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Contents

I Structural Analysis with FEM 1

1 Scope and Goals of the Course 3

1.1 Key Note . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Educational Goals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Scope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Steps of a Finite-Element Analysis . . . . . . . . . . . . . . . . . . . . . . . 5

2 Ancillary Concepts 7

2.1 Variational Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Notations and Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2.1 Summation Convention . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2.2 Kronecker Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.3 Permutation Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . 82.2.4 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2.3 Boundary and Initial Value Problems . . . . . . . . . . . . . . . . . . . . . . 92.4 Partial Integration and Theorems . . . . . . . . . . . . . . . . . . . . . . . . 9

2.5 Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.6 The Variational Symbol . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.7 The Minimum of a Quadratic Functional . . . . . . . . . . . . . . . . . . . . 112.8 Principle of Minimum Potential Energy . . . . . . . . . . . . . . . . . . . . 122.9 Ritz Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3 Governing Equations 15

3.1 Linear Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.1.1 Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3.1.2 Kinematical Equations . . . . . . . . . . . . . . . . . . . . . . . . . . 163.1.3 Generalized Hookes Law . . . . . . . . . . . . . . . . . . . . . . . . 183.1.4 Temperature- and Moisture strains . . . . . . . . . . . . . . . . . . . 193.1.5 Displacement Formulation . . . . . . . . . . . . . . . . . . . . . . . . 193.1.6 Matrix Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.2 Simplified Elasticity Equations . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2.1 Plane Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.2.2 Axial Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.2.3 Generalized Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . 253.2.4 Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.2.5 Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2.6 Plate Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303.2.7 Thick-Walled Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . 323.2.8 Restrictions on Elastic Constants: Isotropic Materials . . . . . . . . 33

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3.3 Heat Conduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4 Numerical Systems of Equations 37

4.1 Differential and Energy Equations . . . . . . . . . . . . . . . . . . . . . . . 374.2 Weak Variational Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.4 Discretization of the Weak Variational Form . . . . . . . . . . . . . . . . . . 39

4.5 Numerical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.5.1 Gauss Quadrature for Line and Quadrilateral Elements . . . . . . . 404.5.2 Integration Rules for Triangular Elements . . . . . . . . . . . . . . . 43

5 Shape Functions and Derivatives 45

5.1 Local and Natural Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 455.2 Isoparametric Finite Elements . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5.3 Convergence Criteria on Shape Functions . . . . . . . . . . . . . . . . . . . 475.4 Higher-Degree Shape Functions and Curvilinear Elements . . . . . . . . . . 475.5 Deriving Polynomial Element Shape Functions . . . . . . . . . . . . . . . . 495.6 Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

5.6.1 Strain Approximation Functions B . . . . . . . . . . . . . . . . . . . 495.6.2 Domain Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

5.6.3 Surface Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.7 Shape Functions for Selected Element Geometries . . . . . . . . . . . . . . . 52

5.7.1 One-Dimensional C0 Element . . . . . . . . . . . . . . . . . . . . . . 525.7.2 Linear Triangular Element . . . . . . . . . . . . . . . . . . . . . . . . 535.7.3 Bilinear Quadrilateral Element . . . . . . . . . . . . . . . . . . . . . 56

5.7.4 Quadratic curvilinear Serendipity Element with Eight Nodes . . . . 585.7.5 Linear Tetrahedral Volume Element . . . . . . . . . . . . . . . . . . 59

6 Element Types 65

6.1 Truss and Bar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 666.2 Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

6.2.1 Shear Deformable Beam Element . . . . . . . . . . . . . . . . . . . . 686.2.2 Kirchhoff Beam Element . . . . . . . . . . . . . . . . . . . . . . . . . 70

6.3 Plane Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 726.3.1 Principle of Virtual Displacements and Variational Form . . . . . . . 726.3.2 Weak Variational Form . . . . . . . . . . . . . . . . . . . . . . . . . 73

6.3.3 Discretization of the Variational Form . . . . . . . . . . . . . . . . . 746.4 Generalized Plane Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 746.5 Mixed Form Robust Solid Elements . . . . . . . . . . . . . . . . . . . . . . . 76

6.6 Plate Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 776.7 Thick-Walled Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

7 Assembly and Solution 83

7.1 Model Organization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 837.1.1 Nodal Point Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . 837.1.2 Connectivity Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

7.1.3 Essential Boundary Conditions Specification . . . . . . . . . . . . . . 84

7.2 Assembly . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 847.3 Essential Boundary Conditions Implementation . . . . . . . . . . . . . . . . 857.4 Bandwidth Minimization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

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7.5 Direct Solution Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 877.5.1 Gaussian Elimination . . . . . . . . . . . . . . . . . . . . . . . . . . 887.5.2 Cholesky Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 89

7.6 Iterative Solution Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . 907.6.1 Jacobi Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 917.6.2 Gauss Seidel Iteration . . . . . . . . . . . . . . . . . . . . . . . . . . 92

7.6.3 Method of Steepest Descend . . . . . . . . . . . . . . . . . . . . . . . 927.6.4 Conjugate Gradient Method . . . . . . . . . . . . . . . . . . . . . . . 947.6.5 Preconditioning . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

7.7 Solution of Eigenvalue Problem by Matrix Iteration . . . . . . . . . . . . . 967.8 Degrees of Freedom Elimination . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.8.1 Substructure Technique . . . . . . . . . . . . . . . . . . . . . . . . . 98

7.8.2 Elastic Foundation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 997.8.3 Complex Finite Elements . . . . . . . . . . . . . . . . . . . . . . . . 101

8 Post Processing 105

8.1 Strain Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1058.1.1 Optimum Points for Strain Calculation . . . . . . . . . . . . . . . . 106

8.2 Stress Calculation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

8.3 Strength Assessments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1108.4 Visual Representation of Numerical Results . . . . . . . . . . . . . . . . . . 110

9 Modeling of Structures 111

9.1 Problem Analysis or What Do I Want to Know? . . . . . . . . . . . . . . . 1119.1.1 How Stiff ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

9.1.2 How Strong? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1129.1.3 For How Long? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

9.2 Modelling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1129.2.1 Earlier FEM Programs . . . . . . . . . . . . . . . . . . . . . . . . . . 1129.2.2 Current FEM Programs . . . . . . . . . . . . . . . . . . . . . . . . . 1139.2.3 Intrinsic Structural Model . . . . . . . . . . . . . . . . . . . . . . . . 113

9.3 Reduction of Computational Effort . . . . . . . . . . . . . . . . . . . . . . . 114

9.3.1 Bandwidth Minimization . . . . . . . . . . . . . . . . . . . . . . . . 1149.3.2 Model Size Reduction by Using Symmetries . . . . . . . . . . . . . . 1149.3.3 Reduction of Degrees-of-Freedom by Structural Elements . . . . . . 1159.3.4 Working with Several Load Cases . . . . . . . . . . . . . . . . . . . . 115

9.4 Model Validation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1159.4.1 Is the Geometry Correctly Defined? . . . . . . . . . . . . . . . . . . 1159.4.2 Is the Geometry Accurately Mapped? . . . . . . . . . . . . . . . . . 116

9.4.3 Solution Precision or Convergence . . . . . . . . . . . . . . . . . . . 1169.5 What Element for What Purpose . . . . . . . . . . . . . . . . . . . . . . . . 1169.6 Combining Different Element Types . . . . . . . . . . . . . . . . . . . . . . 1169.7 Guidelines for Meshing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

10 Element Efficiency 117

10.1 Element Distortion and Mapping Errors . . . . . . . . . . . . . . . . . . . . 117

10.2 Reduced Integration Techniques . . . . . . . . . . . . . . . . . . . . . . . . . 117

10.3 Patch Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

11 Special Issues 119

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12 Sample Problems 121

12.1 Composite Tube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1211 2 . 2 F l y w h e e l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2 1

12.3 Multidirectional Laminates Edge Effect . . . . . . . . . . . . . . . . . . . . 12112.4 Damage Progression Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . 121

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Part I

Structural Analysis with FEM


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Chapter 1

Scope and Goals of the Course

1.1 Key NoteThe finite-element method (FEM) provides the basis for powerful analysis tools and isapplicable to a vast range of continuum mechanics problems including exotic topics suchas stellar life cycle simulations. Its nature is to find approximations to true solutions thatare not accessible to exact analytical solution methods. Supported by todays availabilityof computational power, the FEM is routinely used in engineering practice. The programsallow the performance of very complex analyses within reasonable time spans. Their userinterfaces are increasingly becoming transparent, convenient, and, at least for the user withtheoretical background, self-explanatory. These attributes stimulate the propagation of thecommercial programs and it is the trend that the designing engineer increasingly analyzesthe structural properties of his parts by himself. So the number of FEM users increasesdramatically and the engineering student, even when not majoring in mechanics, must notneglect structural analysis. All of this provokes the question: can the mastery of the userinterfaces of FEM tools may substitute a deepened understanding of structural mechanics?Thinking of the more simple tasks of structural analysis one might be tempted to affirm.However, when it comes to critically loaded structural parts, attaching high responsibilityto the structural analysis, and for problems whose solution is not so obvious to obtain, atheoretical understanding is as essential as in the old days when industrial firms employedtheir own creators of FEM software.

1.2 Educational Goals

The goal of this lecture course is to impart practical and theoretically founded abilitiesneeded by the practicing structural analyst. The students shall combine some theoreticalunderstanding of the workings of the FEM with some degree of routine in solving structuralanalysis problems by using a commercial FEM program. The students shall also have cometo some critical understanding of the method, its limits, and possible modelling pitfalls.In particular, the student shall understand

how the FEM works and to construct finite elements the program Sequence of FEM Software

how to model structures for FEM analysis how to interpret results and assess modelling and other approximation errors


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4 Scope and Goals of the Course

1.3 Scope

The course Structural Analysis with FEM focuses on linear elasticity and combines an

introduction to the FEM method with application of a commercial FEM package. Thematerial collected in this script is not strictly identical with the lecture classes. It coversmore material than the lectures but the material selected for the lectures is presented therewith additional didactic material.

Chapter 2 cites ancillary concepts such as the gradient theorem, the Gauss-Greentheorem, the minimum of a quadratic functional, the principle of the minimum of thepotential energy which are necessary to develop finite elements from continuum mechanicsequations.

The general elasticity equations as well as those relating to the simplifying mechanicalmodels such as plane-stress, truss, beam, or plate are recalled in chapter 3. The models forgeneralized plane strain and for the long and thick-walled cylinder with layered anisotropic

walls are also included in the script as they provide interesting cases, also in terms of theirfinite-element formulations, that are not discussed in many standard FEM textbooks.

The finite-element method is basically a processor for converting a problem with gov-erning equations and boundary conditions prescribed on the boundary of a given domainto a numerical system of equations. Chapter 4 provides the general recipe for obtainingnumerical equations for a given problem, starting with finding the weak variational formfor a governing equations in displacement formulation and concluding with numerical in-tegration of the discrete variational form.

Chapter 5 deals with shape functions for some element geometries as far as these as-pects can be separated from the specific physical meaning of a finite element.

The finite element stiffness matrices, relating to the mechanical models considered in

chapter 3, are discussed in chapter 6.The assembly of the system equations from the contributions of all the finite elements of

the structural model and the implementation of the essential boundary conditions concludethe conversion of the given elasticity problem to its numerical approximate representationat the beginning of chapter 7. The rest of this chapter provides some insight into basicdirect and iterative techniques for resolving the system of equations or finding eigenvectorsand eigenfrequencies.

The post processing of the displacement solution is considered in Chapter 8, which ad-dresses the important question of how to obtain optimum stress (or strain) points withina given finite element.

Chapter 9 imparts some general ideas relevant for structural analysis with FEM. Re-

grading the lecture class, much of this will be transferred to the students through theexercises with a commercial FEM package.


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1.4 Steps of a Finite-Element Analysis 5

1.4 Steps of a Finite-Element Analysis

The following list taken from [1] gives an overview over the steps involved in the finite-

element analysis of a typical problem.1. Discretization (or representation) of the given domain into a collection of preselected

finite elements. (This step can be postponed until after the finite-element formulationof the equation is completed.)

(a) Construct the finite-element mesh of preselected elements

(b) Number the nodes and elements

(c) Generate the geometric properties (e.g., coordinates, cross-sectional areas, etc.)needed for the problem.

2. Derivation of element equations for all typical elements in the mesh.

(a) Construct the variational formulation of the given differential equation over thetypical element.

(b) Assume that a typical dependent variable u is of the form

u =n

i=1 uii

and substitute it into step 2a to obtain element equations in the form

Ku = f

(c) Derive or select, if already available in the literature, element interpolationfunctions Ni and compute element matrices.

3. Assembly of element equations to obtain the equations of the whole problem.

(a) Identify the inter element continuity conditions among the primary variables(relationship between the local degrees of freedom and the global degrees offreedomconnectivity of elements) by relating element nodes to global nodes.

(b) Identify the equilibrium conditions among the secondary variables (relation-ship between the local source or force components and the globally specifiedsource components.

(c) Assemble element equations using steps 3a and 3b and the superposition prop-

erty.

4. Imposition of the boundary conditions to the problem.

(a) Identify the specified global primary degrees of freedom.

(b) Identify the specified global secondary degrees of freedom (if not already donein step 3b).

5. Solution of the assembled equations.

6. Postprocessing of the results.

(a) Compute the gradient of the solution or other desired quantities from the pri-

mary degrees of freedom computed in step 5.(b) Represent the results in tabular and/or graphic form.


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6 Scope and Goals of the Course

The various steps are considered in some detail throughout the script although thesequence of the topics is not always following that of the list.

I have always thought that the construction of a numerical system of equations to

approximate the solution to a given differential equation over a domain of general shape iswhat the FEM is all about and what makes it such a powerful analysis tool. At the coreof this is the construction of an individual finite element or step 2 of the list. You will findthe systematic symbolic derivation of the numerical equations from a given differentialequation in chapter 4. The construction of element interpolation functions is treated as atopic in its own right and you will find material on this in chapter 5.

The material regarding steps 3, 4, and 5 is cramped together in chapter 7 because it isnot so difficult to understand or represent as long as we delegate the details of band-withminimization or equation solving to somebody else.

Step 6 of the list, postprocessing of the results, is considered in chapter 10.Remains step 1 of the list. For the practitioner, mastering this step is the key to

successful analyzes or correct results. A selection of element types from which to built astructural model is discussed in chapter 6. Questions regarding the meshing of a structure,the combination of different element types and ways to control solution quality can be quitetricky and some aspects are given in 9.


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Chapter 2

Ancillary Concepts

2.1 Variational MethodsThe solution of differential equations with variational methods requires two steps [1]:

1. transform a given differential equation into the weak variational form

2. find an approximate solution using a method such as Ritz or Galerkin

A variational form presents a differential equation in an integral form. When the deriva-tives of the dependent variable and of its variation, or test function, are of the same order,the variational form is called a weak form. The weak form prescribes fewer derivations ofthe dependent variable than the original differential equation.

For most linear problems solving the weak form is equivalent to minimizing thequadratic functional of the total potential energy. In analogy to the necessary condi-tion for the minimum of an ordinary function, the first derivative of a functional withrespect to the dependent variables must vanish. The function that minimizes the func-tional is the true or, when only approximate functions are available, the best solution forthe differential equations and boundary or initial conditions.

A variational method approximates the dependent variable of a problem in terms of alinear combination of suitably chosen functions u =

cjj . The free parameters cj are

determined so that the function u minimizes the functional or, equivalently, satisfies theweak variational form.

2.2 Notations and Conventions

The presentation of the conventions in this section follows the textbook by J.N. Reddy [1].

2.2.1 Summation Convention

It is convenient to abbreviate a summation of terms by understanding that a repeatedindex means summation over all values of that index. Thus the summation

x = x1e1 + x2e2 + x3e3 =3

i=1

xiei (2.1)

can be abbreviated to

x = xiei. (2.2)


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8 Ancillary Concepts

2.2.2 Kronecker Symbol

In an orthogonal coordinate system the basis vectors ei are orthogonal to each other and

their lengths are unity:

e1 e1 = 1 e1 e2 = 0 e1 e3 = 0e2 e2 = 1 e2 e3 = 0 e3 e3 = 1 (2.3)

These six relations can be compactly expressed by the relation

ei ej = ij (2.4)

where

ij = 1 i = j0 i = j

(2.5)

The symbol ij is called Kronecker delta. The dot product of two vectors A and B in arectangular cartesian system can now be expressed in the form

A B = (Aiei) (Bj ej)= AiBj (ei ej)= AiBjij

= AiBi

(2.6)

2.2.3 Permutation SymbolIn an orthogonal system, the basis vectors ei also satisfy the following cross-product rela-tions,

e1 e2 = e3, e2 e1 = e3, e2 e3 = e1e3 e1 = e2, and ei ej = 0 if i = j , (2.7)

or in short

ei ej =

ek if i = j = k and i,j,k, permute in cyclic order ek if i = j = k and i,j,k, permute not in a cyclic order

0 if any two of the indices are the same(2.8)

The permutation symbol can be used to write the cross-product of two vectors in anabbreviated form

A B = (Aiei) (Bj ej)= AiBjijk ek

(2.9)

The volume V spanned by three vectors A, B and C as shown in Fig. 2.1 is given by themixed product

V = A B C= (A2B3 B2A3) C1 + (A3B1 B3A1) C2 + (A1B2 B1A2) C3= ijkAiBjCk.

(2.10)

This abbreviation comes in handy when deriving isoparametric finite elements.


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2.3 Boundary and Initial Value Problems 9

BA

A

BC

Figure 2.1: The volume of the parallelepiped spanned by A, B, and C results from thearea (cross-product from A and B) and the height (dot product with C)

2.2.4 Derivatives

Derivatives of functions of one spatial variable are written as:

f(x) = df(x)dx

(2.11)

Time derivatives are written as

f(t) =df(t)

dt, f(t) =

d2f(t)

dt2. (2.12)

Partial derivatives of functions of several spatial variables are identified by the respectivesubscripts:

Fx(x,y ,z) =F(x,y ,z)

xFxy(x,y ,z) =

y(

F(x,y ,z)

x) (2.13)

2.3 Boundary and Initial Value Problems

The objective of structural analyzes is often to determine unknown functions within ageometric domain bounded by . When the dependent variables, or solution functions,depend on a single independent variable, the domain is one-dimensional and the boundaryconsists of the two end points. This is typical for truss and beam problems. Two spatialindependent variables span a two-dimensional domain on a plane or surface curved inspace. The boundary is then a closed curve. With three independent variables the domainis three-dimensional and the boundary is the enclosing surface. A boundary value problemprescribes, along with the differential equation, values of the dependent variables and theirderivatives on the boundary. An initial value problem prescribes values of the independentvariable and possibly also their derivatives at a starting time (t = 0).

2.4 Partial Integration and Theorems

Let us first consider sufficiently often differentiable functions g and f that depend on asingle variable x. For these we obtain, using partial integration, the identityb

a

fgdx = (fg)|ba ba

fgdx (2.14)

Now let us consider functions of several variables. The gradient of such functions gives thepartial derivatives of this function with respect to the respective variables. The gradient

of such a function is symbolically written with the nabla-operator = ex

x+ ey

y+ ez

z(2.15)


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where ei symbolizes unit vectors pointing in the directions of the base coordinates. Thesecond derivative is written with the Laplace operator:

= 2

x2 +

2

y2 +

2

z2 . (2.16)

With these operators one can write the gradient theorem for scalar functions F:

grad(F)d

F d =

n F d. (2.17)

and the divergence theorem for vector functions G:

div(G)d =

Gd =

nGd. (2.18)

The dot indicates the scalar product of two vectors. The symbol n means the unit vector

perpendicular to the boundary . The following identities result, in the same way asthe gradient theorem and the divergence theorem, from equivalent transformations usingpartial integration. They are useful for deriving the weak variational form of a givendifferential equation. In the following F as well as G are scalar functions.

F(G)d =

nF Gd

(F)Gd. (2.19)

2.5 Functionals

When deriving approximations using variational methods, we consider integral expressionsof the form

I(u) =

F(xx, u, u)d. (2.20)

The integrand is a function of the independent spatial coordinates x as well as of thedependent variables u and their derivatives u. The scalar-valued integral is also calleda functional because it depends on functions of independent variables. Mathematically, afunctional is an operator mapping a vector u onto a scalar I(u). A functional is linear if,and only if, for arbitrary scalar values and the following relation holds:

I(u + v) = I(u) + I(v). (2.21)

A functional I(u, v) is bilinear if, and only if, it is linear in each of its arguments.

2.6 The Variational Symbol

Consider the functional I = I(x, u, u). At a fixed position x I depends on u and u. Thedeviation v from u is called the variation of u. It is written as u:

u = v, = const. (2.22)

The operator is called variational symbol. The variation u of a function u representsan admissible change of the function u(x) at a fixed value of the independent variable x.Whenever u is prescribed at some position, that is usually on the boundary, its variationis equal to zero because the prescribed value can not be changed. Therefore the variation

of the function u must satisfy the homogeneous part of the geometric boundary conditionson u. In other words, the variation u of u vanishes at all points where u is prescribedand is arbitrary anywhere else. Thus, u is a virtual change.


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2.7 The Minimum of a Quadratic Functional 11

a bx

v ( x )

u ( x )

u ( x ) + v ( x )

u ( x ) , v ( x )

u ( b )

u ( a )

Figure 2.2: Function u(x) with prescribed end conditions and admissible test function v(x)

2.7 The Minimum of a Quadratic Functional

The minimum functional theoremas presented in J.N. Reddys textbook [2] starts with theequation with linear, symmetric, and positive semi-definite operator A that is characteristicfor linear continuum mechanics problems

Au = f (2.23)

and shows that the solution u minimizes the quadratic functional

Q(u) = (Au,u) 2(f, u). (2.24)

Vice versa, any function minimizing Q solves (2.23). This principle is significant becauseminimizing (2.24) is often easier than solving (2.23), especially when a best solution inthe sense of an approximation of the true solution is sought.

Theorem : Functional (2.24) is minimized by u = u0 if and only if u0 is also a so-lution of 2.23.

Proof If part. Let u0 be the solution of 2.23. Then f = Au0 and we have

Q(u) = +(Au,u) (2Au0, u)= Q(u0) +(A(u u0), u) (Au0, u) +(Au0, u0)= Q(u0) +(A(u u0), u) (Au0, (u u0))= Q(u0) +(A(u u0), u) (Au u0), u0)= Q(u0) +(A(u u0), (u u0))

(2.25)

To perform these transformations, the symmetry and linearity of the operator A have beenutilized in the following form:

(Au,v) = (Av,u) (A(u + v), u) = (Au,u) + (Av,u) (2.26)

Since A is always positive, it holds for all functions u of the solutions space:

(A(u u0), (u u0)) 0 for all u of the solution region= 0 if and only if u u0 = 0. (2.27)


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Thus, the functional is minimized by solution to (2.23) and it assumes larger values forany function u deviating from u0:

Q(u) Q(u0). (2.28)Proof Only If part. The solution u0 is varied by v with a real number and an arbitraryfunction v. From (2.28) follows:

Q(u0 + v) Q(u0). (2.29)

From the definition (2.24) of Q follows:

Q(u0 + v) = (A(u0 + v), (u0 + v)) 2(f(u0 + v))= (Au0, u0) + 2(Au0, v) +

2(Av,v) 2(f, v) 2(f, v). (2.30)

The right side f, the solution u0, and the test function v are fixed elements. Therefore,2.30 is a quadratic function of only. Then, Q(u0 + v) has a minimum at = 0 if itsfirst derivative with respect to vanishes at = 0:

0 =

d

dQ(u0 + v)

=0

= 2((Au0 f), v). (2.31)

Thus, only if

Au0 = f (2.32)

(2.31) is satisfied for any arbitrary test function v. This proves that the function u0 thatminimizes the functional (2.24) is also the solution of the differential equation (2.23).

2.8 Principle of Minimum Potential Energy

Differential equations of the linear elastic theory show the form of (2.23) where A is alinear operator. The potential energy , the deformation energy U, and the work of thevolume forces and of the external forces W are related by

= U W. (2.33)

The balance (2.33) is a functional that differs from (2.24) only by a factor of two.Consider for example a coherent solid with domain whose part u of its surface is

subject to specified displacements u and whose other part of its surface is loaded bysurface tractions . In addition, distributed volume forces f act inside the solid. Theseloads cause, in the domain , a displacement field u, a strain field , and a stress field. These magnitudes fulfill the basic equations of elasticity theory whose derivations arerecalled in section 3.1: the stresses, being in equilibrium, are related to the strains by theconstitutive law and these follow from the displacements through the kinematical relations.Then (2.33) becomes

=1

2

Td

fTud

Tud

u

Tud. (2.34)

In the displacement formulation of an elasticity problem it holds that all other magnitudesdepend correctly, via the basic equations, on the displacement field u. If this minimizesthe total potential energy functional, it must also satisfy the equilibrium conditions and


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2.9 Ritz Method 13

therefore be the best solution. To illustrate this, we substitute the stresses and thestrains in (2.24) by the displacements u and use from now on index notation for tensors,

= 12

12

(ui,j + uj,i)Cijkl 12

(uk,l + ul,k)d

fiuid

iuidu

iuid (2.35)

Taking the variation of the potential energy , that is deriving it with respect to thefunctions u, the contribution of the specified displacements u vanishes:

=

1

2(ui,j + uj,i)Cijkl

1

2(uk,l + ul,k)d

fiuid

iuid (2.36)

The following transformations show that the principle of minimum potential energy re-produces the differential equations of the considered problem. These transformations shiftthe derivative orders, by applying partial integration, so that displacement variation u

appears in the not derivated form, for instance: ui,jCijkluk,ld =

[(uiCijkluk,l),j uiCijkluk,lj] d

= uiCijkluk,lnjd

uiCijkluk,ljd.

(2.37)

After transforming the whole internal energy integral term, the virtual displacements iwhich are constants with respect to the integration, can be factored out. The boundaryterms created by the partial integrations cancel out with those contained in (2.35). Theresulting equation

= ui 1

2Cijkl(uk,l + ul,k),j + fi d (2.38)

appears very simple and setting the value of equal to zero confirms the principleof minimal potential energy: the squared brackets in (2.38) include the so called eulerequations which are identical with the governing differential equations. The variationalform (2.36) is called weak variational form and is directly created by the principle ofvirtual work. The variational form (2.38) is directly created by the principle of virtualdisplacements. Since the two equations are related via identical transformations, they aremechanically equivalent to each other, and so are the two principles.

2.9 Ritz Method

Consider the variational problem to find an approximate solution u so that the equation

B(v, u) = l(v) (2.39)

holds for all sufficiently often differentiable functions v that satisfy the homogeneous formof the essential boundary conditions on u. If the functional B is bilinear and symmetricand l is linear, the problem of solving 2.39 is equivalent to the task of minimizing thequadratic functional

Q(u) =1

2B(u, u) l(u). (2.40)

The Ritz method seeks the approximate solution of 2.40 in terms of a finite series

u =n

j=1

cjj + 0. (2.41)


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The j are shape functions that are weighted with Ritz coefficients cj. The function0 satisfies the inhomogeneous essential boundary conditions as it has the prescribeddisplacements u on the boundary u. Therefore, 0 is not weighted with a factor and

vanishes if all essential boundary conditions are homogeneous. All other shape functionsj must fulfill the homogenous essential boundary conditions on the boundary u (or havezero values there). This guarantees that the series always satisfies the essential boundaryconditions independent of the values of Ritz coefficient cj . With these restrictions one setsv = j and, after the substitution of the displacements in 2.39, one obtains a system ofequations that determines the Ritz coefficients cj :

B

i, n

j=1

cjj + 0

= l(i), i = 1, 2,...,n. (2.42)

If B is bilinear and symmetric, (2.42) can be written as:n

j=1

B(i, j)cj = l(i) B(i, 0). (2.43)

Also, (2.43) can be obtained by combining 2.41 directly with the quadratic form 2.40. Thenecessary condition for the minimum of the functionals 2.41 is the disappearance of allpartial derivatives with respect to the Ritz coefficients:

Q(cj)

c1= 0,

Q(cj)

c2= 0, , Q(cj)

cn= 0 (2.44)

This yields a system of n linear equations to determine the unknown coefficients ci. Sum-marizing, the shape functions i are required to satisfy the following conditions:

1. (a) i should be such that B(i, j) is well defined and nonzero [i.e. sufficientlydifferentiable as required by the bilinear form B(, )].

(b) i must satisfy at least the homogeneous form of the essential boundary con-ditions of the problem.

2. For any n, the set {i}ni=1 of B(i, j) is linearly independent.3. The set of shape functions {i} is complete.

The above requirements on the approximation functions guarantee, for linear problems,convergence of the Ritz solution to the exact solution as the Value of n is increased.


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Chapter 3

Governing Equations

3.1 Linear ElasticityThe summary of selected aspects of linear elasticity in sections 3.1.1, 3.1.2, 3.1.3, and 3.1.5follows the representation in the textbook by D. Frederick [3]. The arguments justifyingthe use of matrix notation (section 3.1.6) in place of tensor index notation, for the purposeof simplified presentation of linear elasticity equations, can be found in the textbook byR.M. Jones [4].

3.1.1 Equations of Motion

The concept of the equations of motions, or dynamic equilibrium, is based on Newtonssecond law

F = ma. (3.1)

It says that the force F, acting upon a mass point, is proportional to the product of itsmass m and its acceleration a in the direction where F is applied. We can use Newtonssecond law (3.1) in deriving the equilibrium equations for stress components when wereplace the mass point by a body having some spatial extension, or domain, boundedby a surface as indicated in Fig. 3.1. The external force applied to each particle in is

y

z

x

0

9

/

d/d9

fi

Ii

ni

9

Figure 3.1: Solid body whose domain is bounded by a surface

transferred by a stress vector i, acting on the part of its surface dS whose unit direction

vector ni is parallel to the direction of the stress. A body force vector fi may arise fromgravity, for instance, and is acting on the volume element dV. These forces are relatedto acceleration, which is the time rate of change of velocity dvi/dt, according to Newtons


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16 Governing Equations

second law. Then we take the integral of the equilibrium of forces over the domain andits boundary , respectively, and write:

inidS+

fidV =

dvidt dV. (3.2)

We can transform the surface integral (left term on the left-hand side of (3.2)) to a volumeintegral by applying the Gauss-Green theorem to it:

ij ,j +fi dvi

dt

dV = 0. (3.3)

The integral equation (3.2) holds for any domain , when the expression in parenthesisvanishes everywhere. This gives us the equations of motions for a small volume element:

ij ,j +fi = dvidt

. (3.4)

In solid mechanics the coordinates X are fixed to the particles (Lagrangian description).Then (3.3) yields the equations of motion that we can also derive from studying theinfinitesimal small volume element shown in Fig. 3.2,

xx,x + yx,y + zx,z + fx = uxxy,x + yy ,y + zy ,z + fy = uyxz ,x + yz ,y + zz ,z + fz = uz.

(3.5)

xI

yI

zI

xyJ

xzJ

yxJ

yzJ

zxJ zy

J

y

Figure 3.2: Stress components acting upon a small octahedral volume element.

3.1.2 Kinematical Equations

When a continuum changes its configuration, material in the vicinity of each point isgenerally displaced and rotated like a rigid body but also strained. The straining of avolume element is the part of the relative motion between two neighboring points thatcannot be ascribed to a rigid body motion. To derive some measure of strain, considertwo neighboring points before (P, Q) and after (P , Q) a deformation, [3]. The points aregiven by the vectors

P : ai

Q : ai + daiP : xi = ai + ui(a, t)

Q : xi + dxi = ai + dai + ui(a + da, t)

(3.6)


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3.1 Linear Elasticity 17

O

Q

Q

P

P

ui(a+d

a,t)

ui(a,t)

d

xi

ai

+d

ai

dai

xi

xi+

dxi

ai

x1

x2

x3

Figure 3.3: Relative motion between two neighboring points in a solid [3].

The differences of the squares of the differential distances da and dx of two neighboringpoints before and after the deformation are

(dx)2 (da)2 = dxidxi daidai. (3.7)If the positions xi of the particles are given as functions of the initial configuration ai, the

vector dxi can be expressed by the chain rule

dxi =

xiaj

P

daj , (3.8)

where the partial derivatives are evaluated at point P. Combining (3.8) and (3.7) yields

(dx)2 (da)2 =

xrai

xraj

ij

daidaj . (3.9)

Following the Lagrangian point of view, the displacements ui are given as differencesbetween the initial and actual positions of a point:

ui = xi ai. (3.10)With this definition we obtain the square strain measure in terms of the displacements

(dx)2 (da)2 =

ai

(ar + ur)

aj(ar + ur) ij

daidaj

=

ri +urai

rj +

uraj

ij

daidaj

=uiaj

+ujai

+ urai

uraj

daidaj .

(3.11)

Obviously, the bracketed quantity is a measure for strain because only when strain existsis the terms (dx)2

(da)2 different from zero. The Lagrangian strain tensor

Lij =1

2

uiaj

+ujai

+urai

uraj

(3.12)


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is invariant with respect to interchanging the indices i and j, or symmetric. In deriving(3.12) no restriction was imposed regarding the magnitude of the displacements or strains.Linear elasticity assumes the displacement gradients so small that the square terms can

be neglected. In practice, one is used to writing the linear strain tensor as

ij =1

2

uixj

+ujxi

. (3.13)

Fig. 3.4 illustrates the meaning of shear strain.

y

z

x

P

R

Q

_y

_z

_x

_P

_R

_Q

u,zdz

u,xdx

u,ydy

undeformed

deformed

Figure 3.4: Mechanism of shearing strain [3].

3.1.3 Generalized Hookes Law

The generalized Hookes law, a form of constitutive equations, for a linear elastic soliddescribes the relationship between the stress tensor ij and the strain tensor kl with atensor of fourth order Cijkl:

ij = Cijklkl, i, j, k, l = 1, 2, 3. (3.14)

The indices refer strictly to the material coordinates of the solid body but these may bereplaced by the cartesian coordinates of the reference system as long as the deformationsare small. The 81 components of the fourth-order tensor Cijkl are not all independent.

If one studies situations where the stress tensor is symmetric (when distributed bodymoments do not occur), the stress tensor is also symmetric. We have already noted thatthe strain tensor is also symmetric. Then, both the strain and the stress tensor have onlysix components and the tensor C can not have more than 36 independent components.From the expression for the specific deformation energy e

e =

ijdij =1

2Cijklklij , (3.15)

we obtain the material law by differentiating with respect to the strains and the mate-rial constants by differentiating twice. However, the sequence of differentiations can beinterchanged, and so can the indices ij and kl. Therefore the tensor Cijkl is symmetric,

Cijkl = Cklij, (3.16)

and has only 21 independent components.


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3.1 Linear Elasticity 19

3.1.4 Temperature- and Moisture strains

Temperature changes T cause volume changes in bodies which are described with the

temperature expansion coefficient ij . In the anisotropic case, temperature changes mayeven change the shape of bodies. Hindering the free volume expansion through essentialboundary conditions, or other restrictions such as heterogeneous material distribution,leads to temperature stresses. Some technical materials can absorb moisture and a changeof moisture content H leads to analogous effects and is described with the swellingcoefficients ij . In order to capture these effects, the generalized Hookes law (3.14) mustcomplemented with the temperature- and moisture strains:

ij = Cijkl (kl klT klH) , i, j, k, l = 1, 2, 3. (3.17)The various strains can be divided in the categories mechanical strains M, total strains T,and initial strains R (in the field of composite materials initial strains are called residual

strains). The mechanical strains are directly related to the stresses via the constitutiveequations (3.17). The total strains follow directly from the displacement derivatives after(3.13). The temperature- and moisture strains can be measured at the free body undergo-ing temperature or moisture-content changes. The relationship between the various straindefinitions is

M = T R. (3.18)

3.1.5 Displacement Formulation

In the displacement formulation as summarized in the textbook by D. Frederick [3], thebody force fi is given everywhere in the domain, and the components of the displacement

vector ui are given everywhere on the outer surface of the domain for all times after theinitial time t = t0. Furthermore, the displacements and their first time derivatives areprescribed everywhere in the domain at time t = t0. The components of the stress tensorij , strain tensor ij , and displacement ui are to be found as functions of time and positioneverywhere in the domain.

To solve this type of problem, we determine the displacements ui first, since the bound-ary conditions are given in terms of these. Hence, we shall express the equations of motionin terms of ui. Substituting the stress-strain relationship for an isotropic, homogeneous,linear, elastic solid, (3.14), into the equations of motion (3.4) we obtain

Cijklkl,j +fi = ui. (3.19)

In order to express all terms (except the body force) in terms of ui, we use the relations(3.13). Hence, (3.19) becomes

1

2Cijkl (uk,lj +ul,kj ) + fi = ui. (3.20)

These are three scalar partial differential equations for the three unknown field functionsu1, u2, and u3. A solution of these equations satisfying the initial and boundary conditionsyields the ui for every points within the elastic region.

The formulation of the preceding problem has been for elastic solids not in equilibrium(elastodynamics). By dropping the inertia term, we obtain the formulation for elasticsolids in equilibrium (elastostatics)

1

2Cijkl (uk,lj +ul,kj ) + fi = 0. (3.21)


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3.1.6 Matrix Notation

The observed symmetries of the tensors ij , the strain tensor ij , and the constitutive

equations tensor Cijkl allow to write down the basic elasticity equations in matrix notation.The stress tensor and strain tensor components are rearranged in vectors with six entriesand the material constants fill a symmetric matrix with six rows and six columns. Table3.1 shows how the entries are organized. The indices 1, 2, 3 can be replaced with x,y ,z.

1 = 11 , 1 = 112 = 22 , 2 = 223 = 33 , 3 = 334 = 23 , 4 = 23 = 2235 = 31 , 5 = 31 = 2316 = 12 , 6 = 12 = 212

Table 3.1: Tensor and matrix notation

The three continuous displacement components ui are now entries of the displacementvector u, the strain and the stress vector entries, i and i, are also grouped in vector and , respectively, and the elasticity constants can can now be written as a matrix C

u =

uxuy

uz

=

123456

=

123456

C =

c11 c12 c13 c14 c15 c16c21 c22 c23 c24 c25 c26c31 c32 c33 c34 c35 c36c41 c42 c34 c44 c45 c46c51 c52 c35 c45 c55 c56c61 c62 c36 c46 c56 c66

. (3.22)

The constitutive equations can now be written as

= C. (3.23)

To find the Cij in terms of the engineering elasticity constants Youngs moduli, shearmoduli, and Poissons ratios, it is convenient to write down the compliance matrix S inprincipal material directions

S =

1E1

12E1

13E1

0 0 0

21E2

1E2

23E2

0 0 0

31E3

32E2

1E3

0 0 0

0 0 0 1G23

0 0

0 0 0 0 1G31

0

0 0 0 0 0 1G12

. (3.24)

and obtain the stiffness matrix C by inverting S. The determinant S of the coupled part

of S, which is of order 3 3, isS = S11S22S33 S11S223 S22S213 S33S212 + 2S12S23S13. (3.25)


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3.2 Simplified Elasticity Equations 21

With this, the entries of the stiffness matric C are

C11 =S22S33S

223

SC12 =

S13S23S12S33S

C13 =S12S23S13S22

S

C22 =S33S11S213

SC23 =

S12S13S23S11S

C33 =S11S22S

212

S

C44 =1

S44C55 =

1S55

C66 =1

S66.

(3.26)

In matrix notation one needs to define a matrix differential operator L to express thestrain-displacement equations

= Lu, L =

/x 0 0

0 /y 0

0 0 /z

0 /z /y

/z 0 /x

/y /x 0

. (3.27)

The same operator L can be used to express the static equilibrium equations in matrixnotation

LTs + f = u. (3.28)

Finally, the equation of motion in terms of the displacement vector u follows from com-bining (3.23), (3.27), and (3.28):

LTCLu + f = u. (3.29)

To capture residual strain effects, (3.29) is simply modified by subtracting the residualstrain vector R from the total strains = Lu:

LTC

Lu R+ f = u. (3.30)3.2 Simplified Elasticity Equations

Simplified equations for describing the mechanical behavior exist for a number of mechan-

ical problems where certain assumptions on the shape and the loading of bodies have beenmade. Truss, beam, plane elasticity, and plates are among the most popular examples.Generalized plane strain, discussed in section 3.2.3, and the one-dimensional model of athick-walled long cylinder, considered by G. Kress for an FEM-type exact solution method[5] and discussed in section 3.2.7, are interesting cases because they are not standard text-book material and the finite element models of these equations are not found in commercialFEM software.

3.2.1 Plane Elasticity

Plane elasticity is, in a way, closest to the general forms presented above and yet much

simplified. Consider a linear solid of uniform thickness h that is bounded by two planeswhose extension is much larger than h. Furthermore, the body is loaded by in-planeforces only, and by distributed body forces only acting in the lateral directions, so that the


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distribution of stresses is constant along its thickness direction and no stresses are appliedon the parallel planes perpendicular to the thickness direction. The lateral directionsare usually denoted by x and y and the out-of-plane, or transverse, direction is z. The

assumptions imply that the stresses in the z direction are negligibly small,

z = yz = xz = 0. (3.31)

For plane strain problems the situation is reversed: the distance between the boundingplanes is much larger than the extension in the lateral directions. Then, the direct strainz in the z direction is zero. When considering isotropic materials, the shear strains zxand yz are also zero but in the anisotropic case they may be not.

For the cases of plane stress and special plane strain displacements exist only in thex-y plane, and only depend on the positions in that plane:

u = u(x, y),v = v(x, y),w = 0.

(3.32)

Thus, derivatives of stresses or strains with respect to the out-of-plane direction z vanishand so the equilibrium equations are reduced to

x,x +xy,y +fx = 0xy,x +y,y +fy = 0

(3.33)

and the strain-displacement relations are

x = u,x y = v,y xy = u,y +v,x . (3.34)

The equilibrium and the strain-displacement equations can again be expressed by a lineardifferential operator L

L =

/x 0

0 /y

/y /x.

. (3.35)

The stress-strain, or constitutive, relations are

x = C11x + C12yy = C12x + C22yxy = C66xy ,

(3.36)

where the Cij are elastic constants. Plane stress and plane strain equations differ only inthe definition of the Cij .

In order to find the appropriate entries for the Cij in terms of the engineering constantsfor the plane stress case, consider the compliance matrix and note that the out-of-planestresses all vanish (3.31). From the stress-strain equations in terms of the compliancematrix S the equations for the out-of-plane strains can therefore simply be dropped:

12

6

=

S11 S12 0S12 S22 00 0 S66

126

(3.37)


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Since the matrix appearing in (3.37) is coupled only in two dimensions, it is easily invertedby hand and after substituting the definitions of the Sij from (3.24) one obtains the reducedplane-stress stiffness matrix for an orthotropic material

C11 =E1

1 1221 C12 =12E1

1 1221 C66 = G12. (3.38)

In the plane strain case one must use the stiffness matrix entries (3.26) obtained frominverting the full compliance matrix (3.24). Since the full compliance matrix is coupledin three-dimensions, the plane-strain stiffness matrix is more complicated, in terms of theengineering constants, than the reduced plane-stress stiffness matrix,

C11 =1 2332

E2E3 C12 =21 + 3123

E2E3 C22 =1 1331

E1E3 C66 = G12, (3.39)

where = 1 1221 2332 3113 2213213

E1E2E3. (3.40)

For the isotropic case things simplify to Ei = E, ij = , and G = E/2(1 + ) and theisotropic stiffness matrices for isotropic plane stress

C11 = C22 =E

1 2 C12 =E

1 2 C66 =E

2(1 + )(3.41)

and plane strain

C11 = C22 =

E(1

)

(1 + )(1 2) C12 =E

(1 + )(1 2) C66 =E

2(1 + ) (3.42)

are readily verified.

3.2.2 Axial Symmetry

The axisymmetric model is also a plane model because it considers only the two displace-ments u and v in the x-y plane and (3.32) describe the full displacement field. However,there is one additional mechanism included which can be understood when partitioningthe axisymmetric part to be modelled into a set of circular hoops. The model domainof the axisymmetric part is a plane intersecting at right angles with all the hoops, see

the illustration at the left-hand side of Fig. 3.5. Just pick one of these hoops and movethe intersecting point of it in the radial direction x by a displacement u. Then the cir-cumferential length of the hoop is increased from 2x to 2x + u(x). The increase incircumferential length then causes, in the circumferential direction , a strain = u/x.So the axisymmetric model contains strains in the out-of-plane direction varying in thex-y plane although only the two displacements u and v appear. The kinematical relationsof the axisymmetric model are therefore an extension of those holding for the plane statesmodel:

= Lu,

=

xy

xy

, L(+) =

x

0

0 y

y

x

1x

0

, u =

u

v

. (3.43)


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x(1)

y(2)

G

ANIN

N

CNOJNO

AOIO

AGIG

Figure 3.5: Plane element in an axisymmetric solid (left) and stresses and strains (right),after Zienkiewicz

The operator L(+) prescribes both, derivatives and a division by the position x of a pointin the undeformed state. The equilibrium conditions in the radial and in axial directions,x and y,

x : x,x + xy,y + x1(x ) + fx = 0

y : xy,x + y,y + x1xy + fy = 0

, (3.44)

can be derived from the information of Figs. 3.6 and 3.5. However, the matrix notation(3.23) can not be used to write these equilibrium equations correctly. However, we canmultiply (3.44) by x to obtain

x : (xx) ,x + xxy,y + xfx = 0y : (xxy) ,x + xy,y + xfy = 0

(3.45)

which can be presented in matrix-operator notation as

L()T (x) + xf = 0. (3.46)

Here the stress vector s and the operator L() are defined as

=

x

yxy

, L() =

x

0

0

yy

x

1x 0

. (3.47)

To be general, we base the constitutive equations on an orthotropic material with itsprincipal axis in-parallel to the directions x, , and r,

xyxy

=

C11 C12 0 C13C12 C22 0 C23

0 0 C66 0C13 C23 0 C33

xyxy

, (3.48)

so that there is no coupling between the shear strain and the direct stresses. This isconsistent with the so-called torsion-free axial symmetric problem that we have already


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@N

@q

@N@qsq

@N@qsq

@NN@@ NN ++ qss

N@N qs

N

@N@ON qtNBNrw=

NB

Figure 3.6: Forces Acting Upon a Volume Element of Axially Symmetric Stressed Solid

assumed in the kinematical relations. The governing equations can now be written inmatrix notation as

L()T

xCL(+)u

+ xf = 0. (3.49)

The body force fx due to rotation is given by

fx = 2x. (3.50)

3.2.3 Generalized Plane Strain

A generalized plane strain problem arises through elastic coupling mechanisms of theconstitutive equations. Consider for example test coupons of a rectangular shape madefrom laminated composite materials. Loading such a laminate in the longitudinal directionx creates a complicated stress situation. The clamping jaws, as Fig. 3.7 illustrates, hinderthe transverse strain determined by Poissons ratio effects. However, at distances large

P

P

Figure 3.7: Tensile test situation and clamping effects

enough away from the clamping jaws, the clamping effects fade away and the strainsand stresses in the undisturbed region of the coupon remain constant in the longitudinaldirection. Specifically, the strain x is constant everywhere. Since the elastic properties are

different from layer to layer, not only a stress x is created but also stresses y and xy arecreated due to the mutual hindrance of the free transverse straining due to Poissons ratios.Furthermore, these in-plane transverse stresses must vanish along the free edges. The


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equilibrium conditions then enforce stresses in the out-of-plane direction z of the laminatethat are called interlaminar stresses. Thus, the stress state is fully three-dimensional as allsix stress components are present. However, the stress vary only over the cross-sectional

area of the laminated coupon and remain constant along the longitudinal direction inthe region where the clamping effects have decayed. The state of strains and stresses inthat region can be modelled perfectly, and much more economically than with a three-dimensional model, by a two-dimensional generalized plane strain model used by G. Kress[6], [7], and [8] to study free-edge influences on stiffness and strength measurements oflaminated test coupons.

Please note that for the following, the spatial directions x,y, and z are defined slightlydifferent than for the preceding sections. Specifically, in the theory of laminated platesit is customary to denote the laminate in-plane directions with x and y and the out-of-plane direction with z. To stay consistent with that convention, the cross-section of atest coupon is parallel to the y-z plane. So our two-dimensional model is in the y-z plane

instead of the x-y plane that has been used for the models in the previous sections.The difference between the special and the generalized plane-strain state is the in the

latter displacements in the out-of-plane direction x occur:

u = u(x,y ,z)v = v(x,y ,z)w = w(x,y ,z).

(3.51)

However, the derivatives of all displacement components with respect to the longitudinal,or out-of-plane, direction vanish. Moreover, the strain x is constant everywhere:

x = u,x = constant

y = v,yz = w,zyz = v,z +w,yxz = u,zxy = u,y .

(3.52)

Since the longitudinal strain x is not a part of the unknown solution function, it must bespecified as a part of the loading data. This makes sense as one can easily understand thatthe before discussed test coupon of a laminate is loaded by extending it in the longitudinaldirection. Also, since the stresses do not change along the x direction, the derivatives withrespect to x are dropped from the equilibrium equations,

yx,y + zx,z = 0yy ,y + zy ,z = 0yz ,y + zz ,z = 0,

(3.53)

where body forces are not considered. Applying the linear operator L for the generalizedplane-strain problem,

L =

0 0 0

0 /y 0

0 0 /z

0 /z /y

/z 0 0

/y 0 0

, (3.54)


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to the displacement vector u yields the strains except for the direct strain in the out-of-plane direction, x. Therefore we can write

Lu = x, x =

x00000

, (3.55)

where e contains the strains listed in (3.52). Taking this into account, we write thekinematics and the equilibrium conditions in concise matrix-operator notation:

= Lu + x and LT = 0. (3.56)

The material law, making the generalization of the plane-strain problem necessary, is

Cgps =

C11 C12 C13 0 0 C16C12 C22 C23 0 0 C26C13 C23 C33 0 0 C36

0 0 0 C44 C45 00 0 0 C45 C55 0

C16 C26 C36 0 0 C66

. (3.57)

With all the definitions in place, the governing equations in matrix notation read

LT (CLu + x) = 0 (3.58)

and we discover that the loading data ex gets lost because of the differential operator Lacting upon it. In fact, the meaningful part of (3.58) is consequently written as

LTCLu = 0 (3.59)

You may feel a little unsettled by the fact that (3.59) does not seem to carry any loadinformation but dont worry: it will be instructive to see in section 6.4 how the informationon the loading strain x re-enters the weak variational form via the boundary conditions.

3.2.4 Truss

The truss belongs to the so-called structural elements, where the term element does not yetrefer to the finite element method. Rather, the truss structural element is based on defini-tions of geometry and restrictions on loading, resulting in great simplifications. Moreover,structural elements have great practical significance. For instance, the complicated struc-ture of a framework, for instance that of a crane, can be described by using the structuralelement truss. The truss has is extended in one direction, the longitudinal direction, muchlarger than in the two transverse directions. It has a cylindrical shape, i.e. its cross-sectionstays constant in the longitudinal directions. It is also understood that the truss carriesloads only along its longitudinal direction and that no bending occurs. Therefore, the

external loads can only be applied in-line with its longitudinal direction and it is also clearthat the truss must be straight. Fig. 3.8 illustrates the truss model and indicates that itcan also be loaded by a distributed force p in the longitudinal direction. The force p is


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related to the body force fx of the general equilibrium equations, and the internal force Fis related to the stress x, via the cross-section area A of the truss

p = Afx, F = Ax. (3.60)

Thus, the integration of the governing equations for the truss is implicitly already per-formed over the cross-sectional area and only the longitudinal direction remains as a one-dimensional domain over which an integration must be carried out. All stress components

L

p

F, u

x

Figure 3.8: Truss Conventions

except for longitudinal direct stress vanish,

y = z = yz = xz = xy = 0, x = x(x), (3.61)

and all displacements except the one in longitudinal direction vanish too:

v = w = 0, u = u(x). (3.62)

The equilibrium equation becomesF,x +p = 0 (3.63)

and the the stress-strain equations reduce to

x = u,x . (3.64)

The linear differential operator L is only one-dimensional

L =

x. (3.65)

With the constitutive law

F = AEx (3.66)

and the other definitions the one governing equation is written in matrix notation

AEL2u +p = Au. (3.67)

The differential equation governing torsion of a circular shaft has the same structure asthat of the truss. By replacing the area value A by the polar moment of inertia Ip of thecircular cross section, Youngs modulus E by the shear modulus G, the longitudinal forceF by the torque Mt, the displacement u by the absolute angle of twist , the strain bythe angle of twist per unit length of the shaft , and the line force p by the distributed

torque per unit length of the shaft m, we arrive at

IpGL2 + m = Ip,tt. (3.68)


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L

x

z

y

q

Q, w

M , >

Figure 3.9: Coordinates Coordinate convention and loads of the intrinsic beam.

3.2.5 Beam

The beam geometry is identical to the truss geometry. The loading restrictions, however,

are different. Whereas the truss is loaded so that no bending can occur, the beam isloaded so that no overall extension along its neutral axis can occur. That is, the beamis loaded by transverse forces and by bending moments as indicated in Fig. 3.9. Weassume a coordinate system x, y z so that the x-axis is aligned with the beam axis andthe other coordinate directions are arranged as Fig. 3.9 indicates. The integrations overthe transverse directions of the beam are implicitly performed when using the momentof inertial Iy and the shear area As. For the time being we consider only bending inthe x-y plane and must therefore assume that a coupling term Iyz does not exist. Wefurthermore expect that in general, the beam may be deformed in pure bending and alsoin shear. Denoting the deflection by w(x), the following relationship exists between thespatial derivative of the deflection line, or slope, w = w,x, the rotation of the beams

cross-section , and the shear angle :

= dwdx

= w. (3.69)

The global equilibrium of an infinitesimal piece of beam is given by

Q + q = 0, M Q = 0. (3.70)The constitutive equations are

Q = GAs, M = EI = EI . (3.71)

Inserting these into the equilibrium equations gives the governing equations

EI + GAs(w + ) = 0 GAs(w

+ ) + q = 0 (3.72)

The equation on the left-hand-side of (3.72) describes the equilibrium in the bendingdirection , so to say, and the other equation describes the equilibrium in the deflectiondirection w. Again we discover a way to write down the equations more concisely byintroducing linear operators L() and L(+):

L() =

0 /x

/x 1

, L(+) =

0 /x

/x 1

. (3.73)

With the abbreviations

u =

w

, =

, =

MQ

, C =

EI 0

0 GAs

, and q =

q0

, (3.74)


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we write the governing equations (3.72)

LT()CL(+)u + q = 0. (3.75)

The above equations are set up for the beam that can deform in shear. A further simpli-fication results from assuming that the beam is very slender so that the deflection due toshear is so much smaller than that due to bending that it can be neglected. This equivalentto saying that the shear deformation vanishes and from the kinematical relation (3.69)follows that the slope depends only on the derivative of the deflection = w and the sameholds for the curvature = w. Then there remains only one displacement function wand from the two equilibrium equations (3.70) one can be eliminated to obtain

EI + q = 0. (3.76)

Substituting the kinematical relation = w

yields the governing equation of the thinbeam

EI w q = 0. (3.77)The governing equation (3.77) is a scalar equation so that writing it down in matrixnotation for the purpose of simplifying things would really be pointless.

3.2.6 Plate Bending

The plate is an extension of the beam into two dimensions. A plate is a body that isbounded by two parallel plates whose lateral dimensions are much larger than the distancebetween them. Geometrically, a plate is similar to two dimensional bodies bodies in plane

stress state; however, plates are loaded transverse to their x-y midplane by a distributedforce of intensity p. It deflects by a function w(x, y) rotates by angles x and y in thex and y directions, respectively. We consider a shear deformable plate where shearingangles x and y appear due to the transverse stress resultants, or line loads, Qx and Qy,respectively. The bending of the plate due to bending moments Mx, My, and the twistingmoment Mxy are related to the curvatures x, y, and xy. The following kinematical

x,u

y,v

z,w

A

B

CD

zc

u0

A

DC

B

w0

b

zcb

Undeformed Cross Section Deformed Cross Section

Figure 3.10: Plate kinematics

relationships between the generalized displacements and strains exist:

x = w,x +x, y = w,y +y, x = x,x , y = y,y , xy = x,y +y,x . (3.78)


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Here it must be noted that the sign conventions, indicated in Fig. 3.10, are different fromthose customarily used in beam theory. The global equilibrium conditions of the plate interms of force and moment resultants are

Qx,x + Qy,y + q = 0

Mx,x + Mxy,y Qx = 0Mxy,x + My,y Qy = 0

(3.79)

The constitutive equations are

Mx = D11x + D12y

My = D12x + D22y

Mxy = D66xy

Qx = A44x

Qy = A55y.

(3.80)

The kinematical relations between the generalized displacements w, x, y and the gen-eralized strains x, y, xy, x, and y can be expressed in matrix notation, e = Lu withthe operator L:

L =

0 /x 0

0 0 /y

0 /y /x/x 1 0/y 0 1

,

=

x

y

xyx

y

, u =

w

xy

(3.81)

The same operator can be used to write the equilibrium equations (3.79)

LT = p, =

MxMy

MxyQxQy

, p =

p00

. (3.82)

The constitutive equations (3.80) are written in matrix notation

= C, C =

D11 D12 0 0 0

D12 D22 0 0 0

0 0 D66 0 0

0 0 0 A44 0

0 0 0 0 A55

. (3.83)

The stiffness values A44, A55, D11, D12, D22, and D66 correspond to a plate made froman orthotropic material. They are calculated from engineering constants where E1 and E2are Youngs moduli along the x and y directions, respectively, 12 and 21 are Poissons


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ratios, G12, G13 and G23 are the shear moduli in the x y, x z, and the y z planes ofthe material axes of the plate, and h is the thickness of the plate.

D11 D12D21 D22

= h3

12(1 1221)

E1 21E121E2 E2

, D21 = D12, (3.84)

A44 = G23h, A55 = G13h, D66 =G12h

3

12. (3.85)

Substituting the generalized stresses into the equilibrium equations (3.82) by the consti-tutive equations (3.83) and substituting the strains by the kinematical equations (3.81)yields the governing equations in terms of the displacements u in matrix-operator notation

LTCLu = p. (3.86)

3.2.7 Thick-Walled Cylinder

As an engineering student, you will be familiar with the thick-walled cylinder problem.The problem has been extended and generalized by G. Kress [5] to describe layered wallsmade from different anisotropic materials which may be found in products such as fiber-wound cylinders or fiber- reinforced pressure vessels. Although stationary heat flux inthe radial direction and shrink fits, as included in the model [5], are not considered, theone-dimensional problem becomes too messy to be handled by using hand formulas andpocket calculator. The problem is not suited for beginning to understand FEM, and itmay not seem so obvious to treat this problem at all because an exact solution exists. It

is however, an interesting problem as its FEM treatment leads to a different result as inthe more typical cases: the FEM reproduces the exact solution but requires less unknownparameters than it and greatly reduces the numerical solving effort as compared with thenumerical equations corresponding to the other exact solution [5].

We assume that the cylinder is thick-walled, infinitely long, and loaded axisymmet-rically without any bending, so that the cylinder remains straight. The problem can bedescribed in a one-dimensional domain as the stresses and strains are constant along thecylinder axis, here denoted by x, as well as in the circumferential direction . The in-dependent spatial variable is the thickness direction r. The one displacement that varieswith r is the displacement w in thickness direction. The longitudinal displacement u islinear due to constant direct strain x in that direction. The circumferential displacement

v is also linearly distributed along the cylinder axis due to a twist angle per unit length. Summarizing, we have the displacement functions,

u =

u = u(x) =

0xdx

v = v(x) =

dx

w = w(r)

, (3.87)

and the kinematical relations

=

x = 0x

= w/r

r = wr

x = r

. (3.88)


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Noting the differences in axis notation, we recognize that we can use the first of the twoequilibrium equations (3.44):

r,r +1

r (r ) + r2 = 0 (3.89)

We have included the body force term and assumed that the body force is created byinertia effects when the cylinder rotates. The following stress-strain equations are used:

x

r

x

=

C11 C12 C13 C16

C12 C22 C23 C26

C13 C23 C33 C36

C16 C26 C36 C66

Tx

T

Tr

Tx

(3.90)

Combining equations (3.88), (3.89), and (3.90) yields the inhomogeneous ordinary differ-ential equation

w,rr +w,r

r s2 w

r2+ P = 0, s2 =

C22C33

. (3.91)

The inhomogeneous part P of (3.91) carries the load information of the predefined totalstrains in P1 and the body forces due to rotation in P2.

P(r) = P1(r) + P2(r), P1(r) =0xr

(C13 C13), P2(r) = r2

C33. (3.92)

The solution of the homogeneous part of the differential equation is given by

wH(r) = ars + brs. (3.93)

Using the method of variation of the constants, the inhomogeneous displacement solutionfollows from

wP(r) =rs

2s

P(r)r1+sdr r

s

2s

P(r)r1sdr. (3.94)

The complete solution is the sum of the homogeneous and the inhomogeneous part

w(r) = ars + brs + wP(r), (3.95)

where the free parameters a and b are used to fit the solution to boundary and interfacecontinuity conditions. The boundary conditions are specified surface tractions of the innerand outer surface of the wall of the cylinder. Equilibrium requires that at the interface

between layers the radial stress r be continuous. Additionally, the displacement plus thespecified mismatch at interfaces between different parts of the wall must be continuous.

3.2.8 Restrictions on Elastic Constants: Isotropic Materials

The compliance matrix for isotropic materials simplifies to

S =

1E E E 0 0 0

E 1E E 0 0 0

E

E1E

0 0 0

0 0 01G 0 0

0 0 0 0 1G

00 0 0 0 0 1

G

. (3.96)


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The determinants of the 3 3 coupled sub-matrix and of the two significant sub-sub-matrices are:

D = 1 32

23

E3, D11 = (1 )(1 + )

E2, D12 = (1 + )

E2. (3.97)

With these, the stiffness components C11 and C12 are easily calculated after Cramers rule:

C11 =D11D

= E1 v

1 2 , C12 =D12D

= Ev

1 2 . (3.98)

In addition, the shear modulus G is defined in terms of the stiffness components, C11 andC12, or in terms of the elastic modulus, E, and Poissons ratio, , as

G =C11 C12

2

=E

2(1 + )

(3.99)

so that the stiffness matrix C for an isotropic material can be written as

C =E(1 )

(1 + )(1 2)

1 1

1 0 0 0

1 1

1 0 0 0

1

1 1 0 0 0

0 0 0 122(1) 0 0

0 0 0 0 122(1)

0

0 0 0 0 0 122(1)

. (3.100)

Thus, in order that E and G always be positive, i.e. that a positive normal stress or shearstress times the respective normal strain or shear strain yield positive work, it must bethat

> 1. (3.101)Another constrained on the value of Poissons ratio comes from the thought experiment ofsubjecting an isotropic body to hydrostatic pressure, p. The strains to hydrostatic pressurefollow from multiplying the pressure vector with the coupled 3 3 part of the compliancematrix (3.96),

xyz

= 1E1

1 1

ppp

= p(1 2)E111

(3.102)The volumetric strain v is defined as the sum of the three extensional or normal strains:

v = x + y + z =p

E3 (1 2)

=p

K. (3.103)

Then, the bulk modulus,

K =E

3(1 2) , (3.104)

is positive only if E is positive and

0 for every nonzero vector x. It is therefore sufficient to store only theupper half of the matrix including the diagonal. In order to preserve the band struc-ture, decompositions of band matrices make only sense without pivot search. Then the

Cholesky decomposition can be applied with advantage because it uses with n3

/6 + O(n2

)asymptotically only half as many floating-point operations than the Gaussian eliminationand about half as much storage. There are two types of Cholesky decomposition, the firstone is A = UTU, the second one is A = UTDU. The second form is numerically morefavorable because it avoids the evaluation of square roots.

First Method

Perform the following steps:

1. Decomposition A = BTB. For each j = 1, n in increments of 1

(a)

bjj =

ajj j1i=1

b2ij

(b) For each k = j + 1, n in steps of 1

bjk =1

bjj

ajk

j1i=1

bikbij

2. Forward elimination a = BTb. For each j = 1, n in increments of 1

bj = aj j1

i=1

bijbi 1bjj

3. Back substitution Bx = b

(a)

xn =bn

ann

(b) For each i = n 1, 1 in increments of 1

xi =1

aii

bi

nk=i+1

aikxk

The determinant of A can now rather elegantly be calculated by

|A| = |BT||B| = (b11b22 bnn)2 .


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90 Assembly and Solution

Second Method

The used matrices and vectors have the following structures:

D =

d1d2

. . .

dn1dn

, B =

1 b12 b121 b23 b2n

. . .. . .

...1 bn1,n

1

, b =

b1b2...

bn

, z =

z1z2...

zn

.

Perform the following steps:

1. Decomposition A = BTDB. For each j = 1, n in increments of 1

(a) For each i = 1, j 1 in increments of 1i. h = aij

ii. bij = h/di

iii. For each k = i + 1, j in increments of 1 akj is replaced with akj hbik(b) dj = ajj

2. Forward elimination a = BTz, Db = z. For each j = 1, n in increments of 1

(a) zj = aj

(b) For each i = 1, j 1 in increments of 1 zj := zj bijzi(c) bj = zj/dj

3. Back substitution Bx = b. For each j = n, 1 in increments of 1(a) xj = bj

(b) For each i = j + 1, n in increments of 1 xj := xj bjixiThe determinant of A is calculated by

|A| = |BT||D||B| = |D| = (d1d22 dn) .

7.6 Iterative Solution Methods

Iterative methods are suited for use with sparse matrices. When a coefficient matrix ispopulated by only a few nonzero entries and these are not densely concentrated within aband along the diagonal, the matrix is called sparse. An example for a sparse matrix is thematrix to the left in Figs. 7.4 and 7.5 corresponding to the unfavorable numbering of thenodes at the top of Fig. 7.3. Sparse matrices often arise for very large models with manyunknowns. The triangular factors of a sparse matrix A may have many more nonzeroelements than A itself. Factoring may be impossible due to limited memory. The iterativesolution methods seek the primary solution in iterations, avoiding the decompositions step.Most iterative methods are memory-efficient and run quickly with sparse matrices.

Proofs of convergence of iterative methods is based on eigenvector and eigenvalueanalysis of a matrix B. An eigenvector v of B is a nonzero vector that does not rotate

when B is applied to it. In other words, there is some scalar constant such that Bv = v.The value is an eigenvalue of B. Iterative methods often depend on applying B to avector over and over again. When B is repeatedly applied to an eigenvector v, one of two


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7.6 Iterative Solution Methods 91

things can happen. If || < 1, then Biv = iv will vanish as i approaches infinity. If|| > 1, then Biv = iv will grow to infinity. Any vector x can be expressed as a linearcombination of eigenvectors x = 1v1 + 2v2 +

+ nvn. On repeated application of

B to x we have Bix = 1i1v1 + 2i2v2 + + ninvn. If the magnitudes of all theeigenvalues are smaller than one, Bix will converge to zero. This why the spectral radius(B) of a matrix B,

(B) = max|i|, (7.14)is so important in numerical analyzes.

7.6.1 Jacobi Iteration

It is assumed that all diagonal elements aii of the matrix A are nonzero. Then, resolvingthe ith row for the unknown xi directly yields an iteration rule

x(+1)i = bi

aii

i1k=1

aikaii

x()k n

k=i+1

aikaii

x(k ) (i = 1, 2, , n) (7.15)

with the iteration index and the start values x(0)1 , x

(0)2 , , x(n)1 . The iteration rule is

called Jacobi method or parallel method, since all components of the new vector x(+1)

are calculated only from x(). The Jacobi method converges for any start vectors u0 if

maxk

ni=1i=k

|aikaii

| < 1 column sum criterion (7.16)

or

maxi

nk=1k=i

|aikaii | < 1 row sum criterion (7.17)

hold.Why, or when, the Jacobi method converges to the exact solution can be understood

by the following consideration, when it is derived by splitting A into two parts: D, whosediagonal entries are iden

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