [D2l], the cross coupling terms disappear and thestiffness matrix reduces to the following form:

KHTi^rfto) (12I7)In addition, it may be noted that [K'Y = [KlY- In this case the {Ad/*} and {Ad/**}terms for each harmonic order can be solved independently, using the samestiffness matrix. This form of uncoupling is called symmetric uncoupling. Thesymmetrically uncoupled stiffness matrix for the Ith order harmonic is:

<]T[D22][B2l.]r darea (12.18)

where: A = 2 i f / = 0 ;A = l i f / * 0 .

The applied incremental loads are formulated as harmonic coefficients ofincremental nodal force using the equations in Appendices XII. 1 to XII.4. Theseharmonic coefficients of nodal force are expressed as vectors {AR1*} and


T

{AR '*} = I R'r Rlz Rl

e \ is the vector of parallel Ith order harmonic coefficients of

incremental force;

{ARl**} = <Rlr Rl

z -RQ\ is the vector of orthogonal 7th order harmonic

coefficients of incremental force;where:

Rj. , Rlz and Rl

9 are the Ith cosine harmonic coefficient of radial, vertical andcircumferential incremental force respectively;

Rlr , Rl

z and Rle are the Ith sine harmonic coefficient of radial, vertical and

circumferential incremental force respectively.

Thus for each harmonic order two sets of system equations can be written:

2 7 ;J (12.19)1 } = [Kl]° { A d 1 }

The displacements can be solved for by inverting the stiffness matrix in asimilar manner to the conventional finite element method. An important feature ofthis linear elastic formulation is the harmonic uncoupling. A consequence of thisuncoupling is that for any harmonic order the solution coefficients ofdisplacements are only non-zero if the applied coefficients of load for the sameorder are non-zero. Thus the number of harmonics required for an analysis is equalto the number of harmonics required to represent the boundary conditions.

12.3.2 Symmetrical loading conditionsA symmetrical function of #,/.(#), has the property that:

/v(#) = / s ( -0 ) (12-20)

and the Fourier series representation of such a function would only consist of thezeroth (i.e. 0th) and cosine harmonic terms. An asymmetrical function of8,fm(ff),has the property that:

fm(O) = -fm(-O) (12.21)and the Fourier series representation of such a function would only consist of sineharmonic terms.

The linear CFSAFEM formulation has divided the solution displacements,{Ad}, into two parts, parallel symmetry displacements, {Ait}, and orthogonalsymmetry displacements, {Act*}. The parallel symmetry displacements consist ofsymmetrical displacements in the r and z coordinate directions and asymmetricaldisplacements in the # direction. Conversely, orthogonal symmetry displacementsconsist of asymmetrical displacements in the r and z coordinate directions and


symmetrical displacements in the 6 direction. The applied loads, {A/?}, areseparated in a similar manner into parallel loads, {A/?*}, and orthogonal loads,{AR"}.

e=o°Pz is the applied vertical force/unit lengthF2 is axial force applied to the pile = 2nrPz

r is the radial coordinate of the node

a) Axial load on pile

Pr=PcosQ Pe=-PsinQ PX=P ; Py =

Fx is lateral force applied to the pile = 2nrP

b) Lateral load on pile

e=o° r ^ ^ e=o°

PZ=P'cosQ My = nrP'Pz is the applied vertical force/unit lengthMy is the applied turning moment

c) Moment load on pile

Figure 12.2: Load components on pileMany boundary value problems have a symmetry about the 6 = 0° direction,

such that the imposed boundary conditions consist of purely parallel symmetry ororthogonal symmetry terms. For instance, the loading conditions applied to a pilemay consist of a combination of axial loading, lateral loading and turning momentabout an axis perpendicular to the direction of lateral loading. These boundaryconditions can be represented using parallel symmetry, if the direction of lateralloading is parallel with the 6 = 0° direction, see Figure 12.2a-c. Hence, all previousimplementations of the CFSAFEM have been constrained to analyse either parallelsymmetry conditions (Winnicki and Zienkiewicz (1979)), or orthogonal symmetryproblems (Griffiths and Lane (1990)). If the applied loads in a problem do notsatisfy any symmetry about the 6 = 0° direction, then a non-symmetrical


formulation is required, consisting of both parallel and orthogonal symmetry terms.Analyses using only parallel or orthogonal symmetry terms benefit from thereduced number of solution coefficients of displacements that have to be solvedfor. This results in large savings in computer resources required for an analysis andsignificantly simplifies the solution algorithms.

The previous implementations of the CFSAFEM have assumed that if theapplied loading conditions can be represented using either parallel or orthogonalsymmetry, then the resulting solution displacements will also satisfy parallel ororthogonal symmetry, respectively. Though this may have been valid for theparticular cases analysed, it is not true in general. In the linear elastic formulationpresented above, this assumption has been shown to be true only for materials witha [D] matrix which has zero off diagonal terms [Dn] and [D2l], i.e an isotropiclinear elastic material. Due to the coupling of parallel and orthogonal terms, theanalysis of an anisotropic elastic material, with non-zero [Dn] and [D2l], wouldrequire the full non-symmetrical formulation, irrespective of any symmetry in theimposed loads.

The term parallel symmetry stems from the fact that it can be used to representa load in a direction parallel to the 6 = 0° direction, see Figure 12.2b. Similarly,orthogonal symmetry can represent loads in a direction perpendicular to the 6 = 0°direction.

12.3.3 Existing formulations for nonlinear behaviourAll past implementations of the CFSAFEM have used the above formulation forlinear elastic behaviour with a constant material stiffness, [D] matrix, in the 6direction. No formulation has been developed for nonlinear material behaviourwith a variable [D] matrix in the 6 direction. However, nonlinear analyses havebeen undertaken using the linear elastic formulation, in combination with asolution strategy which continually adjusts the right hand side of the governingfinite element equations to cater for material nonlinearity. This is akin to using anelastic stiffness matrix to solve an elasto-plastic problem. Considering the threesolution strategies described and compared in Chapter 9, only the visco-plastic andMNR approaches can be used in this way. The tangent stiffness approach cannotbe employed because it involves the use of the elasto-plastic stiffness. The visco-plastic approach was used by Winnicki and Zienkiewicz (1979) and Griffiths andLane (1990). However, as noted in Chapter 9, this method can be problematic forhighly nonlinear constitutive models. A better alternative is the use of the MNRapproach, with the stiffness matrix calculated using only the elastic part of theconstitutive matrix.

There is some scepticism about the validity of the above approach for nonlinearproblems. It is argued that, since the formulation of the CFSAFEM assumes thatthe [D] matrix is constant in the 6 direction, its use for a material which does notsatisfy this criterion is not valid. The popular practice of using purely parallel orpurely orthogonal symmetry terms in an analysis requires that the system stiffness


matrix is symmetrically uncoupled. The criterion for this is that the off diagonalterms in the [D] matrix are zero, as discussed in Section 12.3.1. This criterion is notsatisfied in general for elasto-plastic materials and thus questions the validity of theabove practice for nonlinear analyses.

An important feature of nonlinear CFSAFEM analyses is that harmonicuncoupling no longer exists. The number of harmonics required to represent thesolution displacements is not just a function of the number of harmonics requiredto represent the boundary conditions, but is also dependent on the materialnonlinearity. This feature cannot be reproduced with the linear elastic formulationwhich, as noted above, only yields non-zero harmonics of displacement if a loadof the same harmonic order is applied. The anomaly is overcome by allowing thecorrective loads, applied by the procedure for nonlinear finite element analysis, tohave harmonics other than those associated with the applied boundary conditions.Past practice has been to allow a greater number of harmonics of solutiondisplacement than is required to represent the boundary conditions. Though thishas given reasonable results, there has never been a rational for it. Increasing thenumber of harmonics used in an analysis increases both the amount of computerresources used and the solution accuracy. Thus the number of harmonics used inan analysis is an important parameter which is based on a trade off betweensolution accuracy and computer resources.

1 2 . 3 . 4 New formulation for nonlinear behaviourThe linear formulation for the CFSAFEM assumes that the [D] matrix is constantin the 6 direction. In general, this assumption is not valid for nonlinear materialbehaviour, since the [D] matrix is now stress history dependent and the stressesvary in the 6 direction. A new nonlinear formulation is proposed for theCFSAFEM which incorporates a variation of [D] in the 0 direction.

The linear elastic CFSAFEM formulation of strain, in terms of a series ofproducts of harmonic coefficients of displacement and harmonic [B] matrices, isused to formulate a system stiffness matrix as described by Equation (12.13).However, the [D] matrix is no longer a constant, but varies with 6. This variationcan be represented with a Fourier series for each component of the [D] matrix:

[D] = [D°] + [Dl]cos0+[Dl]sin0+[D2]cos20+[D2]sin20+ ...

[Z>/]cos/6>+[Z)/]sin/<9+ ... (12.22)

where [D°], [Df] and [Dl] are matrices containing the 0th, Ith order cosine and Ith

order sine harmonic coefficients of the components of the [D] matrix respectively.The number of harmonics used to represent the [D] matrix, M, need not be thesame as the number used to represent displacements, L. These harmonic [D]matrices can also be split into 4 parts representing [/>,,], [Dl2], [D2l] and [Z>22]terms. Separating the cosine and sine harmonics of the [D] matrix allows Equation(12.13) to be rewritten as:


-%i=\j=\k = O [B2?]sink0]

k=0

L

k=0

[B\f]cosk0[B2f]sink0

[Blf]sink0-[B2f]cosk0

[Blf]cosk0] M[B2f]sink0j w=c

[B\f]sink0] M

k=0

r d0darea

[B2f]sink0

[B\*]sink0~[B2f]cosk0

[Dj[]cosm0[D?2]cosm0[D!?l]cosm0[D!?2]cosm0

[Dj[]cosm0lD?2]cosm0

[D%\]cosm0[D2z]cosm0

[D21\]sinm0[D2l]sinm0

[Dfl]sinm0[D?2]sinm0

[D™]]smm0[D2}2]sinm0

[Df[]sinm0[D?2]sinm0

[D"\]sinm0[D"2]sinm0

[D^\]smm0[D22]sinm0

1=0

[B\!j]CQSl0[B2I

j]sinl0

[Ul'.]cos/^[B2lj]sinl0

[Bl!j]sinl0-[B2*j]cosl0

{Arfj }

-[B2j]cosl0

{Ad1**}

{Ad1;*}

£<\[B2'j]sinI0\

t [Bl'j)cosl0]

£ -[J?2'.]cos/0j

(12.32)

As with the linear formulation, the dO integral is performed and a large numberof terms in the stiffness matrix are zero due to the orthogonal properties of theFourier series. For the linear formulation this integration was solved using thestandard solutions for the integration of the product of two Fourier series.However, for the nonlinear formulation the corresponding integration is of theproduct of three Fourier series and a new set of solutions has to be derived, seeAppendix XII.5. Thus the incremental internal work done can be written as:

/= 1 ./=1 k=0

k=0

[Bit]'[B2f][Blf]"

[Blf]'[B2t]_

k -

[B2>]_

\[Dtfy [D^fi=o\_[D2{'Y [D22Y

T

££/=o

r nk J ~\()P r T\k J ~\®P1 - 1 1 1 1 -^H 9 1

1 D^l | 1 •LJ'yj |

1 -^^11' I I -D-i -J I

\. 1\ J Y. 10 J

T r -jL\ [DU' ] [Din ]L A/ o kJ o\

[Bl';][B2'j]

\BI'J]Z

[B2lj}_

r n i ' i1 XJ 1 • 1

1 i* ^ • 1

{Arff}

{Ad1;}

{Ad1;-}

{Ad1"} r darea

(12.24)


where:[DkI]p, [DkI]op, [Dkl]po and [Dk'!]° are the [D] matrices resulting from the d6integration that should be used to form the stiffness matrix relating the kih

harmonic coefficients of load with the Ith harmonic coefficients ofdisplacement;[DkJY is the [D] matrix that relates parallel loads to parallel displacements;[Dk!]op is the [D] matrix that relates orthogonal loads to parallel displacements;[DkJY° is the [D] matrix that relates parallel loads to orthogonal displacements;[DkJY is the [D] matrix that relates orthogonal loads to orthogonal

displacements.

These [D] matrices have been split into four parts representing [Du], [D]2], [D2]]and [D22] terms. The components of these matrices can be evaluated from thefollowing equations:

[Dkir =

[Dklf =

[DkJ]"" =

(12.25)

where:a=lifk=l,0=lifk = l = O,± i s + i fA- /> 0,+ is - if A- /> 0,

= V2 otherwise;= V2 otherwise;i s - i f k - / < 1 ;is + i f k - l<\.

The nonlinear CFSAFEM stiffness matrix does not exhibit harmonicuncoupling, i.e. the zero terms in Equation (12.4) are now non-zero. This explainsthe important feature observed in nonlinear problems that solution displacementshave harmonic terms with orders different from that of the applied loads. Theharmonically coupled stiffness matrix from this formulation is very large and alarge amount of computer resources, similar to that for a full 3D analysis, is


required to invert it. The use of the linear CFSAFEM formulation to solve stronglynonlinear problems also requires a large amount of computer resources, since theelastic [D] matrix used in the formulation is very different from the true [Dep]matrix for the material, and a large amount of resources is required for the iterativecorrection process to account for this error. A compromise is to use the nonlinearformulation and omit the harmonically coupled terms, i.e. the k * I terms inEquation (12.24), thus allowing the system equations for each harmonic order tobe solved individually. The error associated with this omission is corrected usingthe nonlinear solution strategy in a similar manner to the error associated with thelinear formulation. Since the correct elasto-plastic stiffness matrix is used to solveeach set of harmonic system equations, the proposed partial nonlinear formulationis expected to require less computer resources for its correction process than thesimple linear elastic formulation.

The nonlinear formulation is now able to yield a more rational criterion for thesymmetrical uncoupling of the system equations. For symmetrical uncoupling tobe valid the [D]op and [D]po matrices must be zero. The four submatrices associatedwith each of these two matrices contain either only the 0th and cosine harmonics,or only sine harmonics. Thus the criterion the [D] matrix must satisfy forsymmetrical uncoupling is that the [Du] and [D22] parts of the [D] matrix aresymmetric functions of 0, and the [D]2] and [D2]] parts are asymmetric functionsof 6. The resulting [D]op and [D]po would then be zero. In a parallel symmetryanalysis the stresses associated with the top part of the [B] matrix, [Bl] (<r;., oz, a0and arz), are symmetrical functions of #, while the stresses associated with thebottom part, [B2] (or0, az0), are asymmetrical functions of 6. The strains satisfy thesame symmetrical conditions. Thus a material can only be legitimately analysedusing purely parallel symmetry terms if it satisfies the [D] matrix criterion whilesubjected to a stress and strain state associated with parallel symmetry. Similarly,for a purely orthogonal symmetry analysis the [D] matrix criterion must besatisfied for an orthogonal symmetry stress and strain state. An orthogonalsymmetry stress and strain state exists when the stresses and strains associated with[Bl] are asymmetrical functions and those associated with [B2] are symmetricalfunctions of 0.The tangent stiffness solution strategy could be implemented using the full

nonlinear CFSAFEM formulation, but the accuracy of the solution would bestrongly influenced by both the size of the solution increment, as discussed inChapter 9, and the number of harmonics used to represent the applied loads, thesolution displacements and the [D] matrices.

It should be noted that the linear CFSAFEM formulation is a particular case ofthe nonlinear CFSAFEM where the [D] matrix does not vary with 6, i.e. only the0th harmonic [D] matrix exists.


Z, V

12.3.5 Formulation for interface elementsA formulation for zero thicknessisoparametric interface elements for theCFSAFEM has been developed based on ^^-the 2D formulation of Day and Potts / ^(1994), see Section 3.6. The coordinate f fsystem for the isoparametric six and fournoded element is presented in Figure12.3. The global displacements aredefined in the same manner as for thesolid elements, using u, v, and wdisplacement components. Localdisplacements for the interface elementshave to be defined such that:

Aw7 is the incremental local tangentialdisplacement in plane r-z\Av/ is the incremental local normaldisplacement in plane r-z\Aw, is the incremental local circumferential tangential displacement.

The relationship between incremental global and local displacements can beexpressed in matrix form as:

(12.26)

For any point on the interface there are 2 sets of displacements, top and bottomdisplacements, each set describing the displacement on one side of the interface,see Section 3.6. The incremental global displacements on each side of an interfaceelement can be represented using isoparametric shape functions and a Fourierseries in the 6 direction, in a similar manner to the solid element formulation:

r, u

Figure 12.3: Coordinate systemfor interface elements

A«,Av, =Aw.

COStf-sina

0

sin acos or

0

001

Awi Av

Aw

Au=±Ni[£ (Uj cos/0+£// sin/0)] (12.27)where:

Nj is the shape function of the fh node defined in the element;Uj is the Ith cosine harmonic coefficient of radial displacement at the /lh node;

UJ is the Ith sine harmonic coefficient of radial displacement at the fh node;n is the number of nodes on each side of the interface;L is the order of the harmonic series used to represent displacement.The interface elements have three components of incremental strain:Ayp is the in plane shear strain;As is the in plane normal strain;Ay0 is the circumferential shear strain.


These can be related to the incremental global and local displacements as:

KAs [ =

We

. b o t topAu™ - AM;

Av,bot - Av,'0"cos a-sina

0

sinacos a

0

ol0 <1

Aub

Av bot- Autop

- Avtop

Awbot - Awtop(12.28)

Using Equation (12.27) for the incremental global displacements gives:

Kl +cosor +sina 0±sina Tcosa 0

0 0 +1£*,{

UJGOsW+UlsmW

V/cos!0+V/sinI0

WlcosI0+wSsml0

(12.29)

where:T is - for a top node and is + for a bottom node;± is + for a top node and is - for a bottom node.

This can be written as:

£ /To -[B2i ] cos/6{A<*} (12.30)

where:and {Ad/**} are respectively the parallel and orthogonal Ph order

harmonic coefficients of incremental displacement for the /lh node;[Blil is the top part of the strain matrix for the /lh node;[B2j] is the bottom part of the strain matrix for the fh node, so that:

[Bl,]=N, +cosa +sina 0±sma 0

0 +1] (12.31)

Similar to solid elements, the elasticity matrix [D] for the interface element canbe split into four parts:

ACT

ATa

' PAs (12.32)

where:Ar is the incremental in plane shear stress;ACT is the incremental in plane normal stress;AT0 is the incremental circumferential shear stress.The interface element stresses and strains have been formulated in such a

manner that the resulting equation for the internal incremental work has the sameform as that for the solid element:


&W=,* [Bl7]smk0

*f0 [[Bl^sinkOl ' |-[J?2,.]cos*0

h [[B2j]smw\ { j * \-lB2j\cosW r&6 darea

(12.33)

This can be simplified to give:

AW =

n\±± 2{Ad?*}T[Bll]T[Du][Blj]{Ad?} +2{Adr}T[B2i]T[D22][B2J]{Ad?*}= 1.7=1

{Arfr}T[B2,]T[i)22][B2;.]{Arff} rdarea

-{^)\B\i]T[Dn][B2j]{Ad1;} +' = 1 . 7 = 1

r darea

(12.34)

Since both interface and solid element equations are harmonically uncoupled,the system equations for each harmonic of a problem domain can be solvedindependently, even if it consists of both types of elements. Similarly, if the [Dn]and [D2l] terms in both solid and interface elements are zero, symmetricaluncoupling is also valid.

The linear elastic interface element formulation can be extended for nonlinearbehaviour using the same approach as for the solid element. The [D] matrix isrepresented using a Fourier series and the equation for virtual work now yields aharmonically coupled set of system equations. The [D] matrix pertinent for eachharmonic combination of equations is the same as those presented in Section 12.3.4for the solid elements.

12.3.6 Bulk pore fluid compressibilitySaturated clays are two phase materials consisting of a compressible solid phase,the soil skeleton, and a highly incompressible fluid phase - the pore water.Undrained behaviour is assumed if these clays are loaded quickly, such that thereis little dissipation of excess pore water pressure. This type of behaviour can bemodelled in a conventional finite element analysis by specifying a bulk pore fluidcompressibility, see Section 3.4. This formulation is now extended for use inCFSAFEM analyses.

For linear elastic material behaviour the principle of effective stress gives, seeEquations (3.2) and (3.5):


} + [Df]{Ae} (12.35)

Noting Equation (3.6), the second term in the above equation can be written as:

[Df]{A£} = {rj}Ke Aev (12.36)

where {J/}T:={ 1 1 1 0 0 0 }, Ke is the equivalent bulk modulus of the pore fluid andAev is the incremental volumetric strain. Combining Equations (12.35) and (12.36)

^ Asv (12.37)Accordingly, the incremental internal work done, AW, can be expressed as:

T T r sv dVol (12.38)The first integral is the work done by the soil skeleton, which was calculated inSections 12.3.1 and 12.3 A. Therefore only the second integral, the work done bythe pore fluid, AWf, has to be evaluated.

Using Equation (12.8), Aev can be expressed as:

/=1 1=0

\T - i

[2H,'](cos/0 { < } + sin/6> {A<//**}) (12.39)

where {//1}T = ( 1, 1, 1,0) . From here AWf can. be written as:

=\)±t (cos/£ {A<}+ sin/0 {M^fYBX1^{rjl} Ke(12.40)

I t {?7l}T[B^](cosk0 {M**}+ sinkO {Ad)"}) r d6dareaj] k0 'j=] k=0

Carrying out the d6 integral gives:

AWf=±± J2TI{A<} T [51 , 0 ]

( (12.41)

Ke {TtiyiBl'jUAdf}} r dareaThis equation gives the contributions to the global stiffness matrix associated withthe pore fluid compressibility. The system equations are obtained by adding thesefluid compressibility terms to the terms associated with the soil skeleton stiffnesscalculated in Sections 12.3.1 and 123 A. The fluid compressibility terms are bothharmonically and symmetrically uncoupled and their contributions to the paralleland orthogonal symmetry stiffness matrices are the same. For the Ith harmonic it is:

t ± 1 [tfljftol} K {rfl}T[Bllf] r darea (12.42)

' • = 1 7 = 1

In the above constant Ke formulation, the pore fluid pressure is a symmetricalfunction in a parallel analysis and an asymmetrical function in an orthogonalanalysis.


In an undrained material Ke is prescribed a very large value, such that thevolumetric strains are inhibited, see Section 3.4. There are two options for definingKe: to assign it a constant value, or to define it as a multiple of the bulk stiffness ofthe soil skeleton. In the former option care must be taken in assigning the value ofKe. It must be large enough to inhibit volumetric strains, but small enough to avoidill conditioning problems associated with a very large number in the stiffnessmatrix. Finding a suitable value for Ke is particularly difficult when the bulkmodulus of the soil skeleton changes significantly during the course of an analysis.Accordingly, the option for defining Ke as a multiple of the bulk modulus of thesoil skeleton is preferred, because these difficulties are avoided. A typical value forKe is one hundred to one thousand times the bulk stiffness of the soil skeleton.

However, considering the Fourier series formulation, Ke could now vary in the6 direction and would have to be expressed as a Fourier series:

e

where:

(12.43)

K°e , Kle and Kl

e are the 0th, Ith order cosine and /th order sine harmoniccoefficients ofKe\M is the number of harmonics used to represent Ke and need not be the sameas the number used to represent displacements, L.

Substituting this into the equation for A Wf gives:

*Wj• = J J t t (cos/<9 {M/*} +sin/0 {MD^Bl^iTjl}--n i = l /=0

(K + Z K cosl0 + Kle sin/0)- (12.44)

/=i

1 t {7l\}T[B\kj}{coske {M?}+&mkO {Ad***})rd0 darea.7=1 k=0

The &6 integral is carried using the solutions for a triple Fourier series, as shownin Appendix XII.5. From there:

; = 1./=1 1-0 k=0ni k

*=o

B\k] {dk"}r<\area(12.45)

where:a = 1 if k = /, =V2 otherwise;p = 1 if k = I = 0, = V2 o therwise ;± i s + i f / - k> 0, i s - i f / - k<\;+ is - i f / - £>0 , is + i f / - k<\.


In general, there is neither harmonic nor symmetric uncoupling. The criterion forsymmetric uncoupling is that Ke is a symmetrical function.

12.3.7 Formulation for coupled consolidationThe behaviour of saturated soils under any loading condition is strongly influencedby the rate at which the generated pore pressures are able to dissipate within thesoil mass. As noted in Chapter 3, the conventional finite element theory, describedin Chapter 2, can deal with either drained soil behaviour, where full pore fluidpressure dissipation occurs, or undrained behaviour where no dissipation occurs.The latter behaviour is achieved by introducing the effective bulk compressibilityof the pore fluid, as described in Section 3.4. Accounting for such behaviour in theCFSAFEM was described in the previous section (i.e. Section 12.3.6). Often soilbehaviour cannot be simplified to being either fully drained or undrained, forexample when partial drainage occurs during a loading stage, followed by longterm consolidation. To account for such behaviour the equations governing theflow of pore fluid and the mechanical behaviour of the soil must be combined. Thefinite element theory behind such a coupled approach was presented in Chapter 10.In this section this theory is extended for use with the CFSAFEM. Initially, onlysoil with a constant permeability is considered, but subsequently the theory isextended to account for soils which have variable permeability.

The pore fluid pressures in the problem domain are described by the elementpore fluid shape functions and the Fourier series of the nodal pore fluid pressuresof the element:

Pf = S Npi ( I plfi codO+pl

f, sin/0i=\ 1=0

where:TV is the element pore fluid shape function for the fh node;

(12.46)

p'fj and p!fi are respectively the Ith cosine and sine coefficients of pore fluid

pressure for the ?h node;m is the number of pore fluid pressure nodes in the element which is notnecessarily equal to the number of displacement nodes in the element, n.The hydraulic gradient, {V/i}, can be defined as:

dpf

drdP/dz

dp/rd6

m L

-ft

P'-coslddr

dz-lN P' s'mW

•pf, + \

dNn

drdNni

-sin/6>

dz -sin/0

-^-cosW

•f

(12.47)


where:yf is the bulk unit weight of the pore fluid;{iG} is the vector parallel to gravity, where

\dh)drdh_

dz_

~d6

and dh

dr dzand

The incremental volumetric strain at any point is defined as:

A£v = {r]}T{A£} (12.48)The incremental strain vector {AE} can be expressed in terms of the Fourier seriescoefficients of incremental displacement as in Equation (12.8):

[Ul,']sin/0 ,„(12.49)

Thus the incremental volumetric strain can be expressed as:

Asv = t t tol}T[*l,-] (cos/0 {A<//*}+ sin/6> {A<*}) (12.50)/=1 /=0

The two governing equations are the equilibrium equation:

| {A s}1 {A a} dVol +1 Asv Apf dVol = external work done (12.51)and the continuity equation:

W (12.52)

where [k] is the matrix of permeabilities and Q represents any sources and/or sinks,see Chapter 10. Q is expressed as a Fourier series and the parallel and orthogonal

components are Q!* = Ql and Q1** = Q1 respectively.The first integral in Equation (12.51) is the same as that for the incremental

internal work done by a solid element without consolidation, see Equation (12.38).The second integral can be expressed in terms of the incremental nodal pore fluidpressure and displacement Fourier series:

\Asv Apf dVol =j]±t (cos/0 { < } + sin/0 {AdDYiBl'f {rj\} •

Ap)} sin&<9) r dOdarea(12.53)

/=1 k=0


where:= I Npiit *]%CQsI0+Apl

fi sin/60i\ 10i=\ 1=0

and Ap'fl and Aplfi are respectively the 7th cosine and sine coefficients of

incremental pore fluid pressures for the fh node.Carrying out the d6 integral and using the orthogonal properties of Fourier seriesgives:

\AsvApf dVol

}T[/?l,°]T{/7l} Npj Ap°£ +i{Ad!T[Blii]T{T7l} Np} A^

1=1(12.54)

r}T[JBl,']T{77l} Npj Ap'g r&area

where [JL^]7 = 2TT J r [51/]T {i/l} NpJ darea ifl=0; = n\r [Biff {J/1} iVotherwise.

This equation gives the terms that relate the applied incremental loads to theincremental pore fluid pressures. These equations are both symmetrically andharmonically uncoupled. The cosine harmonic coefficients of incremental porefluid pressure are only associated with the parallel loads, hence they are called theparallel components of incremental pore fluid pressure. Similarly, the sinecoefficients are the orthogonal components of incremental pore fluid pressure:

A^J" = Ap1} and A^f = Ap1** (12.55)

The first integral in Equation (12.52) can be treated in a similar manner to obtainthe equation:

\Apf AsvdVol=f,f,'=1f (12.56)

£The second integral in Equation (12.52) can be rewritten as:

)T [k] {Vh}dVol =


The [A:] matrix is split into four parts in accordance with the two parts of the [E\matrix and the d6 integral is carried out. The hydraulic gradient has been dividedinto to two parts, consisting of the pore fluid pressure and gravity components.Accordingly, Equation (12.57) can be written in two parts:

1 [k]Vpf dVol =—jSZYf i=lj=l

l; r darea

—IXIL^ r darea

(12.58)and

(Ap%)T[El°]T([ku]{iGl} +[kn]{iG2}) r darea

^{n,} ( 1 2 5 9 )

where {«,} = 2n J [^l,0]7 ([A,,]{/G1} + [h^iki)) r darea.Both equations are harmonically uncoupled, with the gravity term only

affecting the 0th harmonic. This is correct if gravity acts in a fixed direction in ther-z plane (i.e. z direction). If the direction of gravity acts out of this plane and isdependent on r and 6, the equations became more complex. The dashed line inEquation (12.58) separates the components that are symmetrically uncoupled fromthose that are coupled. Inspection of these terms reveals that the condition forsymmetric uncoupling is that the [kl2] and [k2]] components of the [A:] matrix arezero. Equation (12.58) can be written as:

i=lJ=l (12.60)tT m Vpf

where: [ < ] = 2JLj[£i?]T[*11][Ei;]rdan«i

'Ji r'liEi1^ L-[*2il o J W ]

r darea


A simple time stepping approach is used to carry out the integral from time t0to t = to+At, see Chapter 10:

(12.61)

where:pf0 is pf at time t = t0;Apfis the change in pf over At;p is the parameter that defines the average pore water pressure over the timestep, i.e. p/v=pfQ+pApf.

For the simple constant [D] and constant [k] case, all the components ofEquations (12.51)and (12.52) are harmonically uncoupled. If the criterion forsymmetric uncoupling is satisfied by both the [D] and the [A:] matrices, theresulting equations are also symmetrically uncoupled, with the orthogonal andparallel equations having the same stiffness contributions. Thus for any harmonicorder we can combine Equations (12.51) and (12.52) and write them in matrixform. For the parallel 7th harmonic we get:

-j3At[<PG] ]"(12.62)

where «/ic)) = nG if / =0 ; = 0 otherwise.The same equation can be used for the orthogonal Ith harmonic. However, if

either the [D] matrix or the [k] matrix do not satisfy the symmetry criterion, bothorthogonal and parallel coefficients have to be solved simultaneously:

0[K'GY[L'G]0 -

(12.63)

The material permeability may not be a constant and could vary withstress/strain level. Accordingly, [k] could vary in the circumferential direction. Inthis case [k] can be written as a Fourier series:

[A] = [ '] cos0+[k] ']sin2<9+ ... +(12.64)


where [k°], [k1] and [kl] are matrices containing the 0th, /th order cosine and Ith

order sine harmonic coefficients of the components of the [A:] matrix respectively.The solution to Equation (12.52) now involves the integral of the product of

three Fourier series. The solution is harmonically coupled and is similar to that fora variable [D] matrix:

\V{Apf)J[k]VpfdVol =

1=0 1=\ k=\

V/=o k=\

(12.65)

where:

y.r

•,']

a!*,*,"']+>8[*,*,+'] ±«lT«[4']-A4'] «

«[*n"']-A*n+'] ±«[^[4V^2r] «[

±a[**-/] + /3[*1V/] - «

- fl[*"1+^t'1 ±a

+a[Af,-'] + yS[Af1+'] «|

:*iV']-y9[**2+']:4']-A4'i.

4'1]+#*£']4VA4'].

[Af2-'] + /9[4+/]|_/r99 J — p\k^^ \

[in*]

> r darea

r darea

• r darea

r darea

In the above equations:a = 1 if k = I,P = 1 if k = I = 0,±is + i f £ - / > 0,+ i s - if*- /> 0,Similarly, the solution for Equation (12.59) now involves the integral of the

product of two Fourier series, thus the gravity term affects all the harmonics:

=Vi otherwise;= Vi otherwise;is - if^- /< 1;is + if A - l<\.

r darea

(12.66)


The [LG] terms in Equation (12.63) are unchanged since they are not affectedby [k] or [D]. However, the p^ terms now have to be multiplied with all theharmonically coupled [0G] matrices. Since the equations are now harmonicallycoupled, an equation with a form similar to that of Equation (12.63) can be writtento relate any 7th harmonic right hand side with any kthharmonic left hand side:

/=0

7=0

where ((([Lj]))) = [Lj] if / = k ; =0 otherwise.

The general criterion for symmetric uncoupling in this case is that the [ku] and[k22] parts of the [£] matrix are symmetric functions of 6, and the [k]2] and [k2l]parts are asymmetric. As discussed previously regarding the nonlinear materialbehaviour (i.e. elasto-plastic), CFSAFEM analysis of coupled problems can beperformed either using the constant [k] formulation or a compromised variable [k]formulation, in which the harmonically coupled terms are ignored. In both casesappropriate correction to the right hand side of the finite element equations will benecessary if [A:] is variable.

12.4 Implementation of the CFSAFEMMA A IntroductionFrom the formulation provided in the previous sections it is clear that theCFSAFEM involves many extensions and enhancements of the conventional finiteelement theory presented in Chapters 2, 10 and 11. In particular, the CFSAFEMformulation is expressed in terms of Fourier series coefficients, while theconventional finite element formulation is in terms of real values. This clearly hasimplications for the computer code, the boundary conditions and the output froman analysis.

Inclusion of the CFSAFEM into an existing finite element program involves aconsiderable effort. There is a large increase in data storage requirementscompared to a conventional 2D finite element analysis. This is due to the additionalcoordinate direction, the two additional components of stress and strain and the useof a number of harmonic coefficients in a CFSAFEM analysis. However, due to


the harmonic uncoupling this storage requirement is likely to be considerablysmaller than that required for a comparable conventional 3D analysis.

Efficient data management routines are required to store data and to convert itbetween harmonic coefficient and real values. Methods of specifying inputboundary conditions also have to be enhanced to provide a user friendly interfacewith the finite element program. Significant modifications must also be made tothe nonlinear solution algorithm to enable the correct adjustment to be made to theright hand side of the finite element equations. In the following subsections someof these topics are considered in more detail. The material presented is based onthe Authors' experiences with implementing the CFSAFEM into the computercode ICFEP.

12.4.2 Evaluating Fourier series harmonic coefficientsAll the variables used in a CFSAFEM analysis have to be expressed as Fourierseries, because the CFSAFEM is formulated entirely in terms of Fourier seriesharmonic coefficients. This is a key component of the MNR solution strategy fornonlinear CFSAFEM, since it is used to evaluate the right hand side correctiveloads, as described in Section 12.4.3. It is also used to interpret complex boundaryconditions and to formulate the partial nonlinear stiffness matrix, see Section12.3.4. Thus a general approach has to be devised for expressing the distributionof variables in the 6 direction as a Fourier series.

Consider a variable x which is a function of 6. The harmonic coefficients haveto be evaluated such that x can be expressed as:

.. + (12 68)Ar/cos/0+Ar/sin/0 + ...

If there is an explicit expression for x, i.e. x =./(#), then harmonic coefficients canbe evaluated from the equations:

2% _J/

X1 = — f / ( # ) coslO&O (12 69)

X;=-J/(i9)sin/<9d6>

Often there is no explicit expression for x, instead values of x at specific 6values are known, e.g. x} at 6{, x2 at 62, x3 at 03,... , xf at #,, ... , xn at 6n, where nis the number of known values of x. There is no unique solution to this type ofproblem, since an assumption has to be made regarding the value of x between anytwo specified values. Two methods are suggested: (i) the stepwise linear methodand (ii) the fitted method.


12.4.2.1 The stepwise linear methodThis method is based on an approach suggested by Winnicki and Zienkiewicz(1979). Using the known values of x they assume a stepwise linear distributionwith 6, see Figure 12.4. Thus the value of x at any value of 9 is:

(12.70)

(*,,Equations (12.69) are then

integrated numerically to obtain theharmonic coefficients and Bode'sintegration rule is suggested. There isa trade off between the errorassociated with this numericalintegration and the amount ofcomputer resources required toevaluate it. y

This method is improved upon by Figure 12.4: Step wise linear methodcarrying out the integrationanalytically. This has the advantage of both eliminating any integration errors andreducing the computer resources required. The integration is performed inAppendix XII.6 and the solutions are presented below:

Stepwise linear distributionFourier series representation

£ -+1 - cos£6j.)(12.71)

where:Xn+] = X\

The harmonic coefficients obtained from Equation (12.69) would fit anycontinuous function exactly, if an infinite number of harmonic terms were used inexpression (12.68). This is impractical in any numerical algorithm and invariablya finite number of terms is used. Usually a truncated Fourier series is used wherethe higher order terms are ignored and sufficient terms are considered such that theerror associated with ignoring these higher order terms is small. Additionally, eventhough the resulting Fourier series does not fit the stepwise linear distributionexactly, it will still give a good representation of x, which may even be better thanthe stepwise distribution.


12.4.2.2 The fitted methodIn this method it is assumed that x canbe represented by a Fourier series thatpasses through all known values of x,see Figure 12.5. The number ofharmonic terms used in the Fourierseries representation is an importantparameter. If there are more harmonicterms than known values of x, anumber of possible solutions exist. If

Fourier series representation

Figure 12.5: Fitted method

there are less harmonic terms, a solution does not exist in general. A uniquesolution exists if the number of harmonic coefficients is equal to the number ofknown values. This condition is achieved by using a truncated Fourier series. Theharmonic coefficients for the unique solution can be obtained by substituting eachknown value of x and 6 into Equation (12.68). This yields n equations with nunknowns, where n is the number of unknown values specified, and can beexpressed in matrix form as:

x°X7

, , x sinLO]

cos<92 sin<92 .... cosL02 sinLO2

1 cos sin s'mLB3

1 cos<9/;_, sin<9,,_,

s'mOn cosLOn sinL6)n

X2

w - 1X 2

(12.72)

where L is the order of the truncated Fourier series used (equal to Vi(n-\)) and canbe rewritten as:

x = [H]X (12.73)where

x is the vector of known values x>;X is the vector of unknown harmonic coefficients;[H] is the harmonic transformation matrix that is a function of the 6 values atwhich x is known.The harmonic coefficients, X, can now be solved for by inverting [//] and

premultiplying x by [H]~l. If the 6 values at which x is known are unchanged, then[//] is unchanged. Thus a number of AT corresponding to a number of JC can beobtained very easily by premultiplying the various x by the same [//]"'.

Solutions have been derived for the special case of having known x values atequispaced 6 values, see Appendix XII.7. The number of known values, n, isassumed to be odd such that the resulting truncated Fourier series has the same


number of sine and cosine harmonic terms. If the first x value, JC1? is at 6 = 0°, thesubsequent known x values, xh will occur at 6 = (i-l)a, see Figure 12.6.

The harmonic coefficients can then be evaluated using equations:

Xk =- (12.74)

2TT

xk=±£This equispaced fitted approach is

the most efficient method forobtaining harmonic coefficients. It is,however, only applicable for thespecial case when the x values areequispaced and the number ofharmonic terms (2Z+1) is equal to n.If these criteria are not satisfied, therobust stepwise linear approach isrecommended, since this approachplaces no limitation on either thenumber of harmonic terms used, or Fi9ure 126: Location of x valuesthe location of the x values. In the for equispaced fitted methodremainder of this chapter thesemethods will simply be referred to as the fitted and stepwise approach.

12.4.3 The modified Newton-Raphson solution strategy/ 2.4.3.1 In troduc tionThe MNR solution strategy for nonlinear CFSAFEM is an iterative procedure. Forany imposed boundary condition a trial solution is obtained by solving the systemequations associated with the current global stiffness matrix, as described earlierin this chapter. It is acknowledged that there is possibly an error associated withthis trial solution and a corrective right hand side load is evaluated. This correctiveload is then used to obtain the next trial solution. The process is repeated until thecorrective loads are small. This procedure is very similar to the MNR procedurefor conventional nonlinear finite elements, see Chapter 9. However, the CFSAFEMformulation is in terms of Fourier series coefficients. Thus the corrective right handside loads have to be expressed as Fourier series and are evaluated from the trialsolutions which are also expressed as Fourier series. The material constitutive lawsare used in the process of calculating these corrective loads. These laws areexpressed in terms of real values rather than harmonic coefficients. Thus aprocedure has to be developed which enables a harmonic correction load to beevaluated from the harmonic trial solution, using the material constitutive laws.


1 2.4.3.2 High t hand side correc tionThe corrective right hand side load is expressed as Fourier series coefficients ofnodal load. They are evaluated by comparing the coefficients of nodal load,obtained from the imposed boundary conditions, with the coefficients of nodal loadconsistent with the trial solution displacements. The latter loads are obtained bycalculating, at a number of sampling points within the problem domain, the stresschanges associated with the solution displacements. This is performed bysubstituting the harmonic solution displacement into Equation (12.8) to calculatethe real incremental strains at the sampling points. A stress point algorithm is usedto integrate the material constitutive laws along these strain paths to obtain theincremental stresses, see Section 9.6.2. A volume integral has to be carried out toobtain the harmonic coefficients of nodal load consistent with the stress state in theproblem domain. The stresses at the sampling points are used to evaluate thisintegral, the solution of which is presented in Appendix XII.3. The integral isdivided into two parts, a dO integral and a darea integral, where 'area' is the areaon the r-z plane.

The darea integral is carried out using a Gaussian integration technique and,accordingly, the sampling points are located at Gaussian integration points in ther-z plane. To perform the d8 integral it is necessary to know the 8 coordinates ofthe sampling points. It is therefore convenient to situate the sampling points on anumber of constant 6 planes (slices). Their locations within each slice isdetermined by the Gaussian integration order used and the finite element mesh.The d8 integral is carried out analytically, but requires the stresses to be expressedas a Fourier series in the 9 direction. There is no definitive method of obtaining theFourier series representation of stress from the stresses at the sampling points.Similarly, there are no guidelines for the number and location of the slices in the6 direction. A rational approach is suggested below.

The number of slices used in an analysis should be influenced by the order ofthe Fourier series required to represent stresses. If there were the same number ofslices as harmonic coefficients of stress in an analysis, a Fourier seriesrepresentation of stress could be obtained, which gives the appropriate stresses oneach slice. Theoretically, increasing the number of harmonics would not increasethe Fourier series accuracy since the stresses on the slices could already berepresented exactly. Conversely, increasing the number of slices means that theFourier series would not be able to represent the stresses on all the slices exactlyand the benefit of these additional slices cannot be fully realised. The Fourier seriesused to represent stresses in a CFSAFEM analysis should have the same order asthe Fourier series used to represent forces and displacements. Harmoniccoefficients of stress of a higher order would not influence the CFSAFEM analysis,as shown in Appendix XII.3. Thus it is proposed that the number of slices shouldbe the same as the number of harmonic coefficients used to represent forces anddisplacements in a CFSAFEM analysis. It is also suggested that the slices areequispaced around the circumference, in which case the efficient fitted method, seeSection 12.4.2.2, for obtaining harmonics can be used.


MA A Data storageThe data storage required for CFSAFEM analysis differs from that of conventionalfinite element analysis in two important aspects. Firstly, for each nodal variable,e.g. nodal displacement, it is necessary to store a number of harmonic coefficients,as opposed to a single quantity. Likewise, for each variable at an integration point,e.g. stress component, it is necessary to store a number of quantities, either in theform of harmonic coefficients, or as real values one for each slice. At differentstages during the analysis each variable (e.g. nodal displacement or force, stress orstrain component) is needed either in harmonic form or as real value. However, itis both very inefficient and cumbersome to store data in two different forms.Consequently, efficient data management routines are required to store the data inone form and to convert between harmonic coefficients and real values and viceversa, see Section 12.4.2. The only decision then is in which form (real values orharmonic coefficients) should each variable be stored. As the nodal variables in aCFSAFEM analyses are expressed purely in terms of Fourier series harmoniccoefficients, it seems sensible that these values are stored as harmonic coefficientsand converted to real values when required for output purposes. The decision ismore difficult for the state variables stored at integration points, e.g. stresses,strains and hardening parameters. The stress point algorithm requires thesevariables to be expressed as real values, however the calculation of the harmoniccoefficients of nodal force, consistent with the stresses at the sampling points,requires that the stresses are expressed as Fourier series. The data output algorithmsare also likely to require that these variables are expressed as Fourier series, so thattheir values at any 6 value can be exactly calculated. For these reasons, and to beconsistent with the storage of nodal variables, it seems appropriate to store theharmonic coefficients.

An important property of any coefficients obtained using the fitted method isthat real values calculated from these coefficients are exactly equal to the realvalues the coefficients were derived from, see Figure 12.5. This interchangeabilitybetween the harmonic coefficients and real values dispenses of the need to storeboth harmonic coefficients and real values of any variable. Coefficients obtainedusing the stepwise method do not exhibit this property and there is a slight error,8, when they are used to evaluate the real values that they were derived from, seeFigure 12.4. The magnitude of this transformation error reduces as the number ofharmonics used increases and, in practice, should be small.

A situation where the transformation error would be important is in the storageof the soil state variables at sampling points on the predefined slices. The stresspoint algorithm, and in practice the projecting back subalgorithm, in the nonlinearsolution strategy ensure that the state variables are consistent with the prescribedstrain path, see Chapter 9 and Appendix IX. 1. These real values of the statevariables are then expressed and stored as harmonic coefficients, ready for use bythe CFSAFEM algorithms and the data output algorithms. When the stress pointalgorithms next require these variables, it is critical that the exact real values areevaluated, since even a small error could result in an illegal stress state, e.g.


stresses exceeding yield. In this case the stress point and other associatedalgorithms would not be valid and some arbitrary form of correction is required,which could result in large errors. It is for this reason that the fitted method isrecommended in preference to the stepwise method when calculating the loadvector for the right hand side correction to the finite element equations.

12.4.5 Boundary conditionsAs the CFSAFEM is formulated in terms of Fourier series coefficients, allboundary conditions must be specified in this manner. However, for most practicalboundary value problems the boundary conditions are likely to be in terms of realvalues (e.g. a displacement or force in a certain direction). Consequently, beforeany analysis can be undertaken, these real values must be converted to anequivalent Fourier series representation. Three options for specifying boundaryconditions are evident:

The user determines the Fourier series that is equivalent (or a goodapproximation) to the real boundary condition, and inputs the harmoniccoefficients directly.

- The user inputs a set of numbers representing the value of the boundarycondition at equispaced values of 6. The number of values must be equal to thenumber of Fourier coefficients required. The computer program can then usethe fitted method (see Section 12.4.2.2) to obtain the required harmoniccoefficients.The user inputs two sets of numbers. The first set contains values of theboundary condition at a series of 9 values. The second set contains themagnitudes of these 6 values. The computer program then uses the stepwisemethod (see Section 12.4.2.1) to obtain the harmonic coefficients.

In the Authors' experience all three of the above options should be available to theuser of a computer program. Examples of their use are given in Volume 2 of thisbook.

12.4.6 Stiffness matricesThe partial nonlinear formulation, as described in Sections 12.3.4 and 12.3.7,requires that the constitutive matrix [D] and permeability matrix [K\ are expressedas a Fourier series. These harmonic matrices are required for every integrationpoint located in the r-z plane. To achieve this, for each integration point the real[D] and [K] matrices are calculated at the associated circumferential series ofsampling points, i.e. on each slice. The harmonic coefficients for these matrices canthen be obtained using the fitted method.

The partial nonlinear no symmetry CFSAFEM formulation results in a non-symmetric stiffness matrix when the stiffness contribution for a parallel symmetryload on an orthogonal symmetry displacement is not the same as that for thecorresponding orthogonal symmetry load on the corresponding parallel symmetry


displacement. As this situation is generally applicable, the global stiffness matrixmust be inverted using a non-symmetric solver.

In all other situations, using the partial nonlinear CFSAFEM, the symmetry ofthe global stiffness matrix depends on the nature of the constitutive matrix [D] andpermeability matrix [A]. For example, if a non-associated elasto-plastic constitutivemodel is employed, the global stiffness matrix is non-symmetric.

1 2.4.7 Simplifications due to symmetrical boundary conditions12.4. 7.1 IntroductionAs noted in Section 12.3.2, many boundary value problems have a symmetry aboutthe 0 = 0° direction, such that the imposed boundary conditions consist of purelyparallel or orthogonal symmetry terms. In the parallel symmetry case allcomponents of the boundary conditions in the r and z coordinate directions havea symmetric variation with 8, whereas the components in the 6 direction have anasymmetric variation. Consequently, for all the r and z components of theboundary conditions the sine terms in their Fourier series have zero coefficients,and for the 6 components the zeroth and cosine terms have zero coefficients.Conversely, for orthogonal symmetry the components of the boundary conditionsin the r and z direction have an asymmetric variation and those in 6 direction havea symmetric variation with 6. This results in zero values for the coefficients of thezeroth and cosine terms associated with the r and z components of the boundaryconditions, and for sine terms associated with the components in the 6 direction.

Consequently, for either a parallel or orthogonal symmetry problem, almosthalf of the harmonic coefficients are zero. The majority of previousimplementations of the CFSAFEM have capitalised on the above consequences andhave been coded to deal only with either parallel symmetry (e.g. Winnicki andZienkiewicz (1979)), or orthogonal symmetry (e.g. Griffiths and Lane (1990)), byignoring the harmonics associated with the zero coefficients. While suchapproaches restrict the type of boundary value problem that can be analysed, theylead to a large saving in computer resources required for an analysis andsignificantly simplify the solution algorithms. This saving arises due to the fact thatonly half of the Fourier series terms are considered. However, as noted in Section12.3.2, 12.3 A and 12.3.7, this approach of considering only parallel or orthogonalterms is theoretically valid only if two requirements are met. The first requirementis that the boundary conditions satisfy the appropriate symmetries, and the secondrequirement is that the system equations are symmetrically uncoupled. Thecriterion for this second requirement concerns the nature of the constitutive matrix[D] and the permeability matrix [A:], see Sections 12.3.2, 12.3.4 and 12.37. If thesecond condition is not satisfied, it is then incorrect to perform an analysis withonly parallel or orthogonal symmetry terms, even if the boundary conditions satisfythe symmetry requirements. In such a situation a full non-symmetric analysis,accounting for all terms in the Fourier series, must be performed. An examplehighlighting this problem is presented in the next section.


Initial stresses can also present potential problems for parallel and orthogonalsymmetry analysis. Consider the situation in which the initial stresses in the soil donot vary in the 6 direction. Here all the stress components can be represented bya Fourier series when all the harmonic coefficients, except the zeroth harmonic, arezero. However, from Appendix XII.3, zeroth harmonic components of {A<rl} (i.e.A<7,., A<7Z, AT,., and Ac ) yield parallel loads and zeroth harmonic components of{Ac2} (i.e. AT,.# and Arr6,) yield orthogonal loads. This questions the validity of{(xl} initial stresses in an orthogonal analysis and {al} initial stresses in a parallelanalysis. This problem is partly overcome if an incremental form of the CFSAFEMis implemented, which is likely to be the case for analysing nonlinear problems. Atthe beginning of an analysis, the initial stresses are assumed to be in equilibriumwith themselves and with the initial boundary stresses. The incremental form ofCFSAFEM then evaluates the stress changes, which should be consistent with theapplied nodal loads. Thus, irrespective of the actual values of stress, a parallelanalysis is valid if the stress changes do not yield an orthogonal load. Similarly anorthogonal analysis is valid if there are no stress changes giving parallel loads. Ifthese conditions are not satisfied, a no symmetry analysis should be used. A nosymmetry analysis is also required if on first loading from the initial stress state theconstitutive matrix [D] depends on the initial state of stress, see the example givenbelow in Section 12.4.7.2.

In a linear consolidation CFSAFEM analysis the gravity vector, {iG}, gives riseto a zeroth harmonic flow term on the right hand side of the finite elementequations, see Section 12.3.7. This is consistent with the parallel symmetrycriterion for a symmetrical flow term. However, there is a conflict between thegravity zero harmonic terms and the orthogonal symmetry criterion for anasymmetrical flow term. This conflict can be resolved by specifying a hydrostaticinitial pore fluid pressure regime. The flow terms resulting from this regimecounterbalance the gravity flow terms such that there are no net symmetrical righthand side flow terms. More generally, a linear or nonlinear consolidationCFSAFEM requires that the initial pore fluid pressures and {ia} are consistent withthe symmetry of the analysis. If the symmetry criteria cannot be satisfied, a nosymmetry analysis should be undertaken.

It is the Authors' experience that all three options, e.g. parallel, orthogonal andno symmetry, are useful and should be available in a general purpose finite elementcode. Although a no symmetry analysis is considerably more expensive in termsof computer resources than either a parallel or orthogonal analysis, many problemshave to be analysed in this way because they do not satisfy all the requirements foreither a parallel or orthogonal analysis. Also, due to the uncertainties involved withsatisfying the requirements concerning the nature of the constitutive matrix [/>], itis often useful to compare no symmetry and the relevant parallel or orthogonalsymmetry analysis, to confirm that the requirements are satisfied. It should benoted that when analysing problems with parallel or orthogonal symmetry onlyhalf of the geometry (i.e. 0° < 6 < 180°) needs to be considered, while in a nosymmetry analysis the full geometry (i.e. 0° < 6 < 360°) must be used.


12.4.7.2 Examples of problems associated with parallel andorthogonal analysis

To illustrate some of the problems associated with implementing a CFSAFEMformulation for analysing either purely parallel or orthogonal symmetry problems,and to emphasise the importance of having the option for a no symmetry analysis,two examples are presented.

The first example considersthe hollow cylinder problemdescribed by Gens and Potts(1984). This problem involves theanalysis of a hollow cylinder soilsample subjected to a torsionaldisplacement, while maintainingthe inner and outer pressures equaland constant. In addition, the axialstrain is kept to zero and it isassumed that there are no endeffects. As a consequence of thislatter assumption, stresses andstrains are independent of

Hollow cylinder sample Finite element mesh

Figure 12.7: Hollow cylinder samplesubjected to torsioncoordinates z and 9 and it is

therefore possible to analyse a section of the sample as shown in Figure 12.7. Alsoshown in this figure is the finite element mesh used to perform this analysis. Itrepresents a section of the hollow cylinder wall, with an inner radius of 100 mmand an outer radius of 125 mm. The wall is divided into ten eight nodedisoparametric elements of equal radial size. To simulate the test conditions acircumferential displacement is applied along the boundary CD, whereas the samedisplacement component is held to zero along AB. No restriction is imposed on theradial movements, but the vertical movements along AB and CD are set to zero.To ensure strain compatibility, the applied displacements along CD have to beproportional to the value of the r coordinate. Since the inner and outer pressuresremain unchanged, a zero normal pressure change is applied during the analysis toboundaries AD and BC.

Gens and Potts (1984) analysed this problem using a special quasi axi-symmetric finite element formulation. However, the boundary conditions describedabove can be presented using orthogonal symmetry, with only the zeroth harmonicterms. Analyses are therefore performed using both an orthogonal symmetry anda no symmetry CFSAFEM formulation and the results are compared with those ofGens and Potts (1984). In these analyses the modified Cam clay model is used torepresent the soil and the material parameters and initial stress conditions are givenin Table 12.1. Initially, undrained analyses are performed by specifying anequivalent bulk modulus for the pore fluid, Ke, to be 100 times the bulk modulusof the soil skeleton, see Section 3.4. The results of both the CFSAFEM analyses(i.e. orthogonal and no symmetry) are identical and are in agreement with those of


Gens and Potts (1984). This can be seen in Figure 12.8, which shows the variationof the moment, applied to the top of the sample, with angular rotation (i.e. twist).

The analyses are repeated assuming drained soil conditions, i.e. all parametersand boundary conditions the same, except that the effective bulk modulus of thepore fluid, Ke , is set to zero. The result of the two CFSAFEM analyses arecompared with the results from the quasi axi-symmetric analyses in Figure 12.9,again in terms of moment versus rotation.

Figure 12.1: Material properties for hollow cylinder problem

Overconsolidation ratio

Specific volume at unit pressure on virgin consolidation line, vx

Slope of virgin consolidation line in v-\np' space, X

Slope of swelling line in v-lnp' space, K

Slope of critical state line in J-p' space, Mj

Isotropic initial stress

Elastic shear modulus, G

1.0

1.788

0.066

0.0077

0.693

200kPa

18675kPa

It can be seen that the quasi axi-symmetric and no symmetry CFSAFEMresults are identical. These results showa much softer response and a higherultimate load to those from the undrainedanalyses presented in Figure 12.8.However, the orthogonal CFSAFEManalysis gives very different results,which are similar to those from theundrained analyses. The reason for thediscrepancy is that in a drained problemthe zeroth harmonic coefficients of radialdisplacement are non-zero. However,these are defined to be zero in anorthogonal CFSAFEM formulation.Thus, even though the boundaryconditions can be specified usingorthogonal symmetry, the correctdrained solution cannot be obtained. Theresults tend towards the undrainedanalysis results, which, due to theincompressible material behaviour (i.e.

0.1 0.2Rotation (10"3rad)

Figure 12.8: Undrained hollowcylinder test

+ Quasi-axisymmetricOrthogonalNo symmetry

1 2Rotation (10° rad)

Figure 12.9: Drained hollowcylinder test


Pisa* view

Figure 12.10; Schematic view ofhorizontally loaded pile section

high Ke value), have no radial displacements and under predict the correct drainedultimate load by over 40%. This is an example of a problem where a no symmetryanalysis is required due to the material properties, even though the boundaryconditions can be specified using orthogonal symmetry.

The second example considers arigid horizontal disk pushed laterallythrough a soil mass with a circularboundary. This geometry isrepresentative of a horizontal sectionacross a laterally loaded pile, asshown in Figure 12.10. It is assumedthat there are no vertical strains,hence the problem can be analysedusing both the CFSAFEM and a planestrain analysis. The boundaryconditions and typical meshes usedfor both types of analyses are shownin Figure 12.11.

Interface elements are located atthe disk-soil boundary so that variousinterface conditions, e.g. rough,smooth, rough with breakway,smooth with breakway, can beinvestigated.

The disk diameter, D, is 2m andthe boundary diameter is 40m. Onlythe first Fourier harmonic coefficientis required to represent the boundaryconditions for the CFSAFEManalysis. However, in a nonlinear z,vanalysis the solution displacements t ,J), © © = -8 sinehave to be represented using moreharmonics, see Section 12.3 A. For thepresent example ten harmonics areused. If the direction of loading is inthe 0=0° direction, the boundaryconditions imposed have parallelsymmetry, if it is in the 6=K/2direction they have orthogonalsymmetry, and if it is in any otherdirection they have no symmetry. For

a) Boundary conditions and mesh usedfor plane strain pile section analysis

i—1>r, u

u = 6 cos0

I 1 1 I I _L

b) Boundary conditions and mesh usedfor CFSAFEM pile section analysis

Figure 12.11: Boundary conditionsand meshes for pile section

analysesthe present example the problem is displacement controlled, with a uniformdisplacement in the 0=0° direction imposed on the rigid disk. A displacement S isspecified by prescribing the nodal displacements at the disk-soil interface to be:


2000

1500

20 40 60Displacement (mm)

Radial displacement, u= S cosO ;Circumferential displacement, co = -8 sin#.

The reaction force on the disk due to the imposed displacement can beevaluated. The net reaction force in the 0=0° direction can be obtained by addingthe first cosine harmonic coefficients of radial nodal reactions and subtracting thefirst sine harmonic coefficients of circumferential nodal reactions, see AppendixXII.4.

For case 1, the soil is modelled asmodified Cam clay, using the soilproperties given in Table 9.3 and hasan initial isotropic stress state withGf.'=a2'=a0'=200 kPa. It is assumed tobehave undrained during loading andthis is simulated by specifying a highbulk modulus of the pore fluid, Ke, asfor the first example. The soil failsunder plane strain conditions and,since an associated flow rule is used,the Lode's angle at failure is 0°, seeSection 7.12. For these conditions, i.e.the soil properties given in Table 9.3and an initial isotropic stress state of200 kPa, the undrained shear strengthfor the soil is 75.12 kPa, see Appendix VII.4. The shear strength of the interfaceelements is also assigned a value of 75.12 kPa. The analytical limit load for thisproblem is 1 \.94SUD (Randolph and Houlsby (1984)), giving a value of 1794.07kN. The results show that both parallel symmetry and no symmetry CFSAFEManalyses are identical and are within half a percent of the plane strain analysis, asshown in Figure 12.12. The limit load achieved is within one percent of theanalytical solution.

To illustrate the problems associated with the past implementations of theCFSAFEM, wherein only the parallel symmetry terms in the formulation areimplemented, a further analysis, case 2, was performed. In case 2 the soilproperties are the same as case 1, but there is a different initial stress regime: a,.'=av' = o0' = 138.22 kPa and xrB = -69.0/r kPa, where r is the radial coordinate andan anticlockwise positive convention for 6 is used.

The plane strain analyses required the initial stresses to be expressed in theglobal xG and yG directions, thus at any point the stresses defined above had to berotated in accordance with their location. To keep the undrained shear strength andthus the limit load the same as in case 1, the pre-consolidation pressure, po\ wasset to 210 kPa. This also results in the soil adjacent to the pile being normallyconsolidated. The CFSAFEM analyses were carried out using the parallelsymmetry and no symmetry options and the load displacement curves are shownin Figure 12.13. The no symmetry option agreed with the plane strain analysis, but

Figure 12.12: Load-displacementcurves for pile section, easel


1500

Limit loadParallel symmetry: case 2No symmetry: case 2 & 3Parallel symmetry: case 3Plane strain

20 40 60Displacement (mm)

80

Figure 12.13: Load-displacementcurves for pile section,

case 2 and 3

there was a significant error in theparallel symmetry analysis, whichgave a stiffer response initially andreached a higher limit load. The limitloads from the no symmetry andplane strain analyses agreed with theanalytical solution.

Case 3 is similar to case 2, exceptthat the shear stress is now applied inthe opposite direction, i.e. xrB =69.0/rkPa. The load-displacement curve forthe no symmetry analysis is the sameas for case 2. There is still an error inthe parallel symmetry analysis,however, the load deflection curve isnow softer than for the no symmetryanalyses and the same limit load is reached, see Figure 12.13. The errors in bothcase 2 and case 3 parallel symmetry analyses can be attributed to the symmetricallycoupled nature of the soil. This is an example of a problem where a no symmetryanalysis is required due to the material properties, even though the boundaryconditions could be specified using parallel symmetry. These results emphasise theimportance of the implementation of the no symmetry CFSAFEM formulation,which was able to analyse both case 2 and case 3 accurately.

A no symmetry analysis should be undertaken if there is any doubt about thevalidity of the symmetry criterion, discussed in Section 123 A, for the problemconsidered. The symmetry criterion is that the material elasto-plastic [Dep] matrixhas [Du

ep] and [D22ep] terms that are symmetric functions of 6, and [D]2

P] and[D2]

ep] terms that are asymmetric functions of 0, see Section 12.3.1 for thedefinition of [Dep]. Consider, for example, the [Dep] matrix for case 2 of thenormally consolidated material adjacent to the pile, at the beginning of theanalysis:

Du\Dl2

~D~\b~2

This [Dep] matrix is constant in the 6 direction and hence is equal to the zerothharmonic [Depf. Thus the criterion that the [Dl2

ep] and [D2lep] terms are

asymmetrical functions of 6 is not satisfied and a no symmetry CFSAFEM analysisis required. Similarly, for case 3 the [Dep] matrix for the material adjacent to thepile is:

48990

11640

11640

0-5402

0

11640

48990

11640

0-5402

0

11640

11640

48990

0-5402

0

000

18680

00

-5402

-5402

-5402

-5402

1770

0

00000

18680


4899011640116400

54020

1164048990116400

54020

1164011640489900

54020

000

|5402i5402J5402

18680J540200

!l770i

! o i

000008680

29260-8091-8091000

-809129260-8091000

-8091-809129260000

000

1867000

0000

186700

00000

18670

Thus the criterion that the [D]2P] and [D2X

ep] terms are asymmetrical functions of6 is also not satisfied and a no symmetry CFSAFEM analysis is again required.

The [Dep] matrix at the beginning of the case 1 analysis is:

[Dep] = [Dep]° =

Thus before loading the criterion for symmetrical uncoupling is satisfied and aparallel symmetry analysis can be used. The [Dep] matrices for case 2 and case 3also suggest that, during loading of case 1, the symmetry criterion is still satisfied.If the loading in case 1 causes a stress change at a point (r, 0), such that its stressstate was equal to the normally consolidated stress state described in case 2, thestress state at point (r, -0) would be the normally consolidated stress statedescribed in case 3. In this instance, the criteria for symmetrical uncoupling arestill valid, since the [Du

ep] and [D22ep] terms would be symmetric functions of 6 and

the [Dl2ep] and [D2l

cp] terms would be asymmetric functions of 6. Note that thoughthe [Dl2

p] and [D2lep] terms may be non-zero at any particular 6 coordinate, the

zeroth and cosine harmonic coefficients of [Dxsymmetry criterion is satisfied.

epr'] and [D2r] are zero, and the

12.5 The discrete Fourier series aided finite elementmethod

12.5.1 IntroductionThe discrete FSAFEM is similar to the continuous FSAFEM and reduces a large3D global stiffness matrix into a series of smaller uncoupled submatrices.However, the discrete method considers a full 3D finite element mesh and employsa discrete Fourier series to represent the nodal values of force and displacement forthe succession of nodes on each circumference, corresponding to each 2D axi-symmetric node. In contrast, the continuous method considers a 2D axi-symmetricmesh and employs a continuous Fourier series to represent the variation of forceand displacement in the circumferential direction, for each 2D axi-symmetric node.


The first application of the discrete FSAFEM approach in geotechnicalengineering was by Moore and Booker (1982) to develop a boundary element foruse in deep tunnel problems. Subsequently, Lai and Booker (1991) used themethod to study the behaviour of laterally loaded caissons and Runesson andBooker (1982 and 1983) used it in the study of the effects of consolidation and soillayering on soil behaviour.

12.5.2 Description of the discrete FSAFEM methodThe discrete FSAFEM described in this section is for an analysis consisting ofeight node constant strain brick elements. The method can be expanded to considertwenty node linear strain brick elements, and such an extension is presented in Laiand Booker (1991). The usual practice of using complex numbers in theformulation is avoided, in order to simplify the description.

V

a typicalelement

z

H r

a) Elevation view of a typical vertical section

71-2

a typical wedgeof elements plane i

b) Plan view of a typical horizontal section

Figure 12.14: Finite elementdiscretisation for discrete FSAFEM

The problem domain is discretised using a mesh of hexahedral finite elementsconsisting of n evenly spaced wedges around the circular boundary in the 6direction, as shown in Figure 12.14. This is similar to a finite element mesh for a


full 3D analysis. Consider the shaded annular ring of n elements. Let theseelements be numbered from 0 to n-1 anticlockwise and let the vertical planes benumbered from 0 to n-1, such that the two vertical planes enclosing element / areplane / and /+1, see Figure 12.14. Thus the incremental displacements in element/ are defined by the element shape functions and the incremental displacements onplane / (i.e. {Aw,}) and on plane /+1 (i.e. {Aw/+1}). The standard finite elementformulation for linear elastic behaviour, see Chapter 2, expresses the incrementalinternal work, AW, , done by element i, in terms of the incremental nodaldisplacements of the element, {Aw,} and {Aw,+1}, and the stiffness matrix of theelement [K,] such that:

AW, ={{*,? {AuMf}[K,]^^ (12.75)

Since the material properties are assumed not to vary in the 6 direction, thestiffness matrices for each element are identical, i.e. [A,-] = [A/+1]. Thus the internalincremental work done in the whole annular ring is:

;Awj

The stiffness matrix, [K], can be partitioned into four submatrices [X\, [Y], [Y]T

and [Z], in accordance with the two parts of the element displacements {Aw,} and{Aw/+1}. Hence, Equation (12.76) can be rewritten as:

Expanding Equation (12.77) and using the periodicity of the axi-symmetricgeometry, where {Aw,} = {Aw//+,}, gives:

'2AW = '2 {Aw,}T[X]{A«,} + {Aw,_1}T[F]{AW,} + {AM/+1}T[F]T{AW/}/•=o (12.78)

+ {Aw,}T[Z]{Aw7}Equation (12.78) is the equation solved directly in a full 3D analysis. This form ofthe finite element formulation yields a single large global stiffness matrix, [KG],which relates the vector of all the incremental nodal loads, {A/?}, to the vector ofall the incremental nodal displacements, {Ad}, as in Equation (12.3).

In the discrete Fourier series aided finite element method it is assumed that theincremental nodal displacement {Aw;} on plane / can be evaluated from the discreteFourier series which represents the nodal displacement for that pseudo 2D node:

{Aw,} = -jL-2; {Uk}cosika + {Uk}sinika (12.79)


where:{U~k} and {U~k} are respectively vectors of the kth cosine and sine harmonic

coefficients of {Au} expressed as a discrete Fourier series function;a = 2n/n.

Equation (12.79) only describes incremental nodal displacements and cannot beused to evaluate incremental displacements at intermediate values of 0, i.e. it isonly valid for integer values of/. It is for this reason that the method is referred toas the 'discrete' FSAFEM.

Introducing expression (12.79) into Equation (12.78) transforms the problemfrom determining the unknown incremental nodal displacements to that of findingthe Fourier coefficients:

^w= Z| - p Y {Uk) cosika + {uk) si^ta | [X] | -p . '£ {U,} cosila + {U,} sinila |/=o

i ,M — = Y (i »-i --~Y, {Uk} cos(i-\)ka +{Uk} sin(i-l)ka\ [Y] \-j=Y< {U;} cosila + {U,} sinila4ni *=o J {-Jn i=o

( n-\ — = )

yjn k=o J ' K-Jn 1=0 J

—= ^ {Uk} cosika + {Uk} sinika [Z] —= ^ {U/} cosila + {Uj} sinilay/n k=o J {y/n /=o J

(12.80)The orthogonal properties of a discrete Fourier series are:

J cosika sinila = 0/=0n-\X cosika cosila - 0 for / ^ k, - n for / = k (12.81)/=oX sinika sinila = 0 for / ^ k, = nfor I = k/ = 0

Equations (12.81) are used to simplify Equation (12.80) such that:

1=0

{t^}T[y]T{f^}cos/a + {^"}T[F]T{f7jsin/a - {Z^}T[7]T{Z^}sin/a

(12.82)

The internal incremental work equation has been divided into n independentexpressions, one expression for each harmonic order, thus the large global stiffnessmatrix has been uncoupled into n submatrices, as in Equation (12.4), i.e. theequations are harmonically uncoupled. The stiffness matrix [Kj] for any harmonic


order, /, can be obtained by substituting / = / in Equation (12.82). However, foreach harmonic order, both the sine and cosine coefficients of load anddisplacement have to be solved simultaneously. This is similar to the symmetricallycoupled continuous FSAFEM, as shown in Equation (12.16).

Symmetric uncoupling is possible with the discrete method for an isotropiclinear elastic material. This is achieved by first regrouping the terms in Equation(12.82) as:

/=0

+ w,}1 cosla+ [Z])([Y]sinla-[Y]T s'mla) {U,}([Y]smla-[Y]T s'mla) &}

VjJ (12.83)

The incremental nodal displacements, and accordingly the harmoniccoefficients of incremental displacement, have three components, a radial, acircumferential and a vertical component. Thus the two vectors of nodal harmonic

displacements, {Uk} and {Uk} , can each be divided into two sub-vectors, a radialand vertical component sub vector, and a circumferential component sub-vector:

(12.84)

Expanding matrices [X\, [F], and [Z] accordingly, Equation (12.83) can be writtenin the following form:

n-\AW= y

cosla •

cosla +

sin/a -

cosla

cosla

sinla

smla -. )

smlaLt^M

Each set of wedge shaped elements defined in the mesh is geometricallysymmetrical about a constant 6 plane through the middle of it. For an isotropic


elastic material this implies that the element stiffness matrix has a symmetry suchthat:

[X] + [Z] = 2

[F] + [F]T = 2

[Y] - [F]T = 2

~[XU]

0

'[Yu]

0

0

\Y2l]

0

[X22]

0

[Y22]_

~[Y2if0

(12.86)

Equations (12.86) are used to simplify Equation (12.85):

AW = 2%1=0

— i (12.87)

Parallel displacement terms can be obtained by mixing cosine coefficients ofradial and vertical displacement with sine coefficients of circumferentialdisplacement. Similarly, orthogonal displacements are sine coefficients of radialand vertical displacement and cosine coefficients of circumferential displacement.Using these terms Equation (12.87) can be rearranged such that:

, T

n ] + [Fn]cos/a [F21]T sin/a

[F21]sin/a [A

u] + [Yu]cosla [F21fsin/a

[F2]]sin/a [Y22]cosla

==\ (12-88)

The parallel displacement terms have been uncoupled from the orthogonaldisplacement terms and the stiffness matrix for both set of terms is the same. This


is similar to the symmetrically uncoupled continuous FSAFEM stiffness matrixshown in Equation (12.17). Thus the internal incremental work equation nowconsists of 2/i uncoupled vectors of harmonic incremental displacementcoefficients and n stiffness matrices. There are two vectors and one stiffness matrixfor each harmonic order. The stiffness matrix for the fh harmonic order is obtainedby substituting i for / in Equation (12.88).

The externally applied incremental loads can also be expressed as a discreteFourier series. Parallel loads can be obtained by mixing cosine coefficients ofradial and vertical loads with sine coefficients of circumferential load. Similarly,orthogonal loads can be obtained, for example for the kih harmonic:

-{A<}JThe principle of virtual work is then used to obtain the system equations whichconsists of 2n separate equations (two equations for each harmonic order), forexample for the kth harmonic:

u] + [Yu]coska

[F21]sinfe

[Xu]+[Yu]coska(12.89)

Thus the large global stiffness matrix has been harmonically and symmetricallyuncoupled. The procedure outlined above is for a linear elastic analysis. Theprocedure can be extended to consider nonlinear behaviour by combining the linearelastic formulation with a solution strategy which continually adjusts the right handside of the governing finite element equations, e.g. visco-plastic or MNR methods.

12.6 Comparison between the discrete and thecontinuous FSAFEM

A key difference between the discrete and the continuous FSAFEM method is themanner in which the variations in forces and displacements in the circumferentialdirection are described.

The discrete method notionally considers a full three dimensional finite elementmesh defined in a cylindrical coordinate system. Thus the force and displacementdistribution in the circumferential direction is described using the standard finiteelement approximation, i.e. the shape function of the pertinent element. Lai andBooker (1991) state that 'the discrete method gives an exact representation (ofdisplacements) after a finite number of terms'. This statement is misleading sinceit wrongly implies that the correct displacement field can be represented after a


finite number of Fourier series terms. The discrete method uses a discrete Fourierseries to represent nodal displacements and is able to exactly represent anyvariation of nodal displacements in the circumferential direction. However, ingeneral these displacements would not represent the correct displacement fieldexactly, due to the finite element approximation, and the solution accuracy wouldincrease with the number of nodes used.

The continuous FSAFEM uses a 2D axi-symmetric mesh and represents thevariation of nodal displacements and loads in the circumferential direction as acontinuous Fourier series. The accuracy of this representation depends on thenumber of harmonic coefficients used in an analysis. For the special case of alinear elastic continuous FSAFEM, the number of harmonics required in ananalysis is equal to the number of harmonics required to represent the boundaryconditions. In a discrete FSAFEM the distribution of nodal variables in the 6direction is represented using the nodal values and shape functions of theappropriate annular ring of 3D brick elements. These nodal values are expressedas a Fourier series which has the same number of coefficients as there are nodes.Increasing the number of coefficients in a nonlinear discrete FSAFEM analysisincreases the number of nodes in the 6 direction and increases the solutionaccuracy. Thus for both the continuous and discrete nonlinear FSAFEM the choiceregarding the number of harmonics used in an analysis is based on a trade offbetween the solution accuracy desired and the computer resources required.

To assess the accuracy of the two FSAFEM, their ability to represent thedistribution of radial stress around a pile is considered. In particular, therelationship between the number of harmonics or brick elements used and theaccuracy with which the resulting stress distribution can represent the chosendistribution is examined.

The stress distribution around a laterally loaded pile was presented by Williamsand Parry (1984), using measurements from a model pile test. Only the variationof radial stress around the pile (0 direction) is studied and the variation in the radialor vertical direction is not considered, thus the problem is one dimensional. Thestudy compared the continuous FSAFEM, the discrete FSAFEM with eight nodedconstant strain brick elements and the discrete FSAFEM with twenty noded linearstrain brick elements.

As noted in Section 12.4.3 for the continuous FSAFEM, the number ofsampling points in the 6 direction is equal to the number of harmonic coefficientsused in the FSAFEM analysis, and the harmonic coefficients are obtained from thesampling point stresses using the fitted method. The sampling points areequispaced in the 6 direction and are located at 9 coordinates ia, where:

a = 2TT//7V;/ is an integer and 0 < / < ns-1;ns is equal to the number of sampling points.

It is assumed that the stresses at the sampling points are correct, i.e. they are equalto the measured stresses shown in Figure 12.15. Thus the Fourier series


70

Measured stress8-node discrete20-node discreteContinuous

90Azimuth (deg)

135

Figure 12.15: Postulated and measuredstress distribution, n=1O

representation for radial stressis correct at the samplingpoints, but could be in errorbetween sampling points.

In a discrete FSAFEManalysis with constant strainelements, the elements have aconstant width in the 6direction. If the number ofcoefficients is equal to n, thenumber of nodes is also equalto n and they are located at 6coordinates ia. Due to thelinear nature of thedisplacements, the stresses and strains within an element are constant. It is assumedthat these stresses represent the stress state at the centre of the element. This issimilar to the assumption made regarding the sampling point stresses for acontinuous FSAFEM analysis.

In a discrete FSAFEM analysis with linear strain elements, the elements stillhave a constant width and the nodes are still equispaced in the 6 direction.Similarly, if the number of coefficients in a discrete FSAFEM analysis is equal ton, the number of elements is also equal to n. However, there would now be Innodes, since these elements have mid-side nodes. Due to the quadratic nature of thedisplacements, the stresses and strains within an element are linear. It is assumedthat the stresses at the end nodes are correct and there is a linear stress distributionbetween them. The stresses at the mid-side node depend only on the stresses at theend nodes.

The stress distribution from each of the three types of analyses are presentedin Figure 12.15, for n equal to ten. The difference between the desired stress andthe stress from these analyses is a measure of the error associated with theseanalyses. This error, Err, is quantified using the equation:

Err = •I yj(F(O)-f(0))2 dO

•100% (12.90)

where:f{6) = the measured stress distribution;F(6) = the postulated stress distribution.

The value of Err associated with each type of analysis is presented in Table 12.2.For any value of n the continuous method and linear strain discrete method havesimilar accuracies and the constant strain discrete method is substantially lessaccurate. However, there are n unknown harmonic coefficients in the continuousmethod and In unknown nodal values in the linear strain discrete method. Hence,


the stiffness matrix associated with the linear strain discrete method is larger thanfor the corresponding continuous method. This would result in increased computerresources required for an analysis. Thus it is concluded that the continuousFSAFEM is probably the most efficient method.

Table 12.2: Summary of errors in the representation of radial stressaround pile

Number ofharmonic terms

2

5

8

10

Discrete Fourierseries using 8node elements

100.0%

30.9 %

14.5 %

17.5 %

Discrete Fourierseries using 20node elements

48.5 %

21.9%

10.2%

6.8 %

ContinuousFourier series

47.7 %

19.4 %

17.8 %

7.8 %

Err\

-1 0

-1 0

Some authors (e.g. Lai and Booker(1991)) have stated that the continuousFSAFEM could suffer from 'difficultiesassociated with summing a large number ofFourier terms, e.g. the Gibb's phenomenon'.However, statements like this are ambiguoussince the Gibb's phenomenon does notsuggest such a problem. The Gibb'sphenomenon states that if a Fourier series isused to represent a function which has adiscontinuity, the magnitude of the maximumerror, Err, of the Fourier series representationin the vicinity of the discontinuity is largelyindependent of the number of Fourier termsused. This is illustrated in Figure 12.16 for asquare wave. The error Err is approximatelyequal to 0.09 times the magnitude of thediscontinuity. However, while increasing thenumber of Fourier terms does not reduce themagnitude of Err, it does reduce the totalerror in the vicinity of the discontinuity, i.e.the area between the Fourier series and thesquare wave is reduced. The Gibb'sphenomenon would only be relevant in a finite element analysis if a discontinuityin displacements existed in the circumferential direction, and this is not feasible in

Err\ 77 = 32

- 1 0 1 xn is the number of harmonic terms used

in the Fourier series

Figure 12.16: Fourier seriesapproximation of a square

wave


a conventional finite element analysis. It would only be possible if non axi-symmetric interface elements were present, and it is unclear how either the discreteFSAFEM or continuous FSAFEM could incorporate such behaviour. This criticismof the continuous FSAFEM is therefor not valid.

Lai and Booker (1991) have also stated that 'problems concerning conformityof elements is overcome' by using the discrete method rather than the continuousmethod. This statement is not clear: the continuous method should have no elementconformity problems since the displacements at a boundary, calculated from eitherelement sharing that same boundary, are identical, if isoparametric elements areused. However, both methods can be adapted to undertake an analysis consistingof both full 3D elements and pseudo 2D FSAFEM elements, e.g. a core consistingof full 3D elements surrounded by layers of pseudo 2D elements. In this case thedisplacements between nodes at the boundary, which is between the pseudo 2Ddiscrete Fourier elements and the 3D elements, are described by both types ofelements using the finite element approximation, and thus they match. Thedisplacements in the 6 direction for a continuous FSAFEM analysis are describedby a continuous Fourier series. Thus at the boundary between the pseudo 2Dcontinuous Fourier elements and the 3D elements, the displacements are equal atthe nodal points of the 3D mesh, but do not match at intermediate values of 6.

A further comparisonbetween the continuous anddiscrete FSAFEM, and of theinfluence of the number ofFourier coefficients employedin each method, can be madeusing results for the horizontaldisk problem described inSection 12.4.7.2. Ganendra(1993) performed continuousFSAFEM analyses of this " 0 2 4 6 8 10 12 u 16problem using the Authors' Number of hannonic coefficientsfinite element code (ICFEP).He used the same boundary Figure 12.17: Limit load vs. number ofconditions as described in harmonics for a pile sectionSection 12.4.7.2, butmodelledthe soil as a Tresca material, with a shear modulus, G, of 3000 kPa, a Poisson'sratio, //, equal 0.49 and an undrained shear strength, Su, of 30 kPa. Two sets ofanalyses were performed, one assuming a smooth, the other a rough, interfacebetween rigid disk and soil. Each set of analyses consisted of a number of separateanalyses performed with a different number of Fourier harmonics. The results arepresented in Figure 12.17 in the form of normalised ultimate load, Pu I StlD (wherePu is the ultimate load and D is the diameter of the disk), against number ofharmonics used in the analysis. Lai (1989) also analysed this same problem, butused the discrete FSAFEM. His results are also shown in Figure 12.17. The

12

8

6

i\

2

0

0 ©Analytical rough

Analytical smooth

-

M-

1 1 1

D 9?/

>v/1

>

n

- e -- - • - •

1

_ ^ #

H • C

Smooth - CFSAFEMRough - CFSAFEMSmooth - DFSAFEMRough - DFSAFEM

1 1 1 1


analytical solutions for the limit loads for the rough and smooth interface cases are11.945SUD and 9.HSJD respectively (Randolph and Houlsby (1984)). The roughsolution is exact, while the smooth solution is a lower bound. Inspection of Figure12.17 indicates that the continuous FSAFEM is not strongly influenced by thenumber of harmonics used, and reasonable accuracy is obtained if five or moreharmonic terms are used. The discrete method is very sensitive to the number ofharmonics employed, and if less than ten harmonics are used significant errors areobtained, e.g. the error is greater than 65% if 6 harmonics are used. These resultsare indicative of the number of harmonics required to analyse a laterally loadedpile accurately and suggest that (i) the continuous method is more economical interms of number of harmonics required and (ii) large errors can be obtained usingthe discrete method if an insufficient number of harmonics are used.

a) Horizontal load on a circular footingP

12.7 Comparison of CFSAFEM and full 3D analysisTo investigate the benefits of theCFSAFEM over full 3D analysis, arough rigid circular footing underhorizontal and inclined loading has beenanalysed, using both approaches, seeFigure 12.18. The soil is assumed tobehave according to a Tresca yieldcriterion, with E= 10000 kPa, ju=0A5 and$=100 kPa. The mesh for the 3Danalysis is shown in Figures 12.19a and12.19b. As there is a vertical plane ofsymmetry, only one half of the problemrequires analysis. To be consistent, theCFSAFEM analysis used a 2D finiteelement mesh which was the same as the

b) Inclined load on a circular footing

Figure 12.18: Loading schemesfor circular footing

vertical section of the 3D mesh shown in Figure 12.19b. Due to the symmetrymentioned above, the CFSAFEM was run using the parallel symmetry option.

100 m

if&

Figure 12.19a: Horizontal cross-section of a 3D mesh for a circularfooting


Applied load

a>

50 m

<B

Figure 12.19b: Vertical cross-section of a3D mesh for a circular footing; also a

mesh for CFSAFEM analysis

! 3000 -

CFSAFEM -

3D (41 hours)5 harmonics (4.5 hours)10 harmonics (7 hours)20 harmonics (21 hour)

The footing is displacedhorizontally and the horizontalreaction on the foundation noted.The results from the analyses,expressed in terms of thehorizontal force on the footingversus horizontal displacement,are given in Figure 12.20.Results from a full 3D analysisand three CFSAFEM analyses,each with a different number ofharmonics, are given. Also notedon the figure are the run timesfor each analysis. It can be seen Figure 12.20: Load-displacement curvesthat all analyses produce very for a horizontally loaded circular footingsimilar results, with a maximumdifference of only 1%. However, the CFSAFEM have much smaller run times. Forexample, the CFSAFEM with five harmonics required approximately a tenth of thetime of the full 3D analysis.

A second series of analyses was then performed in which the footing wasdisplaced downwards, at an angle of 45° to the horizontal. As the displacement

o.oi 0.02 0.03 0.04 0.05Horizontal displacement (m)

0.06 0.07


increased, both the horizontaland vertical loads on thefoundation were recorded. Theresults are shown in Figure12.21, in the form of horizontalload plotted against horizontaldisplacement and vertical loadplotted against verticaldisplacement. Results are givenfor a full 3D analysis and for aCFSAFEM analysis with 10harmonics. The run times arealso noted on the figure. As withthe horizontal loading, bothanalyses give similar predictions

CFSAFEM (18 hours)

3D (55 hours)

0.01 0.02 0.03 0.04 0.05 0.06 0.07Horizontal, u, and vertical, v, displacement (m)

Figure 12.21: Load-displacement curvesfor a circular footing under inclined

loading

and again the CFSAFEM is much faster. These two examples clearly show that theCFSAFEM is more economical than a full 3D analysis.

12.8 Summary1. The continuous Fourier series aided finite element method (CFSAFEM) has

been described in detail in this chapter. This method provides a way ofreducing the computer resources required to analyse certain 3D problems. Inthis approach a standard finite element discretisation is used in two dimensions,while displacements are assumed to vary according to a continuous Fourierseries in the third dimension. In this chapter the method is applied to problemshaving an axi-symmetric geometry, butnon axi-symmetric distribution of soilproperties and/or loading conditions. A 2D finite element discretisation is usedin the r-z plane, with displacements varying according to a Fourier series in the6 direction.

2. A new formulation for the CFSAFEM for nonlinear material behaviour hasbeen described. A general approach has been adopted and no symmetryconstraints have been imposed. The definitions of, and requirements for,symmetrical and harmonic coupling have been established and the associatedcomputational savings identified.

3. The definitions of parallel and orthogonal symmetry have been given and therequirements for such symmetries to be valid have been discussed. One of therequirements concerns the nature of the constitutive model. It is shown howsubtle changes in the constitutive behaviour can lead to violation of theserequirements and hence to erroneous results. It is advised that facilities toundertake no symmetry, as opposed to parallel or orthogonal symmetryanalysis, are available within the computer code being used.

4. Extension of the CFSAFEM to include interface elements, the bulkcompressibility of the pore fluid and coupled consolidation has been presented.


5. An alternative to the continuous FSAFEM is the discrete FSAFEM. The theorybehind this approach and the difference between the two methods have beendiscussed. The discrete method requires a full three dimensional finite elementmesh defined in cylindrical coordinates. The distribution of nodal variables inthe 6 direction is represented using the nodal values and shape functions of theappropriate annular ring of 3D brick elements. These nodal values areexpressed as a discrete Fourier series which has the same number ofcoefficients as there are nodes.

6. Comparison of the continuous and discrete FSAFEM indicates that the formermethod is likely to be more economical. However, each method has itsadvantages.

7. Comparison of CFSAFEM and full 3D analysis shows that the former requiresconsiderably less computer resources to obtain answers of the same accuracy.For the examples considered, savings in computer time are up to an order ofmagnitude.

Appendix XII-1: Harmonic coefficients of force fromharmonic point loads

The applied incremental load, AF (component of force per unit circumference), atany angle 6 is written in the form of a Fourier series:

AFk cosk0+AFk sinkO (XII.k=l

Let the associated component of incremental nodal displacement, Ad, be describedusing a Fourier series:

Ad = Ad° + S Adk coskO+ Adk s'mkO (XII.2)

Thus the work done is:

AW= ] AdAF r&O (XII.3)- 7 1

where r is the radius of the circle described by the location of the node.Substituting the Fourier series into the work equation gives:

AW= ) \Ad° + j^ Adk cosk0 + Adk sin*0| AF° + £ AFk cosk0+AFk smk0\r d0-n V k=\ J\ k=\ J

(XII.4)

Carrying out the integral and using the orthogonal properties of the Fourier seriesgives:

AW = 2nr Ad°AF° + nrt Adk AFk + Adk AFk

k=\


Using the principle of virtual work, the contributions to the harmonic coefficientsof incremental nodal force, AR, used in the right hand side of the FSAFEMequations, are:

AT?0 = 2nr AF° for the 0th harmonic;

ARk = nr AFk for the kth cosine harmonic; (XII.6)

ARk =nrAFk for the kth sine harmonic.

Appendix XII.2: Obtaining the harmonics of force fromharmonic boundary stresses

Boundary stresses are prescribed over an axi-symmetric surface defined by asection of the finite element mesh boundary. These stresses can be resolved intoglobal r, z and 6 directions to obtain ar, az, a0, rrz, rr0 and rz0 respectively. Thesestresses can be expressed as a Fourier series, and the variation of each harmoniccoefficient along the boundary is described, for example, with a cubic splinefunction. Thus for any component of incremental stress, ACT, at any point in the r-zplane, a Fourier series can be written such that:

ACT = ACT0 + £ Acrk coshO+ Acrk smkO (XII.7)

The displacement over the axi-symmetric surface is described using the shapefunctions of the appropriate finite elements and Fourier series of the nodaldisplacement of the element. Thus for Aa the corresponding component ofincremental displacement, Ad, can be written as:

Ad=t NAAd? + £ Adk cosk0+Adk sinko] (XII.8)

where:Nj is the shape function of the 7th node defined in the mesh;

Adf, Adk and Adk are the 0th, the kth cosine and the kth sine coefficients ofincremental displacement respectively, for the Ith node;n is the number of nodes in the element.

The work done can then be written as:

AW =\ Ad ACT darea = j J Ad ACT rdOdl (XII.9)- 7 t

where d/ is the incremental length in the r-z plane along the boundary. Substitutingthe Fourier series representation in this equation gives:

AW=\ J ]T NA A^° + ]T Adk coskO+Adk sink0\\ ACT0 + £ ACT7 cosl0+Aal sin/0 r d0d/k A /=1 J

(XII. 10)


Carrying out the d6 integral gives:

Ad* Ao* + Arf* ACT* I d/ (XII. 11)J

The d/ integral is carried out numerically, usually using a Gaussian integration rule.Thus using the principle of virtual work the contributions to the harmoniccoefficients of incremental nodal force at the fh node, ARf, used in the right handside of FSAFEM equations, are:

AR° = 2nr\ Ni ACT0 d/ for the 0th harmonic;

AR? = nr\ Ni Acrk d/ for the kth cosine harmonic; (XII. 12)

ARf = nr\ Nf A<jk d/ for the kth sine harmonic.

Appendix XII.3: Obtaining the harmonics of force fromelement stresses

The stress point algorithm evaluates the incremental stresses at integration pointswithin the problem domain. These stresses are expressed as a Fourier series, i.e:

{ACT} = {Aor0} + £ {A^} cos£<9+ {Ao*} smkO (XII. 13)

where:{A<r} is the vector of incremental stress components;

{A of}, [ACT-} and {Aaf} are the vectors of the 0th, the kth cosine and the kth

sine coefficients of stress respectively.

Vector {Aa} can be split into two sub-vectors, {Acl} and {Ao"2}, i.e:

{Ao) = {Acrr Aa= Aa0 Aarz\Aar0 Aa=0}T = {{Acrl} {Acr2}}T (XII. 14)

where:{Aoi} = {Ao-, Aaz Aae Acr,z}T and {Aa2} = {AarB Aa20}T

Accordingly, the Fourier series representation can be divided using these two sub-vectors, i.e:

[{Ao-2°}J *=' [{Ao-2'}cos^J [{Acr2*} sinifc J

The incremental internal work done by these stresses is:

j {A^}T{Acr} dVol=\) {Ad}T[Bf {ACT} r&O&area (XII. 16)


where:{Ad} is the vector of incremental displacements;[B] is the strain matrix.

Equation (XII. 16) can be expressed in a Fourier series form as:-|T r i^

[Bl1] cos/0 ^ T [Bll] sin/0[B2l] sin/0 ~[B2l] cos/0

1 L [{Acrr*}cosA;0]{A<x20} k-' {

/=o I

r dO darea

(XII. 17)Using the orthogonal properties of the Fourier series, the 8 integral is carried out:

AW = % ±[[B2']\ [{A<r2'}J

-2n{Ad°**}T[B2°]T{Acr20} darea

(XII. 18)

The work done by the coefficients of incremental nodal force consistent withthis stress state is:

AW= {Ad°*}{AR°*}+t {Adk*}{ARk*} + {Adk**}{ARk**} (XII. 19)

Thus using the principle of virtual work, the work done by both equations must bethe same and in addition the terms which are pre-multiplied by the samecoefficients of displacement can be equated. From here the contributions to theharmonic coefficients of incremental nodal load, {AR}, used in the right hand sideof the FSAFEM equations, are:

{AR°} = 2TIJf [*lo]]Tf{A<xlo}l\ U \r darea for the 0th harmonic;[-[S2°]J [{A<x2°}J

{AR1*} =' 7t JI U {AZ=} 1 r darea for the /* parallel harmonic;{ [ 1 } [ ' J

{A/?'"} = TCH N ^A^_^ I r darea for the /th orthogonal harmonic.l-[«']J [{A.2'}j (XII.20)


Appendix XII.4: Resolving harmonic coefficients of nodalforce

Horizontal loadsLet Afr represent the incremental radial force per unit circumference on a node.This can be expressed as a Fourier series:

Afr = AFr° + £ AFrk cosk0+ AFr

k smk9 (XII.21)k=\

The resultant incremental force in the #=0° direction, ATr°~°, is given by equation:

ATr0=0 = ] Afrrcos0d0

-n

= ) (AFr° + X AFk cosk0+AFrk sinik^lcos^ r d0 (XII.22)

= nr AFrl = ARl

r

Similarly, the resultant incremental force in the 0=l/2n direction, ATr^Vzn, is:

ATer=*n = ) Afrrsin0 d0

AFr° + t AFk cosk0+AFk sinA^lsin^ r d0.. _ k± j (XII.23)

= nr AFl = <

fff represents the incremental circumferential force per unit circumferenceon a node, the resultant incremental force in the #=0° direction, ATr

fh0, is given by:

= ] -(AF0° + £ AFk cosk0+ AFg sinA:<9]sin(9 r d0-n V k=l )

= - n r AFle = -ARX

O

Similarly, the resultant incremental force in the O^Vin direction, AT,0=V2n, is:

AT?*** = ) Aforcos0 d0- 7 1

= J f AF0° + t A F / cosk0 + AFk sinA:6>jcos6> r d6> (XII.25)-71 V n = l /

= nr AFQ = ARxe

Thus the total incremental horizontal load in the 0=0° direction is:

ATre=0 = ARi-A^ (XII.26)

and the total incremental horizontal load in the 6=l/2K direction is:

°^ ~ (XII.27)


Axial loadsLet Afz represent the incremental vertical force per unit circumference on a node.This can be expressed as a Fourier series:

Afz = AFZ° + t AFzk cosk0+ AFz

k sinkO (XII.28)

The total incremental force in the vertical direction, ATZ, is given by the equation:

AT2 = ] A

= ) (AF2° + t Aff cosk0+ AF? sinker d9 ( x n 29)= 2nrAF? =

Turning momentsAs in the previous case for axial load, let Af2 represent the incremental verticalforce per unit circumference on a node, which can be expressed as a Fourier series.The incremental turning moment about the 0=l/2K axis, AM/2%, is equal to:

AM*" = ] Af2rcos0 rd0

= J \AFZ° + £ AFzk co$k6+AF2

k smkO cos<9 r2 d6 ^XH 30)

= nr2AF2l =rARl

2

Similarly, for incremental turning moments about the #=0° axis, AAf is equal to:

AM°= J Af2rsm6 rd6-n

= 1 ( AFZ° + t AFk cosk9+ AFk smko\sinO r2 dO ( X n 3 ^

= nr2AF2l=rARl

2

Appendix XII.5: Fourier series solutions for integratingthe product of three Fourier series

Three standard integrals are used for the solution of the integral of the product oftwo Fourier series:

J smkOcoslOdO = 0 for all k and /- 7 1

J sin£<9sin/<9d<9 = 0 i f * * / , = n i f k = l*0, = 0 i f k = l = 0 ( X I I . 3 2 )- 7 1

J coskOcosWdO = 0 i f


Using the above equations six standard integrals can be obtained for the solutionof the integral of the product of three Fourier series:

J sinkOsinlOsinmO dO = ] ~(cos(k-l)0-cos(k + l)0) sinmO dO-n -n *•

= 0

] coskOcoslOsinmO dO = f ^(cos(£ -1)0+ cos(& +1)0) sinmO dO

= 0

J sin£<9cos/<9sinw<9 d0= ) ~(sin(£ -1)0+ sin(k +1)0) sinmO dO

sinkOs'mlOcosmO dO= ] ~-(cos(^ ~ l)e~ cos(k +1)°) ^osmO dO

] coskOcoslOcosmO d0=) —(cos(A; -1)0+ cos(£ +1)0) cosmO dO

j sinkOcoslOcosmO dO= J —(s'm(k -1)0+ sin(£ +1)0) cosmO dO-n -n ^

= 0

where: (XII.33)a = 1 if |* -l\ = m = 0 ; = lA if |k -l\ = m * 0 ; = 0 otherwise;P = 1 if \k +/| = jw = 0 ; = V2 if | i +/| = m * 0 ; = 0 otherwise;±is + i f £ - / > 0 ; is - i f * - / < l ;|x| denotes absolute value of x.

Appendix XII.6: Obtaining coefficients for a stepwiselinear distribution

The variable x is described by a stepwise linear distribution, see Figure 12.4, usingn discrete values of xf at locations 0f, e.g. xx at 0x, x2 at 62,..., x, at #,,..., xn at 5W,such that at any location 0 the value of x is:

/) w h e n gj < g < ^+i (XII.34)

The 0th harmonic is evaluated from:

27ttl i


2

1 A */+i+*,-ffl>i , Of 0 0

= THh iAi+l ~ i} (XII.35)

The kth cosine harmonic is evaluated from:

Xk =— J xcoskO dO

/=i

^,, -cosM .v #

I ,-cosA;6!) . , „ . f /1 1^ ^-^ r/+1sin^+1 -x7sin^6!j (XIL36)

Since xn+l = xl and #,7+1 = ^ j , the summation of the last two terms in the expressionis zero, giving:

(XII.37)

The kth sine harmonic is evaluated from:

Xk =—

Y f(^.^)g+^/^.)| sin^ Jsinkd 0cosk0)/+1 ^ \~F" ~ST"J


( x n 3 8 )

The summation of the last two terms in the expression is zero, giving:

k = (XII.39)

Simplifications can be made if x varies in either a symmetric or non-symmetricmanner. If a symmetric function is stipulated, the specified values of x areconstrained to be located in half space 0 < 6 < n. For each value of x specified, asymmetrical value is assumed in the half space -n < 6 < 0, i.e. for each x, at 0f avalue of X; at -6} is assumed, see Figure XII. 1. The above equations can then beused to obtain 0th and cosine harmonic coefficients. Since the assumed distributionis symmetric, the sine coefficients are zero.

Specified

XAssumed

*«-•

M Nx

x2

xx

Specified

\\

,'0

Figure XII. 1: Symmetricalstep wise linear function

Figure XII. 2: Asymmetricalstepwise linear function

Similarly, if an asymmetric function is stipulated, the specified values of x areconstrained to be located in the half space 0 < 6 < it. For each value of x specified,an asymmetrical value is assumed in the half space -n<0 <0, i.e. for each x, at #,a value of -x, at -#, is assumed, see Figure XII.2. The above equations can then beused to obtain the sine harmonic coefficients. Since the assumed distribution isasymmetric, the 0th and cosine coefficients are zero.

Appendix XII.7: Obtaining harmonic coefficients usingthe fitted method

A Fourier series has to be found which passes through n discrete values of x (x, tox,,). In order to have the same number of sine and cosine harmonics, n is


constrained to be an odd number. Thus the order of the Fourier series required, L,is equal to !/£(«-1):

x = X° + £ Xk cos£6>+ Xk sin£<9 (XII.40)k=\

The values of x are equispaced in the region -n < 6 < n, i.e. x}is located at 6 =(j- l)a, see Figure 12.6. From here Equation (XII.40) can be written such that forany integer value j :

x. = X° + £ Xk cosk{j-\)a+Xk smk(j-\)a (XII.41)k=]

The harmonic coefficients for this Fourier series can then be evaluated usingthe equations:

n /=i

Xk =- t x.cosk(i-l)an i=\

Xk = - t xt sink(i - \)a (XII.42)n ,=i

This can be proved by substituting Equations (XII.42) into Equation (XII.41),which gives:

- t x, coska(i - 1) coska(j - 1)

— f; JC, sin*ar(/- 1) sinka(j-l)nn JI]

Noting that cosA cos^ + sin^ sinB = cos(A - B) Equation (XII.43) can be simplified:

JC;. = — i x , . + - i t xi coska(i - j) (XII.44)

Equation (XII.44) can be rewritten so that the summation of A: is carried out first:

x,=-± xl\ + t coska(i - j)) (XII.45)

A parameter y/^ can now be defined such that Equation (XII.45) becomes:

Xj= - S x, iffy (XII.46)

where:¥ij 2 k={

cos a l J


Using the standard solution:

1 L

— +Ya COSAX = -2 k=\ 2 s i n -

and if/ * j then:

T" 2sin^il

Substituting L into the equation for a gives:

_ 2TT = 2TIa n 2L+1

Substituting this into Equation (XII.47) for ^ gives:

w _ smn(j-0 _ n

(XII.47)

(XII.48)

Equation (XII.48) equals zero for / *j because (j~i) is an integer, and the sine ofany multiple of n is zero. \fi=j, then y/;j = (L+V£) since the cosine of zero is 1.

Equation (XII.46) can be written as:

(XII.49)

Thus Equation (XII.42) yields harmonic coefficients for a Fourier series whichsatisfies Equation (XII.41) for all values ofy.

Simplifications can again be made if x varies in either a symmetric or non-symmetric manner. If a symmetric function is stipulated, then values of x are onlyspecified in the half space -7t<#<0, i.e. only x0 to xL is specified and xIM to xn isobtained using the equation x, = x,,_,. This is illustrated in Figure XII.3. Equation(XII.42) can now be used to obtain the zeroth and cosine harmonic coefficients.Since the assumed distribution is symmetric, the sine coefficients are zero.

Assumed Specified

Assumed Specified

Figure XII. 3: Symmetricalfunction

Figure XII. 4: Asymmetricalfunction


Similarly, if an asymmetric function is stipulated, then values of x are onlyspecified in the half space 0<0<n and an asymmetrical value is assumed in the halfspace -7r<#<0, i.e. only xx to xL is specified and xL+l to xn is obtained using theequation x, = -xn.f. x0 is constrained to be zero for an asymmetric function. Thisis illustrated in Figure XII.4. Equation (XII.42) can now be used to obtain the sineharmonic coefficients. Since the assumed distribution is asymmetric, the zeroth andcosine coefficients are zero.