2_2. finite element theory for linear materials

2. Finite element theory forlinear materials

2.1 SynopsisThis chapter introduces the finite element method for linear problems. The basictheory is described and the finite element terminology is introduced. Forsimplicity, discussion is restricted to two dimensional plane strain situations.However, the concepts described have a much wider applicability. Sufficientinformation is given to enable linear elastic analysis to be understood.

2.2 IntroductionThe finite element method has a wide range of engineering applications.Consequently, there are many text books on the subject. Unfortunately, there arefew books that consider specifically the application of the finite element methodin geotechnical engineering. This chapter presents a basic outline of the method,with particular attention to the areas involving approximation. The discussion isrestricted to linear elastic two dimensional plane strain conditions. Only continuumelements are considered and attention is focussed on the 'displacement based'finite element approach. The chapter begins with a brief overview of the mainstages of the method and follows with a detailed discussion of each stage.

2.3 OverviewThe finite element method involves the following steps.

Element discretisationThis is the process of modelling the geometry of the problem underinvestigation by an assemblage of small regions, termed finite elements. Theseelements have nodes defined on the element boundaries, or within the element.

Primary variable approximationA primary variable must be selected (e.g. displacements, stresses etc.) and rulesas to how it should vary over a finite element established. This variation isexpressed in terms of nodal values. In geotechnical engineering it is usual toadopt displacements as the primary variable.

Element equationsUse of an appropriate variational principle (e.g. Minimum potential energy) toderive element equations:

24 / Finite element analysis in geotechnical engineering: Theory

[KE]{AdE} = {ARE} (2.1)

where [KE] is the element stiffness matrix, {AdE}, is the vector of incrementalelement nodal displacements and {ARE} is the vector of incremental elementnodal forces.

Global equationsCombine element equations to form global equations

[KG]{AdG} = {ARG} (2.2)

where [KG] is the global stiffness matrix, {AdG} is the vector of all incrementalnodal displacements and {ARG} is the vector of all incremental nodal forces.

Boundary conditionsFormulate boundary conditions and modify global equations. Loadings (e.g.line and point loads, pressures and body forces) affect {ARG}, while thedisplacements affect {AdG}.

Solve the global equationsThe global Equations (2.2) are in the form of a large number of simultaneousequations. These are solved to obtain the displacements {AdG} at all the nodes.From these nodal displacements secondary quantities, such as stresses andstrains, are evaluated.

2.4 Element discretisationThe geometry of the boundary value problem under investigation must be definedand quantified. Simplifications and approximations may be necessary during thisprocess. This geometry is then replaced by an equivalent/zmte element mesh whichis composed of small regions calledfinite elements. For twodimensional problems, the finiteelements are usually triangular orquadrilateral in shape, see Figure2.1. Their geometry is specified interms of the coordinates of keypoints on the element called nodes.For elements with straight sidesthese nodes are usually located atthe element corners. If the elementshave curved sides then additionalnodes, usually at the midpoint ofeach side, must be introduced. Theset of elements in the completemesh are connected together by the

<—*-\

^—-j

element sidesnodes.

and a number of Figure 2.1: Typical 2D finite elements

Finite element theory for linear materials / 25

In order to refer to the complete finiteelement mesh, the elements and nodes must benumbered in systematic manner. An example ofa numbering scheme for a mesh of 4 nodedquadrilateral elements is shown in Figure 2.2.The nodes are numbered sequentially from left toright and from bottom to top; the elements arenumbered separately in a similar fashion. Todescribe the location of an element in the mesh,an element connectivity list is used. This listcontains the node numbers in the element,usually in an anticlockwise order. For example,the connectivity list of element 2 is 2, 3, 7, 6.

When constructing the finite element meshthe following should be considered.

- The geometry of the boundary value problemmust be approximated as accurately aspossible.

- If there are curved boundaries or curvedmaterial interfaces, the higher orderelements, with mid-side nodes should beused, see Figure 2.3.In many cases geometric discontinuitiessuggest a natural form of subdivision. Forexample, discontinuities in boundarygradient, such as re-entrant corners or cracks,can be modelled by placing nodes at thediscontinuity points. Interfaces betweenmaterials with different properties can beintroduced by element sides, see Figure 2.4.

- Mesh design may also be influenced by theapplied boundary conditions. If there arediscontinuities in loading, or point loads,these can again be introduced by placingnodes at the discontinuity points, see Figure2.5.

In combination with the above factors, thesize and the number of elements depend largelyon the material behaviour, since this influencesthe final solution. For linear material behaviourthe procedure is relatively straightforward andonly the zones where unknowns vary rapidlyneed special attention. In order to obtain accurate

Y

45

1

JO

56

22

Jl -

67

3.3

Figure 2.2: Element andnode numbering

a) Curved boundaries

b) Curved material interface

Figure 2.3: Use of higherorder elements


Wail •

Re-entrant,corner .

1

/ \

Soil

M&ieriai elementsto be excavated

\

\

Sure

.11,1,

Figure 2.4: Geometricdiscontinuities

Figure 2.5: Effect of boundaryconditions

solutions, these zones require a refined mesh of smaller elements. The situation ismore complex for general nonlinear material behaviour, since the final solutionmay depend, for example, on the previous loading history. For such problems themesh design must take into account the boundary conditions, the materialproperties and, in some cases, the geometry, which all vary throughout the solutionprocess. In all cases a mesh of regular shaped elements will give the best results.Elements with widely distorted geometries or long thin elements should be avoided(see Figure 2.6 for example).

36 Elements

111

m\\

^ ^ 30 Elements

a) Ill-conditioned mesh b) Well conditioned mesh

Figure 2.6: Examples of good and bad meshes


2.5 Displacement approximationIn the displacement based finite element method the primary unknown quantity isthe displacement field which varies over the problem domain. Stresses and strainsare treated as secondary quantities which can be found from the displacement fieldonce it has been determined. In two dimensional plane strain situations thedisplacement field is characterised by the two global displacements u and v, in thex and y coordinate directions respectively.

The main approximation in the finite elementmethod is to assume a particular form for the waythese displacement components vary over thedomain under investigation. Clearly, this assumedvariation must satisfy the conditions ofcompatibility. Over each element the displacementcomponents are assumed to have a simplepolynomial form, where the order of the polynomialdepends on the number of nodes in the element. Thedisplacement components are then expressed interms of their values at the nodes. For example, Figure 2.7: Three nodedconsider the displacement equations for the three elementnoded triangular element shown in Figure 2.7:

u = a , + 7 y(2 3)v = bx+b2x+b3y v }

The six constants ax - b3 can be expressed in terms of the nodal displacementsby substituting the nodal coordinates into the above equations, and then solving thetwo sets of three simultaneous equations which arise:

Uj = ax+a2Xj+a3yj (2.4)

(2.5)vm = bl+b2xh

The above simultaneous equations are solved for ax - b3 in terms of the nodaldisplacements w,, u}, um , v,, vy and vm, to give:

U\ = [N}{u{ Uj um v, v, vm}T = [N]H (2.6)J I J nodes

where the matrix [N\ is known as the matrix ofshape functions. The displacementcomponents u and v are now expressed in terms of their values at the nodes. Forthree and four noded elements there is, therefore, a linear variation of displacementacross the element, see Figure 2.8. For the higher order six and eight noded


Element

Figure 2.8: Linear variation ofdisplacements across a 4 noded

element

elements the displacement field variesquadratically across the element.

The accuracy of a finite elementanalysis depends on the size of theelements and the nature of thedisplacement approximation. For theaccuracy to increase as the elementsbecome smaller, the displacementapproximation must satisfy the followingcompatibility conditions.

- Continuity of the displacement field.In order to avoid gaps or overlapsoccurring when the domain is loaded,the displacement components mustvary continuously within eachelement and across each element side.This can be achieved by ensuring thatthe displacements on an element sidedepend only on the displacements ofthe nodes situated on that side, seeFigure 2.9.

- The displacement approximationshould be able to represent rigid bodymovement. Examples of suchmovements are translations androtations. Such displacements do notinduce strains in an element.

- The displacement approximationshould be able to represent constantstrain rates.

The simple polynomial approximationpresented above satisfies thesecompatibility conditions.

The essential feature of the element-wise approximation is that the variationof the unknown displacements within anelement is expressed as a simple functionof the displacements at the nodes. Theproblem of determining the displacement field throughout the finite element meshis, therefore, reduced to determining the displacement components at a finitenumber of nodes. These nodal displacements are referred to as the unknowndegrees of freedom. For two dimensional plane strain problems there are twodegrees of freedom at each node: the u and v displacements.

"igure 2.9: Continuity ofdisplacement field

DisplacementApproximation

Figure 2.10: Variation ofdisplacement across a mesh of

4 noded elements


2.5.1 Isoparametric finite elementsThe choice of the finite elements to be used for two dimensional problems dependslargely on the geometry being modelled and the type of analysis required. Forgeotechnical problems, the main requirement is that the element types should beuseful for all the geometric situations that may arise, including cases wherestructures have curved boundaries or curved material interfaces. Both triangularand quadrilateral types of elements could equally well be used, since both of thesetypes are easily modified, by the addition of mid-side nodes, to accuratelyrepresent curved edges. As the finite element equations are slightly easier toformulate for the quadrilateral family of isoparametric elements, these will be usedfor any of the derivations presented in this book. This does not imply thatquadrilateral elements are in any way superior to triangular elements. In fact, someexperts are of the opinion that triangular elements are superior. For completeness,the derivation of the interpolation and shape functions for triangular elements ispresented in Appendix II. 1 of this chapter.

An 8 noded quadrilateral Tisoparametric element isshown in Figure 2.11. Thiselement is widely used ingeotechnical finite elementsoftware. It can be used as ageneral shaped quadrilateralelement with curved sides.The global element is derivedfrom a parent element whichhas the same number of nodes,but is defined with respect to a natural coordinate system. Figure 2.11 shows theconfiguration of the parent element. The natural coordinates S, 7 for this elementsatisfy the conditions -1 <S< 1, -1 <T< 1.

The basic procedure in the isoparametric finite element formulation is toexpress the element displacements and element geometry in terms of interpolationfunctions using the natural coordinate system. The term isoparametric arises fromthe fact that the parametric description used to describe the variation of theunknown displacements within an element is exactly the same as that used to mapthe geometry of the element from the global axes to the natural axes.

For the element shown in Figure 2.11 the global coordinates of a point in theelement can be expressed by coordinate interpolations of the form:

4

,8

1(-1,-1)

7

5

( I . I )

6

2>

a) Parent Element

Figure 2.11:b) Global Element

8 noded isoparametricelement

x = y = (2.7)

where x,- andj^ are the global coordinates of the eight nodes in the element, and Nhz-l,..,8, are called interpolation functions. For isoparametric elements theinterpolation functions are expressed in terms of the natural coordinates S and Twhich vary from -1 to +1. To construct the interpolation functions for the eight


noded isoparametric element, the following properties are used. Firstly, there is oneinterpolation function corresponding to each node in the element and each functionis quadratic in the natural coordinates S and T. Secondly, the interpolation functionNl is equal to 1 at node 1 (S = - 1 , T= -1) and is equal to zero at the other sevennodes; a similar relation holds for the interpolation functions N2, N3,..., N$. Usingthese properties it can be shown that the interpolation functions take the form:

Mid - side nodes: Corner nodes:

(2.8)

N, = y2(\-S)(\-T2) N4=y4(iSince the element is isoparametric, the displacements within the element are

interpolated in the same way as the geometry. Therefore, the above interpolationfunctions Nh N2, ..., Ns are used as the shape functions in Equation (2.6).Consequently, the displacements within the 8 noded isoparametric element are alsoassumed to be quadratic in S and T.

The main advantage of the isoparametric formulation is that the elementequations need only be evaluated in the parent element coordinate system. Thus,for each element in the mesh the stiffness matrix integrals can be evaluated by astandard procedure. The integrations need only be performed over a square, withS and T varying between -1 and +1. The boundary conditions can be determinedin a similar fashion. If gravity loads are applied, the nodal forces are determinedfrom element integrals, which can again be reduced to integrals over the parentelement. If stress boundary conditions are applied, the nodal forces are determinedfrom integrals over the boundary of the mesh. In this case, the integrals can bereduced to line integrals over the sides of the parent element.

For all types of finite elements the best results are obtained if the elements havereasonable shapes. Wildly distorted elements may lead to different forms ofinaccuracy. For example, in the case of isoparametric elements difficulties may beencountered if the distortions lead to a non-unique mapping between the global andparent elements. For the quadrilateral isoparametric element described above, thefollowing points should be observed.

- To avoid a non-unique global to parent element mapping, all the interiorangles, at the element corners, should be smaller than 180°. In particular, thebest results are obtained if these angles lie between 30° and 150°.If an element becomes thin, the stiffness coefficients corresponding to thetransverse degrees of freedom are considerably larger than those correspondingto the longitudinal displacements. This results in numerical ill-conditioning ofthe system equations which can lead to large errors in the solution. To avoidthis the ratio between the longest and shortest sides of an element should besmaller than 5:1. For anisotropic materials, if the longitudinal direction of theelement is parallel to the stiffest material direction, this ratio may be exceeded.


The location of mid-side nodes greatly affects the uniqueness of the global toparent element mapping. For elements with curved sides, experience has shownthat the best results are obtained if the radius of curvature of each side is largerthan the length of the longest side.

2.6 Element equationsElement equations are those that govern the deformational behaviour of eachelement. They essentially combine the compatibility, equilibrium and constitutiveconditions.

Displacements: As noted above, the chosen displacements are assumed to be

Strains: The strains corresponding to these displacements follow fromEquation (1.2):

A dx

Azz = Ayxz = Ayzy = 0 ; {

=_d(Av)_y dy

^ _ 3(Au) D(Av)xy dy dx

Asy Ayxy A j

Combining Equations (2.9) and (2.10) for an element with n nodes leads to:

Asy

8N2

~dx~dN2

8Nl

dx0

0

dNH

dy

dy0

dx0

dy0

dx0 ...

dy... 0

Aw,Av,AM2

Av,

Aw,,Av,,

(2.11)

or more conveniently:{As}=[B]{Ad}n (2.12)

where the matrix [B] contains only derivatives of the shape functions Nh while{Ad},, contains the list of nodal displacements for a single element.

If isoparametric elements are being used, the shape functions are identical tothe interpolation functions, see Section 2.5.1, and TV, depend only on the naturalcoordinates S and T. Consequently, the global derivatives dNf I dx, dN; I dy inEquation (2.11) cannot be determined directly. However, using the chain rulewhich relates the x, y derivatives to the S, Tderivatives gives:

a/v,. a/dS dT

where [J] is the Jacobian matrix:dy


dxdSdx

_dT

dy'8SdydT_

(2.14)

Hence, on inverting Equation (2.13), the global derivatives of the interpolationfunctions are given by:

dN,dx

dN;dy

1dydTdxdT

dy~dsdxdS_

dNf

dSdNf

~dTwhere | / | is the Jacobian determinant:

dx_dS dT dS dT

I I dx dy dy dx' " ~d~d ~ ~d~d

(2.15)

(2.16)

The coordinate derivatives in Equations (2.15) and (2.16), which arise from theJacobian transformation, can be found by differentiating the isoparametric relationsgiven by Equation (2.7).

Constitutive model: The constitutive behaviour can be written as in Equation(1.3):

{ACT} = [D]{As} (2.17)where {Ao}T = [Aax Aay AT^ Aaz]

For isotropic linear elastic materials the constitutive matrix [D] takes the formgiven in Section 1.5.5 of Chapter 1. The form of the matrix for transverselyisotropic materials is presented in Chapter 5.

To determine the element equations for linear material behaviour, the principleof minimum potential energy is invoked. This principle states that the staticequilibrium position of a loaded linear elastic body is the one which minimises thetotal potential energy. The total potential energy of a body is defined as:

Total potential energy (E) = Strain energy (W) - Work done by the applied loads (L)

The principle of minimum potential energy states that for equilibrium:

5A£ = 5APT-SAZ = 0 (2.18)The strain energy, AW, is defined as:

dVol (2.19)

where the integrations are over the volume of the body.The work done by applied loads, AL, can be divided into contributions from

body forces and surface tractions and can, therefore, be expressed as:

AL =


}T{AF} dVol + \{Ad}T{AT] dSrf (2.20)Sf

{ } { } {Vol Srf

where:{Ad}T = {Aw , Av} = displacements;{AF}T = {AFx,AFy} = body forces;{A7)T = {ATX , AT }̂ = surface tractions (line loads, surcharge pressures);Srf is that part of the boundary of the domain over which the surface tractionsare applied.Combining Equations (2.19) and (2.20) the total potential energy of the body

becomes:A£ = I j{A^}T[/)]{Af} dVol - j{Ad}T{AF} dVol - j{Ad}T{AT} dSrf n.2\)

Vol Vol Srf

The essence of the finite element method is now to discretise the problem intoelements. This has two effects. Firstly, the potential energy is replaced by the sumof the potential energies of the separate elements:

AE = (2.22)

where N = number of elements. Secondly, the variation of the displacements canbe expressed in terms of nodal values using Equation (2.9). Equation (2.21)therefore becomes:

{ dSrf(2 23)

where the volume integral is now over the volume of an element and the surfaceintegral is over that portion of the element boundary over which surface tractionsare specified. The principal unknowns are the incremental nodal displacementsover the whole mesh, {Ad},}. Minimising the potential energy with respect to theseincremental nodal displacements gives:

5A£ = [B]T[D][B] AVol {Ad}n - J[7V]T{AF} dVol-i voi

\ [ N f {AT} dSrf 1 = 0(2 24)

which is equivalent to a set of equations of the form:

where: [KE] = J Vol [B]T[D] [B] d Vol = Element stiffness matrix;{ARE} = JFo/[A]T{AF} dVol+I^N]T{AT} dSrf= Right hand side load

vector.


Thus the problem is reduced to determining and summing the separate elementequilibrium equations:

n = {ARE} (2.26)

The element stiffness matrix for isoparametric elements is evaluated using thenatural coordinate system. The isoparametric coordinate transformation gives:

dVol = tdxdy = t\j\dSdT (2.27)where for plane strain problems the thickness t is unity and the element stiffnessmatrix, [KE], becomes:

[KE]=] ]t[Bf[D][B]\j\dSdT (2.28)

In Equation (2.28) | / | is given by Equation (2.16) and represents the determinantof the Jacobian matrix arising from the mapping between global and parentelements. The explicit evaluation of Equation (2.28) cannot usually be performed,except for very special element shapes. It is therefore convenient to use anumerical integration procedure.

2.6.1 Numerical integrationTo evaluate the element stiffness matrix andright hand side vector, integrations must beperformed. The explicit evaluation of theseelement integrals cannot usually beperformed, except for special cases, andtherefore a numerical integration scheme isemployed. For example, consider the onedimensional integral, J.} J{x)dx, showngraphically in Figure 2.12a. The simplestway to perform numerical integration is tosplit the x range of the integral (i.e. -\<x <1)into a number of equal steps of size a andassume that the area under the curve is equalto the sum of the trapezoidal areas^(Axi)'hAxi-/)y^' Such a procedure is oftenreferred to as the Trapezoidal rule. Thisprocedure can be refined to obtain thegreatest accuracy for the minimum numberof function evaluations, fix,), and severaldifferent procedures are available.

Ax)

-1 xt ia) Integration using Trapezoidal rule

b) Gaussian integration

Figure 2.12: Examples ofnumerical integration

Essentially, the integral of a function is replaced by a weighted sum of the functionevaluated at a number of integration points (sampling points). For example,consider a one dimensional integral with three integration points, Figure 2.12b:


\f(x)dx = (2.29)

0.57

0.57

0.57

X

X

4-L-+ 1

X

X

1,1)

S

(-17-1) IParent Element Global Element

where: W} are weights;fix,) are the values of the function at the three integration points x/=12 3.

The values of the weights, Wh and the location of the integration points, xhdepend on the nature of the integration scheme being used. The number ofintegration points determines the integration order. The higher order of integrationgives the more accurateintegration process .Unfortunately, the numberof function evaluationsalso depends on theintegration order, so thecost of an analysis willincrease when a higherorder integration isemployed. This becomesparticularly important fortwo and three dimensional 0.77integrals where an array of 0 77~integration points isrequired. For example, fora two dimensionalgeneralisation of theintegral discussed above,an array of 3x3 integrationpoints would be required.

a) 2x2 order

t 0.77

X )

x >

X

X

1,1)

S

(-1,-1)Parent Element Global Element

b) 3x3 order

Figure 2.13: Location of Gauss points

For three dimensional element this increases to a 3x3x3 array.The most common numerical integration scheme is Gaussian integration and

the integration points are often referred to as Gauss points. For Gaussianintegration the optimum integration order depends on the type of element beingused and on its shape. Experience has shown that for the 8 noded isoparametricelement either a 2x2 or a 3x3 order should be used. Figure 2.13 illustrates thelocations of the Gauss points in the parent element and an example of theirpositions in a global element for the 2x2 and 3x3 integration orders. 2x2 and 3x3integration orders are often referred to as reduced andfull integration, respectively.

For nonlinear problems (see Chapter 9), the element stiffness matrix at aparticular increment may, in general, depend on the stresses and strains determinedat the previous increment. Since the stiffness matrix is determined by numericalintegration and the element equations are referred to the integration points, it isalso convenient to restrict evaluation of the stresses and strains to these points.Hence, in many programs the output listings provide values of stresses and strainsat integration points.


2.7 Global equationsThe next stage in the formulation of the finite element equations is the assemblyof the separate element equilibrium equations into a set of global equations:

NfML = Kl (2-3°)where: [KG] = the global stiffness matrix;

{Ad}ttG = a vector containing the unknown degrees of freedom (nodaldisplacements) for the entire finite element mesh;

{ARG} = the global right hand side load vector.As each element stiffness matrix is formed according to the procedure

described in Section 2.6, it is then assembled into the overall global stiffnessmatrix. This assembly process is called the direct stiffness method. The terms of theglobal stiffness matrix are obtained by summing the individual elementcontributions whilst taking into account the degrees of freedom which are commonbetween elements. This process is described in more detail in Section 2.7.1. Theterms of the right hand side load vector are obtained in a similar manner bysumming the individual loads acting at each node. Further details of the load vectorare given in Section 2.8 and in Chapter 3.

It is clear from Equation (2.28) that, if the constitutive matrix [D] is symmetric,the element stiffness matrices and hence the global stiffness matrix will also besymmetric. This situation occurs for a wide range of material behaviour, includinglinear elastic material behaviour. The non-zero terms of the global stiffness matrixarise from the connections between the degrees of freedom through the elements.From the geometry of the mesh, each degree of freedom is only connected to asmall number of other degrees of freedom; hence, the global stiffness matrix willcontain many zero terms. In addition, most of the zero terms will be outside adiagonal band, see Section 2.7.1. The assembly, storage and solution schemes usedfor the global stiffness matrix take into account its symmetric and banded structure.

2.7.1 The direct stiffness assembly methodThe essence of the direct stiffness method is to assemble the individual terms of theelement stiffness matrix, [KE], into the overall global stiffness matrix, [KG],according to the global degree of freedom numbering scheme. At the element level,the stiffness matrix terms correspond to the relative stiffness between the degreesof freedom contained in the element. At the mesh level, the stiffness matrix termscorrespond to the relative stiffness between the degrees of freedom over the wholemesh. For this reason the size of the global stiffness matrix will depend on the totalnumber of degrees of freedom and the non-zero terms will occur from theconnections between the degrees of freedom through the elements.

To illustrate the stages of the assembly process it is convenient to consider amesh of 4 noded elements with only one degree of freedom at each node (note: fortwo dimensional analyses there are usually two degrees of freedom at each node).With only one degree of freedom at each node the stiffness matrices are much


simpler and the assembly process is easier to describe. In such a situation 'degreeof freedom' may be associated with 'node number'.

The numbering of the degrees of freedom and the form of the stiffness matrixfor a single 4 noded element is illustrated in Figure 2.14. It is assumed that all thestiffness matrices are symmetric and therefore only the diagonal and upper triangleof terms are indicated.

KA =^ 2

SYM

Figure 2.14: Stiffness matrix for a single element

If this single element becomes part of a mesh of elements then, with respect tothe global degree of freedom numbering, the stiffness matrix will take the formshown in Figure 2.15. The numerical values of the stiffness terms remain the same,but the quantities they represent in global terms become different. For example, inFigure 2.14 the stiffness term Ku refers to element degree of freedom 1, whereasin Figure 2.15 the same degree of freedom becomes global degree of freedom 2;hence the contribution of the element to the global value of K22 is equal to Kx,. Theimportant thing to note here is that each row and column of the element stiffnessmatrix corresponds to each degree of freedom of the element.

9

5

1

10

6

2

6

11

7

3 ,

,7 [*•]-

K22

SYM

K2i K2-

K7

K2t

Figure 2.15: Element stiffness matrix in terms of global degrees offreedom

The assembly process can now be demonstrated using the stiffness matricesexpressed in terms of the global degrees of freedom. Figure 2.16 shows a simplemesh containing two elements and the degree of freedom numbers. The stages ofFigure 2.16 show the complete assembly of the global stiffness matrix for thissimple mesh. It should be noted that there is some re-ordering of the terms due to


differences between the standard element numbering and the degree of freedomnumbering. In addition, the terms in the global stiffness matrix corresponding tothe degrees of freedom which are common to more than one element arise fromsumming the contributions from the respective elements. For degrees of freedomoccurring in only one element there is only one term assembled into the globalstiffness matrix.

2 4 6

a) Two element mesh

Kl K\2 K\4

^43

Stiffness matrix forelement 1

K.

KL

Global stiffness matrixb) Assembly of element 1

" If1 V2 V1 If2 ~A 3 3 A 3 4 A 3 6 A 3 5

^525.

->

Stiffness matrix forelement 2

c>

K\2

SYM

Assembly

^ 1 3

A33 + A33

Global stif

K\4 + KlKl + K2

fness matri>

of element 2

K45

K

2

K2K46

K-56

_

ATnK2

^ 4 S ^ d ^

SYM

d) Final assembled structure of the global stiffness matrixFigure 2.16: Assembly procedure for a simple mesh with two

elements


The structure of the global stiffness matrix becomes particularly important ifefficient use is to be made of the computer storage, as is discussed in Section 2.9.At this stage, a few points can be made concerning the structure of the assembledstiffness matrix shown in Figure 2.16d. The non-zero terms in the global stiffnessmatrix arise only from the connection between degrees of freedom through theelements. Thus for each row in the global stiffness matrix, the last non-zero termcorresponds to the highest degree of freedom to which a particular degree offreedom is connected. This property leads to a global stiffness matrix which isgenerally sparse (i.e. the matrix has many zero terms) and banded (i.e. the non-zeroterms are concentrated along the main diagonal).

2.8 Boundary conditionsThe final stage in setting up the global system of equations is the application of theboundary conditions. These are the load and displacement conditions which fullydefine the boundary value problem being analysed.

Loading conditions, line loads and surcharge pressures, affect the right handside of the global system of equations. If line or point forces are prescribed, thesecan be assembled directly into the right hand side vector {ARG}. If pressureboundary conditions are defined, these must first be expressed as equivalent nodalforces before being added to {ARG}. As with the assembly process for the stiffnessmatrix, the assembly of the right hand side vector is performed with respect to theglobal degree of freedom numbering system. Body forces also contribute to{ARG}, as do the forces from excavated and constructed elements. These boundaryconditions are discussed in greater detail in Chapter 3.

Displacement boundary conditions affect {Ad}nG. The equation correspondingto the prescribed degree of freedom (displacement component) must be effectivelyeliminated during the solution process, see Section 2.9. In all cases sufficientdisplacement conditions must be prescribed in order to retain any rigid body modesof deformation, such as rotations or translations of the whole finite element mesh.If these conditions are not satisfied, the global stiffness matrix will be singular andthe equations cannot be solved. For two dimensional plane strain problems at leasttwo nodes must have a prescribed displacement in the x direction and one node aprescribed displacement in the y direction, or, alternatively, two nodes must havea prescribed displacement in the y direction and one node a prescribeddisplacement in the JC direction.

2.9 Solution of global equationsOnce the global stiffness matrix has been established and the boundary conditionsadded, it mathematically forms a large system of simultaneous equations. Thesehave to be solved to give values for the nodal displacements {Ad}nG. There areseveral different mathematical techniques for solving large systems of equations.Most finite element programs adopt a technique based on Gaussian elimination.


However, for three dimensional problems iterative techniques may be more costeffective, see Chapter 11.

To illustrate the Gaussian elimination technique, one particular method ofsolution is described. In this approach the stiffness matrix is stored as a 'sky lineprofile' and inversion of the stiffness matrix follows a triangular decomposition.It should be noted, however, that other procedures are available (Crisfield (1986)).

2.9.1 Storage of the global stiffness matrixBecause of the sparse and banded nature of the stiffness matrix it is not necessaryto store the complete stiffness matrix. Reductions in storage can be obtained bytaking into account the symmetry of the matrix and only storing the diagonal andupper triangle of terms. This reduces the storage space by almost a half. As anexample, Figure 2.17 shows the structure of the global stiffness matrix and theterms which need to be stored for a simple finite element mesh composed of four4 noded elements. Again, for simplicity, it is assumed that there is only one degreeof freedom at each node.

Q o9

-6

1 2 3

a) Finite element mesh (1 DOF per node)

, . 1 2 3 4 5 6 7 8 9.X X O X X O O O O

X X X X X O 0 0X 0 X X 0 0 0

X X 0 X X 0X X X X X

SYM X 0 X XX X 0

X XX.

b) Zero, Non-zero structure of [K^]

Figure 2.17: Structure of the global stiffness matrix

123456789

1 2X X

\ X

\

SYM

30XX

\

4XX0X

\

5NXXXXX

\

6\XX0XX

\

7

\0XX0

X\

8

\XXXXX

\

9

\XX0Xx_

BANDWIDTH

123456789

1 2 3 4 5 6 7 8 9

X X X X XX 0 X X _

SYM

Figure 2.18: Diagonal bandstructure of [KJ

Figure 2.19: Column-profile structure of [KJ

As pointed out in Section 2.7.1, the global stiffness matrix has a diagonal bandstructure, see Figure 2.18. The last non-zero term in any row corresponds to thehighest degree of freedom number to which a particular degree of freedom is


connected through an element. For example, the last non-zero term in row 4corresponds to degree of freedom 8 and it is seen in Figure 2.17 that degree offreedom 8 is the highest degree of freedom to which 4 is connected. Counting foreach column the number of terms above the diagonal term up to the last non-zeroterm (column height) and taking the maximum value of this quantity over all thecolumns, gives the bandwidth of the matrix. It is clear from Figure 2.18 that thebandwidth measured column-wise is the same as measured row-wise and in thisexample equals 5. An efficient compact storage scheme based on the band structureis shown in Figure 2.18. In this case the stiffness matrix is stored row-wise as a twodimensional array and the number of columns in the array is equal to thebandwidth. Alternatively, the stiffness matrix could be stored column-wise, but thishas no effect on the storage requirements or the equation algorithm.

The band storage shown in Figure 2.18 requires space for a number of zeroterms which are contained within the band. Many of these terms are not requiredby the solution algorithm and remain zero throughout the solution process; hencethey need not be stored. The most efficient storage scheme is based on the column-profile shown in Figure 2.19. The column-profile contains the terms in eachcolumn from the diagonal up to the last non-zero term and may contain zero termswithin the profile. Comparing Figures 2.18 and 2.19, it can be seen that thecolumn-profile scheme is essentially a variable bandwidth storage and the zeroterms between the profile and diagonal band are omitted. The zero terms within theprofile are essential in the solution algorithm and the storage locations normallycontain non-zero terms at the end of the solution process, see Section 2.9.2.

As noted above, the bandwidth for any degree of freedom is the differencebetween the highest degree of freedom connected through an element to thatdegree of freedom, and the degree of freedom itself. This difference thereforedepends on the way the degrees of freedom are numbered. There are severalalgorithms available for renumbering the degrees of freedom to minimise thebandwidth (Cuthill and McKee (1969), Gibbs etal (1976), Everstine (1979)).

2.9.2 Triangular decomposition of the global stiffness matrixThe solution technique associated with the column-profile storage scheme is basedon a triangular decomposition of the global stiffness matrix. Formally, thisdecomposition reduces the stiffness matrix to the following matrix product:

(2.31)

where [L] is a lower triangular matrix of the form:

"lL21 1 0£ , , L32 1

.Aii A12 A3 •-

(2.32)


and [DM] is a diagonal matrix:

M=D2 0

A (2.33)

In this section explanation is given of how the terms of [L] and [DM] arederived from the terms of [KG], and how [DM] and [L]T overwrite the diagonal andoff-diagonal terms of[KG], respectively, within the column-profile storage scheme.To illustrate the process, a full nxn symmetric matrix will be considered:

[*]=Ki2 KuK22 K23 K2l

AC. (2.34)

Using the triangular factors defined by Equat ions (2.32) and (2.33) and formingthe matr ix produc t given by Equa t ion (2 .31) , t ransforms the [K] matr ix into thefol lowing form:

?, L,2A L,3A - KA

(2.35)In order to determine the terms of the triangular factors [DM] and [L], the terms

of the matrix [K] given in Equations (2.34) and (2.35) need to be equated. Theorder in which these equations are then solved characterises a number of differentsolution algorithms. Some algorithms solve the equations by following the termsin the rows of [K], some follow the columns of [K] and others modify all the termsof [A] successively. Essentially, all these algorithms are equivalent and require thesame number of arithmetic operations. However, the column-wise technique hascertain advantages in that a smaller number of intermediate terms need to beevaluated, consequently reducing the time of the calculation. The set of equationsto be solved is listed below:

A = ^ n

= *,2/AA -L u =L23 =

A22 ~ ^12^1

^ 3 / A


A - 3̂3 ~ 3 A ~

K,n/Dx (2.36)

Un = I*** ~ A*A3A " L2nL23D2]/D3

A = Km-LlDl-L22nD2-L2

3nD3-...-L2n_]nDn_l

Each term of [DM] and [L] is obtained in sequence, using values of [K] andvalues of [DM] and [L] which are previously determined. Hence, it is easy to seehow the values of [K] can be overwritten by [DM] and [L]. Finally, the uppertriangle of the stiffness matrix will contain the terms:

AA

kn

A (2.37)

For the maximum efficiency there is some advantage in carefully ordering thecomputations in each column. These are performed in three stages: firstly, the off-diagonal terms are modified by an inner product accumulation; secondly, the off-diagonal terms are divided by a corresponding diagonal term and, thirdly, a newdiagonal term is found. This is shown in the set of Equations (2.36).

2.9.3 Solution of the finite element equationsHaving determined the triangular factors [DM] and [L] of the global stiffnessmatrix [KG], the solution of the finite element equations can be performed in threestages. The process solves Equation (2.30), now written in the form:

L] I Ad) = \AR) (2.38)L JL JL J V ) \ )

where {Ad} is the vector of unknown degrees of freedom and {AR} is the righthand side vector of prescribed nodal forces. Setting

{Ad"} = [DM][Lf{Ad} (2.39)the first stage of the solution process finds {A*/"} as:

{Ad"} = [Z,]~!{AJl} (2.40)by forward substitution. Then, setting

{Arf'j = [L]T{AR) (2.41)

gives: Ad'} = (2.42)


The final stage of the solution process then determines

{Ad} = [L]~T{Ad'} (2.43)by backward substitution. The three stages of the solution technique are performedwith the triangular factors [DM] and [L], which are stored in column-profile formin a one-dimensional array. The decomposed stiffness matrix remains unchangedthroughout the solution process and all the calculations are performed on the righthand side vector. Hence, the same solution process could be repeated with manydifferent load vectors, without having to re-decompose the stiffness matrix. Toillustrate this solution process, each of the three stages is considered in detailbelow.

STAGE 1: Forward substitutionThis stage solves a set of equations of the form:

1L]2L]3

1

L2n

1

0

.. 1

Adi'Ad'2'Ad;' (2.44)

which lead, simply by forward substitution, to:

Ad;' = / ,Ad'' = f2-LnAd'±

(2.45)

k=\

Since the terms of [L] are stored in column-profile form, some of the termsindicated in Equations (2.44) and (2.45) may be outside the column-profile and aretherefore zero. In practice, the forward substitution sums indicated in Equation(2.45) are only evaluated over the number of terms of [L] within the column-profile.

STAGE 2: invert diagonal matrixIn this stage the simple equation is solved:

(2.46)

which gives:

AD2

0A

0

A.

Ad;'Ad'2Ad'

Ad'n

Adi'Ad'2'Ad;'

Ad"


Ad[ =Ad'2 =

Ad['/Dx

Ad'2'/D2

Ad; = Ad3"/D3

Ad'n = AdyDn

STAGE 3: Backward substitutionFinally, a set of equations is solved, of the form:

(2.47)

1A3 ••• LXr \Adx

Ad2<Ad3

[Adn

. = <

Ad[Ad'2Ad'3

'Ad'n

(2.48)

1

which lead, by backward substitution, to the solution:Adn = Ad;,Adn_x = Ad;,_x-Ln_XnAdn

Adn_2 = Ad;}_2 - Ln_2^_xAdtl_x -,

Adx =

(2.49)

k = 2

The backward substitution sums indicated in Equation (2.49) are only evaluatedover the number of terms of [L] which lie within the column profile.

The solution process described above is extremely efficient and has theadvantage that the decomposed stiffness matrix remains unchanged. This fact isparticularly important for nonlinear problems which are solved by iteration. In thiscase successive solution estimates can be obtained by modifying the right hand sideload vector, using a stiffness matrix which needs to be decomposed into triangularfactors only once.

2.9.4 Modification due to displacement boundary conditionsIn most finite element calculations displacement boundary conditions are definedin order to restrict rigid body movements, or to restrain part of the structure. Theseconditions imply that certain degrees of freedom are prescribed and the system ofequations:

[Kc]{Ad}nG = {***} (2-50)needs to be modified in order to enforce these conditions. To see the effect on thesystem of equations when a particular degree of freedom, for example Adp isprescribed, Equation (2.50) is written in the full form (Equation (2.51)). In thiscase, the force term ARj becomes unknown and is usually required in the solutionas a reaction term:


Ku .

Kn .

* » i •

• * v •

• KM •

• K » , •

• * , „

• Kn

Ad

Ad

[Ad

AR

AR, (2.51)

Since Adt is prescribed, they'"1 equation in system (2.51) is not required and canbe replaced by the simpler equation:

(2.52)Ad, = aj

where a; is the prescribed value. In addition, each other equation contains a termKjjAdj, which is also known and can therefore be transferred to the right hand side.Performing these two operations, Equation (2.51) reduces to :

Ku .. 0 .. Kt

o ;: r ;; oKni .. 0 ..

Adx AR,-K]jaj

AR, -,

(2.53)

Thus the prescribed degree of freedom is taken into account by replacing thefh row and column in [KG] by zeros, the diagonal term by 1 and updating the righthand side vector by simple correction terms. This method has the advantage thatthe symmetry of the stiffness matrix is not destroyed and, hence, the modificationscan easily be made within the compact column-profile storage scheme before thesolution process. However, this method, as it stands, has two importantdisadvantages. The first is that the terms of [KG] are overwritten by O's and 1 fs andhence the value of the reaction force ARj cannot be found later by backsubstitution. The second disadvantage occurs for incremental or iterativetechniques where the values of the prescribed degrees of freedom change. In suchcases, the modifications to the right-hand side vector cannot be performed, becausethe required terms of [KG] are not saved, and a complete new global stiffnessmatrix must be formed at each stage.

The above disadvantages are avoided by a simple modification of the standarddecomposition and solution procedures. First, it is assumed that the stiffness matrixis effectively replaced by the matrix shown in Equation (2.53). However, the termsin the / h row and column of [KG] are not actually changed. During thedecomposition of [KG] the terms associated with the/h row and column of [KG] areskipped over during the calculations (since they are assumed to be zero), and theremaining columns are decomposed as before. Prior to the solution process, theright-hand side vector is modified using the prescribed values and the correct termsfrom [KG]. Then finally, during the solution process, the/h row and column of [KG]are again effectively ignored. This technique clearly allows the reaction forces tobe calculated by simply back-substituting the solution into the appropriateequations stored within the decomposed [KG].


2.10 Calculation of stresses and strainsOnce the global equations have been solved and values of the nodal displacementsobtained, secondary quantities, such as stresses and strains, can be evaluated.Strains can be found using Equation (2.11) and these can then be combined withthe constitutive matrix, [D], to obtain stresses, see Equation (2.17).

Line of symmetry

Footing

/ AE= 10000 fcN/m2

LI = 0.4

Figure 2.20: Footing problem

2.11 ExampleTo illustrate the application of thefinite element method, the boundaryvalue problem of a strip footing on anisotropic linear elastic soil, as shownin Figure 2.20, is considered. Thefooting has a width of 12m and thesoil has a Young's modulus E = 10000kN/m2, and a Poisson's ratio ju = 0.4.As there is a vertical plane ofsymmetry through the centre of thefooting, the finite element analysisneeds to consider only half of theproblem.

The finite element mesh used toanalyse this problem is shown inFigure 2.21. It consists of 42 eightnoded isoparametric elements. Onlythe soil is discretised into finiteelements. The foundation isrepresented by appropriate boundaryconditions. As it is necessary toanalyse a finite domain, the meshextends 20m horizontally and 20mvertically.

With finite element analyses, it isnecessary to specify an x (horizontal)and j ; (vertical) boundary condition ateach node on the boundary of themesh. This boundary condition can be either a prescribed nodal displacement ora nodal force. For the strip footing problem, it is assumed that both thedisplacements AM and Av, in the horizontal and vertical directions, are zero for allnodes along the bottom boundary of the mesh, see Figure 2.21. Consequently,nodal reactions (forces) will be generated in both the horizontal and verticaldirections as a result of the analysis. On the vertical side boundaries, the horizontaldisplacement, Aw, and the vertical force, AFy, have been assumed to be zero. TheAFy =0 condition implies that there can be no vertical shear stress on these

A«=AF,=0

&>

&>

Fc

wmotin

|4

\ \ N

AFx=AFK=0

///

Figure 2.21: Finite element meshfor strip footing problem


boundaries and the nodes are, therefore, free to move in the vertical direction.Horizontal reactions will, however, be generated at the nodes as a result of theanalysis. Along the top boundary of the mesh, from the edge of the strip footingto the right hand side boundary, both the vertical, AF^, and horizontal, AFX, nodalforces are assumed to be zero. This implies a stress free surface which is free tomove both vertically and horizontally.

The boundary condition applied to the mesh boundary immediately below theposition of the strip footing depends on whether the footing is rigid or flexibleand/or smooth or rough. Three different alternatives for this boundary conditionare shown in Figure 2.22.

FootingAv = - 1 0 m m AFX = 0 AF=AFy*a0

FootingAv = -10mm AM = 0 AF=AF=0

Au=AF=0

>

FootingAF, = -lOOkPa AFX = 0 AF,=AF=0

Aw=AF=0 Au=AF>,=0

a) Smooth rigid footing b) Rough rigid footing c) Smooth flexible footing

Figure 2.22: Footing boundary conditionsIn Figure 2.22a the footing is assumed to be rigid and smooth. Consequently,

a vertical displacement of 10mm (downward) and AFx=0 is prescribed at each nodebeneath the position of the footing. In Figure 2.22b the same vertical displacementboundary condition is applied, but instead of the horizontal force being zero, thehorizontal displacement Aw = 0. This models a rough rigid footing. In both theabove cases a rigid footing is simulated by applying a uniform verticaldisplacement to the footing. In Figure 2.22c a uniform surcharge pressure isapplied via the vertical nodal forces, AFy, to the nodes beneath the position of thefooting. In addition, the horizontal nodal forces, AFX, are assumed to be zero.These boundary conditions therefore model a smooth flexible footing.

It should be noted that many finite element programs do not require that theuser specifies an x (horizontal) and;; (vertical) boundary condition at each node onthe boundary of the mesh. In such a situation, the program will make an implicitassumption. Usually, if a boundary condition is not prescribed, the program willassume that the appropriate nodal force is zero. For example, if only a verticaldisplacement is specified at a particular node, the program will assume thehorizontal force AFX = 0.

Analyses with all three of the footing boundary conditions given in Figure 2.22have been performed. The surface settlement profiles predicted by these analysesare compared in Figure 2.23. To aid comparison, the vertical settlement, Av, hasbeen normalised by the settlement under the centre of the footing, Avmax. Thesettlement profiles for both the smooth and the rough rigid footings are verysimilar.


The analytical solution for asmooth flexible footing on anisotropic elastic half-space predictsthat the vertical displacements areinfinite. However, finite values arepredicted by elastic theory if theelastic material has a finite depth. Forthe situation analysed above, theanalytical prediction of the verticalsettlement at the edge of the footing is0.057m. This compares favourablywith the finite element prediction of0.054m. Finite element predictions

x - coordinate

Smooth flexible footing

Smooth rigid footing

Rough rigid footing

Figure 2.23: Ground surfacesettlements

closer to the analytical solution can be obtained if a more refined mesh is used.Vectors of ground movement predicted by the finite element analysis of the

rough rigid footing are shown in Figure 2.24. These vectors indicate thedistribution of movements within the elastic soil. The contours of the majorprincipal stress, Acrb are shown in Figure 2.25.

Figure 2.24: Vectors ofground movement

Figure 2.25: Contours ofprincipal stress

2.12 Axi-symmetric finite element analysisIn the preceding discussion a plane strain situation has been assumed whendeveloping the finite element equations. The basic procedures are, however, similarfor plane stress, axi-symmetric and full three dimensional situations.

As discussed in Chapter 1, several geotechnical problems can be idealised asaxi-symmetric, e.g. axially loaded piles and circular foundations. In this type ofproblem it is usual to adopt cylindrical coordinates r (radial direction), z (verticaldirection) and 6 (circumferential direction), Figure 2.26, as opposed to the


Cartesian coordinates used for plane strain analysis.There are four non-zero stresses (ar, az, a0 and zrz),four non-zero strains (sr, ez, s0 and yrz), and twodisplacements (u and v) in the r and z direction,respectively. The development of the finite elementequations is very similar to that presented for planestrain situations.

The problem is discretised into finite elements inthe r-z plane. If isoparametric elements are used, thegeometry of the elements is expressed usingEquation (2.7), with r and z replacing x and yrespectively. As with plane strain problems, the nodal degrees of freedom are thedisplacements u and v. The formulation of the element equations follows theprocedure outlined in Section 2.6, but with Equation (2.10) replaced by:

Figure 2.26: Cylindricalcoordinates

Aer = - d(Au) ,

d(Av)dr

d(Av)dz

AAs0 = Au

rd(Au)dz

(2.54)

Equation (2.11) replaced by:

Ase

dN,dr0

dNx

0

dNx

dzdNx

dN2

dr0

dN2

0

dN2

dzdN2

dz drA o

r

dzN2

dro

dN^dr dz

o Hi.

dz

dro

Aux

Av,Au2

Av,

AM,,Av,,

(2.55)

all references to x and y replaced by r and z, respectively, and the thickness t inEquat ions (2.27) and (2.28) replaced by 2nr. The assembly and solution of theglobal equations follow the procedures described in Sections 2.7 to 2.9.

The application of the finite element method to three dimensional problems isdiscussed in Chapter 11.

2.13 Summary1. The finite element method involves the following steps:

Element discretisation;Selection of nodal displacements as primary variables;Derivation of element equations using minimum potential energy;Assembly of element equations to form global equations;Formulation of boundary conditions (nodal displacements and forces);Solution of global equations.


2. For geotechnical analyses triangular or quadrilateral elements with curved sidesare usually employed.

3. Isoparametric elements are popular. In this formulation, both the elementdisplacements and element geometry are expressed using the same interpolationfunctions which are expressed in terms of natural coordinates.

4. Numerical integration is used to establish the element stiffness matrix.Consequently, stresses and strains are usually output at integration points.

5. The global stiffness matrix is symmetric if the constitutive matrix is symmetric.The global stiffness matrix is also sparse and banded. These properties are usedto develop efficient storage algorithms.

6. Gaussian elimination is usually adopted to solve (invert) the global stiffnessmatrix. This can involve triangulation of the matrix.

7. Care must be taken when dealing with prescribed displacements.8. Stresses and strains are secondary quantities which are calculated from the

nodal displacements.

Appendix 11.1: Triangular finite elements11.1.1 Derivation of area coordinatesDerivation of shape and interpolation functions for triangular elements is shownusing the example of the three noded element, presented in Figure II. 1. Theposition of an interior point P, in global coordinates x and;/, is determined by thevector r\

r=xi +yj =x3i + y3jwhere: F,f = unit vectors in global

coordinate system x, yeue2 = unit vectors in local

coordinate system £ r\£ rj = system of local coordinates

coinciding with two sidesof a triangle. j *

If, instead of £ r\ coordinates, normalizedcoordinates are introduced, such as:

•r]e2 (II.l)

/ hi Figure II. 1: Three nodedtriangular elementwhere /3! and /32 are the lengths of sides 3,1

and 3,2 of the triangle in Figure II . l , thennew coordinates Lx and L2 vary only between 0 and 1. Equation (II. 1) can now bewritten as:

r = xj + yj + l3XLxex + l32L2e2

When / 3 ] ^ and I32e*2 a r e expressed in terms of global coordinates, i.e:

(II.2)


then vector f becomes:l32e2 = (x2-x3)J + (y2 - y 3 ) j

r = [Lxxx + L2x2 + (1 - Lx - L2)x3] i

[Lxyx+ L2y2+(\- Lx- L2)y3]](113)

(114)

The link between coordinates x andj and Z,, /= 1,2,3, is therefore:x = Lxxx + L2x2 + L3x3

y = Lxyx + L2y2 + L3y3

where L3=\~LX-L2.All three coordinates Lh i= 1,2,3, vary between 0 and 1 and represent

interpolation functions, Nf, of a three noded triangle:

(II.5)

The geometric meaning of these coordinates can be obtained by manipulatingEquation (II.4), which can be written in the form:

? X2 X3

1 1 1(11.6)

Inverting Equation (11.6) gives expressions for coordinates Li9 /=1,2,3, in terms ofglobal coordinates:

a2 b2 c2

orA = — (^ + btx + c^) , / = 1,2,3

(117)

(118)

In Equation (II.7), A is the determinant of the system of Equations (II.6) and it canbe shown to represent the area of a triangle in Figure II. 1. Parameters ai7 b} and c,are:

ax = x2y3 - x2y2 a2 = x2yx - xxy3 a3 = xxy2 - x2yxb]=y2-yi b2=y3-yx b3=yx-y2 (II.9)C\ — X3 X2 C-i — Xry X\

It can also be shown that the nominator in Equation (II.8) represents areas of thethree triangles, marked as 1, 2 and 3, in Figure II.2.


Therefore the geometric interpretationof the coordinates Lh /=1,2,3, is that theyrepresent the ratio of the area, Aj, of theappropriate triangle, 1, 2 or 3, and thetotal area of the element, A:

- —L 7 = 1 2 1 (11.10)

Consequently,coordinates'.

they are named 'area

Figure 11.2: Area coordinates of atriangular element

11.1.2 Isoparametric formulationIn the same way as for quadrilateralelements, the isoparametric formulation isalso used for triangular elements. FigureII.3 shows a three noded isoparametricparent triangular element and its globalderivative. The parent element is a rightangled triangle, in natural coordinates Tand S, where 0<T< 1 and O^^S l̂.

Interpolation functions Nf , /= 1,2,3,can now be expressed in terms of naturalcoordinates S and Tas:

3(0,1)

1(0,0) 2(1,0)

Parent element Global element

Figure 11.3: Isoparametrictriangular element

Nx=\-S-TN2 = SN3 = T

(11.11)

For higher order 6 noded triangular elements, Figure II.4, it can be shown thatthe interpolation functions become:

yMid - side nodes:N4=4S(\-S-T)NS=4STN6=4T(\-S-T)

Corner nodes:Nl = 1-5-7*N2=S

(11.12)

Figure 11.4: Six nodedtriangular element

Figure II.5 shows the location of Gauss integration points for a triangularelement for 3 and 7-point integration.


T 1

£,=0.16666S^O.66666

£,=0.10128£2=0.79742

£4=0.47014

5^0.05971£7=0.33333

T2=S{

Figure 11.5: Gauss integration points fortriangular element

2_2. finite element theory for linear materials

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